Prove function is identically 0 on an open set from it being 0 on a subset of the boundary This is a homework question so please don't provide the full proof.
Hi everyone,
Let $f$ be holomorphic on the open square $U = \{z\in \mathbb{C}: |Re \,z| <1, |Im \,z| < 1\}$ and continuous on its closure $\bar{U}$. Suppose $f=0$ on $\{z\in \bar{U}: Re \, z =1\}$, prove $f = 0$ on $U$.
My attempt
Method 1: By Max Modulus Theorem, $\max_{z\in U} |f(z)| = \max_{z\in \partial U} |f(z)|$, so it suffices to show that $f$ is identically $0$ on the four edges of the rectangle. To this end, I am trying to use the Identity Theorem, but I don't see how it can be applied.
Another method: Define $g(z) = f(iz + i)$, then by assumption, $g$ is identically $0$ on $(-1, 1)$, holomorphic on $\{z\in \mathbb{C}: -1< Re \,z <1, -2<Im \,z < 0\}$. Then I am trying to use the Schwarz reflection principle to extend this function to a holomorphic function on the entire plane and then apply the Identity Theorem. But since this function is only holomorphic on a subset of the lower half plane, I don't see how Schwarz reflection principle can be applied.
Some hints would be greatly appreciated. Thanks a lot.
 A: Hint: consider $f(z)f(iz)$. Show that if a product of two analytic functions is $0$ then one of them must be $0$.
A: This answer is based on Kavi Rama Murthy's hint (and also a hint from the instructor of my class, which is similar to Kavi's, actually). Many thanks to Kavi!
First , define $g_1(z) = f(iz)$. It is immediate that $g$ is holomorphic on $U$, continuous on $\overline{U}$ and is $0$ for all $\{z\in \overline{U} \mid Im\, z = -1\}$.
Similarly, if we define $g_2(z) = f(-z), g_3(z) = f(-iz), g_4(z) = f(z)$, then $g = g_1g_2g_3g_4$ is holomorphic on $U$, continuous on $\overline{U}$ and identically $0$ on $\partial U$.
By Max Modulus Theorem, 
\begin{equation}
\max_{z\in U} |g(z)| = \max_{z\in \partial U} |g(z)| = 0,
\end{equation}
proving $g(z) = 0$ for all $z\in \overline{U}$, which implies that for any $z \in \overline{U}$, there exists $i\in \{1, 2, 3, 4\}$ such that $g_i(z)=0$. And this is of course true for any $z\in \overline{B_{1/2}(0)}$ as well.
Since $\overline{B_{1/2}(0)}$ is uncountable, one of the sets $\{z\in \overline{B_{1/2}(0)} \mid g_i(z)=0\}$, $i = 1, 2, 3, 4$, must be uncountable and hence must have a limit point in $\overline{B_{1/2}(0)} \subseteq U$ by compactness of $\overline{B_{1/2}(0)}$.
By Identity Theorem, for this particular $i$, $g_i$ is identically $0$ on $U$.
It is easy to see from the definitions of $g_1, g_2, g_3, g_4$ that this implies $f$ being identically $0$ on $U$ regardless of what $i$ is, proving the result.
