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 $