Let $H=\{z \in \mathbb{C}: \operatorname{Im}(z) >0\}$ and $\overline{H}=\{z \in \mathbb{C}: \operatorname{Im}(z) \geq 0\}$.
Let $f$ be a bounded function s.t. $f:\overline{H} \rightarrow \mathbb{C}$ be continuous and $f:H \rightarrow \mathbb{C}$ holomorphic. Moreover, for all $z \in \mathbb{R}$ we have $|f(z)| \leq 1$.
Define $g_t(z) = \frac{f(z)}{i+tz}$ for $t>0$ which is continuous on $\overline{H}$ and holomorphic on $H$.
I proved that $|g_t(z)| \leq 1$ for $z \in \mathbb{R}$ and $\lim_{|z|\to\infty} |g_t(z)| = 0 $ for $z \in \overline{H}$.
How can I prove now (by maximum modulus principle), that $|g_t(z)| \leq 1$ for all $z \in \overline{H}$? I don't know which connected open bounded set I should consider.
Thanks for any hint.
Best Answer
Hint: Choose $R>0$ such that $|z|\ge R$ implies $|g_t(z)|\le 1.$ Then all you have to worry about is how large $|g_t|$ can be on $\{|z|\le R\}\cap \{\text {Im }z \ge 0\}.$