Cauchy integral formula and Cauchy estimates for Banach-valued holomorphic functions

The filled quotient. Put $h(w)=(F(w)-F(z))\mathrm{/}(w-z)$ for $w\mathrm{\ne }z$ and $h(z)=F^{\mathrm{\prime }}(z)$. Differentiability of $F$ at $z$ makes $h$ continuous on $D(z_0,R)$, and the quotient rule makes it holomorphic away from $z$. The triangle-subdivision argument of Primitive and Cauchy theorem for Banach-valued holomorphic maps on star-shaped domains proves that a holomorphic map has zero integral around every closed triangle in its domain. This also holds for $h$ on triangles containing $z$: split such a triangle into at most three triangles with vertex $z$; in each remove a similar corner triangle of diameter $\eta$. The remaining quadrilateral can be split into triangles avoiding $z$, whose integrals vanish. Its boundary differs from the original by edges of total length $O(\eta )$, and $h$ is bounded near $z$, so the norm of this difference tends to zero by [L2]. Thus every triangle integral of $h$ in the disc vanishes. Degenerate triangles cancel by reversal.

Scalar circle integrals. Parametrizing $\gamma (s)=z_0+r{e}^{is}$ and setting $\rho :=z-z_0$ with $\mathrm{\mid }\rho \mathrm{\mid }<r$, the geometric series $\frac{1}{w-z}=\frac{1}{w-z_0}{\sum }_{n\ge 0}(\frac{z-z_0}{w-z_0}{)}^n={\sum }_{n\ge 0}\frac{(z-z_0{)}^n}{(w-z_0{)}^{n+1}}$ converges uniformly on $C_r$; integrating termwise and using $\frac{1}{2\pi i}{\oint }_{C_r}(w-z_0{)}^{m}dw=1$ for $m=-1$ and $=0$ for integers $m\mathrm{\ne }-1$ (a direct computation from ${\int }_0^{2\pi }{e}^{ikt}dt=2\pi$ for $k=0$ and $0$ otherwise) gives $\frac{1}{2\pi i}{\oint }_{C_r}\frac{dw}{w-z}=1$ and ${\oint }_{C_r}dw=0$.

A primitive for the filled quotient. Define $H(w)={\int }_{[z_0,w]}h(\zeta )d\zeta$ on the disc. The zero triangle integrals in step 1.1 give $H(w+k)-H(w)=k{\int }_0^{1}h(w+tk)dt$ for small $k$. Continuity of $h$ gives $H^{\mathrm{\prime }}=h$, exactly as in the segment-primitive argument of Primitive and Cauchy theorem for Banach-valued holomorphic maps on star-shaped domains. Applying its piecewise chain-rule and fundamental-theorem argument to $H$ along $C_r$ yields ${\oint }_{C_r}h(w)dw=0$. This uses continuity at the exceptional point, without assuming that $h$ is differentiable there.

Cauchy's integral formula. Put $J:={\oint }_{C_r}h(w)dw=0$. Writing $\frac{F(w)}{w-z}=\frac{F(z)}{w-z}+\frac{F(w)-F(z)}{w-z}$ and using [step 2.1] and [step 1.2], $\frac{1}{2\pi i}{\oint }_{C_r}\frac{F(w)}{w-z}dw=\frac{F(z)}{2\pi i}{\oint }_{C_r}\frac{dw}{w-z}+\frac{J}{2\pi i}=F(z)$.

Power series and coefficients. For $\mathrm{\mid }z-z_0\mathrm{\mid }<r$ the kernel expansion of [step 1.2] is uniformly convergent on $C_r$, so termwise integration of the identity of [step 3.1] gives $F(z)={\sum }_{n\ge 0}{a}_n(r)(z-z_0{)}^n$ with $a_n(r):=\frac{1}{2\pi i}{\oint }_{C_r}\frac{F(w)}{(w-z_0{)}^{n+1}}dw$, and $\mathrm{\parallel }{a}_n(r)\mathrm{\parallel }\le M(r)\mathrm{/}{r}^n$ by [L2]. Two radii $r_1<r_2<R$ give two power series with the same sum for every real $\zeta$ with $\mathrm{\mid }\zeta \mathrm{\mid }<r_1$ after the translation $z=z_0+\zeta$; applying [L3] to these series centered at $0$ gives $a_n(r_1)=a_n(r_2)$ for every $n$; writing $a_n$ for the common value, $F$ is represented on $D(z_0,R)$ by the norm-convergent series $\sum a_n(z-z_0{)}^n$, and in particular the coefficient integrals are independent of $r$.

Norm-$C^{\mathrm{\infty }}$ regularity and Cauchy estimates. Since $\mathrm{\parallel }{a}_n\mathrm{\parallel }\le M(r^{\mathrm{\prime }})\mathrm{/}{r}^{\mathrm{\prime }n}$ for every $0<r^{\mathrm{\prime }}<R$, for each $0<\rho <r^{\mathrm{\prime }}$ the differentiated series ${\sum }_{n\ge 1}n{a}_n(z-z_0{)}^{n-1}$ is dominated on $\mathrm{\mid }z-z_0\mathrm{\mid }\le \rho$ by ${\sum }_{n\ge 1}nM(r^{\mathrm{\prime }}){\rho }^{n-1}\mathrm{/}{r}^{\mathrm{\prime }n}=M(r^{\mathrm{\prime }})r^{\mathrm{\prime }-1}(1-\rho \mathrm{/}{r}^{\mathrm{\prime }}{)}^{-2}<\mathrm{\infty }$, so it converges uniformly there; the standard difference-quotient estimate $\mathrm{\mid }\frac{(z+h-z_0{)}^n-(z-z_0{)}^n}{h}-n(z-z_0{)}^{n-1}\mathrm{\mid }\le \frac{n(n-1)}{2}\mathrm{\mid }h\mathrm{\mid }{\sum }_k\left(\genfrac{}{}{0px}{}{n-2}{k}\right)\mathrm{\mid }z-z_0{\mathrm{\mid }}^k\mathrm{\mid }h{\mathrm{\mid }}^{n-2-k}$ together with the same geometric majorant shows that the difference quotients of the sum converge to the differentiated sum, so $F$ is complex-differentiable with $F^{\mathrm{\prime }}=\sum n{a}_n(\cdot -z_0{)}^{n-1}$; iterating gives $F^{(n)}(z_0)=n!a_n$ for every $n$, so $F$ is norm-$C^{\mathrm{\infty }}$ and the estimate $\mathrm{\parallel }{F}^{(n)}(z_0)\mathrm{\parallel }=n!\mathrm{\parallel }{a}_n\mathrm{\parallel }\le n!M(r)\mathrm{/}{r}^n$ follows from $\mathrm{\parallel }{a}_n\mathrm{\parallel }\le M(r)\mathrm{/}{r}^n$ at any $0<r<R$. Together with [step 3.1] this proves the formula, the expansion with radius-independent coefficients, and the Cauchy estimates, and no choice principle beyond Countable Choice was used.