Proving a sequence converges uniformly

Question

The problem is:

Let $f ∈ C(\mathbb R)$. Prove that the sequence $\{f_n\}^∞_{n=1}$ defined by

$f_n(x)=\frac{1}{n}\sum^{n-1}_{k=0}f(x+\frac{k}{n})$

converges uniformly on each finite interval $[a, b]$.

How would you prove this? I'm not that good at analysis so any hints or advice would be great. I was thinking maybe comparing sequences and using triangle inequality? However, the summation is really throwing me off.

Answer 1

This is a Riemann sum so $$f_n(x) \to g(x) = \int_{x}^{x+1} f(t)\;dt .$$

For showing uniform convergence, note that $$ |f_n(x)-g(x)| = \left|\sum_{k=0}^{n-1} \frac{1}{n}f(x+k/n)-\int_{x+k/n}^{x+(k+1)/n}f(t)\;dt\right|\\\le \frac{1}{n}\sum_{k=0}^{n-1} \left|f(x+k/n) -f(x+k/n+c_k^*)\right|$$ where for each $k,$ $c^*_k \in [0,1/n]$ (by the triangle inequality and mean value theorem). Since $f$ is continuous on $[a,b+1],$ it is uniformly continuous on this interval, so for sufficiently large $n,$ we have $$ |f_n(x)-g(x)| \le \frac{1}{n}\sum_{k=0}^{n-1} \epsilon = \epsilon$$ for all $x\in [a,b].$

Answer 2

This Riemann sum converges to $I(x) = \int_0^1 f(x +t) \, dt$.

Note that

$$\tag{*} |f_n(x) - I(x)| = \left| \frac{1}{n}\sum_{k=0}^{n-1}f(x + k/n) - \sum_{k=0}^{n-1}\int_{k/n}^{(k+1)/n} f(x+t) \, dt\right| \\ \leqslant \sum_{k=0}^{n-1}\int_{k/n}^{(k+1)/n}|f(x + t)- f(x+ k/n)| \, dt$$

When $x \in [a,b]$ we have $x+t, x + k/n \in [a,b+1]$ for all $n$ and $0 \leqslant k \leqslant n-1$.

Since $f$ is uniformly continuous on $[a,b+1]$, for any $\epsilon >0$ there exists $\delta(\epsilon) > 0$ such that $|f(t) - f(s)| < \epsilon$ if $|s - t| < \delta(\epsilon)$.

Can you finish by showing that the RHS of (*) can be made smaller than $\epsilon$ if $n > N$ where $N$ is a positive integer that depends only on $\epsilon$ and not on $x$ -- proving uniform convergence?

Hint: Choose $N < 1/\delta(\epsilon)$.