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Let $S_nf(x) = \sum_{m = -n}^{n} \hat{f}(x)e^{2\pi imx}$ be the partial Fourier series of $f$. Let $\sigma_Nf(x) = \frac{1}{N}\sum_{n = 0}^{N-1} S_nf(x)$ be the Cesaro means of the $S_nf$.

What is $\widehat{\sigma_N f}(k)$? Some preliminary work has got me to

\begin{align*} \widehat{\sigma_Nf}(k) &= \int_{\mathbb{T}} (\sigma_Nf)(x)e^{-2\pi ikx}dx\\ &= \int_{\mathbb{T}} (K_N*f)(x)e^{-2\pi i kx}dx\\ &= \int_{\mathbb{T}^2} f(x-y)K_N(y)e^{-2\pi i kx}dydx\\ &= \int_{\mathbb{T}^2} f(t) e^{-2\pi ik(t+y)} K_N(y)dtdy \ \ \ \ \text{by the substitution $x - y = t$ and Fubini}\\ &= \hat{f}(k)\int_{\mathbb{T}} e^{-2\pi i ky}K_N(y)dy, \end{align*}

where $K_N(y) = \frac{1}{N} \sum_{n=0}^{N-1}D_n(y) = \frac{1}{N}\frac{\sin^2(\pi Ny)}{\sin^2(\pi y)}$ are the Fejer kernels and $D_n(y) = \sum_{l = -n}^n e^{2\pi ily} = \frac{\sin(\pi(2n+1)y)}{\sin(\pi y)}$ are the Dirichlet kernels. I am having trouble evaluating this last integral. It is certainly bounded for the same reason that the Fejer kernels (the $K_N$) are good kernels. However, Mathematica seems to be unable to integrate this and I don't see an obvious way to do it either.

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2 Answers 2

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You had a typo in the definition of the Cesaro mean: $n$ should sum from $n=0$ rather than $-N$.

$$\sigma_Nf(x)=\frac1N\sum_{n=0}^N\sum_{m=-n}^n\hat f(m)e^{2\pi imx}.$$ So \begin{align*} \widehat{\sigma_N f}(k)=& \int_{\mathbb{T}} (\sigma_Nf)(x)e^{-2\pi ikx}dx\\ & =\frac1N\sum_{n=0}^N\sum_{m=-n}^n\hat f(m)\int_{\mathbb T} e^{2\pi imx}e^{-2\pi ikx}\,dx\\ &=\frac1N\sum_{n=0}^N\sum_{m=-n}^n\hat f(m)\delta_{km}. \end{align*} Therefore, if $N<|k|$, then $\widehat{\sigma_N f}(k)=0$; if $N\geq|k|$, then $$\widehat{\sigma_N f}(k)=\frac1N\sum_{n=|k|}^N\hat f(k)=\frac{N-|k|}{N}\hat f(k).$$

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  • $\begingroup$ Ah yes, thanks for catching the typo. I'll fix it now. Also this solution is quite nice, thanks for the clear walk-through. $\endgroup$
    – PhysMath
    Apr 11, 2021 at 1:50
  • $\begingroup$ @PhysMath You’re welcome! $\endgroup$
    – Feng
    Apr 11, 2021 at 1:51
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All you have to do is to use definition of $\sigma_Nf(x) = \frac{1}{N}\sum_{n = -N}^N S_nf(x)$ and the fact that Fourier coefficients of $S_n$ are $1$ (or $0$ for $|k| > n$). One gets $\frac{(N-k)}{N}$

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