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Let $f$ be analytic on $H=\{z:\Re(z)\ge 0\}$. Suppose that:

  • There exists $M>0$ such that if $\Re(z)=0$ then $|f(z)|\le M$.
  • There are $z_1,...,z_n$ such that $\Re(z_k)>0$ and $f(z_k)=0$.

Prove that for all $z\in H$: $$|f(z)|\le M\dfrac{\prod_{k=1}^n|z-z_k|}{\prod_{k=1}^n|z+\overline{z_k}|}.$$

This is suppose to be solved with the Maximum Modulus Theorem. But I don't know exactly which one. Since $f$ shouldn't be neccesary constant, the theorem must be this:

M. M. Theorem: Let $U\subseteq\mathbb{C}$ a bounded domain. Let $f$ be continuous on $\overline{U}$ and analytic on $U$. Then: $$\max\{|f(z)|:z\in\overline{U}\}=\max\{|f(z)|:z\in\partial U\}.$$

Since we need a bounded domain, let $U$ be the unit disk and consider the Möbius transformation $T(z)=\dfrac{-z-1}{z-1}$, so $T$ sends $U$ onto the right half plane.

By the M. M. Theorem, the maximum value of $|f\circ T|$ on $\overline{U}$ occurrs on the unit circle $|z|=1$. But if $|z|=1$, by the first point we have $|f(T(z))|\le M$. Therefore $|f(T(z))|\le M$ for all $z\in\overline{U}$.

But isn't it true that $T(\overline{U})=H$ and then $|f(z)|\le M$ for all $z\in H$? This is not the desired inequality. What am I doing wrong?

Thank you.

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    $\begingroup$ This is not true. The function $e^z$ is a counterexample with $n=0$ (if you don't like $n=0$, the function $(1-z)e^z/(1+z)$ is a counterexample with $n=1$.) You're leaving out a crucial hypothesis (or the author left it out). $\endgroup$ – David C. Ullrich Jan 16 '16 at 21:44
  • $\begingroup$ @DavidC.Ullrich Oh, thank you. It seems the autor left it out. If anyone knows what a crucial hypothesis could it be, please let me know. $\endgroup$ – JonSK Jan 16 '16 at 21:54
  • $\begingroup$ Hard to believe. Where is this from? Look again - what you wrote is exactly what it says? $\endgroup$ – David C. Ullrich Jan 16 '16 at 22:31
  • $\begingroup$ It is from here. It is in spanish though posgradomatematicas.unam.mx/contenidoEstatico/archivo/files/pdf/… $\endgroup$ – JonSK Jan 16 '16 at 22:37
  • $\begingroup$ Author's typo. It should be: There exists $M>0$ such that if $\Re(z)\ge 0 $ then $|f(z)|\le M$ $\endgroup$ – ts375_zk26 Jan 16 '16 at 23:37
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The result as stated is not true. The function $e^z$ is a counterexample with $n=0$; this can easily be modified to give a counterexample for $n>0$.

The result becomes true if we assume in addition that $f$ is bounded. We need the following version of MMT:

Theorem. Suppose $f$ is analytic and bounded in the region $\Re z\ge0$. If $|f(z)|\le M$ for all $z$ with $\Re z=0$ then $|f(z)|\le M$ for all $z$ with $\Re z > 0$.

Proof. The simplest proof may be a Phragmen-Lindelofish argument. For $\epsilon>0$ let $$g_\epsilon(z)=f(z)/(1+\epsilon z).$$Then $|g_\epsilon(z)|\le M$ for $\Re z=0$. And also $g_\epsilon\to0$ at infinity, since $f$ is bounded. So the lim sup of $g_\epsilon$ is less than or equal to $M$ at every boundary point in the extended plane, hence a suitable version of MMT shows that $|g_\epsilon|\le M$ in $\Re z>0$. Now let $\epsilon\to0$. QED.

Now to do the problem. Given $f$ as in the problem, assuming $f$ is also bounded, let $g=$???. Applying the theorem to $g$ shows that $|g|\le M$ in the right half plane, and that shows that $f$ satisfies the conclusion. (Saying $|g|\le M$ implies what you need should be a good hint what $g$ should be...)


The Point: When you figure out what function $g$ you need to apply the theorem to to do the problem, you see that it's not obvious that $|g|\le M$ in the entire region, even if we begin by assuming thet $|f|\le M$ in the entire region. But it is clear that $g$ is bounded (by something, we don't know what) in the entire region, and that $|g|\le M$ on the imaginary axis. Hence the theorem shows that $|g|\le M$ everywhere. (In other words, even if we assume $|f|\le M$ everywhere to start we still need that theorem.)

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  • $\begingroup$ Some questions. From what you wrote it is pretty obvious that $g=f/h$ where $h(z)=\frac{\prod_{k=1}^n|z-z_k|}{\prod_{k=1}^n|z+\overline{z_k}|}$. Then $g$ is analytic because $h$ has the zeros $z_k's$ and they are simple zeros. But I don't see why is it $g$ bounded? The theorem you wrote says $g$ must be bounded in order to apply it. Second question is, why $|g|\le M$ in $\Re(z)=0$? Perhaps it is $|h(z)|=1$ for all $\Re (z)=0$? $\endgroup$ – JonSK Jan 17 '16 at 1:25
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    $\begingroup$ First, $h$ is not quite what you say. We want $h$ to be holomorphic; leave out the absolute value signs. Yes, $|h(z)|=1$ for $\Re z=0$; this follows from the fact that $|\overline\alpha|=|\alpha|$. Why is $f/h$ bounded? Well, it's bounded in a neighborhood of each $z_k$ because it's continuous. And if $z$ satisfies $|z-z_k|>\delta>0$ for every $k$ you can show directly that $|1/h(z)|$ is less than or equal to something (something independent of $z$). $\endgroup$ – David C. Ullrich Jan 17 '16 at 2:11
  • $\begingroup$ Thank you! I forgot to wipe out the absolute values, but I understood it now. Thanks again. $\endgroup$ – JonSK Jan 17 '16 at 3:31

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