Dual space and kernel
Having X , a normed vector space and X* = B(X,R) its dual space ( R is the real numbers).
Show that for all f contained in X*, we have that Ker f included in X is a closed subspace.
Knowing that X* is a Banach space, since R is a complete space, does that make X a Banach space too ?
Needing a little help with this one.
real-analysis functional-analysis normed-spaces
edited Nov 25 at 19:53
Omnomnomnom
126k788176
asked Nov 25 at 19:45
mimi
175
add a comment |
Having X , a normed vector space and X* = B(X,R) its dual space ( R is the real numbers).
Show that for all f contained in X*, we have that Ker f included in X is a closed subspace.
Knowing that X* is a Banach space, since R is a complete space, does that make X a Banach space too ?
Needing a little help with this one.
real-analysis functional-analysis normed-spaces
edited Nov 25 at 19:53
Omnomnomnom
126k788176
asked Nov 25 at 19:45
mimi
175
It is not necessary for $X$ to be a Banach space. Indeed, however, $X^*$ will be a Banach space whether or not $X$ is complete (see this post for instance)
– Omnomnomnom
Nov 25 at 19:51
This answer really comes down to applying the definitions. In particular: what does $ker f$ mean, and what does it mean for a subspace to be closed?
– Omnomnomnom
Nov 25 at 19:52
add a comment |
Having X , a normed vector space and X* = B(X,R) its dual space ( R is the real numbers).
Show that for all f contained in X*, we have that Ker f included in X is a closed subspace.
Knowing that X* is a Banach space, since R is a complete space, does that make X a Banach space too ?
Needing a little help with this one.
real-analysis functional-analysis normed-spaces
edited Nov 25 at 19:53
Omnomnomnom
126k788176
asked Nov 25 at 19:45
mimi
175
Having X , a normed vector space and X* = B(X,R) its dual space ( R is the real numbers).
Show that for all f contained in X*, we have that Ker f included in X is a closed subspace.
Knowing that X* is a Banach space, since R is a complete space, does that make X a Banach space too ?
Needing a little help with this one.
real-analysis functional-analysis normed-spaces
real-analysis functional-analysis normed-spaces
edited Nov 25 at 19:53
Omnomnomnom
126k788176
asked Nov 25 at 19:45
mimi
175
edited Nov 25 at 19:53
Omnomnomnom
126k788176
asked Nov 25 at 19:45
mimi
175
edited Nov 25 at 19:53
Omnomnomnom
126k788176
edited Nov 25 at 19:53
Omnomnomnom
126k788176
edited Nov 25 at 19:53
Omnomnomnom
126k788176
126k788176
asked Nov 25 at 19:45
mimi
175
asked Nov 25 at 19:45
mimi
175
asked Nov 25 at 19:45
mimi
175
175
It is not necessary for $X$ to be a Banach space. Indeed, however, $X^*$ will be a Banach space whether or not $X$ is complete (see this post for instance)
– Omnomnomnom
Nov 25 at 19:51
This answer really comes down to applying the definitions. In particular: what does $ker f$ mean, and what does it mean for a subspace to be closed?
– Omnomnomnom
Nov 25 at 19:52
add a comment |
It is not necessary for $X$ to be a Banach space. Indeed, however, $X^*$ will be a Banach space whether or not $X$ is complete (see this post for instance)
– Omnomnomnom
Nov 25 at 19:51
This answer really comes down to applying the definitions. In particular: what does $ker f$ mean, and what does it mean for a subspace to be closed?
– Omnomnomnom
Nov 25 at 19:52
It is not necessary for $X$ to be a Banach space. Indeed, however, $X^*$ will be a Banach space whether or not $X$ is complete (see this post for instance)
– Omnomnomnom
Nov 25 at 19:51
It is not necessary for $X$ to be a Banach space. Indeed, however, $X^*$ will be a Banach space whether or not $X$ is complete (see this post for instance)
– Omnomnomnom
Nov 25 at 19:51
This answer really comes down to applying the definitions. In particular: what does $ker f$ mean, and what does it mean for a subspace to be closed?
– Omnomnomnom
Nov 25 at 19:52
This answer really comes down to applying the definitions. In particular: what does $ker f$ mean, and what does it mean for a subspace to be closed?
– Omnomnomnom
Nov 25 at 19:52
add a comment |
1 Answer
1
active
oldest
votes
That $ker f$ is a linear subspace is evident. To prove that it is closed, notice that $ker f = f^{-1}({0})$, which is the preimage of the closed set ${0}$ under the continuous map $f$.
This has nothing to do with $X$ being or not being a Banach space.
answered Nov 25 at 19:48
MisterRiemann
5,7291624
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%2f3013298%2fdual-space-and-kernel%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
That $ker f$ is a linear subspace is evident. To prove that it is closed, notice that $ker f = f^{-1}({0})$, which is the preimage of the closed set ${0}$ under the continuous map $f$.
This has nothing to do with $X$ being or not being a Banach space.
answered Nov 25 at 19:48
MisterRiemann
5,7291624
add a comment |
That $ker f$ is a linear subspace is evident. To prove that it is closed, notice that $ker f = f^{-1}({0})$, which is the preimage of the closed set ${0}$ under the continuous map $f$.
This has nothing to do with $X$ being or not being a Banach space.
answered Nov 25 at 19:48
MisterRiemann
5,7291624
add a comment |
That $ker f$ is a linear subspace is evident. To prove that it is closed, notice that $ker f = f^{-1}({0})$, which is the preimage of the closed set ${0}$ under the continuous map $f$.
This has nothing to do with $X$ being or not being a Banach space.
answered Nov 25 at 19:48
MisterRiemann
5,7291624
That $ker f$ is a linear subspace is evident. To prove that it is closed, notice that $ker f = f^{-1}({0})$, which is the preimage of the closed set ${0}$ under the continuous map $f$.
This has nothing to do with $X$ being or not being a Banach space.
answered Nov 25 at 19:48
MisterRiemann
5,7291624
answered Nov 25 at 19:48
MisterRiemann
5,7291624
answered Nov 25 at 19:48
MisterRiemann
5,7291624
answered Nov 25 at 19:48
MisterRiemann
5,7291624
5,7291624
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3013298%2fdual-space-and-kernel%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
It is not necessary for $X$ to be a Banach space. Indeed, however, $X^*$ will be a Banach space whether or not $X$ is complete (see this post for instance)
– Omnomnomnom
Nov 25 at 19:51
This answer really comes down to applying the definitions. In particular: what does $ker f$ mean, and what does it mean for a subspace to be closed?
– Omnomnomnom
Nov 25 at 19:52