How to use the completeness here

Question

Suppose that $G\subset \mathbb{C}$ is an open subset and that $X$ is a Banach space. Fix $z_0 \in G$ and let $V:=\{(h,k) \in \mathbb{C}^*\times\mathbb{C}^* : z_0+h \in G\ ; z_0 + k \in G\}$. Let $f: G \to X$ be a function such that $$ \sup_{(h,k) \in V} \frac{1}{|h-k|}\left\| \frac{f(z_0+h)-f(z_0)}{h}-\frac{f(z_0+k)-f(z_0)}{k}\right\| < \infty $$

I have just read that the above implies, by the completeness of $X$ that the limit $$ \lim_{h \to 0} \frac{f(z_0+h)-f(z_0)}{h} $$ exist in X.

At first sight I thought this could be something like the Cauchy criterion, but I can´t see how. I feel it must be a triviality, can anyone here see further than I?

Thanks in advance !

Answer 1

Let $(h_n)$ a sequence in $V$ converging in $\mathbb{C}$.

Let $u_n = \frac{f(z_0+h_n)-f(z_0)}{h_n}$

The bounded $\sup$ condition implies $||u_p - u_q|| \leq M |h_p-h_q|$

Hence $(u_n)$ is a Cauchy sequence in $X$ so it converges in $X$ by completeness.

Therefore $\lim \frac{f(z_0+h_n)-f(z_0)}{h_n}$ exists in X.