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 


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


So


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


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 $



Representation of Hilbert transform by a singular integral
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-](http://www.cims.nyu.edu/~chou/notes/harmonic.pdf) and reverse the argument shown there. An excellent reference is:


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


