Seemingly unrelated is Gauss’ Mean Value Theorem, which is significantly cooler (in my opinion) than the standard mean value theorem of the reals.  We will define it formally below, but it says the follwing: if f is analytic (equivalent to complex differentiable) on some disk D and a is the center point of this disk, then the average of the values about the boundary of D is equal to f(a).  That is, to find the value of f(a), it suffices to integrate around a circle centered at a and divide by 2\pi (the amount of radians we pass through while integrating).  This is really neat to think about since this tells us not only that, given f there exists some point whose value is equal to the average of the sum of the values of f lying on a circle, but, moreover, that this point is actually the center of the circle.  This is intense stuff.

 

Theorem (Gauss’ Mean Value Theorem).  Let f(z) be analytic on some closed disk D which has center a and radius r.  Let C denote the boundary of the disk (that is, C is the circle bounding D).  Then we have that \displaystyle f(a) = \frac{1}{2\pi}\int_{0}^{2\pi}f(a+re^{i\theta})\,d\theta.

 

The proof of this theorem is pretty straight forward and uses the Cauchy integral formula and some easy substitution.

 

Proof.  Note that we have \displaystyle f(a) = \frac{1}{2\pi i}\oint_{C}\frac{f(z)}{z-a}\,dz.  The equation of a circle with radius r and center a is given by z = a + re^{i\theta} where \theta runs from 0 to 2\pi (if you don’t believe me, plot some points!).  Substituting this value into the integral and noting that dz = ire^{i\theta} we have that

\displaystyle f(a) = \frac{1}{2\pi}\int_{0}^{2\pi}\frac{f(a + re^{i\theta})ire^{i\theta}}{re^{i\theta}}\,d\theta = \frac{1}{2\pi}\int_{0}^{2\pi}f(a+re^{i\theta})\,d\theta

as required.  \diamond

 

Why bring up this neat little theorem?  Well, by itself it doesn’t seem to be all that useful — when would we be able to calculate and sum up a whole ton of values of an analytic function surrounding a point, but not be able to find the point itself?  But this little theorem packs some punch as a way of bounding certain values.  In particular, it gives a neat proof of the Maximum Modulus Theorem.  You might have guessed this from the title of this post.

 

First, let’s note something quickly.

 

Lemma.  Given the assumptions in Gauss’ MVT, we have \displaystyle |f(a)|\leq \frac{1}{2\pi}\int_{0}^{2\pi}|f(a+re^{i\theta})|\,d\theta

 

Be careful here in thinking that this should be an equality; we are now looking at the modulus of our value, and the modulus of each point on the circle.  But this lemma comes almost for free:

 

Proof.  We have |f(a)| =\left|\frac{1}{2\pi}\int_{0}^{2\pi}f(a+re^{i\theta})\,d\theta\right| by using Gauss’ MVT and simply taking the norm of both sides.  Note that

\displaystyle\left|\frac{1}{2\pi}\int_{0}^{2\pi}f(a+re^{i\theta})\,d\theta\right| = \frac{1}{2\pi}\left|\int_{0}^{2\pi}f(a+re^{i\theta})\,d\theta\right|

\displaystyle \leq \frac{1}{2\pi}\int_{0}^{2\pi}|f(a+re^{i\theta})|\,d\theta

whence the inequality above.  \diamond

 

This lemma tells us that the value of the center of any circle is bounded by the sum of the modulus of the values of the points of that circle.  We’ll see why this is the crucial bound we’ll need in the MMT’s proof below. 

 

Theorem (Maximum Modulus Theorem).  Given f analytic on some domain D, if f is non-constant on D then the maximum value of |f(z)| for z\in D will occur on the boundary of D.  (Alternatively, if |f(z)| is maximized by some value not on the boundary of D, then f is constant on D.)

 

Proof.  We’ll split this into two steps.  The first step is for the specific case that D is a closed disk and our maximum modulus occurs at the center of this disk.  The second step will be to get some arbitrary space D and construct some closed disks in the interior of D and "piece these together" to show that f is constant on all of D.

Step 1: Let’s suppose that our maximum modulus is at the center point of D, which we will call z_{0}; that is, we are supposing that |f(z)|\leq |f(z_{0})| for every z\in D.  Since z_{0} is an interior point, we have that there is some r-ball about z_{0} (that is, a ball of radius r) which is completely contained in D.  Let the C denote the circle of radius r centered at the point z_{0}.  By our second lemma above we have that

\displaystyle|f(z_{0})|\leq \frac{1}{2\pi}\int_{0}^{2\pi}|f(z_{0}+re^{i\theta})|\,d\theta

BUT, using that |f(z)|\leq |f(z_{0})| for every z\in D we have that

\displaystyle \frac{1}{2\pi}\int_{0}^{2\pi}|f(z_{0}+re^{i\theta})|\,d\theta \leq \frac{1}{2\pi}\int_{0}^{2\pi}|f(z_{0})|\,d\theta = f(z_{0}).

Stringing these inequalities together and suggestively re-writing |f(z_{0})| = \frac{1}{2\pi}\int_{0}^{2\pi}|f(z_{0})|\,dz, we have that

\displaystyle \frac{1}{2\pi}\int_{0}^{2\pi}|f(z_{0})|\,d\theta = \frac{1}{2\pi}\int_{0}^{2\pi}|f(z_{0}+re^{i\theta})|\,d\theta

and by subtracting,

\displaystyle 0 = \frac{1}{2\pi}\int_{0}^{2\pi}|f(z_{0})| -|f(z_{0}+re^{i\theta})| \,dz

but since the integrand is always positive or zero (why?) it must be the case that

\displaystyle |f(z_{0})| -|f(z_{0}+re^{i\theta})| = 0

or, in other words, |f(z_{0})| =|f(z_{0}+re^{i\theta})|.  Since r was arbitrary, we conclude that |f(z_{0})| = |f(z)| for every z\in D.

 

Step 2: Now suppose we have some arbitrary domain D and f is analytic on all of D.  I will hand-wave a bit here, but you can fill in the details.  Note that a domain (in this context) necessarily means open and path-connected (and, in fact, it usually denotes a simply connected open subset of {\mathbb C}).   Suppose that our maximum modulus occurs at some point on the interior of D which we will call z_{0}.  Now, given any other point w\in D we have some path from z_{0} to w which is completely contained in D.  In fact, we can make this path a finite polygonal path; that is, a path made out of a finite number of straight lines piecewise-connected together; we will denote this z_{0}L_{0}z_{1}L_{1}\cdots L_{n-1}w, where the L_{i} is the line with endpoints z_{i-1} and z_{i}.  I will let you work the details out here, but it can be done. 

 

image

 

Now, the polygonal line might be right next to a boundary, and we don’t want to accidentally hit it when we start making balls around points, so let \epsilon denote whichever is smaller: the distance from the polygonal line to the boundary, or 1.  So, if your polygonal line is right next to the boundary, we might need to make \epsilon pretty small; but if not, we can just let it be whatever we want, so we might as well make it 1.  Note that since D is open, no point on the polygonal path should be on the boundary.  Now, let’s break up our polygonal path into another polygonal path z_{0}L_{0}z_{1}L_{1}\cdots L_{n-1}w where each L_{i} has length less than \frac{\epsilon}{4}.  It is clear we can do this just by partitioning each straight line in our original path so that their lengths are appropriately small; note, we still only have a finite number of endpoints z_{i}.  That’s important.

image

 

(In the picture above, I’ve made the original endpoints blue and then partitioned our polygonal path with the new red endpoints to make each line segment less than \frac{\epsilon}{4}.)

Now everything is going to fall pretty quickly, so keep on your toes.  First, make a disk of radius \epsilon (as defined above) around each z_{i} and call it D_{i}.  Now note that, by our previous step, since our maximum modulus occurs at z_{0}, we have f(z) = f(z_{0}) for every point z\in D_{0}.  But z_{1} is in D_{0}

image

(This picture is not drawn to scale because I am not a good artist; this is illustrating z_{1} being inside the circle D_{0}.)

So now z_{1} is also of maximum modulus (since z_{0} was) and so f(z_{1}) = f(z) for every point in D_{1}.  Continue this and we will obtain f(z_{0}) = f(w).  Since w was an arbitrary point, it follows that f(z_{0}) = f(w)  for every w\in D.  Hence, if f attains a maximum modulus on the interior of some set D, then it is constant.  This implies directly that any non-constant analytic function achieves its maximum modulus on the boundary.  \Box