Mean Value Theorem for complex functions?

Question

Let $\varphi(z,t)$ is defined on $\Omega \times X$ where $X$ is any measure space and $\Omega \subset \mathbb{C}$, and $\varphi(z,t)$ is analytic on $\Omega$. Let $K \subset \Omega$ be a compact set.

I am trying to prove that for any fixed $t \in X$, $$\left|\frac{\varphi(z,t) - \varphi(z_0,t)}{z-z_0}\right| \leq \sup_{s \in K} \left|\frac{d}{dz}\varphi(z,t)|_{z=s}\right|$$ for $z, z_0 \in K$.

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?

Answer 1

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 $$\left|{\phi(z)-\phi(z_0)\over z-z_0}\right|\ \leq\ \sup_{\zeta\in K}\bigl|\phi'(\zeta)\bigr|\ .$$

Answer 2

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.

Answer 3

Please see this paper:

http://ejde.math.txstate.edu/Volumes/2012/34/cakmak.pdf

for your information.