In the text "Fourier Analysis and Related Topics", i'm having trouble proving the following Theorem in $(3.5.5)$ utilizing Fourier Methods/Summability Methods. Also i'm not sure how to approach $(ii)$ with the use of Summability methods may I have some hints please ?
$$\text{Theorem} \, (3.35)$$:
$(i)$ For $f \in C_{2 \pi}$, the Abel means $A_{r}[f]$ of the partial sums of the Fourier series converge to $f$ uniformly on $[\pi, \pi]$ as $r \rightarrow 1$.
$(ii)$ For $2 \pi$-periodic $f$, piece-wise continuous on $[-\pi, \pi]$, the Abel means $A_{r}[f]$ remain bounded as $r \rightarrow 1$, and they converge to $f$ uniformly on every closed sub-interval $[a, \beta]$ of an interval $(a,b)$ where $f$ is continuous.
To formally attack $(i)$, I approached as follows in $\text{Lemma} \, (0.1)$:
$$\text{Lemma} \, (0.1)$$:
$$\text{Definition (0.1)}:$$
If $f$ is an integrable function given on an interval $[a,b]$ of length $L$, then the $n^{th}$ $\text{Fourier Coefficient}$ of $f$ is defined by:
$$f(n)= a_{n} = \frac{1}{2\pi} \int_{-\pi}^{\pi}f(\theta)e^{in \theta} d \theta, n \in \mathbb{Z}$$
The $\text{Fourier Series}$ of $f$ is given formally by: $$f(\theta) \sim \sum_{n=-\infty}^{\infty} a_{n}e^{in \theta}$$
$\text{Remark}$: Note the Fourier Series of $f$ can also be formally expressed as follows: $$f(\theta) \sim \sum_{n=-\infty}^{\infty} \frac{1}{2 \pi}\int_{-\pi}^{\pi}f(\theta)e^{-in \theta} d \theta e^{in \theta}$$ $$$$
$$\text{Definition (0.2)}:$$
A series $A(r)= \sum_{}^{}a_{n}r^{n}$is said to Abel Summable to $L$ if $f(r)$ is convergent for all $|r| < L$ and if $f(r)$ converges to some Limit $L$ as $r \rightarrow 1^{-}$: $$A(r)= \sum_{}^{}a_{n}r^{n}$$
$\text{Remark}$: The developments of Abel Summability, expressed within a prior definition can be fully expressed as follows: $$ \lim_{r \rightarrow 1^{-}} A(r)= \lim_{r \rightarrow 1^{-}} \sum_{}^{}a_{n}r^{n} = L$$
Adapting Abel Summability in the context of Fourier Series, one can make the following the observations in $(1.)$
$(1.)$
$$A_{r}(f)(\theta)=f(\theta) \sum_{n=-\infty}^{\infty} r^{|n|}\frac{1}{2 \pi}\int_{-\pi}^{\pi}f(\theta)e^{-in \theta} d \theta e^{in \theta}$$
$$\lim_{r \rightarrow 1^{-}}A_{r}(f)(\theta)=\lim_{r \rightarrow 1^{-}} \sum_{n=-\infty}^{\infty} r^{|n|}\frac{1}{2 \pi}\int_{-\pi}^{\pi}f(\theta)e^{-in \theta} d \theta e^{in \theta} = L$$
$$ \text{Lemma} \, (0.2)$$:
In order, to calculate our Fourier Coefficients, we'll need to formally develop IBP(Integration By Parts) in $(2.)$
$(2.)$: $$\text{Theorem}: \text{Integration by Parts}$$
if $u = u(x)$ and $du=u'(x)dx$, while $v=v(x)$ and $dv=v'(x)dx$, then by IBP:
$$\int_{a}^{b}u(x) v'(x)dx = [u(x)v(x)]_{a}^{b} - \int_{a}^{b} u'(x)v(x)dx = u(b)v(b)-u(a)v(a)- \int_{a}^{b}u'(x)v(x)dx$$
Applying IBP we have the following developments:
$$\frac{1}{2 \pi}[-\frac{\theta}{in}e^{-in \theta}]_{-\pi}^{\pi} + \frac{1}{2 \pi in} \int_{-\pi}^{\pi}e^{-in\theta}d \theta$$
$$=-\frac{1}{2 \pi}[\frac{\pi}{in}e^{-in \pi}-\frac{-\pi}{in}e^{+in \pi}] + \frac{1}{2 \pi in}\cdot \left[ \frac{e^{-in\theta}}{{-in}}\right] ^{\pi}_{-\pi} $$
$$=-\frac{1}{2 \pi}[\frac{\pi}{in}e^{-in \pi}-\frac{-\pi}{in}e^{+in \pi}] + \frac{1}{2 \pi in}\cdot \left[ \frac{e^{-in\pi}}{{-in}}-\frac{e^{+in\pi}}{{-in}}\right] $$ Since $$ e^{n\pi i}= e^{-n\pi i}=(-1)^n$$ The second term cancels and the first term becomes: $$-\frac{(-1)^n}{in}= \frac{(-1)^{n+1}}{in}$$
So our Fourier Series is expressed as follows:
$$\sum_{n \neq 0} r^{|n|}\frac{(-1)^{n+1}}{in}e^{in \theta}$$
$$\text{Lemma (0.3)}$$
$$\lim_{r \rightarrow -1}A_{r}(f)(\theta)=\sum_{n \neq 0} r^{|n|}\frac{(-1)^{n+1}}{in}e^{in \theta}$$
Are the recent developments leading up $\text{Lemma(0.3)}$, initially vaild also can one provide with hints on how to approach $(ii)$ ?
