Mixing
Why does $int_{-infty}^infty f(x-y)g(y)dy$ exists when $int_{-infty}^infty f(x)dx<infty$ and...
$begingroup$
Why does $int_{-infty}^infty f(x-y)g(y)dy$ exists when $int_{-infty}^infty f(x)dx< infty$
$int_{-infty}^infty g(x)dx< infty$ where $f,g$ are maps from $mathbb{R}$ to $mathbb{R}_+$
I was hoping to show that $f$ and $g$ are bounded as a result of the assumptions on the integrability of $f$ and $g$ and thus conclude
$int_{-infty}^infty f(x-y)g(y)dy<infty$. But obviously integrability of $f$ and $g$ doesnt imply boundedness but just finiteness. I have spent quite a bit of time thinking about this. I thought about applying Holder, but since $f^2$ and $g^2$ are not necessarily integrable given an integrable $g$ and $f$,it doesnt make any sense.
Any hints on how could I show this?
real-analysis convolution
asked Jan 21 at 10:58
user3503589user3503589
1,2841821
$endgroup$
add a comment |
$begingroup$
Why does $int_{-infty}^infty f(x-y)g(y)dy$ exists when $int_{-infty}^infty f(x)dx< infty$
$int_{-infty}^infty g(x)dx< infty$ where $f,g$ are maps from $mathbb{R}$ to $mathbb{R}_+$
I was hoping to show that $f$ and $g$ are bounded as a result of the assumptions on the integrability of $f$ and $g$ and thus conclude
$int_{-infty}^infty f(x-y)g(y)dy<infty$. But obviously integrability of $f$ and $g$ doesnt imply boundedness but just finiteness. I have spent quite a bit of time thinking about this. I thought about applying Holder, but since $f^2$ and $g^2$ are not necessarily integrable given an integrable $g$ and $f$,it doesnt make any sense.
Any hints on how could I show this?
real-analysis convolution
asked Jan 21 at 10:58
user3503589user3503589
1,2841821
$endgroup$
$begingroup$
Are you assuming that $int|f|$ and $int|g|$ converge (as assumed in the answer you accepted instantly) or that $int f$ and $int g$ converge (as stated in your question)?
$endgroup$
– Did
Jan 21 at 15:07
$begingroup$
@Did I was a bit imprecise in asking the question but $f$ and $g$ take values in $mathbb{R}_+$. I will be more careful next time.
$endgroup$
– user3503589
Jan 21 at 15:10
add a comment |
$begingroup$
Why does $int_{-infty}^infty f(x-y)g(y)dy$ exists when $int_{-infty}^infty f(x)dx< infty$
$int_{-infty}^infty g(x)dx< infty$ where $f,g$ are maps from $mathbb{R}$ to $mathbb{R}_+$
I was hoping to show that $f$ and $g$ are bounded as a result of the assumptions on the integrability of $f$ and $g$ and thus conclude
$int_{-infty}^infty f(x-y)g(y)dy<infty$. But obviously integrability of $f$ and $g$ doesnt imply boundedness but just finiteness. I have spent quite a bit of time thinking about this. I thought about applying Holder, but since $f^2$ and $g^2$ are not necessarily integrable given an integrable $g$ and $f$,it doesnt make any sense.
Any hints on how could I show this?
real-analysis convolution
asked Jan 21 at 10:58
user3503589user3503589
1,2841821
$endgroup$
Why does $int_{-infty}^infty f(x-y)g(y)dy$ exists when $int_{-infty}^infty f(x)dx< infty$
$int_{-infty}^infty g(x)dx< infty$ where $f,g$ are maps from $mathbb{R}$ to $mathbb{R}_+$
I was hoping to show that $f$ and $g$ are bounded as a result of the assumptions on the integrability of $f$ and $g$ and thus conclude
$int_{-infty}^infty f(x-y)g(y)dy<infty$. But obviously integrability of $f$ and $g$ doesnt imply boundedness but just finiteness. I have spent quite a bit of time thinking about this. I thought about applying Holder, but since $f^2$ and $g^2$ are not necessarily integrable given an integrable $g$ and $f$,it doesnt make any sense.
Any hints on how could I show this?
real-analysis convolution
real-analysis convolution
asked Jan 21 at 10:58
user3503589user3503589
1,2841821
asked Jan 21 at 10:58
user3503589user3503589
1,2841821
edited Jan 21 at 15:12
user3503589
asked Jan 21 at 10:58
user3503589user3503589
1,2841821
asked Jan 21 at 10:58
user3503589user3503589
1,2841821
asked Jan 21 at 10:58
user3503589user3503589
1,2841821
1,2841821
$begingroup$
Are you assuming that $int|f|$ and $int|g|$ converge (as assumed in the answer you accepted instantly) or that $int f$ and $int g$ converge (as stated in your question)?
$endgroup$
– Did
Jan 21 at 15:07
$begingroup$
@Did I was a bit imprecise in asking the question but $f$ and $g$ take values in $mathbb{R}_+$. I will be more careful next time.
$endgroup$
– user3503589
Jan 21 at 15:10
add a comment |
$begingroup$
Are you assuming that $int|f|$ and $int|g|$ converge (as assumed in the answer you accepted instantly) or that $int f$ and $int g$ converge (as stated in your question)?
$endgroup$
– Did
Jan 21 at 15:07
$begingroup$
@Did I was a bit imprecise in asking the question but $f$ and $g$ take values in $mathbb{R}_+$. I will be more careful next time.
$endgroup$
– user3503589
Jan 21 at 15:10
$begingroup$
Are you assuming that $int|f|$ and $int|g|$ converge (as assumed in the answer you accepted instantly) or that $int f$ and $int g$ converge (as stated in your question)?
$endgroup$
– Did
Jan 21 at 15:07
$begingroup$
Are you assuming that $int|f|$ and $int|g|$ converge (as assumed in the answer you accepted instantly) or that $int f$ and $int g$ converge (as stated in your question)?
$endgroup$
– Did
Jan 21 at 15:07
$begingroup$
@Did I was a bit imprecise in asking the question but $f$ and $g$ take values in $mathbb{R}_+$. I will be more careful next time.
$endgroup$
– user3503589
Jan 21 at 15:10
$begingroup$
@Did I was a bit imprecise in asking the question but $f$ and $g$ take values in $mathbb{R}_+$. I will be more careful next time.
$endgroup$
– user3503589
Jan 21 at 15:10
add a comment |
1 Answer
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$begingroup$
First note that if a measurable function $phi:Bbb Rto Bbb Rcup {pminfty}$ is integrable, i.e. $int_{Bbb R} |phi(x)|dx<infty$, then
$$
|phi(x)|<infty
$$ for almost every $xinBbb R$.
Define $phi(x)=int |f(x-y)||g(y)|dy$, which takes its values in $[0,infty]$. By Fubini-Tonelli's theorem, it holds
$$
begin{split}
int phi(x)dx &=intleft[int |f(x-y)||g(y)|dyright]dx \
&=int |g(y)|left[int|f(x-y)|dxright]dy =|f|_1|g|_1<infty.
end{split}
$$ This implies that $phi(x)<infty$ a.e., hence the integral $$(f*g)(x)=int f(x-y)g(y)dy$$ converges absolutely for almost every $xin Bbb R$. This shows almost everywhere finiteness of convolution. Moreover, we can observe that since
$$
|f*g|(x)le phi(x),
$$ it holds
$$
|f*g|_1le |phi|_1le |f|_1|g|_1.
$$ This is a special case of Young's inequality, which is saying that
$$
|f*g|_r le |f|_p|g|_q
$$ for $frac{1}{p}+frac{1}{q}=1+frac{1}{r}$, $p,q,rin [1,infty]$.
edited Jan 21 at 14:59
Daniele Tampieri
2,3872922
answered Jan 21 at 11:32
SongSong
16.8k11043
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
First note that if a measurable function $phi:Bbb Rto Bbb Rcup {pminfty}$ is integrable, i.e. $int_{Bbb R} |phi(x)|dx<infty$, then
$$
|phi(x)|<infty
$$ for almost every $xinBbb R$.
Define $phi(x)=int |f(x-y)||g(y)|dy$, which takes its values in $[0,infty]$. By Fubini-Tonelli's theorem, it holds
$$
begin{split}
int phi(x)dx &=intleft[int |f(x-y)||g(y)|dyright]dx \
&=int |g(y)|left[int|f(x-y)|dxright]dy =|f|_1|g|_1<infty.
end{split}
$$ This implies that $phi(x)<infty$ a.e., hence the integral $$(f*g)(x)=int f(x-y)g(y)dy$$ converges absolutely for almost every $xin Bbb R$. This shows almost everywhere finiteness of convolution. Moreover, we can observe that since
$$
|f*g|(x)le phi(x),
$$ it holds
$$
|f*g|_1le |phi|_1le |f|_1|g|_1.
$$ This is a special case of Young's inequality, which is saying that
$$
|f*g|_r le |f|_p|g|_q
$$ for $frac{1}{p}+frac{1}{q}=1+frac{1}{r}$, $p,q,rin [1,infty]$.
edited Jan 21 at 14:59
Daniele Tampieri
2,3872922
answered Jan 21 at 11:32
SongSong
16.8k11043
$endgroup$
add a comment |
$begingroup$
First note that if a measurable function $phi:Bbb Rto Bbb Rcup {pminfty}$ is integrable, i.e. $int_{Bbb R} |phi(x)|dx<infty$, then
$$
|phi(x)|<infty
$$ for almost every $xinBbb R$.
Define $phi(x)=int |f(x-y)||g(y)|dy$, which takes its values in $[0,infty]$. By Fubini-Tonelli's theorem, it holds
$$
begin{split}
int phi(x)dx &=intleft[int |f(x-y)||g(y)|dyright]dx \
&=int |g(y)|left[int|f(x-y)|dxright]dy =|f|_1|g|_1<infty.
end{split}
$$ This implies that $phi(x)<infty$ a.e., hence the integral $$(f*g)(x)=int f(x-y)g(y)dy$$ converges absolutely for almost every $xin Bbb R$. This shows almost everywhere finiteness of convolution. Moreover, we can observe that since
$$
|f*g|(x)le phi(x),
$$ it holds
$$
|f*g|_1le |phi|_1le |f|_1|g|_1.
$$ This is a special case of Young's inequality, which is saying that
$$
|f*g|_r le |f|_p|g|_q
$$ for $frac{1}{p}+frac{1}{q}=1+frac{1}{r}$, $p,q,rin [1,infty]$.
edited Jan 21 at 14:59
Daniele Tampieri
2,3872922
answered Jan 21 at 11:32
SongSong
16.8k11043
$endgroup$
add a comment |
$begingroup$
First note that if a measurable function $phi:Bbb Rto Bbb Rcup {pminfty}$ is integrable, i.e. $int_{Bbb R} |phi(x)|dx<infty$, then
$$
|phi(x)|<infty
$$ for almost every $xinBbb R$.
Define $phi(x)=int |f(x-y)||g(y)|dy$, which takes its values in $[0,infty]$. By Fubini-Tonelli's theorem, it holds
$$
begin{split}
int phi(x)dx &=intleft[int |f(x-y)||g(y)|dyright]dx \
&=int |g(y)|left[int|f(x-y)|dxright]dy =|f|_1|g|_1<infty.
end{split}
$$ This implies that $phi(x)<infty$ a.e., hence the integral $$(f*g)(x)=int f(x-y)g(y)dy$$ converges absolutely for almost every $xin Bbb R$. This shows almost everywhere finiteness of convolution. Moreover, we can observe that since
$$
|f*g|(x)le phi(x),
$$ it holds
$$
|f*g|_1le |phi|_1le |f|_1|g|_1.
$$ This is a special case of Young's inequality, which is saying that
$$
|f*g|_r le |f|_p|g|_q
$$ for $frac{1}{p}+frac{1}{q}=1+frac{1}{r}$, $p,q,rin [1,infty]$.
edited Jan 21 at 14:59
Daniele Tampieri
2,3872922
answered Jan 21 at 11:32
SongSong
16.8k11043
$endgroup$
First note that if a measurable function $phi:Bbb Rto Bbb Rcup {pminfty}$ is integrable, i.e. $int_{Bbb R} |phi(x)|dx<infty$, then
$$
|phi(x)|<infty
$$ for almost every $xinBbb R$.
Define $phi(x)=int |f(x-y)||g(y)|dy$, which takes its values in $[0,infty]$. By Fubini-Tonelli's theorem, it holds
$$
begin{split}
int phi(x)dx &=intleft[int |f(x-y)||g(y)|dyright]dx \
&=int |g(y)|left[int|f(x-y)|dxright]dy =|f|_1|g|_1<infty.
end{split}
$$ This implies that $phi(x)<infty$ a.e., hence the integral $$(f*g)(x)=int f(x-y)g(y)dy$$ converges absolutely for almost every $xin Bbb R$. This shows almost everywhere finiteness of convolution. Moreover, we can observe that since
$$
|f*g|(x)le phi(x),
$$ it holds
$$
|f*g|_1le |phi|_1le |f|_1|g|_1.
$$ This is a special case of Young's inequality, which is saying that
$$
|f*g|_r le |f|_p|g|_q
$$ for $frac{1}{p}+frac{1}{q}=1+frac{1}{r}$, $p,q,rin [1,infty]$.
edited Jan 21 at 14:59
Daniele Tampieri
2,3872922
answered Jan 21 at 11:32
SongSong
16.8k11043
edited Jan 21 at 14:59
Daniele Tampieri
2,3872922
edited Jan 21 at 14:59
Daniele Tampieri
2,3872922
edited Jan 21 at 14:59
Daniele Tampieri
2,3872922
2,3872922
answered Jan 21 at 11:32
SongSong
16.8k11043
answered Jan 21 at 11:32
SongSong
16.8k11043
answered Jan 21 at 11:32
SongSong
16.8k11043
16.8k11043
add a comment |
add a comment |
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$begingroup$
Are you assuming that $int|f|$ and $int|g|$ converge (as assumed in the answer you accepted instantly) or that $int f$ and $int g$ converge (as stated in your question)?
$endgroup$
– Did
Jan 21 at 15:07
$begingroup$
@Did I was a bit imprecise in asking the question but $f$ and $g$ take values in $mathbb{R}_+$. I will be more careful next time.
$endgroup$
– user3503589
Jan 21 at 15:10