The discussion revolves around proving that certain real-valued functions derived from an analytic function cannot achieve local maxima on an open and connected subset of the complex plane. The specific functions under consideration are the modulus of the function, its real part, and its imaginary part. Participants explore theoretical implications and relevant principles, particularly the Maximum Modulus Principle.
Participants generally agree on the implications of the Maximum Modulus Principle for the modulus of the function and the real part of the function. However, the discussion about the imaginary part remains unresolved, with differing approaches and uncertainty about how to prove its behavior.
Participants express uncertainty regarding the necessity of specific inequalities for applying the Maximum Modulus Principle and the implications of non-constancy of the function. There are also unresolved questions about the approach to the imaginary part of the function.
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?
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.AlexYoucis said:Well, the problem for $|f|$ is a pretty important theorem known as the Maximum Modulus Principle.
Ah okay so $\text{Re}(f(z))$ because $|e^{f(z)}|$ ain't bounded?For $\text{Re}(z)$ note that if $\text{Re}(f(z))$ acheived a maximum then so would $e^z$ since $|e^{f(z)}|=e^{\text{Re}(f(z))}$
But if you suppose that you have such a maximum you get such a desired inequality!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.
Ah okay so $\text{Re}(f(z))$ because $|e^{f(z)}|$ ain't bounded?
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.AlexYoucis said:But if you suppose that you have such a maximum you get such a desired inequality!
Oh yes, so that concludes that $\text{Re}f(z)$ doesn't achieve its maximum.AlexYoucis said:No, because $|e^{f(z)}|$ can't reach a maximum neither can $\ln(|e^{f(z)}|)=\text{Re}(f(z))$.