are linear functionals on C[0, 1] bounded and thus continuous
I'm currently on a theorem showing a general representation of linear functionals on the space of continuous functions on the interval $[0,1]$.
My problem is on the beginning of the proof.
First we define a linear functional $f(x)$ on the space $C[0,1]$.
After that using the Cantor theorem we show that every continuous function on a compact interval is actually bounded and thus $C[0,1]$ is a subspace of $M[0,1]$ - bounded functions on the interval $[0,1]$.
Using the Hahn-Banach theorem we show that our functional $f$ can be extended to a functional $F$ on the whole space $M[0, 1]$ with the same norm.
Maybe I've missed something foundamental, but how do we know that the functional $f$ is bounded, so that it has a norm and it's norm is the same with the extension $F$?
Thanks in advance!
functional-analysis hahn-banach-theorem
asked Jan 4 at 10:53
NikolaNikola
736719
add a comment |
I'm currently on a theorem showing a general representation of linear functionals on the space of continuous functions on the interval $[0,1]$.
My problem is on the beginning of the proof.
First we define a linear functional $f(x)$ on the space $C[0,1]$.
After that using the Cantor theorem we show that every continuous function on a compact interval is actually bounded and thus $C[0,1]$ is a subspace of $M[0,1]$ - bounded functions on the interval $[0,1]$.
Using the Hahn-Banach theorem we show that our functional $f$ can be extended to a functional $F$ on the whole space $M[0, 1]$ with the same norm.
Maybe I've missed something foundamental, but how do we know that the functional $f$ is bounded, so that it has a norm and it's norm is the same with the extension $F$?
Thanks in advance!
functional-analysis hahn-banach-theorem
asked Jan 4 at 10:53
NikolaNikola
736719
add a comment |
I'm currently on a theorem showing a general representation of linear functionals on the space of continuous functions on the interval $[0,1]$.
My problem is on the beginning of the proof.
First we define a linear functional $f(x)$ on the space $C[0,1]$.
After that using the Cantor theorem we show that every continuous function on a compact interval is actually bounded and thus $C[0,1]$ is a subspace of $M[0,1]$ - bounded functions on the interval $[0,1]$.
Using the Hahn-Banach theorem we show that our functional $f$ can be extended to a functional $F$ on the whole space $M[0, 1]$ with the same norm.
Maybe I've missed something foundamental, but how do we know that the functional $f$ is bounded, so that it has a norm and it's norm is the same with the extension $F$?
Thanks in advance!
functional-analysis hahn-banach-theorem
asked Jan 4 at 10:53
NikolaNikola
736719
I'm currently on a theorem showing a general representation of linear functionals on the space of continuous functions on the interval $[0,1]$.
My problem is on the beginning of the proof.
First we define a linear functional $f(x)$ on the space $C[0,1]$.
After that using the Cantor theorem we show that every continuous function on a compact interval is actually bounded and thus $C[0,1]$ is a subspace of $M[0,1]$ - bounded functions on the interval $[0,1]$.
Using the Hahn-Banach theorem we show that our functional $f$ can be extended to a functional $F$ on the whole space $M[0, 1]$ with the same norm.
Maybe I've missed something foundamental, but how do we know that the functional $f$ is bounded, so that it has a norm and it's norm is the same with the extension $F$?
Thanks in advance!
functional-analysis hahn-banach-theorem
functional-analysis hahn-banach-theorem
asked Jan 4 at 10:53
NikolaNikola
736719
asked Jan 4 at 10:53
NikolaNikola
736719
edited Jan 4 at 17:15
Nikola
asked Jan 4 at 10:53
NikolaNikola
736719
asked Jan 4 at 10:53
NikolaNikola
736719
asked Jan 4 at 10:53
NikolaNikola
736719
736719
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
No, there exist linear functionals on $C[0,1]$ that are not bounded.
In fact, for every infinite dimensional space there exists a discontinuous (and therefore unbounded) linear functional, see here.
I would suspect that there is an additional assumption somewhere that the linear functional is bounded.
answered Jan 4 at 11:04
supinfsupinf
6,1041028
I guess it's just presumed. I think this theorem is a partial case of the Riesz representation theorem, in the end it is just shown that every linear functional can be expressed as a Stiltes integral with a diferential a function with bounded variation.
– Nikola
Jan 4 at 11:12
1
@Nikola No, the Riesz theorem does not show what you said! It's not true that every linear functional is given by a Stieltjes integral. Every bounded linear functional is given by a Stieltjes integral.
– David C. Ullrich
Jan 4 at 13:32
Then boundedness of the functional is just presumed in the book I"m using.
– Nikola
Jan 4 at 14:03
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%2f3061518%2fare-linear-functionals-on-c0-1-bounded-and-thus-continuous%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
No, there exist linear functionals on $C[0,1]$ that are not bounded.
In fact, for every infinite dimensional space there exists a discontinuous (and therefore unbounded) linear functional, see here.
I would suspect that there is an additional assumption somewhere that the linear functional is bounded.
answered Jan 4 at 11:04
supinfsupinf
6,1041028
I guess it's just presumed. I think this theorem is a partial case of the Riesz representation theorem, in the end it is just shown that every linear functional can be expressed as a Stiltes integral with a diferential a function with bounded variation.
– Nikola
Jan 4 at 11:12
1
@Nikola No, the Riesz theorem does not show what you said! It's not true that every linear functional is given by a Stieltjes integral. Every bounded linear functional is given by a Stieltjes integral.
– David C. Ullrich
Jan 4 at 13:32
Then boundedness of the functional is just presumed in the book I"m using.
– Nikola
Jan 4 at 14:03
add a comment |
No, there exist linear functionals on $C[0,1]$ that are not bounded.
In fact, for every infinite dimensional space there exists a discontinuous (and therefore unbounded) linear functional, see here.
I would suspect that there is an additional assumption somewhere that the linear functional is bounded.
answered Jan 4 at 11:04
supinfsupinf
6,1041028
I guess it's just presumed. I think this theorem is a partial case of the Riesz representation theorem, in the end it is just shown that every linear functional can be expressed as a Stiltes integral with a diferential a function with bounded variation.
– Nikola
Jan 4 at 11:12
1
@Nikola No, the Riesz theorem does not show what you said! It's not true that every linear functional is given by a Stieltjes integral. Every bounded linear functional is given by a Stieltjes integral.
– David C. Ullrich
Jan 4 at 13:32
Then boundedness of the functional is just presumed in the book I"m using.
– Nikola
Jan 4 at 14:03
add a comment |
No, there exist linear functionals on $C[0,1]$ that are not bounded.
In fact, for every infinite dimensional space there exists a discontinuous (and therefore unbounded) linear functional, see here.
I would suspect that there is an additional assumption somewhere that the linear functional is bounded.
answered Jan 4 at 11:04
supinfsupinf
6,1041028
No, there exist linear functionals on $C[0,1]$ that are not bounded.
In fact, for every infinite dimensional space there exists a discontinuous (and therefore unbounded) linear functional, see here.
I would suspect that there is an additional assumption somewhere that the linear functional is bounded.
answered Jan 4 at 11:04
supinfsupinf
6,1041028
answered Jan 4 at 11:04
supinfsupinf
6,1041028
answered Jan 4 at 11:04
supinfsupinf
6,1041028
answered Jan 4 at 11:04
supinfsupinf
6,1041028
6,1041028
I guess it's just presumed. I think this theorem is a partial case of the Riesz representation theorem, in the end it is just shown that every linear functional can be expressed as a Stiltes integral with a diferential a function with bounded variation.
– Nikola
Jan 4 at 11:12
1
@Nikola No, the Riesz theorem does not show what you said! It's not true that every linear functional is given by a Stieltjes integral. Every bounded linear functional is given by a Stieltjes integral.
– David C. Ullrich
Jan 4 at 13:32
Then boundedness of the functional is just presumed in the book I"m using.
– Nikola
Jan 4 at 14:03
add a comment |
I guess it's just presumed. I think this theorem is a partial case of the Riesz representation theorem, in the end it is just shown that every linear functional can be expressed as a Stiltes integral with a diferential a function with bounded variation.
– Nikola
Jan 4 at 11:12
1
@Nikola No, the Riesz theorem does not show what you said! It's not true that every linear functional is given by a Stieltjes integral. Every bounded linear functional is given by a Stieltjes integral.
– David C. Ullrich
Jan 4 at 13:32
Then boundedness of the functional is just presumed in the book I"m using.
– Nikola
Jan 4 at 14:03
I guess it's just presumed. I think this theorem is a partial case of the Riesz representation theorem, in the end it is just shown that every linear functional can be expressed as a Stiltes integral with a diferential a function with bounded variation.
– Nikola
Jan 4 at 11:12
I guess it's just presumed. I think this theorem is a partial case of the Riesz representation theorem, in the end it is just shown that every linear functional can be expressed as a Stiltes integral with a diferential a function with bounded variation.
– Nikola
Jan 4 at 11:12
1
1
@Nikola No, the Riesz theorem does not show what you said! It's not true that every linear functional is given by a Stieltjes integral. Every bounded linear functional is given by a Stieltjes integral.
– David C. Ullrich
Jan 4 at 13:32
@Nikola No, the Riesz theorem does not show what you said! It's not true that every linear functional is given by a Stieltjes integral. Every bounded linear functional is given by a Stieltjes integral.
– David C. Ullrich
Jan 4 at 13:32
Then boundedness of the functional is just presumed in the book I"m using.
– Nikola
Jan 4 at 14:03
Then boundedness of the functional is just presumed in the book I"m using.
– Nikola
Jan 4 at 14:03
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%2f3061518%2fare-linear-functionals-on-c0-1-bounded-and-thus-continuous%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
