8.5 Bounded complex Hessians and elliptic extensionsEA00781761AE

Proof. On a coordinate ball the Euclidean Laplacian is $\Delta u_ j=4\sum _ i (u_ j)_{i\bar i}$, so it is uniformly bounded. The interior Laplace estimate on two nested balls gives

A finite cover gives the global bound. Weak $W^{2,p}$ limits coincide with the uniform limit. The bounded complex Hessians have weak-star limits in $L^\infty $; testing against compactly supported smooth functions identifies them with $H(u)$. Choose $p$ with $1-2n/p>\alpha $. The compact Sobolev embedding into $C^{1,\alpha }$ gives convergence of subsequences. Every such limit is $u$, hence the whole sequence converges. This estimates the full real Hessian in every finite $L^ p$ space, without asserting that it is bounded. $\square$

Proof. Set $D_ j^ a=\mathrm{dd}^{\mathrm{c}}u_ j^ a$ and $D^ a=\mathrm{dd}^{\mathrm{c}}u^ a$. On a smaller coordinate ball, convolution gives smooth $u_\epsilon ^ a$ whose complex Hessians are uniformly bounded and converge to $H(u^ a)$ in every finite $L^ p$ space. Their coefficient products consequently converge in finite $L^ p$ spaces. In particular the products of the $D^ a$ are closed currents, since their smooth approximants are closed.

Expand the smooth background factors and telescope each remaining product. Every difference has the form

where $T_ j$ is a product of $D_ j^ b$ and $D^ b$ with $b\ne a$. It has uniformly bounded coefficients and is closed. Local convolution, followed by strong convergence of its finite coefficient products, justifies integration by parts:

Here $\xi $ includes the chosen smooth background factors and the test form, so their derivatives are retained. The right-hand side has the asserted uniform bound. A finite partition of unity proves the claim on $X$. This argument also proves that the coefficient product is the distributional limit; it does not identify a separately defined bounded Hessian operator for all nonsmooth admissible functions. $\square$

Proof. In the closed $2$-cone,

Thus $K$ is bounded and closed. Maclaurin gives $\sigma _ j(A)\geq \sigma _ m(A)^{j/m}\geq f_0^ j$ for $j\leq m$, so every matrix in $K$ is in the interior. In a unitary eigenframe,

The lower bound is 8.2.5; the upper bound uses $|\lambda _ i|\leq L$ in the polynomial numerator and its positivity. These bounds give possible values of $\vartheta $ and $\Theta $. Concavity places each tangent plane above $F$ on the cone; the plane at $B$ gives equality when $B\in K$. An infimum of affine functions is concave. Each plane increases under $P\geq 0$ by a number between the two displayed trace bounds; taking infima preserves these bounds. The same estimates give global Lipschitz continuity of $\widehat F$. $\square$

Proof. The first lemma gives $u\in W^{2,p}$ for every finite $p$ and a bounded complex Hessian. The matrices $A_ j=I+g^{-1/2}H(u_ j)g^{-1/2}$ converge weakly in $L^2$ to the analogous matrix $A$ for $u$. Mazur's lemma gives convex combinations of tails converging strongly in $L^2$, hence almost everywhere along a subsequence. The pointwise closed cone is convex, so $A\in C_{n,m}$ almost everywhere. The product lemma passes the equation to the limit as currents, since its right-hand side converges uniformly. Equality of the bounded densities then gives the equation almost everywhere.

The right-hand sides are bounded above and away from zero. The complex Hessian bound gives a uniform upper bound $L$ for $\operatorname {tr}A_ j$. Choose $f_0>0$ below all their normalized roots. Thus every $A_ j$, and almost every $A$, belongs to the compact set $K$ of the preceding lemma. Every $u_ j$ is a classical solution of

For completeness, if a smooth test $q$ touches $u$ from above at an interior point $x_0$, add a small positive quartic term to make the local contact strict. Local maximum points of $u_ j-q$ tend to $x_0$. At these points $D^2u_ j\leq D^2q$, and ellipticity gives $\mathcal G(x,D^2q)\geq r_ j(x)$. Pass to the limit, then remove the quartic term. A test from below gives the reverse inequality. The extension is defined on every real Hessian, so no admissibility of a lower test has been assumed. This proves the viscosity statement. $\square$