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.
Best Answer
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.