Littlewood's Tauberian theorem: Let $a_n=O(1/n)$. (In particular, given any $0<r<1$, the power series $\sum a_nr^n$ converges.) If the function defined by the power series $$f(r)=\sum a_nr^n$$ has a limit when $r$ tends to $1$ from below, then the power series also converges at $r=1$, and $f$ is continuous at $r=1$. That is, the series $\sum a_n$ is convergent, and its sum is $$\sum a_n = \lim_{r\nearrow 1} \sum a_nr^n\mbox{.}$$
I'm reading Stein & Shakarchi's Fourier Analysis. Given the above theorem, it is an exercise to prove the following:
(Problem 3iii, on chapter 2) If $f:\mathbb R\to\mathbb R$ is $2\pi$-periodic, continuous, and satisfies $\hat f(n)=O(1/n)$, then the Fourier series of $f$ converges uniformly to $f$.
Well, using:
- the fact that the Abel means of $f$ are given by the convolution of $f$ with the Poisson kernel $$P_r(x)=\sum_{n\in\mathbb Z}r^{|n|}e^{inx};$$
- the fact that the Poisson kernels form an approximation to the identity; and
- the continuity of $f$,
I can show that the Abel means $$P_r * f$$ converge uniformly to $f$ when $r\nearrow 1$. Therefore, the Abel means also converge pointwise, and then using Littlewood's theorem, I can show that the Fourier series of $f$ also converges pointwise. The problem is that I "lost" the information about the uniform convergence of the Abel means, since Littlewood's theorem is only concerned about the pointwise convergence of the series for each $x$, independently.
I know that the result in Problem 3iii, as stated above, does hold (that is, the exercise is not "mistaken" in any way). However, I can't see how this result could follow from Littlewood's theorem (I believe it must be a simple argument, but I can't see it). Any tips?