# Identity principle for harmonic functions

The question: Show that if $$u$$ is real and harmonic on a connected open set $$U \subseteq \mathbb{C}$$, and $$u = 0$$ on some small disc $$D(P,r) \subseteq U$$, then $$u = 0$$ on all of U.

Now here is a couple of ways I thought of solving this problem: The first way is this: since $$u$$ is harmonic, there exists a holomorphic function $$F$$, on set $$U$$ such that $$Re(F) = u$$ on $$U$$. In my textbook, there is one corollary which says that if a holomorphic function $$F = 0$$ on some disc $$D(P,r)$$ which is a subset of a connected open set $$U$$ then $$F = 0$$ on $$U$$. The problem is that since $$F$$ is holomorphic such that $$Re(F) = u = 0$$ on the disc $$D(P,r)$$, $$F$$ is constant on $$U$$ but not exactly zero.

My second option is to use the mean value property on the disc $$D(P,r)$$. We have $$u(P) = \int_{0}^{2 \pi} u(P + re^{i \theta}) d\theta = 0$$, since $$u = 0$$ on the disc. My idea is the construct a disc $$D(P, r_{0})$$ such that the disc is a subset of U and the boundary of U. Do you think I am on the right track? Thank you for your help!

• how did you get "$F$ is constant on $U$?? do you mean "$F$ is purely imaginary on $U$"? – mathworker21 Aug 11 at 1:19
• Yeah thats what I meant – Overachiever Aug 11 at 1:49
• It is not true that there exists a holomorphic $F$ s.t. $\Re{F}=u$ unless the domain is simply connected (for a counterexample take $\log |z|$ on the punctured unit disc); the result is true only locally in general, or on any simple connected subdomain of U – Conrad Aug 11 at 2:34

Let A the set of $$z_0 \in U$$ for which $$u$$ is zero in a neighborhood of $$z_0$$. A is obviously open, but now if $$z_1 \in U$$ is in the closure of A (in U), let B a small open disc containing $$z_1$$ and included in U; then there is an $$F_B$$ analytic on B, $$\Re{F_B}=u$$ on B; but there is a $$z_0$$ in A also contained in B (as $$z_1$$ is in the closure of A!), so there is a small open set C included in B on which $$u=0$$, hence $$F_B$$ is purely imaginary on C, hence constant on C (by the open mapping property for example), hence constant on B by the identity principle, hence $$z_1$$ has the property that defines A, so it is in A. This means A is open and closed in U, so being non-empty by hypothesis, it is all U, done!