Fejer and Poisson Summability of Fourier Series - Examples

The Fourier coefficients of $e_k$ vanish except at $k$, where the coefficient is $1$. Hence $S_j{e}_k=\{\begin{array}{ll}0,& j<\mathrm{\mid }k\mathrm{\mid },\\ e_k,& j\ge \mathrm{\mid }k\mathrm{\mid }\mathrm{.}\end{array}$

If $\mathrm{\mid }k\mathrm{\mid }>N$, then every term in the Cesaro average from [L1] is $0$, so ${\sigma }_N{e}_k=0$. If $\mathrm{\mid }k\mathrm{\mid }\le N$, then exactly $N+1-\mathrm{\mid }k\mathrm{\mid }$ of the terms equal $e_k$, and therefore ${\sigma }_N{e}_k=\frac{N+1-\mathrm{\mid }k\mathrm{\mid }}{N+1}{e}_k=(1-\frac{\mathrm{\mid }k\mathrm{\mid }}{N+1})e_k\mathrm{.}$ This is the same as the displayed $(\cdot {)}_{+}$ formula, and for $N=0$ it gives ${\sigma }_0{e}_0=e_0$ and ${\sigma }_0{e}_k=0$ for $k\mathrm{\ne }0$.

The Fourier coefficients of $e_k$ vanish except at $k$, where the coefficient is $1$. Substituting into [L1] leaves only one term: $A_r{e}_k(x)=r^{\mathrm{\mid }k\mathrm{\mid }}{e}_k(x)\mathrm{.}$

Since step 1.1 holds for every $x$, it is exactly the asserted identity $A_r{e}_k=r^{\mathrm{\mid }k\mathrm{\mid }}{e}_k$.

Direct integration gives $\widehat{q}(2m)=0$ for every $m$ and $\widehat{q}(2m+1)=\frac{2}{\pi i(2m+1)},\widehat{q}(-(2m+1))=-\frac{2}{\pi i(2m+1)}\mathrm{.}$ Hence $S_{j}q(x)=\frac{4}{\pi }{\sum }_{0\le m,2m+1\le j}\frac{\sin (2\pi (2m+1)x)}{2m+1}\mathrm{.}$

Averaging the partial sums from step 1.1 as in [L1] shows that the odd mode $2m+1$ appears with weight $1-(2m+1)\mathrm{/}(N+1)$ when $2m+1\le N$ and with weight $0$ otherwise. This is exactly the displayed formula for ${\sigma }_{N}q$.

Every term of the finite sine sum in step 2.1 vanishes at $x=0$, so ${\sigma }_{N}q(0)=0$ for every $N$. Also [L1] and [L2] give ${\sigma }_{N}q(x)={\int }_0^{1}q(x-t)F_N(t)dt,$ where the integral is Riemann because $q$ is a step function. Since $\mathrm{\mid }q\mathrm{\mid }\le 1$, positivity and total mass one imply $\mathrm{\mid }{\sigma }_{N}q(x)\mathrm{\mid }\le 1$ for every $x$. Thus the Fejer means average across the jump and do not exhibit a fixed overshoot.

For each $N$, the function $S_{j}f$ is a trigonometric polynomial for every $0\le j\le N$, so [L1] makes ${\sigma }_{N}f$ a trigonometric polynomial as well. In particular, every ${\sigma }_{N}f$ is continuous.

If ${\sigma }_{N}f$ converged uniformly to $f$, then the uniform limit of the continuous functions ${\sigma }_{N}f$ would be continuous. But the chosen $f$ has a jump at $0$, so it is discontinuous. Therefore uniform convergence to $f$ is impossible. This single step function refutes the universal statement.

The ordinary partial sums are $1,0,1,0,\dots ,$ so they do not converge.

For $0\le r<1$, [F1] makes the associated Abel sum $1-r+r^2-r^3+\cdots =\frac{1}{1+r}\mathrm{.}$ As $r\uparrow 1$, this tends to $1\mathrm{/}2$. Thus the series is Abel summable but not ordinarily convergent, refuting the statement.