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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:

$S_{n}(f)$ 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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Let me write $t_k(x) = \hat{f}(k)\,e^{ikx}$ so that $s_n = \sum_{k=-n}^{n} t_k$. Notice that $$\begin{align*} s_n = & & & s_n \\ s_{n+1}= & & t_{-(n+1)} + & s_n + t_{n+1} \\ s_{n+2}= & & t_{-(n+2)} + t_{-(n+1)} + & s_n + t_{n+1} + t_{n+2}\\ s_{n+3}= & & t_{-(n+3)} + t_{-(n+2)} + t_{-(n+1)} + & s_n + t_{n+1} + t_{n+2}+ t_{n+3} \\ \vdots & & & \vdots \end{align*}$$ and collect the terms vertically. You didn't really specify what your notation means, please define the symbols, while widespread $S_n$ and $\sigma_m$ are not completely standard for the Fourier and the Fejér sums. – t.b. Oct 3 '11 at 0:16
My apologies. $S_{n}(f)$ is the truncated Fourier series: $\Sigma_{j=-n}^{n}\widehat{f}(j)e^{ijt}$ and $\sigma_{n}(f)$ is the Cesaro mean of $f$ which is given by $\Sigma_{j=0}^{n-1}S_{j}(f)$ – roo Oct 3 '11 at 0:33
I understood it this way, more precisely the Cesàro means of the truncated Fourier series $\sigma_m = \frac{1}{m} \sum_{n=0}^{m-1} s_n$, no? (that's what I call Fejér sums). But you need not worry about that too much anyway, since you're only interested in the second half of this claim which you should easily obtain by looking at the triangle above. If you work out the sum $s_n + s_{n+1} + s_{n+2} + s_{n+3}$ in the triangle I displayed above, the pattern should emerge. – t.b. Oct 3 '11 at 0:38
Your triangle explains it. Thank you. – roo Oct 3 '11 at 1:27

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