Sequence of real convolution is uniformly convergent. Let $f$ be a $C^0_C(\mathbb R)$ function ($=0$ for all $x$ outside a fixed compact interval and continuous) and $g$ a $C^{\infty}_C(\mathbb R)$ non-negative function that is equal to $0$ outside $[-1,1]$. We define $g_n(x)= \frac{ng(nx)}{\int_{\mathbb R}g}$. Show that $$f_n=(f * g_n)(x)=\int_{-\infty}^{+ \infty}f(t)g_n(x-t)dt$$
converges uniformly to $f$.
I noticed that $g_n$ is equal $0$ outside of $[\frac{-1}n,\frac1n]$, but I don't know how to use it (if it is useful) to show the uniform convergence.
 A: Note $$\int_{-\infty}^\infty f(t)g_n(x-t)dt = \int_{-\infty}^\infty f(x-t)g_n(t)dt = \int_{-\infty}^\infty f(x-t)\frac{ng(nt)}{\int g}dt.$$ Letting $y= nt$ gives $$\int_{-\infty}^\infty f\left(x-\frac{y}{n}\right)\frac{g(y)}{\int g}dy = \int_{-1}^1 f\left(x-\frac{y}{n}\right)\frac{g(y)}{\int g}dy.$$ So, using that $\int_{-\infty}^\infty \frac{g(y)}{\int g}dy = 1$, we have \begin{align*}\left|\int_{-\infty}^\infty f(t)g_n(x-t)dt - f(x)\right| &= \left|\int_{-1}^1 \left[f\left(x-\frac{y}{n}\right)-f(x)\right]\frac{g(y)}{\int g} dy\right| \\ &\le \int_{-1}^1 \left|f\left(x-\frac{y}{n}\right)-f(x)\right|\frac{g(y)}{\int g}dy.\end{align*} Now, since $f$ is continuous and is compactly supported, $f$ is uniformly continuous, so for any $\epsilon > 0$, there's some $\delta > 0$ so that for all $x$, we have $|f(z)-f(x)| < \epsilon$ for all $z$ with $|z-x| < \delta$. Therefore, for fixed $\epsilon > 0$, if we take $n > 1/\delta$, we have $$\int_{-1}^1 \left|f\left(x-\frac{y}{n}\right)-f(x)\right|\frac{g(y)}{\int g}dy \le \int_{-1}^1 \epsilon \frac{g(y)}{\int g}dy = \epsilon.$$ This proves uniform convergence.
