14.22.8 The real symbol and exterior-ball propagation1FF2DD5C252C

To propagate an extremum without a weight, we write the linearized operator in real coordinates. Its symbol is continuous and semipositive. At a point where a ball outside the maximum set touches it, the normal direction must lie in the kernel of the symbol; this forces the maximum set to be invariant under the flows of vector fields with values in the range of the symbol.

In a holomorphic chart $z_ j=x_ j+\mathrm{i} y_ j$, let $C_{\rm cf}$ be the Hermitian matrix characterized by

for every Hermitian matrix $H=(H_{j\bar k})$. Positivity of mixed products makes $C_{\rm cf}$ positive semidefinite. Write $C_{\rm cf}=P_{\rm cf}+\mathrm{i} Q_{\rm cf}$, where $P_{\rm cf}$ is real symmetric and $Q_{\rm cf}$ is real skew-symmetric. In the real coordinate order $(x_1,\ldots ,x_ n,y_1,\ldots ,y_ n)$,

Indeed $\varphi _{j\bar k}=(\varphi _{x_ jx_ k}+\varphi _{y_ jy_ k} +\mathrm{i}\varphi _{x_ jy_ k}-\mathrm{i}\varphi _{y_ jx_ k})/4$; inserting this expression in (14.22.9) proves (14.22.10). For real coordinate vectors $\xi ,\eta \in \mathbb R^ n$, moreover,

Thus $a$ is a continuous positive semidefinite principal symbol, the operator $L$ has no first-order term in holomorphic coordinates, and the range of $a$ is that of $Q^\sharp $.

An exterior-ball argument propagates the maximum for this operator. Let $W\in C^2$ satisfy $LW\geq 0$ in a chart, suppose $W\leq M$, and put $K=\{ W=M\} $. If $z\in K$ is a point of $K$ nearest to a point $y\notin K$ of the chart, then

To prove this, suppose the left side is positive, set $r_0=|y-z|$, and shrink the tangent ball toward $z$: $y'=z+\theta (y-z)$, $R=\theta r_0$, with $0<\theta <1$ small. The closed ball $\overline{B(y',R)}$ lies in the chart and meets $K$ only at $z$. This follows from the triangle inequality; equality in $|k-y|\leq |k-y'|+|y'-y|\leq r_0$, for $k\in K$, forces $k=z$. For large $\alpha >0$, the function

satisfies

in a neighborhood of $z$, because $z-y'$ is a nonzero multiple of $y-z$, so that $(z-y')^ Ta(z)(z-y')>0$ by assumption. Choose a sufficiently small $r<R$, and consider the lens $D=B(z,r)\cap B(y',R)$. On the part of its boundary lying on the sphere $\partial B(y',R)$, we have $\psi =0$ and $W-M\leq 0$. On the remaining compact boundary portion, $W-M<0$, since the tangent ball meets $K$ only at $z$. Hence, for sufficiently small $\varepsilon >0$, $W-M\leq \varepsilon \psi $ on $\partial D$. At a positive interior maximum of $W-M-\varepsilon \psi $, the Hessian would be nonpositive, whereas $L(W-M-\varepsilon \psi )>0$; this is impossible since $a\geq 0$. Hence the inequality $W-M\leq \varepsilon \psi $ holds in $D$. Outside the ball $B(y',R)$ we have $\psi \geq 0$, so this inequality holds in a full neighborhood of $z$. Since equality holds at $z$, the function $W-M-\varepsilon \psi $ has a local maximum at $z$, and therefore

proving (14.22.11).

Since $a\geq 0$, its range is the orthogonal complement of its kernel, and (14.22.11) implies that $y-z$ lies in this kernel. Consequently $(y-z)\cdot Z(z)=0$ for every vector field $Z$ with values in the range of $a$. If $Z$ is locally Lipschitz and $\gamma $ is one of its flow curves starting in $K$, let $d(t)$ denote the squared distance from $\gamma (t)$ to $K$ in a sufficiently small compact chart neighborhood. Shortening the time interval ensures that nearest points $z_ t$ lie in the chart interior. Write $D^+d(t)=\limsup _{h\downarrow 0}(d(t+h)-d(t))/h$ for the upper right derivative. Holding such a nearest point fixed when differentiating the distance from above gives

As $d(0)=0$, Gronwall's inequality gives $d(t)=0$. The same argument for $-Z$, followed by continuation, proves invariance under both signs of time and under finite concatenations of such flows.

For $C^{2,1}$ potentials, the columns of $a$ are locally Lipschitz, so the orbit criterion still applies to them. The bracket criterion requires smooth vector fields.