Uniform convergence for operator of translation
$begingroup$
For $ain R^d$, let $T_af(x)=f(x-a),$ for all $fin L^p(R^d), 1leq p<infty$ and all $xin R^d$. I need example that this operator doesn't converge uniformly when $arightarrow 0$. I know that this operator has strong (norm) limit.
functional-analysis convergence
asked Jan 12 at 10:53
HanaHana
161
$endgroup$
add a comment |
$begingroup$
For $ain R^d$, let $T_af(x)=f(x-a),$ for all $fin L^p(R^d), 1leq p<infty$ and all $xin R^d$. I need example that this operator doesn't converge uniformly when $arightarrow 0$. I know that this operator has strong (norm) limit.
functional-analysis convergence
asked Jan 12 at 10:53
HanaHana
161
$endgroup$
add a comment |
$begingroup$
For $ain R^d$, let $T_af(x)=f(x-a),$ for all $fin L^p(R^d), 1leq p<infty$ and all $xin R^d$. I need example that this operator doesn't converge uniformly when $arightarrow 0$. I know that this operator has strong (norm) limit.
functional-analysis convergence
asked Jan 12 at 10:53
HanaHana
161
$endgroup$
For $ain R^d$, let $T_af(x)=f(x-a),$ for all $fin L^p(R^d), 1leq p<infty$ and all $xin R^d$. I need example that this operator doesn't converge uniformly when $arightarrow 0$. I know that this operator has strong (norm) limit.
functional-analysis convergence
functional-analysis convergence
asked Jan 12 at 10:53
HanaHana
161
asked Jan 12 at 10:53
HanaHana
161
asked Jan 12 at 10:53
HanaHana
161
asked Jan 12 at 10:53
HanaHana
161
asked Jan 12 at 10:53
HanaHana
161
161
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $d=1,p=1$ and $f_n(x)=nx^{n}$ for $0<x<1$, $0$ for $x notin (0,1)$. Then $|f_n| <1$ for all $n$ . Suppose $int_{mathbb R}|f_n(x-a)-f_n(x)|, dx to 0$ uniformly in $n$ as $ato 0$. Then, given $epsilon in (0,e^{-1}-e^{-2}) $ there exists $b>0$ such that $int_{mathbb R}|f_n(x-a)-f_n(x)|, dx <epsilon $ for all $a in (0,b)$ for all $n$. Put $a=frac 1 n$ where $n$ is so large that $frac 1 n <b$. After a little computation we get $frac n {n+1} ((1-frac 1n)^{1+n}-(frac 1 n)^{1+n}-(1-frac 2 n)^{1+n}) <epsilon $. Letting $n to infty$ we get $e^{-1}-e^{-2}<epsilon$. This is a contradiction.
answered Jan 12 at 13:10
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
$begingroup$
Why you said that $sup$ converge to $infty$ for each $ain(0,1)$ when $a$ need to converge to 0?
$endgroup$
– Hana
Jan 15 at 7:25
$begingroup$
If we had uniform convergence then there would be $a_0$ such that $int |f_n(x-a)-f_n(x)|, dx <1$ for all $n$ for all $a<a_0$. In my example you can get a contradiction to this by fixing $a$, say $a=a_0/2$ and letting $n to infty$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 7:28
$begingroup$
Thanks, I understand that now. But I cannot get that limit is infinity: $int_a^{1-a}|n(x-a)^n-nx^n|= frac{n}{n+1}((1 - a)^{1+n} - a^{1 + n}-(1-2a)^{1+n})$ this converge to 0 for $ain(0,1)$ when $nrightarrowinfty$ or I made some mistake.
$endgroup$
– Hana
Jan 15 at 16:03
$begingroup$
@Hana I was a bit careless with details but the sequence ${f_n}$ does work. Please look at my revised answer.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 23:21
$begingroup$
Thanks! Whether the uniform convergence of this operator $T_a$ depends on p? I.e. whether the uniform convergence will be valid for some $p$?
$endgroup$
– Hana
Jan 17 at 19:25
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%2f3070781%2funiform-convergence-for-operator-of-translation%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
$begingroup$
Let $d=1,p=1$ and $f_n(x)=nx^{n}$ for $0<x<1$, $0$ for $x notin (0,1)$. Then $|f_n| <1$ for all $n$ . Suppose $int_{mathbb R}|f_n(x-a)-f_n(x)|, dx to 0$ uniformly in $n$ as $ato 0$. Then, given $epsilon in (0,e^{-1}-e^{-2}) $ there exists $b>0$ such that $int_{mathbb R}|f_n(x-a)-f_n(x)|, dx <epsilon $ for all $a in (0,b)$ for all $n$. Put $a=frac 1 n$ where $n$ is so large that $frac 1 n <b$. After a little computation we get $frac n {n+1} ((1-frac 1n)^{1+n}-(frac 1 n)^{1+n}-(1-frac 2 n)^{1+n}) <epsilon $. Letting $n to infty$ we get $e^{-1}-e^{-2}<epsilon$. This is a contradiction.
answered Jan 12 at 13:10
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
$begingroup$
Why you said that $sup$ converge to $infty$ for each $ain(0,1)$ when $a$ need to converge to 0?
$endgroup$
– Hana
Jan 15 at 7:25
$begingroup$
If we had uniform convergence then there would be $a_0$ such that $int |f_n(x-a)-f_n(x)|, dx <1$ for all $n$ for all $a<a_0$. In my example you can get a contradiction to this by fixing $a$, say $a=a_0/2$ and letting $n to infty$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 7:28
$begingroup$
Thanks, I understand that now. But I cannot get that limit is infinity: $int_a^{1-a}|n(x-a)^n-nx^n|= frac{n}{n+1}((1 - a)^{1+n} - a^{1 + n}-(1-2a)^{1+n})$ this converge to 0 for $ain(0,1)$ when $nrightarrowinfty$ or I made some mistake.
$endgroup$
– Hana
Jan 15 at 16:03
$begingroup$
@Hana I was a bit careless with details but the sequence ${f_n}$ does work. Please look at my revised answer.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 23:21
$begingroup$
Thanks! Whether the uniform convergence of this operator $T_a$ depends on p? I.e. whether the uniform convergence will be valid for some $p$?
$endgroup$
– Hana
Jan 17 at 19:25
add a comment |
$begingroup$
Let $d=1,p=1$ and $f_n(x)=nx^{n}$ for $0<x<1$, $0$ for $x notin (0,1)$. Then $|f_n| <1$ for all $n$ . Suppose $int_{mathbb R}|f_n(x-a)-f_n(x)|, dx to 0$ uniformly in $n$ as $ato 0$. Then, given $epsilon in (0,e^{-1}-e^{-2}) $ there exists $b>0$ such that $int_{mathbb R}|f_n(x-a)-f_n(x)|, dx <epsilon $ for all $a in (0,b)$ for all $n$. Put $a=frac 1 n$ where $n$ is so large that $frac 1 n <b$. After a little computation we get $frac n {n+1} ((1-frac 1n)^{1+n}-(frac 1 n)^{1+n}-(1-frac 2 n)^{1+n}) <epsilon $. Letting $n to infty$ we get $e^{-1}-e^{-2}<epsilon$. This is a contradiction.
answered Jan 12 at 13:10
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
$begingroup$
Why you said that $sup$ converge to $infty$ for each $ain(0,1)$ when $a$ need to converge to 0?
$endgroup$
– Hana
Jan 15 at 7:25
$begingroup$
If we had uniform convergence then there would be $a_0$ such that $int |f_n(x-a)-f_n(x)|, dx <1$ for all $n$ for all $a<a_0$. In my example you can get a contradiction to this by fixing $a$, say $a=a_0/2$ and letting $n to infty$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 7:28
$begingroup$
Thanks, I understand that now. But I cannot get that limit is infinity: $int_a^{1-a}|n(x-a)^n-nx^n|= frac{n}{n+1}((1 - a)^{1+n} - a^{1 + n}-(1-2a)^{1+n})$ this converge to 0 for $ain(0,1)$ when $nrightarrowinfty$ or I made some mistake.
$endgroup$
– Hana
Jan 15 at 16:03
$begingroup$
@Hana I was a bit careless with details but the sequence ${f_n}$ does work. Please look at my revised answer.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 23:21
$begingroup$
Thanks! Whether the uniform convergence of this operator $T_a$ depends on p? I.e. whether the uniform convergence will be valid for some $p$?
$endgroup$
– Hana
Jan 17 at 19:25
add a comment |
$begingroup$
Let $d=1,p=1$ and $f_n(x)=nx^{n}$ for $0<x<1$, $0$ for $x notin (0,1)$. Then $|f_n| <1$ for all $n$ . Suppose $int_{mathbb R}|f_n(x-a)-f_n(x)|, dx to 0$ uniformly in $n$ as $ato 0$. Then, given $epsilon in (0,e^{-1}-e^{-2}) $ there exists $b>0$ such that $int_{mathbb R}|f_n(x-a)-f_n(x)|, dx <epsilon $ for all $a in (0,b)$ for all $n$. Put $a=frac 1 n$ where $n$ is so large that $frac 1 n <b$. After a little computation we get $frac n {n+1} ((1-frac 1n)^{1+n}-(frac 1 n)^{1+n}-(1-frac 2 n)^{1+n}) <epsilon $. Letting $n to infty$ we get $e^{-1}-e^{-2}<epsilon$. This is a contradiction.
answered Jan 12 at 13:10
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
Let $d=1,p=1$ and $f_n(x)=nx^{n}$ for $0<x<1$, $0$ for $x notin (0,1)$. Then $|f_n| <1$ for all $n$ . Suppose $int_{mathbb R}|f_n(x-a)-f_n(x)|, dx to 0$ uniformly in $n$ as $ato 0$. Then, given $epsilon in (0,e^{-1}-e^{-2}) $ there exists $b>0$ such that $int_{mathbb R}|f_n(x-a)-f_n(x)|, dx <epsilon $ for all $a in (0,b)$ for all $n$. Put $a=frac 1 n$ where $n$ is so large that $frac 1 n <b$. After a little computation we get $frac n {n+1} ((1-frac 1n)^{1+n}-(frac 1 n)^{1+n}-(1-frac 2 n)^{1+n}) <epsilon $. Letting $n to infty$ we get $e^{-1}-e^{-2}<epsilon$. This is a contradiction.
answered Jan 12 at 13:10
Kavi Rama MurthyKavi Rama Murthy
56k42158
edited Jan 15 at 23:20
answered Jan 12 at 13:10
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Jan 12 at 13:10
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Jan 12 at 13:10
Kavi Rama MurthyKavi Rama Murthy
56k42158
56k42158
$begingroup$
Why you said that $sup$ converge to $infty$ for each $ain(0,1)$ when $a$ need to converge to 0?
$endgroup$
– Hana
Jan 15 at 7:25
$begingroup$
If we had uniform convergence then there would be $a_0$ such that $int |f_n(x-a)-f_n(x)|, dx <1$ for all $n$ for all $a<a_0$. In my example you can get a contradiction to this by fixing $a$, say $a=a_0/2$ and letting $n to infty$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 7:28
$begingroup$
Thanks, I understand that now. But I cannot get that limit is infinity: $int_a^{1-a}|n(x-a)^n-nx^n|= frac{n}{n+1}((1 - a)^{1+n} - a^{1 + n}-(1-2a)^{1+n})$ this converge to 0 for $ain(0,1)$ when $nrightarrowinfty$ or I made some mistake.
$endgroup$
– Hana
Jan 15 at 16:03
$begingroup$
@Hana I was a bit careless with details but the sequence ${f_n}$ does work. Please look at my revised answer.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 23:21
$begingroup$
Thanks! Whether the uniform convergence of this operator $T_a$ depends on p? I.e. whether the uniform convergence will be valid for some $p$?
$endgroup$
– Hana
Jan 17 at 19:25
add a comment |
$begingroup$
Why you said that $sup$ converge to $infty$ for each $ain(0,1)$ when $a$ need to converge to 0?
$endgroup$
– Hana
Jan 15 at 7:25
$begingroup$
If we had uniform convergence then there would be $a_0$ such that $int |f_n(x-a)-f_n(x)|, dx <1$ for all $n$ for all $a<a_0$. In my example you can get a contradiction to this by fixing $a$, say $a=a_0/2$ and letting $n to infty$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 7:28
$begingroup$
Thanks, I understand that now. But I cannot get that limit is infinity: $int_a^{1-a}|n(x-a)^n-nx^n|= frac{n}{n+1}((1 - a)^{1+n} - a^{1 + n}-(1-2a)^{1+n})$ this converge to 0 for $ain(0,1)$ when $nrightarrowinfty$ or I made some mistake.
$endgroup$
– Hana
Jan 15 at 16:03
$begingroup$
@Hana I was a bit careless with details but the sequence ${f_n}$ does work. Please look at my revised answer.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 23:21
$begingroup$
Thanks! Whether the uniform convergence of this operator $T_a$ depends on p? I.e. whether the uniform convergence will be valid for some $p$?
$endgroup$
– Hana
Jan 17 at 19:25
$begingroup$
Why you said that $sup$ converge to $infty$ for each $ain(0,1)$ when $a$ need to converge to 0?
$endgroup$
– Hana
Jan 15 at 7:25
$begingroup$
Why you said that $sup$ converge to $infty$ for each $ain(0,1)$ when $a$ need to converge to 0?
$endgroup$
– Hana
Jan 15 at 7:25
$begingroup$
If we had uniform convergence then there would be $a_0$ such that $int |f_n(x-a)-f_n(x)|, dx <1$ for all $n$ for all $a<a_0$. In my example you can get a contradiction to this by fixing $a$, say $a=a_0/2$ and letting $n to infty$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 7:28
$begingroup$
If we had uniform convergence then there would be $a_0$ such that $int |f_n(x-a)-f_n(x)|, dx <1$ for all $n$ for all $a<a_0$. In my example you can get a contradiction to this by fixing $a$, say $a=a_0/2$ and letting $n to infty$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 7:28
$begingroup$
Thanks, I understand that now. But I cannot get that limit is infinity: $int_a^{1-a}|n(x-a)^n-nx^n|= frac{n}{n+1}((1 - a)^{1+n} - a^{1 + n}-(1-2a)^{1+n})$ this converge to 0 for $ain(0,1)$ when $nrightarrowinfty$ or I made some mistake.
$endgroup$
– Hana
Jan 15 at 16:03
$begingroup$
Thanks, I understand that now. But I cannot get that limit is infinity: $int_a^{1-a}|n(x-a)^n-nx^n|= frac{n}{n+1}((1 - a)^{1+n} - a^{1 + n}-(1-2a)^{1+n})$ this converge to 0 for $ain(0,1)$ when $nrightarrowinfty$ or I made some mistake.
$endgroup$
– Hana
Jan 15 at 16:03
$begingroup$
@Hana I was a bit careless with details but the sequence ${f_n}$ does work. Please look at my revised answer.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 23:21
$begingroup$
@Hana I was a bit careless with details but the sequence ${f_n}$ does work. Please look at my revised answer.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 23:21
$begingroup$
Thanks! Whether the uniform convergence of this operator $T_a$ depends on p? I.e. whether the uniform convergence will be valid for some $p$?
$endgroup$
– Hana
Jan 17 at 19:25
$begingroup$
Thanks! Whether the uniform convergence of this operator $T_a$ depends on p? I.e. whether the uniform convergence will be valid for some $p$?
$endgroup$
– Hana
Jan 17 at 19:25
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%2f3070781%2funiform-convergence-for-operator-of-translation%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
