Corrections to a paper of Allard and Almgren on the uniqueness of tangent cones

William K. Allard

Supported in part by Los Alamos National Laboratory.

§ INTRODUCTION.

The paper referred to in the title is [2].

Several months ago Francesco Maggi emailed me saying that the inequalities
5.3(4),(5) of [2] were wrong. In fact, as he pointed out,
their incorrectness is immediately apparent if one takes $Z=0$ there.
Maggi and his coauthor wanted to use these inequalities in their paper
[5]. They were able to obtain a version of these inequalities
which suffice for the carrying out the work in [5].

I started writing this paper in order to provide a
version of these inequalities as needed in [AA81].
In thinking about this material I began to realize there were other problems
with the paper. As a result I ended up completely rewriting 5.1-5.4 on pages
243-248 of [2]; this rewrite is the contents of this paper. In addition to many annoying misprints many of the
needed definitions and proofs in 5.1-5.4 are incomplete or absent. For example,
it is not said where $z$ in 5.1(2) comes from; this omission completely
surprised me since I remember doing a lot of work to come up with $z$
when [2] was being written. Also, much of the necessary material about
the reach of a submanifold as in [4] is not provided in [2].

The table of contents can serve as an index. In particular one sees there
where the constants $\epsilon_1$ through $\epsilon_6$ are introduced.

This material is extremely technical. One way to navigate this paper would
be to start looking at Proposition <ref> and work backwards.

§ PRELIMINARIES.

If $X$ is a set we let

$$i_X=\{(x,x):x\in X\};$$

thus $i_X:X\rightarrow X$ and $i_X(x)=x$ for $x\in X$; we call $i_X$ the
identity map of $X$; if $X\subset Y$ the $i_X$ is also called the
inclusion map of $X$ into $Y$.

If $(X,\rho)$ and $(Y,\sigma)$ are metric spaces and $f:X\rightarrow Y$
we let

$$\op{Lip}{f}=\sup\left\{{\sigma(f(x_1),f(x_2))\over\rho(x,y)}:x_i\in X, \ i=1,2, \ \hbox{and} \ x_1\neq x_2\right\}.$$

§.§ Some linear algebra.

be the vector space of linear maps from $V$ to $W$. We let

If $V_1,\ldots,V_N$ and $W$ are vector spaces we let

be the vector space of multilinear maps from $V_1\times\cdots\times V_N$
into $W$.

If $V$ and $W$ are vector spaces we let

If $V$ is a vector space we let

A Euclidean space is a finite dimensional inner product space.
If $E$ is a Euclidean space and $l\in\N$ we let

$$\mG_l(E) \  \hbox{and} \ \mO_l(E)$$

be, respectively, the family of $l$ dimensional linear subspaces of $E$ and the family
of orthogonal projections of $E$ onto a member of $\mG_l(E)$.

If $E_i$, $i=1,\ldots,m$ and $F$ are Euclidean spaces and
$\mu:E_1\times\cdots\times E_m\rightarrow F$ is multilinear we let

$$||\mu||=\sup\{|\mu(v_1,\ldots,v_m)|:v_i\in E_i \ \hbox{and} \ |v_i|\leq 1,
\ i=1,\dots,m\};$$

the vector space of multilinear maps from $E_1\times E_m$ into $F$ has a natural Euclidean structure and the norm with respect to this structure of $\mu$
as above will be larger than $||\cdot||$ unless the $E_i$ are one dimensional.
For example, if $i$ is the identity map of the Euclidean space $E$ then
$||i||=1$ but the Euclidean norm of $i$ is the square root of the dimension of

Suppose  $S\in\mG_m(\R^n)$, $L\in\Lin(S,\R^n)$, $\op{ker}{L}=\{0\}$.
and $T=\op{rng}{L}$. and $\op{dim}{T}=m$.

There are unique $A$ and $O$ such that
$A\in\End(T)$, $O\in\Lin(S,T)$, $A^*=A$, $A(v)\bullet v> 0$ for $v\in T$, $A^2=L\circ L^*$, $O$ is an isometry and $L=A\circ O$.

$L\circ L^*$ is a positive definite and carries $T$
isomorphically onto $T$. Let $A$ be the unique positive definite
square root of $L\circ L^*$ and let $O=A^{-1}\circ L$.

Suppose $S$ and $T$ are finite dimensional inner product spaces.
For $L,M\in\Lin(S,T)$

$$L\bullet M=\op{trace}{L^*\circ M}.$$

This is an inner product on $\Lin(S,T))$. If $u_1,\ldots,u_m$
is an orthonormal basis for $S$ we have

$$L\bullet M=\sum_{i=1}^m L(u_i)\bullet M(u_i).$$

It follows that

$$|L\bullet M|\leq m\,||L||\,||M||;$$

note that if $L$ is an isometry then $||L||=1$ and $|L|=\sqrt{m}$.

§.§ A useful notational convention.

Suppose $T$, $X$ and $Y$ are nonempty sets and $W\subset T\times X$.

If $h:T\times X\rightarrow Y$ and $t\in\cJ$ we let

$$h_t=\{(x,h(t,x)):(t,x)\in W\}.$$

$$h_t:\{x:(t,x)\in W\}\rightarrow Y\quad\hbox{for $t\in\cJ$};$$

of course $h_t=\emptyset$ if $W\cap(\{t\}\times X)=\emptyset$.

§ SUBMANIFOLDS OF EUCLIDEAN SPACES.

Suppose $E$ is a Euclidean space,
$m\in\N^+$ and $m\leq\op{dim}{E}$.
We let

be the family of nonempty subsets $M$ of $E$ such that for each $a\in M$
there are $U,V,\psi,\phi$ such that $V$ is an open subset of $E$; $U$ is an open subset of $\R^n$,
$\psi:V\rightarrow U$, $\psi$ is smooth, $\phi:U\rightarrow V$, $\phi$
is smooth, $a\in V$ and

$$\psi(\phi(u))=u\quad\hbox{for $u\in U$.}$$

This a slight variation of the definition in [FE, 3.1.19(4).
Note that $\phi=(\psi|M)^{-1}$

A member of $\cM_m(E)$ is called a
smooth $m$ dimensional submanifold of $E$.

Evidently, $M\in\cM_{\op{dim}{E}}(E)$ if and only if $M$ is an open subset
of $E$. Also, if $L\subset M$ then $L\in\cM_m(E)$ if and only if $L$
is nonempty and is open relative to $M$.

It is elementary that

\quad\hbox{for $u\in U$.}$$

It follows that $m$ is determined by $M$; we call it the dimension of $M$.
We let

\quad\hbox{and}\quad
\nu:M\rightarrow\mG_{\op{dim}{E}-m}(E)$$

be the tangent map and the normal map of $M$, respectively;
by definition,
$\tau(x)$ is orthogonal projection of $E$ onto $\Tan{M,x}$ and $\nu(x)=\tau(x)^\perp$ for $x\in M$.

We now fix $m,n\in\N^+$ with $m\leq n$.

For the remainder of this Section we fix $M\in\cM_m(\R^n)$
and a normed vector space $Y$.

§.§ Smooth functions on $M$.

A function $f:M\rightarrow Y$
is smooth if $f\circ\phi$ is smooth if whenever $a,U,V,\phi,\psi$ are as in
Definition <ref>.

be the vector space of smooth maps from $M$ into $Y$.

Suppose $Y$ is a Euclidean space, $f:M\rightarrow Y$,
$f$ is smooth and $a\in M$.  There is and only one

such that if $a\in M$ then

\quad\hbox{for $v\in\R^m$}$$

Moreover, if $N$ is a smooth submanifold of some Euclidean space
and $F:N\rightarrow M$ is smooth then $f\circ F$ is smooth and

$$D(f\circ F)(b)=Df(F(b))\circ Df(b)
\quad\hbox{for $b\in N$.}$$

Suppose $a,U,V,\phi,\psi$ are as in
Definition <ref>. Then
$(f\circ F)|F^{-1}[V]=(f\circ\phi)\circ(\psi\circ F)|F^{-1}[V]$.

If $f\in\cE(M,Y)$, $a\in M$ and $v\in E$ we let

$$D_v f(a)=Df(a)(v).$$

If $k\in\N^+$ and $v=(v_1,\ldots,v_k)\in E^k$ we let

$$D_v f(a)=D_{v_1}\cdots D_{v_k}f(a).$$

Suppose $f\in\cE(M,Y)$, $a\in M$, $W$ is an open neighborhood
of $a$ in $\R^n$, $F:W\rightarrow Y$ and $F|M=f|W$. Then
$Df(a)(v)=DF(a)(v)$ whenever $a\in M\cap W$ and $v\in\Tan{M,a}$.

If $U,V,\phi,\psi$ are as in Definition <ref> then $F\circ\phi$ equals $f\circ\phi$
in an open neighborhood of $\psi(a)$.

We leave the straightforward proof of the following Proposition to the reader.
Suppose $E$ is a Euclidean space,
$m\in\N^+$ and $m\leq\op{dim}{E}$, $M$ is a nonempty subset of $E$
and for each $a\in M$ there are $U,V,\psi,\phi$ such that $V$ is an open subset of $E$; $U$ is an open subset of some $m$ dimensional submanifold of some
Euclidean space,
$\psi:V\rightarrow U$, $\psi$ is smooth, $\phi:U\rightarrow V$, $\phi$
is smooth, $a\in V$ and

$$\psi(\phi(u))=u\quad\hbox{for $u\in U$.}$$

Then $M\in\cM_m(E)$.

If $i_M$ is the inclusion map of $M$ in $\R^n$ then $i_M$ is smooth

$$Di_M(x)=\tau(x)\quad\hbox{for $x\in M$.}$$

It follows that $\tau$ is smooth and since $\nu(x)=i_{\R^n}-\tau(x)$
for $x\in M$ we find that $\nu$ is smooth.

For each $k\in\N$ we define

$$D^k f(a)\in\MultiLin(\underbrace{E,\ldots,E}_{\hbox{$k$ times}},Y)$$

by letting $D^0 f=f$, $D^1 f=Df$ and, if $k\geq 2$, requiring that

$$D^k f(a)(D\phi(\psi(a))(v_1),\ldots,D\phi(\psi(a))(v_k))
=D^k (f\circ\phi)(\psi(a))(v_1,\ldots,v_k)$$

whenever $a,U,V,\phi,\psi$ are as in Definition <ref>
and $v_1,\ldots,v_k\in \R^m$
and that

$$D^k f(a)(w_1,\ldots,w_k)=0
\quad\hbox{whenever $w_1,\ldots,w_k\in E$ and
$w_i\in\Nor{M,a}$ for some $i=1,\ldots,m$.}$$

As is well known $D^k f(a)$ is symmetric. We have

$$D^k f(a)(w_1,\ldots,w_k)=D_{w_1}\cdots D_{w_k} f(a)
\quad\hbox{whenever $w_1,\ldots,w_k\in E$.}$$

Suppose $Y$ is a Euclidean space and
$f:M\rightarrow Y$ is smooth. For each $k\in\N$ we let

$$\chi^k(f)=\sup\{||D^k f(x)||:x\in M\}$$

and we let

$$\chi^{[k]}(f)=\max\{\chi^j(f):j\in \N \ \hbox{and} \ j\leq k\}.$$

Any of these numbers may be $\infty$.

We will not use $\chi$ as on page 229 of [2].

§.§ More on differentiation.

Suppose $N\in\N^+$ and, for each $i=1,\ldots,N$, $E_i$ is a Euclidean space.
For each $i\in\{1,\ldots,N\}$ we let

$$\iota_i:E_i\rightarrow E_1\times\cdots\times E_N$$

be such that for each $u\in E_i$ and $j\in\{1,\ldots,N\}$

u&\hbox{if $j=i$,}\cr
0&\hbox{if $j\neq i$.}
\end{cases}$$

For each $j\in\N^+$ we let

$$\iota_i^:E_i^j\rightarrow(E_1\times\cdots\times E_N)^j$$

be such that

\quad\hbox{for $(u_1,\ldots,u_j)\in E_i^j$.}$$

Suppose $Y$ is a normed vector space and
$f:M_1\times\cdots\times M_N\rightarrow Y$ is smooth
and $(k_1,\ldots,k_N)\in\N^N$ and $K>0$ where $K$ is the number
of positive $k_i$, $i=1,\ldots,N$.

Let $l_j$, $j=1,\ldots,K$, be such that

We let

$$D^{k_1,\ldots,k_N}f:M_1\times\cdots\times M_N
\rightarrow\MultiLin(E_{l_1}^{k_{l_1}},\ldots,E_{l_K}^{k_{l_K}},Y)$$

be such that

=D^{\sum_{i=1}^K k_i}f(x_1,\ldots,x_N)

for $(U_1,\ldots,U_K)\in E_{l_1}^{k_{l_1}},\ldots,E_{l_K}^{k_{l_K}}$.
We also let

(x_1,\ldots,x_N)\in M_1\times\cdots\times M_N\}.$$

Supposed $N=3$, $K=2$ and $(k_1,k_2,k_3)=(3,0,2)$ and

$$f:E_1\times E_2\times E_3\rightarrow Y.$$

Then $(l_1,l_2)=(1,3)$
and, if $U_1\in E_1^3$ and $U_2\in E_3^2$ then

=D^5 f(x_2,x_2,x_3)((U_1)_1,(U_1)_2,(U_1)_3,(U_2)_1,(U_2)_2).$$

for $(x_1,x_2,x_3)\in E_1\times E_2\times E_3$.

§.§ Diffeomorphisms.

$$f:X\rightarrow Y.$$

We say $f$ carries
$X$ diffeomorphically onto $Y$ if

(i) there are $m\in\N^+$ and Euclidean spaces $E$ and $F$
such that $X\in\cM_m(E)$ and $Y\in\cM_m(F)$;

(ii) $f$ is smooth and univalent;

(iii) $\op{rng}{f}=Y$ and $f^{-1}$ is smooth.

(i) $m\in\N^+$ and $E$ and $F$ are Euclidean spaces;

(ii) $X\in\cM_m(E)$, $Y\in\cM_m(F)$ and

$$f:X\rightarrow Y;$$

(iii)  $f$ is smooth and univalent;

(iv) $\op{rank}{Df(a)}=m$ for each $a\in M$;

(v) $X$ is compact and $X$ and $Y$ are homeomorphic.

Then $\op{rng}{f}=Y$ and $f$ carries
$X$ diffeomorphically onto $Y$.

This Proposition need not hold if $X$ is not compact.

Let $\cC$ be the family of connected components of $Y$.
$\cC$ is finite since $Y$ is compact.

Suppose $a\in X$. Since $f$ is smooth and $\op{rank}{Df(a)}=m$
is elementary that there are $G$ and $H$ such that $G$ is a relatively
open subset of $X$, $a\in G$, $H$ is an open subset of $F$ and
$f[G]=H\cap\op{rng}{f}$. Let $d=\op{dist}{(a,Y\sim H)}$. Then $d>0$
since $Y\sim H$ is compact and $a\not\in Y\sim H$. So there is an open
subset $W$ of $H$ such that $a\in W$ and $W\cap Y=W\cap f[G]$. Thus $\op{rng}{f}$
is open relative to $Y$. So there is a subfamily $\cB$ of $\cC$
such that $\op{rng}{f}=\bigcup\cB.$ Since $\op{rng}{f}$ is homeomorphic to
$X$ and $X$ is homeomophic to $Y$ we find that $\cB=\cC$ so
$\op{rng}{f}=Y$. That $f^{-1}$ is smooth follows from the Inverse Function
Theorem. So $f$ carries $X$ diffeomorphically onto $Y$

§.§ $\cL_m$, $\cP_m$ and $\cW_m$.

Let $\cL_m=\{L\in\End(\R^n):\op{rank}{L}=m\}$; as is well known,

be such that $\cP_m(L)$ is orthogonal projection of $\R^n$ onto $\op{rng}{L}$;
as is well known, $\cP_m$ is smooth and for each $R\in\R^+$ there
is constant $C_{\cP_m,R}$ such that

\begin{equation}\label{CPmR}
||\cP_m(L)-\cP_m(M)||\leq C_{\cP_m,R}||L-M||.
\end{equation}

whenever $L,M\in\cL_m$ and $\max\{||L||,||M||\}\leq R$.

be such that $\cW_m(L)=|\altdown{m}L|$ for $L\in\cL_m$.
It is elementary that $\cW_m$ is smooth and that

§.§ The second fundamental form of $M$.

We define

by requiring that, for any $a\in M$,

where $U,V,\psi,\psi$ are as in Definition <ref> and $s,t\in\R^m$
and that

$$B(a)(u,v)\bullet w=0\quad\hbox{if either $u\in\Nor{M,a}$ or $v\in\Nor{M,a}$
or $w\in\Tan{M,a}$.}$$

One calls $B$ the second fundamental form of $M$.

Suppose $a\in M$, $u,v\in\Tan{M,a}$. Then

Let $U,V,\phi,\psi$ are as in Definition <ref>,
let $b=\psi(a)$, note that $\phi(b)=a$ and let $s,t\in\R^m$
be such that $u=D\phi(b)(s)$ and $v=D\phi(b)(t)$. Then

\begin{equation}\label{eq}
\nu(\phi(x))(D\phi(x)(t))=0\quad\hbox{for $x\in U$.}
\end{equation}

We differentiate (<ref>) at $x=b$ in the direction $s$ and find that

\begin{align*}
\end{align*}

We define

by requiring that, for any $a\in M$,

$$A(a)(u)(v)\bullet w=u\bullet B(a)(v,w)\quad\hbox{whenever $u,v,w\in\R^n$.}$$

It follows that

$$A(a)(u)(v)\bullet w=A(a)(u)(w)\bullet v\quad\hbox{if $u,v,w\in\R^n$.}$$

We have

$$||B||=\sup\{|B(a)(u,v)\bullet w|:a\in M \ \hbox{and} \ u,v,w\in\Sm\};$$

It is elementary that $||B||<\infty$ if $M$ is compact and that
$||B||=0$ if and only if for each point $a\in M$ there are an open neighborhood
$V$ of $a$ and $S\in\mG_m(\R^n)$ such that $M\cap V=(a+S)\cap V$.
We let

it is elementary that $0<R<\infty$ if $M$ is compact.

§ VECTOR BUNDLES.

$$\cP(l)=l\circ l-l\quad\hbox{for $l\in\mathbbm{Sym}(\R^n)$.}$$

Suppose $p\in\mO_m(\R^n)$. Suppose $l\in\op{ker}{D\cP(p)}$.
We have

$$0=D\cP(p)(l)=l\circ p+l\circ l-l=l\circ p-p^\perp\circ l;$$

this implies that $p\circ l\circ p=0$ and $p^\perp\circ l\circ p^\perp=0$
so $l=p\circ l\circ p^\perp+p^\perp\circ l\circ p$
and since $l^*=l$ we have $p^\perp\circ l\circ p=(p\circ l\circ p^\perp)^*$.

Suppose $l,N$ are integers and $1\leq l\leq N$.

and $\gamma$ is smooth. We let

$$\bE(\gamma)=\{(x,y)\in M\times\R^N:y\in\op{rng}{\gamma(x)}\}$$

and for $r\in\R^+$ we let

$$\bE(\gamma,r)=\{(x,y)\in \bE(\gamma):|y|<r\}.$$

We let

\quad\hbox{and we let}\quad
\Nor{M}=\bE(\nu).$$

Suppose $a\in M$ and $u\in\Tan{M,a}$. Then

$$D\gamma(a)(u)=\gamma(a)\circ D\gamma(a)(u)\circ\gamma(a)^\perp
+\gamma(a)^\perp\circ D\gamma(a)(u)\circ\gamma(a).$$

The first equation follows from $\gamma(x)^*=\gamma(x)$
for all $x\in M$.

Since $\gamma(x)\circ\gamma(x)=\gamma(x)$ for $x\in M$ we have

$$D\gamma(a)(u)\circ\gamma(a)+\gamma(a)\circ D\gamma(a)(u)=D\gamma(a)(u)$$

for $u\in\R^n$.
Premultiplying and postmultiplying this equation by $\gamma(a)$ we
infer that $\gamma(a)\circ D\gamma(a)(u)\circ\gamma(a)=0$.
Premultiplying and postmultiplying this equation by $\gamma(a)^\perp$ we
infer that $\gamma(a)^\perp\circ D\gamma(a)(u)\circ\gamma(a)^\perp=0$.

Moreover, for  any $(x,y)\in\bE(\gamma)$ we have

\begin{equation}\label{tangamma}
\begin {split}
\Tan{\bE(\gamma),(x,y)}&=\{(u,v)\in\Tan{M,x}\times\R^N:
\gamma(x)^\perp(v)=D\gamma(x)(u)(y)\}\\
\end{split}
\end{equation}

We have

\quad\hbox{for $(x,y)\in M\times\R^N$.}$$

Let $F:M\times\R^N\rightarrow\R^N$ be such that

$$F(x,y)=\gamma(x)^\perp(y)\quad\hbox{for $(x,y)\in M\times\R^N$.}$$

Then $\bE(\gamma)=\{(x,y)\in M\times\R^N:F(x,y)=0\}$.

Suppose $(x,y)\in\bE(\gamma)$. We have

$$DF(x,y)(u,0)=-D\gamma(x)(u)(y)\in\op{rng}{\gamma(x)^\perp}\quad\hbox{for $u\in\Tan{M,x}$}$$

as well as

$$DF(x,y)(0,v)=\gamma(x)^\perp(v)\quad\hbox{for $v\in\R^N$};$$

it follows that

\op{dim}{\op{rng}{DF(x,y)}}=N-l;$$

\gamma(x)^\perp(v)=D\gamma(x)(u)(y)\}.$$

The Theorem now follows from the Implicit Function Theorem.

Suppose $Z\in\cE(\gamma)$, $x\in M$ and $u\in\Tan{M,x}$. Then

If $Z\in\cE(\gamma)$ we let

$$D_\gamma Z:M\rightarrow \Lin(\R^n,\R^l)$$

be such that $D_\gamma Z(x)=\gamma(x)\circ DZ(x)$ for $x\in\R^n$.
Evidently, $\op{rng}{D_\gamma Z(x)}\subset\op{rng}{\gamma(x)}$ and
$\Nor{M,x}\subset\op{ker}{D_\gamma Z(x)}$. The preceding Corollary implies that

\begin{equation}\label{Dgamma}
DZ(x)=D_\gamma Z(x)+D\gamma(x)(u)(Z(x))
\quad\hbox{for $x\in M$.}
\end{equation}

If $(x,y)\in\Tan{M}$ then

\Tan{\Tan{M},(x,y)}

and if $(x,y)\in\Nor{M}$ then

Suppose $a\in M$ and $u\in\Tan{M,a}$. Then

\begin{equation}\label{WN}
\tau(a)(DN(a)(u))=-A(a)(N(a))(u)\quad\hbox{if $N\in\cE(\nu)$}
\end{equation}

\begin{equation}\label{WT}
\nu(a)(DT(a)(u))=B(a)(T(a),u)\quad\hbox{if $T\in\cE(\tau)$.}
\end{equation}

Suppose $N\in\cE(\nu)$. Then $\tau(x)(N(x))=0$ for $x\in M$;
differentiating at $x=a$ we find that

Suppose $T\in\cE(\tau)$. Then $\nu(x)(T(x))=0$ for $x\in M$. Differentiating
at $x=a$ we find that

§.§ The mean curvature normal of $M$.

We let

be such that

$$H(x)\bullet u=\op{trace}{A(x)(u)}\quad\hbox{for $x\in M$ and $u\in\Nor{M,x}$.}$$

$H$ is called the mean curvature normal of $M$.

(i) $I$ is an open interval, $0\in I$,

$$h:I\times M\rightarrow \R^n$$

and $h$ is smooth;

(ii) for each $t\in I$, $M_t=h_t[M]\in\cM_m(\R^n)$ and
$h_t$ carries $M$ diffeomorphically onto $M_t$;

(iii) for each $t\in I$, $X_t:M_t\rightarrow\R^n$ is such that

$$X_t(y)={d\over d\epsilon}h_{t+\epsilon}(h_t^{-1}(y))|_{\epsilon=0}$$

for $y\in M_t$.

Then $X_t$ is smooth and

$${d\over dt}\cH^m(h_t[M])=\int X_t\bullet H_t\,d||M_t||$$

for each $t\in I$.

For the proof see the literature.

§ REACH.

Suppose $M\in\cM_m(\R^n)$.

We let

be such that
$\Xi(x,y)=x+y\quad\hbox{for $(x,y)\in\Nor{M}$}$
and we let

$$\br=\sup\left(\{0\}\cup\{r\in(0,\infty):\Xi|\bE(\nu,r) \ \hbox{is univalent.}\}\right)$$

$\Xi$ is smooth.

$\Xi$ is the restriction to $\bE(\nu)$ of the smooth map

$$\R^n\times\R^n\ni(x,y)\mapsto x+y\in\R^n.$$

Suppose $0<r<\infty$.
The following statements are equivalent:

(i) $r\leq\br$;

(ii) $M\cap\ob{a+v}{|v|}=\{a\}$ whenever $(a,v)\in\Nor{M}$
and $0<|v|<r$;

(iii) $|\nu(a)(x-a)|\leq{|x-a|^2\over 2r}$ whenever $a,x\in M$.

Suppose (i) holds, $a\in M$ and $u\in\Nor{M,a}\cap\Sm$.
Let $I=\{s\in(0,r):a+su\not\in M$}. Suppose $I$ were not empty.
Then $0\leq\inf I<r$. We have $\inf I>0$ since $u\not\in\Tan{M,a}$.
Let $t=\inf I/2$. Then $\Xi(a,tu)=a+tu=\Xi(a+2tu,-tu)$ which is impossible.
Suppose $s\in I$ and $x\in M$ is such that
$\op{dist}{(a+su,M)}=|(a+su)-x|$. Then $(a+su)-x\in\Nor{M,x}$ so
$\Xi(x,a+su)=\Xi(a,su)$ so, as $s\leq r\leq\br$ we have
$M\cap\ob{a+su,s}=\{a\}$ which implies that $\ob{a+ru,r}\cap M=\emptyset$.
Thus (ii) holds.

It is a simple matter to verify that (ii) implies (i)
and that (ii) and (iii) are equivalent.

$\br\leq {1\over||B||}$.

Suppose $a\in M$. Suppose
$U,V,\phi,\psi$ are as in Definition <ref> and $u\in\R^m$.
We have

as well as, for $t$ sufficiently small,

$$\phi(tu)=\phi(0)+tD\phi(0)(u)+{t^2\over 2}D^2\phi(0)(u,u)+\rho(t)$$

where $\rho(t)/t^2\rightarrow 0$ as $t\rightarrow 0$.

\begin{align*}
{\nu(a)(\phi(tu)-\phi(0))\over t^2}
&={1\over 2}\nu(a)(D^2\phi(0)(u,u))+{\nu(a)(\rho(t))\over t^2}\\
&={1\over 2}(B(a)(D\phi(0)(u),D\phi(0)(u)+{\nu(a)(\rho(t))\over t^2}.
\end{align*}

$${|\nu(a)(\phi(tu)-\phi(0))|\over t^2}\leq{|\phi(tu)-\phi(0)|^2\over 2r\,t^2}
\rightarrow{|D\phi(0)(u)|^2\over 2r}\quad\hbox{as} \ t\rightarrow 0.$$

Thus $|B(a)(D\phi(0)(u),D\phi(0)(u)|\leq{|D\phi(0)(u)|^2\over r}$.

Suppose $(a,b)\in\Nor{M}$ and $(u,v)\in\Tan{M,a}\times\Nor{M,a}$.

\begin{equation}\label{Xi1}
\end{equation}

\begin{equation}\label{Xi2}
\end{equation}

In particular, if $|b|<1/||B||$, then

\begin{equation}\label{Xi3}
|D\Xi(a,b)(u-A(a)(b)(u),v)|\geq|u-A(a)(b)(u)|\geq 1-||B||\,|v|>0
\end{equation}

and $D\Xi(a,b)$ carries $\Tan{\Nor{M},(a,b)}$ isomorphically onto

$(u-A(a)(b)(u),v)\in\Tan{\Nor{M,a},(a,b)}$ by Corollary <ref>.

Let $F(x,y)=x+y$ for $(x,y)\in\R^n\times\R^n$. Since $\Xi=F|\Nor{M}$,
(<ref>) follows from Corollary <ref>. (<ref>) holds
since $u-A(a)(b)(u)(v)\in\Tan{M,a}$ and $v\in\Nor{M,a}$. If $|b|<1/||B||$
then $|u-A(a)(b)(u)|>1-||B||\,|v|$ and (<ref>) follows from (<ref>).
(<ref>) implies $D\Xi(a,b)|\Tan{\Nor{M},(a,b)}$ is univalent.
This and the fact that $\Nor{M}\in\cM_n(\R^n\times\R^n)$ implies the
final conclusion.

For $r\in(0,\infty]$ we set

Obviously, $N(r)$ is an open subset of $\R^n$.

Suppose $\br>0$. Then $\Xi$ carries $\bE(\nu,\br)$
diffeomorphically onto $N(\br)$.

$\Xi|\bE(\nu,\br)$ is univalent by the definition of $\br$.
If $(a,b)\in\bE(\nu,\br)$ then $D\Xi(a,b)$ carries $\Tan{\Nor{M},(a,b)}$
isomorphically onto $\Tan{N(\br),a+b}=\R^n$ by the preceding Theorem.

$\br=\infty$ if and only if $M$ is a subset
of some member of $\mG_m(\R^n)$.

Suppose $\br=\infty$ and $a\in M$. It follows from (ii) of the
preceding Proposition that $M\subset\Tan{M,a}$.

Suppose $M\subset\Sm$. If $a\in M$ then
$A(a)(u)=u$ for $u\in\Tan{M,a}$; in particular, $||B||\geq 1$
and $\br\leq 1$.

Let $X(x)=x$ for $x\in M$. Then $X\in\cE(\nu)$
and, if $u\in\Tan{M,a}$, $DX(a)(u)=u$ so
so $A(a)(u)=u$ by a (<ref>); also, if $|u|=1$, then
$||B||\geq||A(a)(u)||= 1$.

(i) $Y_i$, $i=1,2$, is a Euclidean space;

(ii) $N_i\in\cM_m(Y_i)$ for $i=1,2$;

(iii) $K$ is a compact subset of $N_1$;
$F:N_1\rightarrow N_2$ is smooth; $F|K$ is univalent and,
for each $x\in K$,
$DF(x)$ carries $\Tan{N_1,x}$ isomorphically onto $\Tan{N_2,F(x)}$;

(iv) $U_\epsilon$, $0<\epsilon<\infty$, is an open subset of $N_1$
such that

(a) if $0<\epsilon_1<\epsilon_2<\infty$ then $U_{\epsilon_1}\subset

(b) if $V$ is an open subset of $N_1$ and $K\subset V$ then
$U_\epsilon\subset V$ for some $\epsilon>0$;

(c) $K=\bigcap_{\epsilon>0}U_\epsilon$.

(v) there is $\epsilon>0$ such that $DF(x)$ carries $\Tan{N_1,x}$ isomorphically onto $\Tan{N_2,F(x)}$
for $x\in U_\epsilon$ and such that $F|U_\epsilon$ is univalent;

(vi) if $\epsilon$ is as in (v) then $F[U_\epsilon]$ is an open neighborhood in $N_2$ of $F[K]$ and $F$ carries $U_\epsilon$ diffeomorphically onto

What follows is a paraphrase of the argument on page 116 of

Let $V_1$ be the set of $x\in N_1$ such that $DF(x)$ carries
$\Tan{N_1,x}$ isomorphically onto $\Tan{N_2,F(x)}$.
Then $V_1$ is an open subset
of $N_1$ so there is by (iii)(b) $\epsilon>0$ such that $U_\epsilon\subset V_1$.

Suppose (v) is false. Let $N_*$ be a positive integer such that the
closure of $K$ of $U_{1/N_*}$ in $N_1$ is compact.
For each integer $N$ such that $N\geq N_*$ there are $a_N,b_N\in U_{1/N}$ such that
$a_N\neq b_N$ and $F(a_N)=F(b_N)$. Since $K$ is compact there are
$a_*$ and $b_*$ in $K$ which are accumulation points
of $\{a_N:N=1,2,\ldots\}$ and $\{b_N:N=1,2,\ldots\}$, respectively.
Then $F(a_*)=F(b_*)$ which implies $\lim_{N\rightarrow\infty}|a_N-b_N|=0$.
Thus $F$ would not be univalent in any open neighborhood
of $a_*$. But as $DF(a_*)$ carries $\Tan{N_1,a_*}$ isomorphically onto
$\Tan{N_2,F(a_*)}$, $F$ must be univalent in some open neighborhood of $a_*$.

(vi) is an elementary consequence of (v) and the smoothness of $F$.

Suppose $M\in\cM_m(\R^n)$ and $M$ is compact. Then

$$Y_1=\R^n\times\R^n;\quad Y_2=\R^n;\quad N_1=\Nor{M};\quad N_2=\R^n;$$

let $F=\Xi$ and,
for $\epsilon>0$, let $U_\epsilon=\bE(\gamma,\epsilon)$; and let
$K=\{(x,0):x\in M\}\subset N_1$. Then (i)-(iv) of Lemma <ref> hold.
So if $\epsilon$ is as in Lemma <ref> then $\reach{M}\geq\epsilon$.

Suppose $M$ is compact.
Since $\Xi$ carries $\bE(\nu,\br)$ diffeomorphically onto $N(\br)$
we may define smooth functions

$$\xi:N(\br)\rightarrow \R^n\quad\hbox{and}\quad\sigma:N(\br)\rightarrow\R^n$$

by requiring that

Note that

\begin{equation}\label{xisigma}
\quad\hbox{for $z\in N(\br)$.}
\end{equation}

Suppose $b\in N(\br)$, $a=\xi(b)$ and $c=\sigma(b)$.

(i) $a\in M$, $c\in\Nor{M,a}$, $b=a+c$ and $|c|=\op{dist}{(b,M)}$;

(ii) $D\xi(b)(u-A(a)(c)(u))=u$ for $u\in\Tan{M,a}$ and

(iii) $D\sigma(b)(u-A(a)(c)(u))=-A(a)(c)(u)$ for $u\in\Tan{M,a}$ and
$D\sigma(b)(v)=v$ for $v\in\Nor{M,b}$;

(iv) $||D\xi(b)-\tau(a)||\leq{||B||\,|c|\over 1-||B||\,|c|}\leq{\br|c|\over \br-|c|}$.

Keep in mind that

$$\Tan{M,a}\ni u\mapsto u-A(a)(c)(u)\in\Tan{M,a}$$

is a linear isomorphism since $||B||\,|c|<1$.

We have $\xi\circ\Xi(x,y)=x$ and $\sigma\circ\Xi(x,y)=y$
for $(x,y)\in\bE(\nu,\br)$. (i)-(iii) now follow from (<ref>)
and the Chain Rule.

Let $l\in\End(\Tan{M,a})$ and $L\in\Lin(\Tan{M,a},\R^n)$ be such that $l(u)=A(a)(c)(u)$
and $L(u)=u-l(u)$ for $u\in\Tan{M,a}$. Then $||l||\leq ||B||\,|c|<1$
and $||L||\geq 1-||l||$ so $L$ is invertible and $||L^-1||\leq(1-||l||)^{-1}$.
Let $i$ be the identity map of $\Tan{M,a}$. (ii) implies that
$D\xi(b)\circ L=i$ so $D\xi_b|\Tan{M,a}=L^{-1}$ which implies that

=||L^{-1}-i||\leq||L^{-1}||\,||i-L||\leq{||l||\over 1-||l||}
\leq{||B||\,|c|\over 1-||B||\,|c|}.$$

(iv) follows since (ii) and (iii) imply that $\Nor{M,a}\subset\op{ker}{D\xi(b)}$.

Suppose $z\in N(\br)$. Then

\begin{equation}\label{xitau}
\end{equation}

\begin{equation}\label{Dxitau}
\end{equation}

(<ref>) follows from (ii) of Theorem <ref>.
Let $C(\epsilon)=(1-\epsilon)z+\epsilon\xi(z)$ for $\epsilon\in(0,1)$.
$\ob{z}{\dist{z,M}}\subset N(r)=\op{dmn}{\xi}$ we find that
$\op{rng}{C}\subset\op{dmn}{\xi}$ so that

%=\int_0^1{d\over d\epsilon}D\xi(C(\epsilon))\,d\epsilon
=\int_0^1 D(D\xi)(C(\epsilon))(C^\prime(\epsilon))\,d\epsilon$$

so, as $C^\prime(\epsilon)=z-\xi(z)$ for $\epsilon\in(0,1)$, (<ref>)

Suppose $f$ is a smooth function on $M$ with values in a
normed space $Y$. Then

$$\op{Lip}{f}\leq \left(1+{1\over\br}\right)\chi^{[1]}(f).$$

Suppose $0<\lambda<1$ and
$|a-b|\leq 2\lambda\br$. Let $C(t)=(1-t)a+tb$ for $t\in(0,1)$.
Then $\op{dist}{(C(t),M)}<\lambda\br$ for $t\in(0,1)$ so $\op{rng}{C}\subset\op{dmn}{\xi}$ and, by Theorem <ref>(ii),

\leq {1\over 1-||B||\op{dist}{(C(t),M)}}
\leq{1\over 1-\lambda}$$

$${d\over dt}(f\circ\xi\circ C)(t)
\leq \chi^1(f)||D\xi(C(t))||\,|C^\prime(t)|
\leq {\chi^1(f)\over 1-\lambda}|a-b|.$$

Integrating this inequality from $0$ to $1$ we infer that

$$|f(a)-f(b)|\leq {\chi^1(f)\over 1-\lambda}|a-b|.$$

On the other hand, if $|a-b|>2\lambda\br$ then

$$|f(a)-f(b)|\leq \op{diam}{\op{rng}{f}}
\leq 2\chi^0(f)
<{\chi^0(f)\over \lambda\br}|a-b|.$$

$$|f(a)-f(b)|\leq\chi^{[1]}\max\left\{{1\over 1-\lambda},{1\over\lambda\br}\right\}|a-b|$$

so we should set $\lambda=1/(1+\br)$.

§.§ $\theta$ and $\iota$.

\quad\hbox{and}\quad
\iota:\Sm\rightarrow\mO_1(\R^n)$$

be the tangent map of $\Sm$ and normal map of $\Sm$, respectively.

Suppose $(x,v)\in\bE(\theta)$. Then

=\{(u,w)\in\Tan{\Sm,x}\times\R^n:w\bullet x=-u\bullet v\}.$$

§ $\BBETA$ AND $\BO$.

We define

\quad\hbox{and}\quad
\bo:\R^n\sim\{0\}\rightarrow\mathbb{G}_{n-1}(\R^n)$$

at $x\in\R^n\sim\{0\}$
by letting $\bbeta(x)=|x|^{-1}x$
and letting $\bo(x)$ be orthogonal projection of $\R^n$
onto $\Tan{\Sm,x}=\{w\in\R^n:w\bullet x=0\}$.

Suppose $x,y\in\Sm\sim\{0\}$. Then

\begin{equation}\label{eta}
|\bbeta(x)-\bbeta(y)|\leq {|x-y|\over\min\{|x|,|y|\}}.
\end{equation}

Let $s,t\in\R^+$ and $u,v\in\Sm$ be such that
$x=su$ and $y=tv$. Since both sides of (<ref>) are symmetric in $x$ and $y$
we may suppose that $0<s=|x|\leq|y|=t$. Then (<ref>) amounts to

$$|u-v|\leq{|su-tv|\over s}=|u-\lambda v|$$

where we have set $\lambda=t/s\geq 1$. We have

$$|u-\lambda v|-|u-v|
=1-2\lambda u\bullet v+\lambda^2-(1-2 u\bullet v+1)
=(\lambda-1)(\lambda+1-2 u\bullet v)
\geq 0.$$

For $x\in\R^n\sim\{0\}$ we have

$$D\bbeta(x)={\bo(x)\over |x|};$$

in particular,

$$D\bbeta(x)(w)=|x|^{-3}(|x|^2 w-(w\bullet x)x)
\quad\hbox{for $w\in\R^n$.}$$

Suppose $w\in\R^n$. Then

$$D\bbeta(x)(w)={d\over d\epsilon}{x+\epsilon w\over |x+\epsilon w|}\big|_{\epsilon=0}
=|x|^{-3}(|x|^2 w-(w\bullet x)x).$$

Suppose $x\in\Sm$ and $v\in\R^n$ and $v\bullet x=0$. Then

We have

\leq|v|^2.$$

§  $F$.

We let

be such that

$$f(x,y)=(1+y\bullet x)\bbeta(x+y)\quad\hbox{for $(x,y)\in\bE(\nu,\br)$.}$$

We let

be such that $\upsilon(z)=z\bullet\xi(z)$ for $z\in N(\br)$.

On page 221 of [2] $f$ there is based on the exponential map
of $\Sm$ applied to $\Xi|\bE(\gamma)$. The $f$ here is as in [FN]
which I believe is easier to work with. The modifications to required to
accomodate this $f$ in [2] are straightforward.

Suppose $z\in N(\br)$ and $x=\xi(z)$. Then

$$z\bullet\xi(z)=(x+(z-x))\bullet x=1+(z-x)\bullet x
\geq 1-|z-x|>1-\br\geq 0.$$

We define the smooth function

by setting

$$\rho(z)={z\over\upsilon(z)}-\xi(z)\quad\hbox{for $z\in N(\br)$.}$$

$f$ is a diffeomorphism with range an open subset
$V$ of $N(\br)$ and
with inverse $(\xi,\rho)|V$. In particular,

\begin{equation}\label{DxifP}
\end{equation}

\begin{equation}\label{DrhofQ}
\end{equation}

for any $(a,b)\in\bE(\nu,\br)$ and $(u,v)\in\Tan{M,a}\times\Nor{M,a}$
and where

$$P(a,b)=a\quad\hbox{and}\quad Q(a,b)=b\quad\hbox{for $(a,b)\in\bE(\nu,\br)$.}$$

Suppose $(x,y)\in\bE(\nu,\br)$ and $z=f(x,y)$. Then

(i) $1+y\bullet x>1-\br\geq 0$ and $|z|=1+y\bullet x$;

(ii) $z-x\in\Nor{M,x}$;

(iii) $z\in\op{dmn}{\xi}$ and $x=\xi(z)=x$;

(iv) $y=\rho(z)$.

$y\bullet x\leq|y|<\br$ and $|z|=1+y\bullet x$ so (i) holds.

$\{x,y\}\subset\Nor{M,x}$ so (ii) holds.

Let $\alpha(r)=(1+r)^2-2{(1+r)^2\over 1-r}+1$ for $r\in[0,1)$. Then
$(1-r)(r^2-\alpha(r))=2(2|y|^2+|y|+1>0$ so $\alpha(|y|)<|y|^2<\br^2$.
We infer from
$1+y\bullet x\leq 1+|y|$ and $|x-y|>1-|y|$ that

\begin{align*}
&=\left|(1+y\bullet x)^2-2(1+y\bullet x){x+y\over|x+y|}\bullet x+1\right|^2\\
&=(1+|y|)^2-2{(1+|y|)^2\over 1-|y|}+1\\
\end{align*}

so that $\xi(z)=x$ and (iii) holds.

(i) implies that

Taking the inner product with $\xi(z)=x$ we find that

$$z\bullet\xi(z)={1+y\bullet x\over|x+y|}={|z|\over|x+y|}$$

so that $|x+y|={|z|\over z\bullet\xi(z)}$ and

so (iv) holds.

We have shown that $(\xi,\rho)\circ f(x,y)=(x,y)$ for $(x,y)\in\bE(\nu,\br)$.
Thus $f$ is univalent and, by the Chain Rule, $Df(x,y)$
carries $\Tan{\bE(\nu,\br),(x,y)}$ isomorphically onto $\R^n$ so
$f$ is an open mapping. The Theorem follows.

§.§ $\gamma$.

$$\gamma(x)=\nu(x)-\iota(x)\quad\hbox{for $x\in M$;}$$

evidently, $\gamma:M\rightarrow\mO_{n-m-1}(\R^n)$, $\gamma$ is smooth
and $\op{rng}{\gamma(x)}=\Nor{M,x}\cap\Tan{\Sm,x}$ for any $x\in M$.

Suppose $a\in M$, $u\in\Tan{M,a}$ and $y\in\Nor{M,a}$. Then

\begin{equation}\label{auy}
\begin{cases}
u&\hbox{if $y=a$,}\cr
0&\hbox{if $y\in\op{rng}{\gamma(a)}$}
\end{cases}
\end{equation}

\begin{equation}\label{DgammaW}
DW(a)=D_\gamma W(a)-A(a)(W(a))\quad\hbox{for $W\in\cE(\gamma)$.}
\end{equation}

\begin{equation}\label{auyTan}
\Tan{\bE(\gamma),(a,v)}
\oplus\{(0,w):w\in\op{rng}{\gamma(a)}\}.
\end{equation}

We have

$$\gamma(x)(y)=\nu(x)(y)-(y\bullet x)x\quad\hbox{for $x\in M$
and $y\in\R^n$.}$$

For $u\in\Tan{M,a}$ we have

$$D\gamma(a)(u)(y)=-A(a)(y)(u)-(y\bullet u)a-(y\bullet a)u.$$

$$(y\bullet u)a+(y\bullet a)u
u&\hbox{if $y=a$,}\cr
0&\hbox{if $y\in\op{rng}{\gamma(a)}$}
\end{cases}$$

(<ref>) implies (<ref>). (<ref>) now follows from (<ref>).

The following Theorem is a Corollary of Theorem <ref>.
$f$ carries $\bE(\gamma,\br)$ diffeomorphically
onto $V\cap\Sm$.

Suppose $x\in M$. Then

$$D\rho(x)(w)=\gamma(x)(w)\quad\hbox{whenever $w\in\Nor{M,x}\cap\Tan{\Sm,x}$.}$$

We have $D\xi(x)=\tau(x)$.

Suppose $w\in\Nor{M,x}\cap\Tan{\Sm,x}$. Then

$$D\upsilon(x)(w)=w\bullet x+x\bullet D\xi(x)(w)=0.$$

so, as $\upsilon(x)=1$,

§.§ $F$.

Whenever $W\in\cE(\gamma,\br)$ we define

by letting $F(W)=f\circ(i_M,W)$; it follows from Theorem <ref> that
$F(W)$ carries $M$ diffeomorphically onto its range; in particular,

\begin{equation}\label{Fprop}
\op{rng}{F(W)}\in\cM_m(\R^n),\quad\xi\circ F=i_M \quad\hbox{and}\quad \rho\circ F(W)=W.
\end{equation}

§ FAMILIES OF SUBMANIFOLDS OF $\SM$.

We suppose throughout this section that $\cJ$ is a Euclidean space.
For each $\epsilon\in\R^+$ and $s\in\cJ$ we let

\quad\hbox{and we let}\quad
\cJ(s,\epsilon)=\{t\in\cJ:|s-t|<\epsilon\}.$$

We suppose

(i) $\cS$ is a open subset of $\cJ$ and $0\in\cS$;

(II) $M\in\cM_m(\R^n)$, $M\subset\Sm$ and $M$ is compact;

(III) $\bz:\cS\times M\rightarrow\Sm$ and $\bz$ is smooth;

(IV) $\bz_s$ is a diffeomorphism for each $s\in \cS$;

(IV) $\bz_0(p)=p$ for $p\in M$.

be the reach of $M$.

$$M_s=\bz_s[M]\quad\hbox{for each $s\in\cS$;}$$

it follows that $M_s\in\cM_m(\R^n)$, $M_s$ is compact
and $M_s\subset\Sm$.

We let

$$T:\cS\times M\rightarrow\cL_m$$

be such that $T(s,p)=\cP_m(D\bz_s(p))$ for $(s,p)\in\cS\times M$.

For each $s\in \cS$ we let

$$\tau_s, \ \nu_s, \ B_s, \ A_s \quad\hbox{be as in Section \ref{subman};}$$

$$ \br_s, \ \Xi_s, \ N_s(r), \ r\in\R^+, \ \xi_s, \ \sigma_s\quad\hbox{be as in Section \ref{reach}};$$

$$\upsilon_s, \ \rho_s, \ f_s, \ V_s, F_s
\quad\hbox{be as in Section \ref{f}}$$

with $M$ replaced by $M_s$.

We have

$$T(s,p)=\tau_s(\bz_s(p))\quad\hbox{for $(s,p)\in\bS\times M$.}$$

§.§ $\epsilon_1$.

We fix

such that the closure of $\cJ(\epsilon_1)$ is a compact subset of $\cS$.

For each $(j,k)\in\N\times\N$ we let

\begin{equation}\label{bound}
C_{j,k}=\chi^{j,k}(\bz|(\cJ(\epsilon_1)\times M))
\quad\hbox{for any $(j,k)\in\N$.}
\end{equation}

and note that $C_{j,k}<\infty$.

There is a constant $C_T$ such that

\begin{equation}\label{CT}
||T(s,p)-T(t,q)||\leq C_T(\epsilon)(|s-t|+|p-q|)
\quad\hbox{whenever $s,t\in\cJ(\epsilon_1)$ and $p,q\in M$}
\end{equation}

Suppose  $s,t\in\cJ(\epsilon_1)$ and $p,q\in M$. By (<ref>)
we have

\int_0^1{d\over d\lambda}T((1-\lambda)t+\lambda s,p)\,d\lambda
=\int_0^1 DT((1-\lambda)t+\lambda s,p)(s-t,0)\,d\lambda

and, by Theorem <ref>,

\leq\left(1+{1\over\br}\right)\chi^{[1]}(T_t)|p-q|.$$

$$\ubr=\inf\{\br_s:s\in \cJ(\epsilon_1)\}
\quad\hbox{and let}\quad
\ubb=\sup\{||B_s||:s\in\cJ(\epsilon_1)\}.$$

We have

$$0<\ubr\leq 1\quad\hbox{and}\quad \ubb\leq{1\over\ubr}.$$

Let $\cH$ be the closure of $\cJ(\epsilon_1)$.

let $F:N_1\rightarrow N_2$
be such that $F((s,p),v)=\bz(s,p)+v$ for $((s,p),v)\in\bE(T)$.
Let $K=\{((s,p),0):(s,p)\in\cH\times M\}$ and for each $\epsilon\in\R^+$

$$U_\epsilon=\{((s,p),v)\in\bE(T):\op{dist}{(s,K)}<\epsilon \ \hbox{and}
\ |v|<\epsilon\}.$$

Keeping in mind Theorem <ref> we find that (i)-(iv)
of Lemma <ref> hold. Let $\epsilon_*$ be as in (v) of
Lemma <ref>. Then
$\epsilon_*\leq\br_s$ for any $s\in\cJ(\epsilon_1)$.

$\ubb\leq 1/\ubr$ since $||B_s||\br_s\leq 1$ for $s\in\cJ$.

For each $(s,t)\in\cJ(\epsilon_1)$ we let

$$\bc_{s,t}:[0,1]\rightarrow \cJ(\epsilon_1)$$

be such that

$$\bc_{s,t}(\lambda)=t+\lambda(s-t)\hbox{for $\lambda\in[0,1]$.}$$

Suppose $s,t\in \cJ(\epsilon_1)$. Then

$$|\bz_s(p)-\bz_t(p)|\leq C_{1,0}|s-t|\quad\hbox{for any $p\in M$.}$$

$$M_t\subset N_s(r)\quad\hbox{if $C_{1,0}|s-t|<r\leq\ubr$.}$$

For any $p\in M$ we have

\left|\int_0^1 {d\over d\lambda}\bz(\bc_{s,t}(\lambda),p)\,d\lambda\right|
=\left|\int_0^1 D\bz(\bc_{s,t}(\lambda),p)(s-t,0)\,d\lambda\right|
\leq C_{1,0}|s-t|.$$

Suppose $z\in M_t$. Let $p\in M$ be such that $\bz_t(p)=z$. Then

$$\op{dist}{(z,M_s)}\leq|\bz_t(p)-\bz_s(p)|\leq C_{1,0}|s-t|.$$

be such that $C_{1,0}\delta_1<\ubr$.

If $s,t\in\cJ(\epsilon_1)$
and $|s-t|<\delta_1$ then

$$M_t\subset N_s(\ubr).$$

We define

$$\mu:\cS\times M\rightarrow\R^+$$

by letting

|D\bz_s(p)(u)|\geq\lambda|u| \ \hbox{for} \ u\in\Sm\cap\Tan{M,p}\}.$$

We let

$$\mmu=\inf\{\mu(s,p):(s,p)\in\cJ(\epsilon_1)\times M\}$$

We leave as an elementary exercise for the reader to prove the following
$\mu$ is positive and continuous and $\mmu$ is positive.

There is a constant

such that if $s,t\in\cJ(\epsilon_1)$ and $|s-t|<\delta_1$ then
$M_t\subset N_s(\ubr)$ and

$$||T(t,p)-\tau_s(\xi_s(\bz_t(p)))||\leq C_\mu|s-t|
\quad\hbox{for $p\in M$.}$$

Suppose $s,t\in\cJ(\epsilon_1)\cap\cJ(\delta_1)$.
From Corollary <ref>
we infer that

$$M_t\subset N_s(\ubr)\subset\op{dmn}{\xi_s}.$$

Suppose $p\in M$. Then

Let $q\in M$ be such that $\xi_s(\bz_t(p))=\bz_s(q)$.

$$c=\left(1+{1\over \ubr}\right)\left({1\over\ul{\mu}}\right)$$

we infer from Theorem <ref> that

\begin{align*}
&\leq\left(1+{1\over \br_s}\right)\left({1\over\ul{\mu}}\right)|\bz_s(p)-\bz_s(q)|\\
&\leq c|\bz_s(p)-\bz_s(q)|\\
&\leq c(|\bz_s(p)-\bz_t(p)|+|\bz_t(p)-\xi_s(\bz_t(p))|\\
&\leq 2c|\bz_s(p)-\bz_t(p)|\\
&\leq 2c C_{1,0}|s-t|.
\end{align*}

\begin{align*}
&\leq C_{1,0}\chi^{1}(T,\cJ(\epsilon_1)\times M)|s-t|+C_T|p-q|\\
&\leq C_\mu|s-t|
\end{align*}

$$C_\mu=\chi^{1}(T,\cJ(\epsilon_1)\times M)|s-t|
+C_T(2c C_{1,0}).$$

Suppose $s\in\cJ(\epsilon_1)$.
There is $\delta\in(0,\epsilon_1-|s|]$
such that if $t\in\cJ(s,\delta)$ then $t\in\cJ(\epsilon_1)$,
$M_t\subset N_s(\ubr)$ and

$$\Lambda_t \ \hbox{carries $M_t$ diffeomorphically onto $M_s$}$$

$$\Lambda:\cJ(s,\delta)\times M\rightarrow M$$

is such that

\quad\hbox{ for $(t,q)\in\cJ(s,\delta)\times M$.}$$

That $s,t\in\cJ(\epsilon_1)$ and $|s-t|<\delta_1$.
Then $M_t\subset N_s(\ubr)$ by Corollary <ref>.

For each $(t,x)\in\cJ(s,\delta_1)\times M_t$ we let
$\cN(t,x)=\{z\in M_t:\xi_s(z)=x\}$ and we let
$N(t,x)$ equal the cardinality of $\cN(t,x)$.

We choose $\delta_2\in(0,\delta_1]$ such that $C_\mu\delta_2<1$
where $C_\mu$ is as in Proposition <ref>. Suppose $p\in M$. By Proposition
<ref> we have

\quad\hbox{ if $t\in\cJ(s,\delta_2)$.}$$

It follows from (ii) of Theorem <ref>
that $\Tan{M_t,\bz_t(p)}$
meets $\op{ker}{D\xi_s(\bz_t(p))}$ transversely so that

carries $\Tan{M_t,\bz_t(p)}$ isomorphically onto $\Tan{M_s,\xi_s(\bz_t)(q)}$.}$$

This implies that there is an open neighborhood of $\bz_t(p)$ such that
$\bz_s$ carries $U\cap M_t$ diffeomorphically onto an open neighborhood of
$\bz_s(\bz_t(p))$ in $M_s$.
Since $M_t$ is compact we find that if $x\in\op{rng}{\xi_s|M_t}$ then
$N_t(x)$ is finite and there are an open neighborhood $U$ of $x$
in $M_s$ and  a map
$\cU$ with domain $\cN_t(x)$ such that $\cU(z)$ is an open neighborhood of
$z$ in $M_t$ which $\xi_s$ carries diffeomorphically onto $U$.
It follows that $\cN(t,x)$ is finite and that $N_t$ is locally constant.

We choose $\delta_3\in(0,\delta_2]$ as follows. Let $\cC$ be the family of
connected components of $M_s$; since $M_s$ is compact $\cC$ is finite.
For $x\in M_s$ we let $C_x\in\cC$ be such that $x\in C_x$.
If $\cC$ has one member we let $\delta_2=\delta_1$. Otherwise we let

$$d=\inf\{|x-y|:x,y\in M_s \ \hbox{and} \ C_x\neq C_y\}.$$

We choose $\delta_3\in(0,\delta_2]$ such that
Suppose $|s-t|<\delta_3$ and $x\in M_s$. There is $z\in M_t$
such that $|x-z|=\op{dist}{(x,M_t)}\leq C_{1,0}|s-t|<d$.
Hence $|\xi_s(z)-x|=|\op{dist}{z,M_s}\leq|x-z|<d$ so $z\in C_x$
and $N_t(\xi_s(z))\geq 1$
Since $N_t$ is locally constant we must have $N_t(x)\geq 1$.
Thus $N_t(x)\geq 1$ for all $x\in M_s$.

Let $\bz_s^t=\bz_t\circ\bz_s^{-1}$ for $t\in\cJ(s,\delta_2)$.

\begin{align*}
\int N_t\,d||M_s||
&\rightarrow \ ||M_s||(\R^n)
\end{align*}

$$\lim_{t\rightarrow s}
\sup\{|\cW_m(D\xi_s(\bz_s^t(x)))\cW_m(D\bz_s^t(x))-1|:x\in M_s\}

Thus there is $\delta\in(0,\delta_3]$ such that

$$N_t(x)=1\quad\hbox{if $t\in\cJ(s,\delta)$ and $x\in M_s$.}$$

It follows that $\Lambda_t$ carries $M_s$ diffeomorphically onto $M_t$
for such $t$.

§.§ $\ul{M}$.

We let

$$\ul{M}=\{(s,z):s\in(0,\epsilon_1) \ \hbox{and} \ z\in M_s\}.$$

Suppose $s\in\cJ(\epsilon_1)$. Let $\delta=\epsilon_1-|s|$. Let

$$\Psi:\cJ(s,\delta)\times N_s(\ubr)\rightarrow\cJ(s,\delta)\times N_s(\ubr)$$

be such that $\Psi(t,z)=(t,\bz_s^{-1}(\xi_s(z)))$ for $(t,z)\in\cJ(s,\delta)\times N_s(\ubr)$ and note that $\Psi$ is smooth.

Let $\delta$ and $\Lambda$ be as in Theorem <ref>;
note that $\Lambda$ is smooth and let

$$\Gamma:\cJ(s,\delta)\times M\rightarrow\cJ(s,\delta)\times M$$

be such that $\Gamma_t=\Lambda_t^{-1}$ for $t\in\cJ(s,\delta)$ and note that
$\Gamma$ is smooth.

$$\Phi:\cJ(s,\delta)\times M\rightarrow\cJ(s,\delta)\times N_s(\ubr)$$

be such that

$$\Phi(t,p)=(t,\bz_t(\Lambda_t^{-1}(p)))\quad\hbox{for $(t,p)\in\cJ(s,\delta)\times M$.}$$

Note that $\Phi$ is smooth.

If $(t,p)\in\cJ(s,\delta)\times M$ then

\begin{align*}
\Psi(\Phi(t,p))
\end{align*}

The Theorem now follows from Proposition <ref>.

§.§ $\protect\underline{\tau}$, $\protect\underline{\nu}$, $\protect\underline{\gamma}$, $\protect\underline{B}$, $\protect\underline{A}$.

We define smooth maps

\ul{\nu}:\ul{M}\rightarrow\mO_{n-m}(\R^n);\quad
\ul{\gamma}:\ul{M}\rightarrow\mO_{n-m-1}(\R^n)$$

by requiring that

$$\ul{\tau}(s,x)(u)=Di_{\ul{M}}(x,s)(u,0)\quad\hbox{for $(s,x)\in\ul{M}$
and $u\in\R^n$}$$

and letting

\ul{\gamma}(s,x)=\ul{\nu}(s,x)-\iota(x)$$

for $(s,x)\in\ul{M}$ and $u\in\R^n$. It follows that

\ul{\nu}(s,x)=\nu_s(x);\quad
\ul{\gamma}(s,x)=\gamma_s(x)$$

for $(x,s)\in\ul{M}$.

We define the smooth maps

\quad\hbox{and}\quad
\ul{A}:\ul{M}\rightarrow\Lin(\R^n,\End(\R^n))$$

by letting

\quad\hbox{whenever $(s,x)\in\ul{M}$ and $u,v\in\R^n$}$$

and requiring that

$$\ul{A}(s,x)(u)(v)\bullet w=u\bullet\ul{B}(s,x)(v,w)
\quad\hbox{whenever $(s,x)\in\ul{M}$ and $u,v,w\in\R^n$.}$$

It follows that

\quad\hbox{and}\quad
\ul{A}(s,x)(u)=A_s(x)(u)$$

whenever $(s,x)\in\ul{M}$ and $u,v\in\R^n$.

§.§ $\protect\underline{N}$, $\protect\underline{\Xi}$, $\protect\underline{\xi}$, $\protect\underline{\sigma}$.

For each $r\in(0,\ubr]$ we let

$$\ul{N}(r)=\{(s,z)\in \cJ(\epsilon_1)\times\R^n:\op{dist}{(z,M_s)}<r\}.$$

We let

be such that

$$\ul{\Xi}((s,x),v)=(s,\Xi_s(x,v))\quad\hbox{for $((s,x),v)\in\bE(\ul{\nu},\ubr)$.}$$

It follows from Theorem <ref> that $\ul{\Xi}$ carries $\bE(\ul{\nu},\ubr)$
diffeomorphically onto $\ul{N}(\ubr)$; in particular, $\ul{N}(r)$
is an open subset of $\cJ\times\R^n$ whenever $0<r\leq\ubr$.
Since $\ul{\Xi}$ carries $\bE(\ul{\nu},\ubr)$ diffeomorphically onto $\ul{N}(\ubr)$
we may define smooth functions

$$\ul{\xi}:\ul{N}(\ubr)\rightarrow \R^n\quad\hbox{and}\quad\ul{\sigma}:\ul{N}(\ubr)\rightarrow\R^n$$

by requiring that

Note that

\begin{equation}\label{ulxisigma}
\quad\hbox{for $(s,z)\in \ul{N}(\ubr)$.}
\end{equation}

§.§ $\protect\underline{f}$, $\protect\underline{\upsilon}$, $\protect\underline{\rho}$, $\protect\underline{V}$, $\protect\underline{g}$.

We define the smooth maps

\ul{\upsilon}:\ul{M}(\ubr)\rightarrow\R^+;\quad
\ul{\rho}:\ul{M}(\ubr)\rightarrow\R^n$$

by letting

\ul{\upsilon}((t,z),v)=(t,\upsilon_t(x,v);\quad
\ul{\rho}((t,z),v)=(t,\rho_t(x,v)$$

for $((t,x),v))\in\bE(\ul{\nu})$.
We also set

$\ul{V}$ is an open subset of $\cJ\times\R^n$ and
$\ul{f}$ carries $\bE(\ul{\nu},\ubr)$ diffeomorphically onto $\ul{V}$
with inverse $(\ul{\xi},\ul{\rho})$.

Moreover, $\ul{f}$ carries $\bE(\ul{\gamma},\ubr)$ diffeomorphically
onto $\ul{V}\cap\Sm$.

This follows from the results of Section <ref>.

For $r\in(0,\ubr)$ we let

(s,z)\in\ul{N}(r) \ \hbox{and} \ (t,z)\in\ul{N}(r)\}.$$

Since $\ul{N}(r)$ is an open subset of $\cJ\times\R^n$ it follows
$\ul{O}(r)$ is an open subset of $\cJ\times\cJ\times\R^n$.

Suppose $s,t\in\cJ(\epsilon_1)$, $q,r\in\R^+$, $q+r\leq\br$
and $(s,t,z)\in\ul{O}(r)$. If $\chi^{1}(\bz)|s-t|\leq q$ then

Since $(s,z)\in\ul{N}(r)$, $\op{dist}{(z,M_s)}<r\leq\ubr\leq\br_s$
so $z\in\op{dmn}{\xi_s}$ which implies that there is $w\in M_s$ such
that $\op{dist}{(z,M_s)}=|z-w|$.
Suppose $\lambda\in[0,1]$ and $u=(1-\lambda)s+\lambda t$.

\leq|z-w|+|\bz_s(p)-\bz_u(p)|
<r+\chi^{1}(\bz)\lambda|t-s|\leq r+q$$

so $z\in\ul{N}(q+r)$. By the same argument with $s$ replaced by $t$
we find that $z\in\ul{N}(q+r)$.

§  $F_S^T$.
Suppose $s\in \cS$ and $W\in\cE(\gamma_s,\br_s)$.

$$F_s(W):M_s\rightarrow V_s$$

be such that

Then $\op{rng}{F_s(W)}$ is a smooth compact submanifold of $\Sm$,
$F_s(W)$ carries $M_s$ diffeomorphically onto

$$\xi_s\circ F_s(W)=i_{M_s}
\quad\hbox{and}\quad
\rho_s\circ F_s(W)=0.$$

Suppose $t\in \cS$, $W_i\in\cE(\gamma_s,\br_s)$, $i=1,2$
and $\op{rng}{F_s(W_1)}=\op{rng}{F_s(W_2)}$. Then $W_1=W_2$.

Suppose $x_1\in M_s$. There is $x_2\in M_s$ such that
$F_s(W_2)(x_2)=F_s(W_1)(x_1)$. Thus

Applying $g_s$ we find that $(x_1,W_1(x_1))=(x_2,W_2(x_2))$.

Suppose $s\in \cS$, $X\in\cM_m(\R^n)$ and $X\subset V_s$.
is $W\in\cE(\gamma_s,\br_s)$ such that $\op{rng}{F_s(W)}=X$ if and only if
$\xi_s|X$ carries $X$ diffeomorphically onto $M_s$.

Suppose $W\in\cE(\gamma_s,\br_s)$ and $\op{rng}{F_s(W)}=X$. Then
$F_s(W)=f_s\circ(i_{M_s},W)$ so

$$(\xi_s\circ F_s(W),\rho_s\circ F_s(W))
=g_s\circ F_s(W)=g_s\circ f_s\circ(i_{M_s},W)=(i_{M_s},W)$$

so $\xi_s\circ F_s(W)=i_{M_s}$. It follows that
$\xi_s|X$ carries $X$ diffeomorphically onto $M_s$.

Suppose $\xi_s|X$ carries $X$ diffeomorphically onto $M_s$. Let
$W\in\cE(\gamma_s,\br_s)$ be such that
$W\circ(\xi_s|X)=\rho_s|X\in\cE(M_s,\gamma_s)$. Then

so $\op{rng}{F_s(W)}=X$.

For $s,t\in \cS$ we let

\op{rng}{F_s(W)}=\op{rng}{F_t(Z)}\}.$$

Suppose $s,t\in \cS$, $W\in\cE(\gamma_s,\br_s)$
and $Z\in\cE(\gamma_t,\br_t)$.
The following statements are equivalent.

(I) $(W,Z)\in F_s^t$;

(II) $\xi_t\circ F_s(W)$ carries $M_s$ diffeomorphically onto $M_t$

\begin{equation}\label{Fsteqn}
Z\circ\xi_t\circ F_s(W)=\rho_t\circ F_s(W).
\end{equation}

Suppose (I) holds. Let $g=F_t(Z)^{-1}\circ F_s(W)$. Then
$g$ carries $M_s$ diffeomorphically onto $M_t$ and
$F_t(Z)\circ g=F_s(W)$ so for any $y\in M_t$ we have

$$(i_{M_t},Z)\circ g=(\xi_t,\rho_t)\circ F_t(Z)\circ g
=(\xi_t,\rho_t)\circ F_s(W)$$

so $g=\xi_t\circ F_s(W)$ and $Z\circ g=\rho_t\circ F_s(W)$
so (II) holds.

Suppose (II) holds. Let $X=\op{rng}{F_s(W)}$. (<ref>) implies that
$Z\circ(\xi_t|X)=\rho_t|X$. So

=f_t\circ g_t\circ i_X

so $\op{rng}{F_t(Z)}=\op{rng}{i_X}=X=\op{rng}{F_s(W)}$.

If $s,t\in\cS$ then $0\in\op{dmn}{F_s^t}$ if and only if
$\xi_t$ carries $M_s$ diffeomorphically onto $M_t$.

Suppose $s\in \cS$ and $W\in\cE(\gamma_s)$. Then

$$\op{Lip}{W}\leq \left(1+{8\over\br_s}\right)\chi^{[1]}(W)
\quad\hbox{and}\quad
\op{Lip}{F_s(W)}\leq 1+\left(1+{8\over\br_s}\right)\chi^{[1]}(W)$$

The first inequality follows from
Proposition <ref>.

Suppose $x_i\in M_s$, $i=1,2$. Using Proposition <ref> we find that

\begin{align*}
\end{align*}

so the second inequality holds.

Suppose $E$ is a Euclidean space, $h:\ul{N}(\ubr)
\rightarrow E$, $h$ is smooth and

$$\Box h:\ul{O}(\br/2)\rightarrow\Lin(\cJ,E)$$

is such that

$$\Box h(s,t,z)(u)=\int_0^1 D^{1,0}h(\bc_{s,t}(\lambda),z)(u)\,d\lambda
\quad\hbox{for $(s,t,z)\in\ul{O}(\ubr/2)$ and $u\in\cJ$.}$$

Then $\Box h$ is smooth and

\begin{equation}\label{hst}
h(s,z)-h(t,z)=\Box h(s,t,z)(s-t)\quad\hbox{for $(s,t,z)\in\ul{O}(\ubr/2).$}
\end{equation}

If $(s,t,z_i)\in\ul{O}(\ubr/2)$ for $i=1,2$ then

\begin{equation}\label{hst2}
||\Box h(s,t,z_1)-\Box h(s,t,z_2)||\leq C_{1,\Box}(h)|z_1-z_2|
\end{equation}

where we have set

Since $\ul{O}(\br)$ is an open subset
of $\cJ(\epsilon_1)\times\cJ(\epsilon_1)\times\R^n$ it is elementary that $\Box h$ is smooth.
(<ref>) follows since

$$h(s,z)-h(t,z)=\int_0^1{d\over d\lambda}h(\bc_{s,t}(\lambda),z)\,d\lambda
=\int D^{1,0}h(\bc_{s,t}(\lambda),z)(s-t)\,d\lambda$$

whenever $(s,t,z)\in\ul{O}(\ubr/2)$.

Suppose$(s,t,z_i)\in\ul{O}(\ubr/2)$ for $i=1,2$.
If $|z_2-z_1|<\ubr/4$ then
Corollary <ref> implies that $(s,t,w)\in\ul{O}(\ubr/2)$ if $w\in\ob{z_1}{\ubr/4}$
so that

$$||\Box h(s,t,z_1)-\Box h(s,t,z_2)||\\
\leq\chi^{1}(\Box h)\,|z_1-z_2|\leq\chi^{2}(h)\,|z_1-z_2|.$$

On the other hand, if $|z_1-z_2|\geq\ubr/4$

$$||\Box h(s,t,z_1)-\Box h(s,t,z_2)||\leq
2\chi^{1}(h)\leq 8{\chi^{1}(h)\over\ubr}|z_1-z_2|.$$

So (<ref>) holds.

§.§ $\epsilon_2$ and $\cG$.

be such that

$$C_{1,\Box}(\ul{\xi})(2\epsilon_2)<{1\over 2}.$$

\left(1+{8\over\ubr}\right)\chi^{[1]}(W)<1 \right\}.

Suppose $(s,W)\in\cG$. Then

$$\chi^0(W)<{\ubr\over 8}, \ \op{Lip}{W}<1 \ \hbox{and} \  \op{Lip}{F_s(W)}<2.$$

$\chi^0(W)\leq\chi^{[1]}(W)<\ubr/8$. The second and third
inequalities follow directly from Proposition <ref>.

Suppose $s,t\in \cJ(\epsilon_2)$ and $(s,W)\in\cG$.

\begin{equation}\label{Ost}
\{(s,t,F_s(W)(x)):x\in M_s\}\subset\ul{O}(\ubr/2);
\end{equation}

in particular, $\op{rng}{F_s(W)}\subset\op{dmn}{\xi_t}$.

Suppose $x\in M_s$. There is $p\in M$ such that $\bz_s(p)=x$.
By Proposition <ref> we have

\leq|\bbeta(x+W(x))-x|
\leq \chi^0(W)
\leq{\ubr\over 4}$$

\leq|\bbeta(x+W(x))-x|+|\bz(s,p)-\bz(t,p)|
\leq \chi^0(W)+C_{1,0}|s-t|\leq{\ubr\over 2}$$

so (<ref>) holds and, since $s,t\in\cJ(\epsilon_2)$, $\op{rng}{F_s(W)}
\subset\op{dmn}{\xi_t}$ since $C_{1,\Box}(\ul{\xi})(2\epsilon_2)<1/2$.

For $(s,W)\in\cG$ and $(t,x)\in \ul{O}(\ubr/2)$ we let

$$A_s^t(W)(x)=\Box \ul{\xi}(s,t,F_s(W)(x))(s-t)$$

and we let

$$B_s^t(W)(x)=\Box \ul{\rho}(s,t,F_s(W)(x))(s-t).$$

Suppose $s,t\in \cJ(\epsilon_2)$ and $(s,W)\in\cG$.

\begin{equation}\label{xist}
\xi_t(F_s(W)(x))%=x-(\xi_s(F_s(W)(x))-\xi_t(F_s(W)(x)))
\end{equation}

\begin{equation}\label{rhost}
\rho_t(F_s(W)(x))%=x-(\rho_s(F_s(W)(x))-\rho_t(F_s(W)(x)))
\end{equation}

\begin{equation}\label{biLip}
\begin{split}
{1\over 2}&|x_1-x_2|\\
&\leq{3\over 2}|x_1-x_2|
\end{split}
\end{equation}

if $x_i\in M_s$, $i=1,2$;

\begin{equation}\label{inFst}
\end{equation}

(<ref>) and (<ref>) follow from (<ref>)
since $\xi_s(F_s(W)(x))=x$ and $\rho_s(F_s(W)(x))=W(x)$ for $x\in M_s$.

Suppose  $x_i\in M_s$, $i=1,2$. From (<ref>) we infer that

\begin{align*}
\ul{\xi}(t,&F_s(W)(x_1))-\ul{\xi}(t,F_s(W)(x_2))\\
-(\Box \ul{\xi}(s,t,F_s(W)(x_1))(s-t)-\Box \ul{\xi}(s,t,F_s(W)(x_2))(s-t))
\end{align*}

and, by Proposition <ref> and Proposition <ref>,

\begin{align*}
|\Box \ul{\xi}(s,t,F_s(W)(x_1))(s-t)&-\Box \ul{\xi}(s,t,F_s(W)(x_2))(s-t)|\\
&\leq C_{1,\Box}(\ul{\xi})|F_s(W)(x_2)-F_s(W)(x_1)||s-t|\\
&\leq C_{1,\Box}(\ul{\xi})2|x_1-x_2||s-t|\\
&\leq {1\over 2}
\end{align*}

It now follows from Proposition <ref> that $W\in\op{dmn}{F_s^t}$.

Suppose $s,t\in\cJ(\epsilon_2)$ and $(s,W)\in\cG$.

\begin{equation}\label{Dxist}
D(\xi_t\circ F_s(W))(x)%=x-(\xi_s(F_s(W)(x))-\xi_t(F_s(W)(x)))
=\tau_s(x)+\Box(D^{0,1}\ul{\xi})(t,s,F_s(W)(x))(t-s)\circ DF_s(W)(x);
\end{equation}

\begin{equation}\label{Drhost}
D(\rho_t\circ F_s(W))(x)
=DW(x)+\Box(D^{0,1}\ul{\rho})(t,s,F_s(W)(x))(t-s)\circ DF_s(W)(x);
\end{equation}

\begin{equation}\label{Fstest}
\chi^0(F_s^t(W))\leq\chi^0(D^{1,0}\ul{\rho})|s-t|+\chi^0(W)
\end{equation}

\begin{equation}\label{DFstest}
\chi^0(DF_s^t(W))\leq
\chi^0(D^{1,1}\ul{\rho})|s-t|+\chi^0(DW).
\end{equation}

Suppose $x\in M_s$. Using (<ref>) we find that

\begin{align*}
D(\xi_t\circ F_s(W))&(x)-D(\xi_s\circ F_s(W))(x)\\
&=(D^{0,1}\ul{\xi}(t,F_s(W)(x))-D^{0,1}\ul{\xi}(s,F_s(W)(x)))\circ DF_s(W)(x)\\
&=(\Box(D^{0,1}\ul{\xi})(t,s,F_s(W))(t-s))\circ DF_s(W)(x).
\end{align*}

Moreover, $\xi_s\circ F_s(W)=i_{M_s}$ so $D(\xi_s\circ F_s(W))(x)=\tau_s(x)$
so (<ref>) holds. With regard to (<ref>) we have
$\rho_s\circ F_s(W)=W$ so a similar analysis yields (<ref>).

By (<ref>) we have

$$F_s^t(W)\circ\xi_t\circ F_s(W)=\rho_t\circ F_s(W).$$

(<ref>) follows from (<ref>). The Chain Rule implies that
for any $x\in M_s$ we have

$$DF_s^t(W)((\xi_t\circ F_s(W)(x)))\circ D(\xi_t\circ F_s(W))(x)
=D(\rho_t\circ F_s)(x);$$

(<ref>) now follows from (<ref>) and (<ref>).

§.§ $\cL_m$, $\cW_m$ and $\epsilon_3$.

Let $\cL_m=\{L\in\End(\R^n):\op{rank}{L}=m\}$ and let

be such that $\cW_m(L)=|\altdown{m}L|$ for $L\in\cL_m$.
It is elementary that if $L\in\cL_m$ and $p$ is orthogonal projection
of $\R^n$ onto the kernel of $L$ then

Suppose $(s,W)\in\cG$ and $(s,t,x)\in\ul{O}(\ubr/2)$.

$$\bc_s^t(W)(x)=\cW_m(D(\xi_t\circ F_s(W))(x))-1.$$

Since $D(\xi_t\circ F_s(W))(x)\in\cL_m$ we infer from (<ref>) and (<ref>) that

\begin{equation}\label{cstest}
\begin{split}
&\leq(2^m-1)||D(\xi_t\circ F_s(W))(x)-D(\xi_s\circ F_s(W))(x)||\\
&\leq C_{\cW_m}|s-t|
\end{split}
\end{equation}

where we have set

be such that
$C_{\cW_m}(2\epsilon_3)<1$. Suppose $s,t\in\cJ(\epsilon_3)$.
Then $-|\bc_s^t(W)|\leq \bc_s^t(W)\leq|\bc_s^t(W)|$ so

\begin{equation}\label{cW1}
\leq1-C_{\cW_m}|s-t|
\leq 1-|\bc_s^t(W)|
\leq 1-\bc_s^t(W)
\leq 1+|\bc_s^t(W)|
\leq 1+C_{\cW_m}|s-t|).
\end{equation}

Suppose $s,t\in\cJ(\epsilon_3)$
and  $(s,W)\in\cG$. Then

$$\{(s,t,F_s(W)(x)):x\in M_s\}\subset\ul{O}(\ubr/2)$$

and, for any $\epsilon\in\R^+$,

\leq (\epsilon+C_{F,1}|s-t|)P_s(W,W)+\left(1+{1\over\epsilon}\right)C_{F,2}|s-t|^2$$

$$C_{F,1}=C_{\cW_m}\quad\hbox{and}\quad C_{F,2}=\chi^0(\Box\ul{\rho})^2||M_s||(\R^n)(1+C_{\cW_m})(2\epsilon_3).$$

Keeping in mind (<ref>) we have that

\begin{align*}
&=\int |F_s^t(W)\circ\xi_t\circ F_s(W)|^2(1+\bc_s^t(W))\,d||M||_s\\
&=\int |\rho_t\circ F_s(W)|^2(1+\bc_s^t(W))\,d||M||_s\\
&=\int |W-B_s^t(W)|^2(1+\bc_s^t(W))\,d||M||_s\\
&=\int (|W|^2-2W\bullet B_s^t(W)+|B_s^t|^2)(1+\bc_s^t(W))\,d||M||_s
\end{align*}

so, using (<ref>), we find that

\begin{align*}|
&=\left|\int \bc_s^t(W)|W|^2+(-2W\bullet B_s^t(W)
\end{align*}

Since $|B_s^t(W)|^2\leq(\chi^0(\Box\ul{\rho}))^2|s-t|^2$ our estimate holds.

§ THE FAMILY OF INTEREST.

Suppose $M\in\cM_m(\R^n)$, $M\subset\Sm$ and

\begin{equation}\label{H}
H(x)=-x\quad\hbox{for $x\in M$}
\end{equation}

where $H$ is the mean curvature normal of $M$. As is well known and is implied
by Proposition <ref>
(<ref>) is equivalent to the statement that

\begin{equation}\label{EH}
{d\over d\epsilon}\cH^m(h_t[M])\big|_{\epsilon=0}=0
\end{equation}

whenever $I$ is an open interval $I$ containing $0$, $h:I\times M\rightarrow\Sm$,
$h$ is smooth and $h_0(x)=x$ for $x\in M$.

Let $\gamma$ be as in <ref> and let $\br$ be the reach of $M$.

§.§ The symmetric quadratic functionals $P,Q,R,S$.

We define symmetric quadratic functionals

$$ P,Q,R,S:\cE(\gamma)\times\cE(\gamma)\rightarrow\R$$

by letting

\begin{align*}
P(X,Y)&=\int X\bullet Y\,d||M||;\\
Q(X,Y)&=\int D_\gamma X\bullet D_\gamma Y\,d||M||;\\
R(X,Y)&=\int A(x)(X(x))\bullet A(x)(Y(x))\,d||M||x;\\
S(X,Y)&=Q(X,Y)-m P(X,Y)-R(X,Y)
\end{align*}

for $X,Y\in\cE(\gamma)$.

§.§ $\Phi$; $\Psi$; $\bL$; and $E_\lambda$, $\lambda\in\R$.

For $Z\in\cE(\gamma,\br)$ we let $F(Z)$ be as in (<ref>)
and we let

We define

by requiring that

$$P(\Psi(Z),W)={d\over d\epsilon}\Phi(Z+\epsilon W)\big|_{\epsilon=0}.$$

(In fact, the right hand side of this formula is an integral over $M$ of linear function of $DW$  and the preceding formula
follows by an integration by parts in the integral to obtain an integral over
$M$ of a linear function of $W$.)

be the linearization at $0$ of $\Psi$; that is,

$$\bL(Z)={d\over d\eta}\Psi(\eta Z)\big|_{\eta=0}
\quad\hbox{for $Z\in\cE(\gamma)$.}$$

As is well known,

\begin{equation}\label{symmetric}
P(\bL(Z),W)=P(Z,\bL(W))=S(Z,W)\quad\hbox{whenever $Z,W\in\cE(\gamma)$.}
\end{equation}

For each $\lambda\in\R$ we let

$$E_\lambda=\{Z\in\cE(\gamma):\bL(Z)=\lambda Z\};$$

so $Z\in E_{\lambda}$ if and only if

$$S(Z,W)=\lambda P(Z,W)\quad\hbox{for $W\in\cE(\gamma)$.}$$

§.§ The Hölder spaces $\cC^{k,\alpha}(\gamma)$.

Let us fix $\alpha\in(0,1)$.
For each $k\in\N$ we let $\cC^{k,\alpha}(\gamma)$ be the vector space
of maps $Z:M\rightarrow\R^n$ such that $\gamma(x)(Z(x))=Z(x)$ for each
$x\in M$; $D^j Z$ exists and is continuous for each $j\in\N$ such that $j\leq k$; and $D^k Z$ is Hölder continuous with exponent $\alpha$. $\cC^{k,\alpha}(\gamma)$ has a natural Banach space structure.
Clearly, $P$ and $R$ have uniqued continuous extensions, still denoted by $P$
and $R$, to $\cC^{0,\alpha}(\gamma)$
and $Q$ and $S$ have natural continuous extensions, still denoted by $Q$
and $S$, to $\cC^{1,\alpha}(\gamma)$. $\bL$ has an unique extension,
still denoted by $\bL$ to a continuous map from $\cC^{2,\alpha}(\gamma)$
to $\cC^{0,\alpha}(\gamma)$. Moreover, (<ref>) holds when

$\Phi$ has a unique continuous extension to $\{Z\in\cC^{1,\alpha}(\gamma):
|Z|<\br\}$ and $\Psi$ has a unique extension to a continuous map
from $\{Z\in\cC^{2,\alpha}(\gamma)\}$ to $\cC^{0,\alpha}(\gamma)$.
These extensions are real analytic.

$$\cJ=\op{ker}{\bL} \ \hbox{and let} \ \bj=\op{dim}{\cJ}.$$

The members of $\cJ$ are called Jacobi fields relative to $\Sm$.
The inner product on $\cJ$ is the restriction to $\cJ\times\cJ$
of $P$. So if $s,t\in\cJ$ then

$$s\bullet t=\int s(x)\bullet t(x)\,d||M||.$$

By standard elliptic theory,

$$\bj \ \hbox{is finite;}$$

for each $\epsilon\in\R^+$ there is a constant $C$ such that

$$\sum_{|\lambda|\leq N}\op{dim}{E_\lambda}\leq(1+N)^{m/2+\epsilon}
\quad\hbox{whenever $N\in\R^+$;}$$

in particular, $\op{dim}{\op{rng}{E_\lambda}}$ is finite for each
$\lambda\in\R$ and $\{\lambda\in\R:E_\lambda\neq 0\}$ is discrete;
for each $s\in\R$ and each $Z\in\cE(\gamma)$,

$$\Sigma_s(Z)=P(E_0(Z),E_0(Z))+\sum_{\lambda\in\R\sim\{0\}}|\lambda|^2 P(E_\lambda(Z),E_\lambda(Z))<\infty$$

and $\{\Sigma_s:s\in\R\}$ is a family of seminorms which define the
strong topology on $\cE(\gamma)$.

§.§ Construction of the family of interest.

For $k=0,2$ let $\cH_k$ be the $P$ orthogonal complement
of $\cJ$ in $\cC^{k,\alpha}(\gamma)$; since $\bj$ is finite $\cH_k$
is closed; let $\Pi_k$ be the
$P$ orthogonal projection of $\cC^{k,\alpha}(\gamma)$
onto $\cJ$ and let $\Pi_k^\perp$ be the complementary projection; thus
$Z=\Pi_k(Z)+\Pi_k^\perp(Z)$ for $Z\in\cC^{k,\alpha}(\gamma)$.
By standard elliptic theory,

$$\hbox{$\bL|\cH_2$ is Banach space isomorphism from $\cH_2$ onto $\cH_0$.}$$

Since $\Psi(0)=0$ and since $\Psi$ is real analytic
it follows from the Implicit Function Theorem that there are open neighborhoods $\cS$ of $0$ in $\cJ$ and of $0$ in $\cV$ of $0$ in $\op{rng}{\Pi_0}^\perp\tilde{\Gamma}$ and a
real analytic function $\Upsilon:\cS\rightarrow \op{rng}{\Pi_0}^\perp$ such that
$\{s+\Upsilon(s):s\in\cS\}\subset \cV$
and such that if $s\in\cS$, $W\in\cJ^\perp$ and $s+W\in \cV$ then

$$\Pi_0^\perp(\Psi(s+W))=0 \ \Leftrightarrow  \ W=\Upsilon(s).$$

We let $\Omega:\cS\rightarrow\cJ$ be such that

$$\Omega(s)=\Pi_0(\Psi(s+\Upsilon(s))\quad\hbox{for $s\in\cS$}$$

and we let

Note that $\Omega$ is a real analytic function so $\cA$ is a real analytic set.

$$\{Z\in\cV:\Psi(Z)=0\}=\{s+\Upsilon(s):s\in\cS \ \hbox{and} \ \Omega(s)=0\}.$$

Since $\Psi(0)=0$ we have

We now assume that

\begin{equation}\label{main}
\Tan{\cA,0}=\cJ;
\end{equation}

owing to the real analyticity of $\cA$ this implies that

$$\Omega \ \hbox{vanishes near $0$ in $\cS$.}$$

That $\Tan{\cA,0}=\cJ$ is clearly implied by the assumption about Jacobi
fields made in (1) of the Introduction of [2]. So we have

We define

$$ \bz:\cS\times M\rightarrow\Sm$$

by letting

\quad\hbox{for $(s,x)\in\cS\times M$.}$$

we find that

$$M_s \ \hbox{is $\Phi$ stationary.}$$

It follows from Proposition <ref> that

\begin{equation}\label{stationary}
\cH^m(M_s)=\cH^m(M)\quad\hbox{for $s$ in the connected component of $0$ in $\cS$.}
\end{equation}

In fact (<ref>) holds without assuming (<ref>)
owing to the fact that the real analyticity of $\cA$ implies that $\cA$
is locally rectifiably arcwise connected.

be the Jacobi fields on $M_s$ relative to $\Sm$.

For $s\in\cS$ we let

be such that $\gamma_s(y)$ equals orthogonal projection of $\R^n$ onto
For each $s\in\cS$ let

be such that

=\gamma_s(\bz(s,x))\left({d\over d\epsilon}\bz(s+\epsilon t,x)\big|_{\epsilon=0}\right)\quad\hbox{for $t\in\cJ$.}$$

Since $M_{s+\epsilon t}$ is stationary for sufficiently small $\epsilon$
we find that

$D\Upsilon(0)=0$ and $\bJ_0(t)=t$ for $t\in\cJ$.

\begin{equation}\label{kerbJ}
s\in\cS \ \Rightarrow \ \op{ker}{\bJ_s}=\{0\}.
\end{equation}

Suppose $s\in\cS$, $t\in\cJ$ and
$W=s+\Upsilon(s)\in\cE(\gamma_s)$. Let $X=t+D\Upsilon(s)(t)\in\cE(\gamma_s)$.
Suppose $x\in M$ and $y=\bz(s,x)$. Using Proposition <ref>) we
find that

\begin{align*}
\bJ_s&(t)(y)\\
{d\over d\epsilon}(F(s+\epsilon t))(x)\big|_{\epsilon=0}\right)\\
&=\gamma_s(y)\left({d\over d\epsilon}\bbeta(x+(s+\epsilon t+\Upsilon(s+\epsilon t))(x))\big|_{\epsilon=0}\right)\\
&=\gamma_s(y)\left(1+|W(x)|)^{-3/2}\left((1+|W(x)|^2)X(x)-(X(x)\bullet W(x))(x+W(x))\right)\right)\\
&=(1+|W(x)|)^{-3/2}\left((1+|W(x)|^2)X(x)-(X(x)\bullet W(x))(W(x))\right).\\
\end{align*}

In particular, if $s=0$ then $W=0$ so that

Since $\bJ_0(t)\in\cJ$ and $D\Upsilon(0)(t)\in\cH_0$
we find that $D\Upsilon(0)(t)=0$ and $\bJ_0(t)=t$.

Now suppose $t\in\op{ker}{\bJ_s}$. Then

$$X(x)={(X(x)\bullet W(x))W(x)\over 1+|W(x)|^2}$$

which implies that

$$|X(x)|\leq|X(x)|{|W(x)|^2\over 1+|W(x)|^2}$$

which in turn implies that $X(x)=0$ which implies that $t(x)=0$.

Suppose $s,t\in\cJ$. Then

$$s\bullet t=P_0(\bJ_0(s),\bJ_0(t)).$$

We have

=\int s(x)\bullet t(x)\,d||M||=s\bullet t.$$

Suppose $t\in\cJ(\epsilon_1)$, $u\in\cJ$ and $x\in M_t$. Then

This is 5.4(7) of [2].

Let $x=\bz(t,p)\in M_t$.
Let $I=\{\epsilon\in\R:t+\epsilon u\in\cJ\}$  and note that $I$ is an open neigborhood of $0$ in $\cS$.
For $\epsilon\in I$ let

$$g(\epsilon)=\ul{f}(t+\epsilon u,(\bz(t+\epsilon u,p),0))
=\bbeta(\bz(t+\epsilon u,p)).$$

We have $g(0)=\bz(t,p)=x$ and, by Proposition <ref>,

since $|\bz(t,p)|=1$ for $(t,p)\in\cS\times M$.
Since $g(\epsilon)\in M_{t+\epsilon u}$ we have

$$0=\ul{\rho}(t+\epsilon u,g(\epsilon))\quad\hbox{for $\epsilon\in I$;}$$

Differentiating at $\epsilon=0$ we find using Proposition <ref> that

\begin{align*}
0&={d\over d\epsilon}\ul{\rho}(t+\epsilon u,g(\epsilon))\big|_{\epsilon=0}\\
\end{align*}

There is a constant $C_{\bJ,0}$ such that

$$|\bJ_s(u)(x)-\bJ_t(u)(\xi_t(x))|\leq C\,|s-t|\,|u|$$

Suppose $s,t\in\cJ(\epsilon_1)$, $u\in\cJ$ and $x\in M_s$. Then

\begin{align*}
\bJ_s(u)&(x)-\bJ_t(u)(\xi_t(x))\\
\end{align*}

Now apply Proposition <ref>.

There is a constant $C_\bJ$ such that

\leq C_\bJ|s-t|\,|u|\,|v|\quad\hbox{for $s,t\in\cJ(\epsilon_1)$
and $u,v\in\cJ$.}$$

Suppose $s,t\in\cJ(\epsilon_1)$ and $u,v\in\cJ$. For each
$x\in M_s$ let

\begin{align*}
\Gamma_1(x)&=\bJ_s(u)(x);\\
\Gamma_2(x)&=\bJ_s(v)(x);\\
\Gamma_3(x)&=\bJ_t(u)(\xi_t(x));\\
\Gamma_4(x)&=\bJ_t(v)(\xi_t(x));\\
\Gamma_5(x)&=\bJ_t(u)(\xi_t(x))\bullet\bJ_t(v)(\xi_t(x))\bc_s^t(0)(x)).
\end{align*}

\begin{align*}
\end{align*}

Now apply the preceding Lemmma and (<ref>).

The following Corollary follows by letting $t=0$ above.
If $u\in\cJ$ and $s\in\cJ(\epsilon_1)$ then

$$P_s(\bJ_s(u),\bJ_s(u))\leq (1+C_\bJ)|s|\,|u|.$$

§.§ Symmetric bilinear functions.

Suppose $V$ is an inner product space, $\bB$ is the vector space of
real value symmetric bilinear functions on $V\times V$. We say $T\in\bB$
is nonnegative if $T(v,v)\geq 0$ for $v\in V$.and $\bB^+$
is the set of those $T\in\bB$ such that $T(v,v)\geq 0$ for $v\in V$.

\begin{equation}\label{expand}
T(v,w)={1\over 2}(T(v+w,v+w)-T(v,v)-T(w,w)
\end{equation}

and, if $T$ is nonnegative,

\begin{equation}\label{CaS}
|T(v,w)|\leq\sqrt{T(v,v)T(w,w)}\leq{1\over 2}(T(v,v)+T(w,w))
\end{equation}

for $v,w\in V$.

Suppose $T_i\in\bB$, $i=1,2$, $U\in\bB$, $U$ is nonnegative

$$|T_1(v,v)-T_2(v,v)|\leq U(v,v)\quad\hbox{for $v\in V$.}$$

$$|T_1(v,w)-T_2(v,w)|\leq{3\over 2}(U(u,u)+U(v,v))\quad\hbox{for $u,v\in V$.}$$

Suppose $v,w\in V$. Then

\begin{align*}
&={1\over 2}|(T_1(v+w,v+w)\\
&={1\over 2}|(T_1(v+w,v+w)\\
&\leq{1\over 2}(U(v+w,v+w)+U(v,v)+U(w,w))\\
&\leq{3\over 2}(U(v,v)+U(w,w)),
\end{align*}

§ $L_S^T$ AND $\EPSILON_4$.

Suppose $s,t\in\cJ(\epsilon_2)$. We define a linear map

as follows. Keeping in mind that Proposition <ref> implies that
$\xi_t$ carries $M_s$ diffeomorphically onto $M_t$ we require that

\begin{equation}\label{Lst1}
$W\in\cE(\gamma_s)$ and $x\in M_s$.}
\end{equation}

This is not the $L_s^t$ of Part Two of the proof of
Lemma 5.4 in [2]. There
we see

$$L_s^t(W)={d\over d\epsilon}F_s^t(\epsilon W)\big|_{\epsilon=0}.$$

This has the pleasant property that $L_s^u=L_t^u\circ L_s^t$ which $L_s^t$
above does not have but it is harder to work with.

We let

be such that $h_L(s,z)=\gamma_s(\xi_s(z))$
for $(s,z)\in \ul{N}(\ubr)$; note that $h_L$ is smooth. We infer from
Proposition <ref> that

=\Box h_L(s,t,z)(s-t)
\quad\hbox{for $(s,t,z)\in\ul{O}(\ubr/2)$.}$$

If $W\in\cE(\gamma_s)$ and $x\in M_s$ we have

\begin{equation}\label{Lst2}
L_s^t(W)(\xi_t(x))=W(x)-\Box h_L(s,t,\xi_t(x))(s-t)(W(x))
\end{equation}

whenever $x\in M_s$ since

$$C_L=\sup\{||\Box h_L(s,t,z)(u)||:(s,t,z)\in \ul{O}(\ubr), \ u\in\cJ
\ \hbox{and} \ |u|\leq 1\}.$$

Then $C_L<\infty$ and

\begin{equation}\label{CL0}
||\gamma_s(\xi_s(z))-\gamma_t(\xi_t(z))||\leq C_L|s-t|
\quad\hbox{whenever $z\in\ul{O}(\ubr/2)$}
\end{equation}

\begin{equation}\label{CL}
\end{equation}

be such that $2C_L(2\epsilon_4)<1$.

The following Theorem is an immediate consequence of (<ref>).
Suppose $s,t\in\cJ(\epsilon_4)$.
Then $L_s^t$ carries $\cE(\gamma_s)$ isomorphically onto $\cE(\gamma_t)$Su

and there is a constant $C_{L,P}$ such that

$$|P_t(L_s^t(X),L_s^t(Y))-P_s(X,Y)|\leq C_{L,P}|s-t|)P_s(X,Y).$$

Suppose $W\in\cE(\gamma_s)$ and $\bc_s^t$ is as in (<ref>).Using
<ref> we find that

\begin{align*}\int|L_s^t(W)|^2&-|W|^2\,d||M_t||\\
&=\int |L_s^t(W)\circ\xi_t|^2\cW_m(D(\xi_t|M_s)) -|W|^2\,d||M_s||\\
&=\int |L_s^t(W)\circ\xi_t|^2 (1+\bc_s^t(0))\,d||M_s||\\
&=\int |W(x)-\Box h_L(s,t,\xi_t(x)(s-t)|^2(1+\bc_s^t(0)(x))-|W(x)|^2\,d||M_s||\\
\end{align*}

Now use (<ref>) and Proposition <ref> to complete the proof.

Suppose $s\in\cJ(\epsilon_1)$ and $W\in\cE(\gamma_s)$. Then

$$|R_s(W,W)|\leq m\ubr^2 P_s(W,W).$$

Suppose $u_1,\ldots,u_n$ is an orthonormal basis for $\R^n$.
For any $x\in M_s$ we have

=\sum_{i=1}^m |A_s(x)(W(x))(u_i)|^2
\leq m\,||B_s(x)||^2\,||W(x)|^2
\leq{m\ubb}^2|W(x)|^2.$$

Suppose $s,t\in\cJ(\epsilon_4)$.
There is a constant $C_{L,R}$ such that

$$|R_t(L_s^t(X),L_s^t(Y))-R_s(X,Y)|\leq C_{L,R}|s-t|)P_s(X,Y)
\quad\hbox{for $X,Y\in\cE(\gamma_s)$.}$$

We let

\quad\hbox{for $(s,z)\in\ul{O}(\ubr)$.}$$

Suppose $W\in\cE(\gamma_s)$.
Since $(s,t,x)\in\ul{O}(\ubr/2)$ we have

as well as

\begin{align*}
\end{align*}

\begin{align*}
&=\int h_A(t,x)(W(x))(1+\bc_s^t(0)(x))-h_A(s,x)(W(x))\,d||M_s||\\
&=\int (h_A(t,x)-h_A(s,x))(W(x))+h_A(t,x)(W(x))\bc_s^t(0)(x))\,d||M_s||x.
\end{align*}

The existence of the constant $C_{L,R}$ now follows from Proposition <ref>,
(<ref>) and Proposition <ref>.

Suppose $s,t\in\cJ(\epsilon_4)$. There is a constant $C_{L,Q}$
such that for $X,Y\in\cE(\gamma_s,\ubr)$ we have

\begin{align*}
\leq C_{L,Q}|s-t|\big((P_t+Q_t)(X,X)+(P_t+Q_t)(Y,Y)\big)
\end{align*}

For $u\in\cJ(\epsilon_4)$ and $U,V\in\cE(\gamma_u,\ubr)$ we set

$$T_u(U,V)=\int DU\bullet DV\,d||M_u||$$

and observe that

\begin{align*}
&=\int (D_\gamma U(y)+A_u(y)(U(y)))\bullet(D_\gamma V(y)+A_u(y)(V(y)))\,d||M_u||y\\
&=\int D_\gamma U(y)\bullet D_\gamma V(y)+A_u(y)(U(y))\bullet A_u(y)(V(y))\,d||M_u||y\\
\end{align*}

Suppose $W\in\cE(\gamma_s,\ubr)$. We calculate

\begin{align*}
&=\int DL_s^t(W)(y)\bullet DL_s^t(W)(y)\,d||M_t||y\\
&=\int DL_s^t(W)(\xi_t(x))\bullet DL_s^t(W)(\xi_t(x))
\end{align*}

Now by (ii) of Theorem <ref> and (<ref>) we have
for any $W\in\cE(\gamma_s,\ubr)$ that

\begin{equation}\label{QT}
\begin{split}
&=DL_s^t(W)(\xi_t(x))\circ D\xi_t(\xi_t(x))(u)\\
&=DW(x)(u)-D^{0,0,1}\Box h_L(s,t,\xi_t(x))(u)(s-t)(W(x))\\
&\quad\quad\quad-\Box h_L(s,t,\xi_t(x))(s-t)(DW(x)(u))\\
\end{split}
\end{equation}

\MultiLin(\End(\R^n)\times\cJ\times\R^n,\End(\R^n)))
\quad i=0,1,2,$$

are such that

\begin{align*}
\Gamma_0(s,t,x)(l,v,w)(u)&=l(u);\\
\Gamma_1(s,t,x)(l,v,w)(u)&=-D^{0,0,1}\Box h_L(s,t,\xi_t(x))(u)(v)(w);\\
\Gamma_2(s,t,x)(l,v,w)(u)&=-\Box h_L(s,t,\xi_t(x))(v)(l(u))
\end{align*}

For $i,j=0,1,2$ we let

be such that

for $i,j=0,1,2$ and $(s,t,x)\in\ul{O}(\ubr/2)$. Then (<ref>) implies that

\begin{align*}
&=\int DL_s^t(W)(\xi_t(x))\bullet DL_s^t(W)(\xi_t(x))(1+\bc_s^t(0))\,d||M_s||x\\
&=\int \sum_{i,j=0}^2 Z_{i,j}(W,W)(s,t,x)(1+\bc_s^t(0)(x))\,d||M_s||x.
\end{align*}

We have

$$|Z_{0,0}(W,W)-|DW|^2|=|\bc_s^t||DW|^2\leq C_{\cW_m}|s-t||DW|^2.$$

If $i,j\in\{0,1,2\}$ and not both $i$ and $j$ are $0$ there is a constant
$c_1$ such that

$$|Z_{i,j}(W,W)(s,t,x)(1+\bc_s^t(x))|\leq c_1|s-t|(|W|^2+|DW|^2).$$

It follows that

Since $Q_s=T_s-R_s$ the Theorem now follows from Theorem <ref>
and Proposition <ref>.

Suppose $s,t\in\cJ(\epsilon_4)$. There is a constant
$C_{L,S}$ such that

$$|S_t(L_s^t(X),L_s^t(Y))-S_s(X,Y)|\leq C_{L,S}|s-t|((P_s+Q_s)(X,X)+(P_s+Q_s)(Y,Y))\quad\hbox{for $X,Y\in\cE(\gamma_s)$.}$$

Suppose $X,Y\in\cE(\gamma_s)$. Then

\begin{align*}
\end{align*}

Now apply Theorems <ref>, <ref> and <ref>.

l b

§ $\EPSILON_5$.

be such that

\begin{equation}\label{epsilon5}
2C_{L,Q}\epsilon_5<{1\over 2}.
\end{equation}

The following Lemma is a replacement for Part Four of Lemma 5.4 of [2]
which is false; to see this let $\lambda$ there equal zero. Here is a corrected version for which the inequalities derived in the proof of Part Four of
Lemma 5.4 of [2] do not suffice to prove the replacement.

There is a constant $C_{L,*}$ such that if $s,t\in\cJ(\epsilon_5)$,
$X\in\cE(\gamma_s)$, $X^*=X-E_{s,0}(X)$ and

$$S_t(L_s^t(X),Y)=\lambda P_t(L_s^t(X),Y)\quad\hbox{for $Y\in\cE(\gamma_t)$}$$

$$\kappa_s P_s(X^*,X^*)
\leq\big(|\lambda|(1+C_{L,*}|s-t|)+C_{L,*}|s-t|\big)P_s(X,X).$$

We have

$$\kappa_s P_s(X^*,X^*)\leq |S_s(X,X^*)|\leq\Gamma_1+\Gamma_2$$

\quad\hbox{and}\quad
\Gamma_2=|S_t(L_s^t(X),L_s^t(X^*))-S_s(X,X^*)|.$$

\begin{align*}
\Gamma_1&=|\lambda||P_t(L_s^t(X),L_s^t(X^*))|\\
\end{align*}

\begin{align*}
\Gamma_2
&\leq C_{L,S}|s-t|\left((P_s+Q_s)(X,X)+(P_s+Q_s)(X^*,X^*)\right)\\
&\leq C_{L,S}|s-t|(2P_s(X,X)+Q_s(X,X)+Q_s(X^*,X^*)).
\end{align*}

We estimate

\begin{align*}
&\leq Q_t(L_s^t(X),L_s^t(X))+C_{L,Q}|s-t|(P_s(X,X)+Q_s(X,X))\\
&=\lambda P_t(L_s^t(X),L_s^t(X)+(mP_t+R_t)(L_s^t(X),L_s^t(X))\\
&\quad\quad+ C_{L,Q}|s-t|(P_s(X,X)+Q_s(X,X))
\end{align*}

so, as $1-C_{L,Q}|s-t|\geq 1-C_{L,Q}(2\epsilon_5)\geq 1/2$ we have

$$Q_s(X,X)\leq 2((|\lambda|+m)(1+C_{L,P}|s-t|)+(1+C_{L,R}|s-t|))P_s(X,X).$$

Since $S_s(X_0,Y)=0$ for any $Y\in\cE(\gamma_s)$ we estimate

\begin{align*}
&\leq Q_s(X,X)+(2m+2m^2\ubb^2)P_s(X,X).
\end{align*}

The existence of $C_{L,*}$ follows.

§.§ $\Pi_s$.

For each $s\in\cJ(\epsilon_5)$ let

be $Pi_s$ orthogonal projection of $\cE(\gamma_s)$ onto $\op{rng}{\bJ_s}$.

There is a constant $C_\Pi$ such that

whenever $s\in\cJ(\epsilon_5)$ and $W\in\cE(\gamma_s)$.

Let $A_s=\bJ_s^*\circ\bJ_s$. Then $\bJ_s\circ A_s^{-1/2}$ is an isometry
from $\cJ$ to $\op{rng}{\bJ_s}$. Indeed, if $u,v\in\cJ$ we have

\begin{equation}\label{Aiso}
\begin{split}
P_s(\bJ_s\circ A_s^{-1/2}(u),&\bullet\bJ_s\circ A_s^{-1/2}(v))\\
&=\bJ_s^*\circ\bJ_s\circ A_s^{-1/2}(u)\bullet v\\
&=A_s\circ A_s^{-1/2}(u)\bullet A_s^{-1/2}(v)\\
&=u\bullet v.
\end{split}
\end{equation}

Suppose $W\in\cE(\gamma_s)$.
If $u_1,\ldots,u_{\bj}$ is an orthonormal basis for $\cJ$ and
$U_{s,i}=\bJ_s^*\circ A^{-1/2}(u_i)\in\cE(\gamma_s)$ then

$$\Pi_s(W)=\sum_{i=1}^\bj P_s(W,U_{s,i})U_{s,i}$$

which implies that

$$\Pi_t(L_s^t(W),L_s^t(W))=\sum_{i=1}^\bj P_t(L_s^t(W),U_{t,i})^2.$$

\begin{align*}
&=\int L_s^t(W)\bullet U_{t,i}\,d||M_t||\\
&=\int L_s^t(W)(\xi_t(x))\bullet U_{t,i}(\xi_t(x))(1+\bc_s^t(x))\,d||M_s||x\\
&=\int \gamma_t(\xi_t(x))(W(x))\bullet U_{t,i}(\xi_t(X))(1+\bc_s^t(x))\,d||M_s||x
\end{align*}

The Lemma now follows from the estimates occurring in the proof of Theorem <ref>.

§.§ $\epsilon_6$.

The following Lemma is, essentially, Part Six of Lemma 5.4 of

Suppose $s,t\in\cJ(\epsilon_5)$, $\Pi_s=E_{s,0}$, $\lambda\in\R$
and $\Pi_t\neq E_{t,\lambda}$. Then

$$\kappa_s(1-C_{L,P}|s-t|)\leq |\lambda|(1+C_{L,*}|s-t|)+C_{L,*}|s-t|.$$

Choose $Y$ such that $Y\in\op{rng}{E_{t,\lambda}}\sim\{0\}$ and
$\Pi_t(Y)=0$; if $\lambda=0$ this is possible since $\op{rng}{\Pi_t}\subset
E_{t,0}$ and if $\lambda\neq 0$ this is possible since $\op{rng}{\Pi_t}\subset
E_{t,0}$ which does not meet $E_{t,\lambda}$. Let $X\in\cE(\gamma_s)\sim\{0\}$ be
such that $Y=L_s^t(X)$; this is possible since $L_s^t$ carries $\cE(\gamma_s)$
isomorphically onto $\cE(\gamma_t)$.
Rescaling $Y$ if necessary may assume that $P_s(X,X)=1$.

We have

\begin{align*}
&\leq C_{L,P}|s-t|P_s(X,X)\\
\end{align*}

so that, as $\op{rng}{\Pi_s}\subset \op{rng}{E_{s,0}}$,

\begin{align*}
\kappa_s(1&-C_{L,P}|s-t|)
\end{align*}

now apply Lemma <ref>.

be such that $\kappa_0>(C_{L,P}+C_{L,*})\epsilon_6$.

$\op{rng}{\bJ_t}=\op{rng}{E_{t,0}}$ for $t\in\cJ(\epsilon_6)$.

Suppose $t\in\cJ(\epsilon_6)$ and $\Pi_t\neq E_{t,0}$.
Applying Lemma <ref> with $\lambda=0$ we would have
$\kappa_0(1-C_{L,P}|t|)\leq C_{L,*}|t|$ which would imply  $\kappa_0\leq(C_{L,S}(1+C)|t|<(C_{L,P}+C_{L,*})\epsilon_6$.

There is a constant $C_\kappa$ such that

$$s,t\in\cJ(\epsilon_6) \ \Rightarrow \ |\kappa_s-\kappa_t|\leq C_\kappa|s-t|.$$

For $u\in\cJ(\epsilon_6)$ let $\Lambda_u=\{\lambda\in\R\sim\{0\}:E_{t,\lambda}\neq 0\}$.

Suppose $u\in\cJ(\epsilon_6)$ and $\lambda\in\Lambda_u$. Since $\Pi_0=E_{0,0}$ and $\Pi_u=E_{u,0}\neq E_{u,\lambda}$  Lemma <ref> implies

Since $\kappa_0=\inf\Lambda_0$ we have

Now suppose $s,t\in\cJ(\epsilon_6)$, $\kappa_t\geq\kappa_s$ and $\lambda\in\Lambda_s$. Since $\Pi_s=E_{s,\lambda}$, Lemma <ref> implies that

Since $\kappa_s=\inf\Lambda_s$ we find that

subtracting $\kappa_s(1-C_{L,P}|s-t|)$ from both sides we obtain

\begin{align*}
&\leq{\kappa_0(1+C_{L,*}|s|)+C_{L,*}|s|\over 1-C_{L,P}|s|}
\end{align*}

so the Corollary holds.

§ PROOF OF A STATEMENT SIMILAR TO 5.4(4) OF  [2].

There is $\delta\in\R^+$ such that if $(s,Z)\in\cG$ and $|s|<\epsilon_6/2$ and $\chi^{[1]}(Z)<\delta$ there is $t\in\cJ(s,\epsilon_6/2)$ such

$$|s-t|^2\leq 2(1+C_\bJ|s|)P_s(E_{s,0}(Z),E_{s,0}(Z))$$

$$ E_{t,0}(F_s^t(Z))=0.$$

§.§ $\zeta_{Z,s}$.

For $(s,Z)\in\cG$ with $|s|<\epsilon_6/2$ we define

by requiring that

$$\zeta_{Z,s}(t)\bullet v
=\int F_s^t(Z)\bullet \bJ_t(v)\,d||M_t||
\quad\hbox{for $t\in \cJ(\epsilon_6)$ and $v\in\cJ$.}$$

Since $F_s^s(Z)=Z$ we infer from Corollary <ref> that

\begin{align*}
|\zeta_{Z,s}(s)\bullet v|
\end{align*}

\begin{equation}\label{gs}
|\zeta_{Z,s}(s)|\leq{\epsilon_6\over 4} \quad\hbox{if}\quad
\chi^0(Z)\left(1+C_\bJ{\epsilon_6\over 2}\right)^{1/2}<{\epsilon_6\over 4}.
\end{equation}

There is a constant $C_{1,\zeta}$ such that

$$||D\zeta_{Z,s}(t)-D\zeta_{0,s}(t)||\leq C_{1,\zeta}\chi^{[1]}(Z).$$

We have

\begin{align*}
\zeta_{Z,s}(t)\bullet v
&=\int F_s^t(Z)(\xi_t(F_s(Z)(x)))\bullet \bJ_t(v)(\xi_t(F_s(Z)(x)))
\cW_m(D(\xi_t\circ F_s(Z))(x))\,d||M_s||\\
&=\int \rho_t(F_s(Z)(x)\bullet \bJ_t(v)(\xi_t(F_s(Z)(x)))
\cW_m(D(\xi_t\circ F_s(Z)\,d||M_s||x\\
&=\int A_s(Z)(t,x)B_s(Z)(t,x)(v)C_s(Z)(t,x)\,d||M_s||x
%&=\int \Gamma_s(t,F_s(Z)(x),D^{0,1}\ul{\xi}(t,F_s(Z)(x))\circ DF_s(Z)(x))(v)\,d||M_s||x
\end{align*}

\begin{align*}
A_s(Z)(t,x)&=\rho_t(F_s(Z)(x))=\ul{\rho}(t,F_s(Z)(x));\\ %\ul{\rho}(t,F_s(Z(x)));\\
B_s(Z)(t,x)(v)&=\bJ_t(v)(\xi_t(F_s(Z)(x)))=-D^{1,0}\ul{\rho}(t,\ul{\xi}(t,F_s(Z)(x)))(v);\\ %\xi_tD^{1,0}\ul{\rho}(t,\xi_t(F_s(Z)(x)))(v);\\
C_s(Z)(t,x)&=\cW_m(D(\xi_t\circ F_s(Z))(x))
=\cW_m(D^{0,1}\ul{\xi}(t,F_s(Z)(x))\circ DF_s(Z)(x));
\end{align*}

\begin{align*}
&=- D(D^{1,0}\ul{\rho})(t,\ul{\xi}(t,F_s(Z)(x)))(u,D^{1,0}\xi(t,F_s(Z)(x))(u))(v);\\
&=D\cW_m(D^{0,1}\ul{\xi}(t,F_s(Z)(x))\circ DF_s(Z)(x)))\\
&\quad\quad\quad(D^{1,0}(D^{0,1}\ul{\xi})(t,F_s(Z)(x)(u)\circ DF_s(Z)(x)))).
\end{align*}

In particular,

\begin{align*}
&=- D(D^{1,0}\ul{\rho})(t,\ul{\xi}(t,x))(u,D^{1,0}\xi(t,x)(u))(v);\\
&=D\cW_m(D^{0,1}\ul{\xi}(t,x)\circ \tau_s(x))\\
&\quad\quad\quad(D^{1,0}(D^{0,1}(\ul{\xi})(t,x)(u)\circ \tau_x(x))).
\end{align*}

There are constants $c_A,c_B,c_C$ such that

$$||D^{1,0}A(Z)(t,x)-D^{1,0}A(0)(t,x)||\leq c_A\chi^{[1]}(Z);$$

$$||D^{1,0}B(Z)(t,x)-D^{1,0}B(0)(t,x)||\leq c_B\chi^{[1]}(Z);$$

$$||D^{1,0}C(Z)(t,x)-D^{1,0}C(0)(t,x)||\leq c_C\chi^{[1]}(Z).$$

Observe that $F_s(Z)(x)=f_s(x,Z(x))$ and $F_s(0)(x)=f_s(x,0)$.
This implies that $DF_s(Z)(x)=Df_s(x,Z(x))\circ(\tau_s(x),DZ(x))$
and $DF_s(0)(x)=Df_s(x,0)\circ(\tau_s(x),0)$.

There is a constant $C_{2,\zeta}$ such that

$$||D\zeta_{0,s}(t)-D\zeta_{0,s}(s)||\leq C_{2,\zeta}|s-t|
\quad\hbox{for $s\in\cJ(\epsilon_6)$ and $t\in\cJ(\epsilon_6)$.}$$

We have

\begin{align*}
\zeta_{0,s}(t)\bullet v
&=\int \rho_t(x)\bullet D^{1,0}\ul{\rho}(t,x)(v)
\cW_m(D\xi_t(x)\circ\tau_s(x))\,d||M_s||.\\
\end{align*}

It follows that

\begin{align*}
D\zeta_{0,s}&(t)(u)\bullet v\\
&=\int D^{1,0}\ul{\rho}(t,x)(u)\bullet D^{1,0}\ul{\rho}(t,x)(v)
\cW_m(D\xi_t(x)\circ\tau_s(x))\\
&\quad +\rho_t(x)\bullet D^{2,0}\ul{\rho}(t,x)(u\odot v)\cW_m(D\xi_t(x)\circ \tau_s(x))\\
&\quad +\rho_t(x)\bullet D^{1,0}\ul{\rho}(t,x)(v)D\cW_m(D\xi_t(x)\circ\tau_s(x))(D^{1,0}\ul{\xi}(t,x)(u)\circ\tau(x)\,d||M_s||x
\end{align*}

Since $\rho_s(x)=0$ and $D\xi_s(x)\circ\tau_s(x)=\tau_s(x)$ for $x\in M_s$ we have

\begin{equation}
D\zeta_{0,s}(s)(u)\bullet v
=\int D^{1,0}\ul{\rho}(s,x)(u)\bullet D^{1,0}\ul{\rho}(s,x)(v)\,d||M_s||x
\end{equation}

If $(t,x)\in\ul{N}(\ubr)$ then $\ul{\rho}(t,x)=-\Box\ul{\rho}(s,t,x)(s-t)$

=D^{1,0}\ul{\rho}(s,x)-\Box (D^{1,0}\ul{\rho})(s,t,x)(s-t).$$

We have

$$||D\zeta_{0,s}(s)-i_{\cJ}||\leq C_\bJ|s|
\quad\hbox{for $s\in\cJ(\epsilon_1)$.}$$

Suppose $s\in\cJ(\epsilon_6)$ and $u,v\in\cJ$. Then

\begin{align*}
D\zeta_{0,s}&(s)(u)\bullet v-u\bullet v\\
&=\int D^{1,0}\ul{\rho}(s,x)(u)\bullet D^{1,0}\ul{\rho}(s,x)(v)\,d||M_s||x
&=\int \bJ_s(u)\bullet \bJ_s(v)\,d||M_s||
\end{align*}

Now apply Corollary <ref>.

§.§ Finding a zero of $\zeta_{Z,s}$.

Let $B=\{u\in\cJ:|u-s|\leq\zeta_6/2\}\subset\cJ(\epsilon_6)$; let $g:B\rightarrow\cJ$
be such that $g(u)=\zeta_{Z,s}(u)$ for $u\in B$; and let

It follows from Lemmas (<ref>), (<ref>) <ref> that

\leq||D\zeta_{Z,s}(t)-D\zeta_{0,s}(s)||+||D\zeta_{0,s}(s)-i_{\cJ}||
\leq C_{1,\zeta}\chi^{[1]}(Z)+C_{2,\zeta}|s-t|+C_\bJ|s|.$$

Suppose $(\ref{gs})$; then

$$|g(s)|<{\epsilon_6\over 4}.$$

\begin{equation}\label{Lipp}
C_{1,\zeta}\chi^{[1]}(Z)+C_{2,\zeta}|s-t|+C_\bJ|s|\leq{1\over 2}.
\end{equation}

$p:B\rightarrow\cJ$ be such that $p(u)=(g(u)-g(s))-(u-s)$
for $u\in B$. Then $p(s)=0$ and $\op{Lip}{p}\leq 1/2$.
Let $C:B\rightarrow\cJ$ be such that
$C(u)=s-g(s)-p(u)$ for $u\in B$. Then $\op{Lip}{C}=\op{Lip}{p}\leq 1/2$.

Suppose $u\in B$. Then

\leq |g(s)|+\op{Lip}{p}|u-s|\leq\epsilon_6-|s|$$

so $C[B]\subset B$. Since $B$ is closed and $\op{Lip}{C}\leq 1/2<1$, $C$ has a unique fixed point $t\in B$. Since
$t=C(t)=s-g(s)-p(t)=s-g(s)-(p(t)-p(s)$ we have

$$|s-t|=|g(s)+(p(t)-p(s))|\leq |g(s)|+{|s-t|\over 2}$$

$$|s-t|\leq 2|g(s)|.$$

Suppose $u\in\cJ$ and $|u|=1$. Then

$$g(s)\bullet u
=\int Z\bullet \bJ_s(u)\,d||M_s||

so by Corollary <ref> we have

\begin{align*}
|g(s)\bullet u|^2&\leq P_s(E_{s,0}(Z),E_{s,0}(Z))P_s(\bJ_s(u),\bJ_s(u))\\
&\leq P_s(E_{s,0}(Z),E_{s,0}(Z))(P_0(\bJ_0(u),\bJ_0(u))+C_\bJ|s|)\\
&\leq P_s(E_{s,0}(Z),E_{s,0}(Z))(1+C_\bJ|s|)
\end{align*}

so that

$$|s-t|^2\leq2|g(s)|\leq P_s(E_{s,0}(Z),E_{s,0}(Z))(1+C_\bJ|s|).$$

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