Mean Value Theorem for complex functions? 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?
 A: 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.  
A: 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
$$\left|{\phi(z)-\phi(z_0)\over z-z_0}\right|\ \leq\ \sup_{\zeta\in K}\bigl|\phi'(\zeta)\bigr|\  .$$
A: Please see this paper:
http://ejde.math.txstate.edu/Volumes/2012/34/cakmak.pdf
for your information.
