Is convolution integrable in $L_1(\mathbb{R}^n)$?

Question

Let $|h(y)|\in L_1(\mathbb{R}^n)$, i.e. $\int\limits_{\mathbb{R}^n}|h(y)|\,dy<+\infty$. Consider the function $F(x)=\int\limits_{\mathbb{R}^n}|h(x-y)|\,dy$. It is known that $F(x)$ be bounded and uniformly continuous. I try to find condtitions when $F(x)\in L_1(\mathbb{R}^n)$, i.e. $$ \int\limits_{\mathbb{R}^n}F(x)\,dx= \int\limits_{\mathbb{R}^n}\biggl(\int\limits_{\mathbb{R}^n}|h(x-y)|\,dy\biggr)\,dx<+\infty. $$ But I can't even find a specific example for that my assumption holds. Is this possible? Because I think $F(x)$ also be an increasing function, so its integral will always divergents.

Answer 1

No. Indeed, $F(x)\equiv \|h\|_{L_1(\mathbb{R}^n)}$, so that $$ \int\limits_{\mathbb{R}^n}F(x)\,dx=\|h\|_{L_1(\mathbb{R}^n)}\int \limits_{\mathbb{R}^n}\,dy=\infty. $$ Thanks for comments to @Chris and @RyszardSzwarc!