# Hilbert transform

Let $\mathscr H$ be the Hilbert transform, that is the Fourier multiplier by $\text{sign } \xi$: $$(\mathscr H u)(x)=\int_{\mathbb R} e^{2iπ x\xi }(\text{sign } \xi) \hat u(\xi) d\xi.$$ The Hilbert transform is a singular integral defined by the convolution $$\lim_{\epsilon\rightarrow 0_+}\int_{\vert x-y\vert>\epsilon}\frac{u(y)}{iπ (x-y)} dy.$$ It is a bounded (and even unitary and selfadjoint) operator on $L^2(\mathbb R)$, bounded on $L^p(\mathbb R)$ for $1<p<+\infty$, not bounded on $L^1$ or $L^\infty$.

My question: are there some boundedness results of $\mathscr H$ from $L^p$ to $L^q$ for some values of $p,q\in (0,1)$?

• I doubt there is such. – T. Amdeberhan Feb 16 '17 at 16:10
• Wouldn't you get boundedness on $L^1$ by interpolation if you had such a property? – Christian Remling Feb 16 '17 at 17:36