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It is known that if a function $f \in \mathcal{L}^1 $, then its Hilbert transform may not map it to the $\mathcal{L}^1$ space. So my question is under what kind of condition, the Hilbert transform of $f$ can still be in the $\mathcal{L}^1$ space?

lioness99a
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Claire Fan
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1 Answers1

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The space of integrable functions whose Hilbert transform is also integrable is the Hardy space $H^1$ (not to be confused with the Sobolev space $H^1$.) A necessary condition (but not sufficient) is that its integral be $0$.

Julián Aguirre
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  • Thank you! Is it also valid for multi-dimensional function? – Claire Fan Mar 08 '17 at 14:28
  • Yes, with the Riesz transforms instead of the Hilbert transform. – Julián Aguirre Mar 08 '17 at 14:30
  • Can you provide some references? Because I saw a definition of real-variable Hardy space. If $f$ is in this space, then $\mathcal{H}f \in L^1$. I'm not clear about the difference between Hardy space and real-variable Hardy space. – Claire Fan Mar 09 '17 at 09:39
  • I learned from the classical books by Stein (Singular integrals...) and Stein&Weiss (Introduction to Harmonic Analysis...) although they deal with $H^p$, $1<p<|infty$. In the late seventies there was a lot of literature about the atomic decomposition of $H^p$ spaces and the real variable characterization. These spaces van also be characterized in terms of non-tangential maximal functions. A search with those terms should provide with references. – Julián Aguirre Mar 09 '17 at 11:58