Convergence of a sequence of functions defined as integrals 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 $x_n<x$ and forget the $n/2$):
$$\int\limits_{x_n-\frac{1}{n}}^{x+\frac{1}{n}}{f(t)\,dt}-\int\limits_{x_n+\frac{1}{n}}^{x+\frac{1}{n}}{f(t)\,dt}$$
Then, using the fact that $f(t)<f(x)+\epsilon$ over each of these intervals, everything whittles down to being $$\le n\cdot |f(x)+\epsilon|\cdot (x-x_n)\,.$$
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? 
 A: Ok, this is very confusing for me to read.  One big problem is that you use the index $n$ to refer to two different things simultaneously.  You use it both for the sequence $f_n$ and for the sequence $x_n$.  With this small change, everything you wrote is correct, and what follows is not much different:
Now, the idea is we fix $k$ and look at $f_k$ and we want to show that $f_k$ is continuous.  As you wrote, 
$$|f_k(x_n)-f_k(x)|=\left|\frac{k}{2}\int\limits_{x_n-\frac{1}{k}}^{x_n+\frac{1}{k}}{f(t)\,dt}-\frac{k}{2}\int\limits_{x-\frac{1}{k}}^{x+\frac{1}{k}}{f(t)\,dt}\right|.$$  The key idea is that if $k$ is fixed, we can choose $\delta $ really really small so that these integrals almost line up entirely and cancel out.  So, assume $x_n<x$ and $|x-x_n|<\delta$ and say $\delta <\frac{1}{k}$.  Then 
$$\left|\frac{k}{2}\int\limits_{x_n-\frac{1}{k}}^{x_n+\frac{1}{k}}{f(t)\,dt}-\frac{k}{2}\int\limits_{x-\frac{1}{k}}^{x+\frac{1}{k}}{f(t)\,dt}\right|= 
\left|\frac{k}{2}\int\limits_{x_n-\frac{1}{k}}^{x-\frac{1}{k}}{f(t)\,dt}+\frac{k}{2}\int\limits_{x_n+\frac{1}{k}}^{x+\frac{1}{k}}{f(t)\,dt}\right|.$$   But these two integrals are over an interval of length $\delta$.  Since $f$ must be bounded on $[-1,2]$ we can choose $M$ such that $M>|f(x)|$ for all $x$.  Then the above term is strictly less than
$$\frac{2\delta kM}{2}=kM\delta. $$   Since both $k$ and $M$ are fixed constants, we can choose $\delta$ small enough so that this is less than a given $\epsilon.$
Hope that helps,
Short Answer: 
Not sure why, but I rewrote the whole proof above.  Here is the short answer to your question:
You had $n |x-x_n| |f(x)+\epsilon|$ as an upper bound.  Replacing the $n$ with a $k$, to distinguish the two sequences, using an upper bound for $f(x)$ on the whole interval and using the fact that $|x-x_n|<\delta$ this upper bound becomes $$k\delta M.$$   Then as $k,M$ are fixed, choose $\delta =\epsilon/kM$ and it is finished.
