# A UCLA Qualifying Complex Analyis Problem , possibly related to Phragmén-Lindelöf Theorem [closed]

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 $\phi(z)=\exp(-z^c)$.

## closed as off-topic by heropup, Claude Leibovici, user147263, J. W. Perry, DidMay 28 '15 at 12:36

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• Not sure why this is getting downvoted. The OP has thought about it some, as the title indicates (Phragmen-Lindelof). Perhaps the OP could explain a little more about what he's tried. – zhw. May 24 '15 at 5:44
• I add a possible backgrond of the problem – Hang May 24 '15 at 14:48

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.

• Thank you! But, the problem states that $f(x) \to 0, x \to 0$along the positive real axis, not that$f(x+iy) \to 0, x \to 0$ – Hang May 25 '15 at 4:51
• @Henry That is all I use. We have a normal family $\mathscr{F}$, so every subsequence $(f_{n_k})$ of $(f_n)$ has a locally uniformly convergent subsequence. We know that $f_n(x) \to 0$ for every $x\in (0,+\infty)$ (the positive real half-axis; that is precisely the given condition), hence the limit function of every convergent subsequence is $0$ (since it is $0$ on the non-discrete set $(0,+\infty)$), hence the full sequence converges locally uniformly to $0$ on the full right half-plane. – Daniel Fischer May 25 '15 at 9:03
• I recently found some problem. We only prove a locally uniform convergence, but the question claim that $f(z) \to 0$ as $z\to 0$ and we must notice that $0$ is a boundary point here. Anyway, I fail to write down a formal proof using $\epsilon - \delta$ language. – Hang Jun 2 '15 at 12:33
• @Henry Given $\varepsilon > 0$, by the locally uniform convergence of the $f_n$, there is an $n_0$ (depending on $\varepsilon$ and $\phi$ of course) such that $\lvert f_n(z)\rvert \leqslant \varepsilon$ for every $n \geqslant n_0$ on the annular sector $S = \{ z : \lvert\arg z\rvert \leqslant \phi, 1 \leqslant \lvert z\rvert \leqslant 2\}$. Looking at how $f_n$ is defined, that is $$\lvert f(z) \rvert \leqslant \varepsilon\text{ for } z \text{ with } \lvert \arg z\rvert \leqslant \phi \text{ and } \lvert z\rvert \leqslant 2^{1-n_0}.$$ Take $\delta = 2^{1-n_0}$. – Daniel Fischer Jun 2 '15 at 12:53
• I think the argument cannot be valid since you must pass to a subsequence of $f_n$. Therefore, your argument only gives $∣f(z)∣\le ε$ for z with $∣\arg z∣\le ϕ$ and for $z \in \cup2^{-n_k}S$ but not $|z| \le 2^{1−n_0}.$ – Hang Jun 9 '15 at 2:20