14.5 Second derivatives and ellipticityCBFCDA7183ED

Proof. The identity $\mathrm{d}\omega _ u=\mathrm{d}\omega $ is equivalent to

together with its complex conjugate; this can also be checked directly, since the third derivatives of $u$ cancel.

For tensors with two holomorphic and one antiholomorphic lower index, we use the norm

Set $A_{ik\bar j}=\nabla _ i h_{k\bar j}$ and $D_{ik\bar j}=T_{ik}^ p h_{p\bar j}$. We first claim

To prove this, start with $\Delta _ hS=h^{i\bar j}g^{k\bar\ell }\nabla _ i\nabla _{\bar j}h_{k\bar\ell }$. Switch $\bar j,\bar\ell $ by the conjugate of (14.5.3), commute $\nabla _ i,\nabla _{\bar\ell }$, and then switch $i,k$ by (14.5.3). The terms with a derivative of $h$ are

The first switch also produces the term $-h^{i\bar j}g^{k\bar\ell }\overline{T_{j\ell }^ q}\nabla _ i h_{k\bar q}$. Applying (14.5.3) to it gives one of the two conjugate derivative terms displayed above, together with

The part of this expression containing $h_{p\bar q}$ is $\| D\| ^2$. Differentiating the logarithmic determinant twice gives

By the antisymmetry of $T$, the two remaining derivative terms combine to $2\operatorname {Re}\langle A,D\rangle $, where $\langle \cdot ,\cdot \rangle $ is the inner product of the norm above. These identities give (14.5.4).

To bound the remainder $E$ explicitly, let

The remainder is the contraction by $h^{i\bar j}g^{k\bar\ell }$ of

The commutator $\mathcal C$ is the curvature of $\nabla $ acting linearly on $h$, as one sees by inserting $\nabla =\partial +\Gamma $ and canceling the derivatives of $h$. In a $g$-unitary frame diagonalizing $h$, every coefficient in this remainder is bounded by $C_ g(1+S)$, so the contraction is bounded by $C_ g(1+S)\sum _ i h^{i\bar i}=C_ g(1+S)B$, as asserted. In particular, no derivative of $h$ remains in $E$.

It remains to control the gradient term introduced by the logarithm. Define

By (14.5.3),

Moreover $\| H\| ^2=|\partial S|_ h^2/S$. Expanding the square therefore gives

Combining this with (14.5.4) and with $\Delta _ h\log S=\Delta _ hS/S-|\partial S|_ h^2/S^2$, we obtain the exact identity

Since $K\geq 0$ and $S\geq 1$, and since $E/S\geq -2C_ gB$, enlarging the constant proves the lemma. $\square$

Proof. Use $h=g+(u_{i\bar j})$, $S=\operatorname {tr}_ g h$, and $B=\operatorname {tr}_ h g$ as in lemma 14.5.1. Write $s=u-\inf _ Xu+1$, so that $1\leq s\leq D+1$, and consider the maximum of

on $X$, where $A\geq 1$ will be fixed below. Let $x$ be a maximum point. If $S(x)<1$, an upper bound for $S$ on $X$ already follows from the oscillation bound and the definition of $Q$, as at the end of the proof. We may therefore assume $S(x)\geq 1$ and use lemma 14.5.1.

Since the eigenvalues of $h$ relative to $g$ are positive, $SB\geq n^2$, and hence $\Delta _ gG/S\geq -KB/n^2$. Moreover, $\partial Q=0$ at $x$ gives

Because $|\tau |_ h^2\leq C_ gB$, the absolute value of the torsion term is bounded by

If $S(x)^2\leq (A+1)^2(D+1)^3$, then $S(x)$ is bounded once $A$ is fixed, and again so is $S$ on $X$. In the remaining case the coefficient of $B$ in the last term is at most $C_ g$.

On the other hand, direct computation gives

The trace inequality and (14.5.6) imply at $x$

after a fixed enlargement of $C_ g$. Choosing $A>C_ g+K/n^2+1$, we conclude that $B(x)$ is bounded. If $\lambda _1,\ldots ,\lambda _ n$ are the eigenvalues of $h$ relative to $g$, then $\prod _ i\lambda _ i=\mathrm{e}^ G$, and for $n\geq 2$

For $n=1$, directly $S=\mathrm{e}^ G$. Thus $S(x)$ is bounded in all cases. For every $y\in X$, the maximality of $Q$ at $x$ gives

This proves $S\leq C$ on $X$. The lower bound for the determinant then gives $\lambda _ i\geq \mathrm{e}^{-K}C^{-(n-1)}$, which proves the uniform equivalence. $\square$

Proof. Subtract $\inf _ Xv$, so that $0\leq v\leq M$, and put $K=\| \mathrm{d}v\| _{L^\infty (g)}$. Fix $p>2n$ and $0<\alpha <1-2n/p$. The fixed operator $\Delta _ g$ is uniformly elliptic on the underlying real manifold. The local $W^{2,p}$ estimates, patched by a finite atlas, and Morrey's inequality give

These are the standard linear elliptic estimates of [Chapter 9, GT01]; the constants depend on the fixed metric and on the chosen $p,\alpha $.

For $0<t\leq t_0$, with $t_0$ fixed by the atlas, integration along a coordinate segment of length $t$ gives

Indeed, the difference quotient is bounded by $2M/t$, and its difference from the derivative at the initial point is bounded by $C_ gt^\alpha \| v\| _{C^{1,\alpha }}$. A finite atlas with smaller charts controls every point and coordinate direction; equivalence with the $g$-norm gives the displayed estimate. Consequently,

Choose $t\leq t_0$ so small that $C_ g^2At^\alpha \leq 1/2$. This choice depends only on $g,A$, including when $A=0$. Absorbing the last term bounds $K$ in terms of $g,A,M$, and the assumed trace estimate then bounds $\| \Delta _ gv\| _\infty $. $\square$

Proof. Theorem 14.5.5 gives uniform ellipticity and a bound for $\Delta _ gu=S-n$. The linear elliptic $W^{2,p}$ estimates for the fixed operator $\Delta _ g$ first bound $u$ in $C^{1,\alpha }$ for every $0<\alpha <1$. The local Evans–Krylov estimate for the concave logarithmic determinant then gives a uniform $C^{2,\beta }$ bound for some $0<\beta <1$. To apply it in its real form, project real symmetric Hessians to their complex Hermitian part and extend the logarithmic determinant concavely from the compact set of positive matrices given by the preceding theorem. The extension is uniformly elliptic on real symmetric matrices because the projection preserves positivity and the trace, so the usual Evans–Krylov theorem applies. Finally, differentiating the equation gives linear uniformly elliptic equations with Hölder coefficients, and Schauder estimates and induction give all higher bounds. No regularity results beyond these standard local elliptic estimates are needed. $\square$