I saw this post and I was wondering about the convolution of an $L^2$ function with and $L^1$ function.
Suppose $f \in L^2(\mathbb{R}^d)$ and $g \in L^1(\mathbb{R}^d)$
Could we say that the convolution $\int f(x-y)g(y)dy$ converges for almost every $x$ based on the information above?
Yes, we can say that. We can write $$ f = f_\infty + f_1 $$ with $$ f_\infty = 1_{\{\vert f \vert \leq 1\}}, f_1= 1_{\{\vert f \vert >1\}}. $$ Then we have $f_1\in L^1(\mathbb{R}^d)$ as $\vert f_1\vert \leq \vert f_1\vert^2 $ and $f_\infty \in L^\infty(\mathbb{R}^d)$ as $\vert f_\infty \vert \leq 1$. However, as $g\in L^1(\mathbb{R}^d)$, we get $$ \int \vert f_\infty(x-y) \vert \cdot \vert g (y) \vert \leq \Vert g \Vert_1. $$ The other part reduces to the convolution of two $L^1(\mathbb{R}^d)$ functions, where the convergence follows from Young's convolution inequality.