# Doubt about integral Cauchy theorem proof

I'm having trouble with the proof of the homological integral Cauchy theorem. I'm studying on Serge Lang, complex analysis, chapter $$4$$, page $$148$$, theorem $$2.5$$.

$$f\colon A \subseteq \mathbb{C} \to \mathbb{C}$$ is holomorphic in $$A$$ open set.

We have the continuous function $$g\colon A \times A \subseteq \mathbb{C}^2 \to \mathbb{C} \mid g(z,w)= \begin{cases} \frac{f(w)-f(z)}{w-z} \text{ if } z \neq w\\ f'(z) \text{ if } z=w\end{cases}$$.

We then define $$h\colon A \to \mathbb{C} \mid h(z)=\oint_{\gamma} g(z,w)\,\text{d}w$$ where $$\gamma\colon [a,b] \to \mathbb{C}$$ is a piecewise smooth closed curve in $$A$$.

We want to show that $$h$$ is continuous on $$A$$.

The textbook says that since $$g$$ is uniformly continuous on every compact subset of $$A \times A$$, then it follows at once that $$h$$ is continuos in $$A$$. Now, I can't see why.



I have tried the following approach:

If $$K \subseteq A \times A$$ is a compact set, then $$g$$ is uniformly continuous on $$K$$, namely:

$$\forall \,\epsilon>0 \quad \exists \,\delta>0 \mid \forall \,(z_0,w_0) \in K \quad \forall \,(z,w) \in B_{\delta}(z_0,w_0) \cap K \quad |g(z,w)-g(z_0,w_0)|<\epsilon$$.

Then:

$$\forall \,z_0 \in A \quad \forall \,\epsilon>0 \quad \exists \,\delta>0 \mid \forall \,w \in \gamma([a,b]) \quad \forall \,z \in B_{\delta}(z_0) \subseteq \overline{B_{\delta}(z_0)} \subseteq A$$

$$|g(z,w)-g(z_0,w)|<\epsilon$$

since:

$$(z_0,w) \in K=\overline{B_{\delta}(z_0)} \times \gamma([a,b]) \subseteq A \times A$$ and $$(z,w) \in B_{\delta}(z_0,w) \cap K$$.

Then:

$$\forall \,z_0 \in A \quad \forall \,\epsilon>0 \quad \exists \,\delta>0 \mid \forall \,z \in B_{\delta}(z_0) \quad |h(z)-h(z_0)|=|\oint_{\gamma} (g(z,w)-g(z_0,w))\,\text{d}w| \le$$

$$\le \text{length}(\gamma)\max_{w \in \gamma([a,b])} |g(z,w)-g(z_0,w)|<\text{length}(\gamma)\epsilon$$.

This shows that $$h$$ is continuous in $$z_0 \in A$$, which is arbitrary, and we have done.

Am I totally wrong? Thank you!

The key results are that $$g$$ is uniformly continuous on compact sets and if $$f$$ is bounded by $$M$$ on the range of $$\gamma$$ then $$|\int_\gamma f(w)dw| \le M l(\gamma)$$, where $$l$$ ls the length of $$\gamma$$.
Note that $$C_2 = \gamma([a,b]) \subset A$$ is compact. Pick some $$z \in A$$ and some $$r>0$$ such that the compact set $$C_1=\overline{B(z,r)} \subset A$$. Note that $$C=C_1 \times C_2 \subset A^2$$ is compact and so $$g$$ is uniformly continuous on $$C$$.
Pick $$\epsilon >0$$, then there is some $$\delta$$ such that if $$\|(z,w)-(z',w')\|_2 < \delta$$ then $$|g(z,w)-g(z',w')| < \epsilon$$. In particular, if $$|z-z'| < \epsilon$$ then $$|g(z,w)-g(z',w)| < \epsilon$$.
Hence $$|h(z)-h(z')| \le | \int_\gamma (g(z,w)-g(z',w))dw| \le \epsilon l(\gamma)$$.