Let $g(x):\mathbb{R} \to \mathbb{R}$ be a function such that $\int_{-\infty}^\infty g(x) dx =1$. Show that for any continuous function $f:\mathbb{R} \to\mathbb{R}$,we ahve the sequence $g_n(x)=n\int_{-\infty}^\infty g(n(x-y))f(y)dx$converges uniformly on the compact sets of $\mathbb{R}$ to f(x)
First, use the substitution $t=n(x-y)$ in the integral (the integration is with respect to $y$). I think we have to assume $f$ bounded, because if $tg(t)$ is not integrable (for example when $g(t)=\frac 1{\pi(1+t^2)}$, and $f(t)=t$, we have some problems.