2
$\begingroup$

Suppose $X$ is a measurable space, $\Omega$ is an open set in the complex plane, $\phi$ is a bounded function on $\Omega\times X$ such that $\phi(z,t)$ is a measurable function of $t$, for each $z\in\Omega$, and $\phi(z,t)$ is holomorphic in $\Omega$, for each $t\in X$.

Prove that to every compact $K\subset \Omega$ there corresponds a constant $M<\infty$ such that $\left|\frac{\phi(z,t)-\phi(z_{0},t)}{z-z_{0}}\right|

This question is really tough to me, and I have no idea how to prove it. Any hints will be appreciated. Thanks

  • 0
    Comment: The $t$ variable is pretty much irrelevant here since the function is bounded.. you can fix $t$ and the bound you get will be independent of $t$. See Willie Wong's hints on how to get such a bound.2011-09-13

1 Answers 1

1

A really rough bound, hint:

  1. The LHS is bounded by $\sup_K |\frac{d}{dz}\phi(\cdot,t)|$.
  2. Since $K$ is compact, there exists some $r$ such that for any point $z$ in $K$, the circle of radius $r$ around $z$ is in $\Omega$.
  3. You can estimate the derivative of a holomorphic function f' by the Cauchy differentiation formula, to control it by a bound on $f$ and a lower-bound on the radius $r$.
  • 0
    Thanks!I am totally new for Complex analysis, so maybe I just need some tricks.2011-09-13