Young's Inequality for Convolutions; when $r = infty$











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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!










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  • This is just the standard Holder inequaity
    – Federico
    Nov 20 at 16:08















up vote
1
down vote

favorite












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!










share|cite|improve this question






















  • This is just the standard Holder inequaity
    – Federico
    Nov 20 at 16:08













up vote
1
down vote

favorite









up vote
1
down vote

favorite











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!










share|cite|improve this question













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






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asked Nov 20 at 16:03









김현빈

525




525












  • This is just the standard Holder inequaity
    – Federico
    Nov 20 at 16:08


















  • This is just the standard Holder inequaity
    – Federico
    Nov 20 at 16:08
















This is just the standard Holder inequaity
– Federico
Nov 20 at 16:08




This is just the standard Holder inequaity
– Federico
Nov 20 at 16:08










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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
$$|f ast g(x)| le int |f(x-y)| |g(y)| , dy le |f_x|_p |g|_q = |f|_p |g|_q.$$






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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
    $$|f ast g(x)| le int |f(x-y)| |g(y)| , dy le |f_x|_p |g|_q = |f|_p |g|_q.$$






    share|cite|improve this answer

























      up vote
      0
      down vote













      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
      $$|f ast g(x)| le int |f(x-y)| |g(y)| , dy le |f_x|_p |g|_q = |f|_p |g|_q.$$






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        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
        $$|f ast g(x)| le int |f(x-y)| |g(y)| , dy le |f_x|_p |g|_q = |f|_p |g|_q.$$






        share|cite|improve this answer












        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
        $$|f ast g(x)| le int |f(x-y)| |g(y)| , dy le |f_x|_p |g|_q = |f|_p |g|_q.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 20 at 16:09









        Umberto P.

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