Convergence in L^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
asked Jan 9 at 11:33
Goal123Goal123
507212
$endgroup$
|
show 1 more comment
$begingroup$
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
asked Jan 9 at 11:33
Goal123Goal123
507212
$endgroup$
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
|
show 1 more comment
$begingroup$
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
asked Jan 9 at 11:33
Goal123Goal123
507212
$endgroup$
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
asked Jan 9 at 11:33
Goal123Goal123
507212
asked Jan 9 at 11:33
Goal123Goal123
507212
edited Jan 9 at 16:37
Goal123
asked Jan 9 at 11:33
Goal123Goal123
507212
asked Jan 9 at 11:33
Goal123Goal123
507212
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
|
show 1 more comment
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
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 9 at 23:21
MindlackMindlack
4,750210
$endgroup$
$begingroup$
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
|
show 1 more comment
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Jan 9 at 23:21
MindlackMindlack
4,750210
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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.
answered Jan 9 at 23:21
MindlackMindlack
4,750210
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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.
answered Jan 9 at 23:21
MindlackMindlack
4,750210
$endgroup$
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.
answered Jan 9 at 23:21
MindlackMindlack
4,750210
answered Jan 9 at 23:21
MindlackMindlack
4,750210
answered Jan 9 at 23:21
MindlackMindlack
4,750210
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
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$begingroup$
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