Interior $W^{2,p}$ estimate for uniformly elliptic equations with continuous coefficients

Frozen estimate on nested balls. Fix $x\in B_R(x_0)$ and $0<\rho <R\mathrm{/}4$ with $B_{2\rho }(x)\subset B_R(x_0)$. Freeze $A$ at $A_0=A(x)$ and choose $\eta \in C_c^{\mathrm{\infty }}(B_{\rho }(x))$ with $0\le \eta \le 1$ and $\eta =1$ on $B_{\rho \mathrm{/}2}(x)$, with $\mathrm{\mid }D\eta \mathrm{\mid }\le C_n{\rho }^{-1}$ and $\mathrm{\mid }{D}^2\eta \mathrm{\mid }\le C_n{\rho }^{-2}$. Set $w=\eta u$, extended by zero. The constant-coefficient estimate [F2] and the product identity [F4] give $\mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_{L^p(B_{\rho \mathrm{/}2}(x))}\le \mathrm{\parallel }{D}^{2}w{\mathrm{\parallel }}_{L^p({\mathbb{R}}^n)}\le C\mathrm{\parallel }{A}_0:D^{2}w{\mathrm{\parallel }}_{L^p({\mathbb{R}}^n)}\le C(\mathrm{\parallel }f{\mathrm{\parallel }}_{L^p(B_{\rho }(x))}+{\omega }_A(\rho )\mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_{L^p(B_{\rho }(x))}+(M_b+\mathrm{\Lambda }{\rho }^{-1})\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p(B_{\rho }(x))}+(M_c+\mathrm{\Lambda }{\rho }^{-2}\left)\mathrm{\parallel }u{\mathrm{\parallel }}_{L^p(B_{\rho }(x))}\right),$ where ${\omega }_A(\rho ):=\sup \{\mathrm{\mid }A(y)-A(z)\mathrm{\mid }:y,z\in {\bar{B}}_R(x_0),\mathrm{\mid }y-z\mathrm{\mid }\le \rho \}$ and the constant $C$ depends only on $n,p,\lambda ,\mathrm{\Lambda }$. Apply the doubled-ball interpolation inequality [F3] to $u$ on $B_{2\rho }(x)$: $\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p(B_{\rho }(x))}\le \epsilon \rho \mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_{L^p(B_{2\rho }(x))}+C_{n,p,\epsilon }{\rho }^{-1}\mathrm{\parallel }u{\mathrm{\parallel }}_{L^p(B_{2\rho }(x))}\mathrm{.}$ Since $\mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_{L^p(B_{\rho })}\le \mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_{L^p(B_{2\rho })}$, choose $\epsilon >0$ small and then ${\rho }_{\ast }>0$ so that for every $0<\rho \le {\rho }_{\ast }$ the coefficient of $\mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_{L^p(B_{2\rho }(x))}$ after substitution is at most any prescribed ${\theta }_0>0$; this is possible because ${\omega }_A(\rho )\to 0$ and $\rho M_b\le R{M}_b$. Absorbing constants in the lower-order term yields the local estimate $\mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_{L^p(B_{\rho \mathrm{/}2}(x))}\le C_0\mathrm{\parallel }f{\mathrm{\parallel }}_{L^p(B_{\rho }(x))}+C_0{\rho }^{-2}\mathrm{\parallel }u{\mathrm{\parallel }}_{L^p(B_{2\rho }(x))}+{\theta }_0\mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_{L^p(B_{2\rho }(x))},$ where $C_0$ depends on $n,p,\lambda ,\mathrm{\Lambda },R{M}_b,R^2{M}_c$ and the modulus of continuity of $A$, but not on $x$ or $\rho \le {\rho }_{\ast }$. The cutoff is identically one on the smaller ball, so the left side is the unweighted Hessian norm there; no division by a vanishing cutoff is used.

Finite-overlap cover and hole filling. Write $M(r):=\mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_{L^p(B_r(x_0))}$ and $U:=\mathrm{\parallel }u{\mathrm{\parallel }}_{L^p(B_R(x_0))}$, $F:=\mathrm{\parallel }f{\mathrm{\parallel }}_{L^p(B_R(x_0))}$. For $r<R$ with $\delta :=s-r>0$ sufficiently small, put $\rho =\delta \mathrm{/}4$ and cover $B_r(x_0)$ by the balls $B_{\rho \mathrm{/}2}(x_j)$ of a cubic lattice of mesh $\rho \mathrm{/}(4\sqrt{n})$; the enlarged balls $B_{2\rho }(x_j)$ lie in $B_s(x_0)$ and have overlap bounded by a constant depending only on $n$. Applying step 1.1 on each patch and taking the $p$-sum, the finite-overlap bounds give $M(r)\le C_{1}F+C_1{\delta }^{-2}U+\theta M(s),$ where $\theta$ can be fixed in advance as small as desired by choosing ${\theta }_0$ small enough relative to the overlap constant, and $C_1$ is independent of $r,s$ (it may depend on ${\rho }_{\ast }^{-1}$ only through the permitted modulus-of-continuity dependence). Choose $\theta <1\mathrm{/}4$. Take ${\delta }_j={\delta }_0{2}^{-j}$ with $0<2{\delta }_0\le R\mathrm{/}4$ and ${\delta }_0\mathrm{/}4\le {\rho }_{\ast }$, and put $r_0=3R\mathrm{/}4$, $r_{j+1}=r_j+{\delta }_j$; then $r_j\uparrow r_{\mathrm{\infty }}\le R$. Iterating gives $M(r_0)\le C_{1}F{\sum }_{j=0}^{N-1}{\theta }^j+C_{1}U{\sum }_{j=0}^{N-1}{\theta }^j{\delta }_j^{-2}+{\theta }^{N}M(r_N)\mathrm{.}$ The first series is bounded, the second converges because ${\delta }_j^{-2}={\delta }_0^{-2}{4}^j$ and $4\theta <1$, and the final term tends to zero since $u\in W^{2,p}(B_R(x_0))$. Therefore $\mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_{L^p(B_{3R\mathrm{/}4})}\le C(F+R^{-2}U)$, with ${\delta }_0^{-2}$ written as $R^{-2}$ times a constant depending on the permitted dimensionless radius ratio. To control first derivatives on $B_{R\mathrm{/}2}$, choose a cubic lattice of mesh $R\mathrm{/}(32\sqrt{n})$ and retain the finitely many centers $x_j\in B_{9R\mathrm{/}16}(x_0)$ whose balls $B_{R\mathrm{/}16}(x_j)$ meet $B_{R\mathrm{/}2}(x_0)$. These inner balls cover $B_{R\mathrm{/}2}$: every point is within $R\mathrm{/}64$ of a lattice point, and such a point lies in $B_{9R\mathrm{/}16}$. Their doubled balls $B_{R\mathrm{/}8}(x_j)$ lie in $B_{11R\mathrm{/}16}(x_0)\subset B_{3R\mathrm{/}4}(x_0)$. Apply the doubled-ball interpolation inequality [F3] with radius $R\mathrm{/}16$ on each patch and take the finite $p$-sum; bounded overlap gives $R^{-1}\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p(B_{R\mathrm{/}2})}\le C_n(\mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_{L^p(B_{3R\mathrm{/}4})}+R^{-2}U)\le C(F+R^{-2}U)\mathrm{.}$ The zeroth-order term satisfies $R^{-2}\mathrm{\parallel }u{\mathrm{\parallel }}_{L^p(B_{R\mathrm{/}2})}\le R^{-2}U$. Combining this with the Hessian bound proves the displayed scale-invariant estimate. The exponent range is $1<p<\mathrm{\infty }$ as in [F1] and [F3], and the constant has exactly the stated dependence.