Primitive and Cauchy theorem for Banach-valued holomorphic maps on star-shaped domains

Triangle subdivision. For a closed nondegenerate triangle $\mathrm{\Delta }\subset U$, put $I(\mathrm{\Delta })={\int }_{\mathrm{\partial }\mathrm{\Delta }}Fdw$. Subdivide into four similar triangles, with matching boundary orientations; internal edges cancel by [L2, L3], so some child has integral norm at least $\mathrm{\parallel }I(\mathrm{\Delta })\mathrm{\parallel }\mathrm{/}4$. Order the four children once and take the first satisfying this inequality at each subdivision. The resulting nested triangles ${\mathrm{\Delta }}_n$ have diameter $2^{-n}d$ and perimeter $2^{-n}p$, where $d,p$ are those of $\mathrm{\Delta }$, and $\mathrm{\parallel }I({\mathrm{\Delta }}_n)\mathrm{\parallel }\ge 4^{-n}\mathrm{\parallel }I(\mathrm{\Delta })\mathrm{\parallel }$. Their intersection is a point $w_0$: a specified vertex of each triangle is a Cauchy sequence in $\mathbb{C}$, its limit lies in every closed triangle, and the diameters tend to zero.

Goursat's estimate. Differentiability at $w_0$ gives $F(w)=F(w_0)+F^{\mathrm{\prime }}(w_0)(w-w_0)+(w-w_0)r(w)$ with $r(w)\to 0$ as $w\to w_0$ and $r(w_0)=0$. The affine part has polynomial primitive $F(w_0)w+F^{\mathrm{\prime }}(w_0)(w-w_0{)}^2\mathrm{/}2$, so its boundary integral vanishes by [L1]. On ${\mathrm{\Delta }}_n$, $\mathrm{\mid }w-w_0\mathrm{\mid }\le 2^{-n}d$ and ${\sup }_{{\mathrm{\Delta }}_n}\mathrm{\parallel }r\mathrm{\parallel }\to 0$; hence [L2] gives $\mathrm{\parallel }I({\mathrm{\Delta }}_n)\mathrm{\parallel }\le 4^{-n}pd{\sup }_{{\mathrm{\Delta }}_n}\mathrm{\parallel }r\mathrm{\parallel }$. Comparing with step 1.1 proves $I(\mathrm{\Delta })=0$. For a degenerate triangle the oriented segment integrals cancel directly.

The segment primitive. The segment integrand defining $G(z)$ is continuous, so [L1] makes it integrable. Fix $z\in U$ and take $h$ sufficiently small that $[z,z+h]\subset U$. Every point of $\mathrm{conv}\{a,z,z+h\}$ is on a segment from $a$ to a point of $[z,z+h]$, so the triangle lies in $U$. Its boundary integral is zero by step 2.1; additivity and reversal therefore give $G(z+h)-G(z)={\int }_{[z,z+h]}Fdw=h{\int }_0^{1}F(z+th)dt$. The norm of the difference between this quotient and $F(z)$ is at most ${\sup }_{0\le t\le 1}\mathrm{\parallel }F(z+th)-F(z)\mathrm{\parallel }$, which tends to zero by continuity. Thus $G^{\mathrm{\prime }}=F$.

Closed contours. On each $C^1$ piece of a contour $\gamma$, the difference-quotient chain rule gives $(G\circ \gamma {)}^{\mathrm{\prime }}=F(\gamma ){\gamma }^{\mathrm{\prime }}$; this derivative is continuous on the closed piece because $F$ and ${\gamma }^{\mathrm{\prime }}$ are continuous. Applying [L1] piecewise and telescoping gives ${\int }_{\gamma }Fdw=G(\gamma (\beta ))-G(\gamma (\alpha ))$. It is zero for a closed contour; concatenating a contour with the reversal of another having the same endpoints gives path independence. The subdivision choices were specified by a finite ordering, so no choice principle beyond Countable Choice was used.