Continuity of Hilbert transform Suppose $f : \mathbb{R} \to \mathbb{R}$, be a non-negative, bounded and continuous function, and its support is a compact interval in $\mathbb{R}$. Moreover, we have that $\int f(x) \, dx =1$. The Hilbert transform of $f$ is defined as:
$$
g(y) := H[f](y) = PV \frac{1}{\pi} \int_{\mathbb{R}} \frac{f(x)}{x-y} \, dx
$$
What can we say about the continuity and boundedness of the function $g(y)$ on $\mathbb{R}$?
If view the Hilbert transform as the convolution with the function $\frac{1}{\pi x}$, maybe we can use this post?
But $\frac{1}{\pi x}$ is un-bounded at $x = 0$!
 A: Assume  $g\in L^2(\mathbb{R})$ and the Fourier transform $\widehat{g}$ is absolutely integrable. By the  Fourier inversion formula we get
$$g(x)={1\over 2\pi}\int\limits_{-\infty}^\infty \widehat{g}(t)e^{itx}\,dt\qquad (*)$$  The formula $(*)$ states that the function $g(-x)$ is equal to $(2\pi)^{-1}$ times the Fourier transform of the function $\widehat{g}.$ It is a classical theorem that the Fourier transform of an absolutely integrable function
is continuous and vanishes at $\pm \infty$ see. Therefore $g$ is continuous.
Consider $f$ as in the question. Then
$$\widehat{Hf}(x)=-i{\rm sgn}(x)\widehat{f}(x)$$ Assume $\widehat{f}$ is absolutely integrable. Then so is $\widehat{Hf}(x)$ as $|\widehat{Hf}(x)|=|\widehat{f}(x)|,\ x\neq 0.$
Hence $Hf$ is continuous and vanishes at $\pm \infty.$
Now we look for additional assumptions sufficient for   absolute integrability of $\widehat{f}$. If the function $f$ is continuously differentiable with compact support, then
$$\widehat{f'}(x)=ix\widehat{f}(x)$$
Therefore $\widehat{f}(x),\ ix \widehat{f}(x)\in L^2(\mathbb{R}).$ Hence the function $(1+x^2)|\widehat{f}(x)|^2$ is integrable.
Then by the Cauchy-Schwarz inequality we get
$$\int\limits_{-\infty}^\infty |\widehat{f}(x)|\,dx \le \left (\int\limits_{-\infty}^\infty {1\over 1+x^2}\,dx\right )^{1/2}\left (\int\limits_{-\infty}^\infty { (1+x^2)}|\widehat{f}(x)|^2\,dx\right )^{1/2}<\infty$$
