Iterated Cauchy formula, power-series expansion and Cauchy estimates for functions continuous and separately holomorphic on a polydisc

Iterated Cauchy formula. We prove clause 1 by induction on $m$. For $m=1$ it is exactly [F1]. Assume it holds for $m-1\ge 1$ and let $f$ be as in the statement. Fix $z\in {\mathrm{\Delta }}_r(a)$ and write $z=(z_1,z^{\mathrm{\prime }})$, $a=(a_1,a^{\mathrm{\prime }})$. For $z^{\mathrm{\prime }}$ fixed, the slice ${\zeta }_1\mapsto f({\zeta }_1,z^{\mathrm{\prime }})$ is holomorphic on the disc $\mathrm{\mid }{z}_1-a_1\mathrm{\mid }<{\rho }_1$ and continuous on its closure as the restriction of a continuous map, so [F1] gives $f(z_1,z^{\mathrm{\prime }})=\frac{1}{2\pi i}{\oint }_{\mathrm{\mid }{\zeta }_1-a_1\mathrm{\mid }=r_1}\frac{f({\zeta }_1,z^{\mathrm{\prime }})}{{\zeta }_1-z_1}d{\zeta }_1$. For each ${\zeta }_1$ on the circle, the function $z^{\mathrm{\prime }}\mapsto f({\zeta }_1,z^{\mathrm{\prime }})$ is continuous and separately holomorphic on ${\mathrm{\Delta }}_{{\rho }^{\mathrm{\prime }}}(a^{\mathrm{\prime }})$, ${\rho }^{\mathrm{\prime }}=({\rho }_2,\dots ,{\rho }_m)$, so the induction hypothesis gives $f({\zeta }_1,z^{\mathrm{\prime }})$ as an $(m-1)$-fold iterated integral over $\mathrm{\mid }{\zeta }_k-a_k\mathrm{\mid }=r_k$, $k\ge 2$, for $z^{\mathrm{\prime }}\in {\mathrm{\Delta }}_{r^{\mathrm{\prime }}}(a^{\mathrm{\prime }})$. Substituting and interchanging the iterated integrals is legitimate by [F4], since after parametrizing each circle by $[0,2\pi ]$ the integrand is continuous on the compact product and the denominators never vanish ($\mathrm{\mid }{z}_k-a_k\mathrm{\mid }<r_k$). This yields the displayed formula with the outer variable ${\zeta }_1$; the argument applies to every order of the variables, either directly or by [F4].

Expansion of the kernel. Fix $z\in {\mathrm{\Delta }}_r(a)$ and $r^{\mathrm{\prime }}$ with $r_k<r_k^{\mathrm{\prime }}<{\rho }_k$, and consider on the distinguished boundary ${\mathrm{\Gamma }}_{r^{\mathrm{\prime }}}(a)$ the product of the geometric series ${\prod }_{k=1}^m\frac{1}{{\zeta }_k-z_k}={\prod }_{k=1}^m{\sum }_{n\ge 0}\frac{(z_k-a_k{)}^n}{({\zeta }_k-a_k{)}^{n+1}}={\sum }_{\alpha \in {\mathbb{N}}^m}\frac{(z-a{)}^{\alpha }}{(\zeta -a{)}^{\alpha +\mathbf{1}}},$ valid because $\mathrm{\mid }{z}_k-a_k\mathrm{\mid }<r_k^{\mathrm{\prime }}=\mathrm{\mid }{\zeta }_k-a_k\mathrm{\mid }$; its partial sums $S_N(\zeta ):={\sum }_{\mathrm{\mid }\alpha {\mathrm{\mid }}_{\mathrm{\infty }}\le N}(\zeta -a{)}^{-\alpha -\mathbf{1}}(z-a{)}^{\alpha }$ converge uniformly on the compact parameter rectangle to the full product, by [F6] and the Weierstrass test with majorant ${\sum }_{\alpha }{\prod }_k(r_k\mathrm{/}{r}_k^{\mathrm{\prime }}{)}^{{\alpha }_k}={\prod }_k(1-r_k\mathrm{/}{r}_k^{\mathrm{\prime }}{)}^{-1}<\mathrm{\infty }$.

Power-series expansion at a fixed contour radius. Fix any polyradius $s$ with $0<s_k<{\rho }_k$ and any $z\in {\mathrm{\Delta }}_s(a)$. Choose an inner polyradius $\widehat{s}$ with $\mathrm{\mid }{z}_k-a_k\mathrm{\mid }<{\widehat{s}}_k<s_k$ for all $k$; the kernel expansion of step 1.2 applies with $\widehat{s}$ as the inner radius and $s$ as the contour radius. Applying clause 1 with contour radius $s$ and substituting that expansion, the integral of $f{S}_N$ converges to that of $f{\prod }_k({\zeta }_k-z_k{)}^{-1}$ by [F4]; by linearity the integral of $f{S}_N$ is the finite sum ${\sum }_{\mathrm{\mid }\alpha {\mathrm{\mid }}_{\mathrm{\infty }}\le N}{c}_{\alpha }(s)(z-a{)}^{\alpha }$, where $c_{\alpha }(s)$ is the displayed iterated contour integral with radii $s$. Hence these partial sums converge to $f(z)$. The coefficient estimate $\mathrm{\mid }{c}_{\alpha }(s)\mathrm{\mid }\le M(s){\prod }_k{s}_k^{-{\alpha }_k}$ follows from [F5], each factor contributing $(2\pi s_k)\mathrm{/}(2\pi )$. By [F6], the resulting product of geometric series proves absolute convergence on ${\mathrm{\Delta }}_s(a)$ and uniform convergence on every compact subset of ${\mathrm{\Delta }}_s(a)$, since the coordinate ratios have suprema strictly below $1$ there.

Independence of the contour radii. Fix a multi-index $\alpha$ and two polyradii $s,\tilde{s}$ with all components strictly between $0$ and the corresponding ${\rho }_k$. Connect them by changing one radius at a time. For a fixed coordinate $k$, hold the other contour radii fixed and integrate first in those other variables, defining $g({\zeta }_k):=\frac{1}{(2\pi i{)}^{m-1}}{\oint }_{{\mathrm{\Gamma }}_{s_{\mathrm{\ne }k}}(a_{\mathrm{\ne }k})}\frac{f(\zeta )}{{\prod }_{\mathrm{\ell }\mathrm{\ne }k}({\zeta }_{\mathrm{\ell }}-a_{\mathrm{\ell }}{)}^{{\alpha }_{\mathrm{\ell }}+1}}{\prod }_{\mathrm{\ell }\mathrm{\ne }k}d{\zeta }_{\mathrm{\ell }}$. For $m=1$ this means $g=f$. The parameter-integral theorem [F3], applied to the scalar integrands of these iterated contour integrals, shows that $g$ is holomorphic on $\mathrm{\mid }{\zeta }_k-a_k\mathrm{\mid }<{\rho }_k$. Differentiating its one-variable Cauchy formula [F1] ${\alpha }_k$ times at $a_k$ under the contour integral, justified by [F3] because the differentiated kernels are jointly continuous on the contour, gives $\frac{1}{2\pi i}{\oint }_{\mathrm{\mid }{\zeta }_k-a_k\mathrm{\mid }=s_k}\frac{g({\zeta }_k)}{({\zeta }_k-a_k{)}^{{\alpha }_k+1}}d{\zeta }_k=\frac{g^{({\alpha }_k)}(a_k)}{{\alpha }_k!}$. The right side is independent of $s_k$. By [F4] the contour order can be chosen so this is the $k$-th integral in $c_{\alpha }(s)$; therefore changing only $s_k$ leaves that coefficient unchanged. Repeating for each coordinate proves $c_{\alpha }(s)=c_{\alpha }(\tilde{s})$. In particular the coefficients do not depend on the admissible $r^{\mathrm{\prime }}$. For any compact subset of ${\mathrm{\Delta }}_{\rho }(a)$, choose a contour polyradius $s$ strictly larger in each coordinate than the maximum coordinate distance of that compact set from $a$; step 2.1 and the established independence give uniform absolute convergence there.

Complex differentiability at the centre. Write the expansion of step 2.1 at the point $a$ as $f(a+v)=c_0+L(v)+q(v)$ with $L(v):={\sum }_k{c}_{e_k}{v}_k$ (a $\mathbb{C}$-linear map) and $q(v):={\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }\ge 2}{c}_{\alpha }{v}^{\alpha }$. Fix a polyradius $r^{\mathrm{\prime }}$ with $0<r_k^{\mathrm{\prime }}<{\rho }_k$ and take $\mathrm{\mid }{v}_k\mathrm{\mid }\le r_k^{\mathrm{\prime }}\mathrm{/}2$; put ${\theta }_k:=\mathrm{\mid }{v}_k\mathrm{\mid }\mathrm{/}{r}_k^{\mathrm{\prime }}\le 1\mathrm{/}2$. The coefficient bounds give $\mathrm{\mid }q(v)\mathrm{\mid }\le M(r^{\mathrm{\prime }}){\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }\ge 2}{\prod }_k{\theta }_k^{{\alpha }_k}\le M(r^{\mathrm{\prime }})C_m{\max }_k{\theta }_k^2\le \frac{M(r^{\mathrm{\prime }})C_m}{({\min }_k{r}_k^{\mathrm{\prime }}{)}^2}\parallel v{\parallel }^2$, where $C_m:={\sum }_{n\ge 2}\left(\genfrac{}{}{0px}{}{n+m-1}{m-1}\right)2^{-(n-2)}<\mathrm{\infty }$. Thus $q(v)=o(\parallel v\parallel )$ as $v\to 0$. Hence $f$ is complex differentiable at $a$ with $Df(a)=L$. Applying the same argument at an arbitrary centre $\zeta \in {\mathrm{\Delta }}_{\rho }(a)$ — the hypotheses hold on any polydisc centred at $\zeta$ contained in ${\mathrm{\Delta }}_{\rho }(a)$ — gives complex differentiability everywhere on ${\mathrm{\Delta }}_{\rho }(a)$, so $f$ is holomorphic there, with ${\mathrm{\partial }}_{z_k}f(a)=c_{e_k}$.

Derivative formula and estimates at the centre. Differentiating under the iterated integral of clause 1, the difference quotient of the kernel identity $\frac{1}{{\zeta }_k-z_k-h}-\frac{1}{{\zeta }_k-z_k}=\frac{h}{({\zeta }_k-z_k{)}^2}+\frac{h^2}{({\zeta }_k-z_k{)}^2({\zeta }_k-z_k-h)}$ and [F4] give, for every $k$, ${\mathrm{\partial }}_{z_k}f(a)=\frac{1}{(2\pi i{)}^m}{\oint }_{{\mathrm{\Gamma }}_r(a)}\frac{f(\zeta )}{{\prod }_{l\mathrm{\ne }k}({\zeta }_l-a_l)}\cdot \frac{d\zeta }{({\zeta }_k-a_k{)}^2}\mathrm{.}$ By [F5] the modulus is at most $M(r){\prod }_l{r}_l(r_l{)}^{-1}\cdot r_k^{-1}\cdot \dots =M(r)\mathrm{/}{r}_k$ (each unmodified factor contributes $r_l\mathrm{/}{r}_l=1$, the squared factor contributes $r_k\mathrm{/}{r}_k^2$), which is the estimate of clause 3 with $s=r$.

Continuity and separate holomorphy of the derivatives. Fix a polydisc ${\mathrm{\Delta }}_s(b)$ with ${\bar{\mathrm{\Delta }}}_s(b)\subseteq {\mathrm{\Delta }}_{\rho }(a)$. Compactness of ${\bar{\mathrm{\Delta }}}_s(b)$ inside the open polydisc lets us choose a polyradius $s^{\mathrm{\prime }}$ with $s_k<s_k^{\mathrm{\prime }}$ for every $k$ and ${\bar{\mathrm{\Delta }}}_{s^{\mathrm{\prime }}}(b)\subseteq {\mathrm{\Delta }}_{\rho }(a)$. Differentiating the iterated Cauchy formula on ${\mathrm{\Delta }}_{s^{\mathrm{\prime }}}(b)$, for $z\in {\mathrm{\Delta }}_s(b)$, ${\mathrm{\partial }}_{z_k}f(z)=\frac{1}{(2\pi i{)}^m}{\oint }_{{\mathrm{\Gamma }}_{s^{\mathrm{\prime }}}(b)}\frac{f(\zeta )}{({\zeta }_k-z_k{)}^2{\prod }_{\mathrm{\ell }\mathrm{\ne }k}({\zeta }_{\mathrm{\ell }}-z_{\mathrm{\ell }})}d{\zeta }_1\cdots d{\zeta }_m$. The integrand is jointly continuous in $z$ and the contour parameters on the compact product, so the integral is continuous in $z$; fixing all but one coordinate of $z$, the integrand is holomorphic in that coordinate and [F3] makes the integral holomorphic in it. Such polydiscs cover ${\mathrm{\Delta }}_{\rho }(a)$, proving the regularity assertion. The Cauchy bound of clause 3 applies at any centre with any polyradius strictly inside its domain.

Conclusion. Clause 1 is step 1.1; clause 2 combines steps 2.1 and 3.1; clause 3 combines steps 3.2, 3.3 and 4.1. No choice principle was used: the induction is over the finite range $1,\dots ,m$, the geometric series are evaluated by [F6], and the limit is a uniformly convergent sequence of finite partial sums.