Multiplicative interior Harnack for geometric Schrödinger solutions

At any coordinate point $p$, replace $u$ by the quadratic polynomial having its value, first derivatives and second derivatives at $p$. The metric trace of the Levi-Civita Hessian uses only these data: in coordinates its Hessian entries are ${\mathrm{\partial }}_{ij}u-{\mathrm{\Gamma }}_{ij}^k{\mathrm{\partial }}_{k}u$. Apply the smooth formulas in [F1] to this polynomial. The coordinate expression $m^{-1}{\mathrm{\partial }}_i(m{h}^{ij}{\mathrm{\partial }}_{j}u)$ also uses only these same derivatives through order two, so the polynomial and $u$ give equal values on both sides at $p$. Thus ${\mathrm{\Delta }}_{h}u=m^{-1}{\mathrm{\partial }}_i(A^{ij}{\mathrm{\partial }}_{j}u)$ throughout the ball at the stated $C^2$ regularity. Multiplying the pointwise equation by $m$ gives ${\mathrm{\partial }}_i(A^{ij}{\mathrm{\partial }}_{j}u)=qu$, with repeated indices summed. Here $A$ is smooth and symmetric.

Set $D=\{x:u(x)>0\}$, an open set, and define $q_{\ast }(x)=m(x){\mathrm{\Delta }}_{h}u(x)\mathrm{/}u(x)$ on $D$ and $q_{\ast }(x)=0$ off $D$. On $D$ this coefficient is continuous and equals $q$ by the given equation; hence $0\le q_{\ast }\le Q$ everywhere. Its zero extension is Borel: on the open set $D$ it is continuous, and on the complementary closed set it is constant, so preimages of open subsets of $\mathbb{R}$ are Borel. At points where $u=0$, the pointwise equation gives ${\mathrm{\Delta }}_{h}u=Vu=0$. Consequently $q_{\ast }u=qu=m{\mathrm{\Delta }}_{h}u$ everywhere, including those points, and $-\mathrm{div}(A\mathrm{\nabla }u)+q_{\ast }u=0$ pointwise. In particular $q_{\ast }u$ is continuous even though $q_{\ast }$ need not be continuous. This construction includes $u\equiv 0$ and makes no assertion that the originally given $V$ is measurable.

Fix $0<s<r$ and set $v(y)=u(T_s(y))$, $\tilde{A}(y)=A(T_s(y))\mathrm{/}\lambda$ and $\tilde{b}(y)=s^2{q}_{\ast }(T_s(y))\mathrm{/}\lambda$ on $B_1\subset {\mathbb{R}}^n$. The chain rule gives ${\mathrm{\partial }}_{y_j}v=s({\mathrm{\partial }}_{x_j}u)\circ T_s$ and ${\mathrm{div}}_y(\tilde{A}{\mathrm{\nabla }}_{y}v)=(s^2\mathrm{/}\lambda )({\mathrm{div}}_x(A{\mathrm{\nabla }}_{x}u))\circ T_s$. Thus $-{\mathrm{div}}_y(\tilde{A}{\mathrm{\nabla }}_{y}v)+\tilde{b}v=0$ pointwise. The matrix $\tilde{A}$ is smooth, symmetric and at least $I$, and $\mathrm{\mid }{\tilde{A}}^{ij}\mathrm{\mid }\le M\mathrm{/}\lambda$. The potential is Borel and satisfies $0\le \tilde{b}\le K:=r^{2}Q\mathrm{/}\lambda$, with this fixed bound independent of $s$. Because $\overline{B_s(x_0)}$ is compactly contained in $B_r(x_0)$, $v$ and $\tilde{A}$ are respectively $C^2$ and smooth on a neighborhood of $\overline{B_1}$.

If $n\ge 2$, put $d=n$, $w=v$, $H=\tilde{A}$ and $b=\tilde{b}$. If $n=1$, put $d=2$ and, on the two-dimensional unit disk, set $w(z_1,z_2)=v(z_1)$, $H(z_1,z_2)=\mathrm{diag}(\tilde{A}(z_1),1)$ and $b(z_1,z_2)=\tilde{b}(z_1)$. In this latter case ${\mathrm{\partial }}_{2}w=0$ and the equation of step 3.1 gives $-{\mathrm{\partial }}_1(\tilde{A}(z_1)v^{\mathrm{\prime }}(z_1))+b(z_1,z_2)w(z_1,z_2)=0$, exactly $-\mathrm{div}(H\mathrm{\nabla }w)+bw=0$ on the disk. In either case $d\ge 2$, $w\ge 0$, $w$ is $C^2$ near the closed disk or ball, $H$ is smooth symmetric with $H\ge I$, and $b$ is Borel with $0\le b\le K$. Since $M\mathrm{/}\lambda \ge 1$, the entrywise bound ${\sum }_{i,j}\mathrm{\mid }{H}^{ij}\mathrm{\mid }\le d^{2}M\mathrm{/}\lambda$ holds in both cases; for the lifted matrix its sum is at most $M\mathrm{/}\lambda +1$. No one-dimensional version of the imported theorem is assumed.

The function $w$ and its classical first derivatives are bounded on $B_1\subset {\mathbb{R}}^d$ and therefore square integrable. For every real $\phi \in C_c^{\mathrm{\infty }}(B_1)$, the field $\phi w{e}_i$ is $C^1$ and vanishes near the boundary. Applying [F3] to its divergence gives $\int w{\mathrm{\partial }}_i\phi =-\int ({\mathrm{\partial }}_{i}w)\phi$, so these classical derivatives are weak derivatives and $w\in W^{1,2}(B_1)$. Likewise the field $\phi H\mathrm{\nabla }w$ is $C^1$ and vanishes near the boundary. Its divergence integrates to zero; the pointwise equation in step 4.1 and the product rule give ${\int }_{B_1}(H\mathrm{\nabla }w\cdot \mathrm{\nabla }\phi +bw\phi )=0$. The products are measurable and integrable; indeed $bw=\mathrm{div}(H\mathrm{\nabla }w)$ is continuous. These are precisely the weak-solution hypotheses of [F4]. Therefore [F4] gives the comparison on $B_{1\mathrm{/}2}\subset {\mathbb{R}}^d$ with the same constant $C_H(d,d^{2}M\mathrm{/}\lambda ,K)$ for every $s$.

Since $w$ is continuous, its essential supremum and infimum on the open half-ball equal its ordinary supremum and infimum: any strict pointwise violation of an essential bound persists on a nonempty open ball, which has positive Lebesgue measure. For $n\ge 2$ this directly gives $v(y)\le Cv(y^{\mathrm{\prime }})$ for all $y,y^{\mathrm{\prime }}\in B_{1\mathrm{/}2}$. For $n=1$, restrict the resulting pointwise comparison to the pairs $(y,0),(y^{\mathrm{\prime }},0)$ in the half-disk; these range over all $y,y^{\mathrm{\prime }}\in (-1\mathrm{/}2,1\mathrm{/}2)$. Thus in all cases $u(x)\le Cu(x^{\mathrm{\prime }})$ for $x,x^{\mathrm{\prime }}\in B_{s\mathrm{/}2}(x_0)$, where $C=\max \{1,C_H(d,d^{2}M\mathrm{/}\lambda ,r^{2}Q\mathrm{/}\lambda )\}$ depends only on the asserted parameters. Every pair in $B_{r\mathrm{/}2}(x_0)$ lies in some such smaller half-ball, so the same pairwise comparison holds throughout $B_{r\mathrm{/}2}(x_0)$. Taking the supremum in $x$ and then the infimum in $x^{\mathrm{\prime }}$ proves the Statement.