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.

  • 2
    $\begingroup$ 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! $\endgroup$ – Alexander Nov 24 '20 at 7:24

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