Projection estimates for a Gaussian collar cutoff

Put $f=\tau -cR$. The gradient characterization in [F3] also holds up to the smooth graph boundary: for any smooth $\varphi$ on the graph, in a boundary chart put $Y^i={\sum }_j{h}^{ij}{\mathrm{\partial }}_j\varphi$, where $(h^{ij})=(h_{ij}{)}^{-1}$. The inverse matrix is smooth up to the boundary by the adjugate formula and positive definiteness. Thus $Y$ is smooth and $h(Y,X)=X\varphi$ for all tangent $X$ there. If two fields satisfy this identity, their difference $Z$ satisfies $h(Z,Z)=0$, so they agree; this also makes the chart formulas agree on overlaps. At the reference slice $c\mathrm{\mid }dR{\mathrm{\mid }}_{h_0}=c<1$; continuity on a compact reference collar permits a common smaller tube and a fixed $c<\gamma <1$ with $c\mathrm{\mid }dR{\mathrm{\mid }}_{h_{\tau }}\le \gamma$. Since $R$ is constant on Gaussian fibres, $g^{-1}(df,df)=-1+c^2\mathrm{\mid }dR{\mathrm{\mid }}_{h_{\tau }}^2\ge -1$. Write $n=vU+z$, with $z\perp U$ and $\mathrm{\mid }z{\mathrm{\mid }}_{h_{\tau }}^2=v^2-1$. As $c\ge 0$ and $v\ge 1$, $df(n)=v-cdR(z)\ge v-\gamma \sqrt{v^2-1}\ge (1-\gamma )v$. Let $A$ be the vector defined by $g(A,\cdot )=df$. Its tangent projection is $A+df(n)n$, which pairs with every tangent $X$ as $Xf$ and is therefore ${\mathrm{\nabla }}_{\mathrm{\Sigma }}(f{\mathrm{\mid }}_{\mathrm{\Sigma }})$. Its squared norm is $g^{-1}(df,df)+df(n{)}^2\ge (1-\gamma {)}^2{v}^2-1$. On $\chi >0$ division by $d^2$ proves the gradient bound with $a=(1-\gamma {)}^2\mathrm{/}{d}^2$ and $b=1\mathrm{/}{d}^2$.

For tangent fields $X,Y$ set $D_{X}Y=({\mathrm{\nabla }}_{X}Y{)}^{\mathrm{\top }}$. Tangential projection preserves the connection rules and metric compatibility; torsion freedom follows from $D_{X}Y-D_{Y}X=({\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_{Y}X{)}^{\mathrm{\top }}=[X,Y]$, because the bracket is tangent. Thus [F4] identifies $D$ with the intrinsic Levi–Civita connection. Differentiating $g(Y,n)=0$ gives $g({\mathrm{\nabla }}_{X}Y,n)=-g(Y,{\mathrm{\nabla }}_{X}n)=k(X,Y)$, hence ${\mathrm{\nabla }}_{X}Y=D_{X}Y-k(X,Y)n$. For $\varphi =f{\mathrm{\mid }}_{\mathrm{\Sigma }}$, differentiating $h({\mathrm{\nabla }}_{\mathrm{\Sigma }}\varphi ,Y)=Y\varphi$ and using metric compatibility yields ${\mathrm{Hess}}_h\varphi (X,Y)=X(Y\varphi )-(D_{X}Y)\varphi$. Substitution into [F1] now gives ${\mathrm{Hess}}_h\varphi (X,Y)={\mathrm{Hess}}_{g}f(X,Y)-k(X,Y)df(n)$. Tracing over a tangent orthonormal basis $(E_i{)}_{i=1}^3$ and using $H=-{\mathrm{tr}}_{h}k$ yields ${\mathrm{\Delta }}_{\mathrm{\Sigma }}\varphi ={\sum }_i{\mathrm{Hess}}_{g}f(E_i,E_i)+Hdf(n)={\sum }_i{\mathrm{Hess}}_{g}f(E_i,E_i)$.

The bound in [F2] and the trace identity imply $\mathrm{\mid }{\mathrm{\Delta }}_{\mathrm{\Sigma }}(f{\mathrm{\mid }}_{\mathrm{\Sigma }})\mathrm{\mid }\le M{\sum }_{i=1}^3\mathrm{\mid }{E}_i{\mathrm{\mid }}_e^2\le 6M{v}^2$. On $\chi >0$ the cutoff is $1+f\mathrm{/}d$, so $\mathrm{\mid }{\mathrm{\Delta }}_{\mathrm{\Sigma }}\chi \mathrm{\mid }\le (6M\mathrm{/}d)v^2$; taking $C=1+6M\mathrm{/}d$ gives a positive constant. This uses only fixed smooth background second derivatives, not graph second-derivative bounds. At the graph boundary the formulas have their smooth one-sided interpretation wherever $\chi >0$.

On the stated graph collar, $-(c+\theta )R\le f\le -(c-\theta )R\le 0$, so $0\le \chi \le 1$. At $R=0$ the cone gives $\tau =0$ and hence $\chi =1$. At $R=r\mathrm{/}2$ the strict inequality $d<(c-\theta )r\mathrm{/}2$ gives $f<-d$ and $\chi =0$. Since $c+\theta >0$, if $R\le d\mathrm{/}[2(c+\theta )]$ then $f\ge -d\mathrm{/}2$ and $\chi \ge 1\mathrm{/}2$. No derivative estimate is asserted across $\chi =0$. Together with the preceding differential estimates these prove every conclusion.