Abel–Poisson summation of Fourier series

Remarks

If $f$ is a Riemann integrable $2\pi$-periodic function and $c_k = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t) e^{-ikt} \, dt$ are its Fourier coefficients, then the $c_k$ are bounded, $A_r(x)$ converges absolutely, and $$A_r(x) = \frac{1}{2\pi} \int_{-\pi}^{\pi} P_r(x - t) f(t) \, dt, \qquad P_r(\theta) = \sum_{k \in \Z} r^{\abs{k}} e^{ik\theta} = \frac{1 - r^2}{1 - 2r\cos\theta + r^2}.$$ Here $P_r$ is the Poisson kernel of the unit disc.

Example: for $f(x) = \cos x$ one has $c_1 = c_{-1} = \frac12$ and $c_k = 0$ otherwise, so $A_r(x) = r \cos x \to \cos x$.

Non-example: for $c_k = 1$ for all $k$ and $x = 0$, one gets $A_r(0) = \sum_{k \in \Z} r^{\abs{k}} = \frac{1 + r}{1 - r} \to +\infty$. So the series $\sum_{k \in \Z} e^{ikx}$ is not Abel–Poisson summable at $0$.