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

Hang
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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 05:44
  • I add a possible backgrond of the problem – Hang May 24 '15 at 14:48

1 Answers1

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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.

Daniel Fischer
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  • 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 04:51
  • @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 09:03
  • 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 02 '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 02 '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 09 '15 at 02:20
  • @Henry The point is that every subsequence has a further subsequence that converges locally uniformly to $0$, therefore the whole sequence converges locally uniformly to $0$. – Daniel Fischer Jun 09 '15 at 07:45
  • I read your referenced question and the general criterion for convergence. I think we should check what topology space is here for the locally uniform convergence. Anyway, for an arbitrary countable normal family, by Montel's theorem, every subsequence of this family must has a further locally uniformly convergent subsequence; however, I think the whole normal family cannot converges in this sense. – Hang Jun 10 '15 at 15:57
  • @Henry We're looking at the space of holomorphic functions on the right half-plane. The topology on that space is the topology of locally uniform convergence [equivalently, since the right half-plane is locally compact, the topology of compact convergence; equivalently, by one of Weierstraß' theorems, the topology of locally uniform convergence of all derivatives]. If it weren't the case that the whole sequence converges locally uniformly to $0$, we could pick a subsequence that had $\lvert f_{n_k}(z_0)\rvert \geqslant \varepsilon$ for some $z_0$ and all $k$. Then that subseqence could not – Daniel Fischer Jun 10 '15 at 19:31
  • have any subsequence converging locally uniformly to $0$. So take a subsequence $\bigl( f_{n_{k_m}}\bigr)_{m\in\mathbb{N}}$ that is locally uniformly convergent. Let $g$ denote its limit function. Then $g\not\equiv 0$, so the zeros of $g$ [if any] are isolated. But that contradicts the behaviour on the real line, where we know that the full sequence $(f_n)$ converges to $0$ pointwise. That contradiction shows that indeed the full sequence converges locally uniformly to $0$ on the whole right half-plane. – Daniel Fischer Jun 10 '15 at 19:31