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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.

xf16
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2 Answers2

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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

$$|a_nr^{|n|}e^{in\theta}|\le \left (\frac{1}{2\pi}\int_0^{2\pi} |f|\right) \cdot r^{|n|}.$$

zhw.
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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.

Daniele Tampieri
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  • Hi Daniele, we only know $a_n$ is uniformly bounded, say $a_n \leq B\ \forall n$, and $e^{in\theta}$ is certainly $\leq 1$, but how did you conclude the limsup is $\leq 1$. I was worried about the $B$ part. Also, for the later part to use Cauchy-Hadamard theorem, why do we need $r < \rho$, we just need $r < R$, right? So as long as we have $r < 1$, it is fine? – xf16 Feb 03 '19 at 19:57
  • I modified my answer to clarify why $\limsup$ is 1. The use of $\rho$ is just to remark that the convergence of $A_r(f)(\theta)$ *is uniform for every $r\in D(0,\rho)$ for every $\rho<1$: it just a matter of notation, and sometimes I am a bit pedantic ;). And we only need $r<R$ since in this case you can nevertheless find a $\rho$ such that $r\le \rho<1$ which assures uniform convergence if $r$ is in the disk $D(0,\rho)$. – Daniele Tampieri Feb 03 '19 at 20:14
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    Oh I think I understand what you mean. For each $r < 1$, we can find a open disk $D(0,\rho)$ such that $r<\rho<1$, so that $\sum_{n=1}^{\infty} c_n r^n$ is uniformly convergent. Is this what you're saying? Thanks again Daniele. – xf16 Feb 03 '19 at 20:22
  • Exactly: uniform convergence is always related to a set of values of the variable, not just one. – Daniele Tampieri Feb 03 '19 at 20:24
  • Thank you so much!! I understand much better now. – xf16 Feb 03 '19 at 20:26
  • You are welcome: I am glad of having been of some help. – Daniele Tampieri Feb 03 '19 at 20:27