$f_n:[0,1]\to [0,1]$ continuous, $f_n\to f$ uniformly, prove: ${\frac1n}\sum_{k=1}^{n}{f_k} \to f$ uniformly I was able to prove, hopefully correctly, that the sum converges uniformly. But, I'm not sure how to show it converges uniformly specifically to $f$.
The way I proved it converges uniformly was by using the Dirichlet convergence test, by taking $a_k(x) = \frac1n$ and $b_k(x) = f_k(x)$. $\sum_{k=1}^n{b_k(x)}$ is uniformly bounded on $[0,1]$ by $n+2$ for instance, and $a_n$ is monotone and uniformly convergent to 0 because it does not depend on the value of $x$.
I got rather stuck after this. I guess I have to use the continuity of the functions and the uniform convergence of both, but I can't figure out how to show that they converge to the same thing. Any help will be greatly appreciated.
 A: Write
$$\left| \frac{1}{n} \sum_{k=1}^n f_k(x) - f(x) \right| \leq  \frac{1}{n} \sum_{k=1}^m |f_k(x)-f(x)| + \frac{1}{n} \sum_{k=m+1}^n |f_k(x)-f(x)| \\ \leq \frac{m}{n} 2M+\frac{n-m}{n} \sup_{k > m} |f_k(x)-f(x)|,$$
where $M$ is a uniform bound on $f_k$ (since all are continuous and converge uniformly).
Take the limsup as $n \to \infty$. Then let $m \to \infty$ to show that the limsup is 0, and hence the limit exists. Since the bounds can be put independent of $x$, we can easily tweak this, by picking $n$'s and $m$'s appropriately, to show that the convergence is uniform.
A: Let $||g||$ denote the $\sup$ norm of $g$ over $[0,1]$
Your goal is to prove that $\displaystyle \left|\left|{\frac1n}\sum_{k=1}^{n}{f_k}-f \right|\right|\to 0$
Note that the LHS is lesser than $\displaystyle \frac1n \sum_{k=1}^n ||f-f_k||$
But since $f_n\to f$ uniformly, the sequence $||f-f_n|| \to 0$
By Cesaro, $\displaystyle \frac1n \sum_{k=1}^n ||f-f_k|| \to 0$
Hence $\displaystyle \left|\left|{\frac1n}\sum_{k=1}^{n}{f_k}-f \right|\right|\to 0$

If you absolutely need to avoid naming Cesaro, just mimic the proof you have of it .
