Let \(f:\mathbb R\to\mathbb R\) be continuous and define \[ f_n(x)=\frac1n\sum_{k=0}^{n-1}f\left(x+\frac{k}{n}\right). \] Prove that \((f_n)\) converges uniformly to a limit on every finite interval \([a,b]\).
Solution
Fix a finite interval \([a,b]\), and define \[ F(x)\coloneqq\int_0^1 f(x+t)\,dt. \]
1. The function \(f\) is uniformly continuous on \([a,b+1]\).
The interval \([a,b+1]\) is compact and \(f\) is continuous on \(\mathbb R\). Therefore the Heine–Cantor theorem applies.
2. Define the modulus\[ \omega(\delta) \coloneqq \sup\left\{ \abs{f(u)-f(v)}: u,v\in[a,b+1], \abs{u-v}\leq\delta \right\}. \]Then\[ \omega(\delta)\longrightarrow0 \qquad(\delta\downarrow0). \]
This is exactly the uniform continuity from step 1, written in terms of the modulus of continuity.
3. For every \(x\in[a,b]\),\[ f_n(x) = \sum_{k=0}^{n-1} \int_{k/n}^{(k+1)/n} f\left(x+\frac{k}{n}\right)\,dt. \]
Each interval of integration has length \(1/n\), so its \(k\)th term is \[ \frac1n f\left(x+\frac{k}{n}\right). \] Summing gives the definition of \(f_n(x)\).
4. For every \(x\in[a,b]\),\[ \abs{f_n(x)-F(x)} \leq \omega\left(\frac1n\right). \]
By step 3 and by splitting the integral defining \(F\) over the same partition, \[ \begin{aligned} \abs{f_n(x)-F(x)} &= \left| \sum_{k=0}^{n-1} \int_{k/n}^{(k+1)/n} \left[ f\left(x+\frac{k}{n}\right)-f(x+t) \right]dt \right|\\ &\leq \sum_{k=0}^{n-1} \int_{k/n}^{(k+1)/n} \left| f\left(x+\frac{k}{n}\right)-f(x+t) \right|dt. \end{aligned} \] If \(t\in[k/n,(k+1)/n]\), then \[ \left| \left(x+\frac{k}{n}\right)-(x+t) \right| \leq \frac1n. \] Moreover both arguments lie in \([a,b+1]\). Thus every integrand is at most \(\omega(1/n)\), and the total length of all subintervals is \(1\). This yields the claimed bound.
5. The sequence \((f_n)\) converges uniformly to \(F\) on \([a,b]\).
Taking the supremum in step 4 gives \[ \sup_{x\in[a,b]}\abs{f_n(x)-F(x)} \leq \omega\left(\frac1n\right). \] By step 2, the right-hand side tends to \(0\).
6. Therefore, on every finite interval,\[ \boxed{ f_n(x)\longrightarrow \int_0^1f(x+t)\,dt } \]uniformly.
The interval \([a,b]\) was arbitrary, and step 5 proves uniform convergence on it.
7. Q.E.D.
Step 6 is the required conclusion.