Home > Complex questions > Complex 3 – Cauchy Ineqs and Schwarz Lemma

Complex 3 – Cauchy Ineqs and Schwarz Lemma

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

Suppose a and b are positive.

(a) Prove that the only entire functions f for which |f(z)|\le a|z|^{1/2}+b for all z are constant.

(b) What can you prove if |f(z)|\le a|z|^{5/2} +b for all z.

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

Suppose you have such an  f in (a). By the Cauchy Inequalities, |f'(z)|\le \frac{M}{R}, where M is the maximum of f(w) on a disk of radius R around z. If |z|=r\ge 0, choose R=r+1, and the given implies that |f'(z)|\le\frac {a (2r+1)^{1/2}+b}{ r+1}  (since 2r+1 is the biggest possible distance from the origin on that circle of radius r+1 around z). Note that the inequality is true for |z|=r, but by the Maximum Modulus Principle, it is also true for all z such that |z|\le r. The RHS of this inequality is bounded above by a constant for 0\le r<\infty (see powers of r), so Liouville’s Theorem implies that f'(z) is a constant. By taking a limit as r\to\infty, we see that that constant must be zero.  Thus, f'(z)=0, and f is a constant function.

Part (b) is similar. Use the same argument, except use the Cauchy inequality for the third derivative: |f'''(z)|\le \frac{(3!)M}{R^3}. Then you get |f'''(z)|\le\frac {6a (2r+1)^{5/2}+6b}{ (r+1)^3} using the same circles. The same argument shows that f'''(z)=0, and so f is a polynomial of degree two.

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Categories: Complex questions
  1. May 30, 2011 at 1:36 pm

    I was asked why to use R=r+1 instead of just R=r. If you do the latter, your bound for |f'(z)| has an r in the denominator, and so your bound goes to \infty as r\to 0. So if instead you do R=r, you would have to do the case |z| \le 1 separately (which you can of course handle since the derivative has to be bounded by a constant on that disk).

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