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.

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).