Convolution is well defined in $L^1$
$begingroup$
I know that this question has been "answered" before, but I am struggling to understand the solutions previously given.
Let $f$ and $g$ be in $L^1(R, L, m)$
a) Show that $f*g$ is well defined for a.e. $x in R$. That is show $f(x-y)g(y)$ is integrable for all x.
b) Show that $ ||f*g||_1 le||f||_1||g||_1$
I have the following so far:
For part a)
$|int f(x-y)g(y)dy|le int |f(x-y)||g(y)|dy le ||f||_infty ||g||_1$ by Holder's Inequality?
$||f*g||_1 = intint f(x-y)g(y)dydx = intint f(x-y)g(y)dxdy$ by Fubini Tonelli
= $int g(y)int f(x-y)dxdy = int g(y)dy int f(x)dx = ||f||_1||g||_1 $ by translation invariance.
First in part b, why is it an inequality instead of an equality? Is this a typo?
I have seen arguments that the proof of b implies a too, or simply saying "apply Tonelli's theorem" to prove part a, but have yet to find anything concrete for an answer.
measure-theory convolution
asked Jan 7 at 19:47
Math LadyMath Lady
1196
$endgroup$
add a comment |
$begingroup$
I know that this question has been "answered" before, but I am struggling to understand the solutions previously given.
Let $f$ and $g$ be in $L^1(R, L, m)$
a) Show that $f*g$ is well defined for a.e. $x in R$. That is show $f(x-y)g(y)$ is integrable for all x.
b) Show that $ ||f*g||_1 le||f||_1||g||_1$
I have the following so far:
For part a)
$|int f(x-y)g(y)dy|le int |f(x-y)||g(y)|dy le ||f||_infty ||g||_1$ by Holder's Inequality?
$||f*g||_1 = intint f(x-y)g(y)dydx = intint f(x-y)g(y)dxdy$ by Fubini Tonelli
= $int g(y)int f(x-y)dxdy = int g(y)dy int f(x)dx = ||f||_1||g||_1 $ by translation invariance.
First in part b, why is it an inequality instead of an equality? Is this a typo?
I have seen arguments that the proof of b implies a too, or simply saying "apply Tonelli's theorem" to prove part a, but have yet to find anything concrete for an answer.
measure-theory convolution
asked Jan 7 at 19:47
Math LadyMath Lady
1196
$endgroup$
$begingroup$
I assume $L^1(R, L, m)$ is supposed to contain $L$-valued functions on $R$ that are measurable w.r.t. $m$. What can $L$ be for you? (i. e. only $mathbb{R}$ or $mathbb{C}$ or an arbitrary Banach space?)
$endgroup$
– 0x539
Jan 7 at 19:57
1
$begingroup$
For $b$, from the definition of $|.|_1$ you should have lots of $|.|$s. Adding these back will show you why its only an inequality, not an equality
$endgroup$
– Calvin Khor
Jan 7 at 20:06
add a comment |
$begingroup$
I know that this question has been "answered" before, but I am struggling to understand the solutions previously given.
Let $f$ and $g$ be in $L^1(R, L, m)$
a) Show that $f*g$ is well defined for a.e. $x in R$. That is show $f(x-y)g(y)$ is integrable for all x.
b) Show that $ ||f*g||_1 le||f||_1||g||_1$
I have the following so far:
For part a)
$|int f(x-y)g(y)dy|le int |f(x-y)||g(y)|dy le ||f||_infty ||g||_1$ by Holder's Inequality?
$||f*g||_1 = intint f(x-y)g(y)dydx = intint f(x-y)g(y)dxdy$ by Fubini Tonelli
= $int g(y)int f(x-y)dxdy = int g(y)dy int f(x)dx = ||f||_1||g||_1 $ by translation invariance.
First in part b, why is it an inequality instead of an equality? Is this a typo?
I have seen arguments that the proof of b implies a too, or simply saying "apply Tonelli's theorem" to prove part a, but have yet to find anything concrete for an answer.
measure-theory convolution
asked Jan 7 at 19:47
Math LadyMath Lady
1196
$endgroup$
I know that this question has been "answered" before, but I am struggling to understand the solutions previously given.
Let $f$ and $g$ be in $L^1(R, L, m)$
a) Show that $f*g$ is well defined for a.e. $x in R$. That is show $f(x-y)g(y)$ is integrable for all x.
b) Show that $ ||f*g||_1 le||f||_1||g||_1$
I have the following so far:
For part a)
$|int f(x-y)g(y)dy|le int |f(x-y)||g(y)|dy le ||f||_infty ||g||_1$ by Holder's Inequality?
$||f*g||_1 = intint f(x-y)g(y)dydx = intint f(x-y)g(y)dxdy$ by Fubini Tonelli
= $int g(y)int f(x-y)dxdy = int g(y)dy int f(x)dx = ||f||_1||g||_1 $ by translation invariance.
First in part b, why is it an inequality instead of an equality? Is this a typo?
I have seen arguments that the proof of b implies a too, or simply saying "apply Tonelli's theorem" to prove part a, but have yet to find anything concrete for an answer.
measure-theory convolution
measure-theory convolution
asked Jan 7 at 19:47
Math LadyMath Lady
1196
asked Jan 7 at 19:47
Math LadyMath Lady
1196
asked Jan 7 at 19:47
Math LadyMath Lady
1196
asked Jan 7 at 19:47
Math LadyMath Lady
1196
asked Jan 7 at 19:47
Math LadyMath Lady
1196
1196
$begingroup$
I assume $L^1(R, L, m)$ is supposed to contain $L$-valued functions on $R$ that are measurable w.r.t. $m$. What can $L$ be for you? (i. e. only $mathbb{R}$ or $mathbb{C}$ or an arbitrary Banach space?)
$endgroup$
– 0x539
Jan 7 at 19:57
1
$begingroup$
For $b$, from the definition of $|.|_1$ you should have lots of $|.|$s. Adding these back will show you why its only an inequality, not an equality
$endgroup$
– Calvin Khor
Jan 7 at 20:06
add a comment |
$begingroup$
I assume $L^1(R, L, m)$ is supposed to contain $L$-valued functions on $R$ that are measurable w.r.t. $m$. What can $L$ be for you? (i. e. only $mathbb{R}$ or $mathbb{C}$ or an arbitrary Banach space?)
$endgroup$
– 0x539
Jan 7 at 19:57
1
$begingroup$
For $b$, from the definition of $|.|_1$ you should have lots of $|.|$s. Adding these back will show you why its only an inequality, not an equality
$endgroup$
– Calvin Khor
Jan 7 at 20:06
$begingroup$
I assume $L^1(R, L, m)$ is supposed to contain $L$-valued functions on $R$ that are measurable w.r.t. $m$. What can $L$ be for you? (i. e. only $mathbb{R}$ or $mathbb{C}$ or an arbitrary Banach space?)
$endgroup$
– 0x539
Jan 7 at 19:57
$begingroup$
I assume $L^1(R, L, m)$ is supposed to contain $L$-valued functions on $R$ that are measurable w.r.t. $m$. What can $L$ be for you? (i. e. only $mathbb{R}$ or $mathbb{C}$ or an arbitrary Banach space?)
$endgroup$
– 0x539
Jan 7 at 19:57
1
1
$begingroup$
For $b$, from the definition of $|.|_1$ you should have lots of $|.|$s. Adding these back will show you why its only an inequality, not an equality
$endgroup$
– Calvin Khor
Jan 7 at 20:06
$begingroup$
For $b$, from the definition of $|.|_1$ you should have lots of $|.|$s. Adding these back will show you why its only an inequality, not an equality
$endgroup$
– Calvin Khor
Jan 7 at 20:06
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Actually, you do prove $b$ before proving $a$. Let $F(x,y)=f(x-y)g(y)$. $F$ is clearly measurable, and $|F|$ is clearly integrable over $mathbb{R}^2$, with integral $|f|_1|g|_1$ (Fubini-Tonelli theorem ensures that all manipulations are legit on the double integral of $|F|$ so you can integrate wrt $x$ first, make the change of variables $x=x’+y$).
Then you use Fubini’s theorem to prove that $x longmapsto int{F(x,y)dy}$ is well-defined and absolutely convergent ae and is in $L^1(mathbb{R})$ with norm at most $|f|_1|g|_1$.
answered Jan 7 at 19:57
MindlackMindlack
2,72717
$endgroup$
add a comment |
$begingroup$
Suppose that $f, gin L^1$. Then
$$int int |f(t)g(x-t)|, dt , dx = int |f(t)| int |g(x - t)|, dx
= int f(t) |g|_1 , dt = |f|_1 |g|_1. $$
answered Jan 7 at 20:01
ncmathsadistncmathsadist
42.4k260103
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
Actually, you do prove $b$ before proving $a$. Let $F(x,y)=f(x-y)g(y)$. $F$ is clearly measurable, and $|F|$ is clearly integrable over $mathbb{R}^2$, with integral $|f|_1|g|_1$ (Fubini-Tonelli theorem ensures that all manipulations are legit on the double integral of $|F|$ so you can integrate wrt $x$ first, make the change of variables $x=x’+y$).
Then you use Fubini’s theorem to prove that $x longmapsto int{F(x,y)dy}$ is well-defined and absolutely convergent ae and is in $L^1(mathbb{R})$ with norm at most $|f|_1|g|_1$.
answered Jan 7 at 19:57
MindlackMindlack
2,72717
$endgroup$
add a comment |
$begingroup$
Actually, you do prove $b$ before proving $a$. Let $F(x,y)=f(x-y)g(y)$. $F$ is clearly measurable, and $|F|$ is clearly integrable over $mathbb{R}^2$, with integral $|f|_1|g|_1$ (Fubini-Tonelli theorem ensures that all manipulations are legit on the double integral of $|F|$ so you can integrate wrt $x$ first, make the change of variables $x=x’+y$).
Then you use Fubini’s theorem to prove that $x longmapsto int{F(x,y)dy}$ is well-defined and absolutely convergent ae and is in $L^1(mathbb{R})$ with norm at most $|f|_1|g|_1$.
answered Jan 7 at 19:57
MindlackMindlack
2,72717
$endgroup$
add a comment |
$begingroup$
Actually, you do prove $b$ before proving $a$. Let $F(x,y)=f(x-y)g(y)$. $F$ is clearly measurable, and $|F|$ is clearly integrable over $mathbb{R}^2$, with integral $|f|_1|g|_1$ (Fubini-Tonelli theorem ensures that all manipulations are legit on the double integral of $|F|$ so you can integrate wrt $x$ first, make the change of variables $x=x’+y$).
Then you use Fubini’s theorem to prove that $x longmapsto int{F(x,y)dy}$ is well-defined and absolutely convergent ae and is in $L^1(mathbb{R})$ with norm at most $|f|_1|g|_1$.
answered Jan 7 at 19:57
MindlackMindlack
2,72717
$endgroup$
Actually, you do prove $b$ before proving $a$. Let $F(x,y)=f(x-y)g(y)$. $F$ is clearly measurable, and $|F|$ is clearly integrable over $mathbb{R}^2$, with integral $|f|_1|g|_1$ (Fubini-Tonelli theorem ensures that all manipulations are legit on the double integral of $|F|$ so you can integrate wrt $x$ first, make the change of variables $x=x’+y$).
Then you use Fubini’s theorem to prove that $x longmapsto int{F(x,y)dy}$ is well-defined and absolutely convergent ae and is in $L^1(mathbb{R})$ with norm at most $|f|_1|g|_1$.
answered Jan 7 at 19:57
MindlackMindlack
2,72717
answered Jan 7 at 19:57
MindlackMindlack
2,72717
answered Jan 7 at 19:57
MindlackMindlack
2,72717
answered Jan 7 at 19:57
MindlackMindlack
2,72717
2,72717
add a comment |
add a comment |
$begingroup$
Suppose that $f, gin L^1$. Then
$$int int |f(t)g(x-t)|, dt , dx = int |f(t)| int |g(x - t)|, dx
= int f(t) |g|_1 , dt = |f|_1 |g|_1. $$
answered Jan 7 at 20:01
ncmathsadistncmathsadist
42.4k260103
$endgroup$
add a comment |
$begingroup$
Suppose that $f, gin L^1$. Then
$$int int |f(t)g(x-t)|, dt , dx = int |f(t)| int |g(x - t)|, dx
= int f(t) |g|_1 , dt = |f|_1 |g|_1. $$
answered Jan 7 at 20:01
ncmathsadistncmathsadist
42.4k260103
$endgroup$
add a comment |
$begingroup$
Suppose that $f, gin L^1$. Then
$$int int |f(t)g(x-t)|, dt , dx = int |f(t)| int |g(x - t)|, dx
= int f(t) |g|_1 , dt = |f|_1 |g|_1. $$
answered Jan 7 at 20:01
ncmathsadistncmathsadist
42.4k260103
$endgroup$
Suppose that $f, gin L^1$. Then
$$int int |f(t)g(x-t)|, dt , dx = int |f(t)| int |g(x - t)|, dx
= int f(t) |g|_1 , dt = |f|_1 |g|_1. $$
answered Jan 7 at 20:01
ncmathsadistncmathsadist
42.4k260103
edited Jan 8 at 0:07
answered Jan 7 at 20:01
ncmathsadistncmathsadist
42.4k260103
answered Jan 7 at 20:01
ncmathsadistncmathsadist
42.4k260103
answered Jan 7 at 20:01
ncmathsadistncmathsadist
42.4k260103
42.4k260103
add a comment |
add a comment |
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$begingroup$
I assume $L^1(R, L, m)$ is supposed to contain $L$-valued functions on $R$ that are measurable w.r.t. $m$. What can $L$ be for you? (i. e. only $mathbb{R}$ or $mathbb{C}$ or an arbitrary Banach space?)
$endgroup$
– 0x539
Jan 7 at 19:57
1
$begingroup$
For $b$, from the definition of $|.|_1$ you should have lots of $|.|$s. Adding these back will show you why its only an inequality, not an equality
$endgroup$
– Calvin Khor
Jan 7 at 20:06