Cesaro and Abel means of a Fourier series

The continuous kernel $P_r$ below is defined by its uniformly convergent series without a choice premise. For the means of an arbitrary integrable function $f$, assume the Axiom of Countable Choice, as in Period-one Fourier coefficients, partial sums, and convolution on the torus, and let $f$ be integrable on one period. For a Riemann-integrable one-periodic representative, the coefficient and kernel integrals in the formulas below may instead be ordinary Riemann integrals; these formulas then need no choice premise and agree with their Lebesgue versions when Countable Choice is assumed.

For $N\ge 0$, the Cesaro mean (or Fejer mean) of order $N$ of the Fourier series of $f$ is

${\sigma }_{N}f(x):=\frac{1}{N+1}{\sum }_{j=0}^N{S}_{j}f(x)\mathrm{.}$

Because Dirichlet and Fejer kernels defines $F_N(t)=\frac{1}{N+1}{\sum }_{j=0}^N{D}_j(t),$ and Period-one Fourier coefficients, partial sums, and convolution on the torus defines convolution and partial sums, one has

${\sigma }_{N}f=f\ast F_N\mathrm{.}$

For $0\le r<1$, the series

$P_r(t):={\sum }_{k\in \mathbb{Z}}{r}^{\mathrm{\mid }k\mathrm{\mid }}{e}_k(t)$

converges absolutely and uniformly on $\mathbb{T}$, because ${\sum }_{k\in \mathbb{Z}}\mid r^{\mathrm{\mid }k\mathrm{\mid }}{e}_k(t)\mid ={\sum }_{k\in \mathbb{Z}}{r}^{\mathrm{\mid }k\mathrm{\mid }}<\mathrm{\infty }\mathrm{.}$ The Abel mean of the Fourier series of $f$ is

$A_{r}f(x):={\sum }_{k\in \mathbb{Z}}{r}^{\mathrm{\mid }k\mathrm{\mid }}\widehat{f}(k)e_k(x)\mathrm{.}$

Because $\mathrm{\mid }\widehat{f}(k)\mathrm{\mid }\le {\int }_0^1\mathrm{\mid }f(t)\mathrm{\mid }dt$, this series also converges absolutely and uniformly in $x$. Therefore termwise integration against the absolutely convergent kernel series gives

$A_{r}f=f\ast P_r\mathrm{.}$