Another question from a theorem in my notes:
Let $f\in BV(\mathbb{T})$. Then for every $x\in\mathbb{R}$, $S_{n}(f)(x)\to \dfrac{f(x+0) + f(x-0)}{2}$ (and to $f(x)$ at every point $x$ where $f$ is continuous).
The proof starts like this:
Definitions:
$S_{n}(f)$ is the truncated Fourier series: $\sum\limits_{j=-n}^{n}\widehat{f}(j)e^{ijt}$
$\sigma_{n}(f)$ is the Cesàro mean of $f$ which is given by $\sum\limits_{j=0}^{n-1}S_{j}(f)$
Let $s_{n} := S_{n}(f)(x)$ and $\sigma_{n} := \sigma_{n}(f)(x)$. Then for $m> n$, we have
$m\sigma_{n} - n\sigma_{n} = s_{n} + ... + s_{m-1} = (m-n)s_{n} + \sum\limits_{n \leq |j| \leq m-1}(m - |j|)\widehat{f}(j)e^{ijx}$.
I can't verify the second half of this last claim. That is, how we got $(m-n)s_{n} + \sum\limits_{n \leq |j| \leq m-1}(m - |j|)\widehat{f}(j)e^{ijx}$.
Any advice?