# Singularity on the boundary of domain of holomorphy

Let $$\phi$$ be a continuous function on the closed upper half-plane $$\{ z\in\mathbb{C}: \operatorname{Im}(z)\ge 0\}$$ and holomorphic in the interior. Suppose that the function $$x\phi(x)$$ is in $$C^1(\mathbb{R})$$. Does it follow that $$\phi$$ is in $$C^1(\mathbb{R})$$, too?

Without the holomorphy, this is false, but maybe holomorphy ''heals'' the singularity on the boundary. Does it?

• Doesn't this follow from the Schwarz reflection theorem? Jun 23 at 17:14
• @Joseph Van Name: Yes, if the function takes real values on $\mathbb R$.
– Echo
Jun 23 at 19:55

## 1 Answer

Write $$\phi=\phi_1+i\phi_2$$. A counterexample is given by $$\phi_2(x)=\begin{cases} 0 & x<0 \\ x & 0\le x\le 1 \end{cases} .$$ We also give $$\phi_2$$ compact support and keep it smooth away from $$x=0$$. We can then set $$\phi(z) = \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{\phi_2(t)\, dt}{t-z}\label{2}\tag{2}$$ for $$\textrm{Im}\: z>0$$ and recover $$\phi_1$$ as the real part of the boundary values of this function. Alternatively, $$\phi_1=H\phi_2$$ (the Hilbert transform).

Thus the only potential problems with $$\phi_1(x)$$ occur at $$x=0$$. A step function has a Hilbert transform that diverges at the discontinuity as $$\log |x|$$, and $$\phi'_2$$ is a step function, so $$\phi_1\simeq x\log|x|$$ near $$x=0$$, which is continuous, and $$x\phi_1\in C^1$$.

Or, easier perhaps, just compute $$\phi_1$$ from \eqref{2}, and cut off the integral at $$t=1$$. The error will be smooth, so doesn't matter for the questions under consideration.