Lemma : Suppose that the Abel means $\displaystyle\sum_{n=1}^{\infty} r^{n}c_{n}$ of the series $\displaystyle\sum_{n=1}^{\infty}c_{n}$ are bounded as $r\rightarrow 1$ for the left. If $c_{n}=O(\frac{1}{n})$ then the partial sum $\displaystyle\sum_{n=1}^{N}c_{n}$ is bounded .
Let $f(\theta)=i(\pi-\theta)$ be odd in $\theta\in(0,\pi)$ and the fourier series of $f$ is $\displaystyle\sum_{n\ne0}\frac{e^{in\theta}}{n}$
Now , apply the lemma to the series $\displaystyle\sum_{n\ne0}\frac{e^{in\theta}}{n}$ and $c_{n}=\frac{1}{n} (e^{in\theta}-e^{-in\theta})$ for $n\ne0$ , so clearly , $c_{n}=O(1/|n|)$ . Finally the Abel means of this series are $(f\ast P_{r})(\theta).$ But $f$ is bounded and $P_{r}$ is a good kernel , so $S_{N}(f)(\theta)$ is uniformly bounded in $N$ and $\theta$
,where $P_{r}$ is the poisson kernel and $S_{N}(f)(\theta) $ is denoted by the $N^{th}$ partial sum of the fourier series of the given $f$ . This content can be found here , page 84~87 of the book .
Q: How come we can conclude that $S_{N}(f)(\theta)$ is uniformly bounded in $N$ and $\theta$ .
By odd extension of the given $f$ , we see that $f$ is continuous at $\theta\in[-\pi,\pi]-\{0\}$ and then $\displaystyle\lim_{ r\rightarrow 1^{-}}(f\ast P_{r})(\theta)=f(\theta)$ for $\theta\in[-\pi,\pi]-\{0\}$ .
At $\theta=0$ is a jump discontinuity of $f$ and $\displaystyle\lim_{ r\rightarrow 1^{-}}(f\ast P_{r})(0)=\frac{f(0^{+})+f(0^{-})}{2}$.
So , now , I think we just can conclude that $$\lim_{r\rightarrow 1^{-}}(f\ast P_{r})(\theta)$$ is bounded for $\theta \in [-\pi,\pi]$, thus , by lemma , we have $S_{N}(f)(\theta)$ is bounded for all $\theta\in [-\pi,\pi]$. But I think I do nothing for the uniformness .
Any comment or suggestion I will be grateful . Thanks for considering my request .
The proof of Abel's theorem actually proves an a priori stronger quantitative version: