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Let $M = \mathbb{R}^3 \setminus \overline{B_1}$ where $\overline{B_1}$ is the closed unit ball.

Let $g$ be an asymptotically metric of the form $g_{ij} = \delta_{ij}+h_{ij}$ in standard coordinates such that $$ \sup_{M} ( |x|^2 |h| + |x|^3 |\partial h| + |x|^4 |\partial^2 h| ) < \epsilon$$ where $\epsilon$ is a small positive number and $|x| = \sqrt{x_1^2+ x_2^2+ x_3^2}$ and $(x_1,x_2,x_3)$ is the standard coordinates on $M$.

Then is it true that the distance function $r(\cdot) := {\rm dist}(\cdot, \partial M)$ from the boundary is differentiable on $M\setminus \partial M$ and hence defines a global foliation of spheres $S_{r_0} := \{ r=r_0 \}$?

It is clear that there exists a neighborhood $U$ of $\partial M$ such that $r$ is differentiable on $U \setminus \partial M$ since $\partial M$ is compact.

Does it suffice to show that if $r$ is differentiable on $B_R = \{x \in M| 0<r(x) < R \}$ for some $R>0$, then the mean curvature $H_r$ of the spheres $S_r$ for $r<R$ is bounded from below? I think that the mean curvature must go to $-\infty$ on $S_R$ if $r$ fails to be differentiable on $S_R$, but I am not sure if that's correct. If so, then we can study the ODE obeyed by the mean curvature $H_r$ and the traceless part of the second fundamental form $\hat A$. If the Ricci curvature falls very quickly, I think it can be shown that $H_r = \frac{1}{r} + O(\frac{1}{r^2})$.

Of course, if $g$ is the euclidean metric, then $r$ is differentiable on $M \setminus \partial M$.

Any help or references is appreciated.

Let $M = \mathbb{R}^3 \setminus \overline{B_1}$ where $\overline{B_1}$ is the closed unit ball.

Let $g$ be an asymptotically flat metric of the form $g_{ij} = \delta_{ij}+h_{ij}$ in standard coordinates such that $$ \sup_{M} ( |x|^2 |h| + |x|^3 |\partial h| + |x|^4 |\partial^2 h| ) < \epsilon$$ where $\epsilon$ is a small positive number and $|x| = \sqrt{x_1^2+ x_2^2+ x_3^2}$ and $(x_1,x_2,x_3)$ is the standard coordinates on $M$.

Then is it true that the distance function $r(\cdot) := {\rm dist}(\cdot, \partial M)$ from the boundary is differentiable on $M\setminus \partial M$ and hence defines a global foliation of spheres $S_{r_0} := \{ r=r_0 \}$?

It is clear that there exists a neighborhood $U$ of $\partial M$ such that $r$ is differentiable on $U \setminus \partial M$ since $\partial M$ is compact.

Does it suffice to show that if $r$ is differentiable on $B_R = \{x \in M| 0<r(x) < R \}$ for some $R>0$, then the mean curvature $H_r$ of the spheres $S_r$ for $r<R$ is bounded from below? I think that the mean curvature must go to $-\infty$ on $S_R$ if $r$ fails to be differentiable on $S_R$, but I am not sure if that's correct. If so, then we can study the ODE obeyed by the mean curvature $H_r$ and the traceless part of the second fundamental form $\hat A$. If the Ricci curvature falls very quickly, I think it can be shown that $H_r = \frac{1}{r} + O(\frac{1}{r^2})$.

Of course, if $g$ is the euclidean metric, then $r$ is differentiable on $M \setminus \partial M$.

Any help or references is appreciated.

Let $M = \mathbb{R}^3 \setminus \overline{B_1}$ where $\overline{B_1}$ is the closed unit ball.

Let $g$ be an asymptotically flat metric satisfying $|Ric| = O(|x|^{-4})$ , where $|x| = \sqrt{x_1^2+ x_2^2+ x_3^2}$ and $(x_1,x_2,x_3)$ is the standard coordinates on $M$.

Then is it true that the distance function $r(\cdot) := {\rm dist}(\cdot, \partial M)$ from the boundary is differentiable on $M\setminus \partial M$ and hence defines a global foliation of spheres $S_{r_0} := \{ r=r_0 \}$? If not, what if we also add a smallness condition on the Ricci: say $\sup_{M} |x|^4|Ric| < \epsilon$?

It is clear that there exists a neighborhood $U$ of $\partial M$ such that $r$ is differentiable on $U \setminus \partial M$ since $\partial M$ is compact.

Does it suffice to show that if $r$ is differentiable on $B_R = \{x \in M| 0<r(x) < R \}$ for some $R>0$, then the mean curvature $H_r$ of the spheres $S_r$ for $r<R$ is bounded from below? I think that the mean curvature must go to $-\infty$ on $S_R$ if $r$ fails to be differentiable on $S_R$, but I am not sure if that's correct. If so, then we can study the ODE obeyed by the mean curvature $H_r$ and the traceless part of the second fundamental form $\hat A$. If the Ricci curvature falls very quickly, I think it can be shown that $H_r = \frac{1}{r} + O(\frac{1}{r^2})$.

Of course, if $g$ is the euclidean metric, then $r$ is differentiable on $M \setminus \partial M$.

Any help or references is appreciated.

Let $M = \mathbb{R}^3 \setminus \overline{B_1}$ where $\overline{B_1}$ is the closed unit ball.

Let $g$ be an asymptotically metric of the form $g_{ij} = \delta_{ij}+h_{ij}$ in standard coordinates such that $$ \sup_{M} ( |x|^2 |h| + |x|^3 |\partial h| + |x|^4 |\partial^2 h| ) < \epsilon$$ where $\epsilon$ is a small positive number and $|x| = \sqrt{x_1^2+ x_2^2+ x_3^2}$ and $(x_1,x_2,x_3)$ is the standard coordinates on $M$.

Then is it true that the distance function $r(\cdot) := {\rm dist}(\cdot, \partial M)$ from the boundary is differentiable on $M\setminus \partial M$ and hence defines a global foliation of spheres $S_{r_0} := \{ r=r_0 \}$?

It is clear that there exists a neighborhood $U$ of $\partial M$ such that $r$ is differentiable on $U \setminus \partial M$ since $\partial M$ is compact.

Does it suffice to show that if $r$ is differentiable on $B_R = \{x \in M| 0<r(x) < R \}$ for some $R>0$, then the mean curvature $H_r$ of the spheres $S_r$ for $r<R$ is bounded from below? I think that the mean curvature must go to $-\infty$ on $S_R$ if $r$ fails to be differentiable on $S_R$, but I am not sure if that's correct. If so, then we can study the ODE obeyed by the mean curvature $H_r$ and the traceless part of the second fundamental form $\hat A$. If the Ricci curvature falls very quickly, I think it can be shown that $H_r = \frac{1}{r} + O(\frac{1}{r^2})$.

Of course, if $g$ is the euclidean metric, then $r$ is differentiable on $M \setminus \partial M$.

Any help or references is appreciated.

Let $M = \mathbb{R}^3 \setminus \overline{B_1}$ where $\overline{B_1}$ is the closed unit ball.

Let $g$ be an asymptotically flat metric satisfying $|Ric| = O(|x|^{-4})$, where $|x| = \sqrt{x_1^2+ x_2^2+ x_3^2}$ and $(x_1,x_2,x_3)$ is the standard coordinates on $M$.

Then is it true that the distance function $r(\cdot) := {\rm dist}(\cdot, \partial M)$ from the boundary is differentiable on $M\setminus \partial M$ and hence defines a global foliation of spheres $S_{r_0} := \{ r=r_0 \}$?

It is clear that there exists a neighborhood $U$ of $\partial M$ such that $r$ is differentiable on $U \setminus \partial M$ since $\partial M$ is compact.

Does it suffice to show that if $r$ is differentiable on $B_R = \{x \in M| 0<r(x) < R \}$ for some $R>0$, then the mean curvature $H_r$ of the spheres $S_r$ for $r<R$ is bounded from below? I think that the mean curvature must go to $-\infty$ on $S_R$ if $r$ fails to be differentiable on $S_R$, but I am not sure if that's correct. If so, then we can study the ODE obeyed by the mean curvature $H_r$ and the traceless part of the second fundamental form $\hat A$. If the Ricci curvature falls very quickly, I think it can be shown that $H_r = \frac{1}{r} + O(\frac{1}{r^2})$.

Of course, if $g$ is the euclidean metric, then $r$ is differentiable on $M \setminus \partial M$.

Any help or references is appreciated.

Let $M = \mathbb{R}^3 \setminus \overline{B_1}$ where $\overline{B_1}$ is the closed unit ball.

Let $g$ be an asymptotically flat metric satisfying $|Ric| = O(|x|^{-4})$  , where $|x| = \sqrt{x_1^2+ x_2^2+ x_3^2}$ and $(x_1,x_2,x_3)$ is the standard coordinates on $M$.

Then is it true that the distance function $r(\cdot) := {\rm dist}(\cdot, \partial M)$ from the boundary is differentiable on $M\setminus \partial M$ and hence defines a global foliation of spheres $S_{r_0} := \{ r=r_0 \}$? If not, what if we also add a smallness condition on the Ricci: say $\sup_{M} |x|^4|Ric| < \epsilon$?

It is clear that there exists a neighborhood $U$ of $\partial M$ such that $r$ is differentiable on $U \setminus \partial M$ since $\partial M$ is compact.

Does it suffice to show that if $r$ is differentiable on $B_R = \{x \in M| 0<r(x) < R \}$ for some $R>0$, then the mean curvature $H_r$ of the spheres $S_r$ for $r<R$ is bounded from below? I think that the mean curvature must go to $-\infty$ on $S_R$ if $r$ fails to be differentiable on $S_R$, but I am not sure if that's correct. If so, then we can study the ODE obeyed by the mean curvature $H_r$ and the traceless part of the second fundamental form $\hat A$. If the Ricci curvature falls very quickly, I think it can be shown that $H_r = \frac{1}{r} + O(\frac{1}{r^2})$.

Of course, if $g$ is the euclidean metric, then $r$ is differentiable on $M \setminus \partial M$.

Any help or references is appreciated.