Why Hahn-Banach theorem is needed for the following theorem?
$begingroup$
One theorem in Rudin's Real and complex analysis says the following:
If $X$ is a normed linear space and if $x_0 in X$, $x_0neq 0$, there is a bounded linear funcitonal $f$ on $X$, of norm 1, so that $f(x_0)=||x_0||$.
He uses the Hahn-Banach theorem to prove it, but why we cannot just say that $f(x)=||x||$ works?
functional-analysis
asked Dec 3 '18 at 14:37
eigenvalueeigenvalue
11
$endgroup$
add a comment |
$begingroup$
One theorem in Rudin's Real and complex analysis says the following:
If $X$ is a normed linear space and if $x_0 in X$, $x_0neq 0$, there is a bounded linear funcitonal $f$ on $X$, of norm 1, so that $f(x_0)=||x_0||$.
He uses the Hahn-Banach theorem to prove it, but why we cannot just say that $f(x)=||x||$ works?
functional-analysis
asked Dec 3 '18 at 14:37
eigenvalueeigenvalue
11
$endgroup$
add a comment |
$begingroup$
One theorem in Rudin's Real and complex analysis says the following:
If $X$ is a normed linear space and if $x_0 in X$, $x_0neq 0$, there is a bounded linear funcitonal $f$ on $X$, of norm 1, so that $f(x_0)=||x_0||$.
He uses the Hahn-Banach theorem to prove it, but why we cannot just say that $f(x)=||x||$ works?
functional-analysis
asked Dec 3 '18 at 14:37
eigenvalueeigenvalue
11
$endgroup$
One theorem in Rudin's Real and complex analysis says the following:
If $X$ is a normed linear space and if $x_0 in X$, $x_0neq 0$, there is a bounded linear funcitonal $f$ on $X$, of norm 1, so that $f(x_0)=||x_0||$.
He uses the Hahn-Banach theorem to prove it, but why we cannot just say that $f(x)=||x||$ works?
functional-analysis
functional-analysis
asked Dec 3 '18 at 14:37
eigenvalueeigenvalue
11
asked Dec 3 '18 at 14:37
eigenvalueeigenvalue
11
asked Dec 3 '18 at 14:37
eigenvalueeigenvalue
11
asked Dec 3 '18 at 14:37
eigenvalueeigenvalue
11
asked Dec 3 '18 at 14:37
eigenvalueeigenvalue
11
11
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, we cannot, because that map is not a linear map.
answered Dec 3 '18 at 14:40
José Carlos SantosJosé Carlos Santos
163k22130234
$endgroup$
add a comment |
$begingroup$
$f(x)=|x|$ is not linear!
There is no guarantee that $f(x+y) = |x+y|$ is equal to $f(x)+f(y) = |x|+|y|$.
In fact this is almost never happens.
answered Dec 3 '18 at 14:41
YankoYanko
7,0381629
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024137%2fwhy-hahn-banach-theorem-is-needed-for-the-following-theorem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, we cannot, because that map is not a linear map.
answered Dec 3 '18 at 14:40
José Carlos SantosJosé Carlos Santos
163k22130234
$endgroup$
add a comment |
$begingroup$
No, we cannot, because that map is not a linear map.
answered Dec 3 '18 at 14:40
José Carlos SantosJosé Carlos Santos
163k22130234
$endgroup$
add a comment |
$begingroup$
No, we cannot, because that map is not a linear map.
answered Dec 3 '18 at 14:40
José Carlos SantosJosé Carlos Santos
163k22130234
$endgroup$
No, we cannot, because that map is not a linear map.
answered Dec 3 '18 at 14:40
José Carlos SantosJosé Carlos Santos
163k22130234
answered Dec 3 '18 at 14:40
José Carlos SantosJosé Carlos Santos
163k22130234
answered Dec 3 '18 at 14:40
José Carlos SantosJosé Carlos Santos
163k22130234
answered Dec 3 '18 at 14:40
José Carlos SantosJosé Carlos Santos
163k22130234
163k22130234
add a comment |
add a comment |
$begingroup$
$f(x)=|x|$ is not linear!
There is no guarantee that $f(x+y) = |x+y|$ is equal to $f(x)+f(y) = |x|+|y|$.
In fact this is almost never happens.
answered Dec 3 '18 at 14:41
YankoYanko
7,0381629
$endgroup$
add a comment |
$begingroup$
$f(x)=|x|$ is not linear!
There is no guarantee that $f(x+y) = |x+y|$ is equal to $f(x)+f(y) = |x|+|y|$.
In fact this is almost never happens.
answered Dec 3 '18 at 14:41
YankoYanko
7,0381629
$endgroup$
add a comment |
$begingroup$
$f(x)=|x|$ is not linear!
There is no guarantee that $f(x+y) = |x+y|$ is equal to $f(x)+f(y) = |x|+|y|$.
In fact this is almost never happens.
answered Dec 3 '18 at 14:41
YankoYanko
7,0381629
$endgroup$
$f(x)=|x|$ is not linear!
There is no guarantee that $f(x+y) = |x+y|$ is equal to $f(x)+f(y) = |x|+|y|$.
In fact this is almost never happens.
answered Dec 3 '18 at 14:41
YankoYanko
7,0381629
answered Dec 3 '18 at 14:41
YankoYanko
7,0381629
answered Dec 3 '18 at 14:41
YankoYanko
7,0381629
answered Dec 3 '18 at 14:41
YankoYanko
7,0381629
7,0381629
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024137%2fwhy-hahn-banach-theorem-is-needed-for-the-following-theorem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
