The transverse lapse is a positive weighted Dirichlet multiplier

The supplied graph, the stationary metric, their differentiated bounds and the range $0<\alpha <1$ satisfy all hypotheses of [F1]. Under [A1], it gives $n^0-1\in C_{-1}^{4,\alpha }$ and $n^i\in C_{-7\mathrm{/}4}^{4,\alpha }$ in $x_q$. Independently [F5] gives $e=g_{00}+1\in C_{-1}^{4,\alpha }$ and $g_{0i}\in C_{-2}^{4,\alpha }$. These are end component estimates; stationarity means restricting a metric component to the graph just evaluates that spatial function.

Where $Y=T=c{\mathrm{\partial }}_{t_q}$, contraction gives $a=-c(g_{00}{n}^0+g_{0i}{n}^i)$ and hence $a-c=c((n^0-1)-e{n}^0-g_{0i}{n}^i)$. By [F7], $e{n}^0=e+e(n^0-1)$ has weight $-1$, and $g_{0i}{n}^i$ has weight $-15\mathrm{/}4$, which includes into weight $-1$. Thus $a-c\in C_{-1}^{4,\alpha }$ on the normalized end. The fixed-$q$ scaling in [F1], equivalently rescaling the annuli of [F6] by $q$, gives the same exponent in $x$. In particular $a\to c>0$ uniformly at infinity.

Smoothness follows from the smooth metric and normal in [F4] and the given smooth field $Y$. Near $S$ the equality $Y=n$ gives $a=-g(n,n)=1$. Everywhere $n$ is future unit timelike and $Y$ is nonzero future timelike, so [F8] gives $a>0$. Step 2.1 supplies a radius outside which $a\ge c\mathrm{/}2$. The remainder is compact by the exterior data. At each point $p$ of that remainder, continuity gives a relative neighborhood on which $a>a(p)\mathrm{/}2>0$. A finite subcover yields a positive lower bound there by taking the minimum of its finitely many positive bounds. Combining this with $c\mathrm{/}2$ proves the global lower bound; an empty remainder needs no compact-part bound.

On the end, the constant $c$ has finite weight-zero norm, and $a-c$ has weight $-1$ by step 2.1, so [F7] puts $a$ in weight zero there. On the fixed compact core and any intervening bounded end region, smoothness on neighborhoods of the compact chart subsets gives finite derivative suprema and fourth-derivative Hölder seminorms. With the norm in [F6], this proves $a\in C_0^{4,\alpha }({\mathrm{\Sigma }}_1)$. Apply [F2] and step 3.1 to obtain $a^{-1}\in C_0^{4,\alpha }$ and inverse bounded multiplication operators on $C_{\beta }^{4,\alpha }$. Boundary restriction is pointwise and continuous by [F3]; since $a$ never vanishes, $w{\mathrm{\mid }}_S=0$ if and only if $(aw){\mathrm{\mid }}_S=0$. Thus both operators preserve the closed Banach domain $X_{\beta }$ from [F3], used under [A1], and restrict to bounded inverses on it.