$|f * g|_ple |f|_1 |g|_p$
Let $f in L^1(lambda^r)$ and $g in L^p(lambda^r)$ with $1le p<infty$ where $lambda^r$ denotes the $r$-dimensional Lebesgue measure.
$h:=f * g$ where $*$ denotes the convolution.
I need to show that $h in L^p(lambda^r)$ and that $|f * g|_ple |f|_1 |g|_p$.
I tried the following:
$Big(int_{mathbb{R}^r} |(f*g)(x)|^pdlambda_r(x)Big)^{1/p} = Big(int_{mathbb{R}^r} Big|int_{mathbb{R}^r} f(x - y)g(y)dlambda_r(y)Big|^p dlambda_r(x)Big)^frac{1}{p} le int_{mathbb{R}^r} Big(int_{mathbb{R}^r} |f(x - y)g(y)|^pdlambda_r(x)Big)^{frac{1}{p}}dlambda_r(y) = int_{mathbb{R}^r} Big(int_{mathbb{R}^r} |f(x - y)|^p dlambda_r(x)Big)^{frac{1}{p}}|g(y)|dlambda_r(y) = |f|_p|g|_1.$
and this also shows that $h in L^p(lambda^r)$.
Is that correct?
real-analysis analysis measure-theory lebesgue-integral
asked Dec 3 '18 at 18:36
conradconrad
757
add a comment |
Let $f in L^1(lambda^r)$ and $g in L^p(lambda^r)$ with $1le p<infty$ where $lambda^r$ denotes the $r$-dimensional Lebesgue measure.
$h:=f * g$ where $*$ denotes the convolution.
I need to show that $h in L^p(lambda^r)$ and that $|f * g|_ple |f|_1 |g|_p$.
I tried the following:
$Big(int_{mathbb{R}^r} |(f*g)(x)|^pdlambda_r(x)Big)^{1/p} = Big(int_{mathbb{R}^r} Big|int_{mathbb{R}^r} f(x - y)g(y)dlambda_r(y)Big|^p dlambda_r(x)Big)^frac{1}{p} le int_{mathbb{R}^r} Big(int_{mathbb{R}^r} |f(x - y)g(y)|^pdlambda_r(x)Big)^{frac{1}{p}}dlambda_r(y) = int_{mathbb{R}^r} Big(int_{mathbb{R}^r} |f(x - y)|^p dlambda_r(x)Big)^{frac{1}{p}}|g(y)|dlambda_r(y) = |f|_p|g|_1.$
and this also shows that $h in L^p(lambda^r)$.
Is that correct?
real-analysis analysis measure-theory lebesgue-integral
asked Dec 3 '18 at 18:36
conradconrad
757
1
Yes. Just one minor point - you wanted to prove that $|f|_1|g|_p$ is the bound and you showed that $|f|_p |g|_1$ is the bound. But you can just use that $fast g = gast f$ or argue a little different.
– Yanko
Dec 3 '18 at 18:49
add a comment |
Let $f in L^1(lambda^r)$ and $g in L^p(lambda^r)$ with $1le p<infty$ where $lambda^r$ denotes the $r$-dimensional Lebesgue measure.
$h:=f * g$ where $*$ denotes the convolution.
I need to show that $h in L^p(lambda^r)$ and that $|f * g|_ple |f|_1 |g|_p$.
I tried the following:
$Big(int_{mathbb{R}^r} |(f*g)(x)|^pdlambda_r(x)Big)^{1/p} = Big(int_{mathbb{R}^r} Big|int_{mathbb{R}^r} f(x - y)g(y)dlambda_r(y)Big|^p dlambda_r(x)Big)^frac{1}{p} le int_{mathbb{R}^r} Big(int_{mathbb{R}^r} |f(x - y)g(y)|^pdlambda_r(x)Big)^{frac{1}{p}}dlambda_r(y) = int_{mathbb{R}^r} Big(int_{mathbb{R}^r} |f(x - y)|^p dlambda_r(x)Big)^{frac{1}{p}}|g(y)|dlambda_r(y) = |f|_p|g|_1.$
and this also shows that $h in L^p(lambda^r)$.
Is that correct?
real-analysis analysis measure-theory lebesgue-integral
asked Dec 3 '18 at 18:36
conradconrad
757
Let $f in L^1(lambda^r)$ and $g in L^p(lambda^r)$ with $1le p<infty$ where $lambda^r$ denotes the $r$-dimensional Lebesgue measure.
$h:=f * g$ where $*$ denotes the convolution.
I need to show that $h in L^p(lambda^r)$ and that $|f * g|_ple |f|_1 |g|_p$.
I tried the following:
$Big(int_{mathbb{R}^r} |(f*g)(x)|^pdlambda_r(x)Big)^{1/p} = Big(int_{mathbb{R}^r} Big|int_{mathbb{R}^r} f(x - y)g(y)dlambda_r(y)Big|^p dlambda_r(x)Big)^frac{1}{p} le int_{mathbb{R}^r} Big(int_{mathbb{R}^r} |f(x - y)g(y)|^pdlambda_r(x)Big)^{frac{1}{p}}dlambda_r(y) = int_{mathbb{R}^r} Big(int_{mathbb{R}^r} |f(x - y)|^p dlambda_r(x)Big)^{frac{1}{p}}|g(y)|dlambda_r(y) = |f|_p|g|_1.$
and this also shows that $h in L^p(lambda^r)$.
Is that correct?
real-analysis analysis measure-theory lebesgue-integral
real-analysis analysis measure-theory lebesgue-integral
asked Dec 3 '18 at 18:36
conradconrad
757
asked Dec 3 '18 at 18:36
conradconrad
757
asked Dec 3 '18 at 18:36
conradconrad
757
asked Dec 3 '18 at 18:36
conradconrad
757
asked Dec 3 '18 at 18:36
conradconrad
757
757
1
Yes. Just one minor point - you wanted to prove that $|f|_1|g|_p$ is the bound and you showed that $|f|_p |g|_1$ is the bound. But you can just use that $fast g = gast f$ or argue a little different.
– Yanko
Dec 3 '18 at 18:49
add a comment |
1
Yes. Just one minor point - you wanted to prove that $|f|_1|g|_p$ is the bound and you showed that $|f|_p |g|_1$ is the bound. But you can just use that $fast g = gast f$ or argue a little different.
– Yanko
Dec 3 '18 at 18:49
1
1
Yes. Just one minor point - you wanted to prove that $|f|_1|g|_p$ is the bound and you showed that $|f|_p |g|_1$ is the bound. But you can just use that $fast g = gast f$ or argue a little different.
– Yanko
Dec 3 '18 at 18:49
Yes. Just one minor point - you wanted to prove that $|f|_1|g|_p$ is the bound and you showed that $|f|_p |g|_1$ is the bound. But you can just use that $fast g = gast f$ or argue a little different.
– Yanko
Dec 3 '18 at 18:49
add a comment |
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Yes. Just one minor point - you wanted to prove that $|f|_1|g|_p$ is the bound and you showed that $|f|_p |g|_1$ is the bound. But you can just use that $fast g = gast f$ or argue a little different.
– Yanko
Dec 3 '18 at 18:49