7.13 Bounded complex Hessians and contact setsFB1768F3514F

Proof. We first prove the chain rule

for a $C^1$ function $\Phi \colon \mathbb R\to \mathbb R$ with bounded derivative; note that $\Phi (g)\in L^1_{\mathrm{loc}}(D)$, since $|\Phi (g)|\leq |\Phi (0)|+\| \Phi '\| _\infty |g|$. Let $V\Subset D$ be open. The mollifications $g_\varepsilon =g*\rho _\varepsilon $ converge to $g$ in $L^1(V)$, together with their first derivatives $\partial _ ag_\varepsilon =(\partial _ ag)*\rho _\varepsilon $, by the remark preceding lemma 2.9.14; passing to a sequence $\varepsilon _ k\to 0$, we may also assume that $g_{\varepsilon _ k}\to g$ almost everywhere on $V$. Since $\Phi $ is Lipschitz, $\Phi (g_{\varepsilon _ k})\to \Phi (g)$ in $L^1(V)$. Moreover,

tends to zero in $L^1(V)$: the first term is bounded by $\| \Phi '\| _\infty |\partial _ ag_{\varepsilon _ k}-\partial _ ag|$, and the second tends to zero by dominated convergence, since it tends to zero almost everywhere, $\Phi '$ being continuous, and is bounded by $2\| \Phi '\| _\infty |\partial _ ag|$. For $\psi \in C^\infty _ c(V)$ the classical chain rule gives $\int \Phi (g_{\varepsilon _ k})\partial _ a\psi =-\int \Phi '(g_{\varepsilon _ k})\partial _ ag_{\varepsilon _ k}\, \psi $, and letting $k\to \infty $ we obtain $\int \Phi (g)\partial _ a\psi =-\int \Phi '(g)\partial _ ag\, \psi $. As $V$ was arbitrary, this proves the chain rule.

Choose a smooth function $\chi $ on $\mathbb R$ supported in $(-2,2)$, with $0\le \chi \le 1$ and $\chi =1$ on $[-1,1]$, and put $\Phi _\varepsilon (t)=\int _0^ t\chi (s/\varepsilon )\, \mathrm ds$ for $\varepsilon >0$. Then $\Phi _\varepsilon $ is smooth with $0\leq \Phi _\varepsilon '\leq 1$, $|\Phi _\varepsilon |\le 2\varepsilon $, and by the chain rule

as $\varepsilon \to 0$, by dominated convergence, since $\chi (g/\varepsilon )\to \mathbf1_{\{ g=0\} }$ pointwise and $|\chi (g/\varepsilon )\nabla g|\leq |\nabla g|$. On the other hand, $\Phi _\varepsilon (g)\to 0$ uniformly, so $\nabla \Phi _\varepsilon (g)\to 0$ in the sense of distributions. Since convergence in $L^1_{\mathrm{loc}}(D)$ implies convergence in the sense of distributions, the locally integrable function $\mathbf1_{\{ g=0\} }\nabla g$ defines the zero distribution, and hence vanishes almost everywhere (proposition 2.5.3). This is the first assertion.

If $g\in W^{2,1}_{\mathrm{loc}}(D)$, then each $\partial _ ag$ belongs to $W^{1,1}_{\mathrm{loc}}(D)$ and, by the first assertion, vanishes almost everywhere on $\{ g=0\} $. Applying the first assertion to $\partial _ ag$, formed with any measurable representative, shows that $\nabla \partial _ ag=0$ almost everywhere on $\{ \partial _ ag=0\} $, a set which contains $\{ g=0\} $ up to a null set. Since there are finitely many derivatives, the union of the exceptional null sets is null. $\square$

Proof. Considering real and imaginary parts separately, we may assume that $f$ is real. Let $f_\varepsilon =f*\rho _\varepsilon $ be the mollifications of $f$. For every compact set $L\subset D$ they are smooth on a neighborhood of $L$ for small $\varepsilon $, and by proposition 2.5.3 and lemma 2.5.1, $f_\varepsilon \to f$ and $\Delta f_\varepsilon =(\Delta f)*\rho _\varepsilon \to \Delta f$ in $L^2(L)$ as $\varepsilon \to 0$.

We first show that the gradients converge. Let $\chi $ be a real test function, and let $h$ be a real smooth function on a neighborhood of $\operatorname{Supp}\chi $. Since $\int \partial _ a(\chi ^2h\, \partial _ ah)=0$ for each $a$, by Fubini's theorem and the fundamental theorem of calculus, we have

where we used $2xy\le x^2/2+2y^2$. Hence

Let $L\subset D$ be compact, and choose $\chi \in C^\infty _ c(D)$ equal to one on $L$ (lemma 2.5.1). Apply the last inequality to $h=f_\varepsilon -f_\delta $. As $\varepsilon ,\delta \to 0$, the norms $\| f_\varepsilon -f_\delta \| _{L^2(\operatorname{Supp}\chi )}$ tend to zero, while the norms $\| \Delta f_\varepsilon -\Delta f_\delta \| _{L^2(\operatorname{Supp}\chi )}$ stay bounded. Hence the gradients $\nabla f_\varepsilon $ form a Cauchy family in $L^2(L)$. For $\psi \in C^\infty _ c(\operatorname{Int}L)$ we have $\int \partial _ af_\varepsilon \, \psi =-\int f_\varepsilon \, \partial _ a\psi \to -\int f\, \partial _ a\psi $, so the $L^2$ limits are the distributional first derivatives of $f$ on $\operatorname{Int}L$. Since $L$ was arbitrary, the first derivatives of $f$ belong to $L^2_{\mathrm{loc}}(D)$, and $\nabla f_\varepsilon \to \nabla f$ in $L^2(L)$ for every compact $L\subset D$.

For the second derivatives, let $L\subset D$ be compact, choose $\chi \in C^\infty _ c(D)$ equal to one on a neighborhood $N$ of $L$, and for small $\varepsilon $ put $v_\varepsilon =\chi f_\varepsilon $, extended by zero to $\mathbb R^ d$. These are smooth functions with compact support, and

is a Cauchy family in $L^2(\mathbb R^ d)$, by the first part applied on $\operatorname{Supp}\chi $. For every real $v\in C^\infty _ c(\mathbb R^ d)$, integrating by parts twice gives

Applying this identity to $v_\varepsilon -v_\delta $, we see that all second derivatives of $v_\varepsilon $ form Cauchy families in $L^2(\mathbb R^ d)$. Testing against compactly supported smooth functions identifies their limits with the distributional second derivatives of $\chi f$, which agree with those of $f$ on $N$, where $\chi f=f$. Thus the second derivatives of $f$ are square-integrable on $L$. Since $L$ was arbitrary, $f\in W^{2,2}_{\mathrm{loc}}(D)$. The last assertion holds because continuous functions and locally bounded measurable functions are locally square-integrable. $\square$

Proof. Apply lemma 7.13.1 to $f-g\in W^{2,1}_{\mathrm{loc}}(\Omega )$, whose zero set, formed with the chosen representatives, is $E$. The complex derivatives $f_{j\bar k}$ and $g_{j\bar k}$ are fixed linear combinations of the real second derivatives, which gives the second assertion. $\square$

Proof. For smooth $u_ j$ we polarize (7.2.4), as in the proof of corollary 7.8.8. Applied to the smooth function $\sum _ jt_ ju_ j$ with $t_1,\ldots ,t_ n\in \mathbb R$, it gives the identity

between exterior powers of smooth forms, and both sides are polynomials in $t_1,\ldots ,t_ n$. We compare the coefficients of $t_1\cdots t_ n$. Since $2$-forms commute, on the left this coefficient is $n!$ times the exterior product $\mathrm{dd}^{\mathrm{c}}u_1\wedge \cdots \wedge \mathrm{dd}^{\mathrm{c}}u_ n$, which is the Bedford–Taylor product by proposition 7.2.2(1); on the right it is $(n!)^2\pi ^{-n}D(A_1,\dots ,A_ n)\, \mathrm dV_{2n}$, by the definition of the mixed discriminant. This proves the first formula for smooth $u_ j$.

For continuous $u_ j$, let $u_{j,\varepsilon }=u_ j*\rho _\varepsilon $ be their radial mollifications, which are smooth and plurisubharmonic on $\Omega _\varepsilon $ (proposition 5.9.2). They converge to $u_ j$ uniformly on compact sets (lemma 2.5.1), so their products converge weakly by proposition 7.2.6. By proposition 2.5.3, the complex Hessian matrix of $u_{j,\varepsilon }$ is the mollification $A_{j,\varepsilon }$ of the matrix $A_ j$ of locally bounded functions. Let $K\subset \Omega $ be compact. For small $\varepsilon $, the entries of $A_{j,\varepsilon }$ are bounded on $K$ by the bounds of the entries of $A_ j$ on a fixed compact neighborhood of $K$, and by lemma 2.5.1 they converge to those of $A_ j$ in $L^1(K)$. Products of functions which are bounded on $K$ uniformly in $\varepsilon $ and converge in $L^1(K)$ converge in $L^1(K)$, because $|ab-a'b'|\leq |a||b-b'|+|b'||a-a'|$. Since $D$ is a polynomial in the matrix entries, it follows that $D(A_{1,\varepsilon },\dots ,A_{n,\varepsilon })\to D(A_1,\dots ,A_ n)$ in $L^1(K)$. By the first part, $(n!/\pi ^ n)D(A_{1,\varepsilon },\dots ,A_{n,\varepsilon })\, \mathrm dV_{2n}$ is the product of the $\mathrm{dd}^{\mathrm{c}}u_{j,\varepsilon }$, so the weak limit of these products is $(n!/\pi ^ n)D(A_1,\dots ,A_ n)\, \mathrm dV_{2n}$. This proves the first formula in general.

By lemma 7.13.2, each $u_ j$ belongs to $W^{2,2}_{\mathrm{loc}}(\Omega )\subset W^{2,1}_{\mathrm{loc}}(\Omega )$, so proposition 7.13.3, applied to $u_ j$ and the chosen representative of $v_ j$, gives $A_ j=B_ j$ almost everywhere on $\{ u_ j=v_ j\} $, hence on $E$. Substituting into the first formula proves the second, and also shows that $\mathbf1_ ED(B_1,\dots ,B_ n)=\mathbf1_ ED(A_1,\dots ,A_ n)$ is locally bounded. $\square$