I have seen a theorem which proves that the Hilbert transform $H$ of a $C^{1}$ function defined on $(-\pi, \pi)$ is equal (in the point $x$) to $$\lim_{ \epsilon \to 0} \frac{1}{2 \pi} \int_{\epsilon < |z| < \pi} f(x-y) \cot(y/2) dy $$ Then it says that we can extend this definition (using principal value) to $L^{2}$ functions. Now, the problem is that in the proof for the case $f \in C^{1}$ it was used that $$ \lim_{\epsilon \to 0} \int_{\epsilon < |z| < \pi} \frac{f(x-y) - f(x+y)}{y} dy$$ was limited. How can I conclude that if I only have $f\in L^{2}$?
Also, while the proof for $f\in C^{1}$ started from the definition (I have defined the Hilbert transform as an operator in $L^2$ that sends a function $f$ in a function $Hf$ whose fourier coefficiensts are $$Hf_k = -i *sgn(k) f_k$$), how can I conclude that the definition of $H$ with the principal value for $f\in L^{2}$ is such that its Fourier coefficients are exactly those in the definition of $H$? I mean, if I define $$Hf(x)=\lim_{ \epsilon \to 0} \frac{1}{2 \pi} \int_{\epsilon < |z| < \pi} f(x-y) \cot(y/2) dy $$ how can I say that the Fourier coefficients of this function and the Fourier coefficients of the definition are the same?