# Is this mean value theorem? (complex analysis)

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!

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

## 1 Answer

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 $$|\lambda | \le 1$$.

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