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