Let $f:[-1,2]\rightarrow \mathbb{R}$ continuous. Define a function sequences: \begin{eqnarray*} f_{n}(x)=\frac{n}{2}\int_{x-\frac{1}{n}}^{x+\frac{1}{n}} f(t)dt \end{eqnarray*} for all $x\in [0,1]$ and for all $n\in \mathbb{N}$.
Show that $(f_{n})$ converges to $f$ uniformly in $[0,1]$.
My work:
By hypotesis $f$ is continuous in $[-1,2]$ so, By first theorem of first fundamental theorem of calculus $f'_{n}$ exist and : \begin{eqnarray*} f'_{n}=\frac{n}{2}(f(x+\frac{1}{n})-f(x-\frac{1}{n})) \end{eqnarray*} so in this case $x\in[0,1]$ and for all $n\in \mathbb{N}$.
So I think to analysis the uniformly convergence by differential theorem and study: \begin{eqnarray*} |f'_{n}(x)-f'(x)|&=&|\frac{n}{2}(f(x+\frac{1}{n})-f(x-\frac{1}{n}))-f'(x)| \end{eqnarray*} But I don't know if exist some point $x_{0}\in(0,1)$ such that $(f_{n}(x_{0}))$ converges and if $f'(x)$ exist.
HINT
$$|f_n(x)-f(x)|=\left|\frac{n}{2}\int_{x-1/n}^{x+1/n}\left(f(t)-f(x)\right)dt\right| \le \frac{n}{2}\int_{x-1/n}^{x+1/n}|f(x)-f(t)|dt$$
Now you can use that $f$ is uniform continuous on $[-1,2]$ to finish.