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\section{The classical differential}
\label{sec-gdq-intro}

\noindent If $X$, $Y$ are normed spaces, $S$ is a subset of $X$, 
$s_*\in S$, and $L$ is a bounded linear map from $X$ to $Y$,
then we say that $L$ is a {\bf\em Fr\'echet differential}---or, simply, 
a {\bf\em differential}---{\bf\em at $s_*$} of a ppd map 
$F:S\brokarrow Y$ if $F(s)$ is defined for all $s$ 
in some relative neighborhood 
$S\cap\bmath_X(s_*,\delta)=\{s\in S:\|s-s_*\|<\delta\}$ of
$s_*$ in $S$, and 
\begin{equation}\label{frechet}
\lim_{s\to s_*,s\in S,s\neq s_*}\frac{\|F(s)-F(s_*)-L\cdot(s-s_*)\|}
{\|s-s_*\|}=0\,.
\end{equation}
We say that $F$ is {\bf\em differentiable  at $s_*$} if there exists
a bounded linear map \mbox{$L:X\mapsto Y$} such that 
$L$ is a differential of $F$ at $s_*$. Clearly, if
$F$ is  differentiable  at $s_*$, then the differential $L$ of $F$
at $s_*$ is uniquely determined on the closed linear subspace
$$
\Sigma(S,s_*)={\rm Clos}_X\Biggl({\rm lin\,span}T_{s_*}^BS\Biggr)\,,
$$
where $T_{s_*}^BS$ is the {\bf\em Bouligand tangent cone} of $S$ at 
$s_*$, i.e., the set of all limits
$v=\lim_{j\to\infty}\frac{s_j-s_*}{h_j}$, for all sequences
$\{s_j\}_{j=1}^\infty$, $\{h_j\}_{j=1}^\infty$, 
such that (i)~$s_j\in S$ and $h_j>0$ for all $j$,
and (ii)~$h_j\to 0$ as $j\to\infty$.   (Indeed,
if $L_1$, $L_2$ are two differentials of $F$ at $s_*$, and 
$v\in T_{s_*}^BS$, then we can find sequences 
$\{s_j\}_{j=1}^\infty$ in $S$, $\{h_j\}_{j=1}^\infty$ in $\,]0,+\infty\,[$,
such that $v=\lim_{j\to\infty}\frac{s_j-s_*}{h_j}$. Then, if $i=1,2$, and
$j$ is large enough, we have
$L_i\cdot(s_j-s_*)=F(s_j)-F(s_*)+o(\|s_j-s_*\|)$, and $s_j-s_*=h_jv+o(h_j)$, 
so $L_i\cdot(s_j-s_*)=F(s_j)-F(s_*)+o(h_j)$, and then
$$
L_i\cdot\Biggl(\frac{s_j-s_*}{h_j}\Biggr)=
\frac{F(s_j)-F(s_*)}{h_j}+o(1)\,\,\,{\rm as}\,\,\,j\to\infty\,,
$$
from which it follows that 
$$
L_i\cdot v=\lim_{j\to\infty}\Biggl(\frac{F(s_j)-F(s_*)}{h_j}\Biggr)\,.
$$
Hence $L_1\cdot v=L_2\cdot v$. So $L_1$ and $L_2$ agree on $T_{S_*}^BS$ and 
then they agree on $\Sigma(S,s_*)$, since they are linear and continuous.)

We use $\mathcal{D}F(s_*)$ to denote the set of all differentials of
$F$ at $s_*$. Then $\mathcal{D}F(s_*)$ consists of a single 
linear map if and only if $F$ is differentiable at $s_*$ and
\mbox{$\Sigma(S,s_*)=X$}. In that case, the unique member of 
$\mathcal{D}F(s_*)$ will be referred to as {\bf\em the 
\mbox{differential} of $F$ at $s_*$}, and will be  represented 
with the notation $DF(s_*)$. 

More generally, if $S\subseteq S'\subseteq X$, $s_*\in S$, and
$F:S'\brokarrow Y$, a {\bf\em differential of $F$ at $s_*$ in the
direction of $S$} is a differential at $s_*$ of the restriction
$F\restr S$ of $F$ to $S$. We use $\mathcal{D}F_S(s_*)$ to denote the set
of all differentials of $F$ at $s_*$ in the direction of
$S$. Naturally, $\mathcal{D}F_S(s_*)$ consists of a single linear map if
and only if $F$ is differentiable at $s_*$ in the direction of $S$ and
$\Sigma(S,s_*)=X$. In that case, the unique member of $\mathcal{D}F_S(s_*)$ 
will be referred to as {\bf\em the differential of $F$ at
$s_*$ in the direction of $S$}, and will be represented with the 
notation $DF_S(s_*)$.


\subsection{The chain rule}
\begin{proposition}\label{chainrule}
Let $X_1$, $X_2$, $X_3$ be normed real linear spaces, let $S_1$, $S_2$
be subsets of $X_1$, $X_2$, let $F_1$, $F_2$ be ppd maps from $S_1$ to $X_2$
and from $S_2$ to $X_3$, respectively,
such that $F_1(S_1)\subseteq S_2$. Let $s_*\in S_1$
be such that $F_1$ is differentiable at $s_*$, so in particular
$\sigma_*\equaldef F_1(s_*)$ exists and belongs to $S_2$. 
Assume that  $F_2$ is differentiable at $\sigma_*$. Let
$L_1\in\mathcal{D}F_1(s_*)$, $L_2\in\mathcal{D}F_2(\sigma_*)$.
Let $F=F_2\compo F_1$, $L=L_2\compo L_1$. Then $L\in\mathcal{D}F(s_*)$.
\end{proposition}
\begin{proof}
We assume, as we clearly may without loss of generality,
that $s_*=0$, $\sigma_*=0$.  Fix positive numbers 
$r_1^{\#}$, $\bar r_2^{\#}$, such that $F_i$ is everywhere defined on 
$U_i^{\#}\cap S_i$ for $i=1,2$, where
$U_i^{\#}=\bmath_{X_i}(0,r_i^{\#})$. Since 
$\|s\|^{-1}\Big(F_1(s)-L_1(s)\Big)$
goes to zero as $s\to 0$ via values in $U_1^{\#}\cap S_1$, 
we may pick a positive $\tilde r_1$ such that $\tilde r_1\le r_1^{\#}$ 
and $\|F_1(s)-L_1(s)\|\le \|s\|$ whenever $s\in S_1$ and
$\|s\|\le\tilde r_1$. Let $\kappa=1+\|L_1\|$. Then
$\|F_1(s)\|\le\kappa\|s\|$ whenever $s\in S_1$ and
$\|s\|\le\tilde r_1$. Pick a positive $r_1$ such that 
$r_1\le\tilde r_1$ and $\kappa mr_1\le r_2^{\#}$. Then 
$F_1(s)\in U_2^{\#}\cap S_2$ whenever $s\in S_1$ and
$\|s\|\le r_1$. Define $r_2=r_2^{\#}$, and  let
$U_i=\bmath_{X_i}(0,r_i)$ for $i=1,2$. Then, for $i=1,2$, 
$F_i$ is everywhere defined on  $U_i\cap S_i$, and in addition
$F_1(U_1\cap S_1)\subseteq U_2\cap S_2$. 

For $i=1,2$, $0<r\le r_i$, let
$$
\theta_i(r)=\sup\Biggl\{\frac{\|F_i(s)-L_i\cdot s\|}{\|s\|}:
s\in S_i,0<\|s\|\le r\Biggr\}\,,
$$
so that $\theta_i(r)\to 0$ as $r\downarrow 0$. Then, if 
$s\in U_1\cap S_1$, and we write $\sigma=F_1(s)$, we have
\begin{eqnarray*}
\|F(s)-L(s)\|&=&\|F_2(F_1(s))-L_2(L_1(s))\|\\
&\le&\|F_2(F_1(s))-L_2(F_1(s))\|+\|L_2(F_1(s))-L_2(L_1(s))\|\\
&\le&\theta_2(\|\sigma\|)+\|L_2\|\,\|F_1(s)-L_1(s)\|\\
&\le&\theta_2(\|\sigma\|)\|\sigma\|+\|L_2\|\theta_1(\|s\|)\|s\|\,.
\end{eqnarray*}
On the other hand, 
$$
\|\sigma\|=\|F_1(s)\|\le\|F_1(s)-L_1(s)\|+\|L_1(s)\|
\le\theta_1(\|s\|)\,\|s\|+\|L_1\|\,\|s\|\,.
$$
Therefore $\|F(s)-L(s)\|\le\Theta(\|s\|)\,\|s\|$, where
$$
\Theta(r)\equaldef 
\theta_2\Big(r(\theta_1(r)+\|L_1\|)\Big)\Big(\theta_1(r)+\|L_1\|\Big)
+\|L_2\|\theta_1(r)\,.
$$
Clearly, $\lim_{r\downarrow 0}\Theta(r)=0$, so 
$\|s\|^{-1}\|F(s)-L(s)\|$ goes to $0$ as $s\to 0$ via values
in $S_1$, and our proof is complete.
\end{proof}

\subsection{Open mapping theorems for classical differentials}
\label{sub-openmap-clarkewarga}

\begin{definition}
Let $X$, $Y$ be topological spaces, let $F:X\mapsto Y$ be a map, and let 
$x\in X$. We say that $F$ is {\bf\em open at $x$} if $F$ maps 
neighborhoods of $x$ to neighborhoods of $F(x)$ (that is, whenever $U$ 
is a subset of $X$ such that $x\in{\rm Int}\,U$ it follows that 
$F(x)\in{\rm Int}\,F(U)$). \proofend
\end{definition}

Recall that a {\bf\em cone} in a real linear space $X$ is a 
nonempty subset $C$ of $X$ such that $r\cdot v\in C$ whenever 
$v\in C$, $r\in\rr$, and $r\ge 0$.  It follows from this definition 
that if $C$ is a cone in $X$ then automatically $0\in C$. 

\begin{definition}\label{def-weak-open-dir}
Let $X$, $Y$ be normed real linear spaces, let $\bar x$, $v$ be points of 
$X$, $Y$, respectively, let $S$ be a subset of 
$X$ such that $\bar x\in S$, and let $F:S\mapsto Y$ be a map. 
We say that $F$ is {\bf\em directionally open at $\bar x$ along $S$ 
in the direction of $v$} if there exists a convex cone $D$ in $Y$ 
such that 
\begin{itemize}
\item[(1)] $v\in{\rm Int}(D)$,
\item[(2)] for every positive number $\alpha$
there exists a positive number $\beta=\beta(\alpha)$ such that 
\begin{itemize}
\item[(2.i)] the image under $F$ of the set 
$\{\,x\in S:\|x-\bar x\|\le\alpha\,\}$ contains the set 
$\{\,F(\bar x)+y:y\in D,\|y\|\le\beta\,\}$. 
\end{itemize}\end{itemize}
If the function $\alpha\mapsto\beta(\alpha)$ can be chosen so
that there exist positive numbers $\alpha^*$, $\mu$ such that
$\beta(\alpha)=\mu\alpha$ for $0<\alpha\le\alpha^*$, then
$F$ is said to be {\bf\em directionally open with linear rate
at $\bar x$ along $S$ 
in the direction of $v$}.\proofend
\end{definition}

\medbreak
\begin{center}
\fbox{\fbox{
\begin{minipage}{12cm}
{\large
\begin{theorem}\label{theoopendirclass} 
{\rm (The $\,$directional $\,$open $\,$mapping $\,$theorem 
$\,$for $\,$the $\,$classical \mbox{differential}).} 
Let $X$, $Y$ be normed real linear spaces such that
$Y$ is finite-dimensional, let $\bar x\in X$, $v\in Y$, let $C$ be
a convex cone in $X$, and let 
$$
S=\bar x+C=\{\,x\in X:x-\bar x\in C\,\}\,.
$$
Let $U$ be a neighborhood of $\bar x$ in $X$, let 
$F:U\cap S\mapsto Y$ be a continuous map, and let
$L:X\mapsto Y$ be a bounded linear map such that
$L\in\mathcal{D}F(\bar s)$. 
Assume that $v\in{\rm Int}(L\cdot C)$.  Then $F$ is directionally
open with linear rate at $\bar x$ along $S$ in the direction of $v$.
\end{theorem}}
\end{minipage}}}
\end{center}

\begin{proof}
We assume, without loss of generality, that $\bar x=0$ and 
$F(\bar x)=0$. We let $m={\rm dim}\,Y$.
If $w\in Y$ and $\rho>0$, we use $\bar\bmath_Y(w,\rho)$ to denote the closed
ball   $\{\,y\in Y:\|y-w\|\le\rho\,\}$.  
We let $(\tilde e_1,\ldots,\tilde e_m)$ be a basis of $Y$, write 
$\tilde e_0=-(\tilde e_1+\ldots+\tilde e_m)$,
and pick an $r$ such that $r>0$ and $v+r\tilde e_j\in L\cdot C$ for
every $j\in\{0,\ldots,m\}$. We then let $e_j=v+r\tilde e_j$, 
and  observe that the vectors $e_0,\ldots,e_m$ are affinely independent, 
i.e., that whenever a linear combination
$a_0e_0+a_1e_1+\ldots+a_me_m$ with $a_0+a_1+\ldots+a_m=0$ is equal to zero, 
it follows that $a_0=a_1=\ldots=a_m=0$. (Indeed, if 
$a_0e_0+a_1e_1+\ldots+a_me_m=0$ and $a_0+a_1+\ldots+a_m=0$, 
we get $(a_0+a_1+\ldots+a_m)v+ra_0\tilde e_0
+ra_1\tilde e_1+\ldots+ra_m\tilde e_m=0$. 
Then $a_0\tilde e_0+a_1\tilde e_1+\ldots+a_m\tilde e_m=0$,
so $(a_1-a_0)\tilde e_1+\ldots+(a_m-a_0)\tilde e_m=0$,
which implies $a_1-a_0=a_2-a_0=\ldots=a_m-a_0=0$,
so $a_0=a_1=a_2=\ldots=a_m$, and then
$0=a_0+a_1+\ldots+a_m=(m+1)a_0$, so 
$a_0=a_1=a_2=\ldots=a_m=0$.) We then 
remark that every  member $y$ of $Y$ can be expressed 
in a unique way as an affine combination (that is, 
a linear combination whose coefficients add up to~$1$) 
$a_0(y)e_0+a_1(y)e_1+\ldots+a_m(y)e_m$ of the $e_j$. 
(Indeed, we can find $b_1,\ldots,b_m$ such that
$r^{-1}(y-v)=b_1\tilde e_1+\ldots+b_m\tilde e_m$, 
since the $\tilde e_j$ form a basis of $Y$. Then we can
let $a_0=(m+1)^{-1}(1-b_1-\ldots-b_m)$ and $a_j=a_0+b_j$
for $j=1,\ldots,m$. It is then easy to verify
that $y=a_0e_0+a_1e_1+\ldots+a_me_m$ and $a_0+a_1+\ldots+a_m=1$.
Uniqueness follows from the fact that $e_0,\ldots,e_m$ are affinely 
independent.) It is clear that the functions
$Y\ni y\mapsto a_j(y)\in\rr$ are continuous.

Since $v=(m+1)^{-1}(e_0+\ldots+e_m)$, we have $a_j(v)=(m+1)^{-1}$
for $j=0,\ldots,m$. Therefore we can fix a $\rho$ such that
$\rho>0$ and $a_j(y)>0$ for all $j\in\{0,\ldots,m\}$ and
all  $y\in\bar\bmath_Y(v,\rho)$. Let $W=\bar\bmath_Y(v,\rho)$. 
Then $W\subseteq L\cdot C$, because 
if $y\in\bar\bmath_Y(v,\rho)$ then 
$y=\sum_{j=0}^ma_j(y)e_j$, the $e_j$ are in $L\cdot C$, 
$\sum_{j=0}^ma_j(y)=1$, and $a_j(y)>0$, so 
$y$ is a convex combination of the $e_j$.

Since $e_j\in L\cdot C$, we can  pick $c_j\in C$ such that
$L\cdot c_j=e_j$. 
Let $M:Y\mapsto X$ be the map that sends $y\in Y$ to 
the affine combination $M\cdot y\equaldef\sum_{j=0}^m a_j(y)c_j$. 
Then $M$ is a continuous (and in fact affine, but not necessarily
linear) map, and $L\circ M={\rm identity}_{Y}$, because
$L(M\cdot y)=\sum_{j=0}^m a_j(y)L\cdot c_j
=\sum_{j=0}^m a_j(y)e_j=y$.   

Let $\kappa=\sup\{\|M\cdot y\|:y\in W\}$. For $s>0$, let 
$$
\theta(s)=\sup\Biggl\{\frac{\|F(x)-L\cdot x\|}{\|x\|}:x\in C,\|x\|
\le s\Biggr\}\,,
$$
so $\theta(s)\to 0$ as $s\downarrow 0$. 

Now fix a $\rho^*$ such that (i)~$0<\rho^*<\rho$, and (ii)~if $v\neq 0$ 
then $\rho^*<\|v\|$. Then define $W^*=\bar\bmath_Y(v,\rho^*)$.  
Also, let $\delta^*$ be such that $\delta^*>0$ and
$\bar\bmath_X(0,\delta^*)\subseteq U$.  Then pick a $\tau^*$ such
that $0<\tau^*$, $\kappa\tau^*\le\delta^*$, and
$\kappa\theta(\tau^*\kappa)\le\rho-\rho^*$. Let
$\alpha^*=\kappa\tau^*$, so $0<\alpha^*\le\delta^*$.  For each
positive number $\alpha$, define
$V_\alpha=\bar\bmath_X(0,\tilde\alpha)$ and
$\tau_\alpha=\frac{\tilde\alpha}{\kappa}$, where
$\tilde\alpha=\min(\alpha,\alpha^*)$.  
Then $V_\alpha\subseteq U$, and 
$$
tM\cdot y\in V_\alpha\cap C\quad
{\rm whenever}\quad\alpha>0\,,y\in W\,,\,0\le t\le\tau^*\,.
$$
We then define 
$$
G_t(y)=\frac{1}{t}\Big(\,F(tM\cdot y)\,\Big)\quad
{\rm for}\quad y\in W\,,\,\,0<t\le\tau^*\,.
$$
Clearly, $G_t$ is a continuous map from $W$ to $Y$, if 
$t\in\,]\,0,\tau^*]$, and 
\begin{eqnarray*}
\|G_t(y)-y\|&=&\frac{1}{t}\Big(\|F(tM\cdot y)-ty\|\Big)\\
&\le&\frac{1}{t}\Big(\|F(tM\cdot y)-\|t(L\circ M)\cdot y)\|\Big)\\
&=&\frac{1}{t}\Big(\|F(tM\cdot y)-\|L(tM\cdot y)\|\Big)\\
&\le&\frac{1}{t}\Big(\theta(t\kappa)t\kappa\Big)\\
&=&\kappa\theta(t\kappa)\,.
\end{eqnarray*}

If $y\in W^*$ and $0<t\le\tau^*$, 
we can consider the map $\mu_{t,y}:W\mapsto Y$
given by 
$$
\mu_{t,y}(z)=z-G_t(z)+y\,.
$$
Then 
$$
\|\mu_{t,y}(z)-v\|=\|z-G_t(z)+y-v\|\le\|z-G_t(z)\|+\|y-v\|
\le\kappa\theta(t\kappa)+\rho^*\le\rho\,,
$$
since $\kappa\theta(\tau^*\kappa)\le\rho-\rho^*$. Therefore
$\mu_{t,y}$ is a continuous map from $W$ to $W$.
By the Brouwer fixed point theorem, there exists, for
each $y\in W^*$ and each $t\in\,]0,t^*]$, a $z=z(t,y)\in W$ 
such that $\mu_{t,y}(z)=z$. Then $z-G_t(z)+y=z$, so $G_t(z)=y$.
But then $F(tM\cdot(z(t,y)))=ty$.  Since this is true for
every $y\in W^*$ and every $t\in\,]0,\tau^*]$, we have shown that
$$
tW^*\subseteq F(tM(W))\,\,\,{\rm whenever}\,\,\,0<t\le\tau^*\,.
$$
Furthermore, the inclusion of the previous line also holds
for $t=0$, since $F(0)=0$. It follows that 
$$
\bigcup_{0\le t\le\tau_\alpha}\,tW^*\subseteq F(V_\alpha\cap C)
\,\,\,{\rm whenever}\,\,\,\alpha>0\,.
$$
Let $D$ be the convex cone generated by $W^*$. Then
$v$ is an interior point of $D$. Furthermore, $D$ is closed. 
(Indeed, if $v=0$ then $D=Y$, so $D$ is closed. Assume that
$v\neq 0$. If $\{y_j\}_{j=1}^\infty$ is a sequence
of members of $D$ that converges to a limit $y$, 
then we can write $y_j=t_jw_j$, $t_j\in[0,\infty\,[\,$, 
$w_j\in W^*$. After passing to a subsequence, if necessary,
we may assume that $w=\lim_{j\to\infty}w_j$ exists and belongs to
$W^*$.  Since $v\neq 0$, the inequality $\rho^*<\|v\|$ implies that
$0\notin W^*$, so $w$ and the $w_j$ are all $\neq 0$. Then
$t_j=\frac{\|y_j\|}{\|w_j\|}$, and $\lim_{j\to\infty}t_j=t=
\frac{\|y\|}{\|w\|}$.  It follows that $y=tw$, so $y\in D$.)

Now fix a constant $\nu$ such that 
$\bmath_Y(0,1)\cap D\subseteq\bigcup_{0\le t\le\nu}tW^*$.
(The existence of $\nu$ is trivial. Indeed, if $v=0$ we can 
pick $\nu=\frac{1}{\rho^*}$. If $v\neq 0$, then 
every $w\in W^*$ satisfies $\|w-v\|\ge\rho^*$, so 
$\|v\|\le\rho^*+\|w\|$, and then 
$\|w\|\ge\|v\|-\rho^*$. Since $\rho^*<\|v\|$, we can take 
$\nu$ to be any nmuber such that $\nu\ge(\|v\|-\rho^*)^{-1}$, and conclude
that if $y\in\bmath_Y(0,1)\cap D$, and we
write $y=tw$, with $w\in W^*$, then 
$t=\frac{\|y\|}{\|w\|}\le\frac{1}{\|v\|-\rho^*}
\le\nu$.)  It follows that 
$\bmath_Y(0,s)\cap D\subseteq\bigcup_{0\le t\le\nu s}tW^*$
for every positive $s$.  

Now, if $\alpha$ is an arbitrary positive number, 
and we let $\beta(\alpha)=\nu^{-1}\tau_\alpha$, we have 
$$
\bmath_Y(0,\beta(\alpha))\cap D\subseteq
\bigcup_{0\le t\le\nu\beta(\alpha)}tW^*=\bigcup_{0\le t\le\tau_\alpha}tW^*
\subseteq F(V_\alpha\cap C)\,.
$$
The proof of Theorem  \ref{theoopendirclass} is thus complete. 
\end{proof}


\medbreak
\begin{center}
\fbox{\fbox{
\begin{minipage}{12cm}
{\large
\begin{theorem}\label{theoopenclass} 
{\rm (The $\,$open $\,$mapping $\,$theorem $\,$for $\,$the 
$\,$classical \mbox{differential}).}
Let $X$, $Y$ be normed real linear spaces such that
$Y$ is finite-dimensional, let $\bar x\in X$, 
let $U$ be a neighborhood of $\bar x$ in $X$, and let 
$F:U\mapsto Y$ be a continuous map. Assume that $F$ is differentiable at
$\bar x$, and let $L=DF(\bar x)$. 
Assume that $L$ is surjective. Then $F$ is open at $\bar x$.
\end{theorem}}
\end{minipage}}}
\end{center}

\begin{proof}
It suffices to apply Theorem  \ref{theoopendirclass} with
$v=0$, $C=X$.
\end{proof}

\subsection{The transversal intersection theorem
for the \mbox{classical} differential}

\begin{definition}
Assume that $Y$ is a normed real linear space, $S$ is a 
subset of $Y$, $s_*\in S$, and $C$ is a convex cone 
in $Y$. We say that $C$ is an {\bf\em approximating cone
to $S$ at $s_*$} if there exist a normed space $X$,
a neighborhood $U$ of $0$ in $X$, a convex cone 
$D$ in $X$, a continuous map $F:U\cap D\mapsto Y$,
and a bounded linear map $L\in{\rm Lin}(X,Y)$, such that
$F(0)=s_*$, $F(U\cap D)\subseteq S$, $LD=C$, 
and $L\in\bmath{D}F(0)$. 

We say that $C$ is a {\bf\em limiting 
approximating cone to $S$ at $s_*$} if there exists
an increasing sequence $\{C_j\}_{j=1}^\infty$ of convex cones
in $Y$ such that $C={\rm Clos}\,\bigcup_{j=1}^\infty C_j$
and each $C_j$ is an approximating cone
to $S$ at $s_*$. \proofend
\end{definition}


\begin{definition}
Let $X$ be a real linear space. We say 
that two convex cones $C^1$, $C^2$ in $X$ are {\bf\em transversal}, and write 
$$
C^1\transv C^2\,\,\,,
$$
if $C^1-C^2=X$.   We say that $C^1$ and $C^2$ in $X$ are {\bf\em strongly
transversal}, and write 
$$
C^1\stransv C^2\,\,\,,
$$
if $C^1\transv C^2$ and $C^1\cap C^2\neq\{0\}$.\proofend
\end{definition}


\begin{definition}
Assume that $Y$ is a normed real linear space, $S^1$ and $S^2$ are
subsets of $Y$, and $s_*\in S^1\cap S^2$.  We say that $S^1$ and $S^2$
are {\bf\em locally separated at $s_*$} if every neighborhood $U$ of
$s_*$ contains a point $s$ of $S^1\cap S^2$ such that $s\neq
s_*$.\proofend
\end{definition}

\begin{definition}
Let $Y$ be a normed real linear space, and let $Y^\dagger$ denote the dual
space of $Y$. Let $C$ be a convex cone
in $Y$. The {\bf\em polar} of $C$ is the set 
\begin{equation}
\quad\quad\qquad\qquad
C^\dagger\,\equaldef\,\{\lambda\in Y^\dagger:\lambda(c)\le 0\,\,{\rm for\,\,
all}\,\,c\in C\}\,.\quad\qquad\qquad\qquad\quad\,\,\,\diamondsuit
\end{equation}
\end{definition}

\begin{definition}
Let $Y$ be a normed real linear space, and let 
$C^1$, $C^2$  be two convex cones in $Y$. We say that 
the cones $C^1$ and $C^2$  are {\bf\em linearly separated}
if \mbox{$(C^1)^\dagger\cap(-C_2)^\dagger\neq\{0\}$}.\proofend
\end{definition}

In other words, {\bf\em $C^1$ and $C^2$  are linearly separated
if and only if there exists a nontrivial continuous linear functional 
$\lambda:Y\mapsto\rr$
such that  $\langle\lambda,v\rangle\le 0$ for all $v\in C^1$ and
$\langle\lambda,v\rangle\ge 0$ for all $v\in C^2$.}

\begin{proposition}\label{mystery}
Let $C^1$, $C^2$ be convex cones in a finite-dimensional 
real linear space $Y$. Then the following three conditions are equivalent:
\ben
\item[(1)] $C^1\transv C^2$;
\item[(2)] $C^1$ and $C^2$ are not linearly separated;
\item[(3)] either
\ben
\item[(a)] $C^1$ are $C^2$ are linear subspaces of $Y$ such that 
$Y=C^1\oplus C^2$, 
\een
or
\ben
\item[(b)] $C^1\stransv C^2$.
\een
\een
\end{proposition}
\begin{proof}
The equivalence of (1) and (2) is very easy to prove: if 
(2) does not hold, then there  exists a  
$\lambda\in Y^\dagger\backslash\{0\}$ such 
that $\langle\lambda,c\rangle\le 0$ for all $c\in C^1$ and
$\langle\lambda,c\rangle\ge 0$ for all $c\in C^2$, so 
that $\langle\lambda,c\rangle\le 0$ for all $c\in C^1-C^2$
and then 
$C^1-C^2$ cannot be the whole space because $\lambda$ is nontrivial,
so (1) does not hold;
conversely, if (1) does not hold, then 
the convex cone $C=C^1-C^22$ is not the whole space, and  then 
the closure $\bar C$ of $C$ is not the whole space either, because
$Y$ is finite-dimensional; therefore the Hahn-Banach theorem 
implies the existence of a nontrivial linear 
functional $\lambda$ which is nonpositive on $C^1-C^2$; 
then $\lambda$ is nonpositive
on $C^1$ and nonnegative on $C^2$, so 
$\lambda\in(C^1)^\dagger\cap(-C^2)^\dagger$ and $\lambda\neq 0$, 
showing that  (2) does  not hold.

The implication (3)$\Rightarrow$(2) is trivial. To prove that 
(2)$\Rightarrow$(3), it suffices to show that if 
$C^1$ and $C^2$ are transversal but not strongly transversal 
then (a) holds. So assume that $C^1\transv C^2$ but 
$C^1$ and $C^2$ are not strongly transversal.
It then follows that $C^1\cap C^2=\{0\}$.  
We will show that
$C^1$ and $C^2$ are linear subspaces. This will then imply---since
$C^1-C^2=Y$ and $C^1\cap C^2=\{0\}$---that $C^1+C^2=Y$ 
and $C^1\cap C^2=\{0\}$, so $C^1\oplus C^2=Y$, establishing (a). 

To prove that $C^1$ and $C^2$ are linear subspaces, we will  just do it for
$C^1$, the other case being similar. Since $C^1$ is a convex cone, it suffices
to show that 
$$
c\in C^1\Longrightarrow -c\in C^1\,.
$$
Pick $c\in C^1$. Using the fact that $C^1\transv C^2$, i.e., that
$C^1-C^2=X$, express $-c$ as a difference $-c=c^1-c^2$, $c^i\in C^i$.  Then
$c^1+c=c^2$. But $c^1+c\in C^1$ and $c^2\in C^2$. Since $C^1\cap
C^2=\{0\}$, we conclude that $c^1+c=0$. So $-c=c^1$, showing that $-c\in
C^1$, and completing our proof.
\end{proof}

\bigbreak
\begin{center}
\fbox{\fbox{
\begin{minipage}{12cm}
{\large
\begin{theorem}\label{theosepbasic-weak}
{\rm (The transversal intersection theorem for classical 
\mbox{differentials}).} Assume that $Y$  is a finite-dimensional
real linear space, $S^1$ and $S^2$ are subsets of $Y$, $s_*\in S^1\cap
S^2$, and $C^1$, $C^2$ are limiting \mbox{approximating} cones to $S^1$,
$S^2$ at $s_*$.  Assume that the cones $C^1$ and $C^2$ are strongly
transversal.  Then $S^1$ and $S^2$ are not locally separated at $s_*$.
\end{theorem}}
\end{minipage}}}
\end{center}

\bigbreak
\begin{proof}
Let $\,\,C^1={\rm Clos}\,\bigcup_{j=1}^\infty C^1_j\,$,
$\,\,C^2={\rm Clos}\,\bigcup_{j=1}^\infty C^2_j\,$,
where each $C_j^i$ is an \mbox{approximating} cone
to $S^i$ at $s_*$ and $C_j^i\subseteq C_{j+1}^i$ for all $i,j$. 
Fix a basis $(e_1,\ldots,e_m)$ of $Y$, and 
then let $e_0=-(e_1+\ldots+e_m)$. Using the fact that $C_1\transv C_2$,
find $e_k^i$, for $i=1,2$, $k=0,\ldots,m$, such that 
$e_k^i\in C^i$ and $e_k^1-e_k^2=e_k$. Then find sequences
$\{e_{k,j}^i\}_{j=1}^\infty$, for each $i$ and each $k$,
such that $e_{k,j}^i\in C_j^i$ for every $i,j,k$, and 
$\lim_{j\to\infty}e_{k,j}^i=e_k^i$. Let 
$e_{k,j}=e_{k,j}^1-e_{k,j}^2$. Then 
$\lim_{j\to\infty}e_{k,j}=e_k$ for $k=0,\ldots,m$.  Since
$0$ is an interior point of the convex hull of 
$e_0,\ldots,e_m$, it is clear that there exists a $j^*$ such 
$0$ is an interior point of the convex hull $\Sigma_j$ of 
$e_{0,j},\ldots,e_{m,j}$ if $j\ge j^*$. Since $\Sigma_j
\subseteq C_j^1-C_j^2$, we see that $C_j^1\transv C_j^2$
if $j\ge j^*$. 

We now claim that 
\begin{itemize}
\item[(\&)] {\em There exists a $j^{\#}$ such that
$j^{\#}\ge j^*$ and $C_j^1\stransv C_j^2$
whenever $j\ge j^{\#}$.}  
\end{itemize}
If (\&) was not true, there would exist an infinite subset $J$ of
\mbox{$\{j\in\nn:j\ge j^*\}$} such that $C_j^1$ is not strongly transversal
to $C_j^2$ for $j\in J$. In view of \mbox{Proposition} \ref{mystery},
\mbox{since} $C_j^1\transv C_j^2$ whenever $j\in J$, 
we can conclude that for every $j\in J$ the cones $C_j^1$ and $C_j^2$
are in fact linear subspaces and $C_j^1\oplus C_j^2=Y$. Since the
sequences $\{C_j^1\}_{j=1}^\infty$, $\{C_j^2\}_{j=1}^\infty$, are
increasing, we have $\bigcup_{j=1}^\infty C^1_j=\bigcup_{j\in
J}^\infty C^1_j$ and \mbox{$\bigcup_{j=1}^\infty C^2_j=\bigcup_{j\in
J}^\infty C^2_j$}.  Furthermore, since the families $\{C_j^i\}_{j\in
J}^\infty$ are increasing \mbox{sequences} of linear \mbox{subspaces},
and $Y$ is finite-dimensional, there exists a $j^\dagger$ such that
$\{C_j^i\}=C_{j^\dagger}^i$ \mbox{whenever} $j\in J$, $j\ge j^\dagger$,
$i=1,2$. But then $\bigcup_{j=1}^\infty
C^1_j=\bigcup_{j\in{}J}C^1_j=C_{j^\dagger}^1$ and
\mbox{$\bigcup_{j=1}^\infty C^2_j=\bigcup_{j\in J}C^2_j=C_{j^\dagger}^2$}.
It follows that $\bigcup_{j=1}^\infty C^1_j$ and $\bigcup_{j=1}^\infty
C^2_j$ are linear subspaces of $Y$. In particular, these two sets are
closed, and then $C^1=\bigcup_{j\in J}C^1_j=C_{j^\dagger}^1$ and
$C^2=\bigcup_{j\in J}C^1_j=C_{j^\dagger}^2$.  But then
$C_{j^\dagger}^1\stransv C_{j^\dagger}^2$, since $C^1\stransv
C^2$. This contradicts the fact that $C_{j^\dagger}^1\oplus
C_{j^\dagger}^2=Y$. This contradiction proves (\&).

If we pick any $j$ such that $j\ge j^\#$, and substitute
$C_j^1$, $C_j^2$ for $C^1$, $C^2$, then 
$C^1$, $C^2$ are approximating cones to $S^1$, $S^2$ 
at $s_*$,  and in addition $C^1\stransv C^2$. 
In other words, we have shown
that we may assume, without loss of generality, that 
\begin{itemize}
\item[(\&\&)] {\em 
$C^1$ and $C^2$ are approximating cones to $S^1$ and $S^2$ at $s_*$.}
\end{itemize}
From now on, we assume that (\&\&) holds, and also that 
$s_*=0$.   

Fix a neighborhood $V$ of $0$ in $Y$. 
For each $i\in\{1,2\}$, choose a normed space $X^i$, a 
\mbox{convex} cone
$D^i$ in $X^i$, a neighborhood $U^i$ of 
$0$ in $X^i$, a continuous map \mbox{$F^i:U^i\cap D^i\mapsto Y$},
and a continuous linear map $L^i:X^i\mapsto Y$, such that
$C^i=L\cdot D^i$, $F(0)=0$, $L\in\mathcal{D}F^i(0)$, 
and $F^i(U^i\cap D^i)\subseteq S^i$.
After shrinking the $U^i$, if necessary, we may assume
that $F^i(U^i\cap D^i)\subseteq V$ for $i=1,2$.

Using the fact that 
$C^1\stransv C^2$, 
choose a linear functional $\mu:Y\mapsto\rr$ 
and a vector $\bar y\in C^1\cap C^2$ such that $\mu(\bar y)>0$. Let
$\alpha=\mu(\bar y)$. 
Write $\bar y=L^1\bar v^1=L^2\bar v^2$, with $\bar v^1\in D^1$,
$\bar v^2\in D^2$. Let $U=U^1\times U^2$, 
$D=D^1\times D^2$. Define a map $F:U\cap D\mapsto Y\times\rr$ by letting 
$$
F(x^1,x^2)=(F^1(x^1)-F^2(x^2),\mu(F^1(x^1)))\,\,\,.{\rm if}
\,\,\,x^1\in U^1\cap D^1\,,\,\,x^2\in U^2\cap D^2\,.
$$
Then $F$ is clearly continuous.  A simple calculation
shows that  $F$ is differentiable at $(0,0)$, and a differential 
of $F$ at $(0,0)$ is the linear map $L:X^1\times X^2\mapsto Y\times\rr$
given by 
$$
L\cdot(v^1,v^2)=(L^1\cdot v^1-L^2\cdot v^2,\mu(L^1\cdot v^1))
$$
for $(v^1,v^2)\in X^1\times X^2$. Given any $k\in\{0,\ldots,m\}$, 
the fact that $C^1\transv C^2$ implies that 
we can find $(\tilde v_k^1,\tilde v_k^2)\in D$ such that 
$L^1\cdot\tilde  v_k^1-L^2\cdot\tilde  v_k^2=e_k$. Then  
$L\cdot(\tilde v_k^1,\tilde v_k^2)=(e_k,r_k)$ for some 
$r_k\in\rr$.  If $r\ge r$, then we can find a pair
$(v_k^1(r),v_k^2(r))\in D$ such that 
$L\cdot(v_k^1(r),v_k^2(r))=(e_k,r)$ by letting 
$v_k^1(r)=\tilde v_k^1+\beta\bar v_1$, $v_k^2=\tilde v_k^2+\beta\bar v_2$, 
where $\beta=\frac{r-r_k}{\alpha}$.  Let $\bar =\max(r_0,\ldots,r_m)$. 
Then we have shown that the vector $(e_k,r)$ belongs to 
$L\cdot D$ if $k\in\{0,\ldots.m\}$ and $r\ge\bar r$. Let 
$\Sigma$ be the convex hull of the set $\{e_0,\ldots,e_m\}$.
Then $\Sigma\times[\bar,\infty\,[\,\subseteq L\cdot D$. In particular,
this implies that the vector $(0,\bar r+1)$  belongs to
the interior ${\rm Int}(L\cdot D)$ of $L\cdot D$. Therefore 
$(0,1)\in{\rm Int}(L\cdot D)$ as well, and then 
Theorem  \ref{theoopendirclass} implies that 
there exists a positive number  $\rho$ such  that 
$(0,r)\in F(U\cap D)$ whenever $0\le r\le\rho$. Pick any $r$ such 
that $0<r\le\rho$. Then $(0,r)\in F(U\cap D)$, so 
there exist $x^1\in U^1\cap D^1$, $x^2\in U^2\cap D^2$, such that 
$F^1(x^1)-F^2(x^2)=0$ and $\mu(F^1(x^1))=r$.  If we let 
$y=F^1(x^1)=F^2(x^2)$, then $y\in S^1\cap S^2$.
Since $\mu(y)=r\neq 0$, $y$ cannot be the origin of $Y$.
Since $y\in F^1(U^1\cap D^1)\subseteq V$, $y$ satisfies
$y\in V$.  Since $V$ is arbitrary, we have shown that 
$S^1\cap S^2$ contains nonzero points arbitrarily close  to the origin,
which is the desired conclusion. 
\end{proof}


\subsection{The linear separation
theorem for the \mbox{classical} differential}

\bigbreak
\begin{center}
\fbox{\fbox{
\begin{minipage}{12cm}
{\large
\begin{theorem}\label{theosepbasic-dual}
{\rm (The linear separation theorem for the
classical  \mbox{differential}).} Let $Y$ be a finite-dimensional real 
normed space.
Let $S^1$, $S^2$ be subsets of $Y$, let $s_*\in S^1\cap S^2$,
and let $C^1$, $C^2$ be limiting approximating cones 
to $S^1$, $S^2$ at $s_*$. Assume that the sets $S^1,S^2$ are 
locally separated at $s_*$. Then either
\ben
\item[(a)] $C^1$ and $C^2$ are linear subspaces and $Y=C^1\oplus C^2$
\een
or 
\ben	
\item[(b)] $C^1$ and $C^2$ are linearly separated.
\een
\end{theorem}} 
\end{minipage}}}
\end{center}

\begin{proof}
Assume that both (a) and (b) are false. Then Proposition 
\ref{mystery} implies that either (a) holds or $C^1\stransv C^2$.
But we are assuming that (a) is false. So $C^1\stransv C^2$.
Then Theorem \ref{theosepbasic-dual}
tells us that $S^1$ and $S^2$ are not
locally separated at $s_*$, and we have reached a contradiction.
\end{proof}
\end{document}

