Mixing
Prove that a pointwise converging sequence, bounded in $L^p$, converge uniformly in $L^q$, using Egorov's...
Let $p > 1$, $E$ be a measurable subset of $mathbb{R}^n$ with finite
Lebesgue measure $mu$, ${f_n}_{nin N}$ be a sequence of measurable
functions $f_n : E → mathbb{R}$, bounded in $L^p(E)$, and pointwise converging to $f : E → mathbb{R}$.
Prove that ${f_n}_{n∈N}$ converges to $f$ in
$L^q(E)$ for all $q ∈(1, p)$.
All the hypothesis of Egorov's theorem are satisfied, then $forallε > 0$ $exists N_ε ⊆ Ω$ measurable such that $mu(N_ε) ≤ ε$ and such that the sequence $f_n$ converges uniformly to $f$ in $Ω setminus N_ε$.
Any hint on how to start the exercise?
functional-analysis convergence
asked Nov 19 '18 at 23:49
sound wave
1598
add a comment |
Let $p > 1$, $E$ be a measurable subset of $mathbb{R}^n$ with finite
Lebesgue measure $mu$, ${f_n}_{nin N}$ be a sequence of measurable
functions $f_n : E → mathbb{R}$, bounded in $L^p(E)$, and pointwise converging to $f : E → mathbb{R}$.
Prove that ${f_n}_{n∈N}$ converges to $f$ in
$L^q(E)$ for all $q ∈(1, p)$.
All the hypothesis of Egorov's theorem are satisfied, then $forallε > 0$ $exists N_ε ⊆ Ω$ measurable such that $mu(N_ε) ≤ ε$ and such that the sequence $f_n$ converges uniformly to $f$ in $Ω setminus N_ε$.
Any hint on how to start the exercise?
functional-analysis convergence
asked Nov 19 '18 at 23:49
sound wave
1598
add a comment |
Let $p > 1$, $E$ be a measurable subset of $mathbb{R}^n$ with finite
Lebesgue measure $mu$, ${f_n}_{nin N}$ be a sequence of measurable
functions $f_n : E → mathbb{R}$, bounded in $L^p(E)$, and pointwise converging to $f : E → mathbb{R}$.
Prove that ${f_n}_{n∈N}$ converges to $f$ in
$L^q(E)$ for all $q ∈(1, p)$.
All the hypothesis of Egorov's theorem are satisfied, then $forallε > 0$ $exists N_ε ⊆ Ω$ measurable such that $mu(N_ε) ≤ ε$ and such that the sequence $f_n$ converges uniformly to $f$ in $Ω setminus N_ε$.
Any hint on how to start the exercise?
functional-analysis convergence
asked Nov 19 '18 at 23:49
sound wave
1598
Let $p > 1$, $E$ be a measurable subset of $mathbb{R}^n$ with finite
Lebesgue measure $mu$, ${f_n}_{nin N}$ be a sequence of measurable
functions $f_n : E → mathbb{R}$, bounded in $L^p(E)$, and pointwise converging to $f : E → mathbb{R}$.
Prove that ${f_n}_{n∈N}$ converges to $f$ in
$L^q(E)$ for all $q ∈(1, p)$.
All the hypothesis of Egorov's theorem are satisfied, then $forallε > 0$ $exists N_ε ⊆ Ω$ measurable such that $mu(N_ε) ≤ ε$ and such that the sequence $f_n$ converges uniformly to $f$ in $Ω setminus N_ε$.
Any hint on how to start the exercise?
functional-analysis convergence
functional-analysis convergence
asked Nov 19 '18 at 23:49
sound wave
1598
asked Nov 19 '18 at 23:49
sound wave
1598
asked Nov 19 '18 at 23:49
sound wave
1598
asked Nov 19 '18 at 23:49
sound wave
1598
asked Nov 19 '18 at 23:49
sound wave
1598
1598
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Here's a hint on how to start. Use Egorov's theorem to split the domain into two parts, one with $epsilon$ measure on which convergence is bad, and one with the majority of the measure on which convergence is good. Then bound the total $q$-norm of the difference of the sequence from the limit by bounding the $q$-norm of the difference on each of these sets.
answered Nov 19 '18 at 23:59
jgon
12.8k21940
Thanks. I would do this: since the sequence converge almost everywhere, take the set $N_epsilon$ of measure zero where the seq does not converge, and $Esetminus N_epsilon$ where the seq converge. Then from what you wrote I got that we have to bound $||f_n-f||_{L^q(E)}$ with $||f_n-f||_{L^q(N_epsilon)}+||f_n-f||_{L^q(Esetminus N_epsilon)}$. But the first norm is infinite because on $N_epsilon$ the sequence does not converge. Do we need to change the choice of the two sets?
– sound wave
Nov 20 '18 at 12:28
Could you say what it the $L^q$ norm of $f_n-f$ on the set $Esetminus N_varepsilon$ ?
– sound wave
Nov 21 '18 at 14:20
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%2f3005711%2fprove-that-a-pointwise-converging-sequence-bounded-in-lp-converge-uniformly%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
Here's a hint on how to start. Use Egorov's theorem to split the domain into two parts, one with $epsilon$ measure on which convergence is bad, and one with the majority of the measure on which convergence is good. Then bound the total $q$-norm of the difference of the sequence from the limit by bounding the $q$-norm of the difference on each of these sets.
answered Nov 19 '18 at 23:59
jgon
12.8k21940
Thanks. I would do this: since the sequence converge almost everywhere, take the set $N_epsilon$ of measure zero where the seq does not converge, and $Esetminus N_epsilon$ where the seq converge. Then from what you wrote I got that we have to bound $||f_n-f||_{L^q(E)}$ with $||f_n-f||_{L^q(N_epsilon)}+||f_n-f||_{L^q(Esetminus N_epsilon)}$. But the first norm is infinite because on $N_epsilon$ the sequence does not converge. Do we need to change the choice of the two sets?
– sound wave
Nov 20 '18 at 12:28
Could you say what it the $L^q$ norm of $f_n-f$ on the set $Esetminus N_varepsilon$ ?
– sound wave
Nov 21 '18 at 14:20
add a comment |
Here's a hint on how to start. Use Egorov's theorem to split the domain into two parts, one with $epsilon$ measure on which convergence is bad, and one with the majority of the measure on which convergence is good. Then bound the total $q$-norm of the difference of the sequence from the limit by bounding the $q$-norm of the difference on each of these sets.
answered Nov 19 '18 at 23:59
jgon
12.8k21940
Thanks. I would do this: since the sequence converge almost everywhere, take the set $N_epsilon$ of measure zero where the seq does not converge, and $Esetminus N_epsilon$ where the seq converge. Then from what you wrote I got that we have to bound $||f_n-f||_{L^q(E)}$ with $||f_n-f||_{L^q(N_epsilon)}+||f_n-f||_{L^q(Esetminus N_epsilon)}$. But the first norm is infinite because on $N_epsilon$ the sequence does not converge. Do we need to change the choice of the two sets?
– sound wave
Nov 20 '18 at 12:28
Could you say what it the $L^q$ norm of $f_n-f$ on the set $Esetminus N_varepsilon$ ?
– sound wave
Nov 21 '18 at 14:20
add a comment |
Here's a hint on how to start. Use Egorov's theorem to split the domain into two parts, one with $epsilon$ measure on which convergence is bad, and one with the majority of the measure on which convergence is good. Then bound the total $q$-norm of the difference of the sequence from the limit by bounding the $q$-norm of the difference on each of these sets.
answered Nov 19 '18 at 23:59
jgon
12.8k21940
Here's a hint on how to start. Use Egorov's theorem to split the domain into two parts, one with $epsilon$ measure on which convergence is bad, and one with the majority of the measure on which convergence is good. Then bound the total $q$-norm of the difference of the sequence from the limit by bounding the $q$-norm of the difference on each of these sets.
answered Nov 19 '18 at 23:59
jgon
12.8k21940
answered Nov 19 '18 at 23:59
jgon
12.8k21940
answered Nov 19 '18 at 23:59
jgon
12.8k21940
answered Nov 19 '18 at 23:59
jgon
12.8k21940
12.8k21940
Thanks. I would do this: since the sequence converge almost everywhere, take the set $N_epsilon$ of measure zero where the seq does not converge, and $Esetminus N_epsilon$ where the seq converge. Then from what you wrote I got that we have to bound $||f_n-f||_{L^q(E)}$ with $||f_n-f||_{L^q(N_epsilon)}+||f_n-f||_{L^q(Esetminus N_epsilon)}$. But the first norm is infinite because on $N_epsilon$ the sequence does not converge. Do we need to change the choice of the two sets?
– sound wave
Nov 20 '18 at 12:28
Could you say what it the $L^q$ norm of $f_n-f$ on the set $Esetminus N_varepsilon$ ?
– sound wave
Nov 21 '18 at 14:20
add a comment |
Thanks. I would do this: since the sequence converge almost everywhere, take the set $N_epsilon$ of measure zero where the seq does not converge, and $Esetminus N_epsilon$ where the seq converge. Then from what you wrote I got that we have to bound $||f_n-f||_{L^q(E)}$ with $||f_n-f||_{L^q(N_epsilon)}+||f_n-f||_{L^q(Esetminus N_epsilon)}$. But the first norm is infinite because on $N_epsilon$ the sequence does not converge. Do we need to change the choice of the two sets?
– sound wave
Nov 20 '18 at 12:28
Could you say what it the $L^q$ norm of $f_n-f$ on the set $Esetminus N_varepsilon$ ?
– sound wave
Nov 21 '18 at 14:20
Thanks. I would do this: since the sequence converge almost everywhere, take the set $N_epsilon$ of measure zero where the seq does not converge, and $Esetminus N_epsilon$ where the seq converge. Then from what you wrote I got that we have to bound $||f_n-f||_{L^q(E)}$ with $||f_n-f||_{L^q(N_epsilon)}+||f_n-f||_{L^q(Esetminus N_epsilon)}$. But the first norm is infinite because on $N_epsilon$ the sequence does not converge. Do we need to change the choice of the two sets?
– sound wave
Nov 20 '18 at 12:28
Thanks. I would do this: since the sequence converge almost everywhere, take the set $N_epsilon$ of measure zero where the seq does not converge, and $Esetminus N_epsilon$ where the seq converge. Then from what you wrote I got that we have to bound $||f_n-f||_{L^q(E)}$ with $||f_n-f||_{L^q(N_epsilon)}+||f_n-f||_{L^q(Esetminus N_epsilon)}$. But the first norm is infinite because on $N_epsilon$ the sequence does not converge. Do we need to change the choice of the two sets?
– sound wave
Nov 20 '18 at 12:28
Could you say what it the $L^q$ norm of $f_n-f$ on the set $Esetminus N_varepsilon$ ?
– sound wave
Nov 21 '18 at 14:20
Could you say what it the $L^q$ norm of $f_n-f$ on the set $Esetminus N_varepsilon$ ?
– sound wave
Nov 21 '18 at 14:20
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%2f3005711%2fprove-that-a-pointwise-converging-sequence-bounded-in-lp-converge-uniformly%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
