Let $f\in C (\mathbb{R})$ and $f_n=\frac{1}{n}\sum\limits_{k=0}^{n-1} f(x+\frac{k}{n})$. Prove that $f_n$ converges uniformly on every finite interval. 

Let $f\in C (\mathbb{R})$ and $f_n=\frac{1}{n}\sum\limits_{k=0}^{n-1} f(x+\frac{k}{n})$. Prove that $f_n$ converges uniformly on every finite interval.

Answer

Note that $f_n(x)$ is a Riemann sum and converges to $\int ^{x+1} _{x} f(t) dt$ as $n ? ?$. Let (a, b) be a finite interval. Then
\[ f_n(x) ? \int ^{x+1} _{x} f(t) dt= \sum \limits ^{n?1} _{k=0} \int ^{x+(k+1)/n}_{ x+k/n} (f(x + k/n) ? f(t)) dt. \]
Since f is uniformly continuous on $[a, b + 1]$ we have that $|f(x + k/n)?f(t)| < \epsilon$ for all $x ? [a, b], x + k/n ? t ? x + (k + 1)/n, 0 ? k ? n ? 1$, and for all $n$ sufficiently large. Therefore
\[\sup_{x \in [a,b]}|f_n(x)-\int ^{x+1} _{x} f(t) dt| <\sum \limits ^{n?1} _{k=0} \int ^{x+(k+1)/n}_{ x+k/n}\epsilon=n \times \epsilon/n=\epsilon. \]

Hence $f_n$ converges uniformly on $[a,b]$ to $\int ^{x+1} _{x} f(t) dt$. 

The answer is accepted.