The Dolbeault Complex and Integral Solutions

At each point, the displayed change from $(d{x}_j,d{y}_j)$ to $(d{z}_j,d{\bar{z}}_j)$ is an invertible complex-linear change of cotangent basis. Its increasing wedges therefore form a basis of the complexified alternating cotensors. By [F1] and [F2], every smooth complex-valued form has a unique expansion in this basis, and its coefficient functions are smooth. Grouping the terms by the numbers $p$ and $q$ of holomorphic and antiholomorphic factors gives the stated direct sum.

For a coefficient function $a$, the real-coordinate formula for $da$ and [F4] give $da={\sum }_j({\mathrm{\partial }}_{z_j}a)d{z}_j+({\mathrm{\partial }}_{{\bar{z}}_j}a)d{\bar{z}}_j$. Since $d(d{z}_j)=d(d{\bar{z}}_j)=0$, [F3] applied termwise to $a_{I,J}d{z}^I\wedge d{\bar{z}}^J$ splits $d\eta$ into exactly the two displayed sums. Their bidegrees differ, so projection onto those summands recovers the coefficient formulas and proves $d=\mathrm{\partial }+\bar{\mathrm{\partial }}$.

If $\mathrm{\Phi }$ is a holomorphic coordinate change, [F5] makes its differential complex-linear; hence ${\mathrm{\Phi }}^{\ast }d{z}_j$ is a linear combination of holomorphic differentials and ${\mathrm{\Phi }}^{\ast }d{\bar{z}}_j$ is the conjugate linear combination of antiholomorphic differentials. Thus pullback preserves each bidegree. By [F6], pullback also commutes with $d$; uniqueness of the bidegree decomposition from step 1.1 implies it commutes separately with its two projections $\mathrm{\partial }$ and $\bar{\mathrm{\partial }}$. The definitions are therefore independent of holomorphic coordinates.

Write a complex form $\xi ={\xi }_1+i{\xi }_2$ with real forms ${\xi }_1,{\xi }_2$. The coordinate formula defining $d$ on complex coefficients is the complex-linear extension of the real exterior derivative, so $d^2\xi =d^2{\xi }_1+i{d}^2{\xi }_2=0$ by [F2].

The real graded product rule [F3] extends to complex forms: write each complex form as real part plus $i$ times imaginary part, expand the wedge product by complex bilinearity, and apply [F3] to each real pair. The coordinate definition of $d$ in [F1] is complex-linear, so the resulting identity is the same signed rule for complex forms.

For a pure type form $\eta \in {\mathrm{\Omega }}^{p,q}(U)$, [F1] gives $d\eta =\mathrm{\partial }\eta +\bar{\mathrm{\partial }}\eta$ and hence $0=d^2\eta ={\mathrm{\partial }}^2\eta +(\mathrm{\partial }\bar{\mathrm{\partial }}+\bar{\mathrm{\partial }}\mathrm{\partial })\eta +{\bar{\mathrm{\partial }}}^2\eta$ by step 1.1. These three terms have respective bidegrees $(p+2,q)$, $(p+1,q+1)$, and $(p,q+2)$; the direct sum uniqueness in [F1] forces each component to vanish, including when an endpoint component is zero by convention.

Let $\eta \in {\mathrm{\Omega }}^{p,q}(U)$ and $\theta \in {\mathrm{\Omega }}^{r,s}(U)$. Wedge products add the two bidegrees (and vanish if a repeated differential occurs). In the complex graded-derivation identity from step 1.2, the terms involving $\mathrm{\partial }$ have bidegree $(p+r+1,q+s)$ and those involving $\bar{\mathrm{\partial }}$ have bidegree $(p+r,q+s+1)$. Projecting onto these distinct summands yields the two displayed Leibniz identities.

Every smooth complex form is a finite sum of its bidegree components, and both operators and wedge product are additive. Applying step 2.2 componentwise proves the graded Leibniz rules for all homogeneous forms; applying step 2.1 componentwise proves all three square and anticommutation identities for arbitrary forms.

In standard oriented coordinates set $dV=d{x}_1\wedge \cdots \wedge d{x}_m$. Every complex $(m-1)$-form has a unique expression $\alpha ={\sum }_{j=1}^m(-1{)}^{j-1}{A}_{j}d{x}_1\wedge \cdots \wedge \widehat{d{x}_j}\wedge \cdots \wedge d{x}_m$ with $C^1$ complex coefficients $A_j$ up to the boundary. Directly differentiating these coefficients (the same coordinate formula as [F6], valid here because they are $C^1$) gives $d\alpha =({\sum }_j{\mathrm{\partial }}_j{A}_j)dV=(\mathrm{div}A)dV$. All terms with a repeated $d{x}_i$ vanish, and the sign $(-1{)}^{j-1}$ is canceled by moving $d{x}_j$ into its ordered volume position.

At a boundary point choose a positively oriented orthonormal tangent frame $t_1,\dots ,t_{m-1}$ so that $(\nu ,t_1,\dots ,t_{m-1})$ is positive, where $\nu$ is the outward unit normal. The displayed form is ${\iota }_{A}dV$. Its boundary trace evaluated on that frame is $dV(A,t_1,\dots ,t_{m-1})=(A\cdot \nu )dS(t_1,\dots ,t_{m-1})$, because the tangential component of $A$ repeats a tangent direction in the top-degree volume form. Thus the outward-normal-first boundary trace is exactly $(A\cdot \nu )dS$; this identity is complex-linear in $A$.

By [F1], full AC supplies the weaker ${\mathrm{A}\mathrm{C}}_{\omega }$ hypotheses in [F2] and the surface convention in [F4]. Apply [F3] separately to the real and imaginary vector fields $\mathrm{Re}A$ and $\mathrm{Im}A$. Adding the two equalities and using steps 1.1 and 1.2 gives ${\int }_{D}d\alpha ={\int }_{\mathrm{\partial }D}\alpha$. This is the only use of AC.

For $0<r<\mathrm{dist}(z,\mathrm{\partial }D)$ set $D_r=D\setminus \overline{B_r(z)}$ and $g(\zeta )=f(\zeta )\mathrm{/}(\zeta -z)$ on its closure. Since $1\mathrm{/}(\zeta -z)$ is holomorphic there, [F3] gives ${\mathrm{\partial }}_{\bar{\zeta }}g=({\mathrm{\partial }}_{\bar{\zeta }}f)\mathrm{/}(\zeta -z)$. Writing $dg=({\mathrm{\partial }}_{\zeta }g)d\zeta +({\mathrm{\partial }}_{\bar{\zeta }}g)d\bar{\zeta }$ and using $d(gd\zeta )=dg\wedge d\zeta$, the repeated $d\zeta$ term vanishes, so $d(gd\zeta )=-({\mathrm{\partial }}_{\bar{\zeta }}f)\mathrm{/}(\zeta -z)d\zeta \wedge d\bar{\zeta }$. This is the needed type component of $d$ from [F4].

The continuous derivative ${\mathrm{\partial }}_{\bar{\zeta }}f$ is bounded on the compact $\bar{D}$. Near $z$ the absolute area density is at most $C\mathrm{\mid }\zeta -z{\mathrm{\mid }}^{-1}dA$, whose integral over $B_{ϵ}(z)$ is at most $2\pi Cϵ$; away from $z$ the integrand is bounded on the bounded domain. Thus the area term is absolutely integrable and its integral over the region $D_r$ defined in step 1.1 converges to that over $D$ as $r\downarrow 0$. Parametrizing the positively oriented circle by $\zeta =z+r{e}^{it}$ gives ${\int }_{\mathrm{\partial }{B}_r(z)}f(\zeta )(\zeta -z{)}^{-1}d\zeta =i{\int }_0^{2\pi }f(z+r{e}^{it})dt\to 2\pi if(z)$ by continuity of $f$.

The boundary of $D_r$ from step 1.1 is the disjoint union $\mathrm{\partial }D$ and the negatively oriented circle $-\mathrm{\partial }{B}_r(z)$. The given full AC is the premise in [F1], so [F2] applies to $gd\zeta$ on $D_r$; if $D_r$ is disconnected, each component has C¹ boundary and there are finitely many components because the compact C¹ boundary has a finite graph-chart cover, each chart meeting only one local interior component. Apply [F5] to the components and add. Using step 1.1 gives ${\int }_{\mathrm{\partial }D}f(\zeta )(\zeta -z{)}^{-1}d\zeta -{\int }_{\mathrm{\partial }{B}_r(z)}f(\zeta )(\zeta -z{)}^{-1}d\zeta =-{\int }_{D_r}({\mathrm{\partial }}_{\bar{\zeta }}f)(\zeta -z{)}^{-1}d\zeta \wedge d\bar{\zeta }$.

Letting $r\downarrow 0$ in step 2.2 and using step 2.1 yields ${\int }_{\mathrm{\partial }D}f(\zeta )(\zeta -z{)}^{-1}d\zeta +{\int }_D({\mathrm{\partial }}_{\bar{\zeta }}f)(\zeta -z{)}^{-1}d\zeta \wedge d\bar{\zeta }=2\pi if(z)$. Division by $2\pi i$ proves the first formula, with the plus sign fixed by the inner boundary orientation and the wedge swap in step 1.1.

Since $d\zeta \wedge d\bar{\zeta }=(dx+idy)\wedge (dx-idy)=-2idA$, the area term in step 3.1 equals $-{\pi }^{-1}{\int }_D({\mathrm{\partial }}_{\bar{\zeta }}f)(\zeta -z{)}^{-1}dA$. This proves the equivalent area form.

Apply [F1] on the manifold $\mathbb{C}$ to $K=D_k^{\mathrm{\prime }}$ and $W=D_k$, obtaining $\chi \in C_c^{\mathrm{\infty }}(D_k)$ equal to $1$ on an open neighborhood of $D_k^{\mathrm{\prime }}$. Compact containment lets us choose an open coordinate disc $D_k^{\mathrm{\prime }\mathrm{\prime }}$ containing $D_k^{\mathrm{\prime }}$ with closure inside that neighborhood. For fixed $z^{\mathrm{\prime }}=(z_1,\dots ,{\widehat{z}}_k,\dots ,z_n)$, extend $h(z^{\mathrm{\prime }},\zeta )=\chi (\zeta )g(z^{\mathrm{\prime }},\zeta )$ by zero from $D_k$ to $\mathbb{C}$; the extension is smooth and supported in the fixed compact set $K_0=\mathrm{supp}\chi$.

In the local integral change variables $\eta =\zeta -z_k$ and write $\tilde{h}(z,\eta )=h(z^{\mathrm{\prime }},z_k+\eta )$, so $T_{k}g(z)=\frac{1}{2\pi i}{\int }_{\mathbb{C}}\tilde{h}(z,\eta ){\eta }^{-1}d\eta \wedge d\bar{\eta }$. For any compact parameter set $Q⋐P^{\mathrm{\prime }\mathrm{\prime }}$, the support of $\tilde{h}$ and all its real parameter derivatives lies in a common disk $\mathrm{\mid }\eta \mathrm{\mid }\le R$, since $K_0$ and the $z_k$-projection of $Q$ are compact. Also ${\int }_{\mathrm{\mid }\eta \mathrm{\mid }\le R}\mathrm{\mid }\eta {\mathrm{\mid }}^{-1}dA(\eta )=2\pi R$, so these integrals converge absolutely. For each real-coordinate multi-index $\alpha$ set $F_{\alpha }(z)=\frac{1}{2\pi i}{\int }_{\mathbb{C}}(D_z^{\alpha }\tilde{h})(z,\eta ){\eta }^{-1}d\eta \wedge d\bar{\eta }$. Fix a real parameter coordinate $t$. The fundamental theorem of calculus writes the difference between the difference quotient of $D_z^{\alpha }\tilde{h}$ in $t$ and $D_t{D}_z^{\alpha }\tilde{h}$ as an average of increments of the latter derivative; on $Q\times \{\mathrm{\mid }\eta \mathrm{\mid }\le R\}$ their supremum tends to zero with the increment by uniform continuity on a slightly larger compact set. Thus $\mathrm{\mid }[F_{\alpha }(z+s{e}_t)-F_{\alpha }(z)]\mathrm{/}s-F_{\alpha +e_t}(z)\mathrm{\mid }\le C{\omega }_{\alpha }(\mathrm{\mid }s\mathrm{\mid }){\int }_{\mathrm{\mid }\eta \mathrm{\mid }\le R}\mathrm{\mid }\eta {\mathrm{\mid }}^{-1}dA(\eta )\to 0$, where ${\omega }_{\alpha }(\delta )\to 0$ is that uniform-continuity modulus and $C$ accounts for the fixed form factor. The same estimate gives continuity of each $F_{\alpha }$, and induction proves $D_z^{\alpha }{T}_{k}g=F_{\alpha }$ for every $\alpha$, hence $T_{k}g\in C^{\mathrm{\infty }}(P^{\mathrm{\prime }\mathrm{\prime }})$.

The transformed formula in step 1.2 and [F4] identify ${\mathrm{\partial }}_{{\bar{z}}_k}{T}_{k}g$ with the integral of ${\mathrm{\partial }}_{\bar{\zeta }}h(z^{\mathrm{\prime }},\zeta )\mathrm{/}(\zeta -z_k)$ against $(2\pi i{)}^{-1}d\zeta \wedge d\bar{\zeta }$. For each fixed $z^{\mathrm{\prime }}$, choose a bounded disc $E\subset \mathbb{C}$ containing both $\mathrm{supp}h(z^{\mathrm{\prime }},\cdot )$ and $D_k^{\mathrm{\prime }\mathrm{\prime }}$; the slice is zero near $\mathrm{\partial }E$. Full AC and the Cauchy–Pompeiu premise are both in [F2], so [F3] on $E$ has zero boundary term and gives this integral equal to $h(z^{\mathrm{\prime }},z_k)=g(z)$ on $P^{\mathrm{\prime }\mathrm{\prime }}$, because $\chi =1$ there. Thus ${\mathrm{\partial }}_{{\bar{z}}_k}{T}_{k}g=g$.

For $\mathrm{\ell }\mathrm{\ne }k$, the cutoff $\chi (\zeta )$ is independent of $z_{\mathrm{\ell }}$, so parameter differentiation in step 1.2 and [F4] give ${\mathrm{\partial }}_{{\bar{z}}_{\mathrm{\ell }}}{T}_{k}g=T_k({\mathrm{\partial }}_{{\bar{z}}_{\mathrm{\ell }}}g)$, with the single cutoff factor already included in $T_k$. If ${\mathrm{\partial }}_{{\bar{z}}_{\mathrm{\ell }}}g=0$, this is zero, proving preservation of each selected equation.

If $g\in C_c^{\mathrm{\infty }}({\mathbb{C}}^n)$, omit the cutoff and put $h=g$. On any compact parameter set, compact support of $g$ and boundedness of $z_k$ again place the translated numerator and every derivative in a common bounded $\eta$-disc. The uniform-continuity estimate of step 1.2 therefore proves that the global integral defines a smooth function.

For fixed values of the other variables, the slice $g(z^{\mathrm{\prime }},\cdot )$ is compactly supported. Its translated parameter derivative is $T_k({\mathrm{\partial }}_{{\bar{z}}_k}g)$, and the Cauchy–Pompeiu argument of step 2.1, using the AC premise and formula [F2, F3], gives ${\mathrm{\partial }}_{{\bar{z}}_k}{T}_{k}g=g$. For every $\mathrm{\ell }\mathrm{\ne }k$, the same differentiation calculation as in step 2.2 gives ${\mathrm{\partial }}_{{\bar{z}}_{\mathrm{\ell }}}{T}_{k}g=T_k({\mathrm{\partial }}_{{\bar{z}}_{\mathrm{\ell }}}g)$.

In each summand one $d{\bar{\zeta }}_j$ is omitted and all $n$ factors $d{\zeta }_1,\dots ,d{\zeta }_n$ remain, so every summand has bidegree $(n,n-1)$ by [F1]. Its scalar coefficient is smooth for $\zeta \mathrm{\ne }z$. For $r=\mathrm{\mid }\zeta -z\mathrm{\mid }$, $\mathrm{\mid }\overline{{\zeta }_j-z_j}\mathrm{/}{r}^{2n}\mathrm{\mid }\le r^{1-2n}$. On a compact neighborhood of $z$, a smooth complementary $(0,1)$ form has bounded coefficients; the resulting top-degree density is bounded by a constant times $r^{1-2n}$. Its absolute integral near $z$ is bounded by a constant multiple of ${\int }_0^{ϵ}{r}^{1-2n}{r}^{2n-1}dr=ϵ$. The finite sum is therefore locally integrable.

If $n=1$, the omitted antiholomorphic factor leaves only $d\zeta$, and $\overline{\zeta -z}\mathrm{/}\mathrm{\mid }\zeta -z{\mathrm{\mid }}^2=1\mathrm{/}(\zeta -z)$; since $(n-1)!\mathrm{/}(2\pi i{)}^n=1\mathrm{/}(2\pi i)$, the formula reduces to $d\zeta \mathrm{/}(2\pi i(\zeta -z))$.

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. ∎