14.21 Extrema and density sublevels6267D9D97C43

Proof. Put $g_\delta (x)=g(x)-\delta |x-p|^2$. The gradient map $F=\nabla g_\delta $ is Lipschitz on the ball, so it maps the null set $N=\overline{B(p,r)}\setminus E$ to a null set. Choose $s\in B(0,\delta \rho )\setminus F(N)$ and maximize $g_\delta (x)-s\cdot (x-p)$ on the closed ball. At any point with $|x-p|\geq \rho $, its value is strictly smaller than its value at $p$, because

Thus a maximizer $x$ lies in $B(p,\rho )$. The first-order condition at $x$ gives $F(x)=s$, so $x\notin N$. The second-order Taylor expansion at this upper contact point gives $D^2g(x)\leq 2\delta I$. Finally, sets $E$ as in the statement exist: by Rademacher's theorem, applied to the Lipschitz gradient of $g$, the function $g$ has a second-order Taylor expansion almost everywhere. $\square$

Proof. We first prove strict weak comparison on a relatively compact domain: if $b\in C^{1,1}$, $Lb:=a^{ij}b_{ij}\geq \kappa >0$ almost everywhere, and $b\leq M$ on the boundary, then $b\leq M$ inside. Otherwise an interior maximum point lies in a closed ball on which $a\leq \Lambda I$. Apply 14.21.5 to the set of full measure where both the differential inequality and the Taylor expansion hold. Choosing $2\delta m\Lambda <\kappa $ gives the contradiction $\kappa \leq Lb\leq 2\delta \operatorname {tr}a\leq 2\delta m\Lambda $.

Take the $C^{1,1}$ representative of $z$, write $M=\sup _\Omega z$, and suppose that $K=\{ z=M\} $ is a nonempty proper subset of $\Omega $. Near a relative boundary point of $K$, choose $y\notin K$ so that the ball of radius $r=\operatorname {dist}(y,K)>0$ is compactly contained in $\Omega $ and touches $K$ at $x_0$. On the closed annulus $r/2\leq |x-y|\leq r$, use fixed bounds $\lambda I\leq a\leq \Lambda I$ and set

For sufficiently large $\alpha $,

almost everywhere on the annulus. On its inner sphere $z<M$ uniformly; on its outer sphere $b=0$. Choose $\eta >0$ small enough that $z+\eta b\leq M$ on both spheres. Strict weak comparison, applied to $z+\eta b$, gives this inequality throughout the annulus. At $x_0$, however, $z=M$, $b=0$, $\nabla z=0$, and $b$ increases to first order in the inward radial direction, which contradicts that inequality. Thus $K$ has no relative boundary point in $\Omega $, and connectedness proves the assertion. The argument uses no derivatives of the measurable coefficients. $\square$

of 14.21.3. We first locate the extrema. Fix $s>0$ and consider a maximum point of

For smooth data, the complex Hessian of this function is nonpositive at a maximum point, that is,

Since $sC>0$, this gives $B>0$, and since $A^ n=B^ n$, also $A>0$. In a $B$-unitary frame put $J=B^{-1/2}AB^{-1/2}$, $K=B^{-1/2}CB^{-1/2}$. Since $\det J=1$, arithmetic–geometric mean and (14.21.11) give

Consequently $\operatorname {tr}K\leq n$, $B\geq C/n$, and $\det K\leq (\operatorname {tr}K/n)^ n\leq 1$. In particular $B^ n\geq C^ n$.

For $C^{1,1}$ potentials, fix a maximum point $p$ of (14.21.10) and apply 14.21.5 on a coordinate ball, with radii tending to zero and errors $\delta _ j\downarrow 0$. For the set of full measure in that lemma, take the points where both Hessians are defined and semipositive and have the common determinant given by (14.21.1). With $\beta _0=\mathrm{dd}^{\mathrm{c}}|z|^2$, the resulting points $x_ j\to p$ satisfy

for large $j$. The matrix argument just given yields $f(x_ j)\mathrm{d}V_{x_ j}\geq C_ j(x_ j)^ n$. Since $f$, $\mathrm{d}V$ and $C$ are continuous, letting $j\to \infty $ gives $p\in P_\chi $; no continuity of the Hessians is needed. Thus every maximum point of the perturbed function lies in $P_\chi $. The set $P_\chi $ is closed, by continuity of $f$, $\mathrm{d}V$ and $C$, hence compact, and it is contained in $U$ because $C^ n>0$. Letting $s\downarrow 0$, a limit point of these maximum points is a maximum point of $w$ lying in $P_\chi $. Applying the same argument to $-w$ gives a minimum point of $w$ in $P_\chi $ as well. In particular, $P_\chi $ is nonempty.

On a relatively compact ball in $U$, the determinants of $A$ and $B$ have a positive lower bound, while their coefficients have an essential upper bound. Every eigenvalue is therefore bounded below by the determinant lower bound divided by a fixed power of the upper bound. Thus $U\subset E_{\rm unif}$.

By (14.20.1), $w$ solves a linear equation in nondivergence form with measurable coefficients, whose principal quadratic form is a fixed positive multiple of $Q$. On each relatively compact coordinate ball in $E_{\rm unif}$, these coefficients are uniformly elliptic. 14.21.6, applied to $w$ and to $-w$, makes $w$ constant on the component of $E_{\rm unif}$ containing the maximum point found above, and constant on the component containing the minimum point found above. Continuity extends the two constant values to the closures, proving (14.21.4). Distinct extrema force disjoint closures.

The same argument applies on $U$ itself, since $U$ contains the extremal points found in $P_\chi $ and the cofactor is locally uniformly elliptic on $U\subset E_{\rm unif}$. This proves the assertions with $U$ in place of $E_{\rm unif}$.

The component criteria are immediate except for the dense two-component case. If an open dense subset of connected $X$ has at most two components with disjoint closures, those closures are disjoint nonempty closed sets whose union is $X$, contradicting connectedness. Finally, when $Z=\{ f=0\} $ is a connected embedded real hypersurface, flattening charts assign to each point of $Z$ the unordered pair of components containing its two local sides. This pair is locally constant along $Z$, so connectedness makes it independent of the point; the two labels may coincide or be exchanged along a loop. Every component of $X\setminus Z$ has a nonempty boundary in $Z$, since otherwise it would be both open and closed in connected $X$. Near such a boundary point it contains a local side. Thus the pair accounts for every component, and $U$ has at most two components. It is dense because a hypersurface has empty interior. This proves the last assertion without choosing a global side. $\square$

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. $\square$