Let $f$ be a continuous function on $[-1,2]$. Given $0\le x\le 1$ and $n\ge 1$, define a sequence of functions: $f_n(x)=\frac{n}{2}\int\limits_{x-\frac{1}{n}}^{x+\frac{1}{n}}{f(t)\,dt}\,.$ Show that each $f_n$ is continuous on $[0,1]$ and that $(f_n)$ converges uniformly to $f$ on $[0,1]$.
My work on continuity:
Let $\epsilon>0$. Let $x_n\to x$ in $[0,1]$. Let $\delta$ be the continuity criterion for $f$ (not sure what else to call it); then there exists $N$ such that $|x_n-x|<\delta$ for $n\ge N$. For such $n$, we thus have $|f(x_n)-f(x)|<\epsilon$. So for $t\in (x-\delta, x+\delta)$, we have $|f(t)|<|f(x)|+\epsilon$.
Then
$|f_n(x_n)-f_n(x)|=\left|\frac{n}{2}\int\limits_{x_n-\frac{1}{n}}^{x_n+\frac{1}{n}}{f(t)\,dt}-\frac{n}{2}\int\limits_{x-\frac{1}{n}}^{x+\frac{1}{n}}{f(t)\,dt}\right|$
Now, I don't want to typeset all the madness I have scribbled down, but this difference of integrals becomes (if we assume $x_n
$\int\limits_{x_n-\frac{1}{n}}^{x+\frac{1}{n}}{f(t)\,dt}-\int\limits_{x_n+\frac{1}{n}}^{x+\frac{1}{n}}{f(t)\,dt}$
Then, using the fact that $f(t)
But... uhhh... ? Any help you can give is appreciated
EDIT: What I want to end up with is $|f_n(x_n)-f_n(x)|<\epsilon$, but I don't see how to get there from what I have. Or can I get that from what I have? Am I on the right path?