We are given an analytic function $f$ on an open subset $G\subseteq \mathbb{C}$. I want to show the function $\phi$ on $G \times G$ given by $$(z,w) \mapsto \frac{f(z)-f(w)}{z-w}$$ if $z\neq w$ and $$(z,w) \mapsto f’(w)$$ if $z=w$, is continuous for all points of the form $(v,v)$.
Intuitively, this makes sense and it is easy to see if I first let $w$ approach $v$ and then let $z$ approach $v$. However, when I try to show this formally, I run into trouble.
Let $\varepsilon>0$. We note that for any $z$ and $w$ distinct, $$|f’(v)-\frac{f(z)-f(w)}{z-w}|\leq |f’(v)-f’(w)| +|f’(w)-\frac{f(z)-f(w)}{z-w}|. $$
The first summand of the right hand side can be made smaller than $\varepsilon/2$ by taking $w$ sufficiently close to $v$ because of the continuity of $f’$, say $ |w-v|<\delta$. I can also make the second summand smaller than $\varepsilon/2$ by taking $z$ sufficiently close to $w$, say $|z-w| < \delta_w$.
My problem is that $\delta_w$ depends on $w$, so I cannot simply assert that $|(v,v) - (z,w)| = |v-z|^2 + |v-w|^2 $ be less than the smaller of $\delta$ and $\delta_w$.
Could you provide a means to continue this approach? or provide an alternative approach?
Thanks.