Strong minimum principle from the annular barrier

If $M$ is empty the alternative is vacuous. If $n=\dim M=0$, each chart is a singleton open set because ${\mathbb{R}}^0$ is a singleton; hence $M$ is discrete. Every singleton is then clopen, so connectedness and [F3] force a nonempty $M$ to be a singleton. The alternative follows from nonnegativity at its sole point. In the remaining argument $n\ge 1$. At any point $x_0$ of an interior chart, replace $u$ by its quadratic Taylor polynomial with the same value, first derivatives and second derivatives at $x_0$. The smooth Hessian formula in [F2] uses only these data and the metric connection coefficients at $x_0$. Under a smooth change of coordinates, the chain rule transforms the derivatives through order two using only these same data, so the Hessian coefficients defined for $u$ transform as those of the polynomial, hence as a bilinear tensor at $x_0$. This proves that their metric trace is well-defined for $u\in C^2$ and gives ${\mathrm{\Delta }}_{h}u=h^{ij}{\mathrm{\partial }}_{ij}u-h^{ij}{\mathrm{\Gamma }}^k{}_{ij}{\mathrm{\partial }}_{k}u$. Thus $a^{ij}=h^{ij}$ and $b^k=-h^{ij}{\mathrm{\Gamma }}^k{}_{ij}$ are smooth on the chart. For any nonempty compact chart subset $K$, its coordinate image is closed and bounded by [F5]. Thus $K\times \{\xi \in {\mathbb{R}}^n:\mathrm{\mid }\xi \mathrm{\mid }=1\}$, in coordinates, is a nonempty closed bounded subset of ${\mathbb{R}}^{2n}$ and is compact by [F5]. The continuous strictly positive function $(x,\xi )\mapsto h^{ij}(x){\xi }_i{\xi }_j$ has a strictly positive attained minimum there by [F6]. Scaling $\xi$ gives uniform ellipticity on $K$. Applying [F6] on $K$ to the absolute values of each of the finitely many smooth coefficients and of $V$ gives finite coefficient and potential bounds. For empty $K$ these bounds are vacuous. These are exactly the metric-chart coefficient conditions in [F4].

Suppose $P$ and $Z$ are both nonempty. If $\bar{P}\cap Z$ were empty, then $\bar{P}\subseteq P$, so the open set $P$ would also be closed. It would be a nonempty proper clopen subset of the connected space $M$, contradicting [F3]. Consequently choose $q\in \bar{P}\cap Z$.

Center coordinates at $q$ and choose $\rho >0$ such that the closed Euclidean ball ${\bar{B}}_{2\rho }(0)$ is contained in the chart image. Choose $z\in P$ with $\mathrm{\mid }z\mathrm{\mid }<\rho \mathrm{/}4$, which is possible since $q\in \bar{P}$. The set $E=Z\cap {\bar{B}}_{\rho }(0)$ in these coordinates contains $0$ and is closed and bounded in ${\mathbb{R}}^n$: it is the zero set of the continuous coordinate function $u$ restricted to the closed ball, which lies in the chart. Thus $E$ is nonempty and compact by [F5]. The continuous function $x\mapsto \mathrm{\mid }x-z\mathrm{\mid }$ on $E$ attains its minimum $R$ at some $p\in E$ by [F6]. Since $p\in Z$ and $z\mathrm{\notin }Z$, $R=\mathrm{\mid }p-z\mathrm{\mid }>0$; since $0\in E$, $R\le \mathrm{\mid }z\mathrm{\mid }<\rho \mathrm{/}4$. Thus ${\bar{B}}_R(z)\subset B_{\rho \mathrm{/}2}(0)$, and its open ball contains no point of $Z$, so it lies in $P$. Its closure lies in the interior chart and $p$ is a zero on its sphere.

On this ball, step 1.1 verifies the operator hypotheses of [F4]. All its function hypotheses hold because $u$ is nonnegative and $C^2$ throughout the interior, positive in the ball by step 2.1, zero at $p$, and satisfies $({\mathrm{\Delta }}_h-V)u\le 0$. The barrier therefore gives ${\mathrm{\partial }}_{v}u(p)>0$ for $v=(z-p)\mathrm{/}R$. But $p$ is an interior minimum of $u$, so the differentiable function $g(t)=u(p+tv)$ has a minimum at $0$ for both signs of small $t$. For $t>0$ its difference quotient $(g(t)-g(0))\mathrm{/}t$ is nonnegative, and for $t<0$ it is nonpositive. Both one-sided limits equal $g^{\mathrm{\prime }}(0)$ by differentiability, so $g^{\mathrm{\prime }}(0)\ge 0$ and $g^{\mathrm{\prime }}(0)\le 0$, hence ${\mathrm{\partial }}_{v}u(p)=g^{\mathrm{\prime }}(0)=0$. The contradiction excludes simultaneous nonempty $P,Z$. If $P$ is empty then $u\equiv 0$; if $Z$ is empty then $u>0$ everywhere. These alternatives also cover the empty interior.