Mixing
Convergence in operator norm doesn't imply uniform convergence, examples?
$begingroup$
I am learning about functional analysis and the relation between operator norm convergence and uniform convergence is not so clear to me. I think that uniform convergence imply convergence in operator norm since if $forall epsilon exists N$ such that if $n>N$ then $||T_n(x)-T(x)|| < epsilon$ $forall x$ then $sup_{||x|| le 1} ||T_n(x)-T(x)|| < epsilon$ for $n>N$.
If we just have convergence in norm, it feels like we don't have necessarily uniform convergence but I cannot find counter-examples.
PS: there was this question without a clear answer (to me at least).
functional-analysis
asked Jan 6 at 14:35
roi_saumonroi_saumon
53028
$endgroup$
add a comment |
$begingroup$
I am learning about functional analysis and the relation between operator norm convergence and uniform convergence is not so clear to me. I think that uniform convergence imply convergence in operator norm since if $forall epsilon exists N$ such that if $n>N$ then $||T_n(x)-T(x)|| < epsilon$ $forall x$ then $sup_{||x|| le 1} ||T_n(x)-T(x)|| < epsilon$ for $n>N$.
If we just have convergence in norm, it feels like we don't have necessarily uniform convergence but I cannot find counter-examples.
PS: there was this question without a clear answer (to me at least).
functional-analysis
asked Jan 6 at 14:35
roi_saumonroi_saumon
53028
$endgroup$
add a comment |
$begingroup$
I am learning about functional analysis and the relation between operator norm convergence and uniform convergence is not so clear to me. I think that uniform convergence imply convergence in operator norm since if $forall epsilon exists N$ such that if $n>N$ then $||T_n(x)-T(x)|| < epsilon$ $forall x$ then $sup_{||x|| le 1} ||T_n(x)-T(x)|| < epsilon$ for $n>N$.
If we just have convergence in norm, it feels like we don't have necessarily uniform convergence but I cannot find counter-examples.
PS: there was this question without a clear answer (to me at least).
functional-analysis
asked Jan 6 at 14:35
roi_saumonroi_saumon
53028
$endgroup$
I am learning about functional analysis and the relation between operator norm convergence and uniform convergence is not so clear to me. I think that uniform convergence imply convergence in operator norm since if $forall epsilon exists N$ such that if $n>N$ then $||T_n(x)-T(x)|| < epsilon$ $forall x$ then $sup_{||x|| le 1} ||T_n(x)-T(x)|| < epsilon$ for $n>N$.
If we just have convergence in norm, it feels like we don't have necessarily uniform convergence but I cannot find counter-examples.
PS: there was this question without a clear answer (to me at least).
functional-analysis
functional-analysis
asked Jan 6 at 14:35
roi_saumonroi_saumon
53028
asked Jan 6 at 14:35
roi_saumonroi_saumon
53028
asked Jan 6 at 14:35
roi_saumonroi_saumon
53028
asked Jan 6 at 14:35
roi_saumonroi_saumon
53028
asked Jan 6 at 14:35
roi_saumonroi_saumon
53028
53028
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The following should allow to find counterexamples - it also explains why the notion of uniform convergence doesn't come up in this context:
Exercise. Suppose $U$ and $V$ are normed vector spaces and $T_n:Uto V$ is linear for $n=1,2dots$. If $T_nto T$ uniformly then there exists $N$ such that $T_n=T$ for all $n>N$.
Hint: What does $||T_n(cx)-T(cx)||<epsilon$ tell you about $||T_n(x)-T(x)||$?
answered Jan 6 at 17:12
David C. UllrichDavid C. Ullrich
60k43994
$endgroup$
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%2f3063931%2fconvergence-in-operator-norm-doesnt-imply-uniform-convergence-examples%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
$begingroup$
The following should allow to find counterexamples - it also explains why the notion of uniform convergence doesn't come up in this context:
Exercise. Suppose $U$ and $V$ are normed vector spaces and $T_n:Uto V$ is linear for $n=1,2dots$. If $T_nto T$ uniformly then there exists $N$ such that $T_n=T$ for all $n>N$.
Hint: What does $||T_n(cx)-T(cx)||<epsilon$ tell you about $||T_n(x)-T(x)||$?
answered Jan 6 at 17:12
David C. UllrichDavid C. Ullrich
60k43994
$endgroup$
add a comment |
$begingroup$
The following should allow to find counterexamples - it also explains why the notion of uniform convergence doesn't come up in this context:
Exercise. Suppose $U$ and $V$ are normed vector spaces and $T_n:Uto V$ is linear for $n=1,2dots$. If $T_nto T$ uniformly then there exists $N$ such that $T_n=T$ for all $n>N$.
Hint: What does $||T_n(cx)-T(cx)||<epsilon$ tell you about $||T_n(x)-T(x)||$?
answered Jan 6 at 17:12
David C. UllrichDavid C. Ullrich
60k43994
$endgroup$
add a comment |
$begingroup$
The following should allow to find counterexamples - it also explains why the notion of uniform convergence doesn't come up in this context:
Exercise. Suppose $U$ and $V$ are normed vector spaces and $T_n:Uto V$ is linear for $n=1,2dots$. If $T_nto T$ uniformly then there exists $N$ such that $T_n=T$ for all $n>N$.
Hint: What does $||T_n(cx)-T(cx)||<epsilon$ tell you about $||T_n(x)-T(x)||$?
answered Jan 6 at 17:12
David C. UllrichDavid C. Ullrich
60k43994
$endgroup$
The following should allow to find counterexamples - it also explains why the notion of uniform convergence doesn't come up in this context:
Exercise. Suppose $U$ and $V$ are normed vector spaces and $T_n:Uto V$ is linear for $n=1,2dots$. If $T_nto T$ uniformly then there exists $N$ such that $T_n=T$ for all $n>N$.
Hint: What does $||T_n(cx)-T(cx)||<epsilon$ tell you about $||T_n(x)-T(x)||$?
answered Jan 6 at 17:12
David C. UllrichDavid C. Ullrich
60k43994
answered Jan 6 at 17:12
David C. UllrichDavid C. Ullrich
60k43994
answered Jan 6 at 17:12
David C. UllrichDavid C. Ullrich
60k43994
answered Jan 6 at 17:12
David C. UllrichDavid C. Ullrich
60k43994
60k43994
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%2f3063931%2fconvergence-in-operator-norm-doesnt-imply-uniform-convergence-examples%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
