Interior small-energy gradient control for almost complex maps

Use the pullback Levi-Civita connection on $u^{\ast }TX$ and put $v=u_s$. The equation is equivalent to $u_t=Jv$, because $J^2=-I$. Torsion-freeness gives ${\mathrm{\nabla }}_t{u}_s={\mathrm{\nabla }}_s{u}_t$. Differentiating $u_t=Jv$ in both directions and using the chain rule therefore gives the following exact tension identity, where partial derivatives of $J$ mean derivatives with $x$ held fixed: $\tau :={\mathrm{\nabla }}_s{u}_s+{\mathrm{\nabla }}_t{u}_t=({\mathrm{\nabla }}_{Jv}J)v+J({\mathrm{\nabla }}_{v}J)v+({\mathrm{\partial }}_{t}J+J{\mathrm{\partial }}_{s}J)v\mathrm{.}$ Indeed ${\mathrm{\nabla }}_t(Jv)=({\mathrm{\partial }}_{t}J+{\mathrm{\nabla }}_{u_t}J)v+J{\mathrm{\nabla }}_{t}v$ and ${\mathrm{\nabla }}_{t}v={\mathrm{\nabla }}_s(Jv)=({\mathrm{\partial }}_{s}J+{\mathrm{\nabla }}_{v}J)v+J{\mathrm{\nabla }}_{s}v$; the last term $J^2{\mathrm{\nabla }}_{s}v$ cancels ${\mathrm{\nabla }}_s{u}_s$. Thus $\tau =Q(s,t,u)(v,v)+R(s,t,u)v$, with $Q$ bilinear and $R$ linear, and with uniformly bounded coefficients and first derivatives by the assumed order-two bounds.

Let $H=\mathrm{\mid }\mathrm{\nabla }du\mathrm{\mid }$ and $P=\mathrm{\mid }du\mathrm{\mid }$. Differentiating the last expression for $\tau$ in a source direction gives a uniform bound $\mathrm{\mid }\mathrm{\nabla }\tau \mathrm{\mid }\le c(P^3+P^2+P+(P+1)H)\mathrm{.}$ For example differentiation of $Q(v,v)$ gives terms with coefficient derivatives times $\mathrm{\mid }v{\mathrm{\mid }}^2$, bounded by $c(P+1)P^2$, and terms bounded by $cPH$; differentiation of $Rv$ gives terms bounded by $c(P+1)P+cH$. This accounts for every source derivative, including the explicit domain dependence of $J$. A single $c$ works for all maps because its ingredients are the fixed tensor bounds.

The needed energy identity follows by metric compatibility and commuting covariant derivatives. For source indices $\alpha ,\beta \in \{s,t\}$, first differentiate $e=\frac{1}{2}{\sum }_{\alpha }⟨u_{\alpha },u_{\alpha }⟩$ twice to obtain $\mathrm{\Delta }e=H^2+{\sum }_{\alpha ,\beta }⟨{\mathrm{\nabla }}_{\beta }{\mathrm{\nabla }}_{\beta }{u}_{\alpha },u_{\alpha }⟩$. Torsion-freeness replaces ${\mathrm{\nabla }}_{\beta }{u}_{\alpha }$ by ${\mathrm{\nabla }}_{\alpha }{u}_{\beta }$, and commuting the remaining derivatives replaces their sum by ${\mathrm{\nabla }}_{\alpha }\tau$ plus curvature terms $R^X(u_{\beta },u_{\alpha })u_{\beta }$, with a sign determined by the curvature convention and irrelevant to the estimate. Compactness bounds their absolute contribution by $c_1{P}^4$. Thus $\mathrm{\Delta }e\ge H^2-\mathrm{\mid }\mathrm{\nabla }\tau \mathrm{\mid }P-c_1{P}^4\mathrm{.}$ Using step 2.1 and $P^3\le (P^4+P^2)\mathrm{/}2$, this is bounded below by $H^2-c_2(P^4+P^2)-c_3(P^2+P)H$. Finally $c_3(P^2+P)H\le H^2\mathrm{/}2+c_4(P^4+P^2)$ by the elementary square inequality. Discard the remaining nonnegative $H^2\mathrm{/}2$ and use $P^2=2e$ to obtain $\mathrm{\Delta }e\ge -a{e}^2-be$ for uniform $a,b\ge 0$.

Set $r_0=1\mathrm{/}\sqrt{\max \{b,1\}}$ and ${\epsilon }_0=3\pi \mathrm{/}(128\max \{a,1\})$. Then $r\le r_0$ implies $b{r}^2\le 1$, and $\int e<{\epsilon }_0$ implies $a\int e<3\pi \mathrm{/}64$. Smoothness makes $e$ a nonnegative $C^2$ function on a neighborhood of the closed disk. Apply [F1] to step 3.1 and multiply its conclusion by two, since $\mathrm{\mid }du(z){\mathrm{\mid }}^2=2e(z)$. This proves the gradient bound, with constants independent of the chosen solution and disk.