MHB Bounded Function on Set S: Proving $|f(z)|\le1$ for All $z\in S$

[Markov2]

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.

 

[AlexYoucis]

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.

 

[Markov2]

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

 

[ThePerfectHacker]

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 $1$.

 

[Jose27]

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]

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 $\partial S$.

 

[Markov2]

Okay, what's actually the solution? I'm confused now. :(

 

[ThePerfectHacker]

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.

 

[Markov2]

I understand, I'll wait you then!

 

[Markov2]

TPH, I really need help on this one, don't know how to solve it yet. :(

 

[Jose27]

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 \(e^{\sqrt{x}}\leq e^{\sqrt{|z|}}\).