Convergence in L^2












2












$begingroup$



Define a function $f_varepsilon: mathbb{R} to mathbb{R}$ as
$$begin{equation}
f_varepsilon(x) =
begin{cases}
-1 & text{if}; ,x<-varepsilon \
sin left(frac{pi x}{2 varepsilon}right) & text{if};, |x| le varepsilon\
1 & text{if};, x>varepsilon
end{cases}
end{equation} $$
Then $f_varepsilon(x)$ converges pointwise to $mbox{sgn}(x)$ as $varepsilon to 0$. And it can be easily proven that $ int_{mathbb{R}}|f_varepsilon(x)- mbox{sgn}(x)|^2,dx= Cvarepsilon $ .




My question is given $u in L^1(mathbb{R}) cap L^2(mathbb{R})$ and $varphi in C^infty_c(mathbb{R})$ can we prove $$int_{mathbb{R}}|f_varepsilon(u(x))- mbox{sgn}(u(x))|^2varphi(x),dx le Cvarepsilon^r$$ for some r>0.
I actually tried to mimic the proof but the main issue over here is we cannot apply the change of variable formula. And I'm always ending up in an upper bound like $varepsilon^r+c$, which is not desired.






analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What would $delta$ be?
    $endgroup$
    – Mindlack
    Jan 9 at 11:40










  • $begingroup$
    If the fact you can't apply change of variable formula brothers you, I'd suggest try to prove it first for u smooth and then just approach your general function by smooth functions on the support of $phi$.
    $endgroup$
    – Keen
    Jan 9 at 13:42












  • $begingroup$
    @Mindlack I'm sorry that would be epsilon.
    $endgroup$
    – Goal123
    Jan 9 at 16:37










  • $begingroup$
    What then if $u=0$?
    $endgroup$
    – Mindlack
    Jan 9 at 16:54






  • 1




    $begingroup$
    @Keen $u$ is required to be one-one to apply change of variable, which need not be true here.
    $endgroup$
    – Goal123
    Jan 9 at 17:59
















2












$begingroup$



Define a function $f_varepsilon: mathbb{R} to mathbb{R}$ as
$$begin{equation}
f_varepsilon(x) =
begin{cases}
-1 & text{if}; ,x<-varepsilon \
sin left(frac{pi x}{2 varepsilon}right) & text{if};, |x| le varepsilon\
1 & text{if};, x>varepsilon
end{cases}
end{equation} $$
Then $f_varepsilon(x)$ converges pointwise to $mbox{sgn}(x)$ as $varepsilon to 0$. And it can be easily proven that $ int_{mathbb{R}}|f_varepsilon(x)- mbox{sgn}(x)|^2,dx= Cvarepsilon $ .




My question is given $u in L^1(mathbb{R}) cap L^2(mathbb{R})$ and $varphi in C^infty_c(mathbb{R})$ can we prove $$int_{mathbb{R}}|f_varepsilon(u(x))- mbox{sgn}(u(x))|^2varphi(x),dx le Cvarepsilon^r$$ for some r>0.
I actually tried to mimic the proof but the main issue over here is we cannot apply the change of variable formula. And I'm always ending up in an upper bound like $varepsilon^r+c$, which is not desired.






analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What would $delta$ be?
    $endgroup$
    – Mindlack
    Jan 9 at 11:40










  • $begingroup$
    If the fact you can't apply change of variable formula brothers you, I'd suggest try to prove it first for u smooth and then just approach your general function by smooth functions on the support of $phi$.
    $endgroup$
    – Keen
    Jan 9 at 13:42












  • $begingroup$
    @Mindlack I'm sorry that would be epsilon.
    $endgroup$
    – Goal123
    Jan 9 at 16:37










  • $begingroup$
    What then if $u=0$?
    $endgroup$
    – Mindlack
    Jan 9 at 16:54






  • 1




    $begingroup$
    @Keen $u$ is required to be one-one to apply change of variable, which need not be true here.
    $endgroup$
    – Goal123
    Jan 9 at 17:59














2












2








2





$begingroup$



Define a function $f_varepsilon: mathbb{R} to mathbb{R}$ as
$$begin{equation}
f_varepsilon(x) =
begin{cases}
-1 & text{if}; ,x<-varepsilon \
sin left(frac{pi x}{2 varepsilon}right) & text{if};, |x| le varepsilon\
1 & text{if};, x>varepsilon
end{cases}
end{equation} $$
Then $f_varepsilon(x)$ converges pointwise to $mbox{sgn}(x)$ as $varepsilon to 0$. And it can be easily proven that $ int_{mathbb{R}}|f_varepsilon(x)- mbox{sgn}(x)|^2,dx= Cvarepsilon $ .




My question is given $u in L^1(mathbb{R}) cap L^2(mathbb{R})$ and $varphi in C^infty_c(mathbb{R})$ can we prove $$int_{mathbb{R}}|f_varepsilon(u(x))- mbox{sgn}(u(x))|^2varphi(x),dx le Cvarepsilon^r$$ for some r>0.
I actually tried to mimic the proof but the main issue over here is we cannot apply the change of variable formula. And I'm always ending up in an upper bound like $varepsilon^r+c$, which is not desired.






analysis







share|cite|improve this question











$endgroup$





Define a function $f_varepsilon: mathbb{R} to mathbb{R}$ as
$$begin{equation}
f_varepsilon(x) =
begin{cases}
-1 & text{if}; ,x<-varepsilon \
sin left(frac{pi x}{2 varepsilon}right) & text{if};, |x| le varepsilon\
1 & text{if};, x>varepsilon
end{cases}
end{equation} $$
Then $f_varepsilon(x)$ converges pointwise to $mbox{sgn}(x)$ as $varepsilon to 0$. And it can be easily proven that $ int_{mathbb{R}}|f_varepsilon(x)- mbox{sgn}(x)|^2,dx= Cvarepsilon $ .




My question is given $u in L^1(mathbb{R}) cap L^2(mathbb{R})$ and $varphi in C^infty_c(mathbb{R})$ can we prove $$int_{mathbb{R}}|f_varepsilon(u(x))- mbox{sgn}(u(x))|^2varphi(x),dx le Cvarepsilon^r$$ for some r>0.
I actually tried to mimic the proof but the main issue over here is we cannot apply the change of variable formula. And I'm always ending up in an upper bound like $varepsilon^r+c$, which is not desired.






analysis




analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 16:37







Goal123

















asked Jan 9 at 11:33









Goal123Goal123

507212




507212








  • 1




    $begingroup$
    What would $delta$ be?
    $endgroup$
    – Mindlack
    Jan 9 at 11:40










  • $begingroup$
    If the fact you can't apply change of variable formula brothers you, I'd suggest try to prove it first for u smooth and then just approach your general function by smooth functions on the support of $phi$.
    $endgroup$
    – Keen
    Jan 9 at 13:42












  • $begingroup$
    @Mindlack I'm sorry that would be epsilon.
    $endgroup$
    – Goal123
    Jan 9 at 16:37










  • $begingroup$
    What then if $u=0$?
    $endgroup$
    – Mindlack
    Jan 9 at 16:54






  • 1




    $begingroup$
    @Keen $u$ is required to be one-one to apply change of variable, which need not be true here.
    $endgroup$
    – Goal123
    Jan 9 at 17:59














  • 1




    $begingroup$
    What would $delta$ be?
    $endgroup$
    – Mindlack
    Jan 9 at 11:40










  • $begingroup$
    If the fact you can't apply change of variable formula brothers you, I'd suggest try to prove it first for u smooth and then just approach your general function by smooth functions on the support of $phi$.
    $endgroup$
    – Keen
    Jan 9 at 13:42












  • $begingroup$
    @Mindlack I'm sorry that would be epsilon.
    $endgroup$
    – Goal123
    Jan 9 at 16:37










  • $begingroup$
    What then if $u=0$?
    $endgroup$
    – Mindlack
    Jan 9 at 16:54






  • 1




    $begingroup$
    @Keen $u$ is required to be one-one to apply change of variable, which need not be true here.
    $endgroup$
    – Goal123
    Jan 9 at 17:59








1




1




$begingroup$
What would $delta$ be?
$endgroup$
– Mindlack
Jan 9 at 11:40




$begingroup$
What would $delta$ be?
$endgroup$
– Mindlack
Jan 9 at 11:40












$begingroup$
If the fact you can't apply change of variable formula brothers you, I'd suggest try to prove it first for u smooth and then just approach your general function by smooth functions on the support of $phi$.
$endgroup$
– Keen
Jan 9 at 13:42






$begingroup$
If the fact you can't apply change of variable formula brothers you, I'd suggest try to prove it first for u smooth and then just approach your general function by smooth functions on the support of $phi$.
$endgroup$
– Keen
Jan 9 at 13:42














$begingroup$
@Mindlack I'm sorry that would be epsilon.
$endgroup$
– Goal123
Jan 9 at 16:37




$begingroup$
@Mindlack I'm sorry that would be epsilon.
$endgroup$
– Goal123
Jan 9 at 16:37












$begingroup$
What then if $u=0$?
$endgroup$
– Mindlack
Jan 9 at 16:54




$begingroup$
What then if $u=0$?
$endgroup$
– Mindlack
Jan 9 at 16:54




1




1




$begingroup$
@Keen $u$ is required to be one-one to apply change of variable, which need not be true here.
$endgroup$
– Goal123
Jan 9 at 17:59




$begingroup$
@Keen $u$ is required to be one-one to apply change of variable, which need not be true here.
$endgroup$
– Goal123
Jan 9 at 17:59










1 Answer
1






active

oldest

votes


















2












$begingroup$

Take, for every $0 < t < 1$, $u(t)=e^{-1/t}$ and $u=0$ everywhere else. Take $varphi$ to be a smooth approximation of $1_{[0,1]}$, that is not lower than this function. On $[0,1]$, ${0 < u leq r}$ has measure $|ln{r}|^{-1}$.



Thus your integral is not lower than $int_{0<|u| leq epsilon/2}{|1-sin(pi/4)|^2} geq frac{c}{|ln(epsilon/2)|}$ where $c >0$ is a numeric constant. So the answer to your question is negative.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why is the measure $|log r|^{-1}$?
    $endgroup$
    – Goal123
    Jan 10 at 9:36










  • $begingroup$
    $0 <u(t) leq r$ iff $0 < t < 1$ and $e^{-1/t} leq r$ and you can figure out the rest.
    $endgroup$
    – Mindlack
    Jan 10 at 9:43










  • $begingroup$
    But $e^{-1/t} leq r$ gives $ t ge -1/log r$ if $r >1$ and $t< -1/ log r$ if $r<1$.
    $endgroup$
    – Goal123
    Jan 10 at 9:55






  • 1




    $begingroup$
    But note that $r < 1$ always because $0 leq u leq e^{-1}$.
    $endgroup$
    – Mindlack
    Jan 10 at 10:02






  • 1




    $begingroup$
    When $r>0$ goes to $0$, then for any $a<0$, $|ln(r)| << r^a$, so this integral is $>>epsilon^{-a}$ for any $a <0$.
    $endgroup$
    – Mindlack
    Jan 10 at 10:24













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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Take, for every $0 < t < 1$, $u(t)=e^{-1/t}$ and $u=0$ everywhere else. Take $varphi$ to be a smooth approximation of $1_{[0,1]}$, that is not lower than this function. On $[0,1]$, ${0 < u leq r}$ has measure $|ln{r}|^{-1}$.



Thus your integral is not lower than $int_{0<|u| leq epsilon/2}{|1-sin(pi/4)|^2} geq frac{c}{|ln(epsilon/2)|}$ where $c >0$ is a numeric constant. So the answer to your question is negative.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why is the measure $|log r|^{-1}$?
    $endgroup$
    – Goal123
    Jan 10 at 9:36










  • $begingroup$
    $0 <u(t) leq r$ iff $0 < t < 1$ and $e^{-1/t} leq r$ and you can figure out the rest.
    $endgroup$
    – Mindlack
    Jan 10 at 9:43










  • $begingroup$
    But $e^{-1/t} leq r$ gives $ t ge -1/log r$ if $r >1$ and $t< -1/ log r$ if $r<1$.
    $endgroup$
    – Goal123
    Jan 10 at 9:55






  • 1




    $begingroup$
    But note that $r < 1$ always because $0 leq u leq e^{-1}$.
    $endgroup$
    – Mindlack
    Jan 10 at 10:02






  • 1




    $begingroup$
    When $r>0$ goes to $0$, then for any $a<0$, $|ln(r)| << r^a$, so this integral is $>>epsilon^{-a}$ for any $a <0$.
    $endgroup$
    – Mindlack
    Jan 10 at 10:24


















2












$begingroup$

Take, for every $0 < t < 1$, $u(t)=e^{-1/t}$ and $u=0$ everywhere else. Take $varphi$ to be a smooth approximation of $1_{[0,1]}$, that is not lower than this function. On $[0,1]$, ${0 < u leq r}$ has measure $|ln{r}|^{-1}$.



Thus your integral is not lower than $int_{0<|u| leq epsilon/2}{|1-sin(pi/4)|^2} geq frac{c}{|ln(epsilon/2)|}$ where $c >0$ is a numeric constant. So the answer to your question is negative.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why is the measure $|log r|^{-1}$?
    $endgroup$
    – Goal123
    Jan 10 at 9:36










  • $begingroup$
    $0 <u(t) leq r$ iff $0 < t < 1$ and $e^{-1/t} leq r$ and you can figure out the rest.
    $endgroup$
    – Mindlack
    Jan 10 at 9:43










  • $begingroup$
    But $e^{-1/t} leq r$ gives $ t ge -1/log r$ if $r >1$ and $t< -1/ log r$ if $r<1$.
    $endgroup$
    – Goal123
    Jan 10 at 9:55






  • 1




    $begingroup$
    But note that $r < 1$ always because $0 leq u leq e^{-1}$.
    $endgroup$
    – Mindlack
    Jan 10 at 10:02






  • 1




    $begingroup$
    When $r>0$ goes to $0$, then for any $a<0$, $|ln(r)| << r^a$, so this integral is $>>epsilon^{-a}$ for any $a <0$.
    $endgroup$
    – Mindlack
    Jan 10 at 10:24
















2












2








2





$begingroup$

Take, for every $0 < t < 1$, $u(t)=e^{-1/t}$ and $u=0$ everywhere else. Take $varphi$ to be a smooth approximation of $1_{[0,1]}$, that is not lower than this function. On $[0,1]$, ${0 < u leq r}$ has measure $|ln{r}|^{-1}$.



Thus your integral is not lower than $int_{0<|u| leq epsilon/2}{|1-sin(pi/4)|^2} geq frac{c}{|ln(epsilon/2)|}$ where $c >0$ is a numeric constant. So the answer to your question is negative.






share|cite|improve this answer









$endgroup$



Take, for every $0 < t < 1$, $u(t)=e^{-1/t}$ and $u=0$ everywhere else. Take $varphi$ to be a smooth approximation of $1_{[0,1]}$, that is not lower than this function. On $[0,1]$, ${0 < u leq r}$ has measure $|ln{r}|^{-1}$.



Thus your integral is not lower than $int_{0<|u| leq epsilon/2}{|1-sin(pi/4)|^2} geq frac{c}{|ln(epsilon/2)|}$ where $c >0$ is a numeric constant. So the answer to your question is negative.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 9 at 23:21









MindlackMindlack

4,750210




4,750210












  • $begingroup$
    Why is the measure $|log r|^{-1}$?
    $endgroup$
    – Goal123
    Jan 10 at 9:36










  • $begingroup$
    $0 <u(t) leq r$ iff $0 < t < 1$ and $e^{-1/t} leq r$ and you can figure out the rest.
    $endgroup$
    – Mindlack
    Jan 10 at 9:43










  • $begingroup$
    But $e^{-1/t} leq r$ gives $ t ge -1/log r$ if $r >1$ and $t< -1/ log r$ if $r<1$.
    $endgroup$
    – Goal123
    Jan 10 at 9:55






  • 1




    $begingroup$
    But note that $r < 1$ always because $0 leq u leq e^{-1}$.
    $endgroup$
    – Mindlack
    Jan 10 at 10:02






  • 1




    $begingroup$
    When $r>0$ goes to $0$, then for any $a<0$, $|ln(r)| << r^a$, so this integral is $>>epsilon^{-a}$ for any $a <0$.
    $endgroup$
    – Mindlack
    Jan 10 at 10:24




















  • $begingroup$
    Why is the measure $|log r|^{-1}$?
    $endgroup$
    – Goal123
    Jan 10 at 9:36










  • $begingroup$
    $0 <u(t) leq r$ iff $0 < t < 1$ and $e^{-1/t} leq r$ and you can figure out the rest.
    $endgroup$
    – Mindlack
    Jan 10 at 9:43










  • $begingroup$
    But $e^{-1/t} leq r$ gives $ t ge -1/log r$ if $r >1$ and $t< -1/ log r$ if $r<1$.
    $endgroup$
    – Goal123
    Jan 10 at 9:55






  • 1




    $begingroup$
    But note that $r < 1$ always because $0 leq u leq e^{-1}$.
    $endgroup$
    – Mindlack
    Jan 10 at 10:02






  • 1




    $begingroup$
    When $r>0$ goes to $0$, then for any $a<0$, $|ln(r)| << r^a$, so this integral is $>>epsilon^{-a}$ for any $a <0$.
    $endgroup$
    – Mindlack
    Jan 10 at 10:24


















$begingroup$
Why is the measure $|log r|^{-1}$?
$endgroup$
– Goal123
Jan 10 at 9:36




$begingroup$
Why is the measure $|log r|^{-1}$?
$endgroup$
– Goal123
Jan 10 at 9:36












$begingroup$
$0 <u(t) leq r$ iff $0 < t < 1$ and $e^{-1/t} leq r$ and you can figure out the rest.
$endgroup$
– Mindlack
Jan 10 at 9:43




$begingroup$
$0 <u(t) leq r$ iff $0 < t < 1$ and $e^{-1/t} leq r$ and you can figure out the rest.
$endgroup$
– Mindlack
Jan 10 at 9:43












$begingroup$
But $e^{-1/t} leq r$ gives $ t ge -1/log r$ if $r >1$ and $t< -1/ log r$ if $r<1$.
$endgroup$
– Goal123
Jan 10 at 9:55




$begingroup$
But $e^{-1/t} leq r$ gives $ t ge -1/log r$ if $r >1$ and $t< -1/ log r$ if $r<1$.
$endgroup$
– Goal123
Jan 10 at 9:55




1




1




$begingroup$
But note that $r < 1$ always because $0 leq u leq e^{-1}$.
$endgroup$
– Mindlack
Jan 10 at 10:02




$begingroup$
But note that $r < 1$ always because $0 leq u leq e^{-1}$.
$endgroup$
– Mindlack
Jan 10 at 10:02




1




1




$begingroup$
When $r>0$ goes to $0$, then for any $a<0$, $|ln(r)| << r^a$, so this integral is $>>epsilon^{-a}$ for any $a <0$.
$endgroup$
– Mindlack
Jan 10 at 10:24






$begingroup$
When $r>0$ goes to $0$, then for any $a<0$, $|ln(r)| << r^a$, so this integral is $>>epsilon^{-a}$ for any $a <0$.
$endgroup$
– Mindlack
Jan 10 at 10:24




















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張江高科駅

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張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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