(New page: Define <math>(f\ast g)(x) = \int_{\mathbb{R}^{n}}f(x-y)g(y)dy</math>. Show that <math>L^{p}(\mathbb{R}^{n})\ast L^{q}(\mathbb{R}^{n})</math> <math>\subset L^{r}(\mathbb{R}^{n}), 1+1/r = 1/...) |
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Revision as of 15:55, 27 July 2009
Define $ (f\ast g)(x) = \int_{\mathbb{R}^{n}}f(x-y)g(y)dy $. Show that $ L^{p}(\mathbb{R}^{n})\ast L^{q}(\mathbb{R}^{n}) $ $ \subset L^{r}(\mathbb{R}^{n}), 1+1/r = 1/p + 1/q, 1<p,q,r<\infty $
Proof: Without loss of generality assume $ \|f\|_{p} = \|g\|_{q} = 1 $
$ \int_{\mathbb{R}^{n}}f(x-y)g(y)dy = \int_{\mathbb{R}^{n}}[f(x-y)^{p/r}g(y)^{q/r}]f(x-y)^{1-p/r}g(y)^{1-q/r}dy $
$ \leq \bigg[\int_{\mathbb{R}^{n}}f(x-y)^{p}g(y)^{q}dy\bigg]^{1/r}\cdot \bigg[\int_{\mathbb{R}^{n}}f(x-y)^{(1-p/r)q'}dy\bigg]^{1/q'}\cdot $ $ \bigg[\int_{\mathbb{R}^{n}}g(y)^{(1-q/r)p'}dy\bigg]^{1/p'} $
