Abel-Poisson summation of Fourier series

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Suppose f∈L1(∂D),z∈∂Df \in L^{1}(\partial \mathbf{D}), z \in \partial \mathbf{D}, and ff is continuous at zz. Prove that

lim⁡r↑1(Prf)(z)=f(z) \lim _{r \uparrow 1}\left(\mathcal{P}_{r} f\right)(z) = f(z)
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Remark.
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The result in this exercise differs from 11.18 because here we are assuming continuity only at a single point and we are not even assuming that ff is bounded, as compared to 11.18, which assumed continuity at all points of ∂D\partial \mathrm{D}.

Find a formula (in closed form, not as an infinite sum) for Prf\mathcal{P}_{r} f, where ff is the function in the second bullet point of Example 11.8.

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Remark.
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Example 11.8 (second bullet point): define f:∂D→Rf: \partial D \rightarrow \mathbb {R} by f(z)=1∣3−z∣2f(z) = \frac{1}{\left|3-z\right|^{2}}. Then for z∈∂Dz \in \partial D, f(z)=18∑n=−∞∞zn3∣n∣f(z) = \frac{1}{8} \sum_{n=-\infty }^{\infty } \frac{z^{n}}{3^{\left|n\right|}}, and the Fourier coefficients are f^(n)=18⋅13∣n∣\widehat{f}(n) = \frac{1}{8} \cdot \frac{1}{3^{\left|n\right|}} for all n∈Zn \in \mathbb {Z}.

Define f:∂D→Rf: \partial \mathbf{D} \rightarrow \mathbb {R} by

f(z)={1 if Im⁡z>0−1 if Im⁡z<00 if Im⁡z=0 f(z) = \begin{cases} 1 & \text{ if } \operatorname {Im} z > 0 \\ -1 & \text{ if } \operatorname {Im} z < 0 \\ 0 & \text{ if } \operatorname {Im} z = 0\end{cases}
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(a)

Show that if n∈Zn \in \mathbf{Z}, then

f^(n)={−2inπ if n is odd 0 if n is even  \widehat{f}(n) = \begin{cases} -\frac{2 i}{n \pi } & \text{ if } n \text{ is odd } \\ 0 & \text{ if } n \text{ is even }\end{cases}
(b)

Show that

(Prf)(z)=2πarctan⁡2rIm⁡z1−r2 \left(\mathcal{P}_{r} f\right)(z) = \frac{2}{\pi } \arctan \frac{2 r \operatorname {Im} z}{1-r^{2}}

for every r∈[0,1)r \in [0,1) and every z∈∂Dz \in \partial \mathbf{D}.

(c)

Verify that lim⁡r↑1(Prf)(z)=f(z)\lim_{r \uparrow 1}\left(\mathcal{P}_{r} f\right)(z) = f(z) for every z∈∂Dz \in \partial \mathbf{D}.

(d)

Prove that Prf\mathcal{P}_{r} f does not converge uniformly to ff on ∂D\partial \mathbf{D} as r↑1r \uparrow 1.

Prove that for the function s(x;r)s(x ; r) (see (21)) the formula holds

s(x;r)=12π∫−ππf(t)1−r21−2rcos⁡(t−x)+r2dt s(x ; r) = \frac{1}{2 \pi } \int _{-\pi }^{\pi } f(t) \frac{1-r^{2}}{1-2 r \cos (t-x)+r^{2}} d t

(Poisson).

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Prove that if a 2π2 \pi-periodic function ff, absolutely integrable over the period, has left- and right-hand limits f(x±0)f(x \pm 0) at the point xx, then

lim⁡r→1−0s(x;r)=lim⁡r→1−012π∫−ππf(t)1−r21−2rcos⁡(t−x)+r2dt==f(x+0)+f(x−0)2 \begin{aligned} & \lim _{r \rightarrow 1-0} s(x ; r) = \lim _{r \rightarrow 1-0} \frac{1}{2 \pi } \int _{-\pi }^{\pi } f(t) \frac{1-r^{2}}{1-2 r \cos (t-x)+r^{2}} d t = \\ & = \frac{f(x+0)+f(x-0)}{2} \end{aligned}

i.e., the Fourier series of the function ff is summable by the Poisson-Abel method to the value f(x+0)+f(x−0)2\frac{f(x+0)+f(x-0)}{2}.

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Prove that if ff is a 2π2 \pi-periodic, continuous function, then s(x;r)s(x ; r) tends uniformly, on the entire real line, to the function ff as r→1−0r \rightarrow 1-0.

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References

[Axl19] Axler S. Measure, Integration & Real Analysis. 3rd ed., Springer, 2019.
[ККЧ+03] Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. Интегралы. Ряды. 2-е изд., ФИЗМАТЛИТ, 2003.