Push forward of measures and weak* converence
I have to prove the following:
Proposition:
Let $$mu := mathcal{L}^1 big|_{[0,1]} $$
and $1<p< infty$. Consider the sequence of functions $ {f_h }_{h>0} subset L^p(mathbb{R}, mu)$ where $f_h(x)=f(hx)$ with
$$ f(x) := begin{cases} 1 quad & text{ if } quad 0 le {x} < frac{1}{2} \ -1 quad &text{ if } quad frac{1}{2} le {x} < 1 end{cases}$$
where ${x}$ is the fractional part of $x$. Let $nu:= frac{1}{2} ( delta_1 + delta_{-1})$. Then
- For each $h>0$ it holds
$$ ((f_h)_{#}) (mu) = nu $$
$f_h overset{ast}{rightharpoonup} 0$ as $h to 0$
where $((f_h)_{#}) (mu)(B) = mu ( f_h^{-1}(B))$ for any borel set $B$.
I'm not sure they are true; I suspect it should be $h>1$ and $h to + infty$. Indeed:
$ ((f_h)_{#}) (mu) = nu Leftrightarrow mu ( f_h^{-1}((a,b))= nu((a,b))$ for every $-infty< a <b <+ infty$. If both (or none of) $1$ and $-1$ belong to $(a,b)$ the equality is easy. Suppose $1 in (a,b)$ and $-1 notin (a,b)$, and take for example $h=1/2$ then
$$nu((a,b))= frac{1}{2} $$
and
$$ mu (f_{1/2}^{-1}((a,b))) = mu ({ x in [0,1] mid f_{1/2}(x) = 1 }) = mu ( {x in [0,1] mid 0 le { frac{1}{2} x } < 1/2 } ) = 1$$
On the other hand, if $h ge 1$ I think the equality holds. Am I wrong?If $h < 1/2$ and $varphi in L^q(mathbb{R}, mu)$ with $1/p + 1/q =1$ then
$$ int_{mathbb{R}} f_h varphi d mu = int_{ { x in [0,1] mid 0 le { hx } < 1/2 } } varphi d mathcal{L}^1 = int_{[0,1]} varphi d mathcal{L}^1 = int_{[0,1]} varphi d mu $$
and then I have $f_h overset{ast}{rightharpoonup} 1$ as $h to 0$. I think I can prove that, if $h to + infty$, the set on which $f_h = 1$ becomes union of smaller and smaller disjoint intervals which total measure is always 1/2. The same for the set on which $f_h = -1$. But, even if it is true, I don't know how to prove statement 2 (with $h to + infty$).
measure-theory lp-spaces weak-convergence
asked Nov 29 '18 at 19:03
Bremen000
424210
|
show 1 more comment
I have to prove the following:
Proposition:
Let $$mu := mathcal{L}^1 big|_{[0,1]} $$
and $1<p< infty$. Consider the sequence of functions $ {f_h }_{h>0} subset L^p(mathbb{R}, mu)$ where $f_h(x)=f(hx)$ with
$$ f(x) := begin{cases} 1 quad & text{ if } quad 0 le {x} < frac{1}{2} \ -1 quad &text{ if } quad frac{1}{2} le {x} < 1 end{cases}$$
where ${x}$ is the fractional part of $x$. Let $nu:= frac{1}{2} ( delta_1 + delta_{-1})$. Then
- For each $h>0$ it holds
$$ ((f_h)_{#}) (mu) = nu $$
$f_h overset{ast}{rightharpoonup} 0$ as $h to 0$
where $((f_h)_{#}) (mu)(B) = mu ( f_h^{-1}(B))$ for any borel set $B$.
I'm not sure they are true; I suspect it should be $h>1$ and $h to + infty$. Indeed:
$ ((f_h)_{#}) (mu) = nu Leftrightarrow mu ( f_h^{-1}((a,b))= nu((a,b))$ for every $-infty< a <b <+ infty$. If both (or none of) $1$ and $-1$ belong to $(a,b)$ the equality is easy. Suppose $1 in (a,b)$ and $-1 notin (a,b)$, and take for example $h=1/2$ then
$$nu((a,b))= frac{1}{2} $$
and
$$ mu (f_{1/2}^{-1}((a,b))) = mu ({ x in [0,1] mid f_{1/2}(x) = 1 }) = mu ( {x in [0,1] mid 0 le { frac{1}{2} x } < 1/2 } ) = 1$$
On the other hand, if $h ge 1$ I think the equality holds. Am I wrong?If $h < 1/2$ and $varphi in L^q(mathbb{R}, mu)$ with $1/p + 1/q =1$ then
$$ int_{mathbb{R}} f_h varphi d mu = int_{ { x in [0,1] mid 0 le { hx } < 1/2 } } varphi d mathcal{L}^1 = int_{[0,1]} varphi d mathcal{L}^1 = int_{[0,1]} varphi d mu $$
and then I have $f_h overset{ast}{rightharpoonup} 1$ as $h to 0$. I think I can prove that, if $h to + infty$, the set on which $f_h = 1$ becomes union of smaller and smaller disjoint intervals which total measure is always 1/2. The same for the set on which $f_h = -1$. But, even if it is true, I don't know how to prove statement 2 (with $h to + infty$).
measure-theory lp-spaces weak-convergence
asked Nov 29 '18 at 19:03
Bremen000
424210
Yes for 1 you need $hinmathbb N$
– Federico
Nov 29 '18 at 19:34
Thank you. What do you think about the second part of the "proposition"?
– Bremen000
Nov 29 '18 at 20:36
Yes for 2 you need $htoinfty$ to conclude $f_hrightharpoonup 0$
– Federico
Nov 29 '18 at 20:40
Otherwise you have $f_h(x)tomathrm{sign}(x)$ if $hto0$.
– Federico
Nov 29 '18 at 20:42
I'll try to work out the details for $h to +infty$. By now, I can't see how to prove it. If $h to 0$ don't you think I'll get 1, being the support of the measure $[0,1]$?
– Bremen000
Nov 29 '18 at 20:50
|
show 1 more comment
I have to prove the following:
Proposition:
Let $$mu := mathcal{L}^1 big|_{[0,1]} $$
and $1<p< infty$. Consider the sequence of functions $ {f_h }_{h>0} subset L^p(mathbb{R}, mu)$ where $f_h(x)=f(hx)$ with
$$ f(x) := begin{cases} 1 quad & text{ if } quad 0 le {x} < frac{1}{2} \ -1 quad &text{ if } quad frac{1}{2} le {x} < 1 end{cases}$$
where ${x}$ is the fractional part of $x$. Let $nu:= frac{1}{2} ( delta_1 + delta_{-1})$. Then
- For each $h>0$ it holds
$$ ((f_h)_{#}) (mu) = nu $$
$f_h overset{ast}{rightharpoonup} 0$ as $h to 0$
where $((f_h)_{#}) (mu)(B) = mu ( f_h^{-1}(B))$ for any borel set $B$.
I'm not sure they are true; I suspect it should be $h>1$ and $h to + infty$. Indeed:
$ ((f_h)_{#}) (mu) = nu Leftrightarrow mu ( f_h^{-1}((a,b))= nu((a,b))$ for every $-infty< a <b <+ infty$. If both (or none of) $1$ and $-1$ belong to $(a,b)$ the equality is easy. Suppose $1 in (a,b)$ and $-1 notin (a,b)$, and take for example $h=1/2$ then
$$nu((a,b))= frac{1}{2} $$
and
$$ mu (f_{1/2}^{-1}((a,b))) = mu ({ x in [0,1] mid f_{1/2}(x) = 1 }) = mu ( {x in [0,1] mid 0 le { frac{1}{2} x } < 1/2 } ) = 1$$
On the other hand, if $h ge 1$ I think the equality holds. Am I wrong?If $h < 1/2$ and $varphi in L^q(mathbb{R}, mu)$ with $1/p + 1/q =1$ then
$$ int_{mathbb{R}} f_h varphi d mu = int_{ { x in [0,1] mid 0 le { hx } < 1/2 } } varphi d mathcal{L}^1 = int_{[0,1]} varphi d mathcal{L}^1 = int_{[0,1]} varphi d mu $$
and then I have $f_h overset{ast}{rightharpoonup} 1$ as $h to 0$. I think I can prove that, if $h to + infty$, the set on which $f_h = 1$ becomes union of smaller and smaller disjoint intervals which total measure is always 1/2. The same for the set on which $f_h = -1$. But, even if it is true, I don't know how to prove statement 2 (with $h to + infty$).
measure-theory lp-spaces weak-convergence
asked Nov 29 '18 at 19:03
Bremen000
424210
I have to prove the following:
Proposition:
Let $$mu := mathcal{L}^1 big|_{[0,1]} $$
and $1<p< infty$. Consider the sequence of functions $ {f_h }_{h>0} subset L^p(mathbb{R}, mu)$ where $f_h(x)=f(hx)$ with
$$ f(x) := begin{cases} 1 quad & text{ if } quad 0 le {x} < frac{1}{2} \ -1 quad &text{ if } quad frac{1}{2} le {x} < 1 end{cases}$$
where ${x}$ is the fractional part of $x$. Let $nu:= frac{1}{2} ( delta_1 + delta_{-1})$. Then
- For each $h>0$ it holds
$$ ((f_h)_{#}) (mu) = nu $$
$f_h overset{ast}{rightharpoonup} 0$ as $h to 0$
where $((f_h)_{#}) (mu)(B) = mu ( f_h^{-1}(B))$ for any borel set $B$.
I'm not sure they are true; I suspect it should be $h>1$ and $h to + infty$. Indeed:
$ ((f_h)_{#}) (mu) = nu Leftrightarrow mu ( f_h^{-1}((a,b))= nu((a,b))$ for every $-infty< a <b <+ infty$. If both (or none of) $1$ and $-1$ belong to $(a,b)$ the equality is easy. Suppose $1 in (a,b)$ and $-1 notin (a,b)$, and take for example $h=1/2$ then
$$nu((a,b))= frac{1}{2} $$
and
$$ mu (f_{1/2}^{-1}((a,b))) = mu ({ x in [0,1] mid f_{1/2}(x) = 1 }) = mu ( {x in [0,1] mid 0 le { frac{1}{2} x } < 1/2 } ) = 1$$
On the other hand, if $h ge 1$ I think the equality holds. Am I wrong?If $h < 1/2$ and $varphi in L^q(mathbb{R}, mu)$ with $1/p + 1/q =1$ then
$$ int_{mathbb{R}} f_h varphi d mu = int_{ { x in [0,1] mid 0 le { hx } < 1/2 } } varphi d mathcal{L}^1 = int_{[0,1]} varphi d mathcal{L}^1 = int_{[0,1]} varphi d mu $$
and then I have $f_h overset{ast}{rightharpoonup} 1$ as $h to 0$. I think I can prove that, if $h to + infty$, the set on which $f_h = 1$ becomes union of smaller and smaller disjoint intervals which total measure is always 1/2. The same for the set on which $f_h = -1$. But, even if it is true, I don't know how to prove statement 2 (with $h to + infty$).
measure-theory lp-spaces weak-convergence
measure-theory lp-spaces weak-convergence
asked Nov 29 '18 at 19:03
Bremen000
424210
asked Nov 29 '18 at 19:03
Bremen000
424210
asked Nov 29 '18 at 19:03
Bremen000
424210
asked Nov 29 '18 at 19:03
Bremen000
424210
asked Nov 29 '18 at 19:03
Bremen000
424210
424210
Yes for 1 you need $hinmathbb N$
– Federico
Nov 29 '18 at 19:34
Thank you. What do you think about the second part of the "proposition"?
– Bremen000
Nov 29 '18 at 20:36
Yes for 2 you need $htoinfty$ to conclude $f_hrightharpoonup 0$
– Federico
Nov 29 '18 at 20:40
Otherwise you have $f_h(x)tomathrm{sign}(x)$ if $hto0$.
– Federico
Nov 29 '18 at 20:42
I'll try to work out the details for $h to +infty$. By now, I can't see how to prove it. If $h to 0$ don't you think I'll get 1, being the support of the measure $[0,1]$?
– Bremen000
Nov 29 '18 at 20:50
|
show 1 more comment
Yes for 1 you need $hinmathbb N$
– Federico
Nov 29 '18 at 19:34
Thank you. What do you think about the second part of the "proposition"?
– Bremen000
Nov 29 '18 at 20:36
Yes for 2 you need $htoinfty$ to conclude $f_hrightharpoonup 0$
– Federico
Nov 29 '18 at 20:40
Otherwise you have $f_h(x)tomathrm{sign}(x)$ if $hto0$.
– Federico
Nov 29 '18 at 20:42
I'll try to work out the details for $h to +infty$. By now, I can't see how to prove it. If $h to 0$ don't you think I'll get 1, being the support of the measure $[0,1]$?
– Bremen000
Nov 29 '18 at 20:50
Yes for 1 you need $hinmathbb N$
– Federico
Nov 29 '18 at 19:34
Yes for 1 you need $hinmathbb N$
– Federico
Nov 29 '18 at 19:34
Thank you. What do you think about the second part of the "proposition"?
– Bremen000
Nov 29 '18 at 20:36
Thank you. What do you think about the second part of the "proposition"?
– Bremen000
Nov 29 '18 at 20:36
Yes for 2 you need $htoinfty$ to conclude $f_hrightharpoonup 0$
– Federico
Nov 29 '18 at 20:40
Yes for 2 you need $htoinfty$ to conclude $f_hrightharpoonup 0$
– Federico
Nov 29 '18 at 20:40
Otherwise you have $f_h(x)tomathrm{sign}(x)$ if $hto0$.
– Federico
Nov 29 '18 at 20:42
Otherwise you have $f_h(x)tomathrm{sign}(x)$ if $hto0$.
– Federico
Nov 29 '18 at 20:42
I'll try to work out the details for $h to +infty$. By now, I can't see how to prove it. If $h to 0$ don't you think I'll get 1, being the support of the measure $[0,1]$?
– Bremen000
Nov 29 '18 at 20:50
I'll try to work out the details for $h to +infty$. By now, I can't see how to prove it. If $h to 0$ don't you think I'll get 1, being the support of the measure $[0,1]$?
– Bremen000
Nov 29 '18 at 20:50
|
show 1 more comment
0
active
oldest
votes
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%2f3019045%2fpush-forward-of-measures-and-weak-converence%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3019045%2fpush-forward-of-measures-and-weak-converence%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
Yes for 1 you need $hinmathbb N$
– Federico
Nov 29 '18 at 19:34
Thank you. What do you think about the second part of the "proposition"?
– Bremen000
Nov 29 '18 at 20:36
Yes for 2 you need $htoinfty$ to conclude $f_hrightharpoonup 0$
– Federico
Nov 29 '18 at 20:40
Otherwise you have $f_h(x)tomathrm{sign}(x)$ if $hto0$.
– Federico
Nov 29 '18 at 20:42
I'll try to work out the details for $h to +infty$. By now, I can't see how to prove it. If $h to 0$ don't you think I'll get 1, being the support of the measure $[0,1]$?
– Bremen000
Nov 29 '18 at 20:50