Mixing
Young's Inequality for Convolutions; when $r = infty$
I've been trying to prove what seems to be a little generalized version of the Young's Inequality for Convolutions. Here's the statement of the Theorem:
Let $1leq p, q leq infty$, $frac{1}{p}+frac{1}{q}geq 1$, and $frac{1}{r}=frac{1}{p}+frac{1}{q}-1$. If $f in L^{p}, g in L^{q}$, then $fast g in L^{r}$ and $||fast g||_{r} leq ||f||_{p}||g||_{q}$.
First we may assume that $f,g$ are non-negative, for we can replace $f,g$ with $|f|,|g|$ if necessary.
In the case where $p,q,r leq infty$, I used the Generalized Hölder's Inequality for three functions with $frac{1}{r}+frac{1}{p_1}+frac{1}{p_2} = 1$, where $frac{1}{p_1}=frac{1}{p}-frac{1}{r}$, and $frac{1}{p_2}=frac{1}{q}-frac{1}{r}$, and got my desired result.
However, I'm stuck trying to prove the case when $r=infty$; that is when $frac{1}{p}+frac{1}{q}=1$, $||fast g||_{infty} leq ||f||_{p}||g||_{q}$ . My guess is that I need to consider the case when either $p$ or $q$ in infinite (so that the other equals 1), and when neither are infinite.
Any help would be appreciated, thanks!
real-analysis convolution
asked Nov 20 '18 at 16:03
김현빈
525
add a comment |
I've been trying to prove what seems to be a little generalized version of the Young's Inequality for Convolutions. Here's the statement of the Theorem:
Let $1leq p, q leq infty$, $frac{1}{p}+frac{1}{q}geq 1$, and $frac{1}{r}=frac{1}{p}+frac{1}{q}-1$. If $f in L^{p}, g in L^{q}$, then $fast g in L^{r}$ and $||fast g||_{r} leq ||f||_{p}||g||_{q}$.
First we may assume that $f,g$ are non-negative, for we can replace $f,g$ with $|f|,|g|$ if necessary.
In the case where $p,q,r leq infty$, I used the Generalized Hölder's Inequality for three functions with $frac{1}{r}+frac{1}{p_1}+frac{1}{p_2} = 1$, where $frac{1}{p_1}=frac{1}{p}-frac{1}{r}$, and $frac{1}{p_2}=frac{1}{q}-frac{1}{r}$, and got my desired result.
However, I'm stuck trying to prove the case when $r=infty$; that is when $frac{1}{p}+frac{1}{q}=1$, $||fast g||_{infty} leq ||f||_{p}||g||_{q}$ . My guess is that I need to consider the case when either $p$ or $q$ in infinite (so that the other equals 1), and when neither are infinite.
Any help would be appreciated, thanks!
real-analysis convolution
asked Nov 20 '18 at 16:03
김현빈
525
This is just the standard Holder inequaity
– Federico
Nov 20 '18 at 16:08
add a comment |
I've been trying to prove what seems to be a little generalized version of the Young's Inequality for Convolutions. Here's the statement of the Theorem:
Let $1leq p, q leq infty$, $frac{1}{p}+frac{1}{q}geq 1$, and $frac{1}{r}=frac{1}{p}+frac{1}{q}-1$. If $f in L^{p}, g in L^{q}$, then $fast g in L^{r}$ and $||fast g||_{r} leq ||f||_{p}||g||_{q}$.
First we may assume that $f,g$ are non-negative, for we can replace $f,g$ with $|f|,|g|$ if necessary.
In the case where $p,q,r leq infty$, I used the Generalized Hölder's Inequality for three functions with $frac{1}{r}+frac{1}{p_1}+frac{1}{p_2} = 1$, where $frac{1}{p_1}=frac{1}{p}-frac{1}{r}$, and $frac{1}{p_2}=frac{1}{q}-frac{1}{r}$, and got my desired result.
However, I'm stuck trying to prove the case when $r=infty$; that is when $frac{1}{p}+frac{1}{q}=1$, $||fast g||_{infty} leq ||f||_{p}||g||_{q}$ . My guess is that I need to consider the case when either $p$ or $q$ in infinite (so that the other equals 1), and when neither are infinite.
Any help would be appreciated, thanks!
real-analysis convolution
asked Nov 20 '18 at 16:03
김현빈
525
I've been trying to prove what seems to be a little generalized version of the Young's Inequality for Convolutions. Here's the statement of the Theorem:
Let $1leq p, q leq infty$, $frac{1}{p}+frac{1}{q}geq 1$, and $frac{1}{r}=frac{1}{p}+frac{1}{q}-1$. If $f in L^{p}, g in L^{q}$, then $fast g in L^{r}$ and $||fast g||_{r} leq ||f||_{p}||g||_{q}$.
First we may assume that $f,g$ are non-negative, for we can replace $f,g$ with $|f|,|g|$ if necessary.
In the case where $p,q,r leq infty$, I used the Generalized Hölder's Inequality for three functions with $frac{1}{r}+frac{1}{p_1}+frac{1}{p_2} = 1$, where $frac{1}{p_1}=frac{1}{p}-frac{1}{r}$, and $frac{1}{p_2}=frac{1}{q}-frac{1}{r}$, and got my desired result.
However, I'm stuck trying to prove the case when $r=infty$; that is when $frac{1}{p}+frac{1}{q}=1$, $||fast g||_{infty} leq ||f||_{p}||g||_{q}$ . My guess is that I need to consider the case when either $p$ or $q$ in infinite (so that the other equals 1), and when neither are infinite.
Any help would be appreciated, thanks!
real-analysis convolution
real-analysis convolution
asked Nov 20 '18 at 16:03
김현빈
525
asked Nov 20 '18 at 16:03
김현빈
525
asked Nov 20 '18 at 16:03
김현빈
525
asked Nov 20 '18 at 16:03
김현빈
525
asked Nov 20 '18 at 16:03
김현빈
525
525
This is just the standard Holder inequaity
– Federico
Nov 20 '18 at 16:08
add a comment |
This is just the standard Holder inequaity
– Federico
Nov 20 '18 at 16:08
This is just the standard Holder inequaity
– Federico
Nov 20 '18 at 16:08
This is just the standard Holder inequaity
– Federico
Nov 20 '18 at 16:08
add a comment |
1 Answer
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It's really just Hölder's inequality. Fix $x$ and write $f_x(y) = f(x-y)$. The translation invariance of the integral gives you $|f_x|_p = |f|_p$ for all $1 le p le infty$ so that
$$|f ast g(x)| le int |f(x-y)| |g(y)| , dy le |f_x|_p |g|_q = |f|_p |g|_q.$$
answered Nov 20 '18 at 16:09
Umberto P.
38.5k13064
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1 Answer
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It's really just Hölder's inequality. Fix $x$ and write $f_x(y) = f(x-y)$. The translation invariance of the integral gives you $|f_x|_p = |f|_p$ for all $1 le p le infty$ so that
$$|f ast g(x)| le int |f(x-y)| |g(y)| , dy le |f_x|_p |g|_q = |f|_p |g|_q.$$
answered Nov 20 '18 at 16:09
Umberto P.
38.5k13064
add a comment |
It's really just Hölder's inequality. Fix $x$ and write $f_x(y) = f(x-y)$. The translation invariance of the integral gives you $|f_x|_p = |f|_p$ for all $1 le p le infty$ so that
$$|f ast g(x)| le int |f(x-y)| |g(y)| , dy le |f_x|_p |g|_q = |f|_p |g|_q.$$
answered Nov 20 '18 at 16:09
Umberto P.
38.5k13064
add a comment |
It's really just Hölder's inequality. Fix $x$ and write $f_x(y) = f(x-y)$. The translation invariance of the integral gives you $|f_x|_p = |f|_p$ for all $1 le p le infty$ so that
$$|f ast g(x)| le int |f(x-y)| |g(y)| , dy le |f_x|_p |g|_q = |f|_p |g|_q.$$
answered Nov 20 '18 at 16:09
Umberto P.
38.5k13064
It's really just Hölder's inequality. Fix $x$ and write $f_x(y) = f(x-y)$. The translation invariance of the integral gives you $|f_x|_p = |f|_p$ for all $1 le p le infty$ so that
$$|f ast g(x)| le int |f(x-y)| |g(y)| , dy le |f_x|_p |g|_q = |f|_p |g|_q.$$
answered Nov 20 '18 at 16:09
Umberto P.
38.5k13064
answered Nov 20 '18 at 16:09
Umberto P.
38.5k13064
answered Nov 20 '18 at 16:09
Umberto P.
38.5k13064
answered Nov 20 '18 at 16:09
Umberto P.
38.5k13064
38.5k13064
add a comment |
add a comment |
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This is just the standard Holder inequaity
– Federico
Nov 20 '18 at 16:08