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
520213
$endgroup$
|
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
520213
$endgroup$
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
520213
$endgroup$
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
520213
asked Jan 9 at 11:33
Goal123Goal123
520213
edited Jan 9 at 16:37
Goal123
asked Jan 9 at 11:33
Goal123Goal123
520213
asked Jan 9 at 11:33
Goal123Goal123
520213
asked Jan 9 at 11:33
Goal123Goal123
520213
520213
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
4,910211
$endgroup$
$begingroup$
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
Your Answer
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%2f3067346%2fconvergence-in-l2%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
$begingroup$
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
4,910211
$endgroup$
$begingroup$
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
4,910211
$endgroup$
$begingroup$
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
4,910211
$endgroup$
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
4,910211
answered Jan 9 at 23:21
MindlackMindlack
4,910211
answered Jan 9 at 23:21
MindlackMindlack
4,910211
answered Jan 9 at 23:21
MindlackMindlack
4,910211
4,910211
$begingroup$
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
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%2f3067346%2fconvergence-in-l2%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
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