TITLE: A UCLA Qualifying Complex Analyis Problem , possibly related to Phragmén-Lindelöf Theorem
QUESTION [5 upvotes]: Let $f$ be a bounded analytic function on the open right half plane such that $f(x) \to 0, x\to 0$ along the positive real axis. Suppose $0<\phi<\pi/2$. Prove that $f(z) \to 0, z \to 0$ uniformly in the sector $|\arg z|\le|\phi|$.
Remark: I guess it cannot be proved just by Montel's theorem as in one of the answer.
 I am reading Chapter VI GTM 11, Functions of a Complex Variable. And a corollary of Phragmén-Lindelöf Theorem (cf page 139) is similar to my question. The corollary states that 
Corollary Suppose f is analytic on $G=\{z:|\arg z|\le\pi/2a\}$ and there is a constant such that $\limsup_{z\to w}|f(z)|\le M$ for all $w\in \partial G$. If there are positive constants $P$ and $b<a$ such that $$|f(z)|\le P \exp(|z|^b)$$ then $|f(z)|\le M$ on $G$.
The proof of the corollary is just using the Phragmén-Lindelöf Theorem with $\phi(z)=\exp(-z^c)$.

REPLY [4 votes]: Phragmén-Lindelöf is not the right tool here. The person to turn to is Paul Montel.
Since $f$ is bounded, the family $\mathscr{F} = \bigl\{ f_n \colon z \mapsto f(2^{-n}\cdot z) \,\;\big\vert\;\, n\in \mathbb{N}\bigr\}$ is normal. Since $(f_n)$ converges (locally uniformly) to $0$ on $(0,+\infty)$, it follows by normality that $f_n \to 0$ locally uniformly in the right half-plane. Looking the annular sector
$$\{ z : 1 \leqslant \lvert z\rvert \leqslant 2, \lvert\arg z\rvert \leqslant \phi\}$$
gives the desired convergence.