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.


1 Answer 1


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.

  • $\begingroup$ Why integrate???? $\endgroup$
    – Rob Arthan
    Sep 27, 2015 at 0:21
  • $\begingroup$ Because if you divide by $z-w$ it is easy to tell the result is continuous as desired. $\endgroup$
    – zhw.
    Sep 27, 2015 at 4:45
  • $\begingroup$ 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 $w$? $\endgroup$
    – Yrogirg
    Sep 27, 2015 at 17:08
  • $\begingroup$ I added some more details. $\endgroup$
    – zhw.
    Sep 27, 2015 at 17:48

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