Function is pointwise limit of integrals This is a question from some old masters exams.

Let $\phi_{n}$ be a sequence of continuous, real functions on $\mathbb{R}$ such that
$\phi_{n}(x) = 0$ for all $|x|\geq 1/n$ and $\phi_{n}(x)\geq 0$ for all $x\in\mathbb{R}$, and
$$\int\limits_{-1}^{1}{\phi_{n}(x)dx}=1$$
For each continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$, let 
$$f_{n}(x)=\int\limits_{-\infty}^{\infty}{\phi_{n}(x-y)f(y)dy}$$
Prove that $f_{n}(x)$ converges pointwise to $f(x)$ and prove that if $f(x)=0$ for $|x|\geq 10$, then $f_{n}$ converges uniformly to $f$.

 A: For pointwise convergence, note that it suffices to show that
$$\int_{-\infty}^\infty \phi_n(x-y)(f(y)-f(x))dy\to 0$$
for all $x$. Fix some $x$. Since $f$ is continuous, for any $\epsilon>0$ we have some $\delta>0$ such that $$|x-y|<\delta\implies |f(x)-f(y)|<\epsilon.$$
Choose $N\in\mathbb N$ such that $N> 1/\delta$. Then for $n\geq N$ we have
$$\begin{align}
\left|\int_{-\infty}^\infty \phi_n(x-y)(f(y)-f(x))dy\right| &=\left|\int_{x-1/n}^{x+1/n} \phi_n(x-y)(f(y)-f(x))dy\right|\\
&\leq \int_{x-1/n}^{x+1/n} \phi_n(x-y)|f(y)-f(x)|dy\\
&< \epsilon \int_{x-1/n}^{x+1/n} \phi_n(x-y)dy = \epsilon
\end{align}$$
hence $f_n\to f$ pointwise. For uniform convergence, note that if $f(x)=0$ when $|x|\geq 10$ then $f$ has compact support, hence we can choose this $\epsilon$ independently of $x$ hence the convergence is uniform in $x$.
A: Here is a hint on how to do the problem. Because $\phi_n(x) = 0$ for $|x| \geq 1/n$, you can say 
$$\int_{-1}^1 \phi_n(x) dx = \int_{-\infty}^{\infty} \phi_n(x) dx.$$
Now choose $x \in\Bbb{R}$. Then
$$\begin{eqnarray*} |f_n(x) - f(x) | &=& \left|\lim_{N \to \infty} \int_{-N}^N \phi_n(x-y)(f(y) - f(x)) dy    \right| \\
&=&\lim_{N \to \infty} \left| \int_{-N}^N \phi_n(x-y)(f(y) - f(x)) dy    \right| \\
&\leq&\lim_{N \to \infty} \int_{-N}^N |\phi_n(x-y)||f(y) - f(x)| dy \\
&=& \lim_{N \to \infty} \int_{x-N}^{x+N} |\phi_n(u)||f(x) - f(x-u)| du \\
&=&  \int_{x-\frac{1}{n}}^{x + \frac{1}{n}} |\phi_n(u)||f(x) - f(x-u)| du. \end{eqnarray*}$$
Facts I used above: The absolute value function is continuous, and the integrals above are continuous functions in $x$. Also, I used that the $\phi_n(x)$ are zero for $n$ sufficiently large, and there is a change of variables above. We can now complete your problem as follows. By continuity of $f$ we know that given any $\epsilon > 0$ , there is $\delta > 0$ such that $|u| < \delta$ will imply that $|f(x) - f(x-u)|  < \epsilon$. Choose $N$ so large that $1/N < \delta$. Then for all $n\geq N$, we would have that $|u| < \frac{1}{n}$ will imply that $|f(x) - f(x-u)| < \epsilon$. The integral in the last line is now less than
$$ \epsilon \int_{x-\frac{1}{n}}^{x + \frac{1}{n}} |\phi_n(u)| du = C\epsilon$$
for some constant $C$. It follows that $f_n \to f$ pointwise.
