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 at 19:03
Bremen000
424110
|
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 at 19:03
Bremen000
424110
Yes for 1 you need $hinmathbb N$
– Federico
Nov 29 at 19:34
Thank you. What do you think about the second part of the "proposition"?
– Bremen000
Nov 29 at 20:36
Yes for 2 you need $htoinfty$ to conclude $f_hrightharpoonup 0$
– Federico
Nov 29 at 20:40
Otherwise you have $f_h(x)tomathrm{sign}(x)$ if $hto0$.
– Federico
Nov 29 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 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 at 19:03
Bremen000
424110
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 at 19:03
Bremen000
424110
asked Nov 29 at 19:03
Bremen000
424110
asked Nov 29 at 19:03
Bremen000
424110
asked Nov 29 at 19:03
Bremen000
424110
asked Nov 29 at 19:03
Bremen000
424110
424110
Yes for 1 you need $hinmathbb N$
– Federico
Nov 29 at 19:34
Thank you. What do you think about the second part of the "proposition"?
– Bremen000
Nov 29 at 20:36
Yes for 2 you need $htoinfty$ to conclude $f_hrightharpoonup 0$
– Federico
Nov 29 at 20:40
Otherwise you have $f_h(x)tomathrm{sign}(x)$ if $hto0$.
– Federico
Nov 29 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 at 20:50
|
show 1 more comment
Yes for 1 you need $hinmathbb N$
– Federico
Nov 29 at 19:34
Thank you. What do you think about the second part of the "proposition"?
– Bremen000
Nov 29 at 20:36
Yes for 2 you need $htoinfty$ to conclude $f_hrightharpoonup 0$
– Federico
Nov 29 at 20:40
Otherwise you have $f_h(x)tomathrm{sign}(x)$ if $hto0$.
– Federico
Nov 29 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 at 20:50
Yes for 1 you need $hinmathbb N$
– Federico
Nov 29 at 19:34
Yes for 1 you need $hinmathbb N$
– Federico
Nov 29 at 19:34
Thank you. What do you think about the second part of the "proposition"?
– Bremen000
Nov 29 at 20:36
Thank you. What do you think about the second part of the "proposition"?
– Bremen000
Nov 29 at 20:36
Yes for 2 you need $htoinfty$ to conclude $f_hrightharpoonup 0$
– Federico
Nov 29 at 20:40
Yes for 2 you need $htoinfty$ to conclude $f_hrightharpoonup 0$
– Federico
Nov 29 at 20:40
Otherwise you have $f_h(x)tomathrm{sign}(x)$ if $hto0$.
– Federico
Nov 29 at 20:42
Otherwise you have $f_h(x)tomathrm{sign}(x)$ if $hto0$.
– Federico
Nov 29 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 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 at 20:50
|
show 1 more comment
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Yes for 1 you need $hinmathbb N$
– Federico
Nov 29 at 19:34
Thank you. What do you think about the second part of the "proposition"?
– Bremen000
Nov 29 at 20:36
Yes for 2 you need $htoinfty$ to conclude $f_hrightharpoonup 0$
– Federico
Nov 29 at 20:40
Otherwise you have $f_h(x)tomathrm{sign}(x)$ if $hto0$.
– Federico
Nov 29 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 at 20:50