The Convergence of Convolution of $f$ and $g$
$begingroup$
I am dealing with a question requiring me to prove that if $f,gin L_{2}$, then the convolution is defined everywhere, bounded, and continuous. Moreover, it will converge to $0$ as $|x|rightarrowinfty$.
I have nearly proved everything except for the last one. Here is my proof for being defined everywhere, boundedness, and continuity.
Firstly, recall the convolution of $f$ and $g$, $$f*g(x):=int f(x-y)g(y)dy$$ Then as $fin L_{2}$ and $gin L_{2}$, and $1/2+1/2=1$, by Holder, we know that $fgin L_{1}$, and $$|f*g(x)|leqint|f(x-y)g(y)|dyleq|f(x-y)|_{2}|g(y)|_{2}=|f(y)|_{2}|g(y)|_{2}<infty$$
The last equality is by translation invariance and reflection invariance.
Therefore, $|f*g(x)|$ is defined everywhere and bounded.
Now for continuity, let $x,zinmathbb{R}$, then consider $$|f*g(x)-f*g(z)|=Big|int f(x-y)g(y)dy-int f(z-y)g(y)dyBig|$$ $$leqint|g(y)(f(x-y)-f(z-y))|dy$$ $$text{now by Holder and by translation invariance and reflection invariance}$$ $$leq|g|_{2}Big(int|f(x-y)-f(z-y)|^{2}Big)^{1/2}=|g|_{2}Big(int|f(-y)-f((z-x)-y)|^{2}Big)^{1/2}$$ $$=|g|_{2}Big(int|f(y)-f(y-(z-x))|^{2}Big)^{1/2}=|g|_{2}|f_{h}(y)-f(y)|_{2} text{where} h=z-x$$
Therefore, by the translation continuity property, we know that $$|f_{h}(y)-f(y)|_{2}rightarrow 0 text{as} zrightarrow x$$
It follows immediately that $$|f*g(x)-f*g(z)|rightarrow 0 text{as} xrightarrow z$$.
I am now having no idea about how to show the convergence to $0$. It seems that I cannot extract the information of $x$.
Any idea will be really appreciated!
measure-theory convolution holder-inequality
asked Dec 18 '18 at 23:12
JacobsonRadicalJacobsonRadical
502111
$endgroup$
add a comment |
$begingroup$
I am dealing with a question requiring me to prove that if $f,gin L_{2}$, then the convolution is defined everywhere, bounded, and continuous. Moreover, it will converge to $0$ as $|x|rightarrowinfty$.
I have nearly proved everything except for the last one. Here is my proof for being defined everywhere, boundedness, and continuity.
Firstly, recall the convolution of $f$ and $g$, $$f*g(x):=int f(x-y)g(y)dy$$ Then as $fin L_{2}$ and $gin L_{2}$, and $1/2+1/2=1$, by Holder, we know that $fgin L_{1}$, and $$|f*g(x)|leqint|f(x-y)g(y)|dyleq|f(x-y)|_{2}|g(y)|_{2}=|f(y)|_{2}|g(y)|_{2}<infty$$
The last equality is by translation invariance and reflection invariance.
Therefore, $|f*g(x)|$ is defined everywhere and bounded.
Now for continuity, let $x,zinmathbb{R}$, then consider $$|f*g(x)-f*g(z)|=Big|int f(x-y)g(y)dy-int f(z-y)g(y)dyBig|$$ $$leqint|g(y)(f(x-y)-f(z-y))|dy$$ $$text{now by Holder and by translation invariance and reflection invariance}$$ $$leq|g|_{2}Big(int|f(x-y)-f(z-y)|^{2}Big)^{1/2}=|g|_{2}Big(int|f(-y)-f((z-x)-y)|^{2}Big)^{1/2}$$ $$=|g|_{2}Big(int|f(y)-f(y-(z-x))|^{2}Big)^{1/2}=|g|_{2}|f_{h}(y)-f(y)|_{2} text{where} h=z-x$$
Therefore, by the translation continuity property, we know that $$|f_{h}(y)-f(y)|_{2}rightarrow 0 text{as} zrightarrow x$$
It follows immediately that $$|f*g(x)-f*g(z)|rightarrow 0 text{as} xrightarrow z$$.
I am now having no idea about how to show the convergence to $0$. It seems that I cannot extract the information of $x$.
Any idea will be really appreciated!
measure-theory convolution holder-inequality
asked Dec 18 '18 at 23:12
JacobsonRadicalJacobsonRadical
502111
$endgroup$
add a comment |
$begingroup$
I am dealing with a question requiring me to prove that if $f,gin L_{2}$, then the convolution is defined everywhere, bounded, and continuous. Moreover, it will converge to $0$ as $|x|rightarrowinfty$.
I have nearly proved everything except for the last one. Here is my proof for being defined everywhere, boundedness, and continuity.
Firstly, recall the convolution of $f$ and $g$, $$f*g(x):=int f(x-y)g(y)dy$$ Then as $fin L_{2}$ and $gin L_{2}$, and $1/2+1/2=1$, by Holder, we know that $fgin L_{1}$, and $$|f*g(x)|leqint|f(x-y)g(y)|dyleq|f(x-y)|_{2}|g(y)|_{2}=|f(y)|_{2}|g(y)|_{2}<infty$$
The last equality is by translation invariance and reflection invariance.
Therefore, $|f*g(x)|$ is defined everywhere and bounded.
Now for continuity, let $x,zinmathbb{R}$, then consider $$|f*g(x)-f*g(z)|=Big|int f(x-y)g(y)dy-int f(z-y)g(y)dyBig|$$ $$leqint|g(y)(f(x-y)-f(z-y))|dy$$ $$text{now by Holder and by translation invariance and reflection invariance}$$ $$leq|g|_{2}Big(int|f(x-y)-f(z-y)|^{2}Big)^{1/2}=|g|_{2}Big(int|f(-y)-f((z-x)-y)|^{2}Big)^{1/2}$$ $$=|g|_{2}Big(int|f(y)-f(y-(z-x))|^{2}Big)^{1/2}=|g|_{2}|f_{h}(y)-f(y)|_{2} text{where} h=z-x$$
Therefore, by the translation continuity property, we know that $$|f_{h}(y)-f(y)|_{2}rightarrow 0 text{as} zrightarrow x$$
It follows immediately that $$|f*g(x)-f*g(z)|rightarrow 0 text{as} xrightarrow z$$.
I am now having no idea about how to show the convergence to $0$. It seems that I cannot extract the information of $x$.
Any idea will be really appreciated!
measure-theory convolution holder-inequality
asked Dec 18 '18 at 23:12
JacobsonRadicalJacobsonRadical
502111
$endgroup$
I am dealing with a question requiring me to prove that if $f,gin L_{2}$, then the convolution is defined everywhere, bounded, and continuous. Moreover, it will converge to $0$ as $|x|rightarrowinfty$.
I have nearly proved everything except for the last one. Here is my proof for being defined everywhere, boundedness, and continuity.
Firstly, recall the convolution of $f$ and $g$, $$f*g(x):=int f(x-y)g(y)dy$$ Then as $fin L_{2}$ and $gin L_{2}$, and $1/2+1/2=1$, by Holder, we know that $fgin L_{1}$, and $$|f*g(x)|leqint|f(x-y)g(y)|dyleq|f(x-y)|_{2}|g(y)|_{2}=|f(y)|_{2}|g(y)|_{2}<infty$$
The last equality is by translation invariance and reflection invariance.
Therefore, $|f*g(x)|$ is defined everywhere and bounded.
Now for continuity, let $x,zinmathbb{R}$, then consider $$|f*g(x)-f*g(z)|=Big|int f(x-y)g(y)dy-int f(z-y)g(y)dyBig|$$ $$leqint|g(y)(f(x-y)-f(z-y))|dy$$ $$text{now by Holder and by translation invariance and reflection invariance}$$ $$leq|g|_{2}Big(int|f(x-y)-f(z-y)|^{2}Big)^{1/2}=|g|_{2}Big(int|f(-y)-f((z-x)-y)|^{2}Big)^{1/2}$$ $$=|g|_{2}Big(int|f(y)-f(y-(z-x))|^{2}Big)^{1/2}=|g|_{2}|f_{h}(y)-f(y)|_{2} text{where} h=z-x$$
Therefore, by the translation continuity property, we know that $$|f_{h}(y)-f(y)|_{2}rightarrow 0 text{as} zrightarrow x$$
It follows immediately that $$|f*g(x)-f*g(z)|rightarrow 0 text{as} xrightarrow z$$.
I am now having no idea about how to show the convergence to $0$. It seems that I cannot extract the information of $x$.
Any idea will be really appreciated!
measure-theory convolution holder-inequality
measure-theory convolution holder-inequality
asked Dec 18 '18 at 23:12
JacobsonRadicalJacobsonRadical
502111
asked Dec 18 '18 at 23:12
JacobsonRadicalJacobsonRadical
502111
asked Dec 18 '18 at 23:12
JacobsonRadicalJacobsonRadical
502111
asked Dec 18 '18 at 23:12
JacobsonRadicalJacobsonRadical
502111
asked Dec 18 '18 at 23:12
JacobsonRadicalJacobsonRadical
502111
502111
add a comment |
add a comment |
1 Answer
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There exists a continuous function $h$ with compact support such that $|f-h|_2 <epsilon$. We have $|(f*g)(x)| leq |((f-h)*g)(x)|+|(h*g)(x)|$. The first term is $leq epsilon |g|_2$. Now consider $int h(x-y)g(y)dy$. Choose a continuous function $phi$ with compact suport such that $|g-phi|_2 <epsilon /M$ where $M=|h|_2$. Then $|int h(x-y)g(y)dy|<epsilon +|int h(x-y)phi(y)dy|$. Can you now show that $int h(x-y)phi(y)dy to 0$ as $ xto infty$ and complete the proof? [Hint: you only have to integrate over a bounded interval so $|x-y|$ can be made large by making $x$ large].
answered Dec 18 '18 at 23:21
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
$endgroup$
$begingroup$
Hi! All the parts before the last sentence make great sense to me. For the last part, the point is that when I make |x-y| large, what will I get? Currently, I can only use uniform continuity, so I cannot connect this with $xrightarrowinfty$
$endgroup$
– JacobsonRadical
Dec 18 '18 at 23:52
$begingroup$
@JacobsonRadical $int h(x-y)phi (y) dy=0$ whenever $|x|$ is sufficiently large because the integrand becomes $0$ for all $y$!.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 23:53
$begingroup$
Oh! I got it, I was thinking too much. Thank you!
$endgroup$
– JacobsonRadical
Dec 19 '18 at 0:36
add a comment |
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$begingroup$
There exists a continuous function $h$ with compact support such that $|f-h|_2 <epsilon$. We have $|(f*g)(x)| leq |((f-h)*g)(x)|+|(h*g)(x)|$. The first term is $leq epsilon |g|_2$. Now consider $int h(x-y)g(y)dy$. Choose a continuous function $phi$ with compact suport such that $|g-phi|_2 <epsilon /M$ where $M=|h|_2$. Then $|int h(x-y)g(y)dy|<epsilon +|int h(x-y)phi(y)dy|$. Can you now show that $int h(x-y)phi(y)dy to 0$ as $ xto infty$ and complete the proof? [Hint: you only have to integrate over a bounded interval so $|x-y|$ can be made large by making $x$ large].
answered Dec 18 '18 at 23:21
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
$endgroup$
$begingroup$
Hi! All the parts before the last sentence make great sense to me. For the last part, the point is that when I make |x-y| large, what will I get? Currently, I can only use uniform continuity, so I cannot connect this with $xrightarrowinfty$
$endgroup$
– JacobsonRadical
Dec 18 '18 at 23:52
$begingroup$
@JacobsonRadical $int h(x-y)phi (y) dy=0$ whenever $|x|$ is sufficiently large because the integrand becomes $0$ for all $y$!.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 23:53
$begingroup$
Oh! I got it, I was thinking too much. Thank you!
$endgroup$
– JacobsonRadical
Dec 19 '18 at 0:36
add a comment |
$begingroup$
There exists a continuous function $h$ with compact support such that $|f-h|_2 <epsilon$. We have $|(f*g)(x)| leq |((f-h)*g)(x)|+|(h*g)(x)|$. The first term is $leq epsilon |g|_2$. Now consider $int h(x-y)g(y)dy$. Choose a continuous function $phi$ with compact suport such that $|g-phi|_2 <epsilon /M$ where $M=|h|_2$. Then $|int h(x-y)g(y)dy|<epsilon +|int h(x-y)phi(y)dy|$. Can you now show that $int h(x-y)phi(y)dy to 0$ as $ xto infty$ and complete the proof? [Hint: you only have to integrate over a bounded interval so $|x-y|$ can be made large by making $x$ large].
answered Dec 18 '18 at 23:21
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
$endgroup$
$begingroup$
Hi! All the parts before the last sentence make great sense to me. For the last part, the point is that when I make |x-y| large, what will I get? Currently, I can only use uniform continuity, so I cannot connect this with $xrightarrowinfty$
$endgroup$
– JacobsonRadical
Dec 18 '18 at 23:52
$begingroup$
@JacobsonRadical $int h(x-y)phi (y) dy=0$ whenever $|x|$ is sufficiently large because the integrand becomes $0$ for all $y$!.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 23:53
$begingroup$
Oh! I got it, I was thinking too much. Thank you!
$endgroup$
– JacobsonRadical
Dec 19 '18 at 0:36
add a comment |
$begingroup$
There exists a continuous function $h$ with compact support such that $|f-h|_2 <epsilon$. We have $|(f*g)(x)| leq |((f-h)*g)(x)|+|(h*g)(x)|$. The first term is $leq epsilon |g|_2$. Now consider $int h(x-y)g(y)dy$. Choose a continuous function $phi$ with compact suport such that $|g-phi|_2 <epsilon /M$ where $M=|h|_2$. Then $|int h(x-y)g(y)dy|<epsilon +|int h(x-y)phi(y)dy|$. Can you now show that $int h(x-y)phi(y)dy to 0$ as $ xto infty$ and complete the proof? [Hint: you only have to integrate over a bounded interval so $|x-y|$ can be made large by making $x$ large].
answered Dec 18 '18 at 23:21
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
$endgroup$
There exists a continuous function $h$ with compact support such that $|f-h|_2 <epsilon$. We have $|(f*g)(x)| leq |((f-h)*g)(x)|+|(h*g)(x)|$. The first term is $leq epsilon |g|_2$. Now consider $int h(x-y)g(y)dy$. Choose a continuous function $phi$ with compact suport such that $|g-phi|_2 <epsilon /M$ where $M=|h|_2$. Then $|int h(x-y)g(y)dy|<epsilon +|int h(x-y)phi(y)dy|$. Can you now show that $int h(x-y)phi(y)dy to 0$ as $ xto infty$ and complete the proof? [Hint: you only have to integrate over a bounded interval so $|x-y|$ can be made large by making $x$ large].
answered Dec 18 '18 at 23:21
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
answered Dec 18 '18 at 23:21
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
answered Dec 18 '18 at 23:21
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
answered Dec 18 '18 at 23:21
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
60.3k42161
$begingroup$
Hi! All the parts before the last sentence make great sense to me. For the last part, the point is that when I make |x-y| large, what will I get? Currently, I can only use uniform continuity, so I cannot connect this with $xrightarrowinfty$
$endgroup$
– JacobsonRadical
Dec 18 '18 at 23:52
$begingroup$
@JacobsonRadical $int h(x-y)phi (y) dy=0$ whenever $|x|$ is sufficiently large because the integrand becomes $0$ for all $y$!.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 23:53
$begingroup$
Oh! I got it, I was thinking too much. Thank you!
$endgroup$
– JacobsonRadical
Dec 19 '18 at 0:36
add a comment |
$begingroup$
Hi! All the parts before the last sentence make great sense to me. For the last part, the point is that when I make |x-y| large, what will I get? Currently, I can only use uniform continuity, so I cannot connect this with $xrightarrowinfty$
$endgroup$
– JacobsonRadical
Dec 18 '18 at 23:52
$begingroup$
@JacobsonRadical $int h(x-y)phi (y) dy=0$ whenever $|x|$ is sufficiently large because the integrand becomes $0$ for all $y$!.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 23:53
$begingroup$
Oh! I got it, I was thinking too much. Thank you!
$endgroup$
– JacobsonRadical
Dec 19 '18 at 0:36
$begingroup$
Hi! All the parts before the last sentence make great sense to me. For the last part, the point is that when I make |x-y| large, what will I get? Currently, I can only use uniform continuity, so I cannot connect this with $xrightarrowinfty$
$endgroup$
– JacobsonRadical
Dec 18 '18 at 23:52
$begingroup$
Hi! All the parts before the last sentence make great sense to me. For the last part, the point is that when I make |x-y| large, what will I get? Currently, I can only use uniform continuity, so I cannot connect this with $xrightarrowinfty$
$endgroup$
– JacobsonRadical
Dec 18 '18 at 23:52
$begingroup$
@JacobsonRadical $int h(x-y)phi (y) dy=0$ whenever $|x|$ is sufficiently large because the integrand becomes $0$ for all $y$!.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 23:53
$begingroup$
@JacobsonRadical $int h(x-y)phi (y) dy=0$ whenever $|x|$ is sufficiently large because the integrand becomes $0$ for all $y$!.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 23:53
$begingroup$
Oh! I got it, I was thinking too much. Thank you!
$endgroup$
– JacobsonRadical
Dec 19 '18 at 0:36
$begingroup$
Oh! I got it, I was thinking too much. Thank you!
$endgroup$
– JacobsonRadical
Dec 19 '18 at 0:36
add a comment |
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