# The boundary value of a bounded analytic function cannot vanish identically on any open subset of the boundary

Suppose $g$ is a bounded function in the region $S=\{z\in\mathbb{C};Imz\le0\}$, analytic in the region $S _0=\{z\in\mathbb{C};Imz\lt0\}$ and continuos on $S$. Let $f$ be the boundary value of this function(i.e $g$ restricted to the real axis).

I need to conclude if $f$ vanishes identically on any non-empty open subset of $\mathbb{R}$,the $f$ is the trivial $0$ function.

I am not able to apply identity theorem because $g$ is not analytic on the real axis. How should i conclude this?

I am basically going through the first part of proposition 3.1 in this paper http://www.ams.org/journals/proc/1983-087-04/S0002-9939-1983-0687648-4/S0002-9939-1983-0687648-4.pdf

There are other things going on here but i believe i have posted the question with all requirements.

If $f$ is zero on an open interval $(a, b)$ then it follows from the Schwarz reflection principle that $g$ can be continued analytically to $$\{z\in\mathbb{C} : \text{Im} \, z\lt0\} \cup (a, b) \cup \{z\in\mathbb{C} :\text{Im} \, z >0\}$$ Now apply the identity theorem to the extended function.
Use the Schwarz reflection principle. $\overline{g(\overline{z})}$ is analytic in the lower half plane, and has the same boundary value $f$ wherever $f$ is real. By Morera's theorem, the combined function $$\cases{g(z) & for \text{Im}(z) > 0\cr \overline{g(\overline{z})} & for \text{Im}(z) < 0\cr 0 & for \text{Im}(z) = 0\cr}$$ would be analytic, and therefore identically $0$.