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!
Best Answer
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 $||f\ast g||_{\infty}\leq ||f||_1||g||_{\infty}$.