f is bounded

LegendZKiller · **Joined:** Mon, 6 Feb 2012 06:18:16 UTC **Posts:** 25

Consider the set $S=\left\{ z\in \mathbb{C}:\text{Re}(z)>0,\text{ }\arg (z)\in \left( -\dfrac{\pi }{4},\dfrac{\pi }{4} \right) \right\},$ and a function $f\in H(S)\cap C(\overline S)$ so that for each $z\in\partial S$ is $|f(z)|\le1$ and for all $z=x+yi\in S$ is $|f(z)|\le e^{\sqrt x}.$ Prove that for all $z\in S$ is $|f(z)|\le1.$

I don't think if I can solve this, since the condition $|f(z)|\le e^{\sqrt x}$ is false if I take $f(z)=e^{z^2}.$ Could this problem be solved? Or do we have to change it?

outermeasure · **Posted:** Mon, 6 Feb 2012 06:32:56 UTC

LegendZKiller wrote:

Consider the set $S=\left\{ z\in \mathbb{C}:\text{Re}(z)>0,\text{ }\arg (z)\in \left( -\dfrac{\pi }{4},\dfrac{\pi }{4} \right) \right\},$ and a function $f\in H(S)\cap C(\overline S)$ so that for each $z\in\partial S$ is $|f(z)|\le1$ and for all $z=x+yi\in S$ is $|f(z)|\le e^{\sqrt x}.$ Prove that for all $z\in S$ is $|f(z)|\le1.$

I don't think if I can solve this, since the condition $|f(z)|\le e^{\sqrt x}$ is false if I take $f(z)=e^{z^2}.$ Could this problem be solved? Or do we have to change it?

What is H(S)?

LegendZKiller · **Joined:** Mon, 6 Feb 2012 06:18:16 UTC **Posts:** 25

Holomorphic function.

outermeasure · **Posted:** Mon, 6 Feb 2012 14:47:42 UTC

It doesn't work for $f(z)=e^{z^2}$ --- so what? Obviously you don't want $f(z)=e^{z^2}$ since it is unbounded in S (consider the positive real axis).

LegendZKiller · **Joined:** Mon, 6 Feb 2012 06:18:16 UTC **Posts:** 25

Yes well so it can't be solved right? I mean, how could we formule the problem in order to solve it?

Shadow · **Posted:** Tue, 7 Feb 2012 02:39:58 UTC

LegendZKiller wrote:

Yes well so it can't be solved right? I mean, how could we formule the problem in order to solve it?

What do you mean? What is your actual goal to understand?

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f is bounded

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