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Lemma 3.24 at page 87 of Alexander Grigor'yan's "Heat Kernel and Analysis on Manifolds" says that (I reformulate it slightly)

For any point $p \in M$ and chart $(U', h)$ around $p$ there exist a $U \subseteq U'$ and a constant $C \ge 1$ such that for all $x,y \in U$ we have

 

$$\frac 1 C \| h(x) - h(y) \| \le d(x,y) \le C \| h(x) - h(y) \| \ .$$

Taking $h = \exp_p ^{-1}$ and $v \in S_r$ you get

$$\frac 1 C \| \exp_p ^{-1} (u) - \exp_p ^{-1} (v) \| \le d(u,v) \le C \| \exp_p ^{-1} (u) - \exp_p ^{-1} (v) \| \ ,$$

whence by taking $\sup _{v \in S_r}$ one gets

$$\frac 1 C \| \exp_p ^{-1} (u) - \exp_p ^{-1} (S_r) \| \le d(u,S_r) \le C \| \exp_p ^{-1} (u) - \exp_p ^{-1} (S_r) \| \ ,$$

the first inequality being what you are asking for.

By looking at Grigor'yan's proof, it seems to me that for small enough $U$ and $r$ one may choose $C$ independent of $p$ and $r$.

Lemma 3.24 at page 87 of Alexander Grigor'yan's "Heat Kernel and Analysis on Manifolds" says that (I reformulate it slightly)

For any point $p \in M$ and chart $(U', h)$ around $p$ there exist a $U \subseteq U'$ and a constant $C \ge 1$ such that for all $x,y \in U$ we have

 

$$\frac 1 C \| h(x) - h(y) \| \le d(x,y) \le C \| h(x) - h(y) \| \ .$$

Taking $h = \exp_p ^{-1}$ and $v \in S_r$ you get

$$\frac 1 C \| \exp_p ^{-1} (u) - \exp_p ^{-1} (v) \| \le d(u,v) \le C \| \exp_p ^{-1} (u) - \exp_p ^{-1} (v) \| \ ,$$

whence by taking $\sup _{v \in S_r}$ one gets

$$\frac 1 C \| \exp_p ^{-1} (u) - \exp_p ^{-1} (S_r) \| \le d(u,S_r) \le C \| \exp_p ^{-1} (u) - \exp_p ^{-1} (S_r) \| \ ,$$

the first inequality being what you are asking for.

By looking at Grigor'yan's proof, it seems to me that for small enough $U$ and $r$ one may choose $C$ independent of $p$ and $r$.

Lemma 3.24 at page 87 of Alexander Grigor'yan's "Heat Kernel and Analysis on Manifolds" says that (I reformulate it slightly)

For any point $p \in M$ and chart $(U', h)$ around $p$ there exist a $U \subseteq U'$ and a constant $C \ge 1$ such that for all $x,y \in U$ we have

$$\frac 1 C \| h(x) - h(y) \| \le d(x,y) \le C \| h(x) - h(y) \| \ .$$

Taking $h = \exp_p ^{-1}$ and $v \in S_r$ you get

$$\frac 1 C \| \exp_p ^{-1} (u) - \exp_p ^{-1} (v) \| \le d(u,v) \le C \| \exp_p ^{-1} (u) - \exp_p ^{-1} (v) \| \ ,$$

whence by taking $\sup _{v \in S_r}$ one gets

$$\frac 1 C \| \exp_p ^{-1} (u) - \exp_p ^{-1} (S_r) \| \le d(u,S_r) \le C \| \exp_p ^{-1} (u) - \exp_p ^{-1} (S_r) \| \ ,$$

the first inequality being what you are asking for.

By looking at Grigor'yan's proof, it seems to me that for small enough $U$ and $r$ one may choose $C$ independent of $p$ and $r$.

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Alex M.
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Lemma 3.24 at page 87 of Alexander Grigor'yan's "Heat Kernel and Analysis on Manifolds" says that (I reformulate it slightly)

For any point $p \in M$ and chart $(U', h)$ around $p$ there exist a $U \subseteq U'$ and a constant $C \ge 1$ such that for all $x,y \in U$ we have

$$\frac 1 C \| h(x) - h(y) \| \le d(x,y) \le C \| h(x) - h(y) \| \ .$$

Taking $h = \exp_p ^{-1}$ and $v \in S_r$ you get

$$\frac 1 C \| \exp_p ^{-1} (u) - \exp_p ^{-1} (v) \| \le d(u,v) \le C \| \exp_p ^{-1} (u) - \exp_p ^{-1} (v) \| \ ,$$

whence by taking $\sup _{v \in S_r}$ one gets

$$\frac 1 C \| \exp_p ^{-1} (u) - \exp_p ^{-1} (S_r) \| \le d(u,S_r) \le C \| \exp_p ^{-1} (u) - \exp_p ^{-1} (S_r) \| \ ,$$

the first inequality being what you are asking for.

By looking at Grigor'yan's proof, it seems to me that for small enough $U$ and $r$ one may choose $C$ independent of $p$ and $r$.