Bounded operator but not Compact
$begingroup$
Let $T : (C([-1,1]),||.||_{infty}) rightarrow (C([-1,1]),||.||_{infty}) $
Such as : $(Tf)(x)=frac{1}{2}(f(x)+f(-x))$ . For all $fin C([-1,1])$
Why $T$ isn't Compact ?
I tried to use the sequence $f_n(x)=x^n$. For $xin [-1,1]$.
But I couldn't prove that $(T(f_n))_n$ has no convergent subsequence.
functional-analysis operator-theory compact-operators unbounded-operators
asked Dec 2 '18 at 17:10
Anas BOUALIIAnas BOUALII
1337
$endgroup$
add a comment |
$begingroup$
Let $T : (C([-1,1]),||.||_{infty}) rightarrow (C([-1,1]),||.||_{infty}) $
Such as : $(Tf)(x)=frac{1}{2}(f(x)+f(-x))$ . For all $fin C([-1,1])$
Why $T$ isn't Compact ?
I tried to use the sequence $f_n(x)=x^n$. For $xin [-1,1]$.
But I couldn't prove that $(T(f_n))_n$ has no convergent subsequence.
functional-analysis operator-theory compact-operators unbounded-operators
asked Dec 2 '18 at 17:10
Anas BOUALIIAnas BOUALII
1337
$endgroup$
1
$begingroup$
It has a convergent subsequence: $lim_{ntoinfty}T(f_{2n-1})=0$.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 17:15
add a comment |
$begingroup$
Let $T : (C([-1,1]),||.||_{infty}) rightarrow (C([-1,1]),||.||_{infty}) $
Such as : $(Tf)(x)=frac{1}{2}(f(x)+f(-x))$ . For all $fin C([-1,1])$
Why $T$ isn't Compact ?
I tried to use the sequence $f_n(x)=x^n$. For $xin [-1,1]$.
But I couldn't prove that $(T(f_n))_n$ has no convergent subsequence.
functional-analysis operator-theory compact-operators unbounded-operators
asked Dec 2 '18 at 17:10
Anas BOUALIIAnas BOUALII
1337
$endgroup$
Let $T : (C([-1,1]),||.||_{infty}) rightarrow (C([-1,1]),||.||_{infty}) $
Such as : $(Tf)(x)=frac{1}{2}(f(x)+f(-x))$ . For all $fin C([-1,1])$
Why $T$ isn't Compact ?
I tried to use the sequence $f_n(x)=x^n$. For $xin [-1,1]$.
But I couldn't prove that $(T(f_n))_n$ has no convergent subsequence.
functional-analysis operator-theory compact-operators unbounded-operators
functional-analysis operator-theory compact-operators unbounded-operators
asked Dec 2 '18 at 17:10
Anas BOUALIIAnas BOUALII
1337
asked Dec 2 '18 at 17:10
Anas BOUALIIAnas BOUALII
1337
asked Dec 2 '18 at 17:10
Anas BOUALIIAnas BOUALII
1337
asked Dec 2 '18 at 17:10
Anas BOUALIIAnas BOUALII
1337
asked Dec 2 '18 at 17:10
Anas BOUALIIAnas BOUALII
1337
1337
1
$begingroup$
It has a convergent subsequence: $lim_{ntoinfty}T(f_{2n-1})=0$.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 17:15
add a comment |
1
$begingroup$
It has a convergent subsequence: $lim_{ntoinfty}T(f_{2n-1})=0$.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 17:15
1
1
$begingroup$
It has a convergent subsequence: $lim_{ntoinfty}T(f_{2n-1})=0$.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 17:15
$begingroup$
It has a convergent subsequence: $lim_{ntoinfty}T(f_{2n-1})=0$.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 17:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: Instead, try
$$ f_n(x) = x^{2n}, $$
and observe that any subsequence of $Tf_n=f_n$ converges pointwise to a discontinuous function.
answered Dec 2 '18 at 17:13
MisterRiemannMisterRiemann
5,8451624
$endgroup$
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%2f3022907%2fbounded-operator-but-not-compact%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
$begingroup$
Hint: Instead, try
$$ f_n(x) = x^{2n}, $$
and observe that any subsequence of $Tf_n=f_n$ converges pointwise to a discontinuous function.
answered Dec 2 '18 at 17:13
MisterRiemannMisterRiemann
5,8451624
$endgroup$
add a comment |
$begingroup$
Hint: Instead, try
$$ f_n(x) = x^{2n}, $$
and observe that any subsequence of $Tf_n=f_n$ converges pointwise to a discontinuous function.
answered Dec 2 '18 at 17:13
MisterRiemannMisterRiemann
5,8451624
$endgroup$
add a comment |
$begingroup$
Hint: Instead, try
$$ f_n(x) = x^{2n}, $$
and observe that any subsequence of $Tf_n=f_n$ converges pointwise to a discontinuous function.
answered Dec 2 '18 at 17:13
MisterRiemannMisterRiemann
5,8451624
$endgroup$
Hint: Instead, try
$$ f_n(x) = x^{2n}, $$
and observe that any subsequence of $Tf_n=f_n$ converges pointwise to a discontinuous function.
answered Dec 2 '18 at 17:13
MisterRiemannMisterRiemann
5,8451624
answered Dec 2 '18 at 17:13
MisterRiemannMisterRiemann
5,8451624
answered Dec 2 '18 at 17:13
MisterRiemannMisterRiemann
5,8451624
answered Dec 2 '18 at 17:13
MisterRiemannMisterRiemann
5,8451624
5,8451624
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%2f3022907%2fbounded-operator-but-not-compact%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
1
$begingroup$
It has a convergent subsequence: $lim_{ntoinfty}T(f_{2n-1})=0$.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 17:15