Let $\infty >p\geq 2$, then for $f\in L^p(\mathbb{T})$ (here $\mathbb{T}=[0,1)$), show that for any real-valued trigonometric polynomial $f$, we have $H(f^2-(Hf)^2)=2fHf$.
The hint is to use the identity $(u+iv)^2=u^2-v^2-2uvi$. I m bit rusty on Hilbert transform Thanks
By the properties of the Hilbert transform, $f+i\,Hf$ is the boundary value of an analytic function $F$; $F^2$ is also analytic, and its boundary value is $(f+i\,Hf)^2$. Then $$ (f+i\,Hf)^2=f^2-(Hf)^2+i\,2\,f\,Hf. $$