Let $f,g: \mathbb{R}^n \to R$. Let $|| \cdot ||_T$ be a translation invariant norm on functions on $\mathbb{R}^n$. How can I prove that $||f*g||_T \leq ||f||_1 ||g||_T$ (where $f*g$ means convolution and $|| \cdot ||_1$ means $L^1$ norm).
I suppose I should specify the class of functions $f,g$ are of, but just take them to be functions such that the norms are well defined. In fact I will settle for a formal proof of the inequality. Any help is appreciated!
Best Answer
Applying the definition of convolution, where I stressed the fact that the norm is in terms of $x$, and $y$ is a dummy variable \begin{align*}\|f\ast g(x)\|_T &=\|\int_{\mathbb{R}^n}f(y)g(x-y)dy\|_T\\ & \leq\int_{\mathbb{R}^n}\|f(y)g(x-y)\|_Tdy\\ & = \int_{\mathbb{R}^n} |f(y)|\|g(x-y)\|_Tdy\\ & =\int_{\mathbb{R}^n}|f(y)|\|g(x)\|_Tdy\\ \\ & =\int_{\mathbb{R}^n}|f(y)|dy \cdot \|g(x)\|_T\\ & \overset{def}{=} \|f\|_{L^1}\|g\|_T \end{align*} where in the second step I have used homogenity and in the third the invariance of the $T$ norm.