Let $f$ be analytic on the open set $G\subset \mathbb{C}$. What's the best way of showing that the function $\phi:G \times G \to \mathbb{C}$ defined by $\phi(z,w)=[f(z)-f(w)]/(z-w)$ for $z \neq w$ and \phi(z,z)=f'(z) is continuous?
This is obvious for $z\neq w$, but I'm having some trouble on the points $(z,z)$. I did it one way but I'm not convinced this is good.
Edit: Rudin's Proof (for the continuity on the diagonal)
Let $z_0\in G$. Since $f$ is analytic, there exists $r>0$ such that $B(a,r)\subset G$ and |f'(\zeta)-f'(z_0)|<\epsilon for all $\zeta \in B(a,r)$. Getting $z,w\in B(a,r)$, than $\zeta(t)=(1-t)z+tw \in B(a,r)$ for $0\le t\le 1$. Now just use that \phi(z,w)-\phi(z_0,z_0)=\int_0^1[f'(\zeta(t))-f'(z_0)]dt and the $\epsilon$-bound to get the desired continuity.