Difference between revisions of "Number one, ye scurvy dogs!" - Rhea

Latest revision as of 04:56, 11 June 2013

Problem #7.1, MA598R, Summer 2009, Weigel

Back to The_Pirate's_Booty

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'} $ (Holder's inequality for three functions)

where $ \frac{1}{r} +\frac{1}{p'} +\frac{1}{q'} = \frac{1}{r}+\bigg(1-\frac{1}{p}\bigg)+\bigg(1-\frac{1}{q}\bigg)=1 $

and one can check that $ (1-\frac{p}{r})q'=p $ and $ (1-\frac{q}{r})p'=q $

therefore, we have (so far) that $ (f\ast g)\leq \bigg[\int_{\mathbb{R}^{n}}f(x-y)^{p}g(y)^{q}dy\bigg]^{1/r}\cdot \|f\|_{p}^{p/q} $ $ \cdot \|g\|_{q}^{q/p}=\bigg[\int_{\mathbb{R}^{n}}f(x-y)^{p}g(y)^{q}dy\bigg]^{1/r}\cdot 1\cdot 1 $

Or, in other words, $ (f\ast g)^{r}\leq f^{p}\ast g^{q} $

We just need to show that $ \|f^{p}\ast g^{q}\|_{1} = \|f^{p}\|_{1}\|g^{q}\|_{1}=\|f\|_{p}^{p}\|g\|_{q}^{q} =1 $ and we are done since

$ \int(f\ast g)^{r}dx\leq \|f^{p}\ast g^{q}\|_{1}<\infty $ by the previous line

To prove the claim, we just need to show that $ \|f\ast g\|_{1}=\|f\|_1\cdot\|g\|_1 $

Since $ (f\ast g) $ is a measurable function in $ \mathbb{R}^{2n} $ we can apply Fubini's Theorem

Thus, $ \int (f\ast g)(x)dx =\int dx\int f(x-y)g(y)dy = \int g(y)dy\cdot\int f(x-y)dx =\|f\|_1\cdot\|g\|_1 $

Q.E.D --Rlalvare 02:15, 28 July 2009 (UTC)

Back to the Pirate's Booty

Back to Assignment 7

Back to MA598R Summer 2009