Is the function $F(w) = \int_{\mathbb{R}^2}g(z)f(w-z)dz$ continuous?

Question

Let $F(w) = \int_{\mathbb{R}^2}g(z)f(w-z)dz$, where $f$ is continuous and $f\cdot g$ is integrable over $\mathbb{R}^2$. The claim is that $F$ is continuous. Is that true? If $f$ was uniformly continuous then I believe I could prove it, but that doesn't seem to be the case, as the domain isn't compact nor bounded. Edit: I now see uniform continuity isn't needed, but I still need $g$ to be bounded at least... I can't think of any counter example tough

Answer 1

not an answer
Let $w_n \to w_0$. We want to show $F(w_n) \to F(w_0)$.
For each fixed $z$, we have $w_n-z \to w_0-z$, hence by continuity $f(w_n-z) \to f(w_0-z)$, hence $g(z)f(w_n-z) \to g(z)f(w_0-z)$.

Then we need some convergence theorem to get $\int g(z)f(w_n-z)\;dz \to \int f(z)f(w_0-z)\;dz$.

If we knew that $f$ is bounded, say by $M$, then we have $|g(z)f(w_n-z)| \le M|g(z)|$ so the dominated convergence theorem applies. (I think Julian's two links will help only if at least one factor is bounded.) But I do not see how to proceed if $f$ is not bounded.