$L^2$ boundedness of cut-off version of Hilbert transform

Question

I know the Hilbert transform is bounded $L^2(\mathbb R)\to L^2(\mathbb R)$: $$f(x)\mapsto-\int_{-\infty}^\infty\frac{f(x+y)-f(x)}{y}dy.$$ I'm wondering if this still true if you "cut off" the integration: $$f(x)\mapsto-\int_{-1}^1\frac{f(x+y)-f(x)}{y}dy.$$ I think this should be true, since you still have the same cancellation around $y=0$.

Answer 1

Maybe this is a little bit lazy, but it seems to prove $L^2$ boundedness:

Neglecting constants, your operator is just convolution with the kernel $\mathbf1_{|y|\leq1}/y$ so its symbol should be given by the Fourier transform of this. According to Mathematica, the Fourier transform is the sine integral of $\xi$. This symbol is bounded so therefore your operator is bounded.