$f(x_0)=0 forall f in X^*$ ($X^*$ is topological dual) then $x_0=0$.
$begingroup$
Let X be a normed vector space of infinite dimension (possibly) and $x_0 in X$.
I would like to show that if $f(x_0)=0 forall f in X^*$ (the topological dual) then $x_0=0$.
I asked this question here but I forgot to mention that the functions were in the topological dual and not the algebraic dual, so I had an explanation for why this is true in the case of algebraic dual which was still very interesting.
The suggestion for when $f$ are in the topological dual was to use Hahn-Banach theorem. I have the version that says that a functional of $Y^*$ (where $Y$ is a subspace of $X$) can be extended to a functional of $X^*$ (again here we consider topological duals). But here I am not sure how this helps me. What should I consider for the subspace $Y$?
functional-analysis
$endgroup$
add a comment |
$begingroup$
Let X be a normed vector space of infinite dimension (possibly) and $x_0 in X$.
I would like to show that if $f(x_0)=0 forall f in X^*$ (the topological dual) then $x_0=0$.
I asked this question here but I forgot to mention that the functions were in the topological dual and not the algebraic dual, so I had an explanation for why this is true in the case of algebraic dual which was still very interesting.
The suggestion for when $f$ are in the topological dual was to use Hahn-Banach theorem. I have the version that says that a functional of $Y^*$ (where $Y$ is a subspace of $X$) can be extended to a functional of $X^*$ (again here we consider topological duals). But here I am not sure how this helps me. What should I consider for the subspace $Y$?
functional-analysis
$endgroup$
3
$begingroup$
Possible duplicate of Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$
$endgroup$
– José Carlos Santos
Jan 12 at 17:52
1
$begingroup$
This is in whatever book you learned the H-B theorem from. Assume $x_0ne0$. Let $Y$ be the span of $x_0$. Take $Lambdain Y^*$ with $Lambda(x_0)ne0$, for example define $Lambda(cx_0)=c$.
$endgroup$
– David C. Ullrich
Jan 12 at 17:56
$begingroup$
Okay, thanks! I am not familiar with those kind of reasoning yet...
$endgroup$
– roi_saumon
Jan 12 at 18:14
add a comment |
$begingroup$
Let X be a normed vector space of infinite dimension (possibly) and $x_0 in X$.
I would like to show that if $f(x_0)=0 forall f in X^*$ (the topological dual) then $x_0=0$.
I asked this question here but I forgot to mention that the functions were in the topological dual and not the algebraic dual, so I had an explanation for why this is true in the case of algebraic dual which was still very interesting.
The suggestion for when $f$ are in the topological dual was to use Hahn-Banach theorem. I have the version that says that a functional of $Y^*$ (where $Y$ is a subspace of $X$) can be extended to a functional of $X^*$ (again here we consider topological duals). But here I am not sure how this helps me. What should I consider for the subspace $Y$?
functional-analysis
$endgroup$
Let X be a normed vector space of infinite dimension (possibly) and $x_0 in X$.
I would like to show that if $f(x_0)=0 forall f in X^*$ (the topological dual) then $x_0=0$.
I asked this question here but I forgot to mention that the functions were in the topological dual and not the algebraic dual, so I had an explanation for why this is true in the case of algebraic dual which was still very interesting.
The suggestion for when $f$ are in the topological dual was to use Hahn-Banach theorem. I have the version that says that a functional of $Y^*$ (where $Y$ is a subspace of $X$) can be extended to a functional of $X^*$ (again here we consider topological duals). But here I am not sure how this helps me. What should I consider for the subspace $Y$?
functional-analysis
functional-analysis
edited Jan 12 at 17:58
David C. Ullrich
60.1k43994
60.1k43994
asked Jan 12 at 17:48
roi_saumonroi_saumon
54538
54538
3
$begingroup$
Possible duplicate of Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$
$endgroup$
– José Carlos Santos
Jan 12 at 17:52
1
$begingroup$
This is in whatever book you learned the H-B theorem from. Assume $x_0ne0$. Let $Y$ be the span of $x_0$. Take $Lambdain Y^*$ with $Lambda(x_0)ne0$, for example define $Lambda(cx_0)=c$.
$endgroup$
– David C. Ullrich
Jan 12 at 17:56
$begingroup$
Okay, thanks! I am not familiar with those kind of reasoning yet...
$endgroup$
– roi_saumon
Jan 12 at 18:14
add a comment |
3
$begingroup$
Possible duplicate of Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$
$endgroup$
– José Carlos Santos
Jan 12 at 17:52
1
$begingroup$
This is in whatever book you learned the H-B theorem from. Assume $x_0ne0$. Let $Y$ be the span of $x_0$. Take $Lambdain Y^*$ with $Lambda(x_0)ne0$, for example define $Lambda(cx_0)=c$.
$endgroup$
– David C. Ullrich
Jan 12 at 17:56
$begingroup$
Okay, thanks! I am not familiar with those kind of reasoning yet...
$endgroup$
– roi_saumon
Jan 12 at 18:14
3
3
$begingroup$
Possible duplicate of Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$
$endgroup$
– José Carlos Santos
Jan 12 at 17:52
$begingroup$
Possible duplicate of Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$
$endgroup$
– José Carlos Santos
Jan 12 at 17:52
1
1
$begingroup$
This is in whatever book you learned the H-B theorem from. Assume $x_0ne0$. Let $Y$ be the span of $x_0$. Take $Lambdain Y^*$ with $Lambda(x_0)ne0$, for example define $Lambda(cx_0)=c$.
$endgroup$
– David C. Ullrich
Jan 12 at 17:56
$begingroup$
This is in whatever book you learned the H-B theorem from. Assume $x_0ne0$. Let $Y$ be the span of $x_0$. Take $Lambdain Y^*$ with $Lambda(x_0)ne0$, for example define $Lambda(cx_0)=c$.
$endgroup$
– David C. Ullrich
Jan 12 at 17:56
$begingroup$
Okay, thanks! I am not familiar with those kind of reasoning yet...
$endgroup$
– roi_saumon
Jan 12 at 18:14
$begingroup$
Okay, thanks! I am not familiar with those kind of reasoning yet...
$endgroup$
– roi_saumon
Jan 12 at 18:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $x_0$ is not the zero vector then consider the functional $f$
which is defined on the space that $x_0$ spans as follows:
$f(a x_0)=a|x_0|$.
It follows from the Hahn-Banach theorem that there is a functional $Fin X^*$ with $F(x)=f(x) forall xin text{span}(x_0)$ and of course $F(x_0)$ is not equal to $0$.
$endgroup$
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%2f3071186%2ffx-0-0-forall-f-in-x-x-is-topological-dual-then-x-0-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $x_0$ is not the zero vector then consider the functional $f$
which is defined on the space that $x_0$ spans as follows:
$f(a x_0)=a|x_0|$.
It follows from the Hahn-Banach theorem that there is a functional $Fin X^*$ with $F(x)=f(x) forall xin text{span}(x_0)$ and of course $F(x_0)$ is not equal to $0$.
$endgroup$
add a comment |
$begingroup$
If $x_0$ is not the zero vector then consider the functional $f$
which is defined on the space that $x_0$ spans as follows:
$f(a x_0)=a|x_0|$.
It follows from the Hahn-Banach theorem that there is a functional $Fin X^*$ with $F(x)=f(x) forall xin text{span}(x_0)$ and of course $F(x_0)$ is not equal to $0$.
$endgroup$
add a comment |
$begingroup$
If $x_0$ is not the zero vector then consider the functional $f$
which is defined on the space that $x_0$ spans as follows:
$f(a x_0)=a|x_0|$.
It follows from the Hahn-Banach theorem that there is a functional $Fin X^*$ with $F(x)=f(x) forall xin text{span}(x_0)$ and of course $F(x_0)$ is not equal to $0$.
$endgroup$
If $x_0$ is not the zero vector then consider the functional $f$
which is defined on the space that $x_0$ spans as follows:
$f(a x_0)=a|x_0|$.
It follows from the Hahn-Banach theorem that there is a functional $Fin X^*$ with $F(x)=f(x) forall xin text{span}(x_0)$ and of course $F(x_0)$ is not equal to $0$.
edited Jan 12 at 19:07
answered Jan 12 at 19:00
NikiforosNikiforos
362
362
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%2f3071186%2ffx-0-0-forall-f-in-x-x-is-topological-dual-then-x-0-0%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
3
$begingroup$
Possible duplicate of Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$
$endgroup$
– José Carlos Santos
Jan 12 at 17:52
1
$begingroup$
This is in whatever book you learned the H-B theorem from. Assume $x_0ne0$. Let $Y$ be the span of $x_0$. Take $Lambdain Y^*$ with $Lambda(x_0)ne0$, for example define $Lambda(cx_0)=c$.
$endgroup$
– David C. Ullrich
Jan 12 at 17:56
$begingroup$
Okay, thanks! I am not familiar with those kind of reasoning yet...
$endgroup$
– roi_saumon
Jan 12 at 18:14