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