$\sum_{j=1}^{n} f\left(\frac{j}{n}\right) \cdot \chi _{\left[\frac{j-1}{n},\frac{j}{n}\right)} \longrightarrow f$ converges uniformly

Question

Let $f:[0,1]\longrightarrow \mathbb{R}$ be a continuous function

and $\displaystyle f_n=\sum_{j=1}^{n} f\left(\frac{j}{n}\right) \cdot \chi _{\left[\frac{j-1}{n},\frac{j}{n}\right)}$

where $\chi _{\left[\frac{j-1}{n},\frac{j}{n}\right)}$ is the characteristic function of ${\left[\frac{j-1}{n},\frac{j}{n}\right)}$

How can we prove that $f_n \longrightarrow f $ uniformly ?

Any hints would be appreciated.

Answer 1

Sure. $f$ is uniformly continuous on $[0,1]$, so, given $\epsilon>0$, there is ... (text hidden)

$\delta>0$ so that whenever $|x-x'|<\delta$, we have $|f(x)-f(x')|<\epsilon$. Just pick $N$ so that $1/N<\delta$. Then it should follow that whenever $n\ge N$, we have $|f_n(x)-f(x)|<\epsilon$ for every $x\in [0,1]$. The point is that $f_n(x)=f(j/n)$ where $x\in [\frac{j-1}n,\frac jn)$, and $|x-j/n|<\delta$.