The convolution of two functions is L1

Question

I have to proof the following corollary:

$ \text{Let } 1 \leq p \leq \infty \text{, } f \in L^{1} \text{, } g\in L^{p} \text{. Then } f \ast g \in L^p \text{ and } \Vert f \ast g \Vert _{L^{p}} \leq \Vert f \Vert _{L^{1}} \Vert g \Vert _{L^{p}}$.

I understood the proof for the case $p < \infty$, and in my lecture notes it is only written "for the case $p = \infty$ the result follows directly from Hölder's inequality. Now my problem is: the Hölder's inequality yields a result for $fg$, not for $f \ast g$, so I don't really see the link.

Thank you!

Answer 1

If $f\in L^1$ and $g\in L^{\infty}$, then $$ |(f\ast g)(x)|=\Big|\int_{\mathbb{R}}f(x-y)g(y)\;dy\Big|\leq \int_{\mathbb{R}}|f(x-y)g(y)|\;dy\leq ||g||_{\infty}\int_{\mathbb{R}}|f(y)|\;dy=||f||_1||g||_{\infty}$$ using the translation-invariance of Lebesgue measure. Therefore $||f\ast g||_{\infty}\leq ||f||_1||g||_{\infty}$.

Answer 2

By Hölder's inequality, you get $$ |f\ast g(x)| \leq \int_{\mathbb{R}^{n}} |f(y)g(x-y)|\;{\rm d}y = \|f g(x-\cdot)\|_1\leq \|g\|_\infty \|f\|_1 $$ since $|g(x-y)|\leq \|g\|_\infty$ for almost every $x\in\mathbb{R}^{n}$ since $\|g(x-\cdot)\|_\infty=\|g\|_\infty$. This implies $$ \|f\ast g\|_\infty \leq \|f\|_1\|g\|_\infty $$