Abel Means of an Integrable Function is Uniformly Convergent 
I am studying Stein Shakarchi's book on Fourier Analysis. This page mentioned that the Abel means are absolutely convergent and uniformly convergent, just because $f$ is integrable. I did figure out why it is absolutely convergent, and also tried to use Weierstrass M-test for the uniform convergence but had no progress.
$a_n$ is the $n$-th fourier coefficient of $f$.
Can anyone share some insights? Much appreciated.
 A: It's uniformly convergent for $\theta \in [0,2\pi]$ for each $r,0\le r<1.$ That follows from the Weierstrass M test: For each $n$ we have 
$$|a_nr^{|n|}e^{in\theta}|\le \left (\frac{1}{2\pi}\int_0^{2\pi} |f|\right) \cdot r^{|n|}.$$
A: The result is a consequence of the Cauchy-Hadamard theorem which incidentally I just used today in order to answer to another question: precisely, since $a_n$ is uniformly bounded, say $|a_n|\le B$ for some real constant $B>0$, then 
$$
\begin{split}
\limsup_{n\to \infty} \sqrt[n]{c_n}&=\limsup_{n\to \infty} \sqrt[n]{\left|a_ne^{in\theta}+a_{-n}e^{-in\theta}\right|}\\
&\le\limsup_{n\to \infty} \sqrt[n]{B}\cdot\sqrt[n]{\left|e^{in\theta}+e^{-in\theta}\right|}\\
&\le\limsup_{n\to \infty} \sqrt[n]{2B}= 1
\end{split}
$$
thus the power series $\sum\limits_{n=0}^{\infty}c_n z^n$ has convergence radius $R\ge 1$, and by Cauchy-Hadamard converges absolutely and uniformly for each $0\le\rho<1$. Now for every $r$ such that $0\le r< \rho<1$, we have 
$$
|A_r(f)(\theta)|=\left|\sum\limits_{n=-\infty}^{\infty} r^{|n|} a_n e^{in\theta}\right|\le\sum\limits_{n=0}^\infty|c_n|r^n\le\sum\limits_{n=0}^\infty|c_n|\rho^n<\infty 
$$
thus the same conclusion holds for this real variable complex valued power series.
