Let $f \in L^2(\mathbb{R})$. Then
(1) Verify that there exists a unique function $f_H \in L^2(\mathbb{R})$ such that $$\mathcal{F}(f_H)(y) = -i\ \mbox{sgn}(y)\mathcal{F}(f)(y)$$ where $\mathcal{F}$ is the Fourier transform of $L^2$ function (Fourier-Plancherel transform).
Call this $f_H$ the $\textbf{Hilbert transform}$ of $f$.
(2) Show that $(f_H)_H = -f$
(Double Hilbert transform give back function with negative sign)
(3) Prove that $f_H = \frac{x}{x^2 + 1}$ where $f = \frac{1}{1+x^2}$
I really have no ideas how I should construct the function $f_H$ in (1). Is it something similar to Fourier trasnform in $L^2$ ? It is constrcuted as a limit of a sequence of functions (has no practical formula).
Any suggestions please ? :(
The Fourier transform is a unitary on $L^2$, so you can define $$ f_H(x)=-i\, \mathcal F^{-1}(\text{sgn}\,(y)\mathcal F(f)(y))(x). $$ This gives existence and uniqueness, since $\mathcal F$ is bijective.
We have, since $(\text{sgn}(y))^2=1$, \begin{align} (f_H)_H(z)&=-i\, \mathcal F^{-1}(\text{sgn}\,(y)\mathcal F(f_H)(y))(z)\\ \ \\ &=-i\,\mathcal F^{-1}(\text{sgn}(y)\,[-i\ \mbox{sgn}(y)\mathcal{F}(f)(y)])(z)\\ \ \\ &=-\mathcal F^{-1}(\mathcal F(f))(z)\\ \ \\ &=-f(z). \end{align}