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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?

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    Your triangle e$x$plains it. Thank you.2011-10-03

1 Answers 1

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There are a couple of minor misprints which may explain the subsequent question? For the Cesaro mean a factor of $1/n$ is missing and later on an index $n$ should be $m$. You have

$ \sigma_n(f) = \frac{1}{n} \sum_{j=0}^{n-1} S_j(f)$ and then since for $p\geq n : S_p = S_n + \sum_{n\leq |j|< p} \widehat{f}(j)$ we get: $ m\sigma_m - n\sigma_n = \sum_{p=0}^{m-1} S_p-\sum_{p=0}^{n-1} S_p= \sum_{p=n}^{m-1} S_p= (m-n) S_n + \sum_{p=n}^{m-1} \sum_{n\leq |j|< p} \widehat{f}_j e^{ijt} $ Interchanging sums in the last expression: $ m\sigma_m - n\sigma_n = (m-n) S_n + \sum_{n\leq |j|< m} (m-|j|)\widehat{f}(j) e^{ijt} $