Exercise 31 from chapter 4 ("Hilbert Spaces: An Introduction") of Stein & Shakarchi's "Real Analysis" The following problem shows that $\{e^{inx}\}_{n \in \mathbb{Z}}$ is an orthonormal basis of $L^2([-\pi, \pi])$. It is taken from [1] (exercise 31 of chapter 4: "Hilbert Spaces: An Introduction", pp. 200-201). I would appreciate answers, hints or references for any of the four parts. Thank you in advance.

Consider a version of the sawtooth function defined on $[-\pi,\pi)$ by 
  $$
K(x) = i(\mbox{sgn}(x)\pi - x)
$$
  (The symbol $\mbox{sgn}(x)$ denotes the sign function: it equals $1$ or $-1$ if $x$ is positive or negative respectively, and $0$ if $x = 0$.) and extend to $\mathbb{R}$ with period $2\pi$. Suppose $f \in L^1([-\pi, \pi])$ is extended to $\mathbb{R}$ with period $2\pi$, and define
  $$
\begin{aligned}
Tf(x) & = \frac{1}{2\pi}\int_{-\pi}^\pi K(x - y)f(y)\, dy \\
& = \frac{1}{2\pi}\int_{-\pi}^\pi K(y)f(x - y)\, dy
\end{aligned}
$$
(a) Show that $F(x) = Tf(x)$ is absolutely continuous, and if $\int_{-\pi}^\pi f(y)\, dy = 0$, then $F'(x) = if(x)$ a.e. $x$.
(b) Show that the mapping $f \mapsto Tf$ is compact and symmetric on $L^2([-\pi,\pi])$.
(c) Prove that $\varphi(x) \in L^2([-\pi,\pi])$ is an eigenfunction for $T$ if and only if $\varphi(x)$ is (up to a constant multiple) equal to $e^{inx}$ for some integer $n \neq 0$ with eigenvalue $1/n$, or $\varphi(x) = 1$ with eigenvalue $0$.
(d) Show as a result that $\{e^{inx}\}_{n \in \mathbb{Z}}$ is an orthonormal basis of $L^2([-\pi, \pi])$.
Note that in Book I, Chapter 2, Exercise 8, it is shown that the Fourier Series of $K$ is
  $$
K(x) \sim \sum_{n \neq 0} \frac{e^{inx}}{n}
$$


References
[1] Stein, Elias M. and Shakarchi, Rami. Real Analysis: Measure Theory, Integration, and Hilbert Spaces. Princeton University Press (2005)
 A: You are given that
$$
         (-2\pi i)Tf = \int_{-\pi}^{\pi}K(x-y)f(y)\,dy \\
     =\int_{-\pi}^{\pi}(\pi\,\mbox{sgn}(x-y)-(x-y))f(y)\,dy\\
     =\int_{-\pi}^{x}(\pi\,-(x-y))f(y)\,dy\\
     +\int_{x}^{\pi}(-\pi-(x-y))f(y)\,dy\\
     =(\pi-x)\int_{-\pi}^{x}f(y)\,dy+\int_{-\pi}^{x}yf(y)\,dy \\
     -(\pi+x)\int_{x}^{\pi}f(y)\,dy+\int_{x}^{\pi}yf(y)\,dy.
$$
The final expression is clearly an absolutely continuous function of $x$ whose derivative is
$$
\begin{align}
      (-2\pi i)(Tf)' & = (\pi-x)f(x)-\int_{-\pi}^{x}f(y)\,dy+xf(x) \\
            & +(\pi+x)f(x)-\int_{x}^{\pi}f(y)\,dy-xf(x), \\
            & = 2\pi f(x) -\int_{-\pi}^{\pi}f(y)\,dy.
\end{align}
$$
Therefore,
$$
             (Tf)' = if +\frac{1}{2\pi i}\int_{-\pi}^{\pi}f(y)\,dy.
$$
You have shown that $T$ is compact and symmetric. Therefore $T$ has an orthonormal basis of eigenfunctions. The eigenfunctions with different eigenvalues are automatically orthogonal.
Part (c). The constant function $1$ is an eigenfunction of $T$ because $T1$ is absolutely continuous with
$$
         (T1)' = i1 +\frac{1}{2\pi i}(2\pi)=0 \implies T1 = \lambda 1.
$$
The eigenvalue is found by evaluating $T1$ at $x=-\pi$, which I leave to you. If $f$ is another eigenvector with eigenvalue $\lambda$ which is orthogonal to $1$, or if $f$ is an eigenvector with eigenvalue $\mu \ne \lambda$, then $(f,1)=0$ can be assumed, which is equivalent to the assumption that
$$
                   \int_{-\pi}^{\pi}f(x)\,dx = 0.
$$
Therefore, for such a non-zero $f$, you have $Tf = \mu f$ and
$$
              \mu f' = (Tf)' = if.
$$
You can see right away there cannot be an eigenfunction with eigenvalue $0$ which is orthogonal to $1$ because the above would give $f\equiv 0$.
Knowing that $Tf$ is periodic $Tf(-\pi)=Tf(\pi)$ gives you the rest of the eigenfunctions and eigenvalues. Added: To see this, use an integrating factor. If $f$ is absolutely continuous and $\mu f'= if$ for some $\mu \ne 0$, then $e^{it/\mu} f(t)$ is also absolutely continuous with
$$
\begin{align}
    \frac{d}{dt}\left(e^{-it/\mu}f(t)\right)
       & =-e^{-it/\mu}\frac{i}{\mu}f(t)+e^{it/\mu}f'(t) \\
       & = \frac{1}{\mu}e^{-it/\mu}\left(if(t)-\mu f'(t)\right) = 0.
\end{align}
$$
Therefore it is necessary that $h(t)=e^{-it/\mu}f(t)$ equal a constant; this is because $h$ is absolutely with $0$ derivative, which leads to
$$
         h(t)-h(0)=\int_{0}^{t}h'(s)\,ds = 0,\;\;\; -\pi \le t \le \pi.
$$
In other words, it is necessary that $f(t)=Ce^{it/\mu}$ for some constant $C$.
The only question is which such exponentials are periodic with period $2\pi$. That is, for which $\mu$ do you have $e^{i\pi/\mu}=e^{-i\pi/\mu}$ or $e^{2\pi i/\mu}=1$? The solutions of this equation are $\mu=\pm 1,\pm 1/2,\pm 1/3,\cdots$. To be careful, plug these back into the operator $T$ and directly verify that all such cases are in fact eigenfunctions of $T$. I'll leave that to you. So, we've found all the eigenvalues and eigenvectors; up to multiplicative factors, they are
$$
         1,e^{\pm it},e^{\pm 2it},e^{\pm 3it},\cdots.
$$
The corresponding eigenvalues are
$$
         0,\pm 1,\pm \frac{1}{2},\pm \frac{1}{3},\cdots.
$$
Part (d): Knowing all the eigenfunctions and eigenvalues of $T$ allows you to construct a complete orthonormal basis because $T$ is selfadjoint and compact.
