5
$\begingroup$

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

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Claude Leibovici, Community, J. W. Perry, Did
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 4
    $\begingroup$ 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. $\endgroup$ – zhw. May 24 '15 at 5:44
  • $\begingroup$ I add a possible backgrond of the problem $\endgroup$ – Hang May 24 '15 at 14:48
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you! But, the problem states that $f(x) \to 0, x \to 0$along the positive real axis, not that$f(x+iy) \to 0, x \to 0$ $\endgroup$ – Hang May 25 '15 at 4:51
  • $\begingroup$ @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. $\endgroup$ – Daniel Fischer May 25 '15 at 9:03
  • $\begingroup$ 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. $\endgroup$ – Hang Jun 2 '15 at 12:33
  • $\begingroup$ @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 $\delta = 2^{1-n_0}$. $\endgroup$ – Daniel Fischer Jun 2 '15 at 12:53
  • $\begingroup$ 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 $|z| \le 2^{1−n_0}.$ $\endgroup$ – Hang Jun 9 '15 at 2:20

Not the answer you're looking for? Browse other questions tagged or ask your own question.