TITLE: Isolated zeros on closure of a domain
QUESTION [10 upvotes]: Let $f$ be an analytic function on the open unit disk domain $D$.  Suppose also that $f$ is bounded.
Since $f$ is bounded I believe that $f$ can be continuously extended to the closed unit disk.
I know that the zeros of $f$ in the open disk $D$ are isolated.   Are the zeros of $f$ in the closed unit disk also necessarily isolated?

REPLY [5 votes]: Jonas Meyer's answer goes much deeper, but let me say the following as well.
Even if your (nonzero -- you didn't say that, but of course you meant it!) function does extend continuously to the closed unit disk it need not have isolated zeros.  Indeed, suppose that $f$ has infinitely many zeros in the open unit disk.  Then the zero set of the (putative) extension of $f$ to the closed unit disk is an infinite subset of a compact space, so must have an accumulation point: i.e., there will be at least one point on the boundary which is a nonisolated zero.
In fact, by the Weierstrass Factorization Theorem, the zero set of an analytic function on the open unit disk can be any discrete subset without accumulation points in the open disk.  You can choose such a set to have the entire boundary $|z| = 1$ of the disk contained in its closure.  We conclude that there exists a nonzero analytic function $f$ on the open unit disk such that -- if it admits a continuous extension to the closed unit disk -- this extension is identically zero on the boundary.
I would be interested to know if this construction can be made unconditional, i.e., whether any analytic function with such a zero set can be extended continuously to the boundary.