Bounding a function using maximum principle

Question

Let $$V=\left\{r\mathrm{e}^{i\theta}:r>0,\theta\in\left(\frac{-\pi}9,\frac\pi9\right)\right\},$$ $f:\overline{V}\to\mathbb{C}$ be continuous and holomorphic on $V$. Show that if $|f(z)|\leq\exp\left(|z|^2\right)$ for $z\in V$ and $|f(z)|\leq 1$ for $\in\overline V\setminus V$, then $|f(z)|\leq 1$ for $z\in\overline V$.

This problem comes with the following hint:

for arbitrary $a>0$ consider function $F_a(z)=f(z)\exp(-az^3)$, show that $$|F_a(z)|\leq|f(z)|\exp\left(-a|z|^3\cos\frac\pi 3\right)$$ and for sufficiently large $R$ if $|z|\geq R$, then $|F_a(z)|\leq1$, and then use the maximum modulus principle.

I don't quite see how to apply the hint, any suggestions?

Answer 1

This is an example of the Phragmén–Lindelöf principle. The proof goes as follows:

• First you show that for every $a > 0$ there is a $R > 0$ such that $|z| \ge R$ implies $|F_a(z)| \le 1$.
• Then use the maximum modulus principle to conclude that $|F_a(z)| \le 1$ for all $z \in \overline V$ and all $a > 0$.
• Finally, write this estimate as $$ |f(z)| \le \exp\left(a|z|^3\cos\frac\pi 3\right) $$ and consider for fixed $z \in \overline V$ the limit $a \to 0$ to conclude that $|f(z)| \le 1$.