Abel means of function $f$ at a jump discontinuity Fourier Analysis An Introduction Chapter 2, Exercise 17 (a):
Abel means of $f$ converge to $f$ whenever $f$ is continuous at $\theta$:
$$\lim_{r\to1}A_{r}(f)(\theta)=\lim_{r\to1}(P_{r}*f)(\theta)=f(\theta),\mbox { with}\ 0 < r< 1.$$
An integrable function is said to have a jump discontinuity at $\theta$ if the two limits 
$$\lim_{h\to0\\h>0}f(\theta+h)=f(\theta^{+})$$ and $$\lim_{h\to0\\h>0}f(\theta-h)=f(\theta^{-})$$
exist.
Prove that if $f$ has a jump discontinuity at $\theta$, then
$$\lim_{r\to1}A_{r}(f)(\theta)=\frac{f(\theta^{+})+f(\theta^{-})}{2},\mbox{ with } 0\le r< 1.$$
[Hint: $\frac{1}{2\pi}\int_{-\pi}^{0}P_{r}(\theta)d\theta = \frac{1}{2\pi}\int_{0}^{\pi}P_{r}(\theta)d\theta=\frac{1}{2}.$]
$$P_{r}(\theta)=\sum_{n=-\infty}^{\infty}r^{|n|}e^{in\theta}=\frac{1-r^{2}}{1-2r\cos\theta+r^{2}},\ with\ 0\le r<1.$$
$$A_{r}(f)(\theta)=\sum_{n=-\infty}^{\infty}r^{|n|}a_{n}e^{in\theta}$$
$$=\sum_{n=-\infty}^{\infty}r^{|n|}(\frac{1}{2\pi}\int_{-\pi}^{\pi}f(\varphi)e^{-in(\varphi-\theta)}d\varphi)e^{in\theta}$$
$$=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(\varphi)(\sum_{n=-\infty}^{\infty}r^{|n|}e^{-in(\varphi-\theta)})d\varphi$$
I tried to prove when $f$ has a jump discontinuity at $\theta$, and I got this:
$$\lim_{r\to1}A_{r}(f)(\theta)=\lim_{r\to1}(\lim_{w\to\theta^{-}}(\frac{1}{2\pi}\int_{-\pi}^{w}f(\varphi)(\sum_{n=-\infty}^{\infty}r^{|n|}e^{-in(\varphi-\theta)})d\varphi)+\lim_{w\to\theta^{+}}(\frac{1}{2\pi}\int_{w}^{\pi}f(\varphi)(\sum_{n=-\infty}^{\infty}r^{|n|}e^{-in(\varphi-\theta)})d\varphi))$$
$$=\lim_{r\to1}(\lim_{w\to0^{-}}(\frac{1}{2\pi}\int_{-\pi}^{w}f(\varphi-\theta)(\sum_{n=-\infty}^{\infty}r^{|n|}e^{-in\varphi})d\varphi)+\lim_{w\to0^{+}}(\frac{1}{2\pi}\int_{w}^{\pi}f(\varphi-\theta)(\sum_{n=-\infty}^{\infty}r^{|n|}e^{-in\varphi})d\varphi))$$
But I still don't know how to prove the formula.
 A: This question is from Stein–Shakarchi and they give the following hint:

Hint: Explain why $\frac{1}{2\pi} \int_{-\pi}^0 P_r(\theta)\,d\theta$
  = $\frac{1}{2\pi}\int_0^{\pi} P_r(\theta) = \frac{1}{2}$, then modify the
  proof given in the text.

The proof they're referring to is that of Theorem 4.1 in chapter 2:

Theorem 4.1. Let $\{K_n\}_{n=1}^{\infty}$ be a family of good kernels, 
  and $f$ an integrable function on the circle. 
  Then $$\lim_{n \to \infty} (f * K_n)(x) = f(x)$$ whenever
  $f$ is continuous at $x$. If $f$ is continuous everywhere,
  then the above limit is uniform.

In particular, since $P_r(\theta)$ is even and nonnegative with $\frac{1}{2\pi}\int_{-\pi}^{\pi} P_r(\theta) = 1$, we have the identity given in the hint. Then
\begin{align}
A_r(f)(\theta) - \frac{f(\theta^+) + f(\theta^-)}{2} 
&= \frac{1}{2\pi}\int_{-\pi}^{\pi} f(\theta-t)P_r(t)\,dt 
- \frac{f(\theta^+) + f(\theta^-)}{2} \\
&= \frac{1}{2\pi}\int_{-\pi}^0 f(\theta - t)P_r(t)\,dt
+ \frac{1}{2\pi}\int_0^{\pi} f(\theta - t)P_r(t)\,dt
- \frac{f(\theta^+) + f(\theta^-)}{2} \\
&= \frac{1}{2\pi}\int_0^{\pi} f(\theta + t)P_r(t)\,dt
+ \frac{1}{2\pi}\int_0^{\pi} f(\theta - t)P_r(t)\,dt
- \frac{f(\theta^+) + f(\theta^-)}{2} \\
&= \frac{1}{2\pi}\int_0^{\pi}[f(\theta + t) - f(\theta^+)]P_r(t)\,dt
+ \frac{1}{2\pi}\int_0^{\pi}[f(\theta-t)-f(\theta^-)]P_r(t)\,dt.
\end{align}
From here you can continue similarly to the proof in the text,
using the left- and right-continuity of $f$ at $\theta$ to 
bound the integrals near $0$ and using the properties of 
good kernels to bound them away from $0$.
