# f is bounded

#### Markov

##### Member
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 $|f(z)|\le1.$

Any ideas? Don't know how to start.

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#### AlexYoucis

##### New member
インテグラルキラー;488 said:
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 $|f(z)|\le1.$

Any ideas? Don't know how to start.
Well, what are some of your initial thoughts.

#### Markov

##### Member
Don't know really, I wish I knew how to use the arg stuff and the closure stuff.

#### ThePerfectHacker

##### Well-known member
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 $|f(z)|\le1.$

Any ideas? Don't know how to start.
The second inequality is entirely not necessary (unless you meant to ask something else). As the problem stands it is entirely trivial consequence of the maximum-modulos principle. If by approaching the boundary the modulos stays bounded by $1$ then the function everywhere inside the domain must also stay bounded by $1$.

• Markov

#### Jose27

##### New member
The second inequality is entirely not necessary (unless you meant to ask something else). As the problem stands it is entirely trivial consequence of the maximum-modulos principle. If by approaching the boundary the modulos stays bounded by $1$ then the function everywhere inside the domain must also stay bounded by $1$.
This is not true: Take $$f(z)=e^{z^2}$$ then $$f=1 on \partial S$$ but it's unbounded. This is an application of the Phragmén-Lindelöf theorem, it shouldn't be too hard to find a good text with a proof.

• ThePerfectHacker

#### ThePerfectHacker

##### Well-known member
This is not true: Take $$f(z)=e^{z^2}$$ then $$f=1 on \partial S$$ but it's unbounded. This is an application of the Phragmén-Lindelöf theorem, it shouldn't be too hard to find a good text with a proof.
I think my mistake is interesting enough to explain. This is what I did. As $\limsup_{z\to \zeta} | f(z) | \leq 1$ for all $\zeta \in \partial S$, I concluded that (by MM) that it follows that $|f(z)|\leq 1$. But I ignored an important case! Here, $\partial S$ includes $\infty$ also (if we view the boundary as belonging to the Riemann sphere topology). Thus, that condition is not true for all points on $\partial S$.

• Markov

#### Markov

##### Member
Okay, what's actually the solution? I'm confused now. #### ThePerfectHacker

##### Well-known member
Okay, what's actually the solution? I'm confused now. Hold on Markov. I will try to get back to you. I am really sorry because I have limited time and writing out these solutions takes up time.
It is not like I do not want to help you as much as I can it is sometimes hard to.

• Markov

#### Markov

##### Member
I understand, I'll wait you then!

#### Markov

##### Member
TPH, I really need help on this one, don't know how to solve it yet. #### Jose27

##### New member
One question: Do you know about the Phragmén-Lindelöf principle (look it up in Wikipedia, if you're not familiar with the name)? If so, this is just a special case since $$e^{\sqrt{x}}\leq e^{\sqrt{|z|}}$$.