Let $\varphi_t$ be an analytic function on an open domain $\Omega\subseteq\mathbb{C}$.

Let $K \subset \Omega$ be a compact set.

I am trying to prove that for any fixed parameter and fixed values: $$\left|\frac{\varphi_t(b) - \varphi_t(a)}{b-a}\right| \leq \sup_{z \in K} \left|\frac{d}{dz}\varphi_t(z)\right|$$

For a second, I thought mean value theorem might work here, but then I realized that MVT does not exist for complex functions.

Any ideas for proving the statement?

  • $\begingroup$ Is this question related to generators of semigroups e.g. time evolutions? $\endgroup$ Commented Dec 8, 2014 at 15:35

3 Answers 3


As mrf has shown there is no general inequality of the conjectured kind. But you can argue as follows: $$\bigl|\phi(z)-\phi(z_0)\bigr|=\left| \int_\gamma \phi'(\zeta)\ d\zeta\right| \leq \int_\gamma \bigl|\phi'(\zeta)\bigr|\ |d\zeta|\leq \sup_{\zeta\in K}\bigl|\phi'(\zeta)\bigr|\ L(\gamma)$$ for any curve $\gamma$ connecting $z_0$ with $z$ within $K$. When $K$ happens to be convex you can take for $\gamma$ the segment connecting $z_0$ with $z$. It has length $|z-z_0|$, so in this case you indeed have an inequality of the form $$\left|{\phi(z)-\phi(z_0)\over z-z_0}\right|\ \leq\ \sup_{\zeta\in K}\bigl|\phi'(\zeta)\bigr|\ .$$

  • 1
    $\begingroup$ Hey, thanks that was a lot simpler that I thought! Your help is greatly appreciated. $\endgroup$
    – aviness
    Commented Oct 7, 2012 at 9:51

This is not true. Look at an arc $K$ of radius $R$ with center at the origin starting close to (but just above) the negative real axis and ending just below the negative real axis. Let $\phi(z) = \log z$ (taken as the principal branch) and let $\Omega$ be a small open neighborhood of the arc, chosen so small that it doesn't contain any points on the negative real axis.

Then with $z = Re^{it}$ and $z_0 = Re^{-it}$ with $t$ close to $\pi$, we get $$|f(z)-f(z_0)| \approx 2\pi$$ and the left hand side of your inequality is very large (since $|z-z_0| \approx 0$). On the other hand, the right hand side is roughly $1/R$ which is very small if $R$ is large.

  • $\begingroup$ If you're interested, you are heartly welcome to post this example on: Complex Mean Value Theorem: Counterexamples $\endgroup$ Commented Dec 8, 2014 at 16:38
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    $\begingroup$ @Freeze_S Duplicated answers is about as bad as duplicated questions! $\endgroup$
    – mrf
    Commented Dec 8, 2014 at 20:57
  • $\begingroup$ Ok, I will close the other thread anyway. Hope you don't mind taking your example. $\endgroup$ Commented Dec 8, 2014 at 20:59

Please see this paper:


for your information.


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