Showing uniform continuity of function giving radius of convergence

Question

Let $f$ be an analytic function on an open disk $D$ and let $R(z)$ denote the radius of convergence of the power series of $f$ about a point $z$. Is there an easy way to show that $|R(z_1) - R(z_2)| \leq |z_1-z_2|$?

Answer 1

The statemet doesn't seem to be true: Take $f(z) = 1/(1-z^2)$ on the unit disk, and take $z_1=1/2$ and $z_2=-1/2$, so that $R(z_1)=r(z_2)=1/2$. Indeed, in general the statement cannot be true if there are any two distinct points $z_1,z_2$ such that $R(z_1)=R(z_2)$. (But this isn't the only failure: still take $f(z) = 1/(1-z^2)$ but now $z_1=3/5$ and $z_2=-4/5$, etc.)