# Fourier coefficients of Cesaro means of partial fourier series

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.

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 $$\widehat{\sigma_N f}(k)=\frac1N\sum_{n=|k|}^N\hat f(k)=\frac{N-|k|}{N}\hat f(k).$$
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 $$\frac{(N-k)}{N}$$