# Show Sequence of Functions Converges Uniformly

I'm working on a Real Analysis book and am stuck on the following problem:

Let $$f$$ be continuous on $$\mathbb R$$ and let $$f_n(x)=\frac1n\sum_{k=0}^{n-1}f\left(x+\frac kn\right).$$ Prove that $$f_n(x)$$ converges uniformly to a limit on every finite interval $$[a,b]$$.

How would I rigorously solve this?

The candidate for the limit is $$g(x) := \displaystyle{\int_x^{x+1}} f(t)dt$$.
For any $$x$$, you can bound the error between the integral and its approximation with the Riemann sums in $$f_n$$:
\begin{align*} |g(x)-f_n(x)| & = \Big| \sum \limits_{k=0}^{n-1} \displaystyle{\int_{x+\frac{k}{n}}^{x+\frac{k+1}{n}}} f(t) - f\big(x+\frac{k}{n}\big)dt\Big| \\ & \le \sum \limits_{k=0}^{n-1} \displaystyle{\int_{x+\frac{k}{n}}^{x+\frac{k+1}{n}}} \big|f(t) - f\big(x+\frac{k}{n}\big)\big|dt \end{align*}
$$f$$ is continuous on $$[a,b]$$, so it is uniformly continuous on that interval. For $$\varepsilon>0$$, we know there exists $$\eta>0$$ such that $$x,y \in [a,b], |x-y|<\eta \implies |f(x)-f(y)| \le \varepsilon$$. Now if you take $$n$$ large enough to have $$\frac{1}{n} \le \eta$$, for $$k \in [\![0,n-1]\!]$$ and $$t \in \big[x+\frac{k}{n}, x+\frac{k+1}{n}\big]$$, $$\big|t-x-\frac{k}{n}\big| \le \eta$$ so $$\big|f(t)-f\big(x+\frac{k}{n}\big)\big| \le \varepsilon$$.
Hence $$|g(x)-f_n(x)| \le \sum \limits_{k=0}^{n-1} \displaystyle{\int_{x+\frac{k}{n}}^{x+\frac{k+1}{n}}} \varepsilon dt = \varepsilon$$, for all $$x \in [a,b]$$, for $$n$$ large enough.
• Ok, I figured out $g$ pretty soon after asking because I realized that the sum was pretty similar to a Riemann integrable, but I had no idea how the proof would work, thank you very much! – Alexander Nov 24 '20 at 7:24