Pome95F071B4092FA nonnegative $C^{1,1}$ density grows at most quadratically away from each of its zeros. If its zero set separates a small ball, a family of parallel segments crossing the zero set therefore contributes at least a constant times $\sqrt t$ to the volume of $\{ f\leq t\} $ for small $t>0$. Excluding this lower bound makes $\{ f>0\} $ connected, and 14.21.3 then gives uniqueness.
Write $\operatorname {Vol}(E)=\int _ E\mathrm{d}V$ for Borel sets $E$.
Theorem 14.21.13.61851D05B072 Let $u,v\in C^2(X)$ satisfy $A,B\geq 0$ and $A^ n=B^ n=f\, \mathrm{d}V$, where $f\geq 0$ is $C^{1,1}$. If
B72083672909
14.21.14
then $\{ f>0\} $ is connected and $u-v$ is constant. In particular this holds when $f^{-1/2}\in L^1(\mathrm{d}V)$, with $f^{-1/2}=+\infty $ on $\{ f=0\} $. The exponent $1/2$ is sharp in every dimension $n\geq 3$.
Proof. Write $Z=\{ f=0\} $. Condition (14.21.14) first implies $\operatorname {Vol}(Z)=0$. We show that the complement of $Z$ in every sufficiently small coordinate ball is connected. Fix such a ball with convex Euclidean image. If the complement of $Z$ in it were disconnected, choose points $a,b$ in different components and small closed disks centered at $a$ and $b$, perpendicular to the segment $[a,b]$ and contained in those respective components. By making the disks small, all parallel segments between corresponding points stay in the ball. Each segment meets $Z$, since otherwise it would connect the two components.
On a fixed coordinate region containing the closed ball, the gradient of $f$ is Lipschitz with some constant $K>0$. Since $f\geq 0$, we have $\nabla f(z)=0$ at every zero $z$, hence
whenever the segment from $z$ to $x$ stays in that region. The endpoint disks are compact and disjoint from $Z$, so the parts of the segments near them contain no zeros. Thus every parallel segment contains a zero at a fixed positive distance from its endpoints. For small $t>0$, (14.21.15) gives on that segment an interval of length $2\sqrt{2t/K}$ contained in $\{ f\leq t\} $. Integrating over the transverse disk, which has positive measure, and using that $\mathrm{d}V$ is comparable to coordinate volume, we obtain
for all sufficiently small $t$, contradicting (14.21.14). This proves the local connectedness assertion.
For each such ball, the complement of $Z$ in it is nonempty, since $Z$ has zero volume, and it lies in one component of $U=X\setminus Z$. Two overlapping balls give the same component, because their open overlap contains a point outside $Z$. Since $X$ is connected, all these balls give the same component. Hence $U$ is connected, and 14.21.3 shows that $u-v$ is constant.
If $f^{-1/2}\in L^1$, then, outside the null zero set,
by dominated convergence. This proves the sufficient integrability condition. The sharpness assertion follows from the examples in 14.29, as quantified at the end of that section. $\square$
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