{"id":35283,"date":"2024-12-20T13:00:01","date_gmt":"2024-12-20T13:00:01","guid":{"rendered":"https:\/\/toposuranos.com\/material\/?p=35283"},"modified":"2025-12-11T17:08:02","modified_gmt":"2025-12-11T17:08:02","slug":"theoreme-de-weierstrass-des-valeurs-extremes","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/fr\/theoreme-de-weierstrass-des-valeurs-extremes\/","title":{"rendered":"Th\u00e9or\u00e8me de Weierstrass des Valeurs Extr\u00eames"},"content":{"rendered":"<style>\np, ul, ol{\ntext-align: justify;\n}\nh1{\ntext-align:center;\ntext-transform: uppercase;\n}\nh2{\ntext-align:center;\ntext-transform: uppercase;\nfont-size:24pt;\n}\nh3 { \n    text-align: center;\n    text-transform: uppercase;\n    font-size: 24px !important;\n}\n<\/style>\n<h1>Th\u00e9or\u00e8me de Weierstrass des Valeurs Extr\u00eames<\/h1>\n<p style=\"text-align:center;\"><em>Pourquoi, dans tant de probl\u00e8mes d\u2019optimisation, consid\u00e8re-t-on presque comme \u00e9vident que \u00ab le maximum existe \u00bb ou qu\u2019\u00ab il y a toujours un minimum \u00bb sur un certain intervalle, alors qu\u2019en r\u00e9alit\u00e9 rien n\u2019obliga a priori \u00e0 ce que cela se produise? Le <strong>Th\u00e9or\u00e8me de Weierstrass<\/strong> est la pi\u00e8ce manquante de ce puzzle: il garantit qu\u2019une fonction continue d\u00e9finie sur un intervalle ferm\u00e9 et born\u00e9 non seulement est born\u00e9e, mais atteint effectivement ses valeurs extr\u00eames. Dans cette entr\u00e9e, nous examinons son \u00e9nonc\u00e9, construisons en d\u00e9tail une d\u00e9monstration rigoureuse fond\u00e9e sur la continuit\u00e9 ponctuelle, la compacit\u00e9 et l\u2019axiome du supremum, et commentons son interpr\u00e9tation moderne en termes de fonctions continues sur des ensembles compacts. L\u2019objectif est qu\u2019\u00e0 la fin vous ne reteniez pas seulement le th\u00e9or\u00e8me comme une phrase, mais que vous compreniez pourquoi il est vrai et pourquoi il appara\u00eet de mani\u00e8re r\u00e9currente en analyse, en optimisation et dans des mod\u00e8les appliqu\u00e9s.<\/em><\/p>\n<p style=\"text-align:center;\"><b>Objectifs d\u2019apprentissage<\/b><\/p>\n<ol>\n<li>\n    <strong>Comprendre l\u2019\u00e9nonc\u00e9 du Th\u00e9or\u00e8me de Weierstrass.<\/strong><br \/>\n    Identifier avec pr\u00e9cision les hypoth\u00e8ses du th\u00e9or\u00e8me (fonction continue sur un intervalle ferm\u00e9 et born\u00e9 <span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span>) et ses conclusions principales: bornitude et existence de valeurs maximale et minimale.\n  <\/li>\n<li>\n    <strong>Interpr\u00e9ter le Th\u00e9or\u00e8me de Weierstrass en termes de compacit\u00e9.<\/strong><br \/>\n    Formuler le r\u00e9sultat en langage moderne: les fonctions continues envoient des ensembles compacts vers des ensembles o\u00f9 les valeurs extr\u00eames sont atteintes, reliant le cas de <span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span> au cadre g\u00e9n\u00e9ral de l\u2019analyse r\u00e9elle.\n  <\/li>\n<li>\n    <strong>Relier le Th\u00e9or\u00e8me de Weierstrass aux probl\u00e8mes d\u2019optimisation.<\/strong><br \/>\n    Reconna\u00eetre le r\u00f4le du th\u00e9or\u00e8me comme fondement th\u00e9orique de l\u2019existence de maxima et minima dans de nombreux probl\u00e8mes d\u2019optimisation en une variable, tant dans des contextes th\u00e9oriques qu\u2019appliqu\u00e9s.\n  <\/li>\n<\/ol>\n<p style=\"text-align:center;\"><b><u>INDEX DES CONTENUS<\/u>:<\/b><br \/>\n<a href=\"#1\"><b>Introduction<\/b><\/a><br \/>\n<a href=\"#2\"><b>\u00c9nonc\u00e9 du Th\u00e9or\u00e8me de Weierstrass<\/b><\/a><br \/>\n<a href=\"#3\">D\u00e9monstration<\/a><br \/>\n<a href=\"#4\">\u00c9tape 1: Continuit\u00e9 ponctuelle sur <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/a><br \/>\n<a href=\"#5\">\u00c9tape 2: Recouvrement ouvert associ\u00e9 \u00e0 la continuit\u00e9<\/a><br \/>\n<a href=\"#6\">\u00c9tape 3: Compacit\u00e9 de <span dir=\"ltr\">[a,b]<\/span> et sous-recouvrement fini<\/a><br \/>\n<a href=\"#7\">\u00c9tape 4: Construction d\u2019un <span class=\"katex-eq\" data-katex-display=\"false\">\\delta<\/span> ind\u00e9pendant de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span> (continuit\u00e9 uniforme)<\/a><br \/>\n<a href=\"#8\">\u00c9tape 5: De la continuit\u00e9 uniforme \u00e0 la bornitude de <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> sur <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/a><br \/>\n<a href=\"#9\">\u00c9tape 6: Existence de valeurs maximale et minimale<\/a><br \/>\n<a href=\"#10\"><b>Interpr\u00e9tation en termes de compacit\u00e9 et conclusion<\/b><\/a>\n<\/p>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/N5mSrhJgCds\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><br \/>\n<a name=\"1\"><\/a><\/br><\/p>\n<h2>Introduction<\/h2>\n<p>\nLe <strong>Th\u00e9or\u00e8me de Weierstrass des Valeurs Extr\u00eames<\/strong> est l\u2019un de ces r\u00e9sultats qui, bien qu\u2019il apparaisse g\u00e9n\u00e9ralement dans les premi\u00e8res unit\u00e9s d\u2019Analyse R\u00e9elle, soutient en r\u00e9alit\u00e9 de mani\u00e8re silencieuse une grande partie des math\u00e9matiques appliqu\u00e9es. Chaque fois qu\u2019en physique, en \u00e9conomie ou en statistique nous parlons de \u00ab maximiser \u00bb ou de \u00ab minimiser \u00bb une quantit\u00e9 soumise \u00e0 certaines contraintes, nous utilisons en arri\u00e8re-plan une id\u00e9e tr\u00e8s proche de celle que garantit ce th\u00e9or\u00e8me: une fonction continue d\u00e9finie sur un intervalle ferm\u00e9 et born\u00e9 <strong>non seulement est born\u00e9e, mais atteint effectivement ses valeurs extr\u00eames<\/strong>.\n<\/p>\n<p>\nIntuitivement, il peut sembler \u00ab \u00e9vident \u00bb que si nous tra\u00e7ons une courbe continue sur un segment <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>, alors il doit exister un point le plus \u00e9lev\u00e9 et un point le plus bas. Cependant, il suffit d\u2019effectuer de petits changements dans les hypoth\u00e8ses pour que cette intuition \u00e9choue de mani\u00e8re spectaculaire: si nous ouvrons l\u2019intervalle, si la fonction cesse d\u2019\u00eatre continue ou si le domaine n\u2019est pas born\u00e9, les maxima et minima peuvent simplement dispara\u00eetre. Le Th\u00e9or\u00e8me de Weierstrass met de l\u2019ordre dans cette intuition et nous indique pr\u00e9cis\u00e9ment <em>quand<\/em> nous pouvons nous y fier et <em>pourquoi<\/em>.\n<\/p>\n<p>\nD\u2019un point de vue th\u00e9orique, ce th\u00e9or\u00e8me constitue la premi\u00e8re rencontre s\u00e9rieuse avec l\u2019id\u00e9e de <strong>compacit\u00e9<\/strong>: dans le langage moderne, il affirme qu\u2019une fonction continue transforme des ensembles compacts en ensembles compacts. D\u2019un point de vue pratique, cela se traduit par l\u2019existence de solutions pour de nombreux probl\u00e8mes d\u2019optimisation en une dimension, et sera un \u00e9l\u00e9ment cl\u00e9 pour des r\u00e9sultats ult\u00e9rieurs tels que le <b>Th\u00e9or\u00e8me des Accroissements Finis<\/b> et, en derni\u00e8re instance, pour comprendre sereinement le Th\u00e9or\u00e8me Fondamental du Calcul.\n<\/p>\n<p>\nDans cette section, nous \u00e9noncerons le Th\u00e9or\u00e8me de Weierstrass et d\u00e9velopperons en d\u00e9tail sa d\u00e9monstration, en nous appuyant sur la notion de continuit\u00e9 sur <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> et sur l\u2019axiome du supremum. L\u2019id\u00e9e est que ce texte vous serve de r\u00e9f\u00e9rence solide: aussi bien pour \u00e9tudier le r\u00e9sultat lui-m\u00eame que pour y revenir chaque fois que vous aurez besoin de l\u2019utiliser pour d\u00e9montrer d\u2019autres th\u00e9or\u00e8mes ou pour justifier rigoureusement l\u2019existence de maxima et minima dans des probl\u00e8mes concrets.\n<\/p>\n<p><a name=\"2\"><\/a><\/br><\/p>\n<h2>\u00c9nonc\u00e9 du Th\u00e9or\u00e8me de Weierstrass<\/h2>\n<table>\n<tbody>\n<tr>\n<td style=\"text-align: justify; background-color: #e0e0ff;\">\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=N5mSrhJgCds&amp;t=439s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Toute fonction <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> d\u00e9finie<\/span><\/strong><\/a> et continue sur <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b],<\/span><\/span> est born\u00e9e et poss\u00e8de des valeurs minimale et maximale, <span class=\"katex-eq\" data-katex-display=\"false\">m<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">M<\/span>, telles que si <span class=\"katex-eq\" data-katex-display=\"false\">x\\in[a,b]<\/span>, alors <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)\\in[m,M]<\/span><\/span>.<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><a name=\"3\"><\/a><\/br><\/p>\n<h3>D\u00e9monstration<\/h3>\n<p>\nMontrons que si <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f:[a,b]\\to\\mathbb{R}<\/span><\/span><\/strong> est continue sur l\u2019intervalle ferm\u00e9 et born\u00e9 <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>, alors <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> est born\u00e9e et atteint une valeur maximale et une valeur minimale sur <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>. Nous diviserons la d\u00e9monstration en deux grandes parties:\n<\/p>\n<ul>\n<li>Tout d\u2019abord, nous montrerons que la continuit\u00e9 de <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> sur <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> implique que <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> est <em>uniform\u00e9ment continue<\/em>, et \u00e0 partir de cela nous d\u00e9duirons qu\u2019elle est <strong>born\u00e9e<\/strong>.<\/li>\n<li>Ensuite, en utilisant l\u2019axiome du supremum, nous prouverons que <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> atteint ses valeurs maximale et minimale sur l\u2019intervalle.<\/li>\n<\/ul>\n<p><a name=\"4\"><\/a><\/br><\/p>\n<h4><b>\u00c9tape 1:<\/b> Continuit\u00e9 ponctuelle sur <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/h4>\n<p>\nPar hypoth\u00e8se, <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> est continue en chaque point <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0\\in[a,b]<\/span><\/span><\/strong>. Selon la d\u00e9finition de la continuit\u00e9 en termes de <span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">\\delta<\/span>, cela signifie que:\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n(\\forall x_0\\in[a,b])(\\forall \\epsilon\\gt 0)(\\exists \\delta(x_0)\\gt 0)\n\n\\big(|x-x_0|\\lt\\delta(x_0)\\Rightarrow |f(x)-f(x_0)|\\lt\\epsilon\\big).\n\n<\/span>\n<\/p>\n<p>\n\u00c0 ce stade, le nombre <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta(x_0)<\/span><\/span><\/strong> peut d\u00e9pendre du point <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong>. Notre objectif imm\u00e9diat sera de construire, \u00e0 partir de ces <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta(x_0)<\/span><\/span><\/strong>, un nombre unique <strong><span class=\"katex-eq\" data-katex-display=\"false\">\\delta<\/span><\/strong> qui ne d\u00e9pende pas de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong> et qui fonctionne simultan\u00e9ment pour tous les points de l\u2019intervalle.\n<\/p>\n<p><a name=\"5\"><\/a><\/br><\/p>\n<h4><b>\u00c9tape 2:<\/b> Recouvrement ouvert associ\u00e9 \u00e0 la continuit\u00e9<\/h4>\n<p>\nFixons un <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon\\gt 0<\/span><\/span><\/strong> quelconque. Pour chaque <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0\\in[a,b]<\/span><\/span><\/strong>, la continuit\u00e9 de <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> nous permet de choisir un nombre <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta(x_0)\\gt 0<\/span><\/span><\/strong> tel que\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|x-x_0|\\lt\\delta(x_0)\\Rightarrow |f(x)-f(x_0)|\\lt\\frac{\\epsilon}{2}.\n\n<\/span>\n<\/p>\n<p>\n\u00c0 partir de ces valeurs, nous d\u00e9finissons, pour chaque <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0\\in[a,b]<\/span><\/span><\/strong>, un intervalle ouvert\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nI_{x_0}=\\left(x_0-\\frac{\\delta(x_0)}{2},\\,x_0+\\frac{\\delta(x_0)}{2}\\right).\n\n<\/span>\n<\/p>\n<p>\nChaque <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{x_0}<\/span><\/span><\/strong> est un ensemble ouvert dans <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}<\/span> et, de plus, la famille\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n\\{I_{x_0}\\}_{x_0\\in[a,b]}\n\n<\/span>\n<\/p>\n<p>\nforme un <strong>recouvrement ouvert<\/strong> de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>. En effet, \u00e9tant donn\u00e9 un point quelconque <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y\\in[a,b]<\/span><\/span><\/strong>, il suffit de prendre <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0=y<\/span><\/span><\/strong> ; par construction, <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y\\in I_y<\/span><\/span><\/strong>. Ainsi, chaque point de l\u2019intervalle appartient \u00e0 au moins un des ouverts <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{x_0}<\/span><\/span><\/strong>.\n<\/p>\n<p>\nCette famille d\u2019ouverts est en g\u00e9n\u00e9ral <strong>infinie<\/strong> (il y en a un pour chaque <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0\\in[a,b]<\/span><\/span><\/strong>). C\u2019est ici qu\u2019intervient la compacit\u00e9 de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>.\n<\/p>\n<p><a name=\"6\"><\/a><\/br><\/p>\n<h4><b>\u00c9tape 3:<\/b> Compacit\u00e9 de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span> et sous-recouvrement fini<\/h4>\n<p>\nNous savons, d\u2019apr\u00e8s le Th\u00e9or\u00e8me de Heine\u2013Borel, qu\u2019un sous-ensemble de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}<\/span><\/span> est compact si et seulement s\u2019il est ferm\u00e9 et born\u00e9. L\u2019intervalle <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> est ferm\u00e9 et born\u00e9, donc il est compact. Par d\u00e9finition de la compacit\u00e9, cela signifie que :\n<\/p>\n<p>\nDe <strong>tout<\/strong> recouvrement ouvert de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> (m\u00eame s\u2019il contient une infinit\u00e9 d\u2019ensembles), on peut extraire un <strong>sous-recouvrement fini<\/strong>.\n<\/p>\n<p>\nEn appliquant cette propri\u00e9t\u00e9 au recouvrement ouvert <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{I_{x_0}\\}_{x_0\\in[a,b]}<\/span><\/span><\/strong>, il s\u2019ensuit qu\u2019il existe des points <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_1,\\dots,x_N\\in[a,b]<\/span><\/span><\/strong> tels que les intervalles correspondants\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nI_{x_1},\\, I_{x_2},\\,\\dots,\\,I_{x_N}\n\n<\/span>\n<\/p>\n<p>\nils continuent de recouvrir tout l\u2019intervalle :\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n[a,b]\\subset I_{x_1}\\cup I_{x_2}\\cup\\cdots\\cup I_{x_N}.\n\n<\/span>\n<\/p>\n<p>\nNous sommes ainsi pass\u00e9s d\u2019une famille infinie d\u2019intervalles ouverts \u00e0 un sous-recouvrement contenant seulement un <strong>nombre fini<\/strong> d\u2019intervalles, sans perdre la propri\u00e9t\u00e9 de recouvrir <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>.\n<\/p>\n<p><a name=\"7\"><\/a><\/br><\/p>\n<h4><b>\u00c9tape 4:<\/b> Construction d\u2019un <span class=\"katex-eq\" data-katex-display=\"false\">\\delta<\/span> qui ne d\u00e9pend pas de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span> (continuit\u00e9 uniforme)<\/h4>\n<p>\n\u00c0 partir du sous-recouvrement fini, nous d\u00e9finissons le nombre\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n\\delta=\\min\\left\\{\\frac{\\delta(x_1)}{2},\\frac{\\delta(x_2)}{2},\\dots,\\frac{\\delta(x_N)}{2}\\right\\}.\n\n<\/span>\n<\/p>\n<p>\nComme il s\u2019agit du minimum d\u2019un ensemble fini de nombres positifs, il s\u2019ensuit que <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta\\gt 0<\/span><\/span><\/strong>. Nous allons montrer que ce <strong><span class=\"katex-eq\" data-katex-display=\"false\">\\delta<\/span><\/strong> fonctionne pour <strong>tout<\/strong> point <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0\\in[a,b]<\/span><\/span><\/strong>, c\u2019est-\u00e0-dire qu\u2019il ne d\u00e9pend pas du choix de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong>.\n<\/p>\n<p>\nConsid\u00e9rons maintenant :\n<\/p>\n<ul>\n<li>un point arbitraire <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0\\in[a,b]<\/span><\/span><\/strong>, et<\/li>\n<li>un point <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in[a,b]<\/span><\/span><\/strong> tel que <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|x-x_0|\\lt\\delta<\/span><\/span><\/strong>.<\/li>\n<\/ul>\n<p>\nComme les intervalles <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{x_1},\\dots,I_{x_N}<\/span><\/span><\/strong> recouvrent <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>, le point <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong> appartient \u00e0 au moins l\u2019un d\u2019eux, disons \u00e0 <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{x_j}<\/span><\/span><\/strong> pour un certain <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">j\\in\\{1,\\dots,N\\}<\/span><\/span><\/strong>. Par la d\u00e9finition de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{x_j}<\/span><\/span><\/strong>, cela signifie que\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|x_0-x_j|\\lt\\frac{\\delta(x_j)}{2}.\n\n<\/span>\n<\/p>\n<p>\nDe plus, par la d\u00e9finition de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta<\/span><\/span><\/strong>, nous avons <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta\\le\\frac{\\delta(x_j)}{2}<\/span><\/span><\/strong>, de sorte que de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|x-x_0|\\lt\\delta<\/span><\/span><\/strong> il s\u2019ensuit\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|x-x_0|\\lt\\frac{\\delta(x_j)}{2}.\n\n<\/span>\n<\/p>\n<p>\nEn appliquant l\u2019in\u00e9galit\u00e9 triangulaire,\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|x-x_j|\\le |x-x_0|+|x_0-x_j|\n\n\\lt \\frac{\\delta(x_j)}{2}+\\frac{\\delta(x_j)}{2}\n\n=\\delta(x_j).\n\n<\/span>\n<\/p>\n<p>\nPar le choix de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta(x_j)<\/span><\/span><\/strong> (continuit\u00e9 de <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> en <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_j<\/span><\/span><\/strong> pour la valeur <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon\/2<\/span><\/span><\/strong>), les in\u00e9galit\u00e9s <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|x_0-x_j|\\lt\\delta(x_j)<\/span><\/span><\/strong> et <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|x-x_j|\\lt\\delta(x_j)<\/span><\/span><\/strong> impliquent\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|f(x_0)-f(x_j)|\\lt\\frac{\\epsilon}{2}\n\n\\quad\\text{et}\\quad\n\n|f(x)-f(x_j)|\\lt\\frac{\\epsilon}{2}.\n\n<\/span>\n<\/p>\n<p>\nEn utilisant de nouveau l\u2019in\u00e9galit\u00e9 triangulaire, on obtient\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|f(x)-f(x_0)|\n\n\\le |f(x)-f(x_j)| + |f(x_j)-f(x_0)|\n\n\\lt \\frac{\\epsilon}{2}+\\frac{\\epsilon}{2}\n\n=\\epsilon.\n\n<\/span>\n<\/p>\n<p>\nComme <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong> et <strong><span class=\"katex-eq\" data-katex-display=\"false\">x<\/span><\/strong> \u00e9taient arbitraires, nous avons d\u00e9montr\u00e9 que pour le <strong><span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon<\/span><\/strong> fix\u00e9 au d\u00e9part, il existe un <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta\\gt 0<\/span><\/span><\/strong>, ind\u00e9pendant de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong>, tel que\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n(\\forall x_0\\in[a,b])(\\forall x\\in[a,b])\n\n\\big(|x-x_0|\\lt\\delta\\Rightarrow |f(x)-f(x_0)|\\lt\\epsilon\\big).\n\n<\/span>\n<\/p>\n<p>\nSi l\u2019on renomme <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong> comme <strong><span class=\"katex-eq\" data-katex-display=\"false\">y<\/span><\/strong>, cela s\u2019\u00e9crit :\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n(\\forall \\epsilon\\gt 0)(\\exists \\delta\\gt 0)(\\forall x,y\\in[a,b])\n\n\\big(|x-y|\\lt\\delta\\Rightarrow |f(x)-f(y)|\\lt\\epsilon\\big),\n\n<\/span>\n<\/p>\n<p>\nce qui est pr\u00e9cis\u00e9ment la d\u00e9finition de la <strong>continuit\u00e9 uniforme<\/strong> de <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> sur <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>. Dans ce qui suit, nous n\u2019aurons besoin d\u2019appliquer ce r\u00e9sultat qu\u2019au cas <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon=1<\/span><\/span><\/strong>.\n<\/p>\n<p><a name=\"8\"><\/a><\/br><\/p>\n<h4><b>\u00c9tape 5:<\/b> De la continuit\u00e9 uniforme \u00e0 la bornitude de <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> sur <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/h4>\n<p>\nAppliquons maintenant la continuit\u00e9 uniforme avec <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon=1<\/span><\/span><\/strong>. Il existe un nombre <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta_1\\gt 0<\/span><\/span><\/strong> tel que pour tous <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x,y\\in[a,b]<\/span><\/span><\/strong> on ait\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|x-y|\\lt\\delta_1\\Rightarrow |f(x)-f(y)|\\lt 1.\n\n<\/span>\n<\/p>\n<p>\nNous divisons \u00e0 pr\u00e9sent l\u2019intervalle <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> en une quantit\u00e9 finie de sous-intervalles dont la longueur est inf\u00e9rieure \u00e0 <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta_1<\/span><\/span><\/strong>. Autrement dit, nous choisissons un entier <strong><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/strong> et des points\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\na = x_0 \\lt x_1 \\lt \\cdots \\lt x_n = b\n\n<\/span>\n<\/p>\n<p>\nde sorte que pour chaque <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0,1,\\dots,n-1]<\/span><\/span><\/strong> on ait\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nx_{k+1}-x_k\\lt\\delta_1.\n\n<\/span>\n<\/p>\n<p>\nConsid\u00e9rons maintenant l\u2019ensemble fini de valeurs\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n\\{f(x_0),f(x_1),\\dots,f(x_{n-1})\\}.\n\n<\/span>\n<\/p>\n<p>\nComme il s\u2019agit d\u2019un ensemble fini de r\u00e9els, nous pouvons d\u00e9finir sans difficult\u00e9\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nC = \\max\\{|f(x_k)| \\;|\\; k=0,1,\\dots,n-1\\}.\n\n<\/span>\n<\/p>\n<p>\nNous allons montrer que <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">C+1<\/span><\/span><\/strong> est une borne sup\u00e9rieure en valeur absolue pour <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> sur tout l\u2019intervalle <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>. Soit <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in[a,b]<\/span><\/span><\/strong> un point arbitraire. Il existe alors un indice <strong><span class=\"katex-eq\" data-katex-display=\"false\">k<\/span><\/strong> tel que <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in[x_k,x_{k+1}]<\/span><\/span><\/strong>. En particulier,\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|x-x_k|\\le x_{k+1}-x_k\\lt\\delta_1.\n\n<\/span>\n<\/p>\n<p>\nPar la continuit\u00e9 uniforme avec <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon=1<\/span><\/span><\/strong>, de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|x-x_k|\\lt\\delta_1<\/span><\/span><\/strong> il s\u2019ensuit que\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|f(x)-f(x_k)|\\lt 1.\n\n<\/span>\n<\/p>\n<p>\nEn utilisant l\u2019in\u00e9galit\u00e9 triangulaire :\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|f(x)|\\le |f(x)-f(x_k)| + |f(x_k)| \\lt 1 + |f(x_k)| \\le 1 + C.\n\n<\/span>\n<\/p>\n<p>\nComme <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in[a,b]<\/span><\/span><\/strong> \u00e9tait arbitraire, nous concluons que\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|f(x)|\\le C+1 \\quad \\text{pour tout } x\\in[a,b],\n\n<\/span>\n<\/p>\n<p>\nc\u2019est-\u00e0-dire que la fonction <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> est <strong>born\u00e9e<\/strong> sur <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>.\n<\/p>\n<p><a name=\"9\"><\/a><\/br><\/p>\n<h4><b>\u00c9tape 6:<\/b> Existence de valeurs maximale et minimale<\/h4>\n<p>\nD\u00e9finissons l\u2019ensemble des valeurs prises par la fonction sur l\u2019intervalle :\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nH=\\{f(x)\\;|\\;x\\in[a,b]\\}\\subset\\mathbb{R}.\n\n<\/span>\n<\/p>\n<p>\nNous savons d\u00e9j\u00e0 que <strong><span class=\"katex-eq\" data-katex-display=\"false\">H<\/span><\/strong> n\u2019est pas vide (car <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> ne l\u2019est pas) et qu\u2019il est born\u00e9, donc par l\u2019axiome du supremum il existe des r\u00e9els\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nM=\\sup H,\\qquad m=\\inf H.\n\n<\/span>\n<\/p>\n<p>\nMontrons que <strong><span class=\"katex-eq\" data-katex-display=\"false\">M<\/span><\/strong> est atteint comme valeur de la fonction, c\u2019est-\u00e0-dire qu\u2019il existe <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_1\\in[a,b]<\/span><\/span><\/strong> tel que <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x_1)=M<\/span><\/span><\/strong>. Nous proc\u00e9derons par l\u2019absurde.\n<\/p>\n<p>\nSupposons que <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)<\/span><\/span><\/strong> n\u2019atteigne jamais la valeur <strong><span class=\"katex-eq\" data-katex-display=\"false\">M<\/span><\/strong>, c\u2019est-\u00e0-dire :\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n(\\forall x\\in[a,b])\\big(f(x)\\lt M\\big).\n\n<\/span>\n<\/p>\n<p>\nSous cette hypoth\u00e8se, la fonction\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\ng(x)=\\frac{1}{M-f(x)}\n\n<\/span>\n<\/p>\n<p>\nest bien d\u00e9finie et positive pour tout <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in[a,b]<\/span><\/span><\/strong>, puisque par hypoth\u00e8se <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">M-f(x)\\gt 0<\/span><\/span><\/strong>. De plus, comme <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> est continue et <strong><span class=\"katex-eq\" data-katex-display=\"false\">M<\/span><\/strong> est constante, la fonction <strong><span class=\"katex-eq\" data-katex-display=\"false\">g<\/span><\/strong> l\u2019est \u00e9galement. D\u2019apr\u00e8s la premi\u00e8re partie de la d\u00e9monstration, toute fonction continue sur <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> est born\u00e9e, il existe donc un nombre <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">N\\gt 0<\/span><\/span><\/strong> tel que\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n(\\forall x\\in[a,b])\\big(g(x)\\le N\\big).\n\n<\/span>\n<\/p>\n<p>\nEn particulier, pour tout <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in[a,b]<\/span><\/span><\/strong>, on v\u00e9rifie que\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n\\frac{1}{M-f(x)} = g(x)\\le N,\n\n<\/span>\n<\/p>\n<p>\nce qui \u00e9quivaut \u00e0\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nM-f(x)\\ge \\frac{1}{N}\n\n\\quad\\Rightarrow\\quad\n\nf(x)\\le M-\\frac{1}{N}.\n\n<\/span>\n<\/p>\n<p>\nCela signifie que toutes les valeurs de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)<\/span><\/span><\/strong> sur <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> sont inf\u00e9rieures ou \u00e9gales \u00e0 <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">M-\\frac{1}{N}<\/span><\/span><\/strong>. En particulier, le supremum de <strong><span class=\"katex-eq\" data-katex-display=\"false\">H<\/span><\/strong> satisfait\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n\\sup H\\le M-\\frac{1}{N}\\lt M,\n\n<\/span>\n<\/p>\n<p>\nce qui contredit la d\u00e9finition de <strong><span class=\"katex-eq\" data-katex-display=\"false\">M<\/span><\/strong> comme supremum de <strong><span class=\"katex-eq\" data-katex-display=\"false\">H<\/span><\/strong>. Par cons\u00e9quent, notre hypoth\u00e8se initiale \u00e9tait fausse, et il doit exister un point <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_1\\in[a,b]<\/span><\/span><\/strong> tel que\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nf(x_1)=M.\n\n<\/span>\n<\/p>\n<p>\nUn raisonnement enti\u00e8rement analogue, appliqu\u00e9 \u00e0 l\u2019infimum <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m=\\inf H<\/span><\/span><\/strong> (par exemple, en consid\u00e9rant la fonction <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">h(x)=-f(x)<\/span><\/span><\/strong>), montre qu\u2019il existe un point <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_2\\in[a,b]<\/span><\/span><\/strong> tel que\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nf(x_2)=m.\n\n<\/span>\n<\/p>\n<p><a name=\"10\"><\/a><\/br><\/p>\n<h2>Interpr\u00e9tation en termes de compacit\u00e9 et conclusion<\/h2>\n<p>\nNous avons montr\u00e9 que toute fonction continue <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f:[a,b]\\to\\mathbb{R}<\/span><\/span><\/strong> est born\u00e9e et atteint ses valeurs maximale et minimale sur <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>. Dans le langage moderne de l\u2019analyse, cela s\u2019interpr\u00e8te en disant que, dans <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}<\/span><\/span><\/strong>, les intervalles ferm\u00e9s et born\u00e9s tels que <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> sont des ensembles compacts et que les fonctions continues envoient les ensembles compacts dans des ensembles compacts.\n<\/p>\n<p>\nEn particulier, si <strong><span class=\"katex-eq\" data-katex-display=\"false\">I<\/span><\/strong> est compact et si <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> est continue sur <strong><span class=\"katex-eq\" data-katex-display=\"false\">I<\/span><\/strong>, alors l\u2019image <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(I)<\/span><\/span><\/strong> est un sous-ensemble compact de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}<\/span><\/span><\/strong>. Cela garantit que <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(I)<\/span><\/span><\/strong> est born\u00e9 et qu\u2019il y existe effectivement une valeur maximale et une valeur minimale, ce qui constitue pr\u00e9cis\u00e9ment le contenu du Th\u00e9or\u00e8me de Weierstrass.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Th\u00e9or\u00e8me de Weierstrass des Valeurs Extr\u00eames Pourquoi, dans tant de probl\u00e8mes d\u2019optimisation, consid\u00e8re-t-on presque comme \u00e9vident que \u00ab le maximum existe \u00bb ou qu\u2019\u00ab il y a toujours un minimum \u00bb sur un certain intervalle, alors qu\u2019en r\u00e9alit\u00e9 rien n\u2019obliga a priori \u00e0 ce que cela se produise? Le Th\u00e9or\u00e8me de Weierstrass est la pi\u00e8ce [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":35255,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":4,"footnotes":""},"categories":[866,569],"tags":[],"class_list":["post-35283","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-calcul-differentiel","category-mathematiques"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Th\u00e9or\u00e8me de Weierstrass des Valeurs Extr\u00eames - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Comprenez le Th\u00e9or\u00e8me de Weierstrass depuis z\u00e9ro : continuit\u00e9, compacit\u00e9 et d\u00e9monstration \u00e9tape par \u00e9tape appliqu\u00e9e aux probl\u00e8mes de maxima et minima.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/toposuranos.com\/material\/fr\/theoreme-de-weierstrass-des-valeurs-extremes\/\" \/>\n<meta property=\"og:locale\" content=\"es_ES\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Th\u00e9or\u00e8me de Weierstrass des Valeurs Extr\u00eames\" \/>\n<meta property=\"og:description\" content=\"Comprenez le Th\u00e9or\u00e8me de Weierstrass depuis z\u00e9ro : continuit\u00e9, compacit\u00e9 et d\u00e9monstration \u00e9tape par \u00e9tape appliqu\u00e9e aux probl\u00e8mes de maxima et minima.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/toposuranos.com\/material\/fr\/theoreme-de-weierstrass-des-valeurs-extremes\/\" \/>\n<meta property=\"og:site_name\" content=\"toposuranos.com\/material\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/groups\/toposuranos\" \/>\n<meta property=\"article:published_time\" content=\"2024-12-20T13:00:01+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2025-12-11T17:08:02+00:00\" \/>\n<meta property=\"og:image\" content=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2021\/06\/Weierstrass-1-1024x683.jpg\" \/>\n<meta name=\"author\" content=\"giorgio.reveco\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:title\" content=\"Th\u00e9or\u00e8me de Weierstrass des Valeurs Extr\u00eames\" \/>\n<meta name=\"twitter:description\" content=\"Comprenez le Th\u00e9or\u00e8me de Weierstrass depuis z\u00e9ro : continuit\u00e9, compacit\u00e9 et d\u00e9monstration \u00e9tape par \u00e9tape appliqu\u00e9e aux probl\u00e8mes de maxima et minima.\" \/>\n<meta name=\"twitter:image\" content=\"https:\/\/toposuranos.com\/material\/wp-content\/uploads\/2021\/06\/Weierstrass-1.jpg\" \/>\n<meta name=\"twitter:creator\" content=\"@topuranos\" \/>\n<meta name=\"twitter:site\" content=\"@topuranos\" \/>\n<meta name=\"twitter:label1\" content=\"Escrito por\" \/>\n\t<meta name=\"twitter:data1\" content=\"giorgio.reveco\" \/>\n\t<meta name=\"twitter:label2\" content=\"Tiempo de lectura\" \/>\n\t<meta name=\"twitter:data2\" content=\"10 minutos\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/toposuranos.com\/material\/fr\/theoreme-de-weierstrass-des-valeurs-extremes\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/toposuranos.com\/material\/fr\/theoreme-de-weierstrass-des-valeurs-extremes\/\"},\"author\":{\"name\":\"giorgio.reveco\",\"@id\":\"https:\/\/toposuranos.com\/material\/#\/schema\/person\/e15164361c3f9a2a02cf6c234cf7fdc1\"},\"headline\":\"Th\u00e9or\u00e8me de Weierstrass des Valeurs Extr\u00eames\",\"datePublished\":\"2024-12-20T13:00:01+00:00\",\"dateModified\":\"2025-12-11T17:08:02+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/toposuranos.com\/material\/fr\/theoreme-de-weierstrass-des-valeurs-extremes\/\"},\"wordCount\":3002,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/toposuranos.com\/material\/#organization\"},\"image\":{\"@id\":\"https:\/\/toposuranos.com\/material\/fr\/theoreme-de-weierstrass-des-valeurs-extremes\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/toposuranos.com\/material\/wp-content\/uploads\/2021\/06\/Weierstrass-1.jpg\",\"articleSection\":[\"Calcul Diff\u00e9rentiel\",\"Math\u00e9matiques\"],\"inLanguage\":\"es\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\/\/toposuranos.com\/material\/fr\/theoreme-de-weierstrass-des-valeurs-extremes\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/toposuranos.com\/material\/fr\/theoreme-de-weierstrass-des-valeurs-extremes\/\",\"url\":\"https:\/\/toposuranos.com\/material\/fr\/theoreme-de-weierstrass-des-valeurs-extremes\/\",\"name\":\"Th\u00e9or\u00e8me de Weierstrass des Valeurs Extr\u00eames - 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