Suppose $f(x)$ is a non-negative function such that $\int_{\mathbb{R}}f(x) \ dx=1$. Define $p_{n,k}$ as follows: $$ p_{n,k}= \int_{\frac{k}{n}}^{\frac{k+1}{n}} f(x) \ dx $$
We know that $\displaystyle \sum_{k\in \mathbb{Z}}p_{n,k}=\sum_{k\in \mathbb{Z}}p_{n,k+1}=1$. If we make finer partitions, i.e making $n$ large can we say anything about the following limit: $$ \lim_{n\rightarrow \infty} \sum_{k \in \mathbb{Z}}|p_{n,k+1}-p_{n,k}| $$ I am trying to prove that it converges to zero. Can anyone help me with this ?
The result is not true. Let $f$ be such that $$ f(x)=\frac{1}{x^2}\quad\text{if}\quad x\ge1. $$ Then for $k\ge n$ $$ p_{n,k+1}-p_{n,k}=\int_0^{1/n}\Bigl[\Bigl(x+\frac{k+1}{n}\Bigr)^{-2}-\Bigl(x+\frac{k}{n}\Bigr)^{-2}\Bigr]\,dx=-\frac{2\,n}{k(k+1)(k+2)}. $$