Convergence of a sequence of continuous function Len $g_n:\mathbb{R}\rightarrow\mathbb{R}$ continuous with $g_n(x)=0$ for $|x|\geq 1/n$, $g_n(x)\geq 0$, $\int_{-1}^1 g_n(x)\,dx=1$. Consider $f:\mathbb{R}\rightarrow\mathbb{R}$ continuous and:
$$
f_n(x)=\int_{-\infty}^\infty g_n(x-y)f(y)\,dy
$$
I want to show that $f_n$ converges to $f$ pointwise and if $f(x)=0$ for $|x|\geq a$ for some $a>1$ then the convergence is uniform.
For the first part after changing variable $z=x-y$ I get $f_n(x)=\int_{-1}^1 g_n(z)f(x-z)dz$. How should I proceed?
The second part with the uniform convergence follows from Dini's Theorem since f is compactly supported and so fn(x) will be compactly supported also?
 A: Hints:


*

*$f_n(x) = \int_{-1/n}^{1/n}g_n(z)f(x-z)\,dz$ by the support assumption on $g_n$.

*If $f$ is continuous, then $|f(x-z) - f(x)|$ is very small for $z \in \left(-\dfrac1n,\dfrac1n\right)$. That is, for fixed $x$ (and large enough $n$), the value of $f(x-z)$ in the above integral is more or less $\equiv f(x)$.

*You know $\int_{-1/n}^{1/n} g_n(z) \,dz = 1$.
A note: $g_n$ is known as an "approximation to the identity", a pretty general and important notion in analysis.
A: As you have noted $$\int\limits_{ - \infty }^{ + \infty } {{g_n}(x - y)f(y)dy}  = \int\limits_{ - \infty }^{ + \infty } {{g_n}(z)f(x - z)dz}  = \int\limits_{ - \frac{1}{n}}^{ + \frac{1}{n}} {{g_n}(z)f(x - z)dz} $$Assuming continuity of $f$ and the existence of it's derivative, for large enough $n$ (small enough $z$ in the integrand limits)one could write $f(x - z) = f(x) + O(z)$, so by replacement $$\int\limits_{ - 1}^{ + 1} {{g_n}(z)f(x - z)dz}  = f(x)\int\limits_{ - 1}^{ + 1} {{g_n}(z)dz}  + \int\limits_{ - 1}^{ + 1} {O(z){g_n}(z)dz}  = f(x) + \int\limits_{ - \frac{1}{n}}^{ + \frac{1}{n}} {O(z){g_n}(z)dz} $$Now, it is easy to see that $$\left| {\int\limits_{ - 1}^{ + 1} {{g_n}(z)f(x - z)dz}  - f(x)} \right| \le O(\frac{1}{n})\int\limits_{ - \frac{1}{n}}^{ + \frac{1}{n}} {{g_n}(z)dz}  = O(\frac{1}{n})$$which implies point-wise convergence. 
