# Prove the function is continuous, exercise from Conway's "Functions of One Complex Variable I"

For the first proof of Cauchy's integral formula, Conway in his book "Functions of One Complex Variable" (Chapter IV, section 5.4) uses the following claim:

Let $G$ be an open subset of $\mathbb C$ and $f : G \to \mathbb C$ an analytic function. Then define $\varphi : G \times G \to \mathbb C$ as $$\varphi(z,w) = \begin{cases} \dfrac{f(z)-f(w)}{z-w}, && z\neq w \\ f'(z) && z=w \end{cases}.$$ Then $\varphi$ is continuous on $G \times G$.

Unfortunately, he does not give a proof, leaving it as an exercise (Exercise 1).

I tried to make a direct proof by taking $|z-z_0| + |w-w_0| < \delta$ as a neighbourhood of $(z_0,w_0)$, but it lead to nowhere. I understand, that $z \mapsto \varphi(z,w)$ and $w \mapsto \varphi(z,w)$ are both holomorphic and continuous, but I do not see how one could use it.

I also saw this claim in several lecture notes but it goes unproven there either.

I would appreciate a hint, we have this problem as a homework.

Hint: The only troublesome case is when $(z,w) \to (a,a)$  for some $a\in G.$ Here note

$$f(z) - f(w) = \int_0^1 f'(w +t(z-w))(z-w)\ dt$$

for $z,w$ in any $D(a,r)\subset G.$

Added the following details later: If $z,w \in D(a,r)\subset G,$ then $\varphi(z,w) = \int_0^1f'(w+t(z-w))\ dt,$ including the case where $z=w.$ Note that $|w+t(z-w) - a| \le |z-a|+|w-a|$ for $t\in [0,1].$ Thus as $(z,w) \to (a,a),$ $f'(w+t(z-w)) \to f'(a)$ uniformly on $[0,1].$ That gives the desired convergence of the integral to $f'(a),$ which is what we want.

• Why integrate???? Sep 27, 2015 at 0:21
• Because if you divide by $z-w$ it is easy to tell the result is continuous as desired.
– zhw.
Sep 27, 2015 at 4:45
• The main point of this rewriting as I see is to get a clearly continuous function under the integral sign. But how then to make a point that after the integration is remains continuous in $z$ and $w$? Sep 27, 2015 at 17:08
• I added some more details.
– zhw.
Sep 27, 2015 at 17:48