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Local maximum

Markov

Member
Feb 1, 2012
149
Consider an analytic function $f$ and non-constant defined on a set $\mathcal U\subset\mathbb C$ open and connected. Prove that the real-valued functions $|f|,\,\text{Re}(z),\,\text{Im}(z)$ can't achieve local maximum.

This one looks hard, how to do it?
 

AlexYoucis

New member
Jan 26, 2012
19
インテグラルキラー;489 said:
Consider an analytic function $f$ and non-constant defined on a set $\mathcal U\subset\mathbb C$ open and connected. Prove that the real-valued functions $|f|,\,\text{Re}(z),\,\text{Im}(z)$ can't achieve local maximum.

This one looks hard, how to do it?
Well, the problem for $|f|$ is a pretty important theorem known as the Maximum Modulus Principle. For $\text{Re}(z)$ note that if $\text{Re}(f(z))$ acheived a maximum then so would $e^z$ since $|e^{f(z)}|=e^{\text{Re}(f(z))}$ from where you could apply the previous part. You try the case for the imaginary part.
 

Markov

Member
Feb 1, 2012
149
Well, the problem for $|f|$ is a pretty important theorem known as the Maximum Modulus Principle.
Yes, I know the principle, but it's supposed that I need an inequality for $f$ in order the principle to be applied. Is it true if I say that if $f$ is non-constant, then we have $|f(z)-w|\le\epsilon$ forall $\epsilon>0$ and $z,w\in\mathbb C$ ? But I don't think if helps me.

For $\text{Re}(z)$ note that if $\text{Re}(f(z))$ acheived a maximum then so would $e^z$ since $|e^{f(z)}|=e^{\text{Re}(f(z))}$
Ah okay so $\text{Re}(f(z))$ because $|e^{f(z)}|$ ain't bounded?
 

AlexYoucis

New member
Jan 26, 2012
19
インテグラルキラー;514 said:
Yes, I know the principle, but it's supposed that I need an inequality for $f$ in order the principle to be applied. Is it true if I say that if $f$ is non-constant, then we have $|f(z)-w|\le\epsilon$ forall $\epsilon>0$ and $z,w\in\mathbb C$ ? But I don't think if helps me.
But if you suppose that you have such a maximum you get such a desired inequality!

Ah okay so $\text{Re}(f(z))$ because $|e^{f(z)}|$ ain't bounded?
No, because $|e^{f(z)}|$ can't reach a maximum neither can $\ln(|e^{f(z)}|)=\text{Re}(f(z))$.
 

Markov

Member
Feb 1, 2012
149
But if you suppose that you have such a maximum you get such a desired inequality!
Thanks you but I don't see how to conclude, I'm supposed to prove that $|f|$ can't achieve a maximum. Okay so if that if such maximum exists, then it follows the inequality (why exactly?), but if the inequality doesn't hold, then $f$ is constant, which is a contradiction.

Sorry if this doesn't make any sense, I'm trying to think on this!

No, because $|e^{f(z)}|$ can't reach a maximum neither can $\ln(|e^{f(z)}|)=\text{Re}(f(z))$.
Oh yes, so that concludes that $\text{Re}f(z)$ doesn't achieve its maximum.

Is it my idea or do we have a typo? I wrote $|f|,\,\text{Re}(z),\,\text{Im}(z),$ shouldn't actually be $|f|,\,\text{Re}f(z),\,\text{Im}f(z)$ ?
Thanks for the help!
 

Markov

Member
Feb 1, 2012
149
Okay so I have this: by direct application of Maximum Modulus Principle, $|f|$ can't achieve local maximum. Now since $|e^{f(z)}|=e^{\text{Re}(f(z))},$ then $\text{Re}(f(z))$ can't achieve a local maximum. Is it enough with this? Or do I need to complete it more?

What about for $\text{Im}\,(f(z))$ ?
 

Markov

Member
Feb 1, 2012
149
Let $f=u+iv$ where $u=\text{Re}f,v=\text{Im}f$ are real-valued, then $-if=v-iu.$ Now, if I consider $e^{-if}$ I think I could get the third part, but I don't see it yet, any help?