Hilbert transform defines as follow:

$$ H: L^2(\mathbb R) \to L^2(\mathbb R) $$

$$ H(f)= \mathcal{F}^{-1}[{F(\gamma) \mathrm{sign}(\gamma)]}$$

Where $F(\gamma)= \mathcal{F}(f) (\gamma)= \hat f$ is the fourier transform of $f(x)$, and $\mathcal{F}^{-1}(F(\gamma))$ is the inverse fourier transform.

I want to check that Hilbert transform can be represented as a singular integral as follow:

$$ H(f)= -\frac{1}{i \pi} \,\,\,p.v.\int_{|x-y|>\epsilon} \frac{f(y)}{x-y} \, \mathrm{d}y = \lim_{\epsilon \to 0} \int_{|x-y|>\epsilon} \frac{f(y)}{x-y} \, \mathrm{d}y $$ which the above limit is taken in the $L^2(\mathbb{R})$ space.

My approach:

I define

$$H_{\epsilon}(f)= \frac{-1}{i \pi} \int_{|x-y|>\epsilon} \frac{f(y)}{x-y} \, \mathrm{d}y $$

which using convolution, it can be rewritten as follows:

$$ H_{\epsilon}(f)=\frac{-1}{i \pi} (f * g_{\epsilon}), $$

$$g_{\epsilon}(x)=\begin{cases} \frac{1}{y} & |y|>\epsilon\\ 0 & |y| \leq \epsilon \end{cases}$$

So $$ H_{\epsilon}(f)= \frac{-1}{i \pi} \mathcal{F}^{-1} \big( \hat f \hat g_{\epsilon} \big) $$

Calculating the Fourier transform of $\hat g_{\epsilon}$ we can conclude:

$$ H_{\epsilon}(f) = \frac{1}{\pi} \mathcal{F}^{-1} \bigg( \hat f(\gamma) \int_{|y|>2 \pi \epsilon} \frac{\mathrm{sin}(\gamma y)}{y} \, \mathrm{d}y \bigg) $$

I should now show that \begin{equation} \label{eq} \| H_{\epsilon}(f) - H(f) \|_{L^2(\mathbb{R})} \to 0 \quad \mathrm{as} \quad \epsilon \to 0 \end{equation}

but I only know that

\begin{equation} \label{eq2} \int_{|y|>2 \pi \epsilon} \frac{\mathrm{sin}(\gamma y)}{y} \, \mathrm{d}y \to \pi \,\, \mathrm{sign(\gamma)} \quad \mathrm{pointwise \,\,\, as} \quad \epsilon \to 0 \end{equation}

My Question is this:

How can I show $H_{\epsilon}(f)$ converge to $H(f)$ in the $L^2(\mathbb{R})$ sence.

My attempts is as follow: Using Plancheral Identity

$$ \| H_{\epsilon}(f) - H(f) \|_{L^2(\mathbb{R})}=\| \hat{H_{\epsilon}(f)} - \hat {H(f)} \|_{L^2(\mathbb{R})} $$


$$ \| H_{\epsilon}(f) - H(f) \|_{L^2(\mathbb{R})}= \| \frac{1}{\pi} \bigg( \hat f(\gamma) \int_{|y|>2 \pi \epsilon} \frac{\mathrm{sin}(\gamma y)}{y} \, \mathrm{d}y \bigg)- \hat f(\gamma) \mathrm{sign}(\gamma)\|_{L^2(\mathbb{R})}$$

Hence it suffices to show that $$ \bigg\| \hat f(\gamma) \bigg(\frac{1}{\pi} \int_{|y|>2 \pi \epsilon} \frac{\mathrm{sin}(\gamma y)}{y} \, \mathrm{d}y - \mathrm{sign}(\gamma)\bigg)\bigg\|_{L^2(\mathbb{R})} \to 0 $$ as $ \epsilon \to 0 $

  • 1
    $\begingroup$ Hint: dominated convergence theorem. $\endgroup$
    – Fan Zheng
    Jan 3, 2017 at 17:06
  • $\begingroup$ Fan Zheng: Can you help How Can I use the Dominated Convergence Theorem. $\endgroup$
    – Hheepp
    Jan 3, 2017 at 17:33
  • $\begingroup$ The multiplier is uniformly bounded and converges pointwise. $\endgroup$
    – Fan Zheng
    Jan 3, 2017 at 18:24
  • $\begingroup$ @ Fan Zheng: But domain is unbounded, it seems for me that uniformly boundedness does not say anything $\endgroup$
    – Hheepp
    Jan 3, 2017 at 18:33
  • 1
    $\begingroup$ $\hat f$ times a uniformly bounded multiplier is dominated by $C\hat f$, which is square integrable. $\endgroup$
    – Fan Zheng
    Jan 3, 2017 at 19:00

1 Answer 1


This is covered in many classical references where the Hilbert transform is defined one way and the other derived (and vice-versa). Take a look at this note, page 42- and reverse the argument shown there. An excellent reference is:

Elias Stein, Singular Integrals and Differentiability Properties of Functions, Princeton Univ. Press, 1970.


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