7.7 The homogeneous Dirichlet problem and balayage22EC1FF0BE9D

Proof. Use the Hermitian product $\langle z,a\rangle =\sum z_ j\bar a_ j$, put $s_ a=\sqrt{1-|a|^2}$, and define

This expression depends smoothly on $(a,z)$ in the open set $\{ (a,z):|a|<1,\ |a||z|<1\} $, where the denominator does not vanish, and it is holomorphic in $z$; moreover $T_0$ is the identity. If $U$ is a unitary matrix, then $\langle Uz,Ua\rangle =\langle z,a\rangle $ and $s_{Ua}=s_ a$, so $T_{Ua}(Uz)=UT_ a(z)$. Given $a\ne 0$, choose $U$ with $Ua=(t,0,\ldots ,0)$, $t=|a|$; it therefore suffices to verify the following identities for $a=(t,0,\ldots ,0)$ with $0\le t<1$, and also for $-a=(-t,0,\ldots ,0)$. Put $s=s_ a=\sqrt{1-t^2}$. Then $\langle z,a\rangle =tz_1$, and the first component of the numerator of $T_ a(z)$ is $sz_1+t^2z_1/(1+s)-t=z_1-t$, because $s(1+s)+t^2=s+s^2+t^2=1+s$. Hence

the second formula by the same computation with $t$ replaced by $-t$. In particular $T_ a(a)=0$ and $T_{-a}(0)=a$. If $w=T_ a(z)$, then

so the first component of $T_{-a}(w)$ is $z_1$ and the others are $s\cdot sz_ k/(1-tz_1)\cdot (1-tz_1)/(1-t^2)=z_ k$; thus $T_{-a}\circ T_ a=\mathrm{id}$, and by symmetry $T_ a\circ T_{-a}=\mathrm{id}$. Finally,

because the terms $\mp 2t\operatorname{Re}z_1$ cancel; dividing by $|1-tz_1|^2$ gives

Hence $T_ a$ maps $B$ into $B$ and the sphere $\partial B$ into itself, and the same holds for $T_{-a}$; since $T_{-a}$ is inverse to $T_ a$, the map $T_ a$ is a biholomorphic automorphism of $B$ which maps $\partial B$ onto itself.

Set $S_{a,h}=T_{-(a+h)}\circ T_ a$ for $|a+h|<1$. It maps $a$ to $T_{-(a+h)}(0)=a+h$ and is the identity for $h=0$. The map $T_ a$ is holomorphic on the neighborhood $\{ z:|a||z|<1\} $ of $\overline B$, and, as we have just seen, it maps $\overline B$ into $\overline B$ and $\partial B$ onto $\partial B$; the same holds for $T_{-(a+h)}$. Hence $S_{a,h}$ is holomorphic on the neighborhood $\{ z:|a||z|<1,\ |a+h||T_ a(z)|<1\} $ of $\overline B$, and it maps $\partial B$ onto $\partial B$. Let $K\subset B$ be compact (we may assume $K\neq \varnothing $), put $t_ K=\max _ K|a|<1$, and choose $h_0>0$ with $t_ K+3h_0<1$. On the open set $O$ of all $(a,h,z)$ with $|a|<t_ K+h_0$, $|h|<2h_0$, $|a||z|<1$ and $(t_ K+3h_0)|T_ a(z)|<1$, we have $|a+h|<1$ and $|a+h||T_ a(z)|<1$, so $(a,h,z)\mapsto S_{a,h}(z)$ is defined and smooth on $O$, and holomorphic in $z$. Since $|T_ a(z)|\le 1$ for $|z|\le 1$, the set $O$ contains the compact set $K\times \overline{B(0,h_0)}\times \overline B$, on which the continuous first and second derivatives of $S_{a,h}(z)$ with respect to $h$ are therefore bounded by a constant $C$. $\square$

Proof. Let $H_\varepsilon =H*\rho _\varepsilon $. Fix $a_0\in U$ and $r>0$ with $\overline{B(a_0,4r)}\subset U$, put $K=\overline{B(a_0,3r)}$, and let $M_0=\sup _ K|H|$, which is finite because $H$ is locally bounded. For $0<\varepsilon <r$, the regularization $H_\varepsilon $ is smooth and plurisubharmonic on a neighborhood of $\overline{B(a_0,2r)}$ (proposition 5.9.2), and $|H_\varepsilon |\le M_0$ on $\overline{B(a_0,2r)}$, since its values there are averages of values of $H$ on $K$. For $x\in \overline{B(a_0,2r)}$ and $|h|\le h_ K$,

by (7.7.3), because $x-y\in K$ for $|y|\le \varepsilon $. Replacing $h$ by $th$ with $0<t\le 1$, dividing by $t^2$ and letting $t\to 0$, we get $D^2H_\varepsilon (x)[h,h]\le 2C_ K|h|^2$ for $|h|\le h_ K$, and hence, by homogeneity, for all $h\in \mathbb C^ n=\mathbb R^{2n}$. On the other hand, the complex Hessian of the smooth plurisubharmonic function $H_\varepsilon $ is positive semidefinite (proposition 5.9.2), and $D^2H_\varepsilon [h,h]+D^2H_\varepsilon [\mathrm ih,\mathrm ih]=4\sum _{j,k}(H_\varepsilon )_{j\bar k}h_ j\bar h_ k\ge 0$; hence $D^2H_\varepsilon (x)[h,h]\ge -2C_ K|h|^2$ as well. Thus the second derivatives of the $H_\varepsilon $ are bounded on $\overline{B(a_0,2r)}$ by a constant $\Lambda _0$ independent of $\varepsilon $. For $x\in \overline{B(a_0,r)}$ and a unit vector $e$, Taylor's formula

gives $|\partial _ eH_\varepsilon (x)|\le 2M_0/r+\Lambda _0r=\Lambda $. By the mean value theorem on the convex set $B(a_0,r)$, $|H_\varepsilon (x)-H_\varepsilon (y)|\le \Lambda |x-y|$ for $x,y\in B(a_0,r)$. Since $H_\varepsilon \to H$ pointwise as $\varepsilon \downarrow 0$, by (3.4.13), the same inequality holds for $H$. Hence $H$ is continuous on $B(a_0,r)$, and, as $a_0$ was arbitrary, on $U$.

Moreover, $H_\varepsilon \to H$ in $L^1$ on compact subsets of $U$, by dominated convergence, and hence in the sense of distributions; and every ball $B_0\Subset U$ is covered by finitely many balls $B(a_0,2r)$ as above, so the second derivatives of the $H_\varepsilon $ are bounded on $B_0$ uniformly in small $\varepsilon $. By the converse part of proposition 2.8.5, the distributional real Hessian of $H$ has locally bounded measurable coefficients $A_{ab}$; hence $H\in C^{1,1}_{\mathrm{loc}}(U)$, and almost every $x\in U$ satisfies the second-order expansion of the statement.

The functions $H_{j\bar k}$ are bounded measurable functions on compact subsets of $U$. By proposition 2.5.3, the complex Hessian of $H_\varepsilon $ has the coefficients $(H_{j\bar k})_\varepsilon $; these are bounded on compact sets uniformly in small $\varepsilon $, and they converge to $H_{j\bar k}$ at every Lebesgue point of $H_{j\bar k}$, hence almost everywhere (theorem 2.8.3). In particular, the matrix $(H_{j\bar k})$ is positive semidefinite almost everywhere, as an almost everywhere limit of the positive semidefinite complex Hessians of the smooth plurisubharmonic functions $H_\varepsilon $. Since $H$ is continuous, $H_\varepsilon \to H$ locally uniformly (lemma 2.5.1(1)), so $(\mathrm{dd}^{\mathrm{c}}H_\varepsilon )^ n\to (\mathrm{dd}^{\mathrm{c}}H)^ n$ weakly on every ball $B_0\Subset U$ as $\varepsilon \downarrow 0$ along a sequence, by proposition 7.2.6 applied on $B_0$. By (7.2.4), $(\mathrm{dd}^{\mathrm{c}}H_\varepsilon )^ n=n!\pi ^{-n}\det ((H_{j\bar k})_\varepsilon )\, \mathrm dV_{2n}$, and the determinants converge almost everywhere and are bounded on compact sets uniformly in small $\varepsilon $. Hence, for every test function $\chi $ supported in such a ball $B_0$, dominated convergence gives

By lemma 2.5.2, every test function on $U$ is a finite sum of test functions supported in balls $B_0\Subset U$, so this identity holds for all test functions on $U$, and the two Radon measures coincide by proposition 2.5.3(4). $\square$

Proof. After an affine change of coordinates $z\mapsto z_0+Rz$ with $R>0$, we may assume that $B$ is the unit ball; plurisubharmonicity, the boundary conditions, $C^2$ data and the vanishing of the Monge–Ampère measure are preserved by such a change, by proposition 5.9.3 and proposition 7.2.2(5). We call the functions $v$ in the statement the subsolutions for $\varphi $ of the homogeneous equation. If $v$ is a subsolution for $\varphi $ and $\varphi \le \varphi '+c$ on $\partial B$ with a constant $c$, then $v-c$ is a subsolution for $\varphi '$. Every subsolution is subharmonic on $B$ (theorem 5.9.1), so the comparison principle (theorem 3.4.5 and the remark following it), applied with the harmonic function $P\varphi $, the Poisson integral of $\varphi $ given by theorem 3.1.18, gives $v\le P\varphi $, and $P\varphi \le \max _{\partial B}\varphi $ because the Poisson kernel is positive with integral $1$. Hence, as soon as the class of subsolutions is nonempty, its upper envelope $H$ is defined, and $H\le P\varphi $.

Step 1: the envelope for $C^2$ data. Suppose that $\varphi $ is the restriction of a $C^2$ function on a neighborhood of $\partial B$. Multiplying this function by a cutoff equal to $1$ near $\partial B$, we obtain a $C^2$ function $\tilde\varphi $ with compact support in $\mathbb {C}^ n$ and $\tilde\varphi =\varphi $ on $\partial B$. Its complex Hessian is continuous with compact support, so there is $\lambda >0$ with $\sum _{j,k}\tilde\varphi _{j\bar k}\xi _ j\bar\xi _ k\ge -\lambda |\xi |^2$ everywhere. The function $s=\tilde\varphi +\lambda (|z|^2-1)$ is then of class $C^2$ with positive semidefinite complex Hessian, so it is plurisubharmonic by proposition 5.9.2. As $s=\varphi $ on $\partial B$, it is a subsolution. Hence $s\le H\le P\varphi $. By theorem 5.10.2, the upper semicontinuous regularization $H^*$ is plurisubharmonic, and $s\le H^*\le P\varphi $ since $P\varphi $ is continuous. As $s$ and $P\varphi $ are continuous on $\overline B$ with boundary values $\varphi $, we have $H^*(z)\to \varphi (x)$ as $z\to x\in \partial B$; so $H^*$ is a subsolution, $H^*\le H$, and $H=H^*$ is plurisubharmonic. Moreover $H(z)\to \varphi (x)$ as $z\to x\in \partial B$, and $H$ is bounded.

Step 2: second differences. Let $K\subset B$ be compact, and let $S_{a,h}$, $h_0$ and $C$ be as in lemma 7.7.1. For $x\in \partial B$ and $a\in K$, the function $g(h)=\tilde\varphi (S_{a,h}(x))$, $|h|\le h_0$, is of class $C^2$, and by the chain rule its second derivatives are bounded by a constant $2C_ K$ independent of $a$ and $x$, since the first and second derivatives of $\tilde\varphi $ are bounded and those of $S_{a,h}(x)$ with respect to $h$ are bounded by $C$. Since $g(0)=\varphi (x)$, Taylor's formula with integral remainder gives $g(h)+g(-h)-2g(0)=\int _0^1(1-t)\bigl (D^2g(th)[h,h]+D^2g(-th)[h,h]\bigr )\, \mathrm dt\le 2C_ K|h|^2$. As $S_{a,\pm h}$ maps $\partial B$ to $\partial B$, where $\tilde\varphi =\varphi $, this reads

The function $w=\frac12(H\circ S_{a,h}+H\circ S_{a,-h})-C_ K|h|^2$ is plurisubharmonic on $B$, by proposition 5.9.3, proposition 5.9.5, since $S_{a,\pm h}$ are automorphisms of $B$ and $H$ is bounded. If $y\in \partial B$ and $z\to y$ in $B$, then $S_{a,\pm h}(z)\to S_{a,\pm h}(y)\in \partial B$, so $H(S_{a,\pm h}(z))\to \varphi (S_{a,\pm h}(y))$ by Step 1, and the preceding inequality gives $\varlimsup _{z\to y}w(z)\le \varphi (y)$. Thus $w$ is a subsolution, so $w\le H$. Evaluating at $z=a$, where $S_{a,\pm h}(a)=a\pm h$, we obtain

Since $K$ was arbitrary and the points $a\pm h=S_{a,\pm h}(a)$ lie in $B$, lemma 7.7.2 applies on $U=B$. Thus $H$ is continuous on $B$ and belongs to $C^{1,1}_{\mathrm{loc}}(B)$, and its distributional real Hessian $A$ and complex Hessian $(H_{j\bar k})$ have the properties stated there: the second-order expansion holds almost everywhere, $(H_{j\bar k})$ is positive semidefinite almost everywhere, and $(\mathrm{dd}^{\mathrm{c}}H)^ n=n!\, \pi ^{-n}\det (H_{j\bar k})\, \mathrm dV_{2n}$.

Step 3: the equation. Suppose that $\det (H_{j\bar k})>0$ on a set of positive measure, and choose a point $a$ of this set at which $(H_{j\bar k}(a))$ is positive semidefinite and the second-order expansion holds. The Taylor polynomial $Q(z)=H(a)+\nabla H(a)\cdot (z-a)+\frac12(z-a)^ tA(a)(z-a)$ is a real quadratic polynomial with real Hessian $A(a)$, so its complex Hessian is the constant matrix $(H_{j\bar k}(a))$; this matrix is positive semidefinite with positive determinant, hence positive definite. Let $\eta >0$ be smaller than its smallest eigenvalue; then $Q-\eta |z-a|^2$ has positive semidefinite complex Hessian and is plurisubharmonic (proposition 5.9.2). For sufficiently small $r>0$ we have $\overline{B(a,r)}\subset B$ and, by the expansion, $|H-Q|<\eta r^2/2$ on the sphere $|z-a|=r$. The polynomial $\tilde Q=Q-\eta |z-a|^2+\eta r^2/4$ is then strictly smaller than $H$ on that sphere, since there $\tilde Q<H+\eta r^2/2-3\eta r^2/4$, and $\tilde Q(a)=H(a)+\eta r^2/4>H(a)$. By proposition 5.9.5, applied with $D=B(a,r)$ and $v=\tilde Q$, whose boundary inequality holds because $\tilde Q$ and $H$ are continuous, the function equal to $\max \{ H,\tilde Q\} $ on $B(a,r)$ and to $H$ elsewhere is plurisubharmonic on $B$. It coincides with $H$ near $\partial B$, so it is a subsolution, and it exceeds $H$ at $a$, a contradiction. Hence $\det (H_{j\bar k})=0$ almost everywhere, and $(\mathrm{dd}^{\mathrm{c}}H)^ n=0$. This proves the theorem for $C^2$ data, including the $C^{1,1}$ regularity, except for uniqueness.

Step 4: continuous data and uniqueness. For continuous $\varphi $, the function $z\mapsto \varphi (z/|z|)$ is continuous on the annulus $\{ 1/2<|z|<2\} $, and its mollifications $\varphi _ j$, $j\ge 4$, are smooth on the neighborhood $\{ 3/4<|z|<3/2\} $ of $\partial B$ and converge to $\varphi $ uniformly on $\partial B$ (lemma 2.5.1(1)). Let $H_ j$ be the envelope for $\varphi _ j|_{\partial B}$; by Steps 1–3 it is continuous on $\overline B$ with boundary values $\varphi _ j$, plurisubharmonic, and $(\mathrm{dd}^{\mathrm{c}}H_ j)^ n=0$. Put $c_ j=\max _{\partial B}|\varphi _ j-\varphi |$. Since $H_ j-c_ j$ is a subsolution for $\varphi $, the class of subsolutions for $\varphi $ is nonempty, and its envelope $H$ satisfies $H_ j-c_ j\le H$; symmetrically, $H-c_ j\le H_ j$. Therefore $\sup _ B|H_ j-H|\le c_ j\to 0$, so $H$ is a uniform limit of continuous functions on $B$; it is continuous, extends continuously to $\overline B$ with the boundary values $\varphi $, and is plurisubharmonic, being equal to its regularization $H^*$, which is plurisubharmonic by theorem 5.10.2. By proposition 7.2.6, $(\mathrm{dd}^{\mathrm{c}}H)^ n=\lim _ j(\mathrm{dd}^{\mathrm{c}}H_ j)^ n=0$. Finally, if $H'$ and $H''$ are two solutions, then $H'-H''\to 0$ at $\partial B$, and the last assertion of theorem 7.3.11, applied in both directions, gives $H'=H''$. $\square$

Proof. Let $u_ j=u*\rho _{\varepsilon _ j}$, with $\varepsilon _ j\downarrow 0$ smaller than the distance from $\overline B$ to $\mathbb C^ n\setminus \Omega $, be radial regularizations of $u$; they are smooth plurisubharmonic functions on a neighborhood of $\overline B$ decreasing to $u$ (proposition 5.9.2). Let $H_ j$ be the solution of theorem 7.7.4 with boundary values $u_ j|_{\partial B}$, that is, the upper envelope of the subsolutions for $u_ j|_{\partial B}$. Since $u_ j$ is such a subsolution, $u\le u_ j\le H_ j$ on $B$. Since $u_{j+1}\le u_ j$ on $\partial B$, every subsolution for $u_{j+1}|_{\partial B}$ is one for $u_ j|_{\partial B}$, so the envelopes decrease: $H_{j+1}\le H_ j$. Let $H=\lim _ jH_ j$. The function $H_1$ is subharmonic on $B$ and continuous on $\overline B$ with boundary values $u_1$, so $H_1\le \max _{\partial B}u_1$ by the comparison principle (theorem 3.4.5). Hence $\inf _{\overline B}u\le H\le \max _{\partial B}u_1$, so $H$ is bounded; it is plurisubharmonic by theorem 5.10.1, and theorem 7.3.7 gives $(\mathrm{dd}^{\mathrm{c}}H)^ n=\lim _ j(\mathrm{dd}^{\mathrm{c}}H_ j)^ n=0$. For $x\in \partial B$, since $H\le H_ j$ and $H_ j$ is continuous on $\overline B$ with $H_ j=u_ j$ on $\partial B$,

By proposition 5.9.5, the function $\widehat u$ equal to $\max \{ u,H\} =H$ on $B$ and to $u$ outside $B$ is plurisubharmonic on $\Omega $. It is locally bounded, since $H$ is bounded and $u$ is locally bounded; it equals $u$ outside $B$, satisfies $\widehat u\ge u$ on $B$, and $(\mathrm{dd}^{\mathrm{c}}\widehat u)^ n=(\mathrm{dd}^{\mathrm{c}}H)^ n=0$ on $B$ by locality.

Let $v$ be plurisubharmonic on $B$ with $\varlimsup _{B\ni z\to x}v(z)\le u(x)$ for $x\in \partial B$. Since $u\le u_ j$, the function $v$ is a subsolution for the boundary values $u_ j|_{\partial B}$, so $v\le H_ j$ by theorem 7.7.4. Letting $j\to \infty $ gives $v\le H$. Conversely, $H$ belongs to this class by the displayed estimate, so $H$ is its largest member. If $u\le u'$, the class for $u$ is contained in the class for $u'$, which proves that $\widehat u\le \widehat{u'}$. $\square$

Proof. Suppose that $(\mathrm{dd}^{\mathrm{c}}u)^ n=0$, and let $v$ be plurisubharmonic on $\Omega $ with $v\le u$ outside a compact set $K$. Choose a bounded open set $D$ with $K\subset D\Subset \Omega $. By upper semicontinuity, $v$ is bounded above on $\overline D$; since $u$ is bounded on $\overline D$, we may choose $M$ with $-M<\inf _ Du$. The function $\max \{ v,-M\} $ is bounded and plurisubharmonic on $D$, and it is at most $u$ on $D\setminus K$; since every point of $D$ near $\partial D$ lies in $D\setminus K$, we get $\varliminf _{D\ni z\to x}(u(z)-\max \{ v(z),-M\} )\ge 0$ for $x\in \partial D$. The last assertion of theorem 7.3.11 therefore gives $\max \{ v,-M\} \le u$ on $D$. Hence $v\le u$ everywhere, and $u$ is maximal.

Conversely, suppose that $u$ is maximal, and let $B\Subset \Omega $ be a ball. The function $\widehat u$ of corollary 7.7.5 is plurisubharmonic, satisfies $\widehat u\ge u$, and equals $u$ outside the compact set $\overline B$. Maximality gives $\widehat u\le u$, so $\widehat u=u$ and $(\mathrm{dd}^{\mathrm{c}}u)^ n=(\mathrm{dd}^{\mathrm{c}}\widehat u)^ n=0$ on $B$. Since the balls $B\Subset \Omega $ cover $\Omega $, and a distribution vanishing on each member of a family of open sets vanishes on their union (lemma 2.5.2), $(\mathrm{dd}^{\mathrm{c}}u)^ n=0$ on $\Omega $. For the last assertion, let every point of $\Omega $ have an open neighborhood $U$ such that $u|_ U$ is maximal on $U$. By the first part applied on $U$, and by locality of the products (proposition 7.2.2), $(\mathrm{dd}^{\mathrm{c}}u)^ n$ vanishes on each such $U$, hence on $\Omega $, and $u$ is maximal on $\Omega $. Conversely, if $u$ is maximal on $\Omega $, then $(\mathrm{dd}^{\mathrm{c}}u)^ n=0$ on every open set $U\subset \Omega $, so $u|_ U$ is maximal on $U$. $\square$