14.19.47 Gradient bounds for concave equationsF77131C8889A

Proof. If $\lambda (P)\in \Gamma $, then $\mathcal A_ U+tP$ is admissible for $t\geq 0$. The function $t\mapsto F(\mathcal A_ U+tP)$ is positive and concave on $[0,\infty )$, so its derivative at $t=0$ cannot be negative; thus $\operatorname {tr}(BP)\geq 0$. Since $F$ is concave and $F(0)=0$, the tangent inequality at $\mathcal A_ U$, evaluated at the origin, gives $\operatorname {tr}(B\mathcal A_ U)\leq h$. Choose $a_*>0$ with $F(a_*\operatorname {Id})>H_0+1$. Concavity gives

Thus $T\geq a_*^{-1}$. Since $\mathcal L(U-V)=\operatorname {tr}(B\mathcal A_ U)-\operatorname {tr}(B\mathcal A_ V)$ and $\mathcal L(\underline U-U)=\operatorname {tr}(B\mathcal A_{\underline U})-\operatorname {tr}(B\mathcal A_ U)$, the remaining two bounds follow from the nonnegativity of $\operatorname {tr}(B\mathcal A_ V)$ and $\operatorname {tr}(B\mathcal A_{\underline U})$.

We derive the alternative from the strict gap. Fix $r,H,\delta $ from lemma 14.19.43, and put $\mu =\lambda (\mathcal A_{\underline U})$. Compactness and the gap give one $S>r$ such that

everywhere. We first prove a scalar alternative. At a point of $X$, let $\lambda \in \Gamma $ satisfy $f(\lambda )=\sigma $, where $\sigma =h$ at that point, and consider the set

It is contained in $\Gamma $ because $\mu -r\mathbf1\in \Gamma $, it is closed, and the matrix bound of lemma 14.19.43, applied to diagonal matrices, makes it bounded; thus $K$ is compact. When nonempty, it contains $\mu -r\mathbf1$ and is star-shaped about that point: moving along a segment toward it decreases each coordinate, hence decreases $f$. Thus $K$ is connected. Let $N=\nabla f(\lambda )$, whose entries are positive, and let $\Pi $ be the tangent hyperplane $N\cdot (v-\lambda )=0$.

If $f(\mu -r\mathbf1)>\sigma $, concavity gives $N\cdot (\mu -\lambda )>r\sum _ iN_ i$. Otherwise $\mu -r\mathbf1\in K$. The diagonal ray $\mu -r\mathbf1+t\mathbf1$, $t\geq 0$, meets $\{ f=\sigma \} $, since $f$ diverges along it; at the meeting point $v$, the tangent inequality gives $N\cdot (v-\lambda )\geq 0$. If $\Pi $ misses $K$, connectedness makes $N\cdot (v-\lambda )$ strictly positive throughout $K$, and in particular at $\mu -r\mathbf1$; this again gives $N\cdot (\mu -\lambda )>r\sum _ iN_ i$.

If $v\in \Pi \cap K$, put $z_ i=v-r\mathbf1+Se_ i$. Then $z_ i\geq \mu -2r\mathbf1+Se_ i$, so $f(z_ i)>\sigma $. The tangent inequality at $\lambda $ gives

Thus $N_ i>(r/S)\sum _ jN_ j$ for every $i$. We have proved that $N\cdot (\mu -\lambda )>r\sum _ iN_ i$ or $N_ i>(r/S)\sum _ jN_ j$ for all $i$, at every positive level $\sigma $, with constants independent of $\sigma $.

To pass to matrices, take $\lambda =\lambda (\mathcal A_ U)$, and order $\lambda $ and $\mu $ decreasingly. By concavity and symmetry, the numbers $f_ i(\lambda )$ are then ordered increasingly. In the eigenframe of $\mathcal A_ U$, the diagonal of $\mathcal A_{\underline U}$ is a doubly stochastic average of permutations of $\mu $, so the rearrangement inequality gives

Since $N=(f_ i(\lambda ))_ i=(B_ i)_ i$, taking $\kappa $ smaller than $\min \{ r,r/S\} $ proves the matrix alternative. $\square$

Proof. Write $\mathscr T^ p_{ik}$ and $R^ p_{i\bar jk}$ for the torsion and curvature of the Chern connection. The commutation formula and the differentiated equation read

The product rule for $|\partial U|^2$ gives both squares $\sum B_ i|U_{ki}|^2$ and $\sum B_ i|U_{k\bar i}|^2$, the term $2\operatorname{Re}\sum _ kh_ kU_{\bar k}$, a curvature error bounded by $CT|\partial U|^2$, and the errors from $\nabla \chi $ and torsion. The latter are bounded by

Absorb the last term into half of the mixed square. Since $U_{i\bar j}=\lambda _ i\delta _{ij}-\chi _{i\bar j}$, the remaining half is at least $\frac14\sum _ iB_ i\lambda _ i^2-CT$. Hence

Finally,

In the product rule for $\mathrm e^\xi Q$, combine the cross term coming from the first sum with $|\partial U|^2|\partial \xi |_ B^2$ and with the square $\sum B_ i|U_{ki}|^2$. This gives $\sum B_ i|U_{ki}+U_ k\xi _ i|^2$, while the remaining weight term is $|\partial \xi |_ B^2$. $\square$

Proof. Let $a\geq 1$ and $\beta >1$ be constants, to be chosen below. Assuming $\| U-V\| _\infty \leq \beta ^{-1}$, put

Then $-2\beta ^{-1}\leq w\leq 0$ and

Writing $D=|\partial (U-V)|_ B^2$, the Cauchy–Schwarz inequality gives

For the first line, split $\xi _ i$ as above and use $\sum B_ i|U_ i|^2\leq 2D+2K_ V^2T$; then apply Young's inequality to the resulting three products, each absorbing at most $\frac1{24}\sum B_ i\lambda _ i^2$. For the second line, use $|\partial \xi |_ B\leq aK_ VT^{1/2}+(a+4)D^{1/2}$ and the bound for $\chi $. In both lines, $C_ a$ depends on $a$ but not on $\beta $ or on the derivatives of $V$.

By lemma 14.19.48 and $w\leq 0$,

Choose $a$ so large that $(a\kappa -C_ g-2)\tau >4H_0+2H_1+1$, and then $\beta \geq 2C_ a+2$. Combining the preceding bounds with lemma 14.19.49, we obtain, at every point where $Q\geq C'(1+K_ V^4)$ for a suitable constant $C'$,

Indeed, the coefficient of $D$ before absorption is $(2\beta -C_ a-1/2)Q-C_ aK_ V^2$, which is at least $\beta Q$ once $C'$ is large enough. The other errors satisfy $C_ aK_ V^4\leq Q$ and $C_ a+C_ aK_ VQ^{1/2}\leq Q$, and they account for the additional term $-2T$ in the bracket.

Consider a maximum point of $\mathrm e^\xi Q$ at which $Q\geq C'(1+K_ V^4)$. In the first alternative of lemma 14.19.48, the bracket is positive by the choice of $a$, since $T\geq \tau $. In the second, $D\geq \kappa T|\partial (U-V)|^2$. For $Q\geq 4(1+K_ V^2)$, the triangle inequality gives $|\partial (U-V)|^2\geq c_*Q$, where $c_*=(\sqrt3-1)^2/4$. Since $\mathcal L(\underline U-U)\geq -H_0$, the bracket is then at least

This is positive once $C'$ is large enough, because $T\geq \tau $. In both cases $\mathcal L(\mathrm e^\xi Q)>0$ at a maximum point, which contradicts the second derivative test. Thus $Q\leq C(1+K_ V^4)$ at the maximum point. The weight has uniformly bounded oscillation:

By the potential bound in proposition 14.19.44, this oscillation is bounded independently of $V$, so the estimate at the maximum point gives $Q\leq C(1+K_ V^4)$ everywhere. Taking square roots and setting $\eta =\beta ^{-1}$ proves the assertion. $\square$

Proof. The right-hand sides in the family with $\| \partial h\| _\infty \leq H_1$ form an equicontinuous set. By corollary 14.19.46, the corresponding functions $U$ form a precompact subset of $C^0(X)$. Choose a finite $\eta $-net in this set, consisting of smooth admissible functions $V_1,\ldots ,V_ N$. Each solution is within $\eta $ of some $V_ j$, so proposition 14.19.50 bounds its gradient by $C(1+\max _ j\| \partial V_ j\| _\infty ^2)$. The set of solutions, and hence the chosen finite net, depends only on the fixed data and $H_1$. Finally, $u=2\pi U$, so the same conclusion holds for $\mathrm du$. $\square$