The Bochner–Martinelli formula for C1 functions

Off the diagonal, differentiating the kernel gives $d(f{\mathrm{\Omega }}_n)={\bar{\mathrm{\partial }}}_{\zeta }f\wedge {\mathrm{\Omega }}_n$. [F1, F2, F3, given, algebra] Put $w=\zeta -z$, $r=\mathrm{\mid }w\mathrm{\mid }$, and let ${\mathrm{\Phi }}_j$ denote the omitted-factor wedge form in the $j$th summand of [F1]. Set $\mathrm{\Theta }=d{\bar{\zeta }}_1\wedge d{\zeta }_1\wedge \cdots \wedge d{\bar{\zeta }}_n\wedge d{\zeta }_n$ and $c_j={\bar{w}}_j{r}^{-2n}$. Since $d{\bar{\zeta }}_j\wedge {\mathrm{\Phi }}_j=\mathrm{\Theta }$, ${\bar{\mathrm{\partial }}}_{\zeta }{\mathrm{\Omega }}_n=\frac{(n-1)!}{(2\pi i{)}^n}{\sum }_{j=1}^n\frac{\mathrm{\partial }{c}_j}{\mathrm{\partial }{\bar{\zeta }}_j}\mathrm{\Theta }=0,$ because $\mathrm{\partial }{c}_j\mathrm{/}\mathrm{\partial }{\bar{\zeta }}_j=r^{-2n}-n\mathrm{\mid }{w}_j{\mathrm{\mid }}^2{r}^{-2n-2}$ and the sum is $n{r}^{-2n}-n{r}^2{r}^{-2n-2}=0$. Also ${\mathrm{\partial }}_{\zeta }(f{\mathrm{\Omega }}_n)=0$ because its holomorphic degree is $n$. The product rule [F3], together with $d=\mathrm{\partial }+\bar{\mathrm{\partial }}$, now gives the identity.

Bounded first derivatives and polar integration prove absolute integrability at the diagonal. [F1, F5, F12, F13, F14, given, algebra] Compactness of $\bar{D}$ bounds the first derivatives of $f$, so coefficients of $\bar{\mathrm{\partial }}f\wedge {\mathrm{\Omega }}_n$ are bounded near $z$ by $C{r}^{1-2n}$. Write ${\sigma }_{2n-1}$ for the finite sphere measure in [F14], and set $g(w)=\mathrm{\mid }w{\mathrm{\mid }}^{1-2n}$ for $0<\mathrm{\mid }w\mathrm{\mid }<ϵ$ and $g(w)=0$ otherwise. This is nonnegative Borel. Translation invariance [F12] makes $w\mapsto w+z$ measure preserving, so [F13] and [F14] give ${\int }_{B_{ϵ}(z)}\mathrm{\mid }\zeta -z{\mathrm{\mid }}^{1-2n}d{\lambda }_{2n}(\zeta )={\sigma }_{2n-1}(S^{2n-1}){\int }_0^{ϵ}{r}^{1-2n}{r}^{2n-1}dr={\sigma }_{2n-1}(S^{2n-1})ϵ<\mathrm{\infty }\mathrm{.}$ Thus the interior density is absolutely integrable at $z$; its integral over $D\setminus {\bar{B}}_{ϵ}(z)$ converges to its integral over $D$, and away from $z$ it is continuous on a bounded set.

The normalized kernel has integral one on every positively oriented sphere centered at $z$. [F4, F5, F7, F8, F9, F10, F11, algebra] Let ${\mathrm{\Psi }}_n={\sum }_{j=1}^n{\bar{w}}_j{\mathrm{\Phi }}_j$. On $\mathrm{\partial }{B}_{ϵ}(z)$, ${\mathrm{\Omega }}_n=(n-1)!(2\pi i{)}^{-n}{ϵ}^{-2n}{\mathrm{\Psi }}_n$, and direct differentiation gives $d{\mathrm{\Psi }}_n=n\mathrm{\Theta }$. Since $d{\bar{\zeta }}_j\wedge d{\zeta }_j=2id{x}_j\wedge d{y}_j$, the chosen orientation gives $\mathrm{\Theta }=(2i{)}^{n}dV$. Stokes [F4] on the ball therefore reduces the sphere integral to its real volume. Its closed ball is Jordan by [F8]; [F9] gives content zero to the boundary sphere. Under countable choice, [F10] identifies the closed ball's Lebesgue measure with its Jordan content and makes the sphere Lebesgue null, so the open and closed balls have the same measure. Formula [F7] in real dimension $2n$ and [F11] iterated at $s=1,\dots ,n$ yield ${\int }_{B_{ϵ}(z)}dV=\frac{{\pi }^n{ϵ}^{2n}}{\mathrm{\Gamma }(n+1)}=\frac{{\pi }^n{ϵ}^{2n}}{n!}\mathrm{.}$ Consequently Stokes gives ${\int }_{\mathrm{\partial }{B}_{ϵ}(z)}{\mathrm{\Omega }}_n=\frac{(n-1)!}{(2\pi i{)}^n}{ϵ}^{-2n}{\int }_{B_{ϵ}(z)}n(2i{)}^{n}dV=1.$

If $f$ is holomorphic, the interior term in the formula vanishes. [F6, given, algebra] At every point of $D$, holomorphicity makes $f$ complex differentiable. By [F6], each antiholomorphic Wirtinger derivative vanishes. Reindexing $k=j-1$ converts the library's zero-based coordinates $0\le k<n$ to the formula's $1\le j\le n$, so $\bar{\mathrm{\partial }}f=0$ and the interior term is zero.

For $n>1$, a kernel coefficient is not holomorphic in the parameter $z$. [F1, given, algebra] Choose distinct indices $j,l$ and write $c_j=\overline{{\zeta }_j-z_j}\mathrm{/}\mathrm{\mid }\zeta -z{\mathrm{\mid }}^{2n}$. Differentiating with respect to ${\bar{z}}_l$ gives $\frac{\mathrm{\partial }{c}_j}{\mathrm{\partial }{\bar{z}}_l}=n\overline{w_j}{w}_l\mathrm{\mid }w{\mathrm{\mid }}^{-2n-2}\mathrm{.}$ This is nonzero when both coordinate differences are nonzero, so this kernel coefficient is not holomorphic in $z$.

Stokes on the punctured domain gives the outer-minus-inner boundary identity. [F4, F5, step 1.1, given, algebra] Choose $0<ϵ<\mathrm{dist}(z,\mathrm{\partial }D)$ and set $D_{ϵ}=D\setminus {\bar{B}}_{ϵ}(z)$. Since $D$ is open by [F15], $z$ is an interior point and this distance is positive; choose $ϵ$ small enough that the closed ball lies in $D$. Its boundary is $\mathrm{\partial }D$ and the oppositely oriented sphere $-\mathrm{\partial }{B}_{ϵ}(z)$. If $D_{ϵ}$ is disconnected, a finite cover of its bounded $C^1$ boundary by graph charts, each with one connected interior side, shows it has only finitely many components; each inherits $C^1$ boundary. Apply [F4] to each component, where $f{\mathrm{\Omega }}_n$ is $C^1$ on the closure. The declared AC supplies [F4]'s premise, and step 1.1 supplies the differential identity. Summing gives ${\int }_{\mathrm{\partial }D}f{\mathrm{\Omega }}_n-{\int }_{\mathrm{\partial }{B}_{ϵ}(z)}f{\mathrm{\Omega }}_n={\int }_{D\setminus {\bar{B}}_{ϵ}(z)}{\bar{\mathrm{\partial }}}_{\zeta }f\wedge {\mathrm{\Omega }}_n\mathrm{.}$

Scaling and step 1.3 show that the inner-sphere integral tends to $f(z)$. [F1, step 1.3, given, algebra] On a fixed small ball about $z$, boundedness of $Df$ and the fundamental theorem of calculus along segments give $\mathrm{\mid }f(\zeta )-f(z)\mathrm{\mid }\le Mϵ$ on $\mathrm{\partial }{B}_{ϵ}(z)$. Under $u\mapsto z+ϵu$, the pullback of ${\mathrm{\Omega }}_n(\cdot ,z)$ to the unit sphere is independent of $ϵ$: its coefficient scales by ${ϵ}^{1-2n}$ and its $(2n-1)$ differentials by ${ϵ}^{2n-1}$. Smoothness on the compact unit sphere gives a finite absolute integral, so $\mid {\int }_{\mathrm{\partial }{B}_{ϵ}(z)}(f(\zeta )-f(z)){\mathrm{\Omega }}_n(\zeta ,z)\mid \le Cϵ⟶0.$ Using ${\int }_{\mathrm{\partial }{B}_{ϵ}(z)}{\mathrm{\Omega }}_n=1$ from step 1.3 proves the limit.

Letting the puncture radius tend to zero in step 2.1 proves the asserted formula. [step 1.2, step 2.1, step 2.2, algebra] By step 1.2 the interior integrals converge; by step 2.2 the inner-sphere integral tends to $f(z)$. Thus ${\int }_{\mathrm{\partial }D}f{\mathrm{\Omega }}_n-f(z)={\int }_D{\bar{\mathrm{\partial }}}_{\zeta }f\wedge {\mathrm{\Omega }}_n,$ and rearranging gives the Statement. ∎