I am considering the boundary values of a bounded holomorphic functions. Suppose $w$ is a bounded holomorphic function in upper half plane, with continuous and bounded boundary value $f$ on real axis. $f$ is smooth, for example, in $C^2$, and bounded. We know that for bounded $f$, hilbert transform $Hf\in BMO$, and in general not in $L^{\infty}$.

My question is: With so many properties of $w$, would it be the case that $Hf\in L^{\infty}$? Is there a sufficient and necessary for $Hf\in L^{\infty}$ for $f\in L^{\infty}$?


1 Answer 1


The answer is no, and no amount of smoothness will help. Indeed, consider an unbounded region $D$ bounded by a smooth simple curve and contained in the half-strip $\{ z:|\Re z|<1,\Im z>0 \} $. Let $f=u+iv$ be a conformal map from the upper half-plane onto $D$ such that $f(\infty)=\infty$. Then $u$ is bounded and $v$ is its Hilbert transform. However $v$ is unbounded. You can even make $\partial D$ analytic in which case $u$ will be analytic on the real ine.

  • $\begingroup$ Thanks Alexandre. Yes the Hilbert transform smooth and bounded functions might not be bounded anymore. Is there any (nontrivial) criterion to determine when the hilbert transform is still bounded? $\endgroup$
    – qingtang
    Commented Nov 30, 2015 at 17:54
  • $\begingroup$ @qingtang: If a criterion means "necessary and sufficient condition", probably there is no useful criterion. $\endgroup$ Commented Nov 30, 2015 at 18:54

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