# Exercise on Maximum Principle

This exercise is taken from Gamelin's Complex Analysis, page 89, exercise 4.

Let $f(z)$ be an analytic function on a domain $D$ that has no zeroes on $D$. (a) Show that $|f(z)|$ attains its minimum on $D$, then $f(z)$ is constant. (b) Show that if $D$ is bounded, and if $D$ is bounded, and if $f(z)$ extends continuously to the boundary $\partial D$ of $D$, then $|f(z)|$ attains its minimum on $\partial D$.

I solved part (a), but part (b) is giving me trouble. Namely, I define $g(z)=1/f(z)$, but how do I know that f(z) doesn't have a zero on $\partial D$? I have to make that extra assumption in my proof below... And if it helps any, the book assumes D is simply connected as well.

Heres my progess:

Part $(a):$

Let $f(z)$ be an analytic function on a domain $D$ that has no zeroes on $D$. Suppose $|f(z)|$ attains a minimum on $D$ at $z_0$, denoted $M$. Since $f(z)$ has no zeroes on $D$, and is analytic, $g(z)=1/f(z)$ is analytic. It follows that $|g(z)|=|1/f(z)|\le 1/M$ for all $z\in\mathbb{Z}$ and $|g(z_0)|=1/M$. So, since $g(z)$ is analytic, it is also harmonic, and thus by the Strict Maximum Principle, $g(z)$ is constant, implying $f(z)$ is constant.

Part $(b):$

Let $f(z)$ be an analytic function on a domain $D$ that has no zeroes on $D$ and $\partial D$. Suppose that $D$ is bounded and connected and that $f(z)$ extends continuously to $\partial D$. Then $D\cup\partial D$ is compact, as it contains its boundary and is bounded. Let $g(z)=1/f(z)$, then since $f(z)$ has no zeroes, $g(z)$ is defined over all of $D\cup\partial D$, and since its a continuous real-valued function on a compact set, it attains a maximum, denoted $M$ at $z_0$. Then $g(z)=1/f(z)\le M$, implying $f(z)\ge 1/M$ for all $z\in\mathbb{Z}$, and since $g(z_0)=M$, it follows that $f(z_0)=1/M$. If $f(z)$ obtains a minimum in $D$, it follows from part $(a)$ that $f(z)$ is constant and thus obtains its minimum on $\partial$$D, and thus f(z) always obtains its minimum on \partial D. • g(z) is harmonic is not correct. It is a function of a complex variable. The harmonic functions are functions of two real variables. – PAD Oct 11 '12 at 8:18 • Suppose that there exists z_0\in\partial D such that f(z_0)=0. Then the |f(z_0)|=0, but this is necessarily a minimum for the nonnegative real valued function |f(z)|. In some sense it seems to me you solved the hard case but not the easy one. Or am I making a stupid mistake? – Giovanni De Gaetano Oct 11 '12 at 8:48 • @PantelisDamianou, in general I would say yes, but on page 87 of Gamelin's Complex Analysis defines the property harmonic as follows "a complex valued function whose real and imaginary parts are harmonic...thus any analytic function is harmonic". – Chris Oct 11 '12 at 21:46 • @GiovanniDeGaetano, thank you! I seem to have missed the obvious but grasped its slightly subtler cousin, – Chris Oct 11 '12 at 21:49 • @Chris Thanks! I have never seen this definition before. But it is not relevant here. – PAD Oct 12 '12 at 21:49 ## 2 Answers I realize this is probably in bad form to answer my own question, but the one comment I thought helped me solve the peice of the problem I was missing, is as I said, a comment, and not something which allows me to close this question. Part (a): Let f(z) be an analytic function on a domain D that has no zeroes on D. Suppose |f(z)| attains a minimum on D at z_0, denoted M. Since f(z) has no zeroes on D, and is analytic, g(z)=1/f(z) is analytic. It follows that |g(z)|=|1/f(z)|\le 1/M for all z\in\mathbb{Z} and |g(z_0)|=1/M. So, since g(z) is analytic, it is also harmonic, and thus by the Strict Maximum Principle, g(z) is constant, implying f(z) is constant. Part (b): Let f(z) be an analytic function on a domain D that has no zeroes on D and a zero on \partial D at z_0. Then f(z_0)=0 implying |f(z_0)|=0, which is neccesarily a minimum on \partial D since |f(z)|> 0 for all z\in D as f(z) has no zeroes in D. Let f(z) be an analytic function on a domain D that has no zeroes on D and \partial D. Suppose that D is bounded and connected and that f(z) extends continuously to \partial D. Then D\cup\partial D is compact, as it contains its boundary and is bounded. Let g(z)=1/f(z), then since f(z) has no zeroes, g(z) is defined over all of D\cup\partial D, and since f is continuous, g is continuous and thus it attains a maximum, denoted M at z_0. Then g(z)=1/f(z)\le M, implying f(z)\ge 1/M for all z\in\mathbb{Z}, and since g(z_0)=M, it follows that f(z_0)=1/M. If |f(z)| obtains a minimum in D, it follows from part (a) that f(z) is constant and thus obtains its minimum on \partial$$D$, and thus $|f(z)|$ always obtains its minimum on $\partial D$.

Since $f$ extends continuously to $\bar{D}$, and since $\bar{D}$ is compact, $\vert f \vert$ is bounded there, and attains both a minimum and a maximum.

Assume that $|f|$ does attain a minimum in $D$. Then $|g|$ attains a maximum in $D$, hence both $f$ and $g$ are constants. The continuity of $f$ now assures that $|f|$ attains a minimum on $\partial D$.