## Complex 3 – Cauchy Ineqs and Schwarz Lemma

Hi all – Ken U. asked me about Complex #3. I quote it here:

Suppose and are positive.

(a) Prove that the only entire functions for which for all are constant.

(b) What can you prove if for all .

**Solution:** well, there are many ways to do this. Here’s one way.

Suppose you have such an in (a). By the Cauchy Inequalities, , where is the maximum of on a disk of radius around . If , choose , and the given implies that (since is the biggest possible distance from the origin on that circle of radius around ). **Note that the inequality is true for , but by the Maximum Modulus Principle, it is also true for all such that .** The RHS of this inequality is bounded above by a constant for (see powers of ), so Liouville’s Theorem implies that is a constant. By taking a limit as , we see that that constant must be zero. Thus, , and is a constant function.

Part (b) is similar. Use the same argument, except use the Cauchy inequality for the third derivative: . Then you get using the same circles. The same argument shows that , and so is a polynomial of degree two.

I was asked why to use instead of just . If you do the latter, your bound for has an in the denominator, and so your bound goes to as . So if instead you do , you would have to do the case separately (which you can of course handle since the derivative has to be bounded by a constant on that disk).