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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.

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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!

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1 Answer 1

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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 $|h(z)-h(z')| \le | \int_\gamma (g(z,w)-g(z',w))dw| \le \epsilon l(\gamma)$.

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