## Applying the Maximum Modulus Principle and Cauchy: A Common Question.

### July 29, 2011

I’ve been a bit busy moving and packing my things, but I’ve tried to keep my problem-doing up to practice for my analysis qual.  One problem that seems to come up quite a bit in the complex analysis part of the exam is something like the following:

Question.  Suppose that $f$ is entire which satisfies $|f(z)| \leq A|z|^{k} + B$ for every $z\in {\mathbb C}$ and every $A,B > 0$.  Prove that $f$ is a polynomial of degree at most $k$.

Just for kicks, let’s remember what the maximum modulus principle says:

Theorem (Maximum Modulus Principle): Let $U\subseteq {\mathbb C}$ be a domain (that is, an open connected set) and $f$ an analytic function on $U$.  If there is a point $z_{0}$ such that $|f(z)| \leq |f(z_{0})|$ for all $z\in U$ then $f$ is a constant function (in $U$.

Teasing out the meaning behind this principle, it says that if $f$ is analytic on some domain and there’s a maximum somewhere inside the domain (that is, not on the boundary), then $f$ is a constant function.  If you really think about this, it’s telling of what kinds of functions can be considered analytic; anything that has local maxes all over the place is not going to be an interesting function, since it’s going to be constant.  A little bit upsetting!

I will also assume you know the Cauchy integral formula in its slightly generalized form.  But just for kicks, I’ll write down:

$\displaystyle f^{(n)}(z_{0}) = \frac{n!}{2\pi i}\oint_{C}\frac{f(z)}{(z-z_{0})^{n+1}}\, dz$

and you can fill in all the assumptions and such.

So, now, how can we use these theorems to help us solve this problem?  Well, let’s follow our noses.

Solution.  We first assume that $|f(z)| \leq A|z|^{k} + B$ for some positive $A, B$ and all $z\in {\mathbb C}$.  So far, we just know that $f$ is analytic, so let’s write it out as a series.

$|a_{0} + a_{1}z + a_{2}z^{2} + \cdots | \leq A|z|^{k} + B$

Now, we note the following.  Since we want to prove that this is a polynomial of degree no more than $k$, we could show that the $n$-th derivative is 0 for every $n > k$ (why?).  First, let’s let our curve $C$ by the circle $|z| = R$, since it’s a pretty nice curve (and because we can approximate the integral with this).  By Cauchy, we know

$\displaystyle |f^{(n)}(0)| = \left|\frac{n!}{2\pi i}\oint_{C}\frac{f(z)}{z^{n+1}}\, dz\right| \leq \left|\frac{n!}{2\pi i}\right|\oint_{C}\left|\frac{f(z)}{z^{n+1}}\right|$

Note now by our assumptions above, we have that $|f(z)| \leq AR^{k} + B$ for $|z| = R$, which gives us

$\displaystyle \left|\frac{f(z)}{R^{n+1}}\right| \leq \left|\frac{AR^{k} + B}{R^{n+1}}\right| = \left|\frac{A}{R^{(n+1)-k}} + \frac{B}{R^{(n+1)}}\right|$

which is a nice estimate.  So.  We now have a nice upper bound for our integral on the right-hand-side, above.  We have that:

$\displaystyle\left|\frac{n!}{2\pi i}\right|\oint_{C}\left|\frac{f(z)}{z^{n+1}}\right| \leq \left|\frac{n!}{2\pi i}\left(\frac{A}{R^{(n+1)-k}} + \frac{B}{R^{n+1}}\right)\cdot 2\pi R\right|$

Simply by estimating our integral by $2\pi$ times the maximum value that it could attain.  This looks crazy messy, so let’s simplify.  (Note, the $i$ goes away, since the modulus of $i$ is just 1).

$\displaystyle = \frac{k!AR}{R^{(n+1)-k}} + \frac{k!BR}{R^{n+1}} = k!\left(\frac{A}{R^{n-k}} + \frac{B}{R^{n}}\right)$

As $R\rightarrow \infty$, we have that this is simply going to give us

$|f^{(n)}(0)| \leq 0$

exactly when $n > k$.  A little thinking will imply that, in our Taylor expansion above, we must have that $0 = a_{k+1} = a_{k+2} = \cdots$ (or else what would $f^{(n)}(0)$ be equal to for, say, $n = k+1$?).  This implies that $f$ must be a polynomial, and its degree is at most $k$$\Box$

If you want a nice quick problem to give to undergraduate students who are studying complex analysis, a similar question is (under the same assumptions as the question above) prove that if $|f(z)| \leq A|z|$ for some $A > 0$, then prove that $|f(z)|$ is at most a linear function.

After you give them that, ask them what $|f(z)| \leq A|z|^{0}$ should be (a constant function; that is, a function of degree at most 0), and then note that this is the same as saying $|f(z)| \leq A$.  Then ask them to recite Liouville’s theorem.  (In fact, the proof above that we gave to solve the problem is the standard proof to prove Liouville’s Theorem!)