0
$\begingroup$

Let $f : \mathbb{C} \rightarrow \mathbb{R} $ is analytic inside and on a simple closed curve $C$, then the maximum value of $f(z)$ occurs over $C$?

It's known the Maximum modulus theorem but in this case you take only the function $f$, is it true or not? I can´t prove it.

$\endgroup$
  • 4
    $\begingroup$ If the codomain of your function is indeed $\mathbb{R}$ and analytic means complex-analytic, then $f$ is necessarily constant. If you mean to talk about an analytic function $f\colon\mathbb{C}\rightarrow\mathbb{C}$ and the maximum of its modulus $|f|$, then I don't see how the statement is different from the Maximum modulus principle. $\endgroup$ – Thorgott Dec 16 '19 at 17:01
  • $\begingroup$ @Thorgott How it's different from MMT: Well of course it is a version of MMT. But getting this from, say. "an analytic function cannot have a local maximum" requires some highly non-trivial topology. (If you don't believe that, tell me, what's the definition of "on and inside $C$"?) $\endgroup$ – David C. Ullrich Dec 16 '19 at 17:27
  • $\begingroup$ What is the meaning of : $f$ have maximum value? . I think that is the following definition. $f$ have maximum value in $z_0$ if $\vert f(z) \vert \leq \vert f(z_0) \vert$ for all $z $ $\endgroup$ – Juan Daniel Valdivia Fuentes Dec 18 '19 at 1:51
0
$\begingroup$

It's true (after fixing the typo; as already noted, the $f:\Bbb C\to\Bbb R$ should be $f:\Bbb C\to\Bbb C$. Proving it is non-trivial, because first you need a definition of "inside $C$".

This is exactly why you're more likely to find language like "on and inside $C$" in elementary books like say Brown&Churchill; they're willing to just wave their hands. In more rigorous/"advanced" texts these days they tend to avoid the notion of the inside of a curve... (what needs to be proved after the definitions are in place is that if $V$ is the bounded open set bounded by $C$ then $C$ is in fact the topological boundary of $V$, or equivalently $\overline V = V\cup C$. Once you know that, the version of MMT you know gives the current version.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.