Problem about weak convergence
$begingroup$
I am reading a lecture note about Radon Riesz theorem, the resources is here: https://faculty.etsu.edu/gardnerr/5210/notes/Radon-Riesz.pdf
At page 4, fourth line from the bottom, it says convergence in measure. But if this result is implied by weak convergence of $f_n$. I can’t proof it.
I have tried some methods, but this lead to a question about the definition of weak convergence. In the definition, which of the following one is exactly the real definition of weak convergence?
- for every $Tin X^*$, for all $varepsilon>0$ May related to $T$, there is an $N$, such that for all $n>N$, $d(Tf_n, Tf)<varepsilon$.
- for all $varepsilon>0$, there is an $N$, such that for all $n>N$ and all $T$, $d(Tf_n, Tf)<varepsilon$.
I checked some book but they generally says like
$$Tf_nto Tf,qquad text{for all } Tin X^*$$
This makes curious for me.
Finally I almost forgot, if the definition is the second one I have proved: weak convergence implies convergence in measure, but if the definition is the first one, the prove failed, I also need a prove or a counterexample that weak convergence can’t imply convergence in measure, I also want to know how the lecture note get the result about convergence in measure if there is really an counterexample. THX.
real-analysis functional-analysis weak-convergence strong-convergence
asked Dec 26 '18 at 15:08
Larry EppesLarry Eppes
471311
$endgroup$
add a comment |
$begingroup$
I am reading a lecture note about Radon Riesz theorem, the resources is here: https://faculty.etsu.edu/gardnerr/5210/notes/Radon-Riesz.pdf
At page 4, fourth line from the bottom, it says convergence in measure. But if this result is implied by weak convergence of $f_n$. I can’t proof it.
I have tried some methods, but this lead to a question about the definition of weak convergence. In the definition, which of the following one is exactly the real definition of weak convergence?
- for every $Tin X^*$, for all $varepsilon>0$ May related to $T$, there is an $N$, such that for all $n>N$, $d(Tf_n, Tf)<varepsilon$.
- for all $varepsilon>0$, there is an $N$, such that for all $n>N$ and all $T$, $d(Tf_n, Tf)<varepsilon$.
I checked some book but they generally says like
$$Tf_nto Tf,qquad text{for all } Tin X^*$$
This makes curious for me.
Finally I almost forgot, if the definition is the second one I have proved: weak convergence implies convergence in measure, but if the definition is the first one, the prove failed, I also need a prove or a counterexample that weak convergence can’t imply convergence in measure, I also want to know how the lecture note get the result about convergence in measure if there is really an counterexample. THX.
real-analysis functional-analysis weak-convergence strong-convergence
asked Dec 26 '18 at 15:08
Larry EppesLarry Eppes
471311
$endgroup$
add a comment |
$begingroup$
I am reading a lecture note about Radon Riesz theorem, the resources is here: https://faculty.etsu.edu/gardnerr/5210/notes/Radon-Riesz.pdf
At page 4, fourth line from the bottom, it says convergence in measure. But if this result is implied by weak convergence of $f_n$. I can’t proof it.
I have tried some methods, but this lead to a question about the definition of weak convergence. In the definition, which of the following one is exactly the real definition of weak convergence?
- for every $Tin X^*$, for all $varepsilon>0$ May related to $T$, there is an $N$, such that for all $n>N$, $d(Tf_n, Tf)<varepsilon$.
- for all $varepsilon>0$, there is an $N$, such that for all $n>N$ and all $T$, $d(Tf_n, Tf)<varepsilon$.
I checked some book but they generally says like
$$Tf_nto Tf,qquad text{for all } Tin X^*$$
This makes curious for me.
Finally I almost forgot, if the definition is the second one I have proved: weak convergence implies convergence in measure, but if the definition is the first one, the prove failed, I also need a prove or a counterexample that weak convergence can’t imply convergence in measure, I also want to know how the lecture note get the result about convergence in measure if there is really an counterexample. THX.
real-analysis functional-analysis weak-convergence strong-convergence
asked Dec 26 '18 at 15:08
Larry EppesLarry Eppes
471311
$endgroup$
I am reading a lecture note about Radon Riesz theorem, the resources is here: https://faculty.etsu.edu/gardnerr/5210/notes/Radon-Riesz.pdf
At page 4, fourth line from the bottom, it says convergence in measure. But if this result is implied by weak convergence of $f_n$. I can’t proof it.
I have tried some methods, but this lead to a question about the definition of weak convergence. In the definition, which of the following one is exactly the real definition of weak convergence?
- for every $Tin X^*$, for all $varepsilon>0$ May related to $T$, there is an $N$, such that for all $n>N$, $d(Tf_n, Tf)<varepsilon$.
- for all $varepsilon>0$, there is an $N$, such that for all $n>N$ and all $T$, $d(Tf_n, Tf)<varepsilon$.
I checked some book but they generally says like
$$Tf_nto Tf,qquad text{for all } Tin X^*$$
This makes curious for me.
Finally I almost forgot, if the definition is the second one I have proved: weak convergence implies convergence in measure, but if the definition is the first one, the prove failed, I also need a prove or a counterexample that weak convergence can’t imply convergence in measure, I also want to know how the lecture note get the result about convergence in measure if there is really an counterexample. THX.
real-analysis functional-analysis weak-convergence strong-convergence
real-analysis functional-analysis weak-convergence strong-convergence
asked Dec 26 '18 at 15:08
Larry EppesLarry Eppes
471311
asked Dec 26 '18 at 15:08
Larry EppesLarry Eppes
471311
edited Dec 26 '18 at 15:18
Larry Eppes
asked Dec 26 '18 at 15:08
Larry EppesLarry Eppes
471311
asked Dec 26 '18 at 15:08
Larry EppesLarry Eppes
471311
asked Dec 26 '18 at 15:08
Larry EppesLarry Eppes
471311
471311
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The second one is false. To see this, if $T$ is a (fixed) continuous linear functional, then any constant multiplied by $T$ is also a continuous linear functional. Then the second statement is essentially asking for
$$|cT(f_n) - cT(f)| = c|T(f_n) - T(f)| leq epsilon$$
for any constant $c$.
The first one is (almost) correct, the $epsilon$ is still arbitrary, just now $N$ is allowed to depend on $T$.
In general, weak convergence does not imply convergence in measure, take $f_n(x) = chi_{[n, n+1)}(x)$ which weakly converges to zero in $L^p(mathbb{R})$ for $1<p<infty$, but fails to converge in measure .
answered Dec 26 '18 at 18:47
XiaoXiao
4,82111436
$endgroup$
$begingroup$
I've got it, yes, for $X^*cong L^q$, so for every $gin L^q$, we can define $Tf:=int fg$ which is finite linearity(linear is obvious, finite is ensured by Holder inequality), then by $int_{[n,n+1)}gto 0$ implies weakly converges to zero.
$endgroup$
– Larry Eppes
Dec 27 '18 at 2:25
$begingroup$
I re-checked the notes in the link, and retried to understand the document in the fourth page, so ${f_n}to f$ in measure is false? although by $|f|^{p-2}fin L^q$, so that $Tg:=int |f|^{p-2}fg$, $forall gin L^p$ is a bounded linear functional, and weakly convergence imply $Tf_nto Tf$ directly without converge in measure.
$endgroup$
– Larry Eppes
Dec 27 '18 at 2:39
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%2f3053018%2fproblem-about-weak-convergence%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 second one is false. To see this, if $T$ is a (fixed) continuous linear functional, then any constant multiplied by $T$ is also a continuous linear functional. Then the second statement is essentially asking for
$$|cT(f_n) - cT(f)| = c|T(f_n) - T(f)| leq epsilon$$
for any constant $c$.
The first one is (almost) correct, the $epsilon$ is still arbitrary, just now $N$ is allowed to depend on $T$.
In general, weak convergence does not imply convergence in measure, take $f_n(x) = chi_{[n, n+1)}(x)$ which weakly converges to zero in $L^p(mathbb{R})$ for $1<p<infty$, but fails to converge in measure .
answered Dec 26 '18 at 18:47
XiaoXiao
4,82111436
$endgroup$
$begingroup$
I've got it, yes, for $X^*cong L^q$, so for every $gin L^q$, we can define $Tf:=int fg$ which is finite linearity(linear is obvious, finite is ensured by Holder inequality), then by $int_{[n,n+1)}gto 0$ implies weakly converges to zero.
$endgroup$
– Larry Eppes
Dec 27 '18 at 2:25
$begingroup$
I re-checked the notes in the link, and retried to understand the document in the fourth page, so ${f_n}to f$ in measure is false? although by $|f|^{p-2}fin L^q$, so that $Tg:=int |f|^{p-2}fg$, $forall gin L^p$ is a bounded linear functional, and weakly convergence imply $Tf_nto Tf$ directly without converge in measure.
$endgroup$
– Larry Eppes
Dec 27 '18 at 2:39
add a comment |
$begingroup$
The second one is false. To see this, if $T$ is a (fixed) continuous linear functional, then any constant multiplied by $T$ is also a continuous linear functional. Then the second statement is essentially asking for
$$|cT(f_n) - cT(f)| = c|T(f_n) - T(f)| leq epsilon$$
for any constant $c$.
The first one is (almost) correct, the $epsilon$ is still arbitrary, just now $N$ is allowed to depend on $T$.
In general, weak convergence does not imply convergence in measure, take $f_n(x) = chi_{[n, n+1)}(x)$ which weakly converges to zero in $L^p(mathbb{R})$ for $1<p<infty$, but fails to converge in measure .
answered Dec 26 '18 at 18:47
XiaoXiao
4,82111436
$endgroup$
$begingroup$
I've got it, yes, for $X^*cong L^q$, so for every $gin L^q$, we can define $Tf:=int fg$ which is finite linearity(linear is obvious, finite is ensured by Holder inequality), then by $int_{[n,n+1)}gto 0$ implies weakly converges to zero.
$endgroup$
– Larry Eppes
Dec 27 '18 at 2:25
$begingroup$
I re-checked the notes in the link, and retried to understand the document in the fourth page, so ${f_n}to f$ in measure is false? although by $|f|^{p-2}fin L^q$, so that $Tg:=int |f|^{p-2}fg$, $forall gin L^p$ is a bounded linear functional, and weakly convergence imply $Tf_nto Tf$ directly without converge in measure.
$endgroup$
– Larry Eppes
Dec 27 '18 at 2:39
add a comment |
$begingroup$
The second one is false. To see this, if $T$ is a (fixed) continuous linear functional, then any constant multiplied by $T$ is also a continuous linear functional. Then the second statement is essentially asking for
$$|cT(f_n) - cT(f)| = c|T(f_n) - T(f)| leq epsilon$$
for any constant $c$.
The first one is (almost) correct, the $epsilon$ is still arbitrary, just now $N$ is allowed to depend on $T$.
In general, weak convergence does not imply convergence in measure, take $f_n(x) = chi_{[n, n+1)}(x)$ which weakly converges to zero in $L^p(mathbb{R})$ for $1<p<infty$, but fails to converge in measure .
answered Dec 26 '18 at 18:47
XiaoXiao
4,82111436
$endgroup$
The second one is false. To see this, if $T$ is a (fixed) continuous linear functional, then any constant multiplied by $T$ is also a continuous linear functional. Then the second statement is essentially asking for
$$|cT(f_n) - cT(f)| = c|T(f_n) - T(f)| leq epsilon$$
for any constant $c$.
The first one is (almost) correct, the $epsilon$ is still arbitrary, just now $N$ is allowed to depend on $T$.
In general, weak convergence does not imply convergence in measure, take $f_n(x) = chi_{[n, n+1)}(x)$ which weakly converges to zero in $L^p(mathbb{R})$ for $1<p<infty$, but fails to converge in measure .
answered Dec 26 '18 at 18:47
XiaoXiao
4,82111436
answered Dec 26 '18 at 18:47
XiaoXiao
4,82111436
answered Dec 26 '18 at 18:47
XiaoXiao
4,82111436
answered Dec 26 '18 at 18:47
XiaoXiao
4,82111436
4,82111436
$begingroup$
I've got it, yes, for $X^*cong L^q$, so for every $gin L^q$, we can define $Tf:=int fg$ which is finite linearity(linear is obvious, finite is ensured by Holder inequality), then by $int_{[n,n+1)}gto 0$ implies weakly converges to zero.
$endgroup$
– Larry Eppes
Dec 27 '18 at 2:25
$begingroup$
I re-checked the notes in the link, and retried to understand the document in the fourth page, so ${f_n}to f$ in measure is false? although by $|f|^{p-2}fin L^q$, so that $Tg:=int |f|^{p-2}fg$, $forall gin L^p$ is a bounded linear functional, and weakly convergence imply $Tf_nto Tf$ directly without converge in measure.
$endgroup$
– Larry Eppes
Dec 27 '18 at 2:39
add a comment |
$begingroup$
I've got it, yes, for $X^*cong L^q$, so for every $gin L^q$, we can define $Tf:=int fg$ which is finite linearity(linear is obvious, finite is ensured by Holder inequality), then by $int_{[n,n+1)}gto 0$ implies weakly converges to zero.
$endgroup$
– Larry Eppes
Dec 27 '18 at 2:25
$begingroup$
I re-checked the notes in the link, and retried to understand the document in the fourth page, so ${f_n}to f$ in measure is false? although by $|f|^{p-2}fin L^q$, so that $Tg:=int |f|^{p-2}fg$, $forall gin L^p$ is a bounded linear functional, and weakly convergence imply $Tf_nto Tf$ directly without converge in measure.
$endgroup$
– Larry Eppes
Dec 27 '18 at 2:39
$begingroup$
I've got it, yes, for $X^*cong L^q$, so for every $gin L^q$, we can define $Tf:=int fg$ which is finite linearity(linear is obvious, finite is ensured by Holder inequality), then by $int_{[n,n+1)}gto 0$ implies weakly converges to zero.
$endgroup$
– Larry Eppes
Dec 27 '18 at 2:25
$begingroup$
I've got it, yes, for $X^*cong L^q$, so for every $gin L^q$, we can define $Tf:=int fg$ which is finite linearity(linear is obvious, finite is ensured by Holder inequality), then by $int_{[n,n+1)}gto 0$ implies weakly converges to zero.
$endgroup$
– Larry Eppes
Dec 27 '18 at 2:25
$begingroup$
I re-checked the notes in the link, and retried to understand the document in the fourth page, so ${f_n}to f$ in measure is false? although by $|f|^{p-2}fin L^q$, so that $Tg:=int |f|^{p-2}fg$, $forall gin L^p$ is a bounded linear functional, and weakly convergence imply $Tf_nto Tf$ directly without converge in measure.
$endgroup$
– Larry Eppes
Dec 27 '18 at 2:39
$begingroup$
I re-checked the notes in the link, and retried to understand the document in the fourth page, so ${f_n}to f$ in measure is false? although by $|f|^{p-2}fin L^q$, so that $Tg:=int |f|^{p-2}fg$, $forall gin L^p$ is a bounded linear functional, and weakly convergence imply $Tf_nto Tf$ directly without converge in measure.
$endgroup$
– Larry Eppes
Dec 27 '18 at 2:39
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%2f3053018%2fproblem-about-weak-convergence%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
