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?



Singularity on the boundary of domain of holomorphy
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.


