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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?

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

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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.

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