In Complex Variables and Applications, Brown & Churchill (9th edition), I stumbled upon a chapter which got me somewhat confused.

On page 175 of the book, there is the theorem, which states the following:

If a function $f$ is analytic and not constant in a given domain $D$, then $\left| f(z) \right|$ has no maximum value in $D$. That is, there is no point $z_0$ in the domain such that $\left|f(z) \right|\le \left|f(z_0) \right|$ for all points $z$ in it.

To me, this means that $f$ is unbounded on $D$ if it's not constant.

But now there is a Corollary on page 176 which seems to contradict this theorem, and which goes as follows:

Suppose that a function $f$ is continuous on a closed bounded region $R$ and that it is analytic and not constant in the interior of $R$. Then the maximum value of $\left|f(z) \right|$ in $R$, which is always reached, occurs somewhere on the boundary of $R$ and never in the interior.

And also there's this theorem:

Suppose that $\left| f(z) \right| \le \left| f(z_0) \right|$ at each point $z$ in some neighborhood $\left| z - z_0\right|<\varepsilon$ in which $f$ is analytic. Then $f(z)$ has the constant value $f(z_0)$ throughout that neighborhood.

So, we have three theorems, where one says that $f(z)$ is either constant or unbounded. Another theorem says that there actually does exist a maximum value of $f$ if it's not constant. While the third theorem says that, nevertheless, $f$ must be constant if it's bounded.

I would really appreciate if someone could please clear this confusion of mine.


The issue here is that these theorems apply to different types of sets. In particular, the first theorem applies to domains, which are defined as open connected sets. Implicitly, the third theorem is also about open sets, as it speaks of the neighborhood of a point.

The second theorem applies to a closed bounded region. In particular, it is explicitly mentions that the function does not obtain a maximum in the interior of the region, which is actually necessary given the other theorems

Also, you need to be careful about your use of the word "unbounded" here - none of the theorems say that $f$ is either constant or unbounded. They just say that $f$ can't obtain a maximum. For instance, over the real interval $(-1,1)$ the function $f(x)=x$ does not obtain a maximum, but it is bounded since $|f(x)|\leq 1$.


You are confusing maximum and supremum.

What is meant is that there is no point in $D$ that achieves the maximum value. But the function can still be bounded. Think of, for instance, of the function $f$ defined on $(-1,1)$ by $$ f(x) = x^2 $$ Well, is bounded... it has a supremum. But it does not have a maximum.

  • $\begingroup$ (In particular, in your first theorem, $D$ is (implicitly, I guess) assumed open -- it is the domain. But in the second, $R$ is closed (by assumption).) $\endgroup$ – Clement C. Jun 16 '16 at 1:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.