Test Functions \(\mathcal D(\Omega)\)Fonctions test \(\mathcal D(\Omega)\)
Distribution theory is built by duality: a generalized function is not a rule assigning a value to each point, but a continuous linear functional acting on a carefully chosen space of smooth test functions. This chapter constructs that test space, \(\mathcal D(\Omega)=C_c^\infty(\Omega)\), from the ground up. We develop multi-index notation, compact support, seminorms, convergence in \(\mathcal D\), standard bump functions and mollifiers, smooth cutoffs, partitions of unity, and the LF-space viewpoint that prepares the rigorous definition of a distribution.La thรฉorie des distributions se construit par dualitรฉ : une fonction gรฉnรฉralisรฉe nโest pas une rรจgle qui associe une valeur ร chaque point, mais une forme linรฉaire continue agissant sur un espace soigneusement choisi de fonctions test lisses. Ce chapitre construit cet espace, \(\mathcal D(\Omega)=C_c^\infty(\Omega)\), ร partir de ses fondements. Nous dรฉveloppons la notation multi-indice, le support compact, les semi-normes, la convergence dans \(\mathcal D\), les fonctions bosse et les mollificateurs standards, les fonctions de coupure lisses, les partitions de lโunitรฉ ainsi que le point de vue dโespace LF qui prรฉpare la dรฉfinition rigoureuse dโune distribution.
Visual Investigations
Objectives & Prerequisites
- Specific objective 1. Manipulate the multi-index calculus fluently: \(|\alpha|\), \(x^\alpha\), \(\partial^\alpha\), the multi-index binomial, and the Leibniz rule.Objectif spรฉcifique 1. Maรฎtriser le calcul multi-indice : \(|\alpha|\), \(x^\alpha\), \(\partial^\alpha\), les coefficients binomiaux multi-indices et la rรจgle de Leibniz.
- Specific objective 2. Define the support of a function and \(\mathcal D(\Omega)=C_c^\infty(\Omega)\) precisely, and verify that a candidate belongs to it.Objectif spรฉcifique 2. Dรฉfinir prรฉcisรฉment le support dโune fonction et \(\mathcal D(\Omega)=C_c^\infty(\Omega)\), puis vรฉrifier quโune fonction donnรฉe appartient ร cet espace.
- Specific objective 3. Prove the one-variable lemma on \(e^{-1/t}\) and deduce the existence of the standard bump and the mollifier \(\rho_\varepsilon\).Objectif spรฉcifique 3. Dรฉmontrer le lemme unidimensionnel portant sur \(e^{-1/t}\), puis en dรฉduire lโexistence de la fonction bosse standard et du mollificateur \(\rho_\varepsilon\).
- Specific objective 4. Construct cutoffs (\(0\le\varphi\le1\), \(\varphi\equiv1\) on a compact \(K\)) and smooth partitions of unity by convolution with a mollifier.Objectif spรฉcifique 4. Construire des fonctions de coupure (\(0\le\varphi\le1\), \(\varphi\equiv1\) sur un compact \(K\)) et des partitions de lโunitรฉ lisses par convolution avec un mollificateur.
- Specific objective 5. State and apply the definition of convergence \(\varphi_k\to\varphi\) in \(\mathcal D(\Omega)\), and decide which sequences converge.Objectif spรฉcifique 5. รnoncer et appliquer la dรฉfinition de la convergence \(\varphi_k\to\varphi\) dans \(\mathcal D(\Omega)\), puis dรฉterminer quelles suites convergent.
- Specific objective 6. Prove the sequential continuity of \(\partial^\alpha\) and of multiplication by \(\psi\in C^\infty\), and explain why \(\mathcal D(\Omega)\) is neither normable nor metrizable in its LF topology.Objectif spรฉcifique 6. Dรฉmontrer la continuitรฉ sรฉquentielle de \(\partial^\alpha\) et de la multiplication par \(\psi\in C^\infty\), puis expliquer pourquoi \(\mathcal D(\Omega)\) nโest ni normable ni mรฉtrisable pour sa topologie LF.
- Multivariable calculus: partial derivatives, the chain rule, \(C^k\) and \(C^\infty\) functions on open sets of \(\mathbb R^n\).Calcul diffรฉrentiel ร plusieurs variables : dรฉrivรฉes partielles, rรจgle de la chaรฎne, fonctions \(C^k\) et \(C^\infty\) sur des ouverts de \(\mathbb R^n\).
- Uniform convergence and the fact that a uniform limit of continuous functions is continuous.Convergence uniforme et fait quโune limite uniforme de fonctions continues est continue.
- The convolution \((f*g)(x)=\int f(x-y)g(y)\,dy\) and differentiation under the integral sign.La convolution \((f*g)(x)=\int f(x-y)g(y)\,dy\) et la dรฉrivation sous le signe intรฉgral.
- Open, closed, and compact subsets of \(\mathbb R^n\); the distance \(\operatorname{dist}(x,A)\).Sous-ensembles ouverts, fermรฉs et compacts de \(\mathbb R^n\) ; distance \(\operatorname{dist}(x,A)\).
- Induction on the order of differentiation with a polynomial \(\times\) exponential ansatz.Rรฉcurrence sur lโordre de dรฉrivation avec un ansatz polynรดme \(\times\) exponentielle.
- Growth comparison: \(s^m e^{-s}\to0\) as \(s\to\infty\) for every \(m\).Comparaison de croissance : \(s^m e^{-s}\to0\) lorsque \(s\to\infty\), pour tout \(m\).
- Convolution with \(\rho_\varepsilon\) to smooth an indicator while controlling its support.Convolution avec \(\rho_\varepsilon\) pour rรฉgulariser une indicatrice tout en contrรดlant son support.
- The Leibniz rule and uniform bounds on a fixed compact set.Rรจgle de Leibniz et majorations uniformes sur un compact fixe.
Diagnostic questions
Where this chapter sits
Later dependence. Everything downstream is dual to this chapter. A distribution (Ch. 2) is a linear functional on \(\mathcal D(\Omega)\) that is continuous for the convergence defined here; the delicacy of that convergence is exactly what makes the dual space so rich. Mollifiers reappear as the engine of approximation and of distributional convergence (Ch. 6); cutoffs and partitions of unity localize distributions and define their support (Ch. 5); multiplication by \(C^\infty\) functions and \(\partial^\alpha\), shown continuous here, are transposed to act on distributions (Ch. 4). The preview of the Schwartz space \(\mathcal S\) at the end points to tempered distributions and the Fourier transform.Dรฉpendances ultรฉrieures. Tout ce qui suit repose sur la dualitรฉ avec ce chapitre. Une distribution (ch. 2) est une forme linรฉaire sur \(\mathcal D(\Omega)\), continue pour la convergence dรฉfinie ici ; la finesse de cette convergence explique prรฉcisรฉment la richesse de lโespace dual. Les mollificateurs rรฉapparaissent comme outils dโapproximation et de convergence des distributions (ch. 6) ; les fonctions de coupure et les partitions de lโunitรฉ localisent les distributions et permettent de dรฉfinir leur support (ch. 5) ; la multiplication par des fonctions \(C^\infty\) et les opรฉrateurs \(\partial^\alpha\), dont la continuitรฉ est รฉtablie ici, sont transposรฉs pour agir sur les distributions (ch. 4). Lโaperรงu de lโespace de Schwartz \(\mathcal S\) ร la fin du chapitre conduit aux distributions tempรฉrรฉes et ร la transformรฉe de Fourier.
Core Definitions
A multi-index is a tuple \(\alpha=(\alpha_1,\dots,\alpha_n)\in\mathbb N_0^n\) of nonnegative integers. Its length (or order) is \(|\alpha|=\alpha_1+\cdots+\alpha_n\), and its factorial is \(\alpha!=\alpha_1!\cdots\alpha_n!\). For \(x=(x_1,\dots,x_n)\in\mathbb R^n\) we write the monomial \[x^\alpha=x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha_n},\] and for a sufficiently differentiable \(f\) the partial derivative operator \[\partial^\alpha=\partial_1^{\alpha_1}\cdots\partial_n^{\alpha_n}=\frac{\partial^{|\alpha|}}{\partial x_1^{\alpha_1}\cdots\partial x_n^{\alpha_n}}.\] We order multi-indices by \(\beta\le\alpha\iff\beta_i\le\alpha_i\ \forall i\), and set \(\binom{\alpha}{\beta}=\frac{\alpha!}{\beta!(\alpha-\beta)!}=\prod_i\binom{\alpha_i}{\beta_i}\) for \(\beta\le\alpha\). For \(C^\infty\) functions the mixed partials commute (Schwarz), so \(\partial^\alpha\partial^\beta=\partial^{\alpha+\beta}\) and the order of the one-variable derivatives is immaterial.Un multi-indice est un n-uplet \(\alpha=(\alpha_1,\dots,\alpha_n)\in\mathbb N_0^n\) dโentiers naturels ou nuls. Sa longueur, ou son ordre, est \(|\alpha|=\alpha_1+\cdots+\alpha_n\), et sa factorielle est \(\alpha!=\alpha_1!\cdots\alpha_n!\). Pour \(x=(x_1,\dots,x_n)\in\mathbb R^n\), on note le monรดme \[x^\alpha=x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha_n},\] et, pour une fonction \(f\) suffisamment dรฉrivable, lโopรฉrateur de dรฉrivation partielle \[\partial^\alpha=\partial_1^{\alpha_1}\cdots\partial_n^{\alpha_n}=\frac{\partial^{|\alpha|}}{\partial x_1^{\alpha_1}\cdots\partial x_n^{\alpha_n}}.\] On ordonne les multi-indices par \(\beta\le\alpha\iff\beta_i\le\alpha_i\ \forall i\), et lโon pose \(\binom{\alpha}{\beta}=\frac{\alpha!}{\beta!(\alpha-\beta)!}=\prod_i\binom{\alpha_i}{\beta_i}\) pour \(\beta\le\alpha\). Pour les fonctions \(C^\infty\), les dรฉrivรฉes partielles mixtes commutent (thรฉorรจme de Schwarz), de sorte que \(\partial^\alpha\partial^\beta=\partial^{\alpha+\beta}\) et que lโordre des dรฉrivations unidimensionnelles est indiffรฉrent.
Take \(\alpha=(2,1)\) and \(f(x_1,x_2)=x_1^3x_2^2\). Then \(|\alpha|=3\), \(x^\alpha=x_1^2x_2\), and \[ \partial^\alpha f=\partial_1^2\partial_2(x_1^3x_2^2) =\partial_1^2(2x_1^3x_2)=12x_1x_2. \] This illustrates that the components of \(\alpha\), not only the total order \(|\alpha|\), determine which derivatives are taken.Prenons \(\alpha=(2,1)\) et \(f(x_1,x_2)=x_1^3x_2^2\). Alors \(|\alpha|=3\), \(x^\alpha=x_1^2x_2\), et \[\partial^\alpha f=\partial_1^2\partial_2(x_1^3x_2^2)=\partial_1^2(2x_1^3x_2)=12x_1x_2.\] Cet exemple montre que les composantes de \(\alpha\), et pas seulement lโordre total \(|\alpha|\), dรฉterminent les dรฉrivรฉes ร effectuer.
Let \(\Omega\subseteq\mathbb R^n\) be open and \(f:\Omega\to\mathbb C\) (or \(\mathbb R\)). The support of \(f\) is the closure, taken in \(\Omega\), of the set where \(f\) does not vanish: \[\operatorname{supp} f=\overline{\{x\in\Omega:f(x)\ne0\}}.\] By construction \(\operatorname{supp} f\) is a closed subset of \(\Omega\), and \(f\equiv0\) on the open set \(\Omega\setminus\operatorname{supp} f\). We say \(f\) has compact support (in \(\Omega\)) if \(\operatorname{supp} f\) is a compact subset of \(\Omega\); equivalently, \(f\) vanishes outside some compact \(K\subset\Omega\) and stays away from \(\partial\Omega\).Soit \(\Omega\subseteq\mathbb R^n\) un ouvert et \(f:\Omega\to\mathbb C\) (ou \(\mathbb R\)). Le support de \(f\) est lโadhรฉrence, prise dans \(\Omega\), de lโensemble des points oรน \(f\) ne sโannule pas : \[\operatorname{supp} f=\overline{\{x\in\Omega:f(x)\ne0\}}.\] Par construction, \(\operatorname{supp} f\) est un fermรฉ de \(\Omega\), et \(f\equiv0\) sur lโouvert \(\Omega\setminus\operatorname{supp} f\). On dit que \(f\) est ร support compact dans \(\Omega\) si \(\operatorname{supp} f\) est un compact contenu dans \(\Omega\) ; de maniรจre รฉquivalente, \(f\) sโannule en dehors dโun compact \(K\subset\Omega\) et son support reste ร distance positive de \(\partial\Omega\).
For \(f(x)=\max(0,1-|x|)\) on \(\mathbb R\), the nonzero set is \((-1,1)\). Therefore \[ \operatorname{supp}f=\overline{(-1,1)}=[-1,1]. \] The endpoints belong to the support even though \(f(\pm1)=0\), because every neighbourhood of either endpoint contains points where \(f\neq0\).Pour \(f(x)=\max(0,1-|x|)\) sur \(\mathbb R\), lโensemble oรน \(f\) ne sโannule pas est \((-1,1)\). Ainsi \[\operatorname{supp}f=\overline{(-1,1)}=[-1,1].\] Les extrรฉmitรฉs appartiennent au support bien que \(f(\pm1)=0\), car tout voisinage de lโune ou lโautre extrรฉmitรฉ contient des points oรน \(f\neq0\).
For \(\Omega\subseteq\mathbb R^n\) open, the space of test functions is \[\mathcal D(\Omega)=C_c^\infty(\Omega)=\{\varphi\in C^\infty(\Omega):\operatorname{supp}\varphi\text{ is compact and contained in }\Omega\}.\] Its elements are smooth functions that vanish identically outside some compact set (which may depend on \(\varphi\)) sitting strictly inside \(\Omega\). It is a complex (or real) vector space under pointwise operations. For a fixed compact \(K\subset\Omega\) we write \(\mathcal D_K(\Omega)=\{\varphi\in C^\infty(\Omega):\operatorname{supp}\varphi\subseteq K\}\), so that \(\mathcal D(\Omega)=\bigcup_{K}\mathcal D_K(\Omega)\), the union over all compact \(K\subset\Omega\).Pour un ouvert \(\Omega\subseteq\mathbb R^n\), lโespace des fonctions test est \[\mathcal D(\Omega)=C_c^\infty(\Omega)=\{\varphi\in C^\infty(\Omega):\operatorname{supp}\varphi\text{ est compact et contenu dans }\Omega\}.\] Ses รฉlรฉments sont des fonctions lisses qui sโannulent identiquement hors dโun compact, รฉventuellement dรฉpendant de \(\varphi\), situรฉ strictement ร lโintรฉrieur de \(\Omega\). Il sโagit dโun espace vectoriel complexe (ou rรฉel) pour les opรฉrations ponctuelles. Pour un compact fixรฉ \(K\subset\Omega\), on note \(\mathcal D_K(\Omega)=\{\varphi\in C^\infty(\Omega):\operatorname{supp}\varphi\subseteq K\}\), de sorte que \(\mathcal D(\Omega)=\bigcup_K\mathcal D_K(\Omega)\), lโunion portant sur tous les compacts \(K\subset\Omega\).
The standard bump \[ j(x)= \begin{cases} e^{-1/(1-x^2)},&|x|<1,\\ 0,&|x|\ge1 \end{cases} \] belongs to \(\mathcal D(\mathbb R)\): it is \(C^\infty\) and its support is the compact set \([-1,1]\). Theorem 1.1 proves the only delicate point, smoothness at \(x=\pm1\).La fonction bosse standard \[j(x)=\begin{cases}e^{-1/(1-x^2)},&|x|<1,\\0,&|x|\ge1\end{cases}\] appartient ร \(\mathcal D(\mathbb R)\) : elle est \(C^\infty\) et son support est le compact \([-1,1]\). Le thรฉorรจme 1.1 รฉtablit le seul point dรฉlicat, ร savoir la rรฉgularitรฉ en \(x=\pm1\).
For a compact set \(K\subset\Omega\), an integer \(N\ge0\), and \(\varphi\in C^\infty(\Omega)\), define \[p_{K,N}(\varphi)=\sup_{|\alpha|\le N}\ \sup_{x\in K}\bigl|\partial^\alpha\varphi(x)\bigr|.\] Each \(p_{K,N}\) is a seminorm: it is nonnegative, absolutely homogeneous \(\bigl(p_{K,N}(\lambda\varphi)=|\lambda|\,p_{K,N}(\varphi)\bigr)\), and subadditive \(\bigl(p_{K,N}(\varphi+\psi)\le p_{K,N}(\varphi)+p_{K,N}(\psi)\bigr)\). It measures the size of \(\varphi\) together with all its derivatives up to order \(N\), uniformly on \(K\). Restricted to \(\mathcal D_K(\Omega)\), the countable family \((p_{K,N})_{N\ge0}\) separates points and turns \(\mathcal D_K(\Omega)\) into a Frรฉchet space (a complete metrizable locally convex space).Pour un compact \(K\subset\Omega\), un entier \(N\ge0\) et \(\varphi\in C^\infty(\Omega)\), on dรฉfinit \[p_{K,N}(\varphi)=\sup_{|\alpha|\le N}\ \sup_{x\in K}|\partial^\alpha\varphi(x)|.\] Chaque \(p_{K,N}\) est une semi-norme : elle est positive, absolument homogรจne \(p_{K,N}(\lambda\varphi)=|\lambda|p_{K,N}(\varphi)\), et sous-additive \(p_{K,N}(\varphi+\psi)\le p_{K,N}(\varphi)+p_{K,N}(\psi)\). Elle mesure la taille de \(\varphi\) ainsi que celle de toutes ses dรฉrivรฉes jusquโร lโordre \(N\), uniformรฉment sur \(K\). Restreinte ร \(\mathcal D_K(\Omega)\), la famille dรฉnombrable \((p_{K,N})_{N\ge0}\) sรฉpare les points et munit \(\mathcal D_K(\Omega)\) dโune structure dโespace de Frรฉchet, cโest-ร -dire dโespace localement convexe complet et mรฉtrisable.
If \(K=[-1,1]\subset\mathbb R\) and \(N=1\), then \[ p_{K,1}(\varphi)=\max\!\left\{\sup_{x\in K}|\varphi(x)|,\ \sup_{x\in K}|\varphi'(x)|\right\}. \] Thus \(p_{K,1}\) controls both the height of \(\varphi\) and the height of its first derivative on the same compact set.Si \(K=[-1,1]\subset\mathbb R\) et \(N=1\), alors \[p_{K,1}(\varphi)=\max\!\left\{\sup_{x\in K}|\varphi(x)|,\ \sup_{x\in K}|\varphi'(x)|\right\}.\] Ainsi, \(p_{K,1}\) contrรดle ร la fois lโamplitude de \(\varphi\) et celle de sa premiรจre dรฉrivรฉe sur le mรชme compact.
Let \((\varphi_k)_{k\ge1}\) and \(\varphi\) lie in \(\mathcal D(\Omega)\). We say \(\varphi_k\to\varphi\) in \(\mathcal D(\Omega)\) when both conditions hold:Soient \((\varphi_k)_{k\ge1}\) et \(\varphi\) dans \(\mathcal D(\Omega)\). On dit que \(\varphi_k\to\varphi\) dans \(\mathcal D(\Omega)\) lorsque les deux conditions suivantes sont satisfaites :
- (i) Common compact support. There is a single compact set \(K\subset\Omega\) with \(\operatorname{supp}\varphi\subseteq K\) and \(\operatorname{supp}\varphi_k\subseteq K\) for every \(k\); and(i) Support compact commun. Il existe un compact unique \(K\subset\Omega\) tel que \(\operatorname{supp}\varphi\subseteq K\) et \(\operatorname{supp}\varphi_k\subseteq K\) pour tout \(k\).
- (ii) Uniform convergence of all derivatives. For every multi-index \(\alpha\), \(\ \partial^\alpha\varphi_k\to\partial^\alpha\varphi\) uniformly on \(\Omega\), i.e. \(\sup_{x}|\partial^\alpha\varphi_k(x)-\partial^\alpha\varphi(x)|\to0\).(ii) Convergence uniforme de toutes les dรฉrivรฉes. Pour tout multi-indice \(\alpha\), \(\partial^\alpha\varphi_k\to\partial^\alpha\varphi\) uniformรฉment sur \(\Omega\), cโest-ร -dire \(\sup_x|\partial^\alpha\varphi_k(x)-\partial^\alpha\varphi(x)|\to0\).
Equivalently, some compact \(K\) contains all the supports and \(p_{K,N}(\varphi_k-\varphi)\to0\) for every \(N\). A sequence is Cauchy in \(\mathcal D(\Omega)\) if the supports lie in one \(K\) and \(p_{K,N}(\varphi_j-\varphi_k)\to0\) for every \(N\) as \(j,k\to\infty\).De maniรจre รฉquivalente, il existe un compact \(K\) contenant tous les supports et tel que \(p_{K,N}(\varphi_k-\varphi)\to0\) pour tout \(N\). Une suite est de Cauchy dans \(\mathcal D(\Omega)\) si tous ses supports sont contenus dans un mรชme compact \(K\) et si \(p_{K,N}(\varphi_j-\varphi_k)\to0\) pour tout \(N\) lorsque \(j,k\to\infty\).
Let \(\rho\in\mathcal D(\mathbb R)\) and set \(\varphi_k=\rho/k\). All supports lie in the fixed compact set \(\operatorname{supp}\rho\), and for every \(m\ge0\), \[ \sup_x|\varphi_k^{(m)}(x)|=\frac1k\sup_x|\rho^{(m)}(x)|\longrightarrow0. \] Hence \(\varphi_k\to0\) in \(\mathcal D(\mathbb R)\).Soit \(\rho\in\mathcal D(\mathbb R)\) et posons \(\varphi_k=\rho/k\). Tous les supports sont contenus dans le compact fixe \(\operatorname{supp}\rho\), et pour tout \(m\ge0\), \[\sup_x|\varphi_k^{(m)}(x)|=\frac1k\sup_x|\rho^{(m)}(x)|\longrightarrow0.\] Par consรฉquent, \(\varphi_k\to0\) dans \(\mathcal D(\mathbb R)\).
The standard bump is \[j(x)=\begin{cases}\exp\!\bigl(-\tfrac{1}{1-|x|^2}\bigr),&|x|<1,\\0,&|x|\ge1,\end{cases}\qquad x\in\mathbb R^n,\] where \(|x|^2=x_1^2+\cdots+x_n^2\). Theorem 1.1 shows \(j\in\mathcal D(\mathbb R^n)\) with \(\operatorname{supp} j=\overline{B(0,1)}\). Normalizing, the standard mollifier is \[\rho=\frac{j}{\int_{\mathbb R^n}j},\qquad\text{so } \rho\ge0,\ \operatorname{supp}\rho=\overline{B(0,1)},\ \int_{\mathbb R^n}\rho=1,\] and for \(\varepsilon>0\) the rescaled mollifier is \(\rho_\varepsilon(x)=\varepsilon^{-n}\rho(x/\varepsilon)\), which satisfies \(\rho_\varepsilon\ge0\), \(\operatorname{supp}\rho_\varepsilon=\overline{B(0,\varepsilon)}\), and \(\int_{\mathbb R^n}\rho_\varepsilon=1\).La fonction bosse standard est \[j(x)=\begin{cases}\exp\!\bigl(-\tfrac{1}{1-|x|^2}\bigr),&|x|<1,\\0,&|x|\ge1,\end{cases}\qquad x\in\mathbb R^n,\] oรน \(|x|^2=x_1^2+\cdots+x_n^2\). Le thรฉorรจme 1.1 montre que \(j\in\mathcal D(\mathbb R^n)\) et \(\operatorname{supp}j=\overline{B(0,1)}\). Aprรจs normalisation, le mollificateur standard est \[\rho=\frac{j}{\int_{\mathbb R^n}j},\qquad \rho\ge0,\ \operatorname{supp}\rho=\overline{B(0,1)},\ \int_{\mathbb R^n}\rho=1,\] et, pour \(\varepsilon>0\), le mollificateur redimensionnรฉ est \(\rho_\varepsilon(x)=\varepsilon^{-n}\rho(x/\varepsilon)\). Il vรฉrifie \(\rho_\varepsilon\ge0\), \(\operatorname{supp}\rho_\varepsilon=\overline{B(0,\varepsilon)}\) et \(\int_{\mathbb R^n}\rho_\varepsilon=1\).
In one dimension, for \(\varepsilon=\tfrac12\), \[ \rho_{1/2}(x)=2\rho(2x),\qquad \operatorname{supp}\rho_{1/2}=[-\tfrac12,\tfrac12],\qquad \int_{\mathbb R}\rho_{1/2}(x)\,dx=1. \] The support is half as wide while the height is doubled, so the total mass remains \(1\).En dimension un, pour \(\varepsilon=\tfrac12\), \[\rho_{1/2}(x)=2\rho(2x),\qquad \operatorname{supp}\rho_{1/2}=[-\tfrac12,\tfrac12],\qquad \int_{\mathbb R}\rho_{1/2}(x)\,dx=1.\] Le support est deux fois plus รฉtroit tandis que la hauteur est doublรฉe, de sorte que la masse totale reste รฉgale ร \(1\).
Theorems & Proofs
Define \(f:\mathbb R\to\mathbb R\) by \(f(t)=e^{-1/t}\) for \(t>0\) and \(f(t)=0\) for \(t\le0\). Then \(f\in C^\infty(\mathbb R)\), and for every \(n\ge0\) one has \(f^{(n)}(0)=0\). Moreover, for \(t>0\) there is a polynomial \(P_n\) with \(f^{(n)}(t)=P_n(1/t)\,e^{-1/t}\).Dรฉfinissons \(f:\mathbb R\to\mathbb R\) par \(f(t)=e^{-1/t}\) pour \(t>0\) et \(f(t)=0\) pour \(t\le0\). Alors \(f\in C^\infty(\mathbb R)\) et, pour tout \(n\ge0\), \(f^{(n)}(0)=0\). De plus, pour \(t>0\), il existe un polynรดme \(P_n\) tel que \(f^{(n)}(t)=P_n(1/t)e^{-1/t}\).
Away from \(0\) the function is a composition of smooth maps, so the only issue is at \(t=0\). Compute the derivatives for \(t>0\) by induction, exhibiting the polynomial-in-\(1/t\) form. Then use the growth estimate \(s^m e^{-s}\to0\) as \(s\to+\infty\) to show every such expression tends to \(0\) at the origin, and feed this into the limit definition of the derivative to climb from \(f^{(n)}(0)=0\) to \(f^{(n+1)}(0)=0\).En dehors de \(0\), la fonction est une composition dโapplications lisses ; le seul point ร traiter est donc \(t=0\). Calculer les dรฉrivรฉes pour \(t>0\) par rรฉcurrence en mettant en รฉvidence une expression polynomiale en \(1/t\). Utiliser ensuite lโestimation de croissance \(s^m e^{-s}\to0\) lorsque \(s\to+\infty\) pour montrer que chacune de ces expressions tend vers \(0\) ร lโorigine, puis appliquer la dรฉfinition de la dรฉrivรฉe par limite afin de passer de \(f^{(n)}(0)=0\) ร \(f^{(n+1)}(0)=0\).
Smoothness for \(t\ne0\). On \((0,\infty)\), \(f\) is a composition of \(t\mapsto -1/t\) and \(\exp\), both \(C^\infty\); on \((-\infty,0)\), \(f\equiv0\). So \(f\) is \(C^\infty\) on \(\mathbb R\setminus\{0\}\).Rรฉgularitรฉ pour \(t\ne0\). Sur \((0,\infty)\), \(f\) est la composition de \(t\mapsto-1/t\) et de lโexponentielle, toutes deux \(C^\infty\) ; sur \(( -\infty,0)\), \(f\equiv0\). Ainsi, \(f\) est \(C^\infty\) sur \(\mathbb R\setminus\{0\}\).
The polynomial form. We show by induction that for \(t>0\), \(f^{(n)}(t)=P_n(1/t)e^{-1/t}\) with \(P_n\) a polynomial. For \(n=0\), \(P_0=1\). If it holds for \(n\), then writing \(s=1/t\) so \(\tfrac{d}{dt}=-s^2\tfrac{d}{ds}\), \[f^{(n+1)}(t)=\frac{d}{dt}\bigl[P_n(s)e^{-s}\bigr]=-s^2\bigl[P_n'(s)e^{-s}-P_n(s)e^{-s}\bigr]=\underbrace{s^2\bigl(P_n(s)-P_n'(s)\bigr)}_{=:P_{n+1}(s)}e^{-s},\] again a polynomial in \(s=1/t\) times \(e^{-1/t}\).Forme polynomiale. Montrons par rรฉcurrence que, pour \(t>0\), \(f^{(n)}(t)=P_n(1/t)e^{-1/t}\), oรน \(P_n\) est un polynรดme. Pour \(n=0\), \(P_0=1\). Supposons la propriรฉtรฉ vraie ร lโordre \(n\). En posant \(s=1/t\), de sorte que \(\tfrac d{dt}=-s^2\tfrac d{ds}\), on obtient \[f^{(n+1)}(t)=\frac d{dt}[P_n(s)e^{-s}]=-s^2[P_n'(s)e^{-s}-P_n(s)e^{-s}]=\underbrace{s^2(P_n(s)-P_n'(s))}_{=:P_{n+1}(s)}e^{-s},\] qui est encore un polynรดme en \(s=1/t\) multipliรฉ par \(e^{-1/t}\).
The growth estimate. For every integer \(m\ge0\), \(s^m e^{-s}\to0\) as \(s\to+\infty\) (the exponential dominates any power). Hence for each \(n\), as \(t\to0^+\) (so \(s=1/t\to+\infty\)), \[f^{(n)}(t)=P_n(1/t)e^{-1/t}\longrightarrow0.\]Estimation de croissance. Pour tout entier \(m\ge0\), \(s^m e^{-s}\to0\) lorsque \(s\to+\infty\), car lโexponentielle domine toute puissance. Ainsi, pour chaque \(n\), lorsque \(t\to0^+\) et donc \(s=1/t\to+\infty\), \[f^{(n)}(t)=P_n(1/t)e^{-1/t}\longrightarrow0.\]
Derivatives at the origin vanish, by induction on \(n\). Base case: \(f(0)=0\) and \(f\) is continuous at \(0\) since \(f(t)\to0\) as \(t\to0^+\) and \(f\equiv0\) for \(t\le0\). Inductive step: assume \(f^{(n)}(0)=0\) and \(f^{(n)}\) is continuous at \(0\). The left derivative of \(f^{(n)}\) at \(0\) is \(0\) (as \(f\equiv0\) there). For the right derivative, \[\lim_{t\to0^+}\frac{f^{(n)}(t)-f^{(n)}(0)}{t}=\lim_{t\to0^+}\frac{P_n(1/t)e^{-1/t}}{t}=\lim_{s\to+\infty}s\,P_n(s)e^{-s}=0,\] again by the growth estimate (the polynomial \(sP_n(s)\) is beaten by \(e^{-s}\)). Both one-sided derivatives equal \(0\), so \(f^{(n+1)}(0)=0\); and \(f^{(n+1)}\) is continuous at \(0\) because \(f^{(n+1)}(t)=P_{n+1}(1/t)e^{-1/t}\to0=f^{(n+1)}(0)\). The induction closes, so \(f\in C^\infty(\mathbb R)\) with all derivatives vanishing at \(0\). โLes dรฉrivรฉes ร lโorigine sโannulent, par rรฉcurrence sur \(n\). Initialisation : \(f(0)=0\) et \(f\) est continue en \(0\), car \(f(t)\to0\) lorsque \(t\to0^+\) et \(f\equiv0\) pour \(t\le0\). Hรฉrรฉditรฉ : supposons \(f^{(n)}(0)=0\) et \(f^{(n)}\) continue en \(0\). La dรฉrivรฉe ร gauche de \(f^{(n)}\) en \(0\) vaut \(0\). ร droite, \[\lim_{t\to0^+}\frac{f^{(n)}(t)-f^{(n)}(0)}{t}=\lim_{t\to0^+}\frac{P_n(1/t)e^{-1/t}}{t}=\lim_{s\to+\infty}sP_n(s)e^{-s}=0,\] encore par lโestimation de croissance. Les deux dรฉrivรฉes unilatรฉrales valent donc \(0\), ce qui donne \(f^{(n+1)}(0)=0\). De plus \(f^{(n+1)}\) est continue en \(0\), puisque \(P_{n+1}(1/t)e^{-1/t}\to0=f^{(n+1)}(0)\). La rรฉcurrence est achevรฉe : \(f\in C^\infty(\mathbb R)\) et toutes ses dรฉrivรฉes sโannulent en \(0\). โ
Depends on: elementary growth \(s^m e^{-s}\to0\); the limit definition of the derivative. Used by: Thm. 1.1 (bump), Def. 1.6, Ex. 1.8โ1.9.Dรฉpend de : croissance รฉlรฉmentaire \(s^m e^{-s}\to0\) ; dรฉfinition de la dรฉrivรฉe par limite. Utilisรฉ dans : th. 1.1 (fonction bosse), dรฉf. 1.6, ex. 1.8-1.9.
The function \(j(x)=e^{-1/(1-|x|^2)}\) for \(|x|<1\) and \(j(x)=0\) for \(|x|\ge1\) belongs to \(C_c^\infty(\mathbb R^n)=\mathcal D(\mathbb R^n)\), with \(j>0\) on the open unit ball and \(\operatorname{supp} j=\overline{B(0,1)}\). Consequently \(\rho=j/\int j\) and \(\rho_\varepsilon(x)=\varepsilon^{-n}\rho(x/\varepsilon)\) lie in \(\mathcal D(\mathbb R^n)\) with the properties listed in Definition 1.6.La fonction \(j(x)=e^{-1/(1-|x|^2)}\) pour \(|x|<1\), et \(j(x)=0\) pour \(|x|\ge1\), appartient ร \(C_c^\infty(\mathbb R^n)=\mathcal D(\mathbb R^n)\), avec \(j>0\) sur la boule unitรฉ ouverte et \(\operatorname{supp}j=\overline{B(0,1)}\). Par consรฉquent, \(\rho=j/\int j\) et \(\rho_\varepsilon(x)=\varepsilon^{-n}\rho(x/\varepsilon)\) appartiennent ร \(\mathcal D(\mathbb R^n)\) et vรฉrifient les propriรฉtรฉs รฉnoncรฉes dans la dรฉfinition 1.6.
Write \(j=f\circ g\) with \(g(x)=1-|x|^2\) a polynomial and \(f\) the flatness function of Lemma 1.1: indeed \(1-|x|^2>0\iff|x|<1\), and \(f(1-|x|^2)=e^{-1/(1-|x|^2)}\) there, \(0\) elsewhere. Smoothness of \(j\) is then the chain rule applied to two \(C^\infty\) maps; compactness of the support is boundedness of the closed ball. Rescaling is a change of variables.รcrire \(j=f\circ g\) avec \(g(x)=1-|x|^2\), qui est un polynรดme, et \(f\) la fonction plate du lemme 1.1. En effet, \(1-|x|^2>0\iff|x|<1\), et \(f(1-|x|^2)=e^{-1/(1-|x|^2)}\) dans la boule, tandis que la valeur est nulle ร lโextรฉrieur. La rรฉgularitรฉ de \(j\) rรฉsulte alors de la rรจgle de la chaรฎne appliquรฉe ร deux applications \(C^\infty\). La compacitรฉ du support provient du caractรจre bornรฉ de la boule fermรฉe. Le changement dโรฉchelle se traite par changement de variables.
\(j\) is a composition. The map \(g:\mathbb R^n\to\mathbb R\), \(g(x)=1-|x|^2=1-\sum_i x_i^2\), is a polynomial, hence \(C^\infty\), and \(g(x)>0\) exactly when \(|x|<1\). With \(f\) as in Lemma 1.1 we have, for all \(x\), \[j(x)=f\bigl(g(x)\bigr)=\begin{cases}e^{-1/(1-|x|^2)},&|x|<1\ \ (g(x)>0),\\ 0,&|x|\ge1\ \ (g(x)\le0).\end{cases}\] Since \(f\in C^\infty(\mathbb R)\) (Lemma 1.1) and \(g\in C^\infty(\mathbb R^n)\), the composition \(j=f\circ g\) is \(C^\infty(\mathbb R^n)\) by the chain rule. (No special treatment of the sphere \(|x|=1\) is needed: \(f\) is genuinely smooth across the value \(g=0\), which is the whole point of Lemma 1.1.)\(j\) est une composition. Lโapplication \(g:\mathbb R^n\to\mathbb R\), \(g(x)=1-|x|^2=1-\sum_i x_i^2\), est un polynรดme, donc \(C^\infty\), et \(g(x)>0\) exactement lorsque \(|x|<1\). Avec \(f\) comme dans le lemme 1.1, on a, pour tout \(x\), \[j(x)=f(g(x))=\begin{cases}e^{-1/(1-|x|^2)},&|x|<1\ (g(x)>0),\\0,&|x|\ge1\ (g(x)\le0).\end{cases}\] Comme \(f\in C^\infty(\mathbb R)\) et \(g\in C^\infty(\mathbb R^n)\), la rรจgle de la chaรฎne donne \(j=f\circ g\in C^\infty(\mathbb R^n)\). Aucun traitement sรฉparรฉ de la sphรจre \(|x|=1\) nโest nรฉcessaire : la fonction \(f\) est rรฉellement lisse au passage de la valeur \(g=0\), ce qui constitue prรฉcisรฉment le contenu du lemme 1.1.
Support. Since \(f(s)>0\) iff \(s>0\), we have \(j(x)>0\) iff \(|x|<1\). Thus \(\{j\ne0\}=B(0,1)\), and \(\operatorname{supp} j=\overline{B(0,1)}\), the closed unit ball. This set is closed and bounded in \(\mathbb R^n\), hence compact (HeineโBorel), and it is contained in \(\Omega=\mathbb R^n\). Therefore \(j\in C_c^\infty(\mathbb R^n)\).Support. Comme \(f(s)>0\) si et seulement si \(s>0\), on a \(j(x)>0\) si et seulement si \(|x|<1\). Ainsi \(\{j\ne0\}=B(0,1)\) et \(\operatorname{supp}j=\overline{B(0,1)}\), la boule unitรฉ fermรฉe. Cet ensemble est fermรฉ et bornรฉ dans \(\mathbb R^n\), donc compact par Heine-Borel, et il est contenu dans \(\Omega=\mathbb R^n\). Par consรฉquent, \(j\in C_c^\infty(\mathbb R^n)\).
Normalization and rescaling. Because \(j\ge0\), \(j\not\equiv0\), and \(j\) is continuous with compact support, \(0<\int_{\mathbb R^n}j<\infty\); set \(\rho=j/\int j\), so \(\rho\ge0\), \(\int\rho=1\), \(\operatorname{supp}\rho=\overline{B(0,1)}\). For \(\varepsilon>0\), \(\rho_\varepsilon(x)=\varepsilon^{-n}\rho(x/\varepsilon)\) is smooth, \(\rho_\varepsilon(x)\ne0\iff|x/\varepsilon|<1\iff|x|<\varepsilon\), so \(\operatorname{supp}\rho_\varepsilon=\overline{B(0,\varepsilon)}\); and the substitution \(y=x/\varepsilon\), \(dy=\varepsilon^{-n}dx\) gives \(\int\rho_\varepsilon(x)\,dx=\int\rho(y)\,dy=1\). โNormalisation et changement dโรฉchelle. Comme \(j\ge0\), \(j\not\equiv0\) et \(j\) est continue ร support compact, \(0<\int_{\mathbb R^n}j<\infty\). Posons \(\rho=j/\int j\). Alors \(\rho\ge0\), \(\int\rho=1\) et \(\operatorname{supp}\rho=\overline{B(0,1)}\). Pour \(\varepsilon>0\), \(\rho_\varepsilon(x)=\varepsilon^{-n}\rho(x/\varepsilon)\) est lisse ; \(\rho_\varepsilon(x)\ne0\iff|x|<\varepsilon\), donc \(\operatorname{supp}\rho_\varepsilon=\overline{B(0,\varepsilon)}\). Le changement de variables \(y=x/\varepsilon\), \(dy=\varepsilon^{-n}dx\), donne enfin \(\int\rho_\varepsilon(x)\,dx=\int\rho(y)\,dy=1\). โ
Depends on: Lem. 1.1, chain rule, HeineโBorel. Used by: Thm. 1.2 (cutoffs), Thm. 1.3 (partitions), Ch. 6 (mollification).Dรฉpend de : lem. 1.1, rรจgle de la chaรฎne, Heine-Borel. Utilisรฉ dans : th. 1.2 (fonctions de coupure), th. 1.3 (partitions), ch. 6 (mollification).
Let \(U\subseteq\mathbb R^n\) be open and \(K\subset U\) compact. Then there exists \(\varphi\in\mathcal D(U)\) with \[0\le\varphi\le1,\qquad \varphi\equiv1\text{ on a neighbourhood of }K,\qquad \operatorname{supp}\varphi\subset U.\]Soit \(U\subseteq\mathbb R^n\) un ouvert et \(K\subset U\) un compact. Il existe alors \(\varphi\in\mathcal D(U)\) telle que \[0\le\varphi\le1,\qquad \varphi\equiv1\text{ sur un voisinage de }K,\qquad \operatorname{supp}\varphi\subset U.\]
Fatten \(K\) slightly to a compact set still inside \(U\), take the indicator of that fattened set, and smooth it by convolving with a mollifier \(\rho_\delta\) whose radius \(\delta\) is smaller than the gaps involved. Convolution with a smooth compactly supported kernel produces a smooth function; the radii are chosen so the result is \(1\) on \(K\) and vanishes before reaching \(\partial U\).รpaissir lรฉgรจrement \(K\) en un compact restant contenu dans \(U\), prendre lโindicatrice de cet รฉpaississement puis la rรฉgulariser par convolution avec un mollificateur \(\rho_\delta\), dont le rayon \(\delta\) est choisi plus petit que les distances en jeu. La convolution avec un noyau lisse ร support compact produit une fonction lisse ; les rayons sont choisis de sorte que le rรฉsultat vaille \(1\) sur \(K\) et sโannule avant dโatteindre \(\partial U\).
Let \(d=\operatorname{dist}(K,\partial U)\); since \(K\) is compact and disjoint from the closed set \(\partial U\), \(d>0\) (interpret \(d=+\infty\) and pick \(d=1\) if \(U=\mathbb R^n\)). Fix \(\delta=d/4\) and set \[K_\delta=\{x:\operatorname{dist}(x,K)\le 2\delta\},\qquad \varphi=\mathbf 1_{K_\delta}*\rho_\delta,\quad\text{i.e. } \varphi(x)=\int_{\mathbb R^n}\mathbf 1_{K_\delta}(x-y)\,\rho_\delta(y)\,dy.\] Here \(K_\delta\) is compact and \(K_\delta\subset U\) because \(2\delta=d/2<d\).Posons \(d=\operatorname{dist}(K,\partial U)\). Comme \(K\) est compact et disjoint du fermรฉ \(\partial U\), on a \(d>0\). Si \(U=\mathbb R^n\), on peut interprรฉter \(d=+\infty\) et choisir simplement \(d=1\). Fixons \(\delta=d/4\) et posons \[K_\delta=\{x:\operatorname{dist}(x,K)\le2\delta\},\qquad \varphi=\mathbf1_{K_\delta}*\rho_\delta,\quad \varphi(x)=\int_{\mathbb R^n}\mathbf1_{K_\delta}(x-y)\rho_\delta(y)\,dy.\] Lโensemble \(K_\delta\) est compact et \(K_\delta\subset U\), puisque \(2\delta=d/2<d\).
Smoothness. Write \(\varphi(x)=\int\mathbf 1_{K_\delta}(z)\rho_\delta(x-z)\,dz\). The integrand is \(C^\infty\) in \(x\) with \(\partial^\alpha_x\rho_\delta(x-z)\) dominated, uniformly in \(x\) on any bounded set, by the integrable compactly supported bound \(\sup|\partial^\alpha\rho_\delta|\,\mathbf 1_{K_\delta}(z)\); differentiating under the integral gives \(\partial^\alpha\varphi=\mathbf 1_{K_\delta}*\partial^\alpha\rho_\delta\), so \(\varphi\in C^\infty\).Rรฉgularitรฉ. รcrivons \(\varphi(x)=\int\mathbf1_{K_\delta}(z)\rho_\delta(x-z)\,dz\). Lโintรฉgrande est \(C^\infty\) en \(x\), et \(\partial_x^\alpha\rho_\delta(x-z)\) est dominรฉe, uniformรฉment en \(x\) sur tout ensemble bornรฉ, par la fonction intรฉgrable ร support compact \(\sup|\partial^\alpha\rho_\delta|\,\mathbf1_{K_\delta}(z)\). On peut donc dรฉriver sous le signe intรฉgral : \(\partial^\alpha\varphi=\mathbf1_{K_\delta}*\partial^\alpha\rho_\delta\), dโoรน \(\varphi\in C^\infty\).
Range \(0\le\varphi\le1\). Since \(0\le\mathbf 1_{K_\delta}\le1\) and \(\rho_\delta\ge0\) with \(\int\rho_\delta=1\), we get \(0\le\varphi(x)\le\int\rho_\delta=1\).Encadrement \(0\le\varphi\le1\). Comme \(0\le\mathbf1_{K_\delta}\le1\), \(\rho_\delta\ge0\) et \(\int\rho_\delta=1\), on obtient \(0\le\varphi(x)\le\int\rho_\delta=1\).
\(\varphi\equiv1\) near \(K\). If \(\operatorname{dist}(x,K)\le\delta\) and \(|y|\le\delta\) (so \(\rho_\delta(y)\) may be nonzero), then \(\operatorname{dist}(x-y,K)\le\operatorname{dist}(x,K)+|y|\le2\delta\), hence \(x-y\in K_\delta\) and \(\mathbf 1_{K_\delta}(x-y)=1\). Therefore \(\varphi(x)=\int\rho_\delta(y)\,dy=1\) for all such \(x\): \(\varphi\equiv1\) on the open neighbourhood \(\{\operatorname{dist}(\cdot,K)<\delta\}\supset K\).\(\varphi\equiv1\) au voisinage de \(K\). Si \(\operatorname{dist}(x,K)\le\delta\) et \(|y|\le\delta\), alors \(\operatorname{dist}(x-y,K)\le\operatorname{dist}(x,K)+|y|\le2\delta\). Ainsi \(x-y\in K_\delta\) et \(\mathbf1_{K_\delta}(x-y)=1\). Par consรฉquent \(\varphi(x)=\int\rho_\delta(y)\,dy=1\) pour tout tel \(x\). Donc \(\varphi\equiv1\) sur le voisinage ouvert \(\{\operatorname{dist}(\cdot,K)<\delta\}\supset K\).
Support. If \(\varphi(x)\ne0\) then some \(y\) with \(|y|\le\delta\) has \(x-y\in K_\delta\), so \(\operatorname{dist}(x,K)\le2\delta+\delta=3\delta<d\). Thus \(\operatorname{supp}\varphi\subseteq\{\operatorname{dist}(\cdot,K)\le3\delta\}\), a compact subset of \(U\). Hence \(\varphi\in\mathcal D(U)\) with all the stated properties. โSupport. Si \(\varphi(x)\ne0\), il existe \(y\) avec \(|y|\le\delta\) et \(x-y\in K_\delta\). Alors \(\operatorname{dist}(x,K)\le2\delta+\delta=3\delta<d\). Ainsi \(\operatorname{supp}\varphi\subseteq\{\operatorname{dist}(\cdot,K)\le3\delta\}\), qui est un compact contenu dans \(U\). Par consรฉquent, \(\varphi\in\mathcal D(U)\) et toutes les propriรฉtรฉs annoncรฉes sont vรฉrifiรฉes. โ
Depends on: Thm. 1.1 (\(\rho_\delta\)); differentiation under the integral. Used by: Thm. 1.3, Ch. 5 (support & localization), Ex. 1.11โ1.12.Dรฉpend de : th. 1.1 (\(\rho_\delta\)) ; dรฉrivation sous le signe intรฉgral. Utilisรฉ dans : th. 1.3, ch. 5 (support et localisation), ex. 1.11-1.12.
Let \(K\subset\mathbb R^n\) be compact and let \(U_1,\dots,U_m\) be open sets with \(K\subseteq U_1\cup\cdots\cup U_m\). Then there exist \(\psi_1,\dots,\psi_m\) with \(\psi_i\in\mathcal D(U_i)\), \(0\le\psi_i\le1\), such that \[\sum_{i=1}^m\psi_i\equiv1\text{ on a neighbourhood of }K,\qquad \sum_{i=1}^m\psi_i\le1\text{ everywhere.}\] The family \(\{\psi_i\}\) is called a smooth partition of unity subordinate to \(\{U_i\}\) (over \(K\)).Soit \(K\subset\mathbb R^n\) compact et soient \(U_1,\dots,U_m\) des ouverts tels que \(K\subseteq U_1\cup\cdots\cup U_m\). Il existe alors \(\psi_1,\dots,\psi_m\) avec \(\psi_i\in\mathcal D(U_i)\), \(0\le\psi_i\le1\), telles que \[\sum_{i=1}^m\psi_i\equiv1\text{ sur un voisinage de }K,\qquad \sum_{i=1}^m\psi_i\le1\text{ partout}.\] La famille \(\{\psi_i\}\) est appelรฉe partition de lโunitรฉ lisse subordonnรฉe au recouvrement \(\{U_i\}\), au voisinage de \(K\).
Cover \(K\) by finitely many small closed balls, each contained in some \(U_i\), and put a bump on each. Sum the bumps assigned to a given \(U_i\) to get a nonnegative \(\theta_i\in\mathcal D(U_i)\); the total \(\theta=\sum\theta_i\) is strictly positive on a neighbourhood of \(K\). Normalize by \(\theta\); but cut off with a fixed cutoff first so the quotient is smooth even where \(\theta\) vanishes.Recouvrir \(K\) par un nombre fini de petites boules fermรฉes, chacune contenue dans un certain \(U_i\), et placer une fonction bosse sur chacune. Additionner les bosses associรฉes ร un mรชme \(U_i\) afin dโobtenir une fonction non nรฉgative \(\theta_i\in\mathcal D(U_i)\). La somme \(\theta=\sum\theta_i\) est strictement positive sur un voisinage de \(K\). On normalise ensuite par \(\theta\), aprรจs avoir introduit une fonction de coupure fixe afin que le quotient reste lisse mรชme lร oรน \(\theta\) sโannule.
Local bumps. For each \(x\in K\) choose \(i(x)\) with \(x\in U_{i(x)}\) and a radius \(r_x>0\) with \(\overline{B(x,2r_x)}\subset U_{i(x)}\). The balls \(\{B(x,r_x)\}_{x\in K}\) cover the compact \(K\), so finitely many \(B(x_1,r_{1}),\dots,B(x_p,r_{p})\) do. Apply Theorem 1.2 with the compact set \(\overline{B(x_j,r_j)}\) and the open set \(B(x_j,2r_j)\). This gives \(\chi_j\in\mathcal D(B(x_j,2r_j))\subset\mathcal D(U_{i(x_j)})\) with \(0\le\chi_j\le1\), \(\chi_j\equiv1\) on a neighbourhood of \(\overline{B(x_j,r_j)}\), and \(\operatorname{supp}\chi_j\subset B(x_j,2r_j)\subset U_{i(x_j)}\).Bosses locales. Pour chaque \(x\in K\), choisissons \(i(x)\) tel que \(x\in U_{i(x)}\) et un rayon \(r_x>0\) tel que \(\overline{B(x,2r_x)}\subset U_{i(x)}\). Les boules \(B(x,r_x)\) recouvrent le compact \(K\) ; un sous-recouvrement fini \(B(x_1,r_1),\dots,B(x_p,r_p)\) suffit. Appliquons le thรฉorรจme 1.2 au compact \(\overline{B(x_j,r_j)}\) et ร lโouvert \(B(x_j,2r_j)\). On obtient \(\chi_j\in\mathcal D(B(x_j,2r_j))\subset\mathcal D(U_{i(x_j)})\), avec \(0\le\chi_j\le1\), \(\chi_j\equiv1\) sur un voisinage de \(\overline{B(x_j,r_j)}\), et \(\operatorname{supp}\chi_j\subset B(x_j,2r_j)\subset U_{i(x_j)}\).
Group by index. For \(i=1,\dots,m\) set \(\theta_i=\sum_{j:\,i(x_j)=i}\chi_j\) (empty sum \(=0\)). Then \(\theta_i\in\mathcal D(U_i)\), \(\theta_i\ge0\), and \(\theta:=\sum_{i}\theta_i=\sum_{j}\chi_j\). On \(K\), each point lies in some \(B(x_j,r_j)\), where \(\chi_j=1\), so \(\theta\ge1>0\); by continuity \(\theta>0\) on an open set \(W\supseteq K\).Regroupement par indice. Pour \(i=1,\dots,m\), posons \(\theta_i=\sum_{j:\,i(x_j)=i}\chi_j\), la somme vide valant \(0\). Alors \(\theta_i\in\mathcal D(U_i)\), \(\theta_i\ge0\), et \(\theta:=\sum_i\theta_i=\sum_j\chi_j\). Sur \(K\), chaque point appartient ร une boule \(B(x_j,r_j)\) sur laquelle \(\chi_j=1\). Ainsi \(\theta\ge1>0\) sur \(K\), et, par continuitรฉ, \(\theta>0\) sur un ouvert \(W\supseteq K\).
Normalize. By Theorem 1.2 choose a cutoff \(\zeta\in\mathcal D(W)\) with \(0\le\zeta\le1\) and \(\zeta\equiv1\) on a neighbourhood \(V\) of \(K\). Define \[\psi_i(x)=\begin{cases}\dfrac{\zeta(x)\,\theta_i(x)}{\theta(x)},&x\in W,\\0,&x\notin W.\end{cases}\] On \(W\), \(\theta>0\) so \(\psi_i\) is smooth. Near \(\partial W\) inside \(W\), \(\zeta=0\) (since \(\operatorname{supp}\zeta\) is a compact subset of \(W\)), so \(\psi_i\) extends by \(0\) to a smooth function on \(\mathbb R^n\). Moreover \(\operatorname{supp}\psi_i\subseteq\operatorname{supp}\theta_i\subseteq U_i\), so \(\psi_i\in\mathcal D(U_i)\); and \(0\le\psi_i\le\zeta\,\theta_i/\theta\le1\).Normalisation. Par le thรฉorรจme 1.2, choisissons une fonction de coupure \(\zeta\in\mathcal D(W)\) telle que \(0\le\zeta\le1\) et \(\zeta\equiv1\) sur un voisinage \(V\) de \(K\). Dรฉfinissons \[\psi_i(x)=\begin{cases}\dfrac{\zeta(x)\theta_i(x)}{\theta(x)},&x\in W,\\0,&x\notin W.\end{cases}\] Sur \(W\), \(\theta>0\), donc \(\psi_i\) est lisse. Prรจs de \(\partial W\), ร lโintรฉrieur de \(W\), \(\zeta=0\), puisque son support est un compact contenu dans \(W\). Ainsi \(\psi_i\) se prolonge par \(0\) en une fonction lisse sur \(\mathbb R^n\). De plus, \(\operatorname{supp}\psi_i\subseteq\operatorname{supp}\theta_i\subseteq U_i\), donc \(\psi_i\in\mathcal D(U_i)\), et \(0\le\psi_i\le\zeta\theta_i/\theta\le1\).
They sum to \(1\) near \(K\). On \(W\), \(\sum_i\psi_i=\zeta\cdot\dfrac{\sum_i\theta_i}{\theta}=\zeta\cdot\dfrac{\theta}{\theta}=\zeta\), and \(\zeta\equiv1\) on \(V\supseteq K\). Off \(W\) all \(\psi_i=0\le1\). Everywhere \(\sum_i\psi_i=\zeta\le1\). This is the required partition of unity. โLa somme vaut \(1\) au voisinage de \(K\). Sur \(W\), \(\sum_i\psi_i=\zeta\,\frac{\sum_i\theta_i}{\theta}=\zeta\,\frac\theta\theta=\zeta\), et \(\zeta\equiv1\) sur \(V\supseteq K\). Hors de \(W\), toutes les \(\psi_i\) sont nulles. Partout, \(\sum_i\psi_i=\zeta\le1\). On obtient donc la partition de lโunitรฉ recherchรฉe. โ
Depends on: Thm. 1.2, Thm. 1.1; compactness of \(K\). Used by: Ch. 2 (gluing distributions), Ch. 5 (support), Ex. 1.21โ1.22.Dรฉpend de : th. 1.2, th. 1.1 ; compacitรฉ de \(K\). Utilisรฉ dans : ch. 2 (recollement des distributions), ch. 5 (support), ex. 1.21-1.22.
\(\mathcal D(\Omega)\) is a vector space under pointwise addition and scalar multiplication. Moreover:\(\mathcal D(\Omega)\) est un espace vectoriel pour lโaddition ponctuelle et la multiplication scalaire. De plus :
- (a) For each multi-index \(\alpha\), the map \(\partial^\alpha:\mathcal D(\Omega)\to\mathcal D(\Omega)\) is well-defined and sequentially continuous: if \(\varphi_k\to\varphi\) in \(\mathcal D(\Omega)\), then \(\partial^\alpha\varphi_k\to\partial^\alpha\varphi\) in \(\mathcal D(\Omega)\).(a) Pour tout multi-indice \(\alpha\), lโapplication \(\partial^\alpha:\mathcal D(\Omega)\to\mathcal D(\Omega)\) est bien dรฉfinie et sรฉquentiellement continue : si \(\varphi_k\to\varphi\) dans \(\mathcal D(\Omega)\), alors \(\partial^\alpha\varphi_k\to\partial^\alpha\varphi\) dans \(\mathcal D(\Omega)\).
- (b) For each \(\psi\in C^\infty(\Omega)\), the map \(\varphi\mapsto\psi\varphi\) sends \(\mathcal D(\Omega)\) into \(\mathcal D(\Omega)\) and is sequentially continuous.(b) Pour toute \(\psi\in C^\infty(\Omega)\), lโapplication \(\varphi\mapsto\psi\varphi\) envoie \(\mathcal D(\Omega)\) dans \(\mathcal D(\Omega)\) et est sรฉquentiellement continue.
Both maps preserve "support inside a fixed \(K\)", so condition (i) of Definition 1.5 is automatic. For (ii) note that differentiating a sequence that converges in every derivative again converges in every derivative; one merely relabels multi-indices. For multiplication, expand \(\partial^\beta(\psi\varphi_k)\) by the Leibniz rule and bound each factor uniformly on \(K\), where \(\psi\) and its derivatives are bounded by continuity.Les deux applications prรฉservent la propriรฉtรฉ ยซ support contenu dans un compact fixe \(K\) ยป, de sorte que la condition (i) de la dรฉfinition 1.5 est automatique. Pour (ii), dรฉriver une suite dont toutes les dรฉrivรฉes convergent revient simplement ร rรฉindexer les multi-indices. Pour la multiplication, dรฉvelopper \(\partial^\beta(\psi\varphi_k)\) ร lโaide de la rรจgle de Leibniz et majorer uniformรฉment chaque facteur sur \(K\), oรน \(\psi\) et ses dรฉrivรฉes sont bornรฉes par continuitรฉ.
Vector space. If \(\varphi,\psi\in\mathcal D(\Omega)\) and \(\lambda\) is a scalar, then \(\varphi+\lambda\psi\in C^\infty(\Omega)\) and \(\operatorname{supp}(\varphi+\lambda\psi)\subseteq\operatorname{supp}\varphi\cup\operatorname{supp}\psi\), a compact subset of \(\Omega\); so \(\mathcal D(\Omega)\) is closed under the operations and is a vector space.Espace vectoriel. Si \(\varphi,\psi\in\mathcal D(\Omega)\) et \(\lambda\) est un scalaire, alors \(\varphi+\lambda\psi\in C^\infty(\Omega)\) et \(\operatorname{supp}(\varphi+\lambda\psi)\subseteq\operatorname{supp}\varphi\cup\operatorname{supp}\psi\), qui est un compact contenu dans \(\Omega\). Ainsi \(\mathcal D(\Omega)\) est stable par ces opรฉrations et constitue un espace vectoriel.
(a) Continuity of \(\partial^\alpha\). First, \(\operatorname{supp}(\partial^\alpha\varphi)\subseteq\operatorname{supp}\varphi\): if \(\varphi\equiv0\) on an open set, all its derivatives vanish there. So \(\partial^\alpha\varphi\in\mathcal D(\Omega)\). Now suppose \(\varphi_k\to\varphi\) in \(\mathcal D(\Omega)\): there is a compact \(K\) with all \(\operatorname{supp}\varphi_k\subseteq K\), and \(p_{K,N}(\varphi_k-\varphi)\to0\) for every \(N\). Then \(\operatorname{supp}(\partial^\alpha\varphi_k)\subseteq K\) as well, and for any multi-index \(\beta\), \[\sup_K\bigl|\partial^\beta(\partial^\alpha\varphi_k)-\partial^\beta(\partial^\alpha\varphi)\bigr|=\sup_K\bigl|\partial^{\alpha+\beta}(\varphi_k-\varphi)\bigr|\le p_{K,|\alpha|+|\beta|}(\varphi_k-\varphi)\to0.\] Thus every derivative of \(\partial^\alpha\varphi_k\) converges uniformly to that of \(\partial^\alpha\varphi\), and the supports stay in \(K\): \(\partial^\alpha\varphi_k\to\partial^\alpha\varphi\) in \(\mathcal D(\Omega)\).(a) Continuitรฉ de \(\partial^\alpha\). Dโabord, \(\operatorname{supp}(\partial^\alpha\varphi)\subseteq\operatorname{supp}\varphi\) : si \(\varphi\equiv0\) sur un ouvert, toutes ses dรฉrivรฉes y sont nulles. Donc \(\partial^\alpha\varphi\in\mathcal D(\Omega)\). Supposons maintenant \(\varphi_k\to\varphi\) dans \(\mathcal D(\Omega)\). Il existe un compact \(K\) tel que tous les \(\operatorname{supp}\varphi_k\subseteq K\), et \(p_{K,N}(\varphi_k-\varphi)\to0\) pour tout \(N\). Alors \(\operatorname{supp}(\partial^\alpha\varphi_k)\subseteq K\), et pour tout multi-indice \(\beta\), \[\sup_K|\partial^\beta(\partial^\alpha\varphi_k)-\partial^\beta(\partial^\alpha\varphi)|=\sup_K|\partial^{\alpha+\beta}(\varphi_k-\varphi)|\le p_{K,|\alpha|+|\beta|}(\varphi_k-\varphi)\to0.\] Toutes les dรฉrivรฉes de \(\partial^\alpha\varphi_k\) convergent donc uniformรฉment vers celles de \(\partial^\alpha\varphi\), et les supports restent dans \(K\). Ainsi \(\partial^\alpha\varphi_k\to\partial^\alpha\varphi\) dans \(\mathcal D(\Omega)\).
(b) Continuity of \(\varphi\mapsto\psi\varphi\). For \(\psi\in C^\infty(\Omega)\) and \(\varphi\in\mathcal D(\Omega)\), \(\psi\varphi\in C^\infty(\Omega)\) and \(\operatorname{supp}(\psi\varphi)\subseteq\operatorname{supp}\varphi\), compact; so \(\psi\varphi\in\mathcal D(\Omega)\). Let \(\varphi_k\to\varphi\) in \(\mathcal D(\Omega)\) with supports in a fixed compact \(K\). The Leibniz rule gives, for any \(\beta\), \[\partial^\beta\bigl(\psi(\varphi_k-\varphi)\bigr)=\sum_{\gamma\le\beta}\binom{\beta}{\gamma}\,\partial^\gamma\psi\,\partial^{\beta-\gamma}(\varphi_k-\varphi).\] Since \(\psi\in C^\infty\), each \(\partial^\gamma\psi\) is continuous, hence bounded on the compact \(K\): put \(M_\beta=\max_{\gamma\le\beta}\sup_K|\partial^\gamma\psi|<\infty\). Because \(\psi(\varphi_k-\varphi)\) is supported in \(K\), taking \(\sup_K\) and using subadditivity, \[p_{K,N}\bigl(\psi\varphi_k-\psi\varphi\bigr)\le C_N\,M\,p_{K,N}(\varphi_k-\varphi)\xrightarrow[k\to\infty]{}0,\] where \(C_N=\max_{|\beta|\le N}\sum_{\gamma\le\beta}\binom{\beta}{\gamma}\) and \(M=\max_{|\beta|\le N}M_\beta\) are finite constants. The supports of \(\psi\varphi_k\) lie in \(K\), so \(\psi\varphi_k\to\psi\varphi\) in \(\mathcal D(\Omega)\). โ(b) Continuitรฉ de \(\varphi\mapsto\psi\varphi\). Pour \(\psi\in C^\infty(\Omega)\) et \(\varphi\in\mathcal D(\Omega)\), le produit \(\psi\varphi\) est \(C^\infty\) et \(\operatorname{supp}(\psi\varphi)\subseteq\operatorname{supp}\varphi\), qui est compact ; donc \(\psi\varphi\in\mathcal D(\Omega)\). Soit \(\varphi_k\to\varphi\) dans \(\mathcal D(\Omega)\), avec les supports contenus dans un compact fixe \(K\). La rรจgle de Leibniz donne, pour tout \(\beta\), \[\partial^\beta(\psi(\varphi_k-\varphi))=\sum_{\gamma\le\beta}\binom{\beta}{\gamma}\partial^\gamma\psi\,\partial^{\beta-\gamma}(\varphi_k-\varphi).\] Chaque \(\partial^\gamma\psi\) est continue, donc bornรฉe sur \(K\). En posant \(M_\beta=\max_{\gamma\le\beta}\sup_K|\partial^\gamma\psi|\), puis en prenant les supremums et en utilisant la sous-additivitรฉ, on obtient \[p_{K,N}(\psi\varphi_k-\psi\varphi)\le C_N M\,p_{K,N}(\varphi_k-\varphi)\to0,\] oรน \(C_N\) et \(M\) sont des constantes finies. Les supports de \(\psi\varphi_k\) restent dans \(K\), donc \(\psi\varphi_k\to\psi\varphi\) dans \(\mathcal D(\Omega)\). โ
Depends on: Def. 1.3โ1.5, Leibniz rule (Ex. 1.3). Used by: Ch. 4 (\(\partial^\alpha\) on distributions), Ch. 3 (\(C^\infty\)-module structure).Dรฉpend de : dรฉf. 1.3-1.5, rรจgle de Leibniz (ex. 1.3). Utilisรฉ dans : ch. 4 (\(\partial^\alpha\) sur les distributions), ch. 3 (structure de module sur \(C^\infty\)).
Choose a compact exhaustion \[ K_1\subset \operatorname{int}K_2\subset K_2\subset \operatorname{int}K_3\subset\cdots, \qquad \bigcup_{j=1}^{\infty}K_j=\Omega. \] Equip each \(\mathcal D_{K_j}(\Omega)\) with the Frรฉchet topology generated by the seminorms \((p_{K_j,N})_{N\ge0}\). Then \[ \mathcal D(\Omega)=\varinjlim_{j\to\infty}\mathcal D_{K_j}(\Omega) \] with its standard strict LF-space topology. A subset \(B\subseteq\mathcal D(\Omega)\) is bounded if and only if there is a single compact \(K\subset\Omega\) containing the support of every \(\varphi\in B\), and for every \(N\), \[ \sup_{\varphi\in B}p_{K,N}(\varphi)<\infty. \] Consequently, for every nonempty open \(\Omega\subseteq\mathbb R^n\), \(\mathcal D(\Omega)\) is not normable and is not metrizable in its LF topology. The full topological proof and completeness of the LF-space are deferred to the functional-analysis appendix; the present chapter uses the concrete sequential convergence of Definition 1.5.Choisissons une exhaustion compacte \[K_1\subset\operatorname{int}K_2\subset K_2\subset\operatorname{int}K_3\subset\cdots,\qquad \bigcup_{j=1}^{\infty}K_j=\Omega.\] Munissons chaque \(\mathcal D_{K_j}(\Omega)\) de la topologie de Frรฉchet engendrรฉe par les semi-normes \((p_{K_j,N})_{N\ge0}\). Alors \[\mathcal D(\Omega)=\varinjlim_{j\to\infty}\mathcal D_{K_j}(\Omega)\] pour sa topologie standard dโespace LF strict. Un sous-ensemble \(B\subseteq\mathcal D(\Omega)\) est bornรฉ si et seulement sโil existe un compact unique \(K\subset\Omega\) contenant le support de toute \(\varphi\in B\), et si, pour tout \(N\), \[\sup_{\varphi\in B}p_{K,N}(\varphi)<\infty.\] En consรฉquence, pour tout ouvert non vide \(\Omega\subseteq\mathbb R^n\), lโespace \(\mathcal D(\Omega)\) nโest ni normable ni mรฉtrisable pour sa topologie LF. La dรฉmonstration topologique complรจte et la complรฉtude de lโespace LF sont reportรฉes ร lโappendice dโanalyse fonctionnelle ; le prรฉsent chapitre utilise la convergence sรฉquentielle concrรจte de la dรฉfinition 1.5.
Two independent requirements. Definition 1.5 asks for both a common compact support and uniform convergence of every derivative. Dropping either wrecks it:Deux exigences indรฉpendantes. La dรฉfinition 1.5 impose ร la fois un support compact commun et la convergence uniforme de toutes les dรฉrivรฉes. Supprimer lโune ou lโautre fait รฉchouer la convergence dans \(\mathcal D\) :
- Supports must not escape. Let \(\varphi_k(x)=\tfrac1k\,\rho(x-k)\) on \(\mathbb R\). Then \(\|\varphi_k\|_\infty=\tfrac1k\sup\rho\to0\), so \(\varphi_k\to0\) uniformly; every derivative also \(\to0\) uniformly. Yet \(\operatorname{supp}\varphi_k=\overline{B(k,1)}=[k-1,k+1]\) marches off to \(+\infty\): there is no compact \(K\) containing all of them. So \(\varphi_k\not\to0\) in \(\mathcal D(\mathbb R)\). A bump marching to infinity does not converge in \(\mathcal D\), however small its amplitude.Les supports ne doivent pas sโรฉchapper. Soit \(\varphi_k(x)=\tfrac1k\rho(x-k)\) sur \(\mathbb R\). Alors \(\|\varphi_k\|_\infty=\tfrac1k\sup\rho\to0\), et toutes les dรฉrivรฉes tendent รฉgalement uniformรฉment vers \(0\). Pourtant, \(\operatorname{supp}\varphi_k=[k-1,k+1]\) se dรฉplace vers \(+\infty\) : aucun compact \(K\) ne contient tous les supports. Donc \(\varphi_k\not\to0\) dans \(\mathcal D(\mathbb R)\). Une bosse qui part vers lโinfini ne converge pas dans \(\mathcal D\), aussi petite soit son amplitude.
- Every derivative must converge. Let \(\varphi_k(x)=\tfrac1k\,\rho(kx)\). The supports \([-1/k,1/k]\) all sit in the fixed compact \([-1,1]\), and \(\varphi_k\to0\) uniformly. But \(\varphi_k'(x)=\rho'(kx)\), so \(\sup|\varphi_k'|=\sup|\rho'|\not\to0\): the first derivatives do not converge to \(0\). Hence \(\varphi_k\not\to0\) in \(\mathcal D(\mathbb R)\), even though it does so uniformly.Toutes les dรฉrivรฉes doivent converger. Soit \(\varphi_k(x)=\tfrac1k\rho(kx)\). Les supports \([-1/k,1/k]\) restent dans le compact fixe \([-1,1]\), et \(\varphi_k\to0\) uniformรฉment. Mais \(\varphi_k'(x)=\rho'(kx)\), donc \(\sup|\varphi_k'|=\sup|\rho'|\not\to0\). Les premiรจres dรฉrivรฉes ne convergent pas vers \(0\). Ainsi \(\varphi_k\not\to0\) dans \(\mathcal D(\mathbb R)\), malgrรฉ la convergence uniforme des fonctions.
Convergence in \(\mathcal D\) is strictly stronger than uniform convergence of the functions themselves; that strength is exactly what forces the dual space of distributions to be so large.La convergence dans \(\mathcal D\) est strictement plus forte que la convergence uniforme des fonctions elles-mรชmes ; cette force est prรฉcisรฉment ce qui rend lโespace dual des distributions si riche.
Worked Examples
Problem. Verify that \(j(x)=e^{-1/(1-x^2)}\) for \(|x|<1\), \(j(x)=0\) for \(|x|\ge1\), is a valid test function in \(\mathcal D(\mathbb R)\), and describe the qualitative shape of its graph.Problรจme. Vรฉrifier que \(j(x)=e^{-1/(1-x^2)}\) pour \(|x|<1\), et \(j(x)=0\) pour \(|x|\ge1\), est une fonction test valide de \(\mathcal D(\mathbb R)\), puis dรฉcrire qualitativement la forme de son graphe.
Definitions used. Def. 1.2 (support), Def. 1.3 (\(\mathcal D\)), Lem. 1.1, Thm. 1.1.Dรฉfinitions utilisรฉes. Dรฉf. 1.2 (support), dรฉf. 1.3 (\(\mathcal D\)), lem. 1.1, th. 1.1.
Strategy. Recognize \(j=f\circ g\) with \(g(x)=1-x^2\) and \(f\) the flatness function; read off positivity, support, and boundary flatness.Stratรฉgie. Reconnaรฎtre \(j=f\circ g\) avec \(g(x)=1-x^2\) et \(f\) la fonction plate ; dรฉterminer ensuite la positivitรฉ, le support et la platitude au bord.
Derivation. With \(g(x)=1-x^2\) (a polynomial, \(C^\infty\)) and \(f(t)=e^{-1/t}\) for \(t>0\), \(f(t)=0\) for \(t\le0\) (Lemma 1.1), we have \(j=f\circ g\in C^\infty(\mathbb R)\) by the chain rule. Now \(j(x)>0\iff g(x)>0\iff |x|<1\), so \(\{j\ne0\}=(-1,1)\) and \(\operatorname{supp} j=[-1,1]\), compact. Hence \(j\in\mathcal D(\mathbb R)\). The peak is \(j(0)=e^{-1}\approx0.368\); \(j\) is even, strictly decreasing on \([0,1)\), and by Lemma 1.1 all derivatives vanish at \(x=\pm1\), so the graph meets the axis with infinite-order tangency.Dรฉrivation. Avec \(g(x)=1-x^2\), qui est un polynรดme donc une fonction \(C^\infty\), et \(f(t)=e^{-1/t}\) pour \(t>0\), \(f(t)=0\) pour \(t\le0\) (lemme 1.1), on a \(j=f\circ g\in C^\infty(\mathbb R)\) par la rรจgle de la chaรฎne. De plus, \(j(x)>0\iff g(x)>0\iff|x|<1\). Ainsi \(\{j\ne0\}=(-1,1)\) et \(\operatorname{supp}j=[-1,1]\), qui est compact. Donc \(j\in\mathcal D(\mathbb R)\). Son maximum vaut \(j(0)=e^{-1}\approx0{,}368\) ; \(j\) est paire, strictement dรฉcroissante sur \([0,1)\), et, dโaprรจs le lemme 1.1, toutes ses dรฉrivรฉes sโannulent en \(x=\pm1\). Le graphe rejoint donc lโaxe avec une tangence dโordre infini.
Verification. \(j'(x)=e^{-1/(1-x^2)}\cdot\bigl(-\tfrac{2x}{(1-x^2)^2}\bigr)\) for \(|x|<1\); as \(x\to1^-\) the polynomial factor \(\tfrac{2x}{(1-x^2)^2}\) blows up but is overwhelmed by \(e^{-1/(1-x^2)}\to0\), so \(j'(x)\to0=j'(1)\). The same growth beats every derivative. โVรฉrification. Pour \(|x|<1\), \(j'(x)=e^{-1/(1-x^2)}\bigl(-\tfrac{2x}{(1-x^2)^2}\bigr)\). Lorsque \(x\to1^-\), le facteur rationnel \(\tfrac{2x}{(1-x^2)^2}\) diverge, mais il est dominรฉ par \(e^{-1/(1-x^2)}\to0\). Ainsi \(j'(x)\to0=j'(1)\). Le mรชme mรฉcanisme domine tous les facteurs polynomiaux apparaissant dans les dรฉrivรฉes dโordre supรฉrieur. โ
Interpretation. The formula is invisible to Taylor series at \(\pm1\): all Taylor coefficients there are \(0\), yet \(j\not\equiv0\). This is why no nonzero test function can be real-analytic.Interprรฉtation. La formule est invisible pour la sรฉrie de Taylor en \(\pm1\) : tous les coefficients de Taylor y sont nuls alors que \(j\not\equiv0\). Cโest la raison pour laquelle aucune fonction test non nulle ne peut รชtre rรฉelle analytique.
Common mistake. "There is a corner at \(x=\pm1\), so \(j\) is only continuous." False: the match is \(C^\infty\), not merely \(C^0\); Lemma 1.1 makes every one-sided derivative equal \(0\) there.Erreur frรฉquente. ยซ Il y a un angle en \(x=\pm1\), donc \(j\) nโest que continue. ยป Cโest faux : le raccordement est \(C^\infty\), et pas seulement \(C^0\). Le lemme 1.1 garantit que toutes les dรฉrivรฉes unilatรฉrales y valent \(0\).
Problem. Construct \(P\in\mathcal D(\mathbb R)\) with \(0\le P\le1\), \(P\equiv1\) on \([-1,1]\), and \(\operatorname{supp} P\subseteq[-2,2]\).Problรจme. Construire \(P\in\mathcal D(\mathbb R)\) telle que \(0\le P\le1\), \(P\equiv1\) sur \([-1,1]\), et \(\operatorname{supp}P\subseteq[-2,2]\).
Definitions used. Lem. 1.1, Thm. 1.2 (cutoff), Def. 1.2.Dรฉfinitions utilisรฉes. Lem. 1.1, th. 1.2 (fonction de coupure), dรฉf. 1.2.
Strategy. Build a smooth ramp \(0\to1\) from the flatness function, then reflect it into a symmetric flat-topped plateau. (Equivalently, convolve \(\mathbf 1_{[-3/2,3/2]}\) with \(\rho_{1/2}\), matching Theorem 1.2 with \(K=[-1,1]\), \(U=(-2,2)\).)Stratรฉgie. Construire ร partir de la fonction plate une transition lisse de \(0\) vers \(1\), puis la rรฉflรฉchir pour obtenir un plateau symรฉtrique. De maniรจre รฉquivalente, on peut convoler \(\mathbf1_{[-3/2,3/2]}\) avec \(\rho_{1/2}\), conformรฉment au thรฉorรจme 1.2 avec \(K=[-1,1]\) et \(U=(-2,2)\).
Derivation. With \(f\) as in Lemma 1.1, set the smooth ramp \[h(t)=\frac{f(t)}{f(t)+f(1-t)},\qquad t\in\mathbb R,\] whose denominator is \(>0\) for all \(t\) (for any \(t\), at least one of \(t,1-t\) is positive). Then \(h\in C^\infty(\mathbb R)\), \(h\equiv0\) on \((-\infty,0]\), \(h\equiv1\) on \([1,\infty)\), and \(h\) increases from \(0\) to \(1\) on \([0,1]\). Define \[P(x)=\begin{cases}1,&|x|\le1,\\ 1-h(|x|-1),&1\le|x|\le2,\\ 0,&|x|\ge2.\end{cases}\] At \(|x|=1\), \(h(0)=0\) gives \(P=1\); at \(|x|=2\), \(h(1)=1\) gives \(P=0\); the pieces match to infinite order because \(h\) is constant near \(0\) and near \(1\). Since \(x\mapsto|x|\) is smooth away from \(0\) and \(P\equiv1\) near \(0\), \(P\in C^\infty(\mathbb R)\).Dรฉrivation. Avec \(f\) comme dans le lemme 1.1, posons \[h(t)=\frac{f(t)}{f(t)+f(1-t)},\qquad t\in\mathbb R.\] Le dรฉnominateur est strictement positif pour tout \(t\), car au moins lโun des nombres \(t\) et \(1-t\) est positif. Ainsi \(h\in C^\infty(\mathbb R)\), \(h\equiv0\) sur \(( -\infty,0]\), \(h\equiv1\) sur \([1,\infty)\), et \(h\) croรฎt de \(0\) ร \(1\) sur \([0,1]\). Dรฉfinissons \[P(x)=\begin{cases}1,&|x|\le1,\\1-h(|x|-1),&1\le|x|\le2,\\0,&|x|\ge2.\end{cases}\] Aux points \(|x|=1\), \(h(0)=0\), donc \(P=1\) ; aux points \(|x|=2\), \(h(1)=1\), donc \(P=0\). Les morceaux se raccordent ร tout ordre, car \(h\) est constante au voisinage de \(0\) et de \(1\). Comme \(x\mapsto|x|\) est lisse hors de \(0\) et que \(P\equiv1\) prรจs de \(0\), on obtient \(P\in C^\infty(\mathbb R)\).
Verification. \(0\le h\le1\Rightarrow0\le P\le1\); \(P\equiv1\) on \([-1,1]\); \(P=0\) for \(|x|\ge2\), so \(\operatorname{supp} P\subseteq[-2,2]\). โ Thus \(P\in\mathcal D(\mathbb R)\).Vรฉrification. De \(0\le h\le1\), on dรฉduit \(0\le P\le1\). On a \(P\equiv1\) sur \([-1,1]\) et \(P=0\) pour \(|x|\ge2\), donc \(\operatorname{supp}P\subseteq[-2,2]\). โ Ainsi \(P\in\mathcal D(\mathbb R)\).
Interpretation. \(P\) is a smooth cutoff for the pair \(K=[-1,1]\subset U=(-2,2)\): it is a concrete instance of Theorem 1.2 and the basic tool for restricting attention to a bounded region without losing smoothness.Interprรฉtation. \(P\) est une fonction de coupure lisse associรฉe au couple \(K=[-1,1]\subset U=(-2,2)\). Il sโagit dโun exemple concret du thรฉorรจme 1.2 et dโun outil fondamental pour restreindre lโรฉtude ร une rรฉgion bornรฉe sans perdre la rรฉgularitรฉ.
Common mistake. Using a piecewise-linear "tent" instead of the smooth ramp \(h\). A tent is only \(C^0\); its corners destroy membership in \(C^\infty\), so it is not a test function.Erreur frรฉquente. Remplacer la transition lisse \(h\) par une fonction ยซ tente ยป affine par morceaux. Une telle fonction nโest que \(C^0\) ; ses angles empรชchent lโappartenance ร \(C^\infty\), donc ce nโest pas une fonction test.
Problem. Let \(\rho\in\mathcal D(\mathbb R)\) be the standard mollifier. Show that \(\varphi_k(x)=\tfrac1k\rho(x-k)\) converges to \(0\) uniformly but not in \(\mathcal D(\mathbb R)\), while \(\psi_k(x)=\tfrac1k\rho(x)\) does converge to \(0\) in \(\mathcal D(\mathbb R)\).Problรจme. Soit \(\rho\in\mathcal D(\mathbb R)\) le mollificateur standard. Montrer que \(\varphi_k(x)=\tfrac1k\rho(x-k)\) converge uniformรฉment vers \(0\), mais ne converge pas dans \(\mathcal D(\mathbb R)\), tandis que \(\psi_k(x)=\tfrac1k\rho(x)\) converge bien vers \(0\) dans \(\mathcal D(\mathbb R)\).
Definitions used. Def. 1.5 (convergence in \(\mathcal D\)), Def. 1.2 (support), Def. 1.6.Dรฉfinitions utilisรฉes. Dรฉf. 1.5 (convergence dans \(\mathcal D\)), dรฉf. 1.2 (support), dรฉf. 1.6.
Strategy. Check the two clauses of Definition 1.5 separately: the common-compact-support clause (i) and the all-derivatives clause (ii).Stratรฉgie. Vรฉrifier sรฉparรฉment les deux conditions de la dรฉfinition 1.5 : (i) lโexistence dโun support compact commun ; (ii) la convergence uniforme de toutes les dรฉrivรฉes.
Derivation. The escaping family. \(\|\varphi_k\|_\infty=\tfrac1k\|\rho\|_\infty\to0\), and for each \(m\), \(\|\varphi_k^{(m)}\|_\infty=\tfrac1k\|\rho^{(m)}\|_\infty\to0\); so every derivative tends to \(0\) uniformly. But \(\operatorname{supp}\varphi_k=[k-1,k+1]\), and \(\bigcup_k[k-1,k+1]=[0,\infty)\) is unbounded: no compact \(K\) contains all supports. Clause (i) fails, so \(\varphi_k\not\to0\) in \(\mathcal D(\mathbb R)\). The fixed family. \(\operatorname{supp}\psi_k=[-1,1]=:K\) for all \(k\), so (i) holds with this single \(K\); and \(\|\psi_k^{(m)}\|_\infty=\tfrac1k\|\rho^{(m)}\|_\infty\to0\) for every \(m\), so (ii) holds. Hence \(\psi_k\to0\) in \(\mathcal D(\mathbb R)\).Dรฉrivation. Famille dont le support sโรฉchappe. On a \(\|\varphi_k\|_\infty=\tfrac1k\|\rho\|_\infty\to0\), et pour tout \(m\), \(\|\varphi_k^{(m)}\|_\infty=\tfrac1k\|\rho^{(m)}\|_\infty\to0\). Toutes les dรฉrivรฉes tendent donc uniformรฉment vers \(0\). Cependant, \(\operatorname{supp}\varphi_k=[k-1,k+1]\), et lโunion de ces supports est non bornรฉe. Aucun compact \(K\) ne peut les contenir tous. La condition (i) รฉchoue, donc \(\varphi_k\not\to0\) dans \(\mathcal D(\mathbb R)\). Famille ร support fixe. Pour \(\psi_k\), \(\operatorname{supp}\psi_k=[-1,1]=:K\) pour tout \(k\), donc (i) est satisfaite. De plus, \(\|\psi_k^{(m)}\|_\infty=\tfrac1k\|\rho^{(m)}\|_\infty\to0\) pour tout \(m\), donc (ii) est รฉgalement satisfaite. Ainsi \(\psi_k\to0\) dans \(\mathcal D(\mathbb R)\).
Verification. For \(\psi_k\): given \(N\), \(p_{K,N}(\psi_k-0)=\tfrac1k\max_{m\le N}\|\rho^{(m)}\|_\infty\to0\). For \(\varphi_k\): any candidate compact \(K'\) is bounded, say \(K'\subseteq[-R,R]\); once \(k>R+1\), \(\operatorname{supp}\varphi_k\not\subseteq K'\). โVรฉrification. Pour \(\psi_k\), รฉtant donnรฉ \(N\), \(p_{K,N}(\psi_k)=\tfrac1k\max_{m\le N}\|\rho^{(m)}\|_\infty\to0\). Pour \(\varphi_k\), tout compact candidat \(K'\) est bornรฉ, disons \(K'\subseteq[-R,R]\). Dรจs que \(k>R+1\), \(\operatorname{supp}\varphi_k\not\subseteq K'\). โ
Interpretation. Amplitude decay is not enough. The topology of \(\mathcal D\) sees where the mass lives, not only how tall it is; a test function that wanders off to infinity is not close to \(0\) no matter how flat it becomes.Interprรฉtation. La dรฉcroissance de lโamplitude ne suffit pas. La topologie de \(\mathcal D\) tient compte de lโendroit oรน se trouve la fonction, et pas seulement de sa hauteur. Une fonction test dont le support part vers lโinfini nโest pas proche de \(0\), mรชme si son amplitude et toutes ses dรฉrivรฉes deviennent petites.
Common mistake. Concluding \(\varphi_k\to0\) in \(\mathcal D\) from \(\varphi_k\to0\) uniformly. Uniform (even \(C^\infty\)-uniform) decay is necessary but not sufficient; the fixed-compact-support clause is an independent, and here decisive, requirement.Erreur frรฉquente. Conclure \(\varphi_k\to0\) dans \(\mathcal D\) ร partir de la seule convergence uniforme. La dรฉcroissance uniforme, mรชme pour toutes les dรฉrivรฉes, est nรฉcessaire mais non suffisante. La condition de support compact fixe est indรฉpendante et, ici, dรฉcisive.
Exercises
Thirty exercises progress from recognition to research. Each lists difficulty, prerequisite and concept tags, and expected method, and carries three progressive hints and a complete, self-contained correction. Click a card to expand.Trente exercices progressent de la reconnaissance jusquโร la recherche. Chacun indique la difficultรฉ, les prรฉrequis, les concepts mobilisรฉs et la mรฉthode attendue, puis propose trois indices progressifs et une correction complรจte et autonome. Cliquez sur une carte pour la dรฉvelopper.
Here \(|\alpha|=2+1=3\), \(x^\alpha=x_1^2x_2\), and \(\partial^\alpha=\dfrac{\partial^3}{\partial x_1^2\,\partial x_2}\). Applying it to \(f=x_1^3x_2^2\): \(\partial_1^2(x_1^3)=6x_1\) and \(\partial_2(x_2^2)=2x_2\), so \(\partial^\alpha f=6x_1\cdot2x_2=12\,x_1x_2\).Ici \(|\alpha|=2+1=3\), \(x^\alpha=x_1^2x_2\), et \(\partial^\alpha=\dfrac{\partial^3}{\partial x_1^2\,\partial x_2}\). En lโappliquant ร \(f=x_1^3x_2^2\), on a \(\partial_1^2(x_1^3)=6x_1\) et \(\partial_2(x_2^2)=2x_2\), donc \(\partial^\alpha f=6x_1\cdot2x_2=12x_1x_2\).
Misconception. \(x^\alpha\) is a monomial (a number once \(x\) is fixed), whereas \(\partial^\alpha\) is a differential operator; they share the tuple \(\alpha\) but are different objects. Do not confuse the exponent pattern with the differentiation pattern.Erreur frรฉquente. \(x^\alpha\) est un monรดme, donc un nombre une fois \(x\) fixรฉ, tandis que \(\partial^\alpha\) est un opรฉrateur diffรฉrentiel. Les deux utilisent le mรชme multi-indice \(\alpha\), mais ce sont des objets diffรฉrents. Ne pas confondre le motif des exposants avec celui des dรฉrivations.Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
By definition \(\partial^\alpha\partial^\beta\varphi=\bigl(\partial_1^{\alpha_1}\cdots\partial_n^{\alpha_n}\bigr)\bigl(\partial_1^{\beta_1}\cdots\partial_n^{\beta_n}\bigr)\varphi\). Since \(\varphi\in C^\infty\), all mixed partials are equal regardless of order (Schwarz's theorem), so any two single derivatives \(\partial_i,\partial_j\) commute. Rearranging the string to gather all \(\partial_i\)'s together for each \(i\) turns it into \(\partial_1^{\alpha_1+\beta_1}\cdots\partial_n^{\alpha_n+\beta_n}=\partial^{\alpha+\beta}\varphi\), since \((\alpha+\beta)_i=\alpha_i+\beta_i\).Par dรฉfinition, \(\partial^\alpha\partial^\beta\varphi=(\partial_1^{\alpha_1}\cdots\partial_n^{\alpha_n})(\partial_1^{\beta_1}\cdots\partial_n^{\beta_n})\varphi\). Comme \(\varphi\in C^\infty\), toutes les dรฉrivรฉes partielles mixtes sont รฉgales quel que soit lโordre de dรฉrivation, dโaprรจs le thรฉorรจme de Schwarz. On peut donc permuter les dรฉrivรฉes simples \(\partial_i\) et regrouper celles portant sur une mรชme variable. On obtient \(\partial_1^{\alpha_1+\beta_1}\cdots\partial_n^{\alpha_n+\beta_n}=\partial^{\alpha+\beta}\varphi\), puisque \((\alpha+\beta)_i=\alpha_i+\beta_i\).
Misconception. Commuting derivatives is a theorem, not a definition: it can fail for functions that are merely \(C^1\) but not \(C^2\). For \(\mathcal D(\Omega)\subseteq C^\infty\) it always holds, which is exactly why the multi-index calculus is clean.Erreur frรฉquente. La commutation des dรฉrivรฉes est un thรฉorรจme et non une dรฉfinition. Elle peut รฉchouer pour des fonctions qui ne sont pas assez rรฉguliรจres. Dans \(\mathcal D(\Omega)\subset C^\infty\), elle est toujours valable, ce qui rend le calcul multi-indice cohรฉrent.Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Iterate the one-variable Leibniz rule coordinate by coordinate. Since \(\partial^\alpha=\partial_1^{\alpha_1}\cdots\partial_n^{\alpha_n}\), apply \(\partial_1^{\alpha_1}\) first: \(\partial_1^{\alpha_1}(fg)=\sum_{\beta_1=0}^{\alpha_1}\binom{\alpha_1}{\beta_1}\partial_1^{\beta_1}f\,\partial_1^{\alpha_1-\beta_1}g\). Now apply \(\partial_2^{\alpha_2}\) to each term, again by the one-variable rule, producing an inner sum over \(\beta_2\); the coefficients multiply to \(\binom{\alpha_1}{\beta_1}\binom{\alpha_2}{\beta_2}\). Continuing through all \(n\) variables and collecting, with \(\beta=(\beta_1,\dots,\beta_n)\le\alpha\) and \(\binom{\alpha}{\beta}=\prod_i\binom{\alpha_i}{\beta_i}\), gives \(\partial^\alpha(fg)=\sum_{\beta\le\alpha}\binom{\alpha}{\beta}\partial^\beta f\,\partial^{\alpha-\beta}g\). Smoothness lets us commute the intermediate derivatives (Ex. 1.2).Appliquer successivement la rรจgle de Leibniz unidimensionnelle dans chaque coordonnรฉe. Comme \(\partial^\alpha=\partial_1^{\alpha_1}\cdots\partial_n^{\alpha_n}\), on commence par \(\partial_1^{\alpha_1}\) : \(\partial_1^{\alpha_1}(fg)=\sum_{\beta_1=0}^{\alpha_1}\binom{\alpha_1}{\beta_1}\partial_1^{\beta_1}f\,\partial_1^{\alpha_1-\beta_1}g\). On applique ensuite \(\partial_2^{\alpha_2}\) ร chacun des termes, puis les dรฉrivations suivantes. Les coefficients binomiaux se multiplient. En regroupant avec \(\beta=(\beta_1,\dots,\beta_n)\le\alpha\) et \(\binom{\alpha}{\beta}=\prod_i\binom{\alpha_i}{\beta_i}\), on obtient \(\partial^\alpha(fg)=\sum_{\beta\le\alpha}\binom{\alpha}{\beta}\partial^\beta f\,\partial^{\alpha-\beta}g\). La rรฉgularitรฉ permet de commuter les dรฉrivรฉes intermรฉdiaires.
Misconception. The sum runs over all \(\beta\le\alpha\) coordinatewise, not just \(|\beta|\le|\alpha|\): there are \(\prod_i(\alpha_i+1)\) terms, and the binomial coefficient is a product, not \(\binom{|\alpha|}{|\beta|}\).Erreur frรฉquente. La somme porte sur tous les \(\beta\le\alpha\) coordonnรฉe par coordonnรฉe, et non seulement sur ceux qui vรฉrifient \(|\beta|\le|\alpha|\). Il y a \(\prod_i(\alpha_i+1)\) termes et le coefficient binomial est le produit \(\binom{\alpha}{\beta}\), non \(\binom{|\alpha|}{|\beta|}\).Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Because \(x^\beta=\prod_{i=1}^n x_i^{\beta_i}\) and \(\partial^\alpha=\prod_i\partial_i^{\alpha_i}\) acts on separate variables, \(\partial^\alpha x^\beta=\prod_i\bigl(\partial_i^{\alpha_i}x_i^{\beta_i}\bigr)\). In one variable \(\partial_i^{\alpha_i}x_i^{\beta_i}=\frac{\beta_i!}{(\beta_i-\alpha_i)!}x_i^{\beta_i-\alpha_i}\) when \(\alpha_i\le\beta_i\), and \(0\) when \(\alpha_i>\beta_i\). Multiplying, if \(\alpha\le\beta\) (all \(\alpha_i\le\beta_i\)) we get \(\prod_i\frac{\beta_i!}{(\beta_i-\alpha_i)!}x_i^{\beta_i-\alpha_i}=\frac{\beta!}{(\beta-\alpha)!}x^{\beta-\alpha}\); if some \(\alpha_i>\beta_i\) one factor is \(0\), so the product is \(0\). Taking \(\alpha=\beta\): \(\partial^\alpha x^\alpha=\frac{\alpha!}{0!}x^0=\alpha!\).Comme \(x^\beta=\prod_{i=1}^n x_i^{\beta_i}\) et que \(\partial^\alpha=\prod_i\partial_i^{\alpha_i}\) agit sรฉparรฉment sur chaque variable, \(\partial^\alpha x^\beta=\prod_i(\partial_i^{\alpha_i}x_i^{\beta_i})\). En une variable, \(\partial_i^{\alpha_i}x_i^{\beta_i}=\frac{\beta_i!}{(\beta_i-\alpha_i)!}x_i^{\beta_i-\alpha_i}\) si \(\alpha_i\le\beta_i\), et vaut \(0\) si \(\alpha_i>\beta_i\). En multipliant, si \(\alpha\le\beta\), on obtient \(\frac{\beta!}{(\beta-\alpha)!}x^{\beta-\alpha}\). Si lโune des inรฉgalitรฉs coordonnรฉe par coordonnรฉe รฉchoue, un facteur est nul et tout le produit sโannule. Pour \(\alpha=\beta\), il vient \(\partial^\alpha x^\alpha=\alpha!\).
Misconception. "\(\partial^\alpha x^\beta=0\) whenever \(|\alpha|>|\beta|\)." The correct test is coordinatewise: \(\partial^{(2,0)}(x_1x_2^5)=0\) already, even though \(|\alpha|=2<6=|\beta|\), because \(\alpha_1=2>1=\beta_1\).Erreur frรฉquente. Lโaffirmation ยซ \(\partial^\alpha x^\beta=0\) dรจs que \(|\alpha|>|\beta|\) ยป est insuffisante. Le bon critรจre est coordonnรฉ : \(\partial^{(2,0)}(x_1x_2^5)=0\) alors mรชme que \(|\alpha|=2<6=|\beta|\), car \(\alpha_1=2>1=\beta_1\).Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
\(\operatorname{supp} f=\overline{\{f\ne0\}}\) is the closure of a set, hence closed in \(\Omega\) by definition of closure. Its complement \(\Omega\setminus\operatorname{supp} f\) is therefore open. If \(x\in\Omega\setminus\operatorname{supp} f\), then \(x\notin\overline{\{f\ne0\}}\), so some neighbourhood \(V\) of \(x\) is disjoint from \(\{f\ne0\}\); thus \(f\equiv0\) on \(V\), in particular \(f(x)=0\). Hence \(f\) vanishes on \(\Omega\setminus\operatorname{supp} f\). For the tent \(f(x)=\max(0,1-|x|)\), \(\{f\ne0\}=(-1,1)\), so \(\operatorname{supp} f=[-1,1]\).Par dรฉfinition, \(\operatorname{supp}f=\overline{\{f\ne0\}}\) est une adhรฉrence, donc un fermรฉ de \(\Omega\). Son complรฉmentaire est ouvert. Si \(x\notin\operatorname{supp}f\), il existe un voisinage \(V\) de \(x\) ne rencontrant pas \(\{f\ne0\}\). Ainsi \(f\equiv0\) sur \(V\), et en particulier \(f(x)=0\). Donc \(f\) sโannule sur \(\Omega\setminus\operatorname{supp}f\). Pour la fonction tente \(f(x)=\max(0,1-|x|)\), on a \(\{f\ne0\}=(-1,1)\), dโoรน \(\operatorname{supp}f=[-1,1]\).
Misconception. The support is the closure of \(\{f\ne0\}\), not \(\{f\ne0\}\) itself: the endpoints \(\pm1\), where \(f=0\), still belong to \(\operatorname{supp} f\). (Note the tent is not smooth, so it is not a test function despite having compact support.)Erreur frรฉquente. Le support est lโadhรฉrence de \(\{f\ne0\}\), et non simplement \(\{f\ne0\}\). Les extrรฉmitรฉs \(\pm1\), oรน \(f=0\), appartiennent encore au support. La fonction tente nโest toutefois pas lisse et nโest donc pas une fonction test malgrรฉ son support compact.Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Let \(W=\Omega\setminus(\operatorname{supp} f\cup\operatorname{supp} g)\), an open set on which both \(f\) and \(g\) vanish. Then \(f+g\equiv0\) on \(W\), so \(\{f+g\ne0\}\subseteq\operatorname{supp} f\cup\operatorname{supp} g\); the latter is closed, so taking closures \(\operatorname{supp}(f+g)\subseteq\operatorname{supp} f\cup\operatorname{supp} g\). For the product, wherever \(f=0\) or \(g=0\) we have \(fg=0\), so \(\{fg\ne0\}\subseteq\{f\ne0\}\cap\{g\ne0\}\subseteq\operatorname{supp} f\cap\operatorname{supp} g\) (closed), giving \(\operatorname{supp}(fg)\subseteq\operatorname{supp} f\cap\operatorname{supp} g\). Strictness: take \(f=\rho\) and \(g=-\rho\); then \(\operatorname{supp} f=\operatorname{supp} g=[-1,1]\) but \(f+g\equiv0\), so \(\operatorname{supp}(f+g)=\varnothing\subsetneq[-1,1]\).Posons \(W=\Omega\setminus(\operatorname{supp}f\cup\operatorname{supp}g)\). Sur cet ouvert, \(f\) et \(g\) sont nulles, donc \(f+g\equiv0\). Ainsi \(\{f+g\ne0\}\subseteq\operatorname{supp}f\cup\operatorname{supp}g\), ensemble fermรฉ, et en prenant les adhรฉrences on obtient \(\operatorname{supp}(f+g)\subseteq\operatorname{supp}f\cup\operatorname{supp}g\). Pour le produit, dรจs que \(f=0\) ou \(g=0\), on a \(fg=0\), donc \(\{fg\ne0\}\subseteq\{f\ne0\}\cap\{g\ne0\}\subseteq\operatorname{supp}f\cap\operatorname{supp}g\). Lโinclusion est stricte pour la somme, par exemple avec \(f=\rho\) et \(g=-\rho\) : alors \(f+g\equiv0\), donc son support est vide, tandis que les supports de \(f\) et \(g\) valent \([-1,1]\).
Misconception. These are inclusions, not equalities: cancellation can shrink the support of a sum, and the product's support can be strictly smaller than the intersection (if the factors' nonvanishing sets meet only on their boundaries).Erreur frรฉquente. Il sโagit dโinclusions et non dโรฉgalitรฉs. Des compensations peuvent rรฉduire le support dโune somme, et le support dโun produit peut รชtre strictement plus petit que lโintersection des supports.Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Let \(\varphi,\psi\in\mathcal D(\Omega)\), \(\lambda\) a scalar. Then \(\varphi+\lambda\psi\in C^\infty(\Omega)\) since \(C^\infty(\Omega)\) is a vector space. By Ex. 1.6, \(\operatorname{supp}(\varphi+\lambda\psi)\subseteq\operatorname{supp}\varphi\cup\operatorname{supp}\psi\). Both supports are compact subsets of \(\Omega\), so their union is compact, and \(\operatorname{supp}(\varphi+\lambda\psi)\) is a closed subset of that compact set, hence compact and contained in \(\Omega\). Thus \(\varphi+\lambda\psi\in\mathcal D(\Omega)\). The zero function is in \(\mathcal D(\Omega)\) (empty support), and the vector-space axioms are inherited pointwise. Therefore \(\mathcal D(\Omega)\) is a vector space.Soient \(\varphi,\psi\in\mathcal D(\Omega)\) et \(\lambda\) un scalaire. Comme \(C^\infty(\Omega)\) est un espace vectoriel, \(\varphi+\lambda\psi\in C^\infty(\Omega)\). Dโaprรจs lโexercice 1.6, \(\operatorname{supp}(\varphi+\lambda\psi)\subseteq\operatorname{supp}\varphi\cup\operatorname{supp}\psi\). Les deux supports sont compacts dans \(\Omega\), leur union est donc compacte, et le support de \(\varphi+\lambda\psi\), fermรฉ dans cette union, est รฉgalement compact et contenu dans \(\Omega\). Ainsi \(\varphi+\lambda\psi\in\mathcal D(\Omega)\). La fonction nulle appartient ร \(\mathcal D(\Omega)\), son support รฉtant vide, et les axiomes dโespace vectoriel sont hรฉritรฉs des opรฉrations ponctuelles. Donc \(\mathcal D(\Omega)\) est un espace vectoriel.
Misconception. One must check the support condition, not only smoothness: the sum of two compactly supported functions could a priori have a larger support, but the union of two compact sets is still compact, so closure holds.Erreur frรฉquente. Il faut vรฉrifier la condition sur le support, et pas seulement la rรฉgularitรฉ. La somme de deux fonctions ร support compact peut avoir un support plus grand, mais celui-ci reste contenu dans lโunion de deux compacts, qui est encore compacte.Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Induct on \(n\). Base: \(f(t)=e^{-1/t}=P_0(1/t)e^{-1/t}\) with \(P_0\equiv1\). Step: assume \(f^{(n)}(t)=P_n(s)e^{-s}\) with \(s=1/t\). Since \(\frac{ds}{dt}=-1/t^2=-s^2\), the chain rule gives \(\frac{d}{dt}=-s^2\frac{d}{ds}\), so \[f^{(n+1)}(t)=-s^2\frac{d}{ds}\bigl[P_n(s)e^{-s}\bigr]=-s^2\bigl[P_n'(s)e^{-s}-P_n(s)e^{-s}\bigr]=s^2\bigl(P_n(s)-P_n'(s)\bigr)e^{-s}.\] Thus \(P_{n+1}(s)=s^2\bigl(P_n(s)-P_n'(s)\bigr)\), a polynomial. Computing: \(P_1(s)=s^2(1-0)=s^2\); \(P_1'(s)=2s\), so \(P_2(s)=s^2(s^2-2s)=s^4-2s^3\).Raisonnons par rรฉcurrence sur \(n\). Initialisation : \(f(t)=e^{-1/t}=P_0(1/t)e^{-1/t}\) avec \(P_0\equiv1\). Supposons \(f^{(n)}(t)=P_n(s)e^{-s}\), oรน \(s=1/t\). Comme \(ds/dt=-s^2\), la rรจgle de la chaรฎne donne \(d/dt=-s^2d/ds\). Par consรฉquent \[f^{(n+1)}(t)=-s^2\frac d{ds}[P_n(s)e^{-s}]=s^2(P_n(s)-P_n'(s))e^{-s}.\] Ainsi \(P_{n+1}(s)=s^2(P_n(s)-P_n'(s))\), qui est un polynรดme. On calcule \(P_1(s)=s^2\), puis \(P_2(s)=s^2(s^2-2s)=s^4-2s^3\).
Misconception. The \(P_n\) are polynomials in \(s=1/t\), not in \(t\); as \(t\to0^+\), \(s\to\infty\), and it is the factor \(e^{-s}\) that forces \(f^{(n)}(t)\to0\) despite \(P_n(s)\to\infty\).Erreur frรฉquente. Les \(P_n\) sont des polynรดmes en \(s=1/t\), et non en \(t\). Lorsque \(t\to0^+\), on a \(s\to\infty\), et cโest le facteur \(e^{-s}\) qui impose \(f^{(n)}(t)\to0\) malgrรฉ la croissance de \(P_n(s)\).Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Set \(g(x)=1-|x|^2=1-\sum_i x_i^2\), a polynomial in \(x\), so \(g\in C^\infty(\mathbb R^n)\), with \(g(x)>0\iff|x|<1\). With \(f\) as in Lemma 1.1, \(j(x)=f(g(x))\) equals \(e^{-1/(1-|x|^2)}\) when \(|x|<1\) and \(0\) when \(|x|\ge1\), matching the definition. Since \(f\in C^\infty(\mathbb R)\) and \(g\in C^\infty(\mathbb R^n)\), the chain rule gives \(j=f\circ g\in C^\infty(\mathbb R^n)\); no separate check at \(|x|=1\) is needed, because \(f\) is smooth across the value \(0\). As \(f(s)>0\iff s>0\), \(\{j\ne0\}=\{|x|<1\}=B(0,1)\), so \(\operatorname{supp} j=\overline{B(0,1)}\), which is closed and bounded, hence compact. Therefore \(j\in C_c^\infty(\mathbb R^n)\).Posons \(g(x)=1-|x|^2=1-\sum_i x_i^2\), qui est un polynรดme et donc une fonction \(C^\infty\), avec \(g(x)>0\iff|x|<1\). Si \(f\) est la fonction du lemme 1.1, alors \(j(x)=f(g(x))\) vaut \(e^{-1/(1-|x|^2)}\) pour \(|x|<1\) et \(0\) pour \(|x|\ge1\). Comme \(f\in C^\infty(\mathbb R)\) et \(g\in C^\infty(\mathbb R^n)\), la rรจgle de la chaรฎne donne \(j=f\circ g\in C^\infty(\mathbb R^n)\). Aucun contrรดle sรฉparรฉ sur \(|x|=1\) nโest nรฉcessaire, car \(f\) est lisse au passage de la valeur \(0\). Enfin, \(f(s)>0\iff s>0\), donc \(\{j\ne0\}=B(0,1)\) et \(\operatorname{supp}j=\overline{B(0,1)}\), qui est compact. Ainsi \(j\in C_c^\infty(\mathbb R^n)\).
Misconception. One need not verify smoothness "by hand" at the sphere \(|x|=1\): that would be error-prone. The entire difficulty is quarantined into Lemma 1.1; composition with the polynomial \(g\) then transfers smoothness automatically.Erreur frรฉquente. Il nโest pas nรฉcessaire de vรฉrifier la rรฉgularitรฉ ยซ ร la main ยป sur la sphรจre \(|x|=1\). Toute la difficultรฉ est isolรฉe dans le lemme 1.1 ; la composition avec le polynรดme \(g\) transmet ensuite automatiquement la rรฉgularitรฉ.Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
With \(y=x/\varepsilon\), \(dx=\varepsilon^n dy\), so \(\int\rho_\varepsilon(x)\,dx=\int\varepsilon^{-n}\rho(x/\varepsilon)\,dx=\int\varepsilon^{-n}\rho(y)\,\varepsilon^n\,dy=\int\rho(y)\,dy=1\), using \(\int\rho=1\). Next, \(\rho_\varepsilon(x)=\varepsilon^{-n}\rho(x/\varepsilon)\ne0\) exactly when \(\rho(x/\varepsilon)\ne0\), i.e. \(|x/\varepsilon|<1\), i.e. \(|x|<\varepsilon\); taking closure, \(\operatorname{supp}\rho_\varepsilon=\overline{B(0,\varepsilon)}\). Finally \(\rho_\varepsilon(0)=\varepsilon^{-n}\rho(0)\), which \(\to\infty\) as \(\varepsilon\to0^+\): the peak grows like \(\varepsilon^{-n}\) while the mass stays \(1\).Avec le changement de variables \(y=x/\varepsilon\), on a \(dx=\varepsilon^n dy\). Ainsi \[\int\rho_\varepsilon(x)\,dx=\int\varepsilon^{-n}\rho(x/\varepsilon)\,dx=\int\rho(y)\,dy=1.\] De plus, \(\rho_\varepsilon(x)\ne0\) exactement lorsque \(|x/\varepsilon|<1\), cโest-ร -dire \(|x|<\varepsilon\). En prenant lโadhรฉrence, \(\operatorname{supp}\rho_\varepsilon=\overline{B(0,\varepsilon)}\). Enfin, \(\rho_\varepsilon(0)=\varepsilon^{-n}\rho(0)\), qui tend vers \(+\infty\) lorsque \(\varepsilon\to0^+\). Le pic croรฎt comme \(\varepsilon^{-n}\) tandis que la masse totale reste รฉgale ร \(1\).
Misconception. The prefactor \(\varepsilon^{-n}\) is not cosmetic: without it the integral would scale like \(\varepsilon^n\to0\). The exponent must match the dimension \(n\) for the mass to be preserved.Erreur frรฉquente. Le facteur \(\varepsilon^{-n}\) est essentiel : sans lui, lโintรฉgrale serait multipliรฉe par \(\varepsilon^n\) et tendrait vers \(0\). Lโexposant doit coรฏncider avec la dimension \(n\) pour prรฉserver la masse.Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Let \(h(t)=\dfrac{f(t)}{f(t)+f(1-t)}\) with \(f\) from Lemma 1.1; then \(h\in C^\infty(\mathbb R)\), \(h\equiv0\) on \((-\infty,0]\), \(h\equiv1\) on \([1,\infty)\), increasing on \([0,1]\). Define \[P(x)=\begin{cases}1,&|x|\le a,\\ 1-h\!\Bigl(\dfrac{|x|-a}{b-a}\Bigr),&a\le|x|\le b,\\ 0,&|x|\ge b.\end{cases}\] At \(|x|=a\): \(h(0)=0\Rightarrow P=1\); at \(|x|=b\): \(h(1)=1\Rightarrow P=0\). Because \(h\) is constant near the ends of \([0,1]\), the pieces match to infinite order, and since \(P\equiv1\) near \(0\), \(P\in C^\infty(\mathbb R)\). Clearly \(0\le P\le1\), \(P\equiv1\) on \([-a,a]\), \(\operatorname{supp} P\subseteq[-b,b]\). Hence \(P\in\mathcal D(\mathbb R)\).Soit \(h(t)=\dfrac{f(t)}{f(t)+f(1-t)}\), oรน \(f\) est celle du lemme 1.1. Alors \(h\in C^\infty(\mathbb R)\), \(h\equiv0\) sur \(( -\infty,0]\), \(h\equiv1\) sur \([1,\infty)\), et \(h\) est croissante sur \([0,1]\). Dรฉfinissons \[P(x)=\begin{cases}1,&|x|\le a,\\1-h\!\left(\dfrac{|x|-a}{b-a}\right),&a\le|x|\le b,\\0,&|x|\ge b.\end{cases}\] Aux points \(|x|=a\), on a \(h(0)=0\), donc \(P=1\), et aux points \(|x|=b\), \(h(1)=1\), donc \(P=0\). Comme \(h\) est constante prรจs des extrรฉmitรฉs de \([0,1]\), les morceaux se raccordent ร tout ordre. De plus \(P\equiv1\) prรจs de \(0\), donc \(P\in C^\infty(\mathbb R)\). Clairement \(0\le P\le1\), \(P\equiv1\) sur \([-a,a]\) et \(\operatorname{supp}P\subseteq[-b,b]\). Par consรฉquent \(P\in\mathcal D(\mathbb R)\).
Misconception. The affine argument \((|x|-a)/(b-a)\) is essential to place the transition in \([a,b]\); forgetting to rescale leaves a plateau that either is not \(1\) on all of \([-a,a]\) or spills past \(b\).Erreur frรฉquente. Lโargument affine \((|x|-a)/(b-a)\) est indispensable pour placer la transition sur \([a,b]\). Sans ce changement dโรฉchelle, le plateau ne vaut pas nรฉcessairement \(1\) sur tout \([-a,a]\) ou son support peut dรฉpasser \([-b,b]\).Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Smoothness. Writing \(\varphi(x)=\int\mathbf 1_{K_\delta}(z)\,\rho_\delta(x-z)\,dz\), the integrand is \(C^\infty\) in \(x\), and for each multi-index \(\alpha\) the derivative \(\partial^\alpha_x\rho_\delta(x-z)\) is bounded by \(\sup|\partial^\alpha\rho_\delta|\) and supported in \(z\in x-\overline{B(0,\delta)}\); since \(\mathbf 1_{K_\delta}\) is integrable with compact support, the dominated-convergence hypotheses hold and we may differentiate under the integral: \(\partial^\alpha\varphi=\mathbf 1_{K_\delta}*\partial^\alpha\rho_\delta\), continuous. So \(\varphi\in C^\infty\). Value on \(K\). For \(x\in K\) and \(|y|\le\delta\), \(\operatorname{dist}(x-y,K)\le\operatorname{dist}(x,K)+|y|=0+\delta\le2\delta\), so \(x-y\in K_\delta\) and \(\mathbf 1_{K_\delta}(x-y)=1\). Hence \(\varphi(x)=\int_{|y|\le\delta}\rho_\delta(y)\,dy=\int\rho_\delta=1\).Rรฉgularitรฉ. En รฉcrivant \(\varphi(x)=\int\mathbf1_{K_\delta}(z)\rho_\delta(x-z)\,dz\), lโintรฉgrande est \(C^\infty\) en \(x\). Pour chaque multi-indice \(\alpha\), la dรฉrivรฉe \(\partial_x^\alpha\rho_\delta(x-z)\) est majorรฉe par \(\sup|\partial^\alpha\rho_\delta|\). Comme \(\mathbf1_{K_\delta}\) est intรฉgrable et ร support compact, on peut dรฉriver sous le signe intรฉgral et obtenir \(\partial^\alpha\varphi=\mathbf1_{K_\delta}*\partial^\alpha\rho_\delta\), qui est continue. Donc \(\varphi\in C^\infty\). Valeur sur \(K\). Si \(x\in K\) et \(|y|\le\delta\), alors \(\operatorname{dist}(x-y,K)\le|y|\le\delta\le2\delta\). Ainsi \(x-y\in K_\delta\) et \(\mathbf1_{K_\delta}(x-y)=1\). Par consรฉquent \(\varphi(x)=\int_{|y|\le\delta}\rho_\delta(y)\,dy=1\).
Misconception. Convolution smooths because derivatives fall on the smooth factor \(\rho_\delta\), never on the rough factor \(\mathbf 1_{K_\delta}\); the indicator is only integrated, so its lack of smoothness is harmless.Erreur frรฉquente. La convolution rรฉgularise parce que les dรฉrivรฉes portent sur le facteur lisse \(\rho_\delta\), jamais sur lโindicatrice \(\mathbf1_{K_\delta}\). Celle-ci intervient uniquement sous lโintรฉgrale ; son manque de rรฉgularitรฉ ne pose donc pas de problรจme.Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Yes: \(\varphi_k\to0\) in \(\mathcal D(\mathbb R)\). Clause (i): \(\operatorname{supp}\varphi_k=\operatorname{supp}\rho=[-1,1]=:K\) for all \(k\), so a single compact set contains every support. Clause (ii): for any multi-index (order \(m\)), \(\partial^m\varphi_k=\tfrac1k\partial^m\rho\), so \(\sup_x|\partial^m\varphi_k-0|=\tfrac1k\sup_x|\partial^m\rho|\to0\). Equivalently, for every \(N\), \(p_{K,N}(\varphi_k)=\tfrac1k\max_{m\le N}\|\rho^{(m)}\|_\infty\to0\). Both clauses hold, so \(\varphi_k\to0\).Oui : \(\varphi_k\to0\) dans \(\mathcal D(\mathbb R)\). Condition (i) : \(\operatorname{supp}\varphi_k=\operatorname{supp}\rho=[-1,1]=:K\) pour tout \(k\), donc un mรชme compact contient tous les supports. Condition (ii) : pour tout ordre \(m\), \(\partial^m\varphi_k=\tfrac1k\partial^m\rho\), et \(\sup_x|\partial^m\varphi_k|=\tfrac1k\sup_x|\partial^m\rho|\to0\). De maniรจre รฉquivalente, pour tout \(N\), \(p_{K,N}(\varphi_k)=\tfrac1k\max_{m\le N}\|\rho^{(m)}\|_\infty\to0\). Les deux conditions sont satisfaites, donc \(\varphi_k\to0\).
Misconception. Convergence to \(0\) in \(\mathcal D\) requires checking all derivatives, not just the function values; here every derivative is a fixed function scaled by \(1/k\), so all seminorms vanish together.Erreur frรฉquente. La convergence vers \(0\) dans \(\mathcal D\) exige de contrรดler toutes les dรฉrivรฉes, et pas seulement les valeurs des fonctions. Ici, chaque dรฉrivรฉe est une fonction fixe multipliรฉe par \(1/k\), donc toutes les semi-normes tendent simultanรฉment vers \(0\).Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Every derivative decays uniformly: \(\varphi_k^{(m)}(x)=\tfrac1k\rho^{(m)}(x-k)\), so \(\|\varphi_k^{(m)}\|_\infty=\tfrac1k\|\rho^{(m)}\|_\infty\to0\) for each \(m\). In particular \(\varphi_k\to0\) uniformly and even in every \(C^m\) norm. However \(\operatorname{supp}\varphi_k=k+\operatorname{supp}\rho=[k-1,k+1]\). If some compact \(K'\) contained all supports, then \(K'\subseteq[-R,R]\) for some \(R\); but for \(k>R+1\), \(k-1>R\), so \([k-1,k+1]\not\subseteq[-R,R]\), a contradiction. Clause (i) of Definition 1.5 fails, so \((\varphi_k)\) has no limit in \(\mathcal D(\mathbb R)\) (in particular not \(0\)).Toutes les dรฉrivรฉes dรฉcroissent uniformรฉment : \(\varphi_k^{(m)}(x)=\tfrac1k\rho^{(m)}(x-k)\), donc \(\|\varphi_k^{(m)}\|_\infty=\tfrac1k\|\rho^{(m)}\|_\infty\to0\) pour tout \(m\). En particulier, \(\varphi_k\to0\) uniformรฉment et mรชme pour toutes les normes \(C^m\). Cependant \(\operatorname{supp}\varphi_k=[k-1,k+1]\). Aucun compact ne peut contenir tous ces supports : si \(K'\subseteq[-R,R]\), alors pour \(k>R+1\), \([k-1,k+1]\not\subseteq[-R,R]\). La condition (i) de la dรฉfinition 1.5 รฉchoue. Ainsi \((\varphi_k)\) nโa pas de limite dans \(\mathcal D(\mathbb R)\), et en particulier ne converge pas vers \(0\).
Misconception. "Uniform convergence of the function and all its derivatives implies convergence in \(\mathcal D\)." It does not: the fixed-compact-support clause is independent and here it is violated.Erreur frรฉquente. La convergence uniforme de la fonction et de toutes ses dรฉrivรฉes nโimplique pas ร elle seule la convergence dans \(\mathcal D\). La condition de support compact fixe est indรฉpendante, et elle รฉchoue ici.Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
No. Clause (i) is fine: \(\operatorname{supp}\varphi_k=[-1/k,1/k]\subseteq[-1,1]=:K\) for all \(k\). But clause (ii) fails at the first derivative. By the chain rule \(\varphi_k'(x)=\tfrac1k\cdot k\,\rho'(kx)=\rho'(kx)\), so \(\sup_x|\varphi_k'(x)|=\sup_y|\rho'(y)|=:c>0\), a constant. Thus \(\varphi_k'\not\to0\) uniformly, so \(\varphi_k\not\to0\) in \(\mathcal D(\mathbb R)\). (The function values do go to \(0\) uniformly, \(\|\varphi_k\|_\infty=\tfrac1k\|\rho\|_\infty\to0\), which is not enough.)Non. La condition (i) est satisfaite : \(\operatorname{supp}\varphi_k=[-1/k,1/k]\subseteq[-1,1]=:K\). Mais la condition (ii) รฉchoue dรจs la premiรจre dรฉrivรฉe. Par la rรจgle de la chaรฎne, \(\varphi_k'(x)=\tfrac1k\,k\rho'(kx)=\rho'(kx)\), donc \(\sup_x|\varphi_k'(x)|=\sup_y|\rho'(y)|=:c>0\), constante indรฉpendante de \(k\). Ainsi \(\varphi_k'\not\to0\) uniformรฉment et \(\varphi_k\not\to0\) dans \(\mathcal D(\mathbb R)\). Les fonctions elles-mรชmes convergent pourtant uniformรฉment vers \(0\), ce qui ne suffit pas.
Misconception. Rescaling the argument by \(k\) amplifies derivatives by powers of \(k\): the \(m\)-th derivative scales like \(k^{m-1}\). Shrinking supports do not help if the derivatives blow up.Erreur frรฉquente. Le changement dโรฉchelle de lโargument par \(k\) amplifie les dรฉrivรฉes par des puissances de \(k\). La dรฉrivรฉe dโordre \(m\) se comporte comme \(k^{m-1}\). La contraction des supports ne compense donc pas lโexplosion des dรฉrivรฉes.Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Since \(0<\tfrac1k\le1\), \(\operatorname{supp}\varphi_k=\tfrac1k+[-1,1]=[\tfrac1k-1,\tfrac1k+1]\subseteq[-1,2]=:K\) for every \(k\ge1\); so clause (i) holds with the single compact \(K=[-1,2]\). Translation preserves sup norms, so \(\|\varphi_k^{(m)}\|_\infty=\tfrac1k\|\rho^{(m)}\|_\infty\to0\) for each \(m\), giving clause (ii). Hence \(\varphi_k\to0\) in \(\mathcal D(\mathbb R)\).Comme \(0<1/k\le1\), \(\operatorname{supp}\varphi_k=1/k+[-1,1]=[1/k-1,1/k+1]\subseteq[-1,2]=:K\) pour tout \(k\ge1\). La condition (i) est donc satisfaite avec le compact fixe \([-1,2]\). La translation prรฉserve les normes sup, si bien que \(\|\varphi_k^{(m)}\|_\infty=\tfrac1k\|\rho^{(m)}\|_\infty\to0\) pour tout \(m\). La condition (ii) est satisfaite. Ainsi \(\varphi_k\to0\) dans \(\mathcal D(\mathbb R)\).
Misconception. A moving centre is not automatically fatal: what matters is whether the supports stay in one compact set. Bounded drift (here toward \(0\)) is fine; only unbounded escape (Ex. 1.14) breaks clause (i).Erreur frรฉquente. Un centre mobile nโest pas automatiquement problรฉmatique. Ce qui compte est que tous les supports restent dans un mรชme compact. Un dรฉplacement bornรฉ, ici vers \(0\), convient ; seul un รฉchappement non bornรฉ fait รฉchouer la condition (i).Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Let \(\varphi_k\to\varphi\) in \(\mathcal D(\Omega)\): there is a compact \(K\) with all \(\operatorname{supp}\varphi_k\subseteq K\) and \(p_{K,N}(\varphi_k-\varphi)\to0\) for every \(N\). Since derivatives do not enlarge support, \(\operatorname{supp}(\partial^\alpha\varphi_k)\subseteq\operatorname{supp}\varphi_k\subseteq K\), so clause (i) holds for the images with the same \(K\). For clause (ii), any multi-index \(\beta\) gives \[\sup_K|\partial^\beta(\partial^\alpha\varphi_k)-\partial^\beta(\partial^\alpha\varphi)|=\sup_K|\partial^{\alpha+\beta}(\varphi_k-\varphi)|\le p_{K,|\alpha|+|\beta|}(\varphi_k-\varphi)\to0.\] Hence \(\partial^\alpha\varphi_k\to\partial^\alpha\varphi\) in \(\mathcal D(\Omega)\). (Sequential continuity is all we need, since \(\mathcal D\) is not metrizable; Prop. 1.2.)Soit \(\varphi_k\to\varphi\) dans \(\mathcal D(\Omega)\). Il existe un compact \(K\) tel que tous les supports de \(\varphi_k\) soient contenus dans \(K\) et que \(p_{K,N}(\varphi_k-\varphi)\to0\) pour tout \(N\). Les dรฉrivรฉes nโagrandissent pas le support, donc \(\operatorname{supp}(\partial^\alpha\varphi_k)\subseteq K\), ce qui vรฉrifie (i). Pour (ii), tout multi-indice \(\beta\) donne \[\sup_K|\partial^\beta(\partial^\alpha\varphi_k)-\partial^\beta(\partial^\alpha\varphi)|=\sup_K|\partial^{\alpha+\beta}(\varphi_k-\varphi)|\le p_{K,|\alpha|+|\beta|}(\varphi_k-\varphi)\to0.\] Ainsi \(\partial^\alpha\varphi_k\to\partial^\alpha\varphi\) dans \(\mathcal D(\Omega)\).
Misconception. Continuity here does not follow from a single operator-norm bound: \(\mathcal D\) has no norm. It follows because differentiation merely shifts the multi-index bookkeeping, and every seminorm of the difference already tends to \(0\).Erreur frรฉquente. La continuitรฉ ne provient pas ici dโune seule borne en norme dโopรฉrateur, car \(\mathcal D\) nโest pas normรฉ. Elle rรฉsulte du fait que la dรฉrivation ne fait que dรฉcaler les multi-indices et que chaque semi-norme de la diffรฉrence tend dรฉjร vers \(0\).Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
First, \(\psi\varphi\in C^\infty(\Omega)\) and \(\operatorname{supp}(\psi\varphi)\subseteq\operatorname{supp}\varphi\), so \(\psi\varphi\in\mathcal D(\Omega)\). Let \(\varphi_k\to\varphi\) in \(\mathcal D(\Omega)\) with all supports in a compact \(K\); then \(\operatorname{supp}(\psi\varphi_k)\subseteq K\) too (clause (i)). Put \(u_k=\varphi_k-\varphi\). For \(|\beta|\le N\), Leibniz gives \[\bigl|\partial^\beta(\psi u_k)\bigr|\le\sum_{\gamma\le\beta}\binom{\beta}{\gamma}\bigl|\partial^\gamma\psi\bigr|\,\bigl|\partial^{\beta-\gamma}u_k\bigr|.\] On the compact \(K\), set \(M=\max_{|\gamma|\le N}\sup_K|\partial^\gamma\psi|<\infty\) (finite by continuity) and \(C_N=\max_{|\beta|\le N}\sum_{\gamma\le\beta}\binom{\beta}{\gamma}\). Taking \(\sup_K\) and then the max over \(|\beta|\le N\), \[p_{K,N}(\psi u_k)\le C_N\,M\,p_{K,N}(u_k)\to0.\] Hence \(\psi\varphi_k\to\psi\varphi\) in \(\mathcal D(\Omega)\).Dโabord, \(\psi\varphi\in C^\infty(\Omega)\) et \(\operatorname{supp}(\psi\varphi)\subseteq\operatorname{supp}\varphi\), donc \(\psi\varphi\in\mathcal D(\Omega)\). Soit \(\varphi_k\to\varphi\) dans \(\mathcal D(\Omega)\), avec tous les supports contenus dans un compact \(K\). Alors les supports de \(\psi\varphi_k\) restent eux aussi dans \(K\). Posons \(u_k=\varphi_k-\varphi\). Pour \(|\beta|\le N\), la rรจgle de Leibniz donne \[|\partial^\beta(\psi u_k)|\le\sum_{\gamma\le\beta}\binom{\beta}{\gamma}|\partial^\gamma\psi|\,|\partial^{\beta-\gamma}u_k|.\] Sur le compact \(K\), les dรฉrivรฉes de \(\psi\) sont bornรฉes. On obtient donc \(p_{K,N}(\psi u_k)\le C_N M p_{K,N}(u_k)\to0\), avec des constantes finies \(C_N\) et \(M\). Par consรฉquent \(\psi\varphi_k\to\psi\varphi\) dans \(\mathcal D(\Omega)\).
Misconception. The bound uses that \(\psi\) and its derivatives are controlled only on the fixed \(K\), where they are bounded by continuity. \(\psi\) itself need not be bounded on all of \(\Omega\); compact support of the \(\varphi_k\) confines everything to \(K\).Erreur frรฉquente. La majoration nโutilise les bornes de \(\psi\) et de ses dรฉrivรฉes que sur le compact fixe \(K\), oรน elles sont bornรฉes par continuitรฉ. La fonction \(\psi\) nโa pas besoin dโรชtre bornรฉe sur tout \(\Omega\), car les supports des \(\varphi_k\) localisent le problรจme sur \(K\).Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
By clause (i) there is a compact \(K\) containing all supports, so \(\varphi_k-\varphi\) vanishes off \(K\) and \(|K|<\infty\). By clause (ii), \(\|\varphi_k-\varphi\|_\infty=p_{K,0}(\varphi_k-\varphi)\to0\). Then \[\Bigl|\int_{\mathbb R^n}(\varphi_k-\varphi)\Bigr|\le\int_K|\varphi_k-\varphi|\le|K|\,\|\varphi_k-\varphi\|_\infty\to0,\] so \(\int\varphi_k\to\int\varphi\). For continuous \(g\), \(m:=\sup_K|g|<\infty\), and \(g(\varphi_k-\varphi)\) is supported in \(K\), so \(\bigl|\int g(\varphi_k-\varphi)\bigr|\le|K|\,m\,\|\varphi_k-\varphi\|_\infty\to0\). Hence \(\int g\varphi_k\to\int g\varphi\).La condition (i) fournit un compact \(K\) contenant tous les supports. Ainsi \(\varphi_k-\varphi\) sโannule hors de \(K\), et \(|K|<\infty\). La condition (ii) donne \(\|\varphi_k-\varphi\|_\infty=p_{K,0}(\varphi_k-\varphi)\to0\). Alors \[\left|\int_{\mathbb R^n}(\varphi_k-\varphi)\right|\le\int_K|\varphi_k-\varphi|\le|K|\,\|\varphi_k-\varphi\|_\infty\to0,\] donc \(\int\varphi_k\to\int\varphi\). Si \(g\) est continue, \(m=\sup_K|g|<\infty\), et \(|\int g(\varphi_k-\varphi)|\le|K|m\|\varphi_k-\varphi\|_\infty\to0\). Ainsi \(\int g\varphi_k\to\int g\varphi\).
Misconception. This uses only the order-\(0\) seminorm and the common compact support; without a fixed \(K\) the bound \(|K|\,\|\cdot\|_\infty\) would be unavailable, which is again why clause (i) matters for duality (Ch. 2).Erreur frรฉquente. On utilise seulement la semi-norme dโordre \(0\) et le support compact commun. Sans compact fixe \(K\), la majoration \(|K|\,\|\cdot\|_\infty\) ne serait pas disponible, ce qui montre encore lโimportance de la condition (i) pour la dualitรฉ.Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Let \(V=\Omega\setminus\operatorname{supp}\varphi\), an open set on which \(\varphi\equiv0\). Every partial derivative of the identically-zero function on an open set is zero, so \(\partial^\alpha\varphi\equiv0\) on \(V\); hence \(\{\partial^\alpha\varphi\ne0\}\subseteq\operatorname{supp}\varphi\), and taking closures (\(\operatorname{supp}\varphi\) is already closed), \(\operatorname{supp}(\partial^\alpha\varphi)\subseteq\operatorname{supp}\varphi\). Strictness: on \(\mathbb R\), let \(\varphi=P\) be the plateau of WEx. 1.2, with \(\operatorname{supp} P=[-2,2]\) but \(P\equiv1\) on \([-1,1]\); then \(P'\equiv0\) on \((-1,1)\), so \(\operatorname{supp} P'\subseteq[-2,-1]\cup[1,2]\subsetneq[-2,2]=\operatorname{supp} P\).Soit \(V=\Omega\setminus\operatorname{supp}\varphi\), ouvert sur lequel \(\varphi\equiv0\). Toute dรฉrivรฉe partielle de la fonction identiquement nulle y est encore nulle, donc \(\partial^\alpha\varphi\equiv0\) sur \(V\). Il en rรฉsulte \(\operatorname{supp}(\partial^\alpha\varphi)\subseteq\operatorname{supp}\varphi\). Pour obtenir une inclusion stricte, prenons la fonction plateau \(P\) de lโexemple rรฉsolu 1.2, avec \(\operatorname{supp}P=[-2,2]\) et \(P\equiv1\) sur \([-1,1]\). Alors \(P'\equiv0\) sur \((-1,1)\), donc \(\operatorname{supp}P'\subseteq[-2,-1]\cup[1,2]\subsetneq[-2,2]=\operatorname{supp}P\).
Misconception. Differentiation can only shrink or preserve the support, never enlarge it; a flat region (where \(\varphi\) is locally constant and nonzero) is where the derivative's support genuinely drops out.Erreur frรฉquente. La dรฉrivation ne peut quโรฉventuellement rรฉduire le support, jamais lโagrandir. Une rรฉgion oรน \(\varphi\) est localement constante et non nulle disparaรฎt du support de sa dรฉrivรฉe.Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Define \(\varphi_i=\psi_i\varphi\). Each \(\varphi_i\in C^\infty\) (product of smooth functions) with \(\operatorname{supp}\varphi_i\subseteq\operatorname{supp}\psi_i\cap\operatorname{supp}\varphi\), a closed subset of the compact \(\operatorname{supp}\psi_i\subseteq U_i\); hence \(\varphi_i\in\mathcal D(U_i)\). Since \(\sum_i\psi_i\equiv1\) on a neighbourhood of \(K\supseteq\operatorname{supp}\varphi\), we have, at every \(x\), \(\sum_i\varphi_i(x)=\varphi(x)\sum_i\psi_i(x)=\varphi(x)\) (both sides vanish off \(\operatorname{supp}\varphi\), and on it \(\sum\psi_i=1\)). Thus \(\varphi=\sum_i\varphi_i\), splitting \(\varphi\) into pieces localized in the \(U_i\).Dรฉfinissons \(\varphi_i=\psi_i\varphi\). Chaque \(\varphi_i\) est \(C^\infty\) et \(\operatorname{supp}\varphi_i\subseteq\operatorname{supp}\psi_i\cap\operatorname{supp}\varphi\), compact contenu dans \(U_i\). Ainsi \(\varphi_i\in\mathcal D(U_i)\). Comme \(\sum_i\psi_i\equiv1\) sur un voisinage de \(K\supseteq\operatorname{supp}\varphi\), on a, pour tout \(x\), \(\sum_i\varphi_i(x)=\varphi(x)\sum_i\psi_i(x)=\varphi(x)\). Donc \(\varphi=\sum_i\varphi_i\), dรฉcomposition en morceaux localisรฉs dans les \(U_i\).
Misconception. The decomposition works because \(\sum\psi_i=1\) exactly where \(\varphi\) lives; off \(\operatorname{supp}\varphi\) the identity \(\sum\psi_i=1\) may fail, but there \(\varphi=0\) anyway, so the product \(\varphi\sum\psi_i=\varphi\) still holds.Erreur frรฉquente. La dรฉcomposition fonctionne parce que \(\sum_i\psi_i=1\) exactement lร oรน \(\varphi\) est supportรฉe. Hors de \(\operatorname{supp}\varphi\), cette รฉgalitรฉ peut รฉchouer, mais \(\varphi=0\) de toute faรงon, de sorte que \(\varphi\sum_i\psi_i=\varphi\) reste vraie.Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
For each \(x\in K\) choose \(i(x)\in\{1,2\}\) with \(x\in U_{i(x)}\) and a ball \(\overline{B(x,2r_x)}\subset U_{i(x)}\); finitely many \(B(x_j,r_j)\) cover \(K\). By Theorem 1.2 take \(\chi_j\ge0\), \(\chi_j\equiv1\) on \(\overline{B(x_j,r_j)}\), \(\operatorname{supp}\chi_j\subset B(x_j,2r_j)\subset U_{i(x_j)}\). Set \(\theta_i=\sum_{j:i(x_j)=i}\chi_j\in\mathcal D(U_i)\), \(\theta_i\ge0\); then \(\theta:=\theta_1+\theta_2\ge1\) on \(K\), so \(\theta>0\) on an open \(W\supseteq K\). Choose \(\zeta\in\mathcal D(W)\), \(0\le\zeta\le1\), \(\zeta\equiv1\) near \(K\). Define \(\psi_i=\zeta\theta_i/\theta\) on \(W\) and \(0\) off \(W\); as in Theorem 1.3 these are smooth (near \(\partial W\), \(\zeta=0\)), \(\operatorname{supp}\psi_i\subseteq U_i\), \(0\le\psi_i\), and \(\psi_1+\psi_2=\zeta\cdot\theta/\theta=\zeta\equiv1\) near \(K\).Pour chaque \(x\in K\), choisir \(i(x)\in\{1,2\}\) tel que \(x\in U_{i(x)}\), ainsi quโune boule fermรฉe \(\overline{B(x,2r_x)}\subset U_{i(x)}\). Un nombre fini de boules \(B(x_j,r_j)\) recouvre \(K\). Par le thรฉorรจme 1.2, choisir \(\chi_j\ge0\), รฉgale ร \(1\) sur \(\overline{B(x_j,r_j)}\) et supportรฉe dans \(B(x_j,2r_j)\subset U_{i(x_j)}\). Posons \(\theta_i=\sum_{j:i(x_j)=i}\chi_j\in\mathcal D(U_i)\). Alors \(\theta=\theta_1+\theta_2\ge1\) sur \(K\), donc \(\theta>0\) sur un ouvert \(W\supseteq K\). Choisissons \(\zeta\in\mathcal D(W)\), \(0\le\zeta\le1\), avec \(\zeta\equiv1\) prรจs de \(K\). Dรฉfinissons \(\psi_i=\zeta\theta_i/\theta\) sur \(W\) et \(0\) hors de \(W\). Comme dans le thรฉorรจme 1.3, les \(\psi_i\) sont lisses, supportรฉes dans \(U_i\), non nรฉgatives, et \(\psi_1+\psi_2=\zeta\equiv1\) prรจs de \(K\).
Misconception. One cannot simply set \(\psi_1=\theta_1/\theta\) globally: where \(\theta=0\) this is \(0/0\). The cutoff \(\zeta\), vanishing before \(\theta\) does, is what makes the quotient a genuine smooth compactly supported function.Erreur frรฉquente. On ne peut pas poser globalement \(\psi_1=\theta_1/\theta\), car aux points oรน \(\theta=0\) on obtiendrait \(0/0\). La fonction de coupure \(\zeta\), qui sโannule avant \(\theta\), rend le quotient lisse et ร support compact.Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Take \(\varphi_k(x)=\tfrac1k\rho(x-k)\). For every \(m\), \(\|\varphi_k^{(m)}\|_\infty=\tfrac1k\|\rho^{(m)}\|_\infty\to0\), so \(\varphi_k\to0\) in each \(C^m(\mathbb R)\)-norm; indeed uniformly with all derivatives. But \(\operatorname{supp}\varphi_k=[k-1,k+1]\), and no compact set contains all of these (given \(K\subseteq[-R,R]\), pick \(k>R+1\)). So clause (i) of Definition 1.5 fails and \(\varphi_k\not\to0\) in \(\mathcal D(\mathbb R)\). This shows \(\mathcal D\)-convergence is strictly stronger than uniform convergence of all derivatives.Prenons \(\varphi_k(x)=\tfrac1k\rho(x-k)\). Pour tout \(m\), \(\|\varphi_k^{(m)}\|_\infty=\tfrac1k\|\rho^{(m)}\|_\infty\to0\). Ainsi \(\varphi_k\to0\) pour chaque norme \(C^m\), donc uniformรฉment avec toutes ses dรฉrivรฉes. Cependant \(\operatorname{supp}\varphi_k=[k-1,k+1]\), et aucun compact ne contient tous ces supports. La condition (i) de la dรฉfinition 1.5 รฉchoue, donc \(\varphi_k\not\to0\) dans \(\mathcal D(\mathbb R)\). Cela montre que la convergence dans \(\mathcal D\) est strictement plus forte que la convergence uniforme de toutes les dรฉrivรฉes.
Misconception. The \(C^m\)-norms know nothing about location, only about size; the \(\mathcal D\)-topology additionally pins down the support. That extra rigidity is deliberate; it is what makes so many functionals continuous on \(\mathcal D\).Erreur frรฉquente. Les normes \(C^m\) mesurent la taille, mais pas la localisation. La topologie de \(\mathcal D\) contrรดle aussi le support. Cette rigiditรฉ supplรฉmentaire est volontaire et explique la continuitรฉ dโun grand nombre de formes linรฉaires sur \(\mathcal D\).Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Solution. Suppose a norm \(\|\cdot\|_*\) induced the usual LF topology on \(\mathcal D(\mathbb R)\). Its open unit ball \[ U=\{\varphi\in\mathcal D(\mathbb R):\|\varphi\|_*<1\} \] would be a bounded \(0\)-neighbourhood. By Proposition 1.2, boundedness would force the supports of all functions in \(U\) to lie in one compact set \(K\subset\mathbb R\).Solution. Supposons quโune norme \(\|\cdot\|_*\) induise la topologie LF usuelle sur \(\mathcal D(\mathbb R)\). Sa boule unitรฉ ouverte \[U=\{\varphi\in\mathcal D(\mathbb R):\|\varphi\|_*<1\}\] serait un voisinage bornรฉ de \(0\). Dโaprรจs la proposition 1.2, la bornitude imposerait que les supports de toutes les fonctions de \(U\) soient contenus dans un mรชme compact \(K\subset\mathbb R\).
Choose a nonzero test function \(\psi\) whose compact support lies outside \(K\), for example a sufficiently far translate of the standard bump. Every neighbourhood of \(0\) in a topological vector space is absorbing, so for sufficiently small \(\lambda\neq0\) one has \(\lambda\psi\in U\). But scalar multiplication by a nonzero scalar does not change support: \[ \operatorname{supp}(\lambda\psi)=\operatorname{supp}\psi\not\subset K, \] contradicting the bounded-set characterization. Therefore no norm can induce the topology of \(\mathcal D(\mathbb R)\).Choisissons une fonction test non nulle \(\psi\) dont le support compact est situรฉ hors de \(K\), par exemple une translatรฉe suffisamment รฉloignรฉe de la fonction bosse standard. Tout voisinage de \(0\) dans un espace vectoriel topologique est absorbant. Il existe donc un scalaire \(\lambda\ne0\), de module assez petit, tel que \(\lambda\psi\in U\). Mais la multiplication par un scalaire non nul ne modifie pas le support : \[\operatorname{supp}(\lambda\psi)=\operatorname{supp}\psi\not\subset K.\] Cela contredit la caractรฉrisation des ensembles bornรฉs. Aucune norme ne peut donc induire la topologie de \(\mathcal D(\mathbb R)\).
Key point. The obstruction is not merely the presence of infinitely many derivatives. A normed space has a bounded neighbourhood of \(0\), whereas every \(0\)-neighbourhood in \(\mathcal D(\mathbb R)\) contains suitably small test functions whose supports can be placed arbitrarily far away.Point essentiel. Lโobstruction ne provient pas seulement de la prรฉsence dโune infinitรฉ de dรฉrivรฉes. Un espace normรฉ possรจde un voisinage bornรฉ de \(0\), alors que tout voisinage de \(0\) dans \(\mathcal D(\mathbb R)\) contient des fonctions test dโamplitude suffisamment petite dont les supports peuvent รชtre placรฉs arbitrairement loin.
Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
\(B\) is not bounded. By Proposition 1.2, boundedness demands one compact \(K\) with \(\operatorname{supp}\varphi\subseteq K\) for all \(\varphi\in B\). But \(\operatorname{supp}\rho(\cdot-k)=[k-1,k+1]\), and \(\bigcup_k[k-1,k+1]\) is unbounded, so no such \(K\) exists; the first criterion already fails. In contrast, \(B'=\{\tfrac1k\rho\}\) is bounded: all supports equal \(K=[-1,1]\), and for each \(N\), \(\sup_k p_{K,N}(\tfrac1k\rho)=\sup_k\tfrac1k\,p_{K,N}(\rho)=p_{K,N}(\rho)<\infty\).Lโensemble \(B\) nโest pas bornรฉ. Dโaprรจs la proposition 1.2, la bornitude exige lโexistence dโun compact \(K\) contenant le support de toute \(\varphi\in B\). Or \(\operatorname{supp}\rho(\cdot-k)=[k-1,k+1]\), et lโunion de ces supports est non bornรฉe. Aucun tel compact nโexiste. En revanche, \(B'=\{\tfrac1k\rho\}\) est bornรฉ : tous les supports sont รฉgaux ร \(K=[-1,1]\), et pour chaque \(N\), \(\sup_k p_{K,N}(\tfrac1k\rho)=p_{K,N}(\rho)<\infty\).
Misconception. Bounded seminorms are not sufficient for boundedness in \(\mathcal D\): the family must also be trapped in one compact set. This is exactly the feature that distinguishes an LF-space from a Frรฉchet space like \(\mathcal D_K\).Erreur frรฉquente. La bornitude des semi-normes ne suffit pas pour quโun ensemble soit bornรฉ dans \(\mathcal D\) : tous ses รฉlรฉments doivent aussi รชtre supportรฉs dans un mรชme compact. Cโest une diffรฉrence essentielle entre un espace LF et un espace de Frรฉchet comme \(\mathcal D_K\).Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Let \(g\in C_c(\mathbb R^n)\), \(\operatorname{supp} g\subseteq K_0\) compact. Set \(g_\varepsilon=g*\rho_\varepsilon\). Then \(g_\varepsilon\in C^\infty\) (differentiating under the integral moves \(\partial^\alpha\) onto \(\rho_\varepsilon\)), and \(\operatorname{supp} g_\varepsilon\subseteq K_0+\overline{B(0,\varepsilon)}\subseteq K_1:=K_0+\overline{B(0,1)}\) for \(\varepsilon\le1\), a fixed compact set; so \(g_\varepsilon\in\mathcal D(\mathbb R^n)\) with supports in one \(K_1\). Since \(\int\rho_\varepsilon=1\), \[|g_\varepsilon(x)-g(x)|=\Bigl|\int[g(x-y)-g(x)]\rho_\varepsilon(y)\,dy\Bigr|\le\sup_{|y|\le\varepsilon}|g(x-y)-g(x)|.\] As \(g\) is uniformly continuous (continuous with compact support), the right side, taken over all \(x\), tends to \(0\) as \(\varepsilon\to0\). Hence \(\|g_\varepsilon-g\|_\infty\to0\): test functions with supports in the fixed \(K_1\) approximate \(g\) uniformly.Soit \(g\in C_c(\mathbb R^n)\), avec \(\operatorname{supp}g\subseteq K_0\) compact. Posons \(g_\varepsilon=g*\rho_\varepsilon\). Alors \(g_\varepsilon\in C^\infty\), car la dรฉrivation sous le signe intรฉgral fait porter \(\partial^\alpha\) sur \(\rho_\varepsilon\). De plus, \(\operatorname{supp}g_\varepsilon\subseteq K_0+\overline{B(0,\varepsilon)}\subseteq K_1:=K_0+\overline{B(0,1)}\) pour \(\varepsilon\le1\). Ainsi \(g_\varepsilon\in\mathcal D(\mathbb R^n)\) et tous les supports sont contenus dans le compact fixe \(K_1\). Comme \(\int\rho_\varepsilon=1\), \[|g_\varepsilon(x)-g(x)|\le\sup_{|y|\le\varepsilon}|g(x-y)-g(x)|.\] La fonction \(g\) est uniformรฉment continue, car elle est continue ร support compact. Le membre de droite tend donc uniformรฉment vers \(0\) lorsque \(\varepsilon\to0\). Ainsi \(\|g_\varepsilon-g\|_\infty\to0\).
Misconception. Density here is in the sup norm on \(C_c\), not in the \(\mathcal D\)-topology (the \(g_\varepsilon\) generally do not converge in \(\mathcal D\), since \(g\) need not be smooth and higher derivatives of \(g_\varepsilon\) can blow up as \(\varepsilon\to0\)).Erreur frรฉquente. La densitรฉ considรฉrรฉe ici porte sur la norme sup de \(C_c\), et non sur la topologie de \(\mathcal D\). En gรฉnรฉral, \(g_\varepsilon\) ne converge pas dans \(\mathcal D\), car \(g\) nโest pas nรฉcessairement lisse et les dรฉrivรฉes dโordre รฉlevรฉ de \(g_\varepsilon\) peuvent diverger lorsque \(\varepsilon\to0\).Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Solution. Fix \(1\le p<\infty\), \(f\in L^p(\mathbb R^n)\), and \(\eta>0\).Solution. Fixons \(1\le p<\infty\), \(f\in L^p(\mathbb R^n)\) et \(\eta>0\).
1. Truncate smoothly. Choose \(R>0\) so that \[ \|f\,\mathbf 1_{\mathbb R^n\setminus B_R}\|_{L^p}<\eta/2. \] By the smooth-cutoff theorem, choose \(\chi_R\in\mathcal D(\mathbb R^n)\) with \(0\le\chi_R\le1\), \(\chi_R\equiv1\) on \(\overline{B_R}\), and \(\operatorname{supp}\chi_R\subset B_{2R}\). Put \(f_R=\chi_R f\). Then \(f_R\in L^p\) has compact support and \[ \|f-f_R\|_p\le \|f\,\mathbf 1_{\mathbb R^n\setminus B_R}\|_p<\eta/2. \]1. Troncature lisse. Choisir \(R>0\) tel que \[\|f\mathbf1_{\mathbb R^n\setminus B_R}\|_{L^p}<\eta/2.\] Par le thรฉorรจme des fonctions de coupure lisses, choisir \(\chi_R\in\mathcal D(\mathbb R^n)\) telle que \(0\le\chi_R\le1\), \(\chi_R\equiv1\) sur \(\overline{B_R}\) et \(\operatorname{supp}\chi_R\subset B_{2R}\). Posons \(f_R=\chi_R f\). Alors \(f_R\in L^p\) est ร support compact et \[\|f-f_R\|_p\le\|f\mathbf1_{\mathbb R^n\setminus B_R}\|_p<\eta/2.\]
2. Mollify. Let \(g_\varepsilon=f_R*\rho_\varepsilon\). Then \(g_\varepsilon\in C^\infty\) and \[ \operatorname{supp}g_\varepsilon\subseteq \operatorname{supp}f_R+\overline{B(0,\varepsilon)}, \] which is compact; hence \(g_\varepsilon\in\mathcal D(\mathbb R^n)\). By Minkowski's integral inequality, \[ \|g_\varepsilon-f_R\|_p \le \int \rho_\varepsilon(y)\,\|f_R(\cdot-y)-f_R\|_p\,dy \le \sup_{|y|\le\varepsilon}\|f_R(\cdot-y)-f_R\|_p. \] Translations are continuous in \(L^p\) for \(1\le p<\infty\), so the last quantity tends to \(0\). Choose \(\varepsilon\) so that \(\|g_\varepsilon-f_R\|_p<\eta/2\). Then \[ \|f-g_\varepsilon\|_p<\eta. \] Thus \(\mathcal D(\mathbb R^n)\) is dense in \(L^p(\mathbb R^n)\) for \(1\le p<\infty\).2. Mollification. Posons \(g_\varepsilon=f_R*\rho_\varepsilon\). Alors \(g_\varepsilon\in C^\infty\) et \[\operatorname{supp}g_\varepsilon\subseteq\operatorname{supp}f_R+\overline{B(0,\varepsilon)},\] qui est compact ; ainsi \(g_\varepsilon\in\mathcal D(\mathbb R^n)\). Par lโinรฉgalitรฉ intรฉgrale de Minkowski, \[\|g_\varepsilon-f_R\|_p\le\int\rho_\varepsilon(y)\|f_R(\cdot-y)-f_R\|_p\,dy\le\sup_{|y|\le\varepsilon}\|f_R(\cdot-y)-f_R\|_p.\] Les translations sont continues dans \(L^p\) pour \(1\le p<\infty\), donc cette quantitรฉ tend vers \(0\). Choisir \(\varepsilon\) tel que \(\|g_\varepsilon-f_R\|_p<\eta/2\). Alors \(\|f-g_\varepsilon\|_p<\eta\). On conclut que \(\mathcal D(\mathbb R^n)\) est dense dans \(L^p(\mathbb R^n)\) pour \(1\le p<\infty\).
Why \(p=\infty\) is different. Translation is not continuous on all of \(L^\infty\), and \(\mathcal D(\mathbb R^n)\) is not dense in \(L^\infty(\mathbb R^n)\) in the essential-supremum norm.Pourquoi le cas \(p=\infty\) est diffรฉrent. La translation nโest pas continue sur tout \(L^\infty\), et \(\mathcal D(\mathbb R^n)\) nโest pas dense dans \(L^\infty(\mathbb R^n)\) pour la norme du supremum essentiel.
Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
The product \(\psi\varphi\) is smooth (product of \(C^\infty\) functions). Wherever \(\varphi(x)=0\), \(\psi(x)\varphi(x)=0\), so \(\{\psi\varphi\ne0\}\subseteq\{\varphi\ne0\}\), and taking closures \(\operatorname{supp}(\psi\varphi)\subseteq\operatorname{supp}\varphi\), a compact subset of \(\Omega\); hence \(\psi\varphi\in\mathcal D(\Omega)\). The map \((\psi,\varphi)\mapsto\psi\varphi\) is bilinear over the scalars, satisfies \((\psi_1\psi_2)\varphi=\psi_1(\psi_2\varphi)\) and \(1\cdot\varphi=\varphi\) (all pointwise identities), so \(\mathcal D(\Omega)\) is a module over the ring \(C^\infty(\Omega)\). Proposition 1.1(b) adds that for fixed \(\psi\) the action is sequentially continuous.Le produit \(\psi\varphi\) est lisse. Lร oรน \(\varphi(x)=0\), on a \(\psi(x)\varphi(x)=0\). Ainsi \(\{\psi\varphi\ne0\}\subseteq\{\varphi\ne0\}\), et en prenant les adhรฉrences \(\operatorname{supp}(\psi\varphi)\subseteq\operatorname{supp}\varphi\), compact contenu dans \(\Omega\). Donc \(\psi\varphi\in\mathcal D(\Omega)\). Lโaction vรฉrifie les identitรฉs ponctuelles \((\psi_1\psi_2)\varphi=\psi_1(\psi_2\varphi)\) et \(1\cdot\varphi=\varphi\), ainsi que la distributivitรฉ. Lโespace \(\mathcal D(\Omega)\) est donc un module sur lโanneau \(C^\infty(\Omega)\). La proposition 1.1(b) ajoute que, pour \(\psi\) fixรฉe, cette action est sรฉquentiellement continue.
Misconception. \(C^\infty(\Omega)\) multipliers need not have compact support; nonetheless \(\psi\varphi\) does, because \(\varphi\) confines the product. It is the test function that supplies compact support, not the multiplier.Erreur frรฉquente. Les multiplicateurs de \(C^\infty(\Omega)\) nโont pas besoin dโรชtre ร support compact. Le produit \(\psi\varphi\), lui, lโest parce que \(\varphi\) confine son support. Cโest la fonction test, et non le multiplicateur, qui fournit la compacitรฉ du support.Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Let \(\varphi\in\mathcal D(\mathbb R)\) be real-analytic. Since \(\operatorname{supp}\varphi\) is compact, \(\varphi\equiv0\) on the nonempty open set \(\mathbb R\setminus\operatorname{supp}\varphi\). A real-analytic function on the connected domain \(\mathbb R\) that vanishes on a nonempty open subset vanishes identically (identity theorem: the set where all derivatives vanish is open, closed, and nonempty, hence all of \(\mathbb R\)). Therefore \(\varphi\equiv0\). Consequently, for any nonzero \(\varphi\in\mathcal D(\mathbb R)\) and any boundary point \(a\) of \(\operatorname{supp}\varphi\), we have \(\varphi(a)=0\) and, by continuity of \(\varphi\equiv0\) just outside, \(\varphi^{(m)}(a)=0\) for all \(m\); the Taylor series of \(\varphi\) at \(a\) is identically \(0\), yet \(\varphi\not\equiv0\); so \(\varphi\) cannot equal its Taylor series near \(a\).Soit \(\varphi\in\mathcal D(\mathbb R)\) rรฉelle analytique. Comme son support est compact, \(\varphi\equiv0\) sur lโouvert non vide \(\mathbb R\setminus\operatorname{supp}\varphi\). Par le thรฉorรจme dโidentitรฉ, une fonction rรฉelle analytique sur le domaine connexe \(\mathbb R\) qui sโannule sur un ouvert non vide est identiquement nulle. Donc \(\varphi\equiv0\). En consรฉquence, pour toute fonction test non nulle \(\varphi\) et tout point frontiรจre \(a\) de son support, on a \(\varphi^{(m)}(a)=0\) pour tout \(m\). La sรฉrie de Taylor en \(a\) est donc identiquement nulle alors que \(\varphi\not\equiv0\). Une fonction test non nulle ne peut donc coรฏncider avec sa sรฉrie de Taylor prรจs dโun point frontiรจre de son support.
Misconception. Smooth is far weaker than analytic: the standard bump has a Taylor series \(\equiv0\) at \(\pm1\) yet is nonzero nearby. This gap; the existence of nonanalytic smooth functions; is precisely what makes \(\mathcal D\) nonempty and the whole theory possible.Erreur frรฉquente. รtre lisse est beaucoup plus faible quโรชtre analytique. La fonction bosse standard possรจde une sรฉrie de Taylor identiquement nulle en \(\pm1\), tout en รฉtant non nulle ร proximitรฉ. Lโexistence de fonctions lisses non analytiques est prรฉcisรฉment ce qui rend \(\mathcal D\) non trivial et la thรฉorie possible.Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Fix an exhaustion \(K_1\subset\operatorname{int}K_2\subset K_2\subset\cdots\), \(\bigcup_j K_j=\Omega\); each \(\mathcal D_{K_j}\) is a Frรฉchet space under \((p_{K_j,N})_N\), and \(\mathcal D(\Omega)=\bigcup_j\mathcal D_{K_j}\) with \(\mathcal D_{K_j}\subseteq\mathcal D_{K_{j+1}}\) a closed subspace inheriting its topology (this is what strict inductive limit means). Endow \(\mathcal D(\Omega)\) with the finest locally convex topology making every inclusion \(\mathcal D_{K_j}\hookrightarrow\mathcal D(\Omega)\) continuous; equivalently, a convex set \(U\ni0\) is a neighbourhood of \(0\) iff \(U\cap\mathcal D_{K_j}\) is a \(0\)-neighbourhood in each \(\mathcal D_{K_j}\). Sequences. A standard LF-space theorem states that any bounded set (hence any convergent sequence) is contained and bounded in a single \(\mathcal D_{K_j}\); there the topology is the Frรฉchet one of the seminorms \(p_{K_j,N}\). So \(\varphi_k\to\varphi\) in \(\mathcal D(\Omega)\) iff all \(\operatorname{supp}\varphi_k\) lie in one \(K_j\) and \(p_{K_j,N}(\varphi_k-\varphi)\to0\) for every \(N\); precisely Definition 1.5. Completeness. A strict inductive limit of a sequence of complete spaces is complete, so \(\mathcal D(\Omega)\) is complete. Non-metrizability. A strict inductive limit of a strictly increasing sequence of Frรฉchet spaces is never metrizable: if it were, it would be Frรฉchet, but the Baire category theorem then forces it to coincide with some \(\mathcal D_{K_j}\) (a proper closed subspace has empty interior), contradicting \(\bigcup_j\mathcal D_{K_j}=\mathcal D(\Omega)\) with strict inclusions. This is the rigorous form of Ex. 1.24, For linear functionals on \(\mathcal D(\Omega)\), the sequence criterion of Definition 1.5 is a practical characterization of continuity; the underlying definition remains continuity for the LF topology.Fixons une exhaustion \(K_1\subset\operatorname{int}K_2\subset K_2\subset\cdots\), avec \(\bigcup_jK_j=\Omega\). Chaque \(\mathcal D_{K_j}\) est un espace de Frรฉchet pour les semi-normes \((p_{K_j,N})_N\), et \(\mathcal D(\Omega)=\bigcup_j\mathcal D_{K_j}\), chaque \(\mathcal D_{K_j}\) รฉtant un sous-espace fermรฉ de \(\mathcal D_{K_{j+1}}\) muni de la topologie induite. Cโest le sens de ยซ limite inductive stricte ยป. On munit \(\mathcal D(\Omega)\) de la plus fine topologie localement convexe rendant continues toutes les inclusions \(\mathcal D_{K_j}\hookrightarrow\mathcal D(\Omega)\). Un thรฉorรจme standard sur les espaces LF affirme que tout ensemble bornรฉ, et donc toute suite convergente, est contenu et bornรฉ dans un mรชme \(\mathcal D_{K_j}\). La convergence y est alors celle des semi-normes \(p_{K_j,N}\). Ainsi \(\varphi_k\to\varphi\) dans \(\mathcal D(\Omega)\) si et seulement si les supports sont contenus dans un mรชme \(K_j\) et si \(p_{K_j,N}(\varphi_k-\varphi)\to0\) pour tout \(N\), ce qui est exactement la dรฉfinition 1.5. Une limite inductive stricte dโune suite dโespaces complets est complรจte, donc \(\mathcal D(\Omega)\) est complet. En revanche, une limite inductive stricte dโune suite strictement croissante dโespaces de Frรฉchet nโest pas mรฉtrisable. Si elle lโรฉtait, elle serait de Frรฉchet, et le thรฉorรจme de Baire forcerait lโespace ร coรฏncider avec lโun des \(\mathcal D_{K_j}\), ce qui contredirait les inclusions strictes. Pour les formes linรฉaires sur \(\mathcal D(\Omega)\), le critรจre sรฉquentiel de la dรฉfinition 1.5 fournit une caractรฉrisation pratique de la continuitรฉ, tandis que la dรฉfinition fondamentale reste la continuitรฉ pour la topologie LF.
Misconception. Non-metrizability does not make \(\mathcal D(\Omega)\) pathological or incomplete: it is a complete, barrelled, reflexive space. It simply is not first-countable at \(0\), so the topology cannot be captured by any sequence of balls; one must argue with the seminorm families \(p_{K,N}\) directly.Erreur frรฉquente. La non-mรฉtrisabilitรฉ ne rend pas \(\mathcal D(\Omega)\) pathologique ni incomplet : cโest un espace complet, tonnelรฉ et rรฉflexif. Il nโest simplement pas ร base dรฉnombrable de voisinages en \(0\). Sa topologie ne peut donc pas รชtre dรฉcrite par une suite de boules ; il faut travailler directement avec les familles de semi-normes \(p_{K,N}\).Compare your work line by line with the derivation above. Every requested clause of the exercise should be addressed, every formula should follow from a stated definition or justified calculation, and the final conclusion should answer the original question explicitly.
Chapter Synthesis
Concept map
Theorem dependency summary
Lemma 1.1 (flatness of \(e^{-1/t}\)) is the seed: composed with the polynomial \(1-|x|^2\) it yields the standard bump (Theorem 1.1), and normalizing gives the mollifier \(\rho_\varepsilon\) (Definition 1.6). Convolving an indicator with \(\rho_\delta\) produces smooth cutoffs (Theorem 1.2), and summing localized cutoffs and normalizing yields smooth partitions of unity (Theorem 1.3). Independently, the multi-index calculus (Definition 1.1) and the Leibniz rule support Proposition 1.1, the sequential continuity of \(\partial^\alpha\) and of multiplication by \(C^\infty\) functions, which are exactly the operations transposed to distributions in Chapter 4. Definition 1.5 fixes the convergence, and Proposition 1.2 records its LF-space nature and the correct notion of bounded set. Misconception 1.1 and Worked Example 1.3 isolate the load-bearing distinction: \(\mathcal D\)-convergence is strictly stronger than uniform convergence of all derivatives, because of the fixed-compact-support clause.Le lemme 1.1, portant sur la platitude de \(e^{-1/t}\), est le point de dรฉpart. Composรฉ avec le polynรดme \(1-|x|^2\), il produit la fonction bosse standard du thรฉorรจme 1.1 ; sa normalisation donne le mollificateur \(\rho_\varepsilon\) de la dรฉfinition 1.6. La convolution dโune indicatrice avec \(\rho_\delta\) fournit des fonctions de coupure lisses, et la combinaison puis la normalisation de coupures localisรฉes conduit aux partitions de lโunitรฉ lisses. Indรฉpendamment, le calcul multi-indice et la rรจgle de Leibniz soutiennent la proposition 1.1 sur la continuitรฉ sรฉquentielle de \(\partial^\alpha\) et de la multiplication par les fonctions \(C^\infty\). La dรฉfinition 1.5 fixe la convergence, tandis que la proposition 1.2 en prรฉcise la nature LF et la notion correcte dโensemble bornรฉ. Lโerreur frรฉquente 1.1 et lโexemple rรฉsolu 1.3 isolent la distinction essentielle : la convergence dans \(\mathcal D\) est strictement plus forte que la convergence uniforme de toutes les dรฉrivรฉes, en raison de la condition de support compact fixe.
Notation summary
- \(\alpha\in\mathbb N_0^n\), \(|\alpha|\), \(\alpha!\), \(x^\alpha\), \(\partial^\alpha\); multi-index calculus\(\alpha\in\mathbb N_0^n\), \(|\alpha|\), \(\alpha!\), \(x^\alpha\), \(\partial^\alpha\) ; calcul multi-indice
- \(\operatorname{supp} f=\overline{\{f\ne0\}}\); support\(\operatorname{supp} f=\overline{\{f\ne0\}}\) ; support
- \(\mathcal D(\Omega)=C_c^\infty(\Omega)\), \(\mathcal D_K(\Omega)\); test functions\(\mathcal D(\Omega)=C_c^\infty(\Omega)\), \(\mathcal D_K(\Omega)\) ; fonctions test
- \(p_{K,N}(\varphi)=\sup_{|\alpha|\le N,\,x\in K}|\partial^\alpha\varphi|\); seminorms\(p_{K,N}(\varphi)=\sup_{|\alpha|\le N,\,x\in K}|\partial^\alpha\varphi|\) ; semi-normes
- \(j,\ \rho,\ \rho_\varepsilon(x)=\varepsilon^{-n}\rho(x/\varepsilon)\); bump & mollifier\(j,\ \rho,\ \rho_\varepsilon(x)=\varepsilon^{-n}\rho(x/\varepsilon)\) ; fonction bosse et mollificateur
- \(\varphi_k\to\varphi\) in \(\mathcal D\); common \(K\) + all \(\partial^\alpha\) uniform\(\varphi_k\to\varphi\) dans \(\mathcal D\) ; compact commun \(K\) + convergence uniforme de toutes les \(\partial^\alpha\)
| English | Franรงais |
|---|---|
| test function | fonction test |
| multi-index | multi-indice |
| support; compact support | support ; support compact |
| bump function; mollifier | fonction bosse ; mollificateur |
| cutoff; partition of unity | fonction de coupure ; partition de lโunitรฉ |
| seminorm | semi-norme |
| convergence in \(\mathcal D\) | convergence dans \(\mathcal D\) |
Frequent misconceptions
- Convergence in \(\mathcal D\) is not pointwise or uniform convergence: it also demands a single compact set containing all supports (Ex. 1.14, 1.23).La convergence dans \(\mathcal D\) nโest ni une simple convergence ponctuelle ni une simple convergence uniforme : elle exige aussi un compact unique contenant tous les supports (ex. 1.14, 1.23).
- A bump marching to infinity, \(\tfrac1k\rho(x-k)\), converges to \(0\) uniformly with all derivatives yet not in \(\mathcal D\).Une bosse qui se dรฉplace vers lโinfini, \(\tfrac1k\rho(x-k)\), converge vers \(0\) uniformรฉment avec toutes ses dรฉrivรฉes, mais pas dans \(\mathcal D\).
- Rescaling the argument, \(\tfrac1k\rho(kx)\), fixes the support but blows up the derivatives, so it does not converge in \(\mathcal D\) either (Ex. 1.15).Le changement dโรฉchelle \(\tfrac1k\rho(kx)\) maintient le support dans un compact fixe mais empรชche la convergence des dรฉrivรฉes ; la suite ne converge donc pas dans \(\mathcal D\) (ex. 1.15).
- Smooth is much weaker than analytic: no nonzero test function is real-analytic (Ex. 1.29); the bump's Taylor series vanishes at the boundary.รtre lisse est beaucoup plus faible quโรชtre analytique : aucune fonction test non nulle nโest rรฉelle analytique (ex. 1.29), et la sรฉrie de Taylor dโune fonction bosse sโannule au bord de son support.
- \(\partial^\alpha x^\beta=0\) is a coordinatewise test (\(\alpha_i>\beta_i\) for some \(i\)), not a test on total orders \(|\alpha|>|\beta|\) (Ex. 1.4).La condition \(\partial^\alpha x^\beta=0\) se vรฉrifie coordonnรฉe par coordonnรฉe : il faut \(\alpha_i>\beta_i\) pour au moins un indice \(i\), et non seulement \(|\alpha|>|\beta|\) (ex. 1.4).
- \(\mathcal D(\Omega)\) is not normable or metrizable; bounded means "common compact support + bounded seminorms" (Prop. 1.2, Ex. 1.25).\(\mathcal D(\Omega)\) nโest ni normable ni mรฉtrisable ; รชtre bornรฉ signifie ยซ support compact commun + semi-normes bornรฉes ยป (prop. 1.2, ex. 1.25).
Oral examination questions
- State the flatness lemma for \(e^{-1/t}\) and prove that all derivatives vanish at \(0\); deduce the existence of the standard bump.รnoncer le lemme de platitude pour \(e^{-1/t}\) et dรฉmontrer que toutes les dรฉrivรฉes sโannulent en \(0\) ; en dรฉduire lโexistence de la fonction bosse standard.
- Define convergence in \(\mathcal D(\Omega)\) and give two sequences that converge uniformly to \(0\) but fail to converge in \(\mathcal D\), for two different reasons.Dรฉfinir la convergence dans \(\mathcal D(\Omega)\) et donner deux suites qui convergent uniformรฉment vers \(0\) mais ne convergent pas dans \(\mathcal D\), pour deux raisons diffรฉrentes.
- Construct a smooth cutoff for a compact \(K\) inside an open \(U\); where is each hypothesis used?Construire une fonction de coupure lisse pour un compact \(K\) contenu dans un ouvert \(U\). Indiquer oรน chacune des hypothรจses est utilisรฉe.
- State and prove the existence of a smooth partition of unity subordinate to a finite open cover of a compact set.รnoncer et dรฉmontrer lโexistence dโune partition de lโunitรฉ lisse subordonnรฉe ร un recouvrement ouvert fini dโun compact.
- Prove that \(\partial^\alpha\) and multiplication by \(\psi\in C^\infty\) are sequentially continuous on \(\mathcal D(\Omega)\).Dรฉmontrer que \(\partial^\alpha\) et la multiplication par \(\psi\in C^\infty\) sont sรฉquentiellement continues sur \(\mathcal D(\Omega)\).
- Explain why \(\mathcal D(\Omega)\) is not metrizable and what "bounded" means there; relate this to the LF-space structure.Expliquer pourquoi \(\mathcal D(\Omega)\) nโest pas mรฉtrisable et prรฉciser la notion dโensemble bornรฉ dans cet espace ; relier ces faits ร la structure dโespace LF.
Assemble a self-contained portfolio proving, in order: (1) the flatness lemma and the existence of \(j\in\mathcal D(\mathbb R^n)\) with \(\operatorname{supp} j=\overline{B(0,1)}\) (Lem. 1.1, Thm. 1.1); (2) the smooth cutoff for \(K\subset U\) by convolution with \(\rho_\delta\), and from it a smooth partition of unity subordinate to a finite cover (Thm. 1.2โ1.3); (3) the sequential continuity of \(\partial^\alpha\) and of \(\varphi\mapsto\psi\varphi\), together with the two counterexamples showing \(\mathcal D\)-convergence exceeds uniform convergence (Prop. 1.1, MIS 1.1). These are the structural facts on which the duality of Chapter 2 rests.Constituer un dossier de dรฉmonstrations autonome รฉtablissant, dans lโordre : (1) le lemme de platitude et lโexistence de \(j\in\mathcal D(\mathbb R^n)\) avec \(\operatorname{supp}j=\overline{B(0,1)}\) (lem. 1.1, th. 1.1) ; (2) la fonction de coupure lisse pour \(K\subset U\) obtenue par convolution avec \(\rho_\delta\), puis une partition de lโunitรฉ lisse subordonnรฉe ร un recouvrement fini (th. 1.2-1.3) ; (3) la continuitรฉ sรฉquentielle de \(\partial^\alpha\) et de \(\varphi\mapsto\psi\varphi\), ainsi que les deux contre-exemples montrant que la convergence dans \(\mathcal D\) est plus forte que la convergence uniforme (prop. 1.1, erreur frรฉquente 1.1). Ce sont les faits structurels sur lesquels repose la dualitรฉ du chapitre 2.
Distributions are defined by duality: \(\mathcal D'(\Omega)\) is the space of linear functionals on \(\mathcal D(\Omega)\) continuous for the convergence of Definition 1.5. A general principle governs the trade-off; the smaller and more rigidly controlled the test space, the larger its dual. Because \(\mathcal D(\Omega)\) is so small (smooth, compactly supported) and its convergence so demanding (common compact support plus all derivatives), an enormous variety of functionals qualify as continuous: point evaluations \(\varphi\mapsto\varphi(x_0)\), integration against any locally integrable function or measure, and their derivatives of all orders. Enlarging the test space shrinks the dual. Relaxing "compact support" to "rapid decay" gives the Schwartz space \[\mathcal S(\mathbb R^n)=\{\varphi\in C^\infty:\ \sup_x|x^\beta\partial^\alpha\varphi(x)|<\infty\ \ \forall\alpha,\beta\},\] a Frรฉchet space (metrizable, unlike \(\mathcal D\)) on which every seminorm controls polynomial growth of every derivative. Its dual \(\mathcal S'(\mathbb R^n)\), the tempered distributions, is smaller than \(\mathcal D'\); it excludes objects that grow too fast, such as \(e^{x}\) as a distribution on \(\mathbb R\); but it is exactly the class on which the Fourier transform acts as an isomorphism, because \(\mathcal S\) is Fourier-invariant. The inclusions \(\mathcal D\subset\mathcal S\subset\mathcal E=C^\infty\) dualize to \(\mathcal E'\subset\mathcal S'\subset\mathcal D'\) (compactly supported, tempered, and general distributions). Everything in this chapter; bumps, mollifiers, seminorms, the delicate convergence; is the apparatus that makes these dualities precise.Les distributions sont dรฉfinies par dualitรฉ : \(\mathcal D'(\Omega)\) est lโespace des formes linรฉaires sur \(\mathcal D(\Omega)\) continues pour la convergence de la dรฉfinition 1.5. Un principe gรฉnรฉral gouverne lโรฉquilibre : plus lโespace de fonctions test est petit et fortement contrรดlรฉ, plus son dual est grand. Comme \(\mathcal D(\Omega)\) est trรจs contraint, avec des fonctions lisses ร support compact et une convergence imposant un support compact commun ainsi que toutes les dรฉrivรฉes, une trรจs grande variรฉtรฉ de formes linรฉaires sont continues : รฉvaluations ponctuelles \(\varphi\mapsto\varphi(x_0)\), intรฉgration contre toute fonction localement intรฉgrable ou toute mesure, ainsi que leurs dรฉrivรฉes de tout ordre. Agrandir lโespace de fonctions test rรฉduit le dual. Remplacer le support compact par une dรฉcroissance rapide conduit ร lโespace de Schwartz \[\mathcal S(\mathbb R^n)=\{\varphi\in C^\infty:\ \sup_x|x^\beta\partial^\alpha\varphi(x)|<\infty\ \forall\alpha,\beta\},\] espace de Frรฉchet, donc mรฉtrisable contrairement ร \(\mathcal D\), dont les semi-normes contrรดlent toute croissance polynomiale de toutes les dรฉrivรฉes. Son dual \(\mathcal S'(\mathbb R^n)\), lโespace des distributions tempรฉrรฉes, est plus petit que \(\mathcal D'\) : il exclut les objets croissant trop rapidement, tels que \(e^x\) considรฉrรฉ comme distribution sur \(\mathbb R\). En revanche, il constitue exactement la classe sur laquelle la transformรฉe de Fourier agit comme un isomorphisme, puisque \(\mathcal S\) est stable par transformation de Fourier. Les inclusions \(\mathcal D\subset\mathcal S\subset\mathcal E=C^\infty\) se dualisent en \(\mathcal E'\subset\mathcal S'\subset\mathcal D'\), correspondant respectivement aux distributions ร support compact, tempรฉrรฉes et gรฉnรฉrales. Tout ce chapitre, fonctions bosses, mollificateurs, semi-normes et convergence subtile, fournit lโappareil nรฉcessaire pour rendre ces dualitรฉs rigoureuses.
Connections to later courses
The seminorms \(p_{K,N}\) and Definition 1.5 are the continuity yardstick for distributions (Ch. 2); the order of a distribution is the smallest \(N\) that suffices locally (Ch. 5). Mollifiers reappear as the regularization \(T*\rho_\varepsilon\to T\) that proves \(\mathcal D\) is dense in \(\mathcal D'\) and drives convergence of distributions (Ch. 6). Cutoffs and partitions of unity define restriction, support, and the gluing of locally-defined distributions into global ones (Ch. 5), and underlie the local structure theorems. The continuity of \(\partial^\alpha\) and of multiplication by \(C^\infty\) functions (Prop. 1.1) is transposed to define distributional derivatives and the \(C^\infty\)-module structure of \(\mathcal D'\) (Ch. 4). Beyond this course, the same duality frames Sobolev spaces, fundamental solutions of PDE, and; through the Schwartz space above; the Fourier analysis of tempered distributions.Les semi-normes \(p_{K,N}\) et la dรฉfinition 1.5 constituent le critรจre de continuitรฉ des distributions au chapitre 2 ; lโordre dโune distribution est le plus petit \(N\) suffisant localement. Les mollificateurs rรฉapparaissent dans la rรฉgularisation \(T*\rho_\varepsilon\to T\), qui joue un rรดle central dans lโapproximation et la convergence des distributions. Les fonctions de coupure et les partitions de lโunitรฉ permettent de dรฉfinir la restriction, le support et le recollement des distributions dรฉfinies localement, et sous-tendent les thรฉorรจmes de structure locale. La continuitรฉ de \(\partial^\alpha\) et de la multiplication par des fonctions \(C^\infty\) se transpose pour dรฉfinir les dรฉrivรฉes au sens des distributions et la structure de module sur \(C^\infty\) de \(\mathcal D'\). Au-delร de ce cours, la mรชme dualitรฉ intervient dans les espaces de Sobolev, les solutions fondamentales des EDP et, via lโespace de Schwartz, lโanalyse de Fourier des distributions tempรฉrรฉes.
Readiness self-assessment
If every box is checked, proceed to Chapter 2: Distributions as Continuous Linear Functionals ยท not yet released, where \(\mathcal D(\Omega)\) becomes the domain and its continuous dual \(\mathcal D'(\Omega)\); the distributions; takes centre stage.Si toutes les cases sont cochรฉes, vous pourrez poursuivre avec le chapitre 2, Distributions comme formes linรฉaires continues, qui nโest pas encore publiรฉ. Lโespace \(\mathcal D(\Omega)\) y devient le domaine dโaction, tandis que son dual continu \(\mathcal D'(\Omega)\), lโespace des distributions, occupe le premier plan.