3
$\begingroup$

In complex plane $\mathbb C$, $G$ is a region and $a,b \in G$. $G$ contains the line segment $C$ from $a$ to $b$. Now $f:G \to \mathbb C$ is holomorphic. Prove that there exist $\lambda$ with $|\lambda|\le1$ and $\theta\in C$ such that $f(b)-f(a)=\lambda (b-a)f'(\theta)$

For me, it looks very like something related to mean value theorem but still i have no idea on how to prove that. Can anyone guide me or provide some hints? Thanks!

$\endgroup$
3
  • $\begingroup$ Have you thought of writing $f= \operatorname{Re}(f) + i\operatorname{Im}(f)$? $\endgroup$
    – David P
    Mar 20, 2016 at 3:59
  • $\begingroup$ Set $g : [0,1]\to\mathbb C$, $g(t) = f(tb + (1-t)a)$ and use that $|g(s) - g(t)|\le \left(\max_{\zeta\in [s,t]}|g'(\zeta)|\right)|s-t|$. $\endgroup$ Mar 20, 2016 at 4:02
  • $\begingroup$ Yes, it is true. $\endgroup$ Mar 20, 2016 at 4:18

1 Answer 1

1
$\begingroup$

It is the multivariable version of the mean value theorem.

We have $f(b)-f(a)= \int_0^1 f'(a+t(b-a)) dt (b-a)$, and so ${ f(b)-f(a) \over b-a } = \int_0^1 f'(a+t(b-a)) dt$.

If $f'(z) = 0 $ for all $z \in C$ then any $z \in C$ will suffice, so suppose $f'(z) \neq 0$ for some $z \in C$.

Let $\theta \in C$ be such that $|f'(\theta)| = \max_{t \in [0,1]} |f'(a+t(b-a))|$, and let $\lambda = {\int_0^1 f'(a+t(b-a)) dt \over f'(\theta)}$. It is straightforward to check that $|\lambda | \le 1$.

$\endgroup$
13
  • $\begingroup$ Hmm. Why is $f(z) = \cos(z) + i\sin(z)$ with $a=0$ and $b=2\pi$ no counter example? $\endgroup$
    – user251257
    Mar 20, 2016 at 4:31
  • 2
    $\begingroup$ @user251257: $\lambda = 0$. ${}$ $\endgroup$
    – copper.hat
    Mar 20, 2016 at 4:32
  • $\begingroup$ oh yeah. I am silly. Thx $\endgroup$
    – user251257
    Mar 20, 2016 at 4:33
  • $\begingroup$ @copper.hat Derive the function in the max. ;o) $\endgroup$ Mar 20, 2016 at 4:33
  • $\begingroup$ @FriedrichPhilipp: I have had a glass of wine and I never drink and derive. $\endgroup$
    – copper.hat
    Mar 20, 2016 at 4:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .