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\begin{document}
\def\ftnt{Supported in part by NSF Grant DMS01-03901.}

\def\authcomp{H\'ector J. Sussmann\footnote{\ftnt}\\
Department of Mathematics\\
Rutgers, the State University of New Jersey\\
Hill Center---Busch Campus\\
110 Frelinghuysen Road, Piscataway, NJ\\
08854-8019, USA\\
$\,$\\
{\it E-mail: sussmann@math.rutgers.edu}\\
{\it Web site: http://www.math.rutgers.edu/$\widetilde{\,\,}$sussmann}
}




\title{High-order open mapping theorems}
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%\author{H\'ector J. Sussmann}
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% Department of Mathematics; Rutgers, the State University of New Jersey
% Hill Center---Busch Campus, 110 Frelinghuysen Road, Piscataway, NJ
% 08854-8019, USA
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% Supported in part by NSF Grant DMS01-03901.

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%\chapter{High-order open mapping theorems}
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\section*{Abstract}


The well-known finite-dimensional first-order 
open mapping theorem
says that a \mbox{continuous} map with a finite-dimensional target is
open at a point if its differential at that point exists and is
surjective.  An identical result, due to Graves, is true when the
target is infinite-dimensional, if ``differentiability'' is replaced
by ``strict differentiability.'' We prove general theorems in which
the linear approximations involved in the usual concept of
differentiability is replaced by an approximation by a map which is
homogeneous relative to a one-parameter group of dilations, and the
error bound in the approximation involves a ``homogeneous pseudonorm''
or a ``homogeneous pseudodistance,'' rather than the ordinary norm.
We outline how these results can be used to derive sufficient
conditions for openness involving higher-order derivatives, and carry
this out in detail for the second-order case.


% And so on and so forth.

\section{Introduction}

\noindent Open mapping theorems (abbr. OMTs) are key tools in the derivation
of \mbox{necessary}
conditions for an optimum in optimization theory, and in particular
in optimal 
\mbox{control}. For example, the usual Lagrange multipliers rule is 
a trivial corollary of the following OMT.

\begin{theorem}\label{linearfindim}
Let $X$, $Y$ be normed spaces, let $\Omega$
be an open subset of $X$, and let $F:\Omega\mapsto Y$ be a continuous
map. Let $x_*\in\Omega$ be such that $F$ is Fr\'echet differentiable
at $x_*$ and the differential $DF(x_*)$ is a surjective linear
map from $X$ to $Y$. Assume that $Y$ is finite-dimensional.
Then $F$ is open at $x_*$, that is, $F$ maps neighborhoods of $x_*$ to 
neighborhoods of $F(x_*)$.
\end{theorem}

\noindent(To deduce the Lagrange rule from Theorem \ref{linearfindim},
let us consider the \mbox{problem} of minimizing the quantity $f_0(x)$ 
subject to finitely many equality constraints $f_1(x)=\cdots=f_m(x)=0$, 
where $f_0,f_1,\ldots,f_m$ are continuous real-valued \mbox{functions}
on an open subset $\Omega$ of a normed space $X$, and 
assume that $x_*$ is a \mbox{solution}. We then define 
$F:\Omega\mapsto\rr^{m+1}$ by letting
$$
F(x)=(f_0(x),f_1(x),\ldots,f_m(x))\,\,\,\,{\rm for}\,\,\,\,x\in\Omega\,,
$$
and observe 
that, if we let $y_*(\eps)=(f_0(x_*)-\eps,0,\ldots,0)\in\rr^{m+1}$
for $\eps\in\rr$, then $F(x_*)=y_*(0)$, and 
$y_*(\eps)\notin F(\Omega)$ whenever $\eps>0$, so $F$ is not
open at $x_*$. If $f_0,f_1,\ldots,f_m$ are Fr\'echet differentiable
at $x_*$, then $F$ is  Fr\'echet differentiable at $x_*$, so 
Theorem \ref{linearfindim} implies that  $DF(x_*)$ is not surjective.
Therefore the $m+1$ vectors $\nabla f_0(x_*),\nabla f_1(x_*),\ldots,
\nabla f_m(x_*)$ are not \mbox{linearly} \mbox{independent}. 
It follows that there exist
numbers $\lambda_0,\lambda_1,\ldots,\lambda_m$ that are not all
zero and satisfy $\sum_{i=0}^m\lambda_i\nabla f_i(x_*)=0$.)


Theorem \ref{linearfindim} is only valid if $Y$ is finite-dimensional.
(Precisely: if $Y$ is any normed space, then $Y$ is finite-dimensional
if and only if it has the property that whenever $F:Y\mapsto Y$ is a
continuous map such that $F(0)=0$, $F$ is Fr\'echet differentiable at
$0$, and the differential $DF(0)$ is the identity map of $Y$, 
it follows that $F$
is open at $0$.)  On the other hand, L.M.~Graves proved in \cite{graves}
(cf. also Dontchev \cite{dontchev}) 
the following infinite-dimensional result. 

\begin{theorem}\label{lineargraves}
Let $X$, $Y$ be Banach spaces, let $\Omega$
be an open subset of $X$, and let $F:\Omega\mapsto Y$ be a continuous
map. Let $x_*\in\Omega$ be such that $F$ is strictly differentiable
at $x_*$ and the differential $DF(x_*)$ is a surjective linear
map from $X$ to $Y$. Then $F$ is open at $x_*$. 
\end{theorem}

\noindent (Graves actually proved a stronger result, in which 
strict differentiability with a surjective differential
is replaced by the weaker condition that there exist
a surjective bounded linear map $L:X\mapsto Y$ such that 
\begin{equation}\label{limofgraves}
\limsup_{x\to x_*,x'\to x_*,x\neq x'}
\frac{\|F(x)-F(x')-L(x-x')\|}{\|x-x'\|}<\|L^{-1}\|^{-1}\,,
\end{equation}
where $\|L^{-1}\|$ is the infimum of the numbers $C$ such that
for every $y\in Y$ there exists $x\in X$ such that
$L(x)=y$ and $\|x\|\le C\|y\|$. 
The definition of ``strict differentiability'' is as follows:
$F$ is strictly differentiable at $x_*$  with differential 
$L$ if the left-hand side of (\ref{limofgraves}) vanishes.)

The purpose of this note is to present ``higher-order'' sufficient
conditions for openness at a point, generalizing
Theorems \ref{linearfindim} and \ref{lineargraves}.

\begin{remark}
{\rm Stronger results can also be proved, in which 
\begin{itemize}
\item[1.] in the infinite-dimensional case, the analogue of strict 
differentiability is replaced by the analogue of condition 
(\ref{limofgraves}); 
\item[2.] in the finite-dimensional case, the analogue of 
differentiability is replaced by the analogue of the condition 
that 
$$
\limsup_{x\to x_*,x\neq x_*}
\frac{\|F(x)-F(x_*)-L(x-x_*)\|}{\|x-x_*\|}<\|L^{-1}\|^{-1}\,;
$$
\item[3.] $F$ is only required to be defined on a ``conic 
neighborhood'' of $x_*$, i.e., a set $S$  of the form 
$$
S=S_X(x_*,\eps,C)\equaldef\{x:\|x-x_*\|<\eps\wedge x-x_*\in C\}\,,
$$
where $\eps>0$ and  $C$ is a convex cone in $X$ with nonempty interior such
that $0\in C\,$;
\item[4.] the conclusion says that 
the map $F$ is ``directionally  open'' at $x_*$ in the \mbox{direction} of a
\mbox{given} vector $v_*\in Y$, provided that $v_*\in{\rm int}\,LC$.
(\mbox{Precisely}: for every positive $\eps$ there exists a
positive $\delta$ such that the image $F(S_X(x_*,\eps,C))$ \mbox{contains}
the set $S_Y(F(x_*),\delta,D)$ for some convex cone $D$ in $Y$ such that
$v_*\in{\rm int}\,D$.)
\end{itemize} 
In this paper, however, we will only carry out the simpler task 
of generalizing the non-directional theorems \ref{linearfindim} and 
\ref{lineargraves}.}\proofend
\end{remark}

Naturally, ``higher-order'' means ``involving higher-oder derivatives.''
A map $F$ of class $C^m$ has a Taylor approximation 
$$
F(x_*+h)\sim F(x_*)+P_1(h)+\frac{1}{2}P_2(h,h)+\cdots
+\frac{1}{m!}P_m(h,h,\ldots,h)\,,
$$
where $P_j=D^{j}F(x_*)$ for $j=1,\ldots,m$, so 
each $P_j$ is a $Y$-valued, continuous,  \mbox{symmetric}, multilinear 
map  defined on the Cartesian product $X^j$ of $j$ copies of $X$.
\mbox{Openness} at $x_*$ follows from the first-order theorems
if the linear map $P_1$ is surjective. When $P_1$ is not surjective,
the high-order results will give openness if the missing directions
in the image $P_1X$ can somehow be realized as image directions
using higher-order derivatives. On the other hand, 
if the ``first-order effect'' $P_1(h)$ of a particular direction
$h$ is nonzero, then this effect will dominate whatever contributions
$h$ may make through higher-order terms.  This suggests that 
one should only look at the second-order effects $P_2(h,h)$ of vectors
$h$ belonging to the kernel $K_2$ of $P_1$.  So, if $m=2$, we 
take $K_1=X$, $K_2={\rm ker}\,P_1$, and consider the approximation
\begin{eqnarray*}
F(x_*+h_1+h_2)&\sim&F(x_*)+P_1(h_1)+P_1(h_2)\\
&\,&\,\,+\frac{1}{2}P_2(h_1,h_1)+\frac{1}{2}P_2(h_2,h_2)+
P_2(h_1,h_2)\,,
\end{eqnarray*}
where $h_1\in K_1$, $h_2\in K_2$. The remainder is clearly 
$o(\|h_1\|^2+\|h_2\|^2)$, which is in particular 
$o(\|h_1\|+\|h_2\|^2)$. Furthermore, the terms 
$\frac{1}{2}P_2(h_1,h_1)$ and $P_2(h_1,h_2)$ are 
$o(\|h_1\|)$, so they are $o(\|h_1\|+\|h_2\|^2)$ as well,
and can be absorbed 
into the remainder. Finally, $P_1(h_2)$ vanishes, because 
$h_2\in{\rm ker}\,P_1$, and we end up with the approximation
\begin{eqnarray}
F(x_*+h_1+h_2)&=&F(x_*)+P_1(h_1)+\frac{1}{2}P_2(h_2,h_2)+
o(\|h_1\|+\|h_2\|^2)\nonumber\\
&\,&\,\,\,\qquad{\rm as}\,\,\,h_1\to 0,\,
h_2\to 0,\,h_1\in K_1,\,h_2\in K_2\,.\label{secondap}
\end{eqnarray}
Such an approximation is valid under more general situations,
even if $F$ is not of class $C^2$ near $x_*$. (For example, 
if $F:\rr^3\mapsto\rr$ is given by 
$$
F(x,y,z)=x+|x|^{3/2}(1+|y|^{1/2})+y^2-z^2\,,
$$ 
then $F(x,y,z)=x+y^2-z^2+o(|x|+y^2+z^2)$, 
so (\ref{secondap}) holds with $K_1=\rr^3$, $K_2=\{0\}\times\rr^2$, and 
obvious choices of $P_1$ and $P_2$.) If we define
$\cx=K_1\times K_2$, and write
\begin{eqnarray*}
\cf(h_1,h_2)&=&F(x_*+h_1+h_2)-F(x_*)\,,\\
\cg(h_1,h_2)&=&P_1(h_1)+\frac{1}{2}P_2(h_2,h_2)\,,\\
\nu(h_1,h_2)&=&\|h_1\|+\|h_2\|^2\,, 
\end{eqnarray*}
for $h_1\in K_1$, $h_2\in K_2$, then (\ref{secondap}) says that
\begin{equation}\label{secondapap}
\cf(\xi)=\cg(\xi)+o(\nu(\xi)) 
\,\,\,\qquad{\rm as}\,\,\,\xi\to 0\,,\,\,\,\,\,\xi\in\cx\,.
\end{equation}
Moreover, if we let 
$$
\delta_t(h_1,h_2)=(th_1, t^{1/2}h_2)\qquad{\rm for}\qquad
h_1\in K_1,\,h_2\in K_2,\,t>0\,,
$$
then
\begin{itemize}\item[]\begin{itemize}
\item[(HA.1)] {\em $\pmb{\delta}=\{\delta_t\}_{t>0}$ is a continuous 
one-parameter group of dilations on $\cx$,}
\item[(HA.2)] {\em $\cg$ is $\pmb{\delta}$-homogeneous, in  the sense
that $\cg(\delta_t(\xi))=t\cg(\xi)$ whenever $\xi\in\cx$ and
$t>0$,}
\item[(HA.3)] {\em $\nu$ is $\pmb{\delta}$-homogeneous, in  the sense
that $\nu(\delta_t(\xi))=t\nu(\xi)$ whenever $\xi\in\cx$ and
$t>0$,}
\item[(HA.4)] {\em $\nu$ is a ``pseudonorm on $\cx$,'' that is,
a continuous nonnegative real-valued \mbox{function} 
that satisfies the conditions 
(i)~$\nu(\xi)=0\Longleftrightarrow \xi=0$ and
(ii)~$\lim_{\|\xi\|\to+\infty}\nu(\xi)=+\infty$.}
\end{itemize}\end{itemize}
So, when we rewrite (\ref{secondap}) as (\ref{secondapap}), we find that 
we are really dealing with an {\em approximation of a map $\cf$
by another map $\cg$ which is homogeneous with 
\mbox{respect} to a 
continuous one-parameter group of dilations}. Furthermore,
{\em the approximation is of exactly the same kind as the approximation
by a linear map involved in the usual concept of differentiability,
except that the ordinary norm on $\cx$ is replaced 
by a dilation-homogeneous pseudonorm}.

Similar considerations apply to higher-order approximations. 
For example, if $F\in C^3$, and we choose three linear subspaces
$K_1$, $K_2$, $K_3$ of $X$, then 
\begin{eqnarray*}
F(x_*+h_1+h_2+h_3)&=&F(x_*)+P_1(h_1)+P_1(h_2)+P_1(h_3)\\
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!
+\frac{1}{2}\Big(P_2(h_1,h_1)+P_2(h_2,h_2)+P_2(h_3,h_3)\Big)\\
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!
+P_2(h_1,h_2)+P_2(h_1,h_3)+P_2(h_2,h_3)+P_3(h_1,h_2,h_3)\\
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!
+\frac{1}{6}\Big(P_3(h_1,h_1,h_1)+P_3(h_2,h_2,h_2)+P_3(h_3,h_3,h_3)\Big)\\
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!
+\frac{1}{2}\Big(P_3(h_1,h_2,h_2)+P_3(h_1,h_1,h_2)
+P_3(h_1,h_3,h_3)\\
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\qquad+P_3(h_1,h_1,h_3)+P_3(h_2,h_3,h_3)+P_3(h_2,h_2,h_3)\Big)\\
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!
+o(\|h_1\|^3+\|h_2\|^3+\|h_3\|^3)\\
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
{\rm as}\,\,\,h_1\to 0,\,
h_2\to 0,\,h_3\to 0,\,h_1\in K_1,\,h_2\in K_2,\,h_3\in K_3\,.
\end{eqnarray*}
%\label{thirdap}
If we choose the $K_i$ such that $K_1=X$, and $K_2$ and $K_3$ are
subsets of ${\rm ker}\,P_1$
(so that $P_1(h_2)=P_1(h_3)=0$), and absorb into the remainder
all the terms that are obviously $o(\|h_1\|+\|h_2\|^2+\|h_3\|^3))$, we find 
\begin{eqnarray*}
F(x_*+h_1+h_2+h_3)&=&F(x_*)+P_1(h_1)+
\frac{1}{2}\Big(P_2(h_2,h_2)+P_2(h_3,h_3)\Big)\\
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
+P_2(h_2,h_3)+\frac{1}{6}P_3(h_3,h_3,h_3)
+o(\|h_1\|+\|h_2\|^2+\|h_3\|^3)\,.
\end{eqnarray*}
(The term $P_3(h_2,h_3,h_3)$ is eliminated because
\begin{eqnarray*}
\|P_3(h_2,h_3,h_3)\|&\le&\|P_3\|\,\|h_2\|\,\|h_3\|^2
\le\|P_3\|(\|h_2\|^{7/3}+(\|h_3\|^2)^{7/4})\\
&=&\|P_3\|(\|h_2\|^{7/3}+\|h_3\|^{7/2})=o(\|h_2\|^2+\|h_3\|^3)\,,
\end{eqnarray*}
using the fact that the inequality $ab\le a^p+b^q$ holds 
if $a\ge 0$, $b\ge 0$,
$p>1$, $q>1$, and $(1/p)+(1/q)=1$, and taking $p=7/3$, $q=7/4$.)

If we require in addition that ``$K_3\subseteq{\rm ker}\,P_2$,'' in the 
precise sense that every $h\in K_3$ must satisfy $P_2(h,h')=0$ 
whenever $h'\in K_2$, then we find 
\begin{eqnarray*}
F(x_*+h_1+h_2+h_3)&=&F(x_*)+P_1(h_1)+
\frac{1}{2}P_2(h_2,h_2)\\
&\,&\,\,+\frac{1}{6}P_3(h_3,h_3,h_3)+o(\|h_1\|+\|h_2\|^2+\|h_3\|^3)\\
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
{\rm as}\,\,\,h_1\to 0,\,
h_2\to 0,\,h_3\to 0,\,h_1\in K_1,\,h_2\in K_2,\,h_3\in K_3\,.
\end{eqnarray*}
So, once again, we find that (\ref{secondapap}) holds, if we 
define $\cx=K_1\times K_2\times K_3$, and let 
\begin{eqnarray*}
\cf(h_1,h_2,h_3)&=&F(x_*+h_1+h_2+h_3)-F(x_*)\,,\\
\cg(h_1,h_2,h_3)&=&P_1(h_1)+\frac{1}{2}P_2(h_2,h_2)
+\frac{1}{6}P_3(h_3,h_3,h_3)\,,\\
\nu(h_1,h_2,h_3)&=&\|h_1\|+\|h_2\|^2+\|h_3\|^3\,, 
\end{eqnarray*}
for $h_1\in K_1$, $h_2\in K_2$,  $h_3\in K_3$. In addition, 
if we let 
$$
\delta_t(h_1,h_2,h_3)=(th_1, t^{1/2}h_2,t^{1/3}h_3)\quad{\rm for}\quad
h_1\in K_1,\,h_2\in K_2,\,h_3\in K_3,\,t>0\,,
$$
then the four homogeneous approximation conditions HA.1-4 hold.

Our higher-order generalization of Theorem \ref{linearfindim} 
involves a condition closely \mbox{resembling} the usual formula 
$\cf(\xi)=\cl(\xi)+o(\|\xi\|)$ that \mbox{characterizes} ordinary 
differentiability of $\cf$ at $0$. The crucial difference
is that  (a)~the linear map $\cl$ is  \mbox{replaced}
by a map $\cg$ which is homogeneous with respect
to some suitable group of dilations, and (b)~a  pseudonorm
$\nu$ is substituted 
for the ordinary norm. The requirement that $\cl$ be surjective,
which occurs in the first-order theorems, will be replaced by the condition
that $\cg$ have a surjective differential at some point $\xi_*\in\cx$
such that $\cg(\xi_*)=0$. 
We will show that these conditions imply that $\cf$ is open at $0$ if 
the target space $\cy$ of $\cf$ is finite-dimensional, generalizing 
Theorem \ref{linearfindim}. To get the generalization of 
Theorem \ref{lineargraves}, in which 
$\cy$ is allowed to be infinite-dimensional, 
we  will have use the appropriate ``strict'' 
analogues of differentiability: the map $\cg$ will have to
be assumed to be strictly differentiable at $\xi_*$, and the approximation
formula (\ref{secondapap}) will have to be replaced by 
\begin{equation}\label{secondapapstrict}
\ce(\xi)-\ce(\xi')=o(\nu(\xi,\xi')) 
\,\,\,\qquad{\rm as}\,\,\,\xi\to 0\,,\,\xi'\to 0\,,
\,\,\,\,\,\xi\in\cx\,,\,\xi'\in\cx\,,
\end{equation}
where the error $\ce$ is defined by $\ce(\xi)=\cf(\xi)-\cg(\xi)$, and
$\nu$ is a ``homogeneous Lipschitz-bounded pseudodistance,''
in a sense to be defined below.

From the general theorems involving homogeneous approximations, it will
then  be possible to derive high-order open mapping theorems
by means of the construction sketched above, using  the obvious 
fact that, when $\cf$ is defined in terms of $F$ as we have done
in our second- and third-order examples, then if $\cf$ is open at $0$
it follows that $F$ is open at $x_*$.

This will be done here for the second-order case in \S\ref{secseco},
where it will be shown that our sufficient conditions for openness
apply in particular when the Hessian of the map has a regular zero,
and in  the cases considered by Avakov in \cite{avakov-a,avakov-b}. 


\section{Preliminaries}\label{prelims}

\noindent{\bf Metric spaces, normed spaces, balls, openness at a
point.} If $X$ is a metric space with distance $d$, $x$ is a point of
$X$, and $r\ge 0$, we will use $\mathbb{B}_X(x,r)$ to denote the
closed $r$-ball with center $x$, i.e., the set 
$\{x'\in X:d(x',x)\le r\}$.

We use the word ``neighborhood'' in the usual sense of
point set topology: if $X$ is a topological space, and
$x_*\in X$, a {\em neighborhood} of $x_*$ in $X$ is 
a subset $U$ of $X$ such that $x_*$ is an interior
point of $U$.
\begin{definition}
{\rm If $X$, $Y$ are topological spaces, $x_*\in X$, and 
$F:X\mapsto Y$ is a map, then
$F$ is said to be {\em open at $x_*$} if
whenever $U$ is a  neighborhood of $x_*$ in $X$, 
it follows that the image $F(U)$ is a  neighborhood of $0$ in 
$Y$.} \proofend
\end{definition}

\noindent All linear spaces will be over $\rr$, the field of real numbers.  If
$X$ is a normed linear space, then we will
write $\mathbb{B}_X(r)$  instead of $\mathbb{B}_X(0,r)$.
If $X$, $Y$ are normed linear spaces, then ${\rm Lin}(X,Y)$
will \mbox{denote} the space of all bounded linear maps from $X$ to $Y$. If
$A\in{\rm Lin}(X,Y)$ and $x\in X$, then we will use interchangeably
the expressions $Ax$, $A\cdot x$ and $A(x)$ to denote the value of $A$
at $x$.  Convergence in ${\rm Lin}(X,Y)$ is uniform
\mbox{convergence}, i.e., convergence relative to the operator norm
\mbox{${\rm Lin}(X,Y)\ni A\mapsto\|A\|\,\equaldef\,\sup\{ \|Ax\|:x\in
X,\|x\|\le 1\}$.}

\medbreak\noindent{\bf Differentiability and strict differentiability.}
The word ``differentiable'' will only be used 
to refer to maps between normed spaces, and will mean ``Fr\'echet 
differentiable.'' That is, if we assume that
\begin{itemize}
\item[(A)] {\em $X$ and $Y$ are normed spaces, $\Omega$ is an open subset 
of $X$, $x_*\!\in\!\Omega$, \mbox{$F:\Omega\!\mapsto\!Y$} is a map, and
$L:X\mapsto Y$ is a bounded linear map,}
\end{itemize}
\noindent then we say that {\em $F$ is differentiable at $x_*$ with 
differential $L$} if
\begin{equation}\label{wecantake-ax}
\lim_{x\to x_*, x\neq x_*}\frac{\|F(x)-F(x_*)-L.(x-x*)\|}
{\|x-x_*\|}=0\,.
\end{equation}
A stronger concept of differentiability is 
``strict differentiability,'' defined as follows.

\begin{definition}\label{defstrict}
{\rm Let $X$, $Y$, $\Omega$, $x_*$, $F$, $L$ be such that (A) above holds.
We say that {\em $F$ is strictly differentiable at $x_*$ with 
strict differential $L$} if  the equality
\begin{equation}\label{wecantake}
\lim_{x\to x_*,x'\to x_*,x\neq x'}\frac{\|F(x)-F(x')-L.(x-x')\|}
{\|x-x'\|}=0\,
\end{equation}
holds.} \proofend
\end{definition}
Clearly, if $F$ is strictly differentiable at $x_*$ with 
strict differential $L$, then $F$ is differentiable at $x_*$ with 
differential $L$, since we can obtain (\ref{wecantake-ax})
by taking $x'=x_*$ in (\ref{wecantake}).

\medbreak\noindent{\bf Dilations, pseudonorms, pseudodistances,
homogeneous maps.} 
\begin{definition}\label{defdila}
{\rm  Assume that $\cx$ is a normed real linear space. 
A {\em continuous one-parameter group of dilations} on $\cx$ is a family
$\pmb{\delta}=\{\delta_t\}_{t>0}$ of bounded linear maps 
$\delta_t:\cx\mapsto\cx$ such that 
\begin{itemize}
\item[D1.] $\delta_1$ is the identity map $\id_{\cx}$ of $\cx$,
\item[D2.] $\delta_t\circ\delta_s=\delta_{ts}$ 
whenever $t>0$ and $s>0$, 
\item[D3.] the map $]\,0,+\infty[\,\ni t\mapsto\delta_t\in{\rm Lin}(\cx,\cx)$ 
is continuous with respect to the uniform operator norm on  
${\rm Lin}(\cx,\cx)$ (that is, $\lim_{t\to s}\|\delta_t-\delta_s\|=0$
whenever $s>0$), 
\item[D4.] $\lim_{t\downarrow 0}\|\delta_t\|=0\,$.\proofend
\end{itemize}}
\end{definition}

\begin{lemma}\label{lemdiltoinf}
If $\cx$ is a normed linear space and $\pmb{\delta}=\{\delta_t\}_{t>0}$ 
is a continuous one-parameter group of dilations on $\cx$, then
\begin{equation}\label{diltoinf}
\lim_{t\to+\infty}\|\delta_t(x)\|=+\infty\qquad{\rm for\,\,\,
all}\qquad x\in\cx\backslash\{0\}\,,
\end{equation}
and 
\begin{equation}\label{diltoinf-a}
{\rm the\,\,map}\,\,\,]\,0,+\infty\,[\,\ni t\mapsto\delta_t(x)\in\cx
\,\,\text{\rm is\,\,one-to-one\,\,whenever}\,\,x\in 
X\backslash\{0\}\,.
\end{equation}
\end{lemma}

\begin{proof}
If $x\neq 0$ and it is not true that 
$\lim_{t\to+\infty}\|\delta_t(x)\|=+\infty$, then 
there exists a sequence $\{t_j\}_{j=1}^\infty$ such that
$\lim_{j\to\infty}t_j=+\infty$ while, on the other hand,
the set $S=\{\delta_{t_j}(x):j\in\nn\}$
is bounded. Then, if we let $\xi_j=\delta_{t_j}(x)$, we can pick a constant
$C$ such that $\|\xi_j\|\le C$ for all $j$, and conclude that 
$x=\delta_{t_j^{-1}}(\xi_j)$, so $\|x\|\le C\|\delta_{t_j^{-1}}\|$.
Since $\|\delta_{t_j^{-1}}\|\to 0$---because $t_j^{-1}\to 0$---it follows
that $x=0$. This completes the proof of (\ref{diltoinf}).

To prove (\ref{diltoinf-a}), we observe that 
if $x\neq 0$, $\delta_t(x)=\delta_s(x)$, and $0<t<s$, then
$\delta_\tau(x)=x$, if $\tau=s/t$. Then $\tau>1$, and 
$\delta_{\tau^k}(x)=x$ for $k=1,2,\ldots$, so 
the norm $\|\delta_{\tau^k}(x)\|$ does not go to $+\infty$ as $k\to\infty$ but
$\tau^k\to+\infty$ as $k\to\infty$, contradicting (\ref{diltoinf}).
\end{proof}


\begin{definition}
{\rm Assume that $\cx$, $\pmb{\delta}$ are as in Definition \ref{defdila}.
A {\em $\pmb{\delta}$-pseudonorm} is a continuous map 
$\nu:\cx\mapsto\mathbb{R}$ such that
(1)~$\nu(x)\ge 0$ for all $x\in\cx$, 
\mbox{(2)~$\nu(x)=0 \Longleftrightarrow x=0$}, 
(3)~$\lim_{\|x\|\to+\infty}\nu(x)=+\infty$,
and (4)~$\nu(\delta_t(x))=t\nu(x)$ whenever $x\in\cx$, $t\in\rr$, 
and $t>0$.}\proofend
\end{definition}

\begin{definition}
{\rm Assume that the space $\cx$ and the dilation group
$\pmb{\delta}$ are as in Definition \ref{defdila}.
A {\em $\pmb{\delta}$-pseudodistance} is a continuous map 
$\nu:\cx\times\cx\mapsto\mathbb{R}_+$  such that 
(1)~$\nu(x,x')=\nu(x',x)\ge 0$ for all $x,x'\in\cx$, 
(2)~$\nu(x,x')=0 \Longleftrightarrow x=x'$, 
(3)~$\lim_{\|x\|\to+\infty}\nu(x,0)=+\infty$,
and (4)~$\nu(\delta_t(x),\delta_t(x'))=t\nu(x,x')$ 
whenever $x,x'\in\cx$, $t\in\rr$, and $t>0$.}\proofend
\end{definition}

If  $\nu:\cx\times\cx\mapsto\rr$ is a $\pmb{\delta}$-pseudodistance on $\cx$,
we define, for each positive number $R$, 
\begin{equation}
\kappa_\nu(R)\,\,\equaldef\,\,\sup
\Biggl\{\,\frac{\nu(x,x')}{\|x-x'\|}:x,x'\in
\mathbb{B}_{\cx}(R)\,,\,x\neq x'\,\Biggr\}\,,\label{chcr-aaa-yyy}
\end{equation}
Then 
\begin{equation}\label{alliffsome}
\kappa_\nu(R)<\infty\,\,\text{\rm for some positive}\,\,R
\Longleftrightarrow
\kappa_\nu(R)<\infty\,\,\text{\rm whenever}\,\,R>0.
\end{equation}
(Indeed, if $R_0$ is such that $R_0>0$ and $\kappa_\nu(R_0)<\infty$, then
there exists a positive $\bar t$ such that $R_0\|\delta_{\bar t}\|<R$, since 
$\lim_{t\downarrow 0}\|\delta_t\|=0$. Then 
$\delta_{\bar t}(\mathbb{B}_\cx(R))\subseteq
\mathbb{B}_\cx(R_0)$. If $x,x'\in\mathbb{B}_\cx(R)$, then
\begin{eqnarray*}
\nu(x,x')&=&\bar t^{-1}\nu(\delta_{\bar t}(x),\delta_{\bar t}(x'))\\
&\le&\bar t^{-1}\kappa_\nu(R_0)\|\delta_{\bar t}(x)-\delta_{\bar t}(x')\|\\
&\le&\bar t^{-1}\kappa_\nu(R_0)\|\delta_{\bar t}\|\,\|x-x'\|\,,
\end{eqnarray*}
so $\kappa_\nu(R)\le\bar t^{-1}\kappa_\nu(R_0)\|\delta_{\bar t}\|<+\infty\,$.)
\begin{definition}
{\rm A pseudodistance $\nu:\cx\times\cx\mapsto\rr$ is
{\em Lipschitz-bounded} if 
$\kappa_\nu(R)<\infty$ for some (and hence every) positive
number $R$.}\proofend
\end{definition}
\begin{definition}\label{defhomog}
{\rm Assume that $\cx$, $\pmb{\delta}$ are as in Definition \ref{defdila},
$\cy$ is a normed \mbox{linear} space, and $\cg:\cx\mapsto\cy$ is a map.
We say that $\cg$ is {\em $\pmb{\delta}$-homogeneous} if
\mbox{$\cg(\delta_t(x))=t\,\cg(x)$} whenever
$x\in\cx$ and $t\ge 0$.}\proofend
\end{definition}

\medbreak\noindent{\bf Regular zeros.}
If $X$, $Y$ are normed spaces, $\Omega$ is open in $X$,
and $F$ is a map from $\Omega$ to $Y$, then a {\em regular zero}
(resp. a {\em strictly regular zero})
of $F$ is a point $x_*\in\Omega$ such that 
$F(x_*)=0$, $F$ is Fr\'echet differentiable 
(resp. strictly Fr\'echet differentiable) at $x_*$, and
the differential $DF(x_*):X\mapsto Y$ is surjective.

\bigbreak

\section{The finite-dimensional theorem}

\begin{theorem}\label{mainfin}
Assume that $\cx$ and $\cy$ are real linear normed spaces, $\cy$ is
finite-dimensional, $\pmb{\delta}=\{\delta_t\}_{t>0}$ is a
\mbox{continuous} one-parameter group of dilations of $\cx$, 
$\nu:\cx\mapsto\mathbb{R}_+$ is a 
\mbox{$\pmb{\delta}$-pseudonorm}, and 
$\cg:\cx\mapsto\cy$ is a \mbox{continuous} 
\mbox{$\pmb{\delta}$-homogeneous}
map. Let $\Omega$ be an open
subset of $\cx$ such that $0\in\Omega$, and let $\cf:\Omega\mapsto\cy$
be a continuous map such that $\cf(0)=0$ and
\begin{equation}\label{limisfin-ux-a}
\lim_{\xi\to 0}\frac{\|\cf(\xi)-\cg(\xi)\|}{\nu(\xi)}=0\,.
\end{equation}
Assume that $\cg$ has a regular zero.  Then $\cf$ is open at $0$.
\end{theorem} 

\begin{proof}
Let $V$ be a neighborhood of $0$ in $\cx$ such that $V\subseteq\Omega$.
Let $R$ be such that $R>0$ and $\mathbb{B}_{\cx}(2R)\subseteq V$.
%and $4\sigma\nu(x_*)<1$, where 
%$$
%\sigma\equaldef\sup\Big\{\,\frac{\|\cf(x)-\cg(x)\|}{\nu(x)}:x\in
%\mathbb{B}_{\cx}(2R)\,\}\,.
%$$
Let $\xi_*$ be a regular zero of $\cg$. For $t>0$, 
let $\xi_{*,t}=\delta_t(\xi_*)$.  
Let $\bar t$ be such that $\bar t>0$ and
the inequalities
$$
\|\delta_t\|\le 1\,,\qquad \|\xi_{*,t}\|\le R\,,
$$
hold  whenever $0<t\le\bar t$.  Let $\cl=D\cg(\xi_*)$, so 
$\cl$ is surjective.  Let $\cm:\cy\mapsto\cx$ be a linear 
map such that $\cl\circ\cm=\id_{\cy}$. Then $\cm$ is bounded,
because $\cy$ is finite-dimensional. Let $r$ be such that
$0<r<R$ and
$$
\|\cg(\xi)-\cg(\xi_*)-\cl(\xi-\xi_*)\|\le\frac{\|\xi-\xi_*\|}{4\|\cm\|}\qquad
{\rm whenever}\qquad \|\xi-\xi_*\|\le r\,.
$$
Let $s$ be such that $s>0$, $s\|\cm\|\le r$, and $\bar\nu<\infty$, where
$$
\bar\nu=\sup\Big\{\,\nu(\xi):\xi\in\mathbb{B}_{\cx}(\xi_*;s\|\cm\|)\,\Big\}\,.
$$
Let $\Psi$ be the map from $\mathbb{B}_{\cy}(s)$
to $\cy$ given by $\Psi(y)=\cg(\xi_*+\cm\cdot y)$. Then,
if $y\in\mathbb{B}_{\cy}(s)$, $\cm(y)$ belongs to $\mathbb{B}_\cx(r)$,
so
\begin{eqnarray*}
\|\Psi(y)-y\|&=&\|\cg(\xi_*+\cm\cdot y)-\cl(\cm(y))\|\\
&=&\|\cg(\xi_*+\cm\cdot y)-\cg(\xi_*)-\cl(\cm(y))\|\\
&\le&\frac{1}{4\|\cm\|}\|\cm\|\cdot\|y\|\\
&\le&\frac{s}{4}\,\,,
\end{eqnarray*}
using the fact that $\cg(\xi_*)=0$. Next, we define
$W_t=\delta_t\Big(\mathbb{B}_{\cx}(\xi_*,s\|\cm\|)\Big)$,
and observe that 
$W_t\subseteq\mathbb{B}_\cx(\,\|\xi_{*,t}\|+s\|\delta_t\|\,\|\cm\|\,)\,$.
In particular,
$$
\lim_{t\downarrow 0}\Big(\sup\{\|\xi\|:\xi\in W_t\}\Big)=0\,,
$$
$$
0<t\le\bar t\,\,\Longrightarrow\,\,W_t\subseteq\mathbb{B}_\cx(2R)\,,
$$
and 
$$
\lim_{t\downarrow 0}\lambda_t=0\,,
$$
where
$$
\lambda_t\equaldef
\sup\Big\{\,\frac{\|\cf(\xi)-\cg(\xi)\|}{\nu(\xi)}:\xi\in W_t\,\Big\}\,.
$$
Define
$$
\Phi_t(y)=t^{-1}(\cf\circ\delta_{t})(\xi_*+\cm\cdot y)\qquad
{\rm for}\qquad y\in\mathbb{B}_{\cy}(s)\,,\quad 0<t\le\bar t\,.
$$
(The definition is possible because, if $y\in\bmath_{\cy}(s)$ 
and $0<t\le\bar t$, then 
$$
\|\delta_t(\xi_*+\cm\cdot y)\|=
\|\xi_{*,t}+\delta_t(\cm\cdot y)\|\le\|\xi_{*,t}\|
+s\|\delta_t\|\,\|\cm\|\le R+r<2R\,,
$$
so $\delta_{t}(\xi_*+\cm\cdot y)\in\mathbb{B}_{\cx}(2R)$ and 
then $(\cf\circ\delta_{t})(\xi_*+\cm\cdot y)$ is well defined.)

Then, if we let
\begin{eqnarray*}
\mu(t,y)&=&\|(\cf\circ\delta_{t})(\xi_*+\cm\cdot y)
-(\cg\circ\delta_{t})(\xi_*+\cm\cdot y)\|\,,\\
\tilde\mu(t,y)&=&\frac{\mu(t,y)}{\nu(\delta_t(\xi_*+\cm\cdot y))}\,,\\
\mu_t&=&\sup\Big\{\,\mu(t,y):y\in\mathbb{B}_{\cy}(s)\,\Big\}\,,
\end{eqnarray*}
we have 
$$
\tilde\mu(t,y)\le\sup\Big\{\,\frac{\|\cf(\xi)-\cg(\xi)\|}
{\nu(\xi)}:\xi\in W_t\,\Big\}=\lambda_t\,,
$$
so that
\begin{eqnarray*}
\mu_t&=&\sup\Big\{\,\tilde\mu(t,y)
\cdot\nu(\delta_t(\xi_*+\cm\cdot y)):y\in\mathbb{B}_{\cy}(s)\,\Big\}\\
&\le&\lambda_t
\sup\Big\{\,\nu(\delta_t(\xi)):\xi\in\mathbb{B}_{\cx}(\xi_*;s\|\cm\|)\,\Big\}\\
&=&\lambda_t\,
\sup\Big\{\,t\nu(\xi):\xi\in\mathbb{B}_{\cx}(\xi_*;s\|\cm\|)\,\Big\}\\
&=&t\,\lambda_t\,
\sup\Big\{\,\nu(\xi):\xi\in\mathbb{B}_{\cx}(\xi_*;s\|\cm\|)\,\Big\}\\
&=&t\,\lambda_t\,\bar\nu\,.
\end{eqnarray*}
Furthermore, 
\begin{eqnarray*}
\|\Phi_t(y)-y\|&\le&t^{-1}\mu(t,y)
+\|t^{-1}\cg(\delta_t(\xi_*+\cm\cdot y))-y\|\\
&\le&t^{-1}\mu(t,y)+\|\cg(\xi_*+\cm\cdot y)-y\|\\
&\le&t^{-1}\mu_t+\|\Psi(y)-y\|\\
&\le&\lambda_t\,\bar\nu+\frac{s}{4}\,.
\end{eqnarray*}
Now choose $t$ such that $\bar\nu\lambda_t\le\frac{s}{4}$. Then
$\|\Phi_t(y)-y\|\le\frac{s}{2}$.  Let $B=\mathbb{B}_{\cy}(\frac{s}{2})$,
$\tilde B=\mathbb{B}_{\cy}(s)$. If $y\in B$, define a map
$\zeta_y:\tilde B\mapsto\cy$ by letting $\zeta_y(y')=y'-\Phi_t(y')+y$
if $y'\in\tilde B$. Then, if $y'\in\tilde B$,
the inequalities 
$$
\|\zeta_y(y')\|\le\|y'-\Phi_t(y')\|+\|y\|
\le\frac{s}{2}+\frac{s}{2}=s
$$
imply that $\zeta_y(y')\in\tilde B$ as well. Hence $\zeta_y$ is a continuous
map from $\tilde B$ to $\tilde B$, so $\zeta_y$ has a fixed point
by Brouwer's theorem.  If $\bar y'$ is a fixed point of $\zeta_y$, then
$\bar y'-\Phi_t(\bar y')+y=\zeta_y(\bar y')=\bar y'$, so
$\Phi_t(\bar y')=y$. Then 
$$
t^{-1}\cf\Big(\delta_{t}(\xi_*+\cm\cdot\bar y')\Big)=y\,,
$$
and $\delta_{t}(\xi_*+\cm\cdot\bar y')\in W_t$.
So, if we let $z=\delta_{t}(\xi_*+\cm\cdot\bar y')$, 
it follows that $z\in V$ (because 
$W_t\subseteq\mathbb{B}_{\cx}(2R)\subseteq V$)
and $\cf(z)=ty$.
Since $y$ is an arbitrary member of $B$, we have shown that
$$
tB\subseteq\cf(V)\,.
$$
Therefore $tB$ is a neighborhood of $0$ in
$\cy$, and $tB\subseteq\cf(V)$. This completes our proof.
\end{proof}

\bigbreak

\section{The infinite-dimensional theorem}

\begin{theorem}\label{maininfin}
Assume that $\cx$ and $\cy$ are real Banach spaces,
$\pmb{\delta}=\{\delta_t\}_{t>0}$ is a
\mbox{continuous} one-parameter group of dilations of $\cx$, 
$\nu:\cx\times\cx\mapsto\mathbb{R}_+$ is a 
Lipschitz-bounded \mbox{$\pmb{\delta}$-pseudodistance}, and 
$\cg:\cx\mapsto\cy$ is a \mbox{continuous} 
\mbox{$\pmb{\delta}$-homogeneous}
map. Let $\Omega$ be an open
subset of $\cx$ such that $0\in\Omega$, and let $\cf:\Omega\mapsto\cy$
be a continuous map such that $\cf(0)=0$ and
\begin{equation}\label{limisinfin-ux-a}
\lim_{\xi\to 0,\xi'\to 0,\xi\neq\xi'}\frac{\Big\|\Big(\cf(\xi)-\cg(\xi)\Big)
-\Big(\cf(\xi')-\cg(\xi'\Big)\Big\|}{\nu(\xi,\xi')}=0\,.
\end{equation}
Assume that $\cg$ has a strictly regular zero.  Then $\cf$ is open at $0$.
\end{theorem} 

\begin{proof}
We define the error functions $\ce:\Omega\mapsto\cy$, 
$\tilde\ce:\cx\times\cx\mapsto\cy$, by letting
\begin{eqnarray*}
\ce(\xi)=\cf(\xi)-\cg(\xi)\,\,\,{\rm for}\,\,\,\xi\in\Omega\,,\\
\tilde\ce(\xi,\xi')=\cg(\xi)-\cg(\xi')-\cl(\xi-\xi')\,\,\,{\rm for}
\,\,\,\xi,\xi'\in\cx\,.
\end{eqnarray*}
Let $V$ be a neighborhood of $0$ in $\cx$ such that $V\subseteq\Omega$.
Let $R$ be such that $R>0$ and $\mathbb{B}_{\cx}(2R)\subseteq V$.
%and $4\sigma\nu(x_*)<1$, where 
%$$
%\sigma\equaldef\sup\Big\{\,\frac{\|\cf(x)-\cg(x)\|}{\nu(x)}:x\in
%\mathbb{B}_{\cx}(2R)\,\}\,.
%$$
Let $\xi_*$ be a strictly regular zero of $\cg$. For $t>0$, 
let $\xi_{*,t}=\delta_t(\xi_*)$.  
Let $\bar t$ be such that $\bar t>0$ and
the inequalities
$$
\|\delta_t\|\le 1\,,\qquad \|\xi_{*,t}\|\le R\,,
$$
hold  whenever $0<t\le\bar t$.  Let $\cl=D\cg(\xi_*)$, so 
$\cl$ is a bounded, surjective linear map from
$\cx$ to $\cy$.  Then the Banach open mapping theorem implies
that there is a positive constant $C$ such that 
\begin{itemize}
\item[(\#)] {\em for every $y\in\cx$ there exists a $\xi\in\cx$ such that
$\cl(\xi)=y$ and $\|\xi\|\le C\|y\|$.}
\end{itemize}

Using the strict differentiability of $\cg$ at $\xi_*$, we choose 
$r$ such that \mbox{$0<r<R$} and
$$
\|\tilde\ce(\xi,\xi')\|\le\frac{\|\xi-\xi'\|}{4 C}\,\,
{\rm whenever}\,\,\|\xi-\xi_*\|\le r\,\,{\rm and}
\,\,\|\xi'-\xi_*\|\le r\,.
$$
Let $s$ be such that $s>0$, $sC\le r$, and $\bar\nu<\infty$, where
$$
\bar\nu=\sup\Big\{\,\nu(\xi,0):\xi\in\mathbb{B}_{\cx}(\xi_*;sC)\,\Big\}\,.
$$
Define 
\begin{eqnarray*}
W_t&=&\delta_t\Big(\mathbb{B}_{\cx}(\xi_*,sC)\Big)\,,\\
\omega_t&=&\sup\Big\{\|\xi\|:\xi\in W_t\Big\}\,,
\end{eqnarray*}
and observe that 
$\omega_t\le\|\xi_{*,t}\|+s\|\delta_t\|\,C\,$ for every $t$.
In particular,
$$
\lim_{t\downarrow 0}\omega_t=0\,,
$$
$$
0<t\le\bar t\,\,\Longrightarrow\,\,W_t\subseteq\mathbb{B}_\cx(\omega_t)
\subseteq\mathbb{B}_\cx(2R)\,,
$$
and 
$$
\lim_{t\downarrow 0}\lambda_t=0\,,
$$
where
$$
\lambda_t\equaldef
\sup\Biggl\{\,\frac{\Big\|\ce(\xi)-\ce(\xi')\Big\|}
{\nu(\xi,\xi')}:\xi\in\mathbb{B}_\cx(\omega_t),\,
\xi'\in\mathbb{B}_\cx(\omega_t),\,\xi\neq\xi'\Biggr\}\,.
$$

The Lipschitz-boundedness of $\nu$ implies that the constant
$\bar\kappa\equaldef\kappa_\nu(R+\|\xi_*\|)$ is finite. 
For each $t$, we let 
$$
\alpha_t\equaldef\frac{1}{4}+ C\lambda_t\bar\kappa\,.
$$
We now fix a $t$ such that 
\begin{equation}
0<t\le\bar t\,,\qquad\,\,\,4\lambda_t\bar\nu\le s\,,\,\,\,\qquad
{\rm and}\,\,\,\qquad2\alpha_t\le 1\,, 
\end{equation}
write $\tau=1/t$, and prove that the ball $\bmath_{\cy}(t\lambda_t\bar\nu)$
is contained in $\cf(W_t)$. For this purpose, we
fix a $y\in\cy$ such that $\|y\|\le t\lambda_t\bar\nu$,
and construct a $\xi\in W_t$ such that
$\cf(\xi)=y$. This $\xi$ will be the limit of a sequence
$\{\xi^j\}_{j=0}^\infty$ of points of $W_t$. 

To begin with, we let $\xi^0=\xi_{*,t}$, and observe that 
$\xi^0\in W_t$, and the error
$e^0\equaldef\cf(\xi^0)-y$ satisfies
$$
e^0=\cf(\xi_{*,t})-\cg(\xi_{*,t})-(\cf(0)-\cg(0))-y\,,
$$
since $\cg(\xi_{*,t})=\cg(0)=\cf(0)=0$. So
\begin{eqnarray*}
\|e^0\|&\le&\|\cf(\xi_{*,t})-\cg(\xi_{*,t})-(\cf(0)-\cg(0))\|+\|y\|\\
&\le&\lambda_t\nu(\xi_{*,t},0)+\|y\|\\
&=&\lambda_t\nu(\delta_t(\xi_{*}),\delta_t(0))+\|y\|\\
&=&t\lambda_t\nu(\xi_{*},0)+\|y\|\\
&\le& t\lambda_t\bar\nu+\|y\|\\
&=&2t\lambda_t\bar\nu\,.
\end{eqnarray*}
We then choose $\zeta^1\in\cx$ such that 
$\cl(\zeta^1)=-t^{-1}e^0$ and
$\|\zeta^1\|\le t^{-1}C\|e^0\|\,$,
and \mbox{define} $\eta^1=\delta_t(\zeta^1)$, $\xi^1=\xi^0+\eta^1
=\delta_t(\xi_*+\zeta^1)$. Then 
$t\cl(\delta_{\tau}(\eta^1))=-e^0$, and 
\mbox{$\|\zeta^1\|\le 2C\lambda_t\bar\nu\,$}. Therefore 
$\|\delta_{\tau}(\xi^{1})-\xi_*\|=\|\zeta^1\|\le 2C\lambda_t\bar\nu
\le 4C\lambda_t\bar\nu\le sC\,$,
from which it follows that $\delta_{\tau}(\xi^{1})\in\bmath_\cx(\xi_*,sC)$, 
so that $\xi^{1}\in W_t$. 

We then let $e^1=\cf(\xi^1)-y$. It follows that
\begin{eqnarray*}
e^1&=&\ce(\xi^1)+\Big(\cg(\xi^1)-y\Big)\\
&=&\Big(\ce(\xi^1)-\ce(\xi^0)\Big)+
\Big(\cf(\xi^0)-\cg(\xi^0)\Big)+\Big(\cg(\xi^1)-y\Big)\\
&=&\ce(\xi^1)-\ce(\xi^0)+
\Big(\cf(\xi^0)-y\Big)+\Big(\cg(\xi^1)-\cg(\xi^0)\Big)\\
&=&\ce(\xi^1)-\ce(\xi^0)+e^0+
t\Big(\cg(\delta_{\tau}(\xi^1))-\cg(\delta_{\tau}(\xi^0))\Big)\\
&=&\ce(\xi^1)-\ce(\xi^0)+e^0+t\tilde\ce(\delta_{\tau}(\xi^1),\delta_{\tau}
(\xi^0))+t\cl(\delta_{\tau}(\eta^1))\\
&=&\ce(\xi^1)-\ce(\xi^0)+
t\tilde\ce(\delta_{\tau}(\xi^1),\delta_{\tau}(\xi^0))\,.
\end{eqnarray*}
On the other hand,
\begin{eqnarray*}
\|\ce(\xi^1)-\ce(\xi^0)\|&\le&\lambda_t\nu(\xi^1,\xi^0)\\
&=&t\lambda_t\nu(\delta_{\tau}(\xi^1),\delta_{\tau}(\xi^0))\\
&\le&t\lambda_t\bar\kappa\|\delta_{\tau}(\xi^1)-\delta_{\tau}(\xi^0))\|\\
&=&t\lambda_t\bar\kappa\|\delta_{\tau}(\eta^1)\|\\
&=&t\lambda_t\bar\kappa\|\zeta^1\|\\
&\le&2Ct\lambda_t^2\bar\kappa\bar\nu\,,
\end{eqnarray*}
and
$$
\|\tilde\ce(\delta_{\tau}(\xi^1),\delta_{\tau}(\xi^0))\|\le
\frac{\|\delta_{\tau}(\xi^1)-\delta_{\tau}(\xi^0)\|}{4C}
=\frac{\|\zeta^1\|}{4C}\le\frac{2C\lambda_t\bar\nu}{4C}
= \frac{1}{2}\lambda_t\bar\nu\,.
$$
Therefore
$$
\|e^1\|\le2Ct\lambda_t^2\bar\kappa\bar\nu+\frac{t}{2}\lambda_t\bar\nu=
t\lambda_t\bar\nu\Big(2C\lambda_t\bar\kappa+\frac{1}{2}\Big)=
2t\lambda_t\bar\nu\alpha_t\,.
$$
Next, we choose $\zeta^2\in\cx$ such that 
$\cl(\zeta^2)=-t^{-1}e^1$ and $\|\zeta^2\|\le t^{-1}C\|e^1\|\,$,
and define $\eta^2=\delta_t(\zeta^2)$, $\xi^2=\xi^1+\eta^2
=\xi^1+\delta_t(\zeta^2)=\delta_t(\xi_*+\zeta^1+\zeta^2)$. Then 
$t\cl(\delta_{\tau}(\eta^2))=-e^1$, and 
$\|\zeta^2\|\le 2C\lambda_t\bar\nu\alpha_t\,$.  Therefore 
$$
\|\delta_{\tau}(\xi^{2})-\xi_*\|=
\|\zeta^1+\zeta^{2}\|\le
2C\lambda_t\bar\nu(1+\alpha_t)
\le 4C\lambda_t\bar\nu\le sC\,,
$$
from which it follows that $\delta_{\tau}(\xi^{2})\in\bmath_\cx(\xi_*,sC)$, 
so that $\xi^{2}\in W_t$. 

We then let $e^2=\cf(\xi^2)-y$. It follows that
\begin{eqnarray*}
e^2&=&\ce(\xi^2)+\Big(\cg(\xi^2)-y\Big)\\
&=&\Big(\ce(\xi^2)-\ce(\xi^1)\Big)+
\Big(\cf(\xi^1)-\cg(\xi^1)\Big)+\Big(\cg(\xi^2)-y\Big)\\
&=&\ce(\xi^2)-\ce(\xi^1)+
\Big(\cf(\xi^1)-y\Big)+\Big(\cg(\xi^2)-\cg(\xi^1)\Big)\\
&=&\ce(\xi^2)-\ce(\xi^1)+e^1+
t\Big(\cg(\delta_{\tau}(\xi^2))-\cg(\delta_{\tau}(\xi^1))\Big)\\
&=&\ce(\xi^2)-\ce(\xi^1)+e^1+t\tilde\ce(\delta_{\tau}(\xi^2),\delta_{\tau}
(\xi^1))+t\cl(\delta_{\tau}(\eta^2))\\
&=&\ce(\xi^2)-\ce(\xi^1)+
t\tilde\ce(\delta_{\tau}(\xi^2),\delta_{\tau}(\xi^1))\,.
\end{eqnarray*}
On the other hand,
\begin{eqnarray*}
\|\ce(\xi^2)-\ce(\xi^1)\|&\le&\lambda_t\nu(\xi^2,\xi^1)\\
&=&t\lambda_t\nu(\delta_{\tau}(\xi^2),\delta_{\tau}(\xi^1))\\
&\le&t\lambda_t\bar\kappa\|\delta_{\tau}(\xi^2)-\delta_{\tau}(\xi^1))\|\\
&=&t\lambda_t\bar\kappa\|\delta_{\tau}(\eta^2)\|\\
&=&t\lambda_t\bar\kappa\|\zeta^2\|\\
&\le&t\lambda_t\bar\kappa(2C\lambda_t\bar\nu\alpha_t)\\
&\le&2Ct\lambda_t^2\bar\kappa\bar\nu\alpha_t\,,
\end{eqnarray*}
and
$$
\|\tilde\ce(\delta_{\tau}(\xi^2),\delta_{\tau}(\xi^1))\|\le
\frac{\|\delta_{\tau}(\xi^2)-\delta_{\tau}(\xi^1)\|}{4C}
=\frac{\|\zeta^2\|}{4C}
\le\frac{2C\lambda_t\bar\nu\alpha_t}{4C}
=\frac{1}{2}\lambda_t\bar\nu\alpha_t\,.
$$
Therefore
$$
\|e^2\|\le2Ct\lambda_t^2\bar\kappa\bar\nu\alpha_t+
\frac{t}{2}\lambda_t\bar\nu\alpha_t=
2t\lambda_t\bar\nu\Big(C\lambda_t\bar\kappa+\frac{1}{4}\Big)\alpha_t
=2t\lambda_t\bar\nu\alpha_t^2\,.
$$

We continue this construction inductively. Suppose we have defined
$$
\xi^0,\ldots,\xi^k\in W_t\,,\,\,\,\,e^0,\ldots,e^k\in\cy\,, 
\,\,\,\,\zeta^1,\ldots,\zeta^k\in\cx\,,\,\,\,\,\eta^1,\ldots,\eta^k\in\cy\,,
$$
such that 
\begin{eqnarray*}
\xi^{j}&=&\xi^{j-1}+\eta^j\phantom{-e^{j-1}\cf(\xi^j)-y}{\rm for}
\,\,j=1,\ldots,k\,,\\
e^{j}&=&\cf(\xi^j)-y\phantom{-e^{j-1}\xi^{j-1}+\eta^j}{\rm for}\,\,
j=0,\ldots,k\,,\\
t\cl(\zeta^j)&=&-e^{j-1}\phantom{\xi^{j-1}+\eta^j\xi^{j-1}+\eta^j}\,
{\rm for}\,\,j=1,\ldots,k\,,\\
\eta^j&=&\delta_t(\zeta_j)\phantom{\xi^{j-1}+\eta^j\xi^{j-1}+\eta^j}\,
\,\,{\rm for}\,\,j=1,\ldots,k\,,\\
\|\zeta^j\|&\le&2C\lambda_t\bar\nu\alpha_t^{j-1}
\phantom{\xi^{j-1}+\eta^j\xi^{j-1}}\,\,
{\rm for}\,\,j=1,\ldots,k\,,\\
\|e^j\|&\le&2t\lambda_t\bar\nu\alpha_t^{j}
\phantom{\xi^{j-1}+\eta^j\xi^{j-1}aai}\,\,
{\rm for}\,\,j=0,\ldots,k\,.
\end{eqnarray*}
We then choose $\zeta^{k+1}\in\cx$ such that 
$$
\cl(\zeta^{k+1})=-t^{-1}e^k\,\,\,{\rm and}\,\,\,
\|\zeta^{k+1}\|\le t^{-1}C\|e^k\|\,,
$$
and define 
\begin{eqnarray*}
\eta^{k+1}&=&\delta_t(\zeta^{k+1})\,,\\
\xi^{k+1}&=&\xi^k+\eta^{k+1}=\xi^k+\delta_t(\zeta^{k+1})=
\delta_t(\xi_*+\zeta^1+\ldots+\zeta^{k+1})\,.
\end{eqnarray*}
Then $t\cl(\delta_{\tau}(\eta^{k+1}))=-e^k$, and 
$$
\|\zeta^{k+1}\|\le t^{-1}C\|e^k\|\le 2C\lambda_t\bar\nu\alpha_t^k\,.
$$
Therefore 
$$
\|\delta_{\tau}(\xi^{k+1})-\xi_*\|=
\|\zeta^1+\ldots+\zeta^{k+1}\|\le
2C\lambda_t\bar\nu\sum_{j=0}^k\alpha_t^j
=\frac{2C\lambda_t\bar\nu}{1-\alpha_t}
\le 4C\lambda_t\bar\nu\,,
$$
so 
$\|\delta_{\tau}(\xi^{k+1})-\xi_*\|\le sC$, from which it follows
that $\delta_{\tau}(\xi^{k+1})\in\bmath_\cx(\xi_*,sC)$, 
so that $\xi^{k+1}\in W_t$. 

We then let $e^{k+1}=\cf(\xi^{k+1})-y$. It follows that
\begin{eqnarray*}
e^{k+1}&=&\ce(\xi^{k+1})+\Big(\cg(\xi^{k+1})-y\Big)\\
&=&\Big(\ce(\xi^{k+1})-\ce(\xi^k)\Big)+
\Big(\cf(\xi^k)-\cg(\xi^k)\Big)+\Big(\cg(\xi^{k+1})-y\Big)\\
&=&\ce(\xi^{k+1})-\ce(\xi^k)+
\Big(\cf(\xi^k)-y\Big)+\Big(\cg(\xi^{k+1})-\cg(\xi^k)\Big)\\
&=&\ce(\xi^{k+1})-\ce(\xi^k)+e^k+
t\Big(\cg(\delta_{\tau}(\xi^{k+1}))-\cg(\delta_{\tau}(\xi^k))\Big)\\
&=&\ce(\xi^{k+1})-\ce(\xi^k)+e^1+
t\tilde\ce(\delta_{\tau}(\xi^{k+1}),\delta_{\tau}
(\xi^k))+t\cl(\delta_{\tau}(\eta^{k+1}))\\
&=&\ce(\xi^{k+1})-\ce(\xi^k)+
t\tilde\ce(\delta_{\tau}(\xi^{k+1}),\delta_{\tau}(\xi^k))\,.
\end{eqnarray*}
On the other hand,
\begin{eqnarray*}
\|\ce(\xi^{k+1})-\ce(\xi^k)\|&\le&\lambda_t\nu(\xi^{k+1},\xi^k)\\
&=&t\lambda_t\nu(\delta_{\tau}(\xi^{k+1}),\delta_{\tau}(\xi^k))\\
&\le&t\lambda_t\bar\kappa\|\delta_{\tau}(\xi^{k+1})-\delta_{\tau}(\xi^k))\|\\
&=&t\lambda_t\bar\kappa\|\delta_{\tau}(\eta^{k+1})\|\\
&=&t\lambda_t\bar\kappa\|\zeta^{k+1}\|\\
&\le&t\lambda_t\bar\kappa(2C\lambda_t\bar\nu\alpha_t)\\
&\le&2Ct\lambda_t^2\bar\kappa\bar\nu\alpha_t^k\,,
\end{eqnarray*}
and
$$
\|\tilde\ce(\delta_{\tau}(\xi^{k+1}),\delta_{\tau}(\xi^k))\|\le
\frac{\|\delta_{\tau}(\xi^{k+1})-\delta_{\tau}(\xi^k)\|}{4C}
=\frac{\|\zeta^{k+1}\|}{4C}
%\le\frac{2C\lambda_t\bar\nu\alpha_t^k}{4C}
\le\frac{1}{2}\lambda_t\bar\nu\alpha_t^k\,.
$$
Therefore
$$
\|e^{k+1}\|\le2Ct\lambda_t^2\bar\kappa\bar\nu\alpha_t^k+
\frac{t}{2}\lambda_t\bar\nu\alpha_t^k=
2t\lambda_t\bar\nu\Big(C\lambda_t\bar\kappa+\frac{1}{4}\Big)\alpha_t^k
=2t\lambda_t\bar\nu\alpha_t^{k+1}\,,
$$
and our inductive construction is complete.

The bound $\|\zeta^j\|\le 2C\lambda_t\bar\nu\alpha_t^{j-1}$ implies---since
$2\alpha_t\le 1$---that the series $\sum_{j=1}^\infty\zeta^j$ 
converges, and the sum $\zeta$ of the series 
satisfies $\|\zeta\|\le 4C\lambda_t\bar\nu\le sC$. Therefore
the limit $\Xi=\lim_{j\to\infty}\delta_{\tau}(\xi^j)$
exists and satisfies $\Xi\in\bmath_\cx(\xi_*,sC)$. So 
the limit $\xi=\delta_t(\Xi)=\lim_{j\to\infty}\xi^j$
exists and belongs to $W_t$. Furthermore, 
$\cf(\xi)-y=\lim_{j\to\infty}(\cf(\xi^j)-y)
=\lim_{j\to\infty}e^j=0$, because of the bound
$\|e^j\|\le2t\lambda_t\bar\nu\alpha_t^{j}$. Therefore
$\cf(\xi)=y$, and our proof is complete.
\end{proof}

\section{Second-order open mapping theorems}\label{secseco}
If $X$ and $Y$ are real linear spaces, and
\mbox{$X\times X\ni(x,x')\mapsto B(x,x')\in Y$} is a symmetric
\mbox{bilinear} map,
we write $Q_B$ to denote the quadratic map 
associated with $B$, i.e., the map 
$X\ni x\mapsto B(x,x)\equaldef Q_B(x)\in Y\,$.
It is well known that $B$ is completely determined by $Q_B$,
since
\begin{equation}\label{bilfromquad}
B(x,y)=\frac{1}{4}\Big(Q_B(x+y)-Q_B(x-y)\Big)\,.
\end{equation}

A {\em quadratic map} from $X$ to $Y$ is a
map $Q$ such that $Q=Q_B$ for some (unique) bilinear symmetric map
\mbox{$B:X\times X\mapsto Y$}. If $Q$ is a quadratic map, then we will
use $B^Q$ to denote the corresponding symmetric bilinear map.

If $X$, $Y$ are normed, then a bilinear map $B:X\times X\mapsto Y$ 
is continuous if and only if it is bounded, in the sense that
there exists a constant $C$ such that $\|B(x,x')\|\le C\|x\|\,\|x'\|$
for all $x,x'\in X$. It follows from (\ref{bilfromquad}) that
a quadratic map $Q:X\to Y$ is continuous if and only if the bilinear
map $B^Q$ is continuous.

\begin{definition}\label{defoflqa}
{\rm If $X$, $Y$ are normed spaces, $\Omega$ is open in $X$,
$x_*\in\Omega$, and $F$ is a map from $\Omega$ to $Y$, then
a {\em linear-quadratic approximation} (abbreviated
LQA) of $F$ at $x_*$ is a triple $A=(L,K,Q)$ such that
\begin{itemize}\item[]\begin{itemize}
\item[LQA1.] $L$ is a bounded linear map from $X$ to $Y$, 
\item[LQA2.] $K$ is a closed linear subspace of $X$,
\item[LQA3.] $Q$ is a continuous quadratic map from $K$ to $Y$.
\item[LQA4.] $F(x_*+x+k)=F(x_*)+Lx+\frac{1}{2}Q(k)+o(\|k\|^2+\|x\|)$
as $(x,k)$ goes to zero via values in $X\times K$.\proofend
\end{itemize}\end{itemize}}
\end{definition}

\medbreak\noindent If $(L,K,Q)$ is a LQA of $F$ at $x_*$, then 
it is clear that $F$ is Fr\'echet differentiable at $x_*$ and $DF(x_*)=L$,
$K\subseteq{\rm ker}\,L$, and the quadratic map $Q$ is completely 
determined by $K$, by means of the formula.
$$
Q(k)=2\,\lim_{t\downarrow 0}t^{-2}(F(x_*+tk)-F(x_*))\,\,
\qquad{\rm for}\,\,\qquad x\in K\,.
$$ 
The error bound of LQA4 is important because it gives an estimate 
of the the error in terms of the function 
$X\times K\ni(x,k)\mapsto\|x\|+\|k\|^2$, which is
positively homogeneous of degree $1$ relative to an appropriate 
group of dilations on $X\times K$. We make this observation precise
in the following statement, whose proof is trivial.
\begin{lemma}\label{itisapseudonorm}
Assume that (a)~$X$, $Y$, $\Omega$, $x_*$, $F$ are as in
Definition \ref{defoflqa},  and
\mbox{(b)~$A=(L,K,Q)$} is a linear-quadratic approximation
of $F$ at $x_*$. Let 
$$
\cx\,\equaldef\,X\times K\,,\qquad\tilde\Omega=\Big\{\,(x,k)\in
\cx:x_*+x+k\in\Omega\,\Big\}\,, 
$$
and define maps
$\nu:\cx\mapsto\mathbb{R}$, $\delta_t:\cx\mapsto\cx$ for $t>0$,  
$\cf:\tilde\Omega\mapsto Y$, $\cg:\cx\mapsto\cy$, 
$\ce:\tilde\Omega\mapsto Y$, by
\begin{eqnarray*}
\nu(x,k)&=&\|x\|+\|k\|^2\phantom{affaaaaaaaaaaaaaaaaaaaa}
{\rm for}\quad(x,k)\in\cx\,,\label{defnusimple}\\
\delta_t(x,k)\!&=&\!(tx,\sqrt{t}k)\phantom{aaffaaaaaaaaaaaaaaaaaaaaa}
{\rm for}\quad
(x,k)\in\cx\,,\,\,t>0\,,\label{defdilsimple}\\
\cf(x,k)&=&F(x_*+x+k)-F(x_*)\phantom{aaaaaaaaaaaaai}\!
{\rm for}\quad(x,k)\in\tilde\Omega\,,\label{defoffnewsimple}\\
\cg_A(x,k)&=&Lx+\frac{1}{2}Q(k)\phantom{aaaaaaaaaaaaaiaaaaaaaaa}\!
{\rm for}\quad(x,k)\in\cx\,,\label{defofgasimple}\\
\ce(x,k)&=&F(x_*+x+k)-F(x_*)-\cg_A(x,k)\nonumber\\
&=&\cf(x,k)-\cg_A(x,k)\phantom{aaaaaaaaaaaaaaaaa}
{\rm for}\quad(x,k)\in\tilde\Omega\,.\label{errorquad}
\end{eqnarray*}
Then $\pmb{\delta}=\{\delta_t\}_{t>0}$ is a continuous
one-parameter group of dilations of $\cx$, 
$\cg_A$ is $\pmb{\delta}$-homogeneous, 
$\nu$ is a $\pmb{\delta}$-pseudonorm, and the error bound
\begin{equation}\label{errorbound}
\lim_{x\to 0,\,k\to 0}\frac{\|\ce(x,k)\|}{\nu(x,k)}=0
\end{equation}
is satisfied. \proofend
\end{lemma}
\begin{definition}
{\rm Assume that $X$, $Y$, $\Omega$, $x_*$, $F$ are as in
Definition \ref{defoflqa}.
A {\em \mbox{strict} linear-quadratic approximation} (abbreviated
SLQA) of $F$ at $x_*$ is a triple \mbox{$A=(L,K,Q)$} that
satisfies conditions LQA1,2,3 of Definition \ref{defoflqa}
and is such that
\begin{itemize}\item[]\begin{itemize}
\item[{\rm (SLQA)}] $\Big(F(x_*+x+k)-Lx-\frac{1}{2}Q(k)\Big)
-\Big(F(x_*+x'+k')-Lx'-\frac{1}{2}Q(k')\Big)$
$\,\,=o\Big(\,
\|x-x'\|+(\sqrt{\|x\|+\|x'\|}+\|k\|+\|k'\|).\|k-k'\|\,\Big)$
as $(x,k,x',k')$ goes to zero via values in $X\times K
\times X\times K$.\proofend
\end{itemize}\end{itemize}}
\end{definition}

\medbreak\noindent If $A=(L,K,Q)$ is a SLQA of $F$ at $x_*$, then 
$A$ is a LQA of $F$ at $x_*\,$, 
$F$ is strictly Fr\'echet differentiable at $x_*$, and 
$DF(x_*)=L\,$.

As in the case of (non-strict) LQAs, the error bound of condition 
\mbox{(SLQA)} is \mbox{important}
because it estimates the error in terms of a function which is 
\mbox{positively}
\mbox{homogeneous} of degree $1$ relative to an appropriate group of 
\mbox{dilations}.
We make this precise in the following statement, whose proof is trivial.
\begin{lemma}\label{itisapseudodis}
Let $X$, $Y$, $\Omega$, $x_*$, $F$, $A$, 
$L$, $K$, $Q$, $\cx$, $\tilde\Omega$,
$\cf$, $\cg_A$, $\pmb{\delta}$, $\ce$ be as in the statement of
Lemma \ref{itisapseudonorm}. 
Assume that $A$ is a strict linear-quadratic approximation
of $F$ at $x_*$. Define a map $\nu:\cx\times \cx\mapsto\mathbb{R}$ by 
letting
\begin{equation}\label{defnu}
\nu(x,k,x',k')\!=\!
\|x\!-\!x'\|\!+\!\Big(\sqrt{\|x\|\!+\!\|x'\|}\!+\!\|k\|\!+\!\|k'\|\Big).
\|k\!-\!k'\|
\end{equation}
for $(x,k,x',k')\in\cx\times\cx\,$. Then 
\begin{itemize}
\item[1.] $\pmb{\delta}$ is a continuous one-parameter group of 
\mbox{dilations} of $\cx$, 

\vup\vup\vup
\item[2.] $\cg_A$ is $\pmb{\delta}$-homogeneous, 

\vup\vup\vup
\item[3.] $\nu$ is a Lipschitz-bounded $\pmb{\delta}$-pseudodistance, 

\vup\vup\vup
\item[4.] the error bound
\begin{equation}\label{errorforsecondor}
\lim_{x\to 0,x'\to 0,x'\neq x}
\frac{\|\ce(x,k)-\ce(x',k')\|}{\nu(x,k,x',k')}=0
\end{equation}
is satisfied.\proofend
\end{itemize}
\end{lemma}

\medbreak An important class of maps that necessarily admit strict 
linear-quadratic \mbox{approximations} consists of the 
{\em maps of class $C^1$ with a differentiable derivative.} Precisely,
let us assume that 
\begin{itemize}
\item[(\#)] {\em $X$, $Y$ are normed spaces, $\Omega$ is an open
subset of $X$, $F:\Omega\mapsto Y$ is a map of class $C^1$,
$x_*\in\Omega$, and the map
$\Omega\ni x\mapsto DF(x)\in{\rm Lin}(X,Y)$ is differentiable
at $x_*\,$.}
\end{itemize}
Then $DF$ is a continuous map from $\Omega$ to the space 
${\rm Lin}(X,Y)$ of bounded linear maps from $X$ to $Y$, 
and the second derivative $D(DF)(x_*)=D^2F(x_*)$ is a bounded 
linear map from $X$ to ${\rm Lin}(X,{\rm Lin}(X,Y))$. 
Furthermore, {\em this map is symmetric}, i.e., 
$D^2F(x_*)(x)\cdot x'=D^2F(x_*)(x')\cdot x$ for all
$x,x'\in X\,$.
(This is true because of the identity 
\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!D^2F(x_*)(x)\cdot x'=\\
&\,&\lim_{\alpha\downarrow 0}\alpha^{-2}\Big(
F(x_*+\alpha x+\alpha x')-F(x_*+\alpha x)-F(x_*+\alpha x')+F(x_*)\Big)\,,
\end{eqnarray*}
whose right-hand side is clearly symmetric under the interchange
of $x$ and $x'$.) 

It follows that we can regard $D^2F(x_*)$ as a bounded symmetric bilinear map 
\mbox{$B:X\times X\mapsto Y$}, given by $B(x,x')=D^2F(x_*)(x)\cdot x'$. Then
$B=B^\cq$, where $\cq:X\mapsto Y$ is the quadratic map
given by $\cq(x)=B(x,x)$, so that $\cq=Q_B$ and $B=B^\cq$. Even more
important for us will be the restriction $Q$ of $\cq$ to the kernel 
$K={\rm ker}\,L$, where $L=DF(x_*)$.

It turns out that the triple $A=(L,K,Q)$ is a strict linear-quadratic 
\mbox{approximation} of $F$ at $x_*$, as we now show. 

First, we write 
\begin{eqnarray*}
M(x)&=&L\cdot x+\frac{1}{2}\cq(x)\,\,,\\
\cg_A(x,k)&=&L\cdot x+\frac{1}{2}Q(k)\,\,,\\
E(x)&=&{}F(x_*+x)-F(x_*)-M(x)\,\,,\\
\ce(x,k)&=&{}F(x_*+x+k)-F(x_*)-\cg_A(x,k)\,\,,
\end{eqnarray*}
for $x\in X$, $k\in K$.  
\begin{lemma}\label{isatsrict}
Let $X$, $Y$, $\Omega$, $F$, $L$, $K$, $\cq$, $Q$, $M$, $\cg_A$, 
$A$, $E$, $\ce$, be as above. Then
\begin{equation}\label{ourconclusion}
\lim_{x\to 0,x'\to 0,x'\neq x}
\frac{\|E(x)-E(x')\|}{(\|x\|+\|x'\|)\|x-x'\|}=0\,,
\end{equation}
and $A$ is a strict linear-quadratic approximation of $F$ at $x_*$.
\end{lemma}
\begin{proof}
Without loss of generality, we assume that $x_*=0$ and
$F(x_*)=0$. We fix a positive $R$ such 
that $\mathbb{B}_{{}X}(R)\subseteq\Omega$. 
For $0<r\le R$, let
$$
\theta(r)=\sup\Biggl\{\frac{\|D{}F(x)-\cl-B^\cq(x,\,\cdot\,)\|}{\|x\|}:
0<\|x\|\le r\,\Biggr\}\,.
$$
Then $\theta$ is monotonically nondecreasing, and
the differentiability assumption implies that 
$\lim_{r\downarrow 0}\theta(r)=0$. Clearly, the bound 
\begin{equation}\label{clearly-aaa}
\|D{}F(\xi).v-\cl.v-B^\cq(\xi,v)\|\le\theta(r)\,\|\xi\|\,\|v\|
\end{equation}
is satisfied whenever $0<r\le R$, $\xi\in\mathbb{B}_{{}X}(r)$,
and $v\in{}X$.

Then, if $x,x'\in\mathbb{B}_{{}X}(R)$, and we write $v=x-x'$
$\xi_s=x'+sv$, we have 
\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!
E(x)-E(x')={}F(x)-{}F(x')-\cl\cdot{v}
-\frac{1}{2}\Big(\cq(x)-\cq(x')\Big)\\  
&\,&\!\!\!\!\!\!\!\!\!\!
=\Big(\int_0^1D{}F(\xi_s)\,ds\Big)\cdot{v}
-\cl\cdot{v}-\frac{1}{2}\Big(\cq(x)-\cq(x')\Big)\\
&\,&\!\!\!\!\!\!\!\!\!\!
=\Big(\int_0^1(D{}F(\xi_s)-\cl)\,ds\Big)\cdot{v}
-\frac{1}{2}\Big(\cq(x)-\cq(x')\Big)\\
&\,&\!\!\!\!\!\!\!\!\!\!
=\int_0^1\Big((D{}F(\xi_s)-\cl).{v}-B^\cq(\xi_s,v)\Big)\,ds\\
&\,&\!\!\!\!\!\!\!\!\!\!\,\,\quad
+\int_0^1B^\cq(\xi_s,v)\,ds-\frac{1}{2}\Big(B^\cq(x,x)-B^\cq(x',x')\Big)\\
&\,&\!\!\!\!\!\!\!\!\!\!
=\int_0^1\Big((D{}F(\xi_s)-\cl).{v}-B^\cq(\xi_s,v)\Big)\,ds\\
&\,&\!\!\!\!\!\!\!\!\!\!\,\,\quad
+B^\cq(x',v)+\frac{1}{2}B^\cq(v,v)
-\frac{1}{2}\Big(B^\cq(x,x)-B^\cq(x',x')\Big)\\
&\,&\!\!\!\!\!\!\!\!\!\!=\int_0^1\Big((D{}F(\xi_s)-\cl).{v}
-B^\cq(\xi_s,v)\Big)\,ds\,,
\end{eqnarray*}
using the identities 
$B^\cq(x',v)+\frac{1}{2}B^\cq(v,v)-\frac{1}{2}\Big(B^\cq(x,x)-
B^\cq(x',x')\Big)=0$ and $\int_0^1B^\cq(\xi_s,v)\,ds=
B^\cq(x',v)+\frac{1}{2}B^\cq(v,v)$. 
Then (\ref{clearly-aaa}), with $\xi=\xi_s$, yields
\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\|(D{}F(\xi_s)-\cl).{v}-B^\cq(\xi_s,v)\|
\le \theta(\|x\|+\|x'\|)\cdot(\|x\|+\|x'\|)\cdot\|x-x'\|\,,
\end{eqnarray*}
since $\|\xi_s\|\le\|x\|+\|x'\|$ whenever
$0\le s\le 1$.  Integrating this inequality, we get the bound
\begin{equation}\label{newbo-ixixi}
\|E(x)-E(x')\|\le\theta(\|x\|+\|x'\|)\cdot(\|x\|+\|x'\|)\cdot\|x-x'\|\,,
\end{equation}
and (\ref{ourconclusion}) follows.

To prove that $A$ is a strict linear-quadratic approximation
of $F$ at $0$, we have to show that 
$$
\ce(x,k)-\ce(x',k')=
o\Big(\|x-x'\|+(\sqrt{\|x\|+\|x'\|}+\|k\|+\|k'\|).\|k-k'\|\Big)
$$
as $(x,x',k,k')\to(0,0,0,0)$.  But
\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!
\ce(x,k)-\ce(x',k')=(F(x+k)-\cg_A(x,k))-(F(x'+k')-\cg_A(x',k'))\\
&\,&=\Big(F(x+k)-L\cdot x-\frac{1}{2}Q(k)\Big)-\Big(F(x'+k')-L\cdot x'-
\frac{1}{2}Q(k')\Big)\\
&\,&=\Big(F(x+k)-L\cdot(x+k)-\frac{1}{2}\cq(x+k)\Big)-
\Big(F(x'+k')-L\cdot(x'+k')\\
&\,&\,\,\,\,\,\,-\frac{1}{2}\cq(x'+k')\Big)
+\frac{1}{2}\Big(\cq(x+k)-\cq(x'+k')-\cq(k)+\cq(k')\Big)\\
&\,&=E(x+k)-E(x'+k')
+\frac{1}{2}\Big(\cq(x+k)-\cq(x'+k')-\cq(k)+\cq(k')\Big)\\
&\,&=E(x+k)-E(x'+k')
+\frac{1}{2}\Big(\cq(x)-\cq(x')\Big)+B^\cq(x,k)+B^\cq(x',k')\\
&\,&=E(x+k)-E(x'+k')\\
&\,&\,\,\,\,\,\,
+\frac{1}{2}\cq(x-x')+B^\cq(x',x'-x)+B^\cq(x-x',k)+B^\cq(x',k-k')\,.\\
\end{eqnarray*}
Therefore (\ref{newbo-ixixi}) implies, if we write $\kappa=\|B^\cq\|$, 
$\Theta(s)=s\theta(s)$, that 
\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!
\|\ce(x,k)-\ce(x',k')\|\le
\Theta\Big(\|x\|+\|x'\|+\|k\|+\|k'\|)\Big)\cdot\Big(\|x-x'\|+\|k-k'\|\Big)\\
&\,&\,\,\,\,+\frac{\kappa}{2}\|x-x'\|^2+\kappa\|x'\|.\|x-x'\|+
\kappa\|x-x'\|.\|k\|+\kappa\|x'\|.\|k-k'\|\,,
\end{eqnarray*}
which is clearly 
$o\Big(\|x-x'\|+(\sqrt{\|x\|+\|x'\|}+\|k\|+\|k'\|).\|k-k'\|\Big)$.
\end{proof}

We can then apply Theorems \ref{mainfin} and \ref{maininfin}
and obtain a number of open mapping theorems. The crucial
condition in all of them is, of course, the existence of a regular 
zero $(\xi_*,k_*)$ of the map $\cg_A$ 
associated to the LQA $A=(L,K,Q)$. (Such a point will then automatically 
be a strictly regular zero, because $\cg_A$ 
is a polynomial map.) It turns out that this condition is equivalent 
to a statement in terms of the ``Hessian'' $H_A$ of $A$, as we now explain.

Let $Z$ be the quotient space $Y/LX$, and let 
$\pi$ be the canonical projection from $Y$ onto $Z$.
The {\em Hessian} $H_A$ is the quadratic map 
$$
K\ni k\mapsto \pi(Q(k,k))\,\,\equaldef\,\,H_A(k)\in Z\,.
$$
\begin{definition}\label{regzeoalg}
{\rm A {\em regular zero} of $H_A$ is a $k_*\in K$ such that 
$H_A(k_*,k_*)=0$ and the map 
$K\ni k\mapsto \pi(B^Q(k_*,k))\in Z$ is surjective.}\proofend
\end{definition}
\begin{remark}
{\rm If $LX$ is a closed subspace of $Y$ then $Z$ is a normed 
space and the quadratic map $H_A:K\mapsto Z$ is continuous, and hence
smooth. In that case, $k_*$ is a regular zero of $H_A$ in the sense
defined above if and  only if it is a regular zero in the sense defined
earlier in \S\ref{prelims}. Here, however, {\em we are not requiring
$LX$ to be closed}, and the concept of a ``regular zero''
of $H_A$ has to be understood in the purely algebraic sense
of Definition \ref{regzeoalg}.}\proofend
\end{remark}
\begin{lemma}
If $A=(L,K,Q)$ is a LQA of a map $F$ at a point $x_*$, and $k_*\in K$, then
$k_*$ is a regular zero of $H_A$ if and only if 
there exists $\xi_*\in X$ such that 
$(\xi_*,k_*)$ is a regular zero of $\cg_A$.
In particular, $H_A$ has a regular zero if and only if  $\cg_A$ has a
regular zero. \proofend
\end{lemma}
\begin{proof}
Recall that a regular zero of $\cg_A$ is a pair 
$(\xi_*,k_*)\in X\times K$ such that
$L\,\xi_*+\frac{1}{2}Q(k_*)=0$ and the linear map $X\times
K\ni(x,k)\mapsto L\,x+B^Q(k_*,k)\in Y$ is surjective. 


If $k_*$ is a regular zero of the Hessian
$H_A$, then $\pi(Q(k_*))=0$, 
so $Q(k_*)$ belongs to $LX$, and then there exists $\xi_*\in X$ such that 
\mbox{$\frac{1}{2}Q(k_*)+L\cdot\xi_*=0$}. Therefore $\cg_A(\xi_*,k_*)=0$,
and $D\cg_A(\xi_*,k_*)(x,k)=L\cdot x+B^Q(k_*,k)$ if $x\in X$, $k\in K$.
If $y\in Y$, then the surjectivity of the map
\mbox{$K\ni k\mapsto\pi(B^Q(k_*,k))\in Z$} implies that 
there exists a $k\in K$ such that $\pi(B^Q(k_*,k))=\pi(y)$, i.e., that 
\mbox{$y-B^Q(k_*,k)$} belongs to $LX$. It follows that 
there exists $x\in X$ such 
that \mbox{$y=B^Q(k_*,k)+Lx=D\cg_A(h_*,k_*)(x,k)$}. 
Since $y$ is an arbitrary member of $Y$, we have shown that
the linear map $D\cg_A(\xi_*,k_*):X\times K\mapsto Y$ is surjective,
so $(\xi_*,k_*)$ is a regular zero of $\cg_A$. 

Conversely, if $(\xi_*,k_*)$ is a regular zero of $\cg_A$, then 
$\frac{1}{2}Q(k_*)+L\cdot\xi_*=0$, so 
$Q(k_*)\in LX$, and then $H_A(k_*)=0$. 
Furthermore, if $v\in Z$, and $y\in Y$ is such that
$\pi(y)=v$, then we can write
$y=D\cg_A(\xi_*,k_*)(x,k)=Lx+B^Q(k_*,k)$ for some 
$x\in X$, $k\in K$. But then $v=\pi(y)=\pi(B^Q(k_*,k))$.
Therefore the map $K\ni k\mapsto\pi(B^Q(k_*,k))\in Z$
is surjective, and we have shown that $k_*$ is a regular zero of $H_A$.
\end{proof}

\begin{theorem}
Assume that $X$, $Y$ are normed spaces, $Y$ is finite-dimensional,
$\Omega$ is open in $X$,
$x_*\in\Omega$, $F:\Omega\mapsto Y$ is a continuous map,
and $A=(L,K,Q)$ is a linear-quadratic approximation 
of $F$ at $x_*$ such that the Hessian $H_A$ has a regular zero.  
Then $F$ is open at $x_*$. 
\end{theorem}
\begin{theorem}
Assume that $X$, $Y$ are Banach spaces,
$\Omega$ is open in $X$,
$x_*\in\Omega$, $F:\Omega\mapsto Y$ is a continuous map
and $A=(L,K,Q)$ is a strict linear-quadratic approximation 
of $F$ at $x_*$ such that the Hessian $H_A$ has a regular zero.  
Then $F$ is open at $x_*$. 
\end{theorem}
In particular, we can take $F$ to be a map of the kind considered
in Lemma \ref{isatsrict}. In that case, the {\em Hessian of $F$
at $x_*$} is the Hessian of the strict LQA $(L,K,Q)$,
where $L=DF(x_*)$, $K={\rm ker}\,L$, and 
$Q$ is the quadratic map \mbox{$K\ni k\mapsto D^2F(x_*)(k,k)$}.
\begin{theorem}\label{likeavakov}
Assume that $X$, $Y$ are Banach spaces,
$\Omega$ is open in $X$,
$x_*\in\Omega$, $F:\Omega\mapsto Y$ is a map of class $C^1$ 
such that the derivative $DF$ is differentiable at $x_*$ 
and the Hessian $H$ of $F$ at $x_*$  has a regular zero.  
Then $F$ is open at $x_*$. 
\end{theorem}

\begin{remark}
{\rm Theorem \ref{likeavakov} is very similar to the result proved by Avakov
(cf. \cite{avakov-a,avakov-b}). Avakov's sufficient condition for 
openness---called ``2-regularity'' by some authors, e.g., 
Ledzewicz and Sch\"attler, cf. \cite{ls,ls2}---is fomulated
in slightly different terms, but is easily seen to be 
equivalent to the existence of a regular zero of the Hessian.

Precisely, the algebraic part of Avakov's condition says---using 
$$
L=DF(x_*)(k_*,k_*)\,,\,\,\,\,\,\,K={\rm ker}\,L\,,
$$
and writing $\cq$ for the map $X\ni x\mapsto D^2F(x_*)(x,x)\in Y$, 
and $Q$ for the restriction of $\cq$ to $K$---that 
\begin{itemize}
\item[Av1.] $Lk_*=0$, 
\item[Av2.] $\cq(k_*)\in LX$, 
\end{itemize}
and
\begin{itemize}
\item[Av3.] {\em the map $X\ni x
\mapsto\ca(x)\,\equaldef\,(Lx,\pi(B^\cq(k_*,x)))
\in LX\times Z$ is surjective}. 
\end{itemize}

\begin{lemma} The algebraic part of Avakov's condition holds if and only 
if $k_*$ is a regular zero of the Hessian.
\end{lemma}
\begin{proof}
If Avakov's condition holds, then 
of course $k_*\in K$, so $Q(k_*)\in LX$, and then 
$H(k_*)=0$. Furthermore, the surjectivity of  $\ca$ implies, in particular,
that given any $z\in Z$ there exists 
a $k\in X$ such that $(Lk,\pi(B^\cq(k_*,k)))=(0,z)$. But then
$Lk=0$, so $k\in K$, and $\pi(B^Q(k_*,k))=z$. This shows that 
the map $K\ni k\mapsto\pi(B^Q(k_*,k))\in Z$ 
is surjective, so $k_*$ is a regular zero of $H$.

Conversely, suppose that $k_*$ is a regular zero of $H$.
Then $\pi(Q(k_*))=0$, so $\cq(k_*)\in LX$. 
We now show that $\ca$ is surjective.
Pick $(y,z)\in LX\times Z$. Write $y=Lx$, $x\in X$.
Let $v'=B^\cq(k_*,x)$, $z'=\pi(v')$. Then the fact that 
the map $K\ni k\mapsto\pi(B^Q(k_*,k))\in Z$ is surjective implies that
there exists a $k\in K$ such that $\pi(B^Q(k_*,k))=z-z'$.
Then $\pi(B^\cq(k_*,x+k))=z'+(z-z')=z$, and  
$L(x+k)=Lx=y$, since $Lk=0$. So $\ca(x+k)=(y,z)$. Hence 
$\ca$ is surjective, and our proof is complete.
\end{proof}

\medbreak\noindent We point out, however, that in the work of Avakov
and Ledzewicz-Sch\"attler it also required, in addition to the
algebraic condition described above, that the space $LX$ be closed,
whereas we do not need to make that extra 
\mbox{requirement}, because
in our framework the purely algebraic condition on the Hessian
\mbox{suffices} to obtain the openness theorem.} \proofend
\end{remark}

%\begin{proof}

\bibliographystyle{amsalpha}
\begin{thebibliography}{A}

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Zh. Vychisl. Mat. Mat. Fiz.} {\bf 25}, No. 5, 1985, pp.~680-693.
Eng. translation in {\it U.S.S.R. Comput. Maths. Math. Phys.}
{\bf 25}, No. 3 (Pergamon Press), 1985, pp.~24-32.} %QQQ

\bibitem{avakov-b}{Avakov,~E.R., ``Necessary conditions for an extremum
for smooth 
\mbox{abnormal} problems with  constraints of equality and inequality
type.'' (Russian) {\it Matematicheskie Zametki} {\bf 45}, No. 6, 1989,
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1990, pp.~425-432.}%QQQ

\bibitem{dontchev}{Dontchev,~A.L., \textit{The 
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\bibitem{graves}{Graves,~L.~M., \textit{Some mapping theorems.} 
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\bibitem{ls} {Ledzewicz, U., and H. Sch\"attler, ``An extended maximum 
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\bibitem{ls2} {Ledzewicz, U., and H. Sch\"attler, ``A high-order
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{\bf 38}, 2000, pp. 823-854.}

\end{thebibliography}

\end{document}

