5.1: The Bochner-Martinelli Kernel
- Page ID
- 74241
A generalization of Cauchy’s formula to several variables is called the Bochner–Martinelli integral formula, which reduces to Cauchy’s (Cauchy–Pompeiu) formula when \(n=1\). As for Cauchy’s formula, we will prove the formula for all smooth functions via Stokes’ theorem. First, let us define the Bochner-Martinelli kernel:
\[\omega(\zeta, z)\overset{\text{def}}{=}\frac{(n-1)!}{(2\pi i)^{n}}\sum\limits_{j=1}^n \frac{\bar{\zeta}_{j}-\bar{z}_{j}}{||\zeta -z||^{2n}}\: d\bar{\zeta}_{1}\wedge d\zeta_{1}\wedge\cdots\wedge\widehat{d\bar{\zeta}_{j}}\wedge d\zeta_{j}\wedge\cdots\wedge d\bar{\zeta}_{n}\wedge d\zeta_{n}.\]
The notation \(\widehat{ d\bar{\zeta}_j }\) means that this term is simply left out.
Let \(U \subset \mathbb{C}^n\) be a bounded open set with smooth boundary and let \(f \colon \overline{U} \to \mathbb{C}\) be a smooth function. Then for \(z \in U\), \[f(z) = \int_{\partial U} f(\zeta) \omega(\zeta,z) - \int_{U} \bar{\partial} f(\zeta) \wedge \omega(\zeta,z) .\] In particular, if \(f \in \mathcal{O}(U)\), then \[f(z) = \int_{\partial U} f(\zeta) \omega(\zeta,z) .\]
- Proof
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The structure of the proof is essentially the same as that of the Cauchy–Pompeiu theorem for \(n=1\), although some of the formulas are somewhat more involved.
Let \(z \in U\) be fixed. Suppose \(r > 0\) is small enough so that \(\overline{B_r(z)} \subset U\). Orient both \(\partial U\) and \(\partial B_r(z)\) positively. As \(f(\zeta) \omega(\zeta,z)\) contains all the holomorphic \(d\zeta_j\),
\[\begin{align}\begin{aligned} d\left( f(\zeta )\omega (\zeta ,z) \right)&=\overline{\partial}\left(f(\zeta )\omega (\zeta ,z)\right) \\ &=\overline{\partial}f(\zeta )\wedge \omega (\zeta ,z) \\ &\quad +f(\zeta )\frac{(n-1)!}{(2\pi i)^{n}}\sum\limits_{j=1}^n \frac{\partial}{\partial\overline{\zeta}_{j}} \left[\frac{\bar{\zeta}_{j}-\bar{z}_{j}}{||\zeta -z||^{2n}}\right]d\bar{\zeta}_{1}\wedge d\zeta_{1}\wedge\cdots\wedge d\bar{\zeta}_{n}\wedge d\zeta_{n}\end{aligned}\end{align}\]
We compute \[\sum_{j=1}^n \frac{\partial}{\partial \bar{\zeta}_j} \left[ \frac{\bar{\zeta}_j-\bar{z}_j}{\norm{\zeta-z}^{2n}} \right] = \sum_{j=1}^n \left( \frac{1}{\norm{\zeta-z}^{2n}} -n \frac{|\zeta_j-z_j|^2}{\norm{\zeta-z}^{2n+2}} \right) = 0 .\] Therefore, \(d \bigl( f(\zeta) \omega(\zeta,z) \bigr) = \bar{\partial} f(\zeta) \wedge \omega(\zeta,z)\).
We apply Stokes: \[ \int_{\partial U} f(\zeta) \omega(\zeta,z) - \int_{\partial B_r(z)} f(\zeta) \omega(\zeta,z) = \int_{U \setminus \overline{B_r(z)}} d \bigl( f(\zeta) \omega(\zeta,z) \bigr) = \int_{U \setminus \overline{B_r(z)}} \bar{\partial} f(\zeta) \wedge \omega(\zeta,z) .\] Again, due to the integrability, which you showed in an exercise above, the right-hand side converges to the integral over \(U\) as \(r \to 0\). Just as for the Cauchy–Pompeiu formula, we now need to show that the integral over \(\partial B_r(z)\) goes to \(f(z)\) as \(r \to 0\).
So \[\int_{\partial B_r(z)} f(\zeta) \omega(\zeta,z) = f(z) \int_{\partial B_r(z)} \omega(\zeta,z) + \int_{\partial B_r(z)} \bigl(f(\zeta)-f(z)\bigr) \omega(\zeta,z) .\] To finish the proof, we will show that \(\int_{\partial B_r(z)} \omega(\zeta,z) = 1\), and that the second term goes to zero. We apply Stokes again and note that the volume of \(B_r(z)\) is \(\frac{\pi^n}{n!}r^{2n}\).
\[\begin{align}\begin{aligned} \int_{\partial B_{r}(z)}\omega (\zeta ,z)&=\int_{\partial B_{r}(z)}\frac{(n-1)!}{(2\pi i)^{n}}\sum\limits_{j=1}^n \frac{\bar{\zeta}_{j}-\bar{z}_{j}}{||\zeta -z||^{2n}}d\bar{\zeta}_{1}\wedge d\zeta_{1}\wedge\cdots\wedge\widehat{d\bar{\zeta}_{j}}\wedge d\zeta_{j}\wedge\cdots\wedge d\bar{\zeta}_{n}\wedge d\zeta_{n} \\ &=\frac{(n-1)!}{(2\pi i)^{n}}\frac{1}{r^{2n}}\int_{\partial B_{r}(z)}\sum\limits_{j=1}^n (\bar{\zeta}_{j}-\bar{z}_{j})d\bar{\zeta}_{1}\wedge d\zeta_{1}\wedge\cdots\wedge\widehat{d\bar{\zeta}_{j}}\wedge d\zeta_{j}\wedge\cdots\wedge d\bar{\zeta}_{n}\wedge d\zeta_{n} \\ &=\frac{(n-1)!}{(2\pi i)^{n}}\frac{1}{r^{2n}}\int_{B_{r}(z)}d\left( \sum\limits_{j=1}^n (\bar{\zeta}_{j}-\bar{z}_{j})d\bar{\zeta}_{1}\wedge d\zeta_{1}\wedge\cdots\wedge\widehat{d\bar{\zeta}_{j}}\wedge d\zeta_{j}\wedge\cdots\wedge d\bar{\zeta}_{n}\wedge d\zeta_{n}\right) \\ &=\frac{(n-1)!}{(2\pi i)^{n}}\frac{1}{r^{2n}}\int_{B_{r}(z)} nd\bar{\zeta}_{1}\wedge d\zeta_{1}\wedge\cdots\wedge d\bar{\zeta}_{n}\wedge d\zeta_{n}\\ &=\frac{(n-1)!}{(2\pi i)^{n}}\frac{1}{r^{2n}}\int_{B_{r}(z)} n(2i)^{n}dV=1 \end{aligned}\end{align}\]
Next, we tackle the second term. Via the same computation as above we find
\[\begin{align}\begin{aligned}\int_{\partial B_{r}(z)}(f(\zeta )-f(z))\omega (\zeta ,z)&=\frac{(n-1)!}{(2\pi i)^{n}}\frac{1}{r^{2n}}\left( \int_{B_{r}(z)}(f(\zeta )-f(z))n\:d\bar{\zeta}_{1}\wedge d\zeta_{1}\wedge\cdots\wedge d\bar{\zeta}_{n}\wedge d\zeta_{n}\right. \\ &\quad \left. +\int_{B_{r}(z)}\sum\limits_{j=1}^n \frac{\partial f}{\partial\bar{\zeta}_{j}}(\zeta )(\bar{\zeta}_{j}-\bar{z}_{j})\: d\bar{\zeta}_{1}\wedge d\zeta_{1}\wedge\cdots\wedge d\bar{\zeta}_{n}\wedge d\zeta_{n}\right).\end{aligned}\end{align}\]
As \(U\) is bounded, \(|f(\zeta)-f(z)| \leq M ||\zeta-z||\) and \(\left|\frac{\partial f}{\partial \bar{\zeta}_j}(\zeta) (\bar{\zeta}_j-\bar{z}_j)\right| \leq M ||\zeta-z||\) for some \(M\). So for all \(\zeta \in \partial B_r(z)\), we have \(|f(\zeta)-f(z)| \leq Mr\) and \(\left| \frac{\partial f}{\partial \bar{\zeta}_j}(\zeta) (\bar{\zeta}_j-\bar{z}_j)\right| \leq Mr\). Hence
\[\left| \int_{\partial B_{r}(z)} (f(\zeta )-f(z))\omega (\zeta, z)\right| \leq\frac{(n-1)!}{(2\pi )^{n}}\frac{1}{r^{2n}}\left(\int_{B_{r}(z)} n2^{n}Mr\:dV\:+\int_{B_{r}(z)} n2^{n}Mr\:dV\right)=2Mr.\] Therefore, this term goes to zero as \(r \to 0\).
Recall that if \(\zeta = x+iy\) are the coordinates in \(\mathbb{C}^n\), the orientation that we assigned to \(\mathbb{C}^n\) in this book\(^{1}\) is the one corresponding to the volume form \[dV = dx_1 \wedge dy_1 \wedge dx_2 \wedge dy_2 \wedge \cdots \wedge dx_n \wedge dy_n .\] With this orientation, \[d\zeta_1 \wedge d\bar{\zeta}_1 \wedge d\zeta_2 \wedge d\bar{\zeta}_2 \wedge \cdots \wedge d\zeta_n \wedge d\bar{\zeta}_n = {(-2i)}^n dV ,\] and hence \[d\bar{\zeta}_1 \wedge d\zeta_1 \wedge d\bar{\zeta}_2 \wedge d\zeta_2 \wedge \cdots \wedge d\bar{\zeta}_n \wedge d\zeta_n = {(2i)}^n dV .\]
Similarly to the Cauchy–Pompeiu formula, note the singularity in the second term of the Bochner–Martinelli formula, and prove that the integral still makes sense (the function is integrable).
Check that for \(n=1\), the Bochner–Martinelli formula reduces to the standard Cauchy–Pompeiu formula.
Recall the definition of \(\partial\) and \(\bar{\partial}\) from Section 4.4, and recall that \(d \eta = \partial \eta + \bar{\partial} \eta\).
One drawback of the Bochner–Martinelli formula \(\int_{\partial U} f(\zeta) \omega(\zeta,z)\) is that the kernel is not holomorphic in \(z\) unless \(n=1\). It does not simply produce holomorphic functions. If we differentiate in \(\bar{z}\) underneath the \(\partial U\) integral, we do not necessarily obtain zero. On the other hand, we have an explicit formula and this formula does not depend on \(U\). This is not the case for the Bergman and Szegö kernels, which we will see next, although those are holomorphic in the right way.
Prove that if \(z \notin \overline{U}\), then rather than \(f(z)\) in the formula you obtain \[\int_{\partial U} f(\zeta) \omega(\zeta,z) - \int_{U} \bar{\partial} f(\zeta) \wedge \omega(\zeta,z) = 0 .\]
Suppose \(f\) is holomorphic on a neighborhood of \(\overline{B_r(z)}\).
- Using the Bochner–Martinelli formula, prove that \[f(z) = \frac{1}{V\bigl(B_r(z)\bigr)} \int_{B_r(z)} f(\zeta) \, dV(\zeta) ,\] where \(V\bigl(B_r(z)\bigr)\) is the volume of \(B_r(z)\).
- Use part a) to prove the maximum principle for holomorphic functions.
Use Bochner–Martinelli for the solution of \(\bar{\partial}\) with compact support. That is, suppose \(g = g_1 d\bar{z}_1 + \cdots + g_n d\bar{z}_n\) is a smooth compactly supported \((0,1)\)-form on \(\mathbb{C}^n\), \(n \geq 2\), and \(\frac{\partial g_k}{\partial \bar{z}_\ell} = \frac{\partial g_\ell}{\partial \bar{z}_k}\) for all \(k, \ell\). Prove that \[\psi(z) = - \int_{\mathbb{C}^n} g(\zeta) \wedge \omega(\zeta,z)\] is a compactly supported smooth solution to \(\bar{\partial} \psi = g\). Hint: Look at the previous proof.
Footnotes
[1] Again, there is no canonical orientation of \(\mathbb{C}^n\), and not all authors follow this (perhaps more prevalent) convention.