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 ? :(

  • $\begingroup$ Existence follows from the theory of fourier transforms. If $f\in L^2(\mathbb{R})$ then it has a fourier transform $\mathcal{F}(f)$ that's also in $L^2$. By definition of $L^2$, the absolute value must be integrable, which gets rid of the sgn, completing existence. Uniqueness follows in a similar fashion $\endgroup$ – Alex R. May 15 '18 at 22:35
  • $\begingroup$ Do you know the Plancherel theorem? $\endgroup$ – David C. Ullrich May 15 '18 at 22:42

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}


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