Krdytkyu

This is a homework question so please don't provide the full proof.

Hi everyone,

Let $f$ be holomorphic on the open square $U = {zin mathbb{C}: |Re ,z| <1, |Im ,z| < 1}$ and continuous on its closure $bar{U}$. Suppose $f=0$ on ${zin bar{U}: Re , z =1}$, prove $f = 0$ on $U$.

My attempt

Method 1: By Max Modulus Theorem, $max_{zin U} |f(z)| = max_{zin partial U} |f(z)|$, 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 ${zin mathbb{C}: -1< Re ,z <1, -2<Im ,z < 0}$. 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.

Hint: consider $f(z)f(iz)$. Show that if a product of two analytic functions is $0$ then one of them must be $0$.

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 ${zin overline{U} mid Im, z = -1}$.

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_{zin U} |g(z)| = max_{zin partial U} |g(z)| = 0,
end{equation}
proving $g(z) = 0$ for all $zin overline{U}$, which implies that for any $z in overline{U}$, there exists $iin {1, 2, 3, 4}$ such that $g_i(z)=0$. And this is of course true for any $zin overline{B_{1/2}(0)}$ as well.

Since $overline{B_{1/2}(0)}$ is uncountable, one of the sets ${zin overline{B_{1/2}(0)} mid g_i(z)=0}$, $i = 1, 2, 3, 4$, 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.