Strong Maximum Principle proof (for harmonic functions in the plane)

Question

This is a question from Section 2.4 of "Complex Variables (Harmonic and Analytic functions)" by Francis J. Flanigan

Strong Maximum Principle: A nonconstant harmonic function in a domain $\Omega$ does not assume its maximum or it minumum on $\Omega$

Proof: Assume the harmonic function $u$ assumes its maximum at $\zeta\in\Omega$, $u(\zeta)=c$. If $u$ is nonconstant in $\Omega$, then the set of $z\in\Omega$ such that $u(z)\lt c$ is nonempty and open (by continuity). Consequently the set of points $z$ for which $u(z)=c$ is not open also, or else $\Omega$ would be disconnected

Thus there is a point $\zeta_1\in\Omega$ such that $u(\zeta_1)=c$ and a disc $\bar D(\zeta_1;r)$ contained in $\Omega$ with boundary $C(\zeta_1;r)$ containing at least one point $z$ with $u(z)\lt c$. Since $u$ is continuous, however, there exists an arc on the circle $C(\zeta_1;r)$ such that $u(z)\lt c$ for all the points in this arc. It follows that (on the polar coordinates $r$, $\theta$ centered at $\zeta_1$) $$\frac{1}{2\pi}\int_0^{2\pi}u(r,\theta)\,d\theta\lt\frac{1}{2\pi}\int_0^{2\pi}u(\zeta_1)\,d\theta =u(\zeta_1)$$ but this is a contradiction (circumferential mean-value property), since $u$ is harmonic (the same sort of proof works in the case of a minimum).

Sorry for posting the whole proof but there are many things I don't understand:

• I get that he set of points $z$ for which $u(z)=c$ is not open, but isn't it safe to assume that it is closed? It is assumed (in the book) that we are in $\Bbb R^2$ so such set is either a singleton or a countably infinte; aren't those always closed?
• Why assume the existence of some $\zeta_1$, possibly different from $\zeta$, when the latter is already a maximum?
• When doing the line integral along $\partial\bar D(\zeta_1;r)=C(\zeta_1;r)$ is it obvious to assume that $u(r,\theta)\le c$ for every $\theta$ in order to write the inequality?
• The fact that the set $\Omega_1=\{z\in\Omega\,;\,u(z)\lt c\}$ is nonempty and open (by continuity) seems kind of obvious to me, but why is it true? Here is a little attempt of a proof: $u$ is nonconstant so there is some $z\in\Omega$ such that $u(z)\ne c$ so $u(z)\lt c$ and $z\in\Omega_1\ne\emptyset$. Now let $z\in\Omega_1$, $(u(z)-\epsilon,u(z)+\epsilon)$ is open (given $\epsilon\lt\vert c-u(z)\vert$ and that the interval is a subset of the range of $u$) so its inverse image in $\Omega_1$ is open and thus $\Omega_1$ is open. Is it right?

Thanks in advance

Answer 1

• Yes, it is closed. But what does that have to do with anything? What he wants is a circle around $\zeta_1$ where $u < c$ somewhere. For that he only needs it to not be open.
• He could have just said "therefore we can choose $\zeta$ such that ...". Personally I think that would have been cleaner. But he argues from the view that $\zeta$ is some definite point that may not have the condition that he needs. So he moves to another point, calling it $\zeta_1$ that does have the condition (the existance of this circle in the domain but extending outside $u^{-1}(c)$).
• $c$ is the maximum value of $u$ everywhere by assumption, so yes, it is quite obvious that it will be an upper bound on the curve.
• That works, though it is easier just to realize that $u$ is continuous, so $\Omega_1 = u^{-1}((-\infty, c))$ is open. Or even the argument you made in the first point: $u^{-1}(c)$ is closed, so its complement is open.