An important class of examples of regular maps is \mbox{provided} by the
following two results, which will be proved together.
\begin{theorem}\label{kakucon}
Assume that $K$ is a compact metric 
\mbox{space},  $Y$ is a normed 
space, and  $C$ is a convex subset of $Y$. Let $\Phi:K\tto C$ be a 
set-valued map such that the graph of $\Phi$ is compact and the value 
$\Phi(x)$ is a nonempty convex set for every $x\in K$. Then $\Phi$ is regular
as a map from $K$ to $C$.
\end{theorem}
\begin{theorem}\label{kakucon-nocomp}
Assume that $X$ is a  metric space, $Y$ is a normed 
space, and $C$ is a convex subset of $Y$. Let $\Phi:K\tto C$ be an 
upper semicontinuous set-valued map with nonempty compact
convex values.  Then $\Phi\in\regu{X}{C}$.
\end{theorem}
\begin{proof}
Theorem \ref{kakucon-nocomp} follows from 
Theorem \ref{kakucon} because, if $\Phi$ satisfies the hypotheses
of Theorem \ref{kakucon-nocomp}, and $K$ is a compact subset
of $X$, then $G\restr K$ is an upper semicontinuous set-valued
map from $K$ to $Y$ with compact values, so 
$\gr(G\restr K)$ is compact. Since 
$G\restr K$ has convex nonempty values, Theorem \ref{kakucon} 
says that $G\restr K\in\regu{K}{C}$. Since this is true for an arbitrary
compact subset $K$ of $X$, Theorem \ref{kakucon-nocomp} follows.

We now prove Theorem \ref{kakucon}.
For $x\in K$, $r>0$, let $B(x,r)$ denote the open $r$-ball in $K$ with center
$x$. For each positive number $\eps$, find a finite subset $F_\eps$ of 
$K$ such that $K=\cup_{\xi\in F_\eps}B(\xi,\eps)$. 
For $x\in K$, $\xi\in F_\eps$,  let 
$$
\psi_{\xi,\eps}(x)=\max\Big(0,\eps-{\rm dist}(x,\xi)\Big)\,, 
$$
so $\psi_{\xi,\eps}:K\to\rr$ is continuous and nonnegative 
and $\psi_{\xi,\eps}(x)>0$ if and only if $x\in B(\xi,\eps)$. 

Define 
$$
\varphi_{\xi,\eps}(x)=\frac{\psi_{\xi,\eps}(x)}{\sum_{\xi'\in
F_\eps}\psi_{\xi',\eps}(x)}\,.
$$
The the $\varphi_{\xi,\eps}$ are continuous nonnegative real-valued functions
on $K$ such that 
$$
\sum_{\xi\in F_\eps}\varphi_{\xi,\eps}(x)=1\quad{\rm for\,\,\,all}\quad x\in
K\,.
$$  
Using the fact that the sets $\Phi(x)$ are nonempty, 
pick a $y_{\xi,\eps}\in\Phi(\xi)$ for each $\xi\in F_\eps$. Define
$H_\eps:K\to Y$ by letting 
\begin{equation}\label{theflaf}
H_\eps(x)\,\,=\,\,\sum_{\xi\in F_\eps}\varphi_{\xi,\eps}(x)y_{\xi,\eps}\,.
\end{equation}
Then each $H_\eps$ is a continuous map from $K$ to $Y$. Moreover, 
the facts that (a)~the $y_{\xi,\eps}$ belong to $\Phi(K)$, which is a subset 
of $C$, and (b)~$C$ is convex, imply that $H_\eps$ takes values in $C$. 

Now let $\{\eps_j\}$ be a sequence of positive numbers
that converges to zero.  We claim that the
graphs $\mathcal{ H}_j$ of the $H_{\eps_j}$ inward-converge to the graph $\gr(\Phi)$ 
of $\Phi$.
To see this, we let 
$$
\alpha_j=\sup\{{\rm dist}(q,\gr(\Phi)):q\in\mathcal{ H}_j\}\,,
$$
and prove by contradiction that $\alpha_j\to 0$. 

Assume that $\{\alpha_j\}$ does not go to zero. Then we may pass to
a subsequence and assume that the $\alpha_j$ are bounded below by a fixed
strictly positive $\alpha$.   Pick a $\beta$ such that 
$0<\beta<\alpha$.  Pick $q_j\in\mathcal{ H}_j$ such that 
\begin{equation}\label{forcontra}
{\rm dist}(q_j,\gr(\Phi))\ge\beta\,.
\end{equation}
Write $q_j=(x_j,y_j)$.  
Then the $x_j$ belong to $K$, so we may assume, after passing to a subsequence,
that the limit $\bar x=\lim_{j\to\infty}x_j$ exists.  

Fix a $\gamma$ be such that $0<\gamma$ and $2\gamma<\beta$. 

Pick a positive  $\delta$  
such  that ${\rm dist}(z,\Phi(\bar x))<\gamma$ whenever 
$w\in K$, $z\in\Phi(w)$, and ${\rm dist}(\bar x,w)\le\delta$. 
(The existence of such a $\delta$ is easily proved: 
suppose, by contradiction, that there exist sequences 
$\{w_k\}$, $\{z_k\}$ such that 
$z_k\in\Phi(w_k)$, $w_k\to\bar x$ as $k\to\infty$, and
${\rm dist}(z_k,\Phi(\bar x))\ge\gamma$; since the graph of 
$\Phi$ is compact we may assume, after passing to a subsequence, that the
sequence  $\{z_k\}$ converges to a limit $z$;  since $z_k\in\Phi(w_k)$, 
and $w_k\to\bar x$, the compactness of $\gr(\Phi)$ also implies
that $z\in\Phi(\bar x)$; since $z_k\to z$, we see that 
${\rm dist}(z_k,\Phi(\bar x))\to 0$, and we have derived a contradiction.)

Now let $j^*$ be such that 
\begin{equation}\label{inbiu}
2\,\eps_j\le\delta\quad{\rm and}\quad
{\rm dist}(x_j,\bar x)\le\min\Big(\gamma,\frac{\delta}{2}\Big)
\end{equation}
whenever $j\ge j^*$.  
If $j\ge j^*$,  $x=x_j$, and $\eps=\eps_j$, then all the terms in the summation
of the right-hand side of (\ref{theflaf}) for which 
${\rm dist}(\xi,\bar x)\ge\delta$ vanish, because
${\rm dist}(\xi,\bar x)\ge\delta$ implies 
${\rm dist}(x_j,\xi)\ge\frac{\delta}{2}\ge\eps_j$ in view of (\ref{inbiu}), 
so $\varphi_{\xi,\eps_j}(x_j)=0$. 
Therefore 
\begin{equation}\label{theflaffa}
y_j\,=\,H_{\eps_j}(x_j)\,\,=\,\,\sum_{\xi\in\hat F_{\eps_j,\bar x}}
\varphi_{\xi,\eps_j}(x_j)y_{\xi,\eps_j}\,,
\end{equation}
where
$$
\hat F_{\eps_j,\bar x}\,\,=\,\,\big\{\xi\in F_{\eps_j}:{\rm
  dist}(\xi,\bar x)<\delta\Big\}\,. 
$$
For every $\xi\in\hat F_{\eps_j,\bar x}$, the point $y_{\xi,\eps_j}$ is in 
$\Phi(\xi)$, so its distance to $\Phi(\bar x)$ is strictly smaller than 
$\gamma$. Therefore we may pick $\tilde y_{\xi,\eps_j}\in\Phi(\bar x)$ 
such that $\|y_{\xi,\eps_j}-\tilde y_{\xi,\eps_j}\|\le\gamma$. 
If we let 
\begin{equation}\label{andcomp}
\tilde y_j=\sum_{\xi\in\hat F_{\eps_j,\bar x}}
\varphi_{\xi,\eps_j}(x_j)\tilde y_{\xi,\eps_j}\,,
\end{equation}
and compare (\ref{andcomp}) with (\ref{theflaffa}), we find
$$
\|\tilde y_j-y_j\|\le\sum_{\xi\in\hat F_{\eps_j,x}}
\varphi_{\xi,\eps_j}(x_j)\|\tilde y_{\xi,\eps_j}
-y_{\xi,\eps_j}\|\le\gamma\,.
$$
On the other hand, $\tilde y_j$ is a convex combination of points
of $\Phi(\bar x)$, so $\tilde y_j\in\Phi(\bar x)$.  Since
$\|y_j-\tilde y_j\|\le\gamma$ and ${\rm dist}(x_j,\bar x)\le\gamma$
for $j\ge j^*$, and $\tilde q_j=(\bar x,\tilde y_j)\in\gr(\Phi)$, 
we can conclude that 
\begin{equation}\label{forcontranew}
{\rm dist}(q_j,\gr(\Phi))\le 2\,\gamma<\beta\quad{\rm if}\quad
j\ge j^*\,.
\end{equation}
With formulas (\ref{forcontra}) and (\ref{forcontranew}), we have
established 
that the assumption that $\alpha_j$ does not go zero leads to a
contradiction. So $\alpha_j\to 0$, and the proof is complete. 
\end{proof}


****************************************************************************

The following example shows that a directional derivate container 
in the restricted sense need not be one in the extended sense. 

We choose $M=N=\rr^2=\comp$, the complex plane, and let $\Omega$
denote the open set $\rr^2\backslash\Big(]\!-\infty,0]\times\{0\}\Big)$,
i.e. the complex plane with the closed negative real axis removed.  We
use $\disc$ to denote the closed unit disc in $\comp$. We identify
${\rm Lin}(\rr^2,\rr^2)$ with $\rr^{2\times 2}$, the space of all 
$2\times 2$ real matrices.

Fix a number $\alpha$ such that $0<\alpha\le\frac{\pi}{2}$, and let $C$ 
be the sector 
$$
C=\{z\in\comp:z=|z|e^{i\theta}\,,\,|\theta|\le\alpha\}\,. 
$$
Let $C'$ denote the interior of $C$, and let $C_0=C\backslash\{0\}$.
Then $C$ is a closed convex cone in $X$,  and $C_0\subseteq\Omega$.  

We use ${\rm Arg}$ to denote the principal branch of the ``argument
function.''  For $z\in\Omega$, ${\rm Arg}(z)$ is the unique real number
$\theta$ such that $|\theta|<\pi$ and $z=|z|e^{i\theta}$.  We then use
${\rm Ln}$ to denote the principal branch ---also defined on $\Omega$---
of the natural logarithm function. Then ${\rm Ln}(z)={\rm ln}\,|z|+
i{\rm Arg}(z)$.  For  $w\in\comp$, $z\in\Omega$, we write 
$z^w=e^{w\,{\rm Ln}(z)}$.  Then for each fixed $w$ the function
$z\to z^w$ is holomorphic on $\Omega$  and its derivative 
is the function $z\to wz^{w-1}$.    We let $f(z)=z^{1+i}$, $g(z)=z^i$,
so both $f$ and $g$ are holomorphic on $\Omega$ and $\frac{df}{dz}=
(1+i)g$.   An easy calculation shows that $g$ is bounded, since
$g(z)=e^{i\,{\rm ln}\,|z|-{\rm Arg}(z)}$ so $|g(z)|=e^{-{\rm Arg}(z)}\le
e^\pi$.   Therefore the derivative of $f$ is globally bounded on $\Omega$,
so $f$ is globally Lipschitz on any convex subset of $\Omega$. In 
particular, $f$ is globally Lipschitz on $C_0$. Since  
$|f(z)|= |z|.|g(z)|\le\kappa|z|$ for some constant $\kappa$, we can extend 
$f$ to a Lipschitz function on $C$ by setting $f(0)=0$. 

Let ${\bf M}$ be the 2-dimensional subspace of $\rr^{2\times 2}$ that
consists of those matrices $\{m_{ij}\}_{1\le i,j\le 2}$ such that
$m_{11}=m_{22}$ and $m_{12}=-m_{21}$.  Then, if $h$ is a map of class
$C^1$ from an open subset $U$ of $\rr^2$ to $\rr^2$, it is clear that
the assertion that $Dh(p)\in{\bf M}$ for all $p\in U$ is exactly
equivalent to the Cauchy-Riemann equations.  In particular, $h$ is
holomorphic iff $Dh(p)\in{\bf M}$ for all $p\in U$.

There is an obvious bijection between $\comp$ and ${\bf M}$, that assigns
to each $z=x+iy\in\comp$ the matrix $L_z$ with components  $m_{ij}$ given by
$m_{11}=m_{22}=x$, $m_{21}=-m_{12}=y$.   Then, if $h$ is holomorphic and 
$h'$ is its derivative in the usual sense of  complex function theory, 
the differential $Dh(z)$ is just $L_{h'(z)}$.

We now compute a weak derivate container for $f$ at $0$ in the
direction of $C$. For this purpose, we let $A$ denote the annulus 
$$
A=\{z\in\comp:a\le|z|\le b\}\,,
$$
where $a=\sqrt{2}e^{-\alpha}$ and $b=\sqrt{2}e^\alpha$, 
and use ${\bf D}$ to denote the set of all matrices  $L_z$ for all $z\in A$.
Then ${\bf D}$ is a compact subset of $\rr^{2\times 2}$.  It is then clear that
$(1+i)z^i\in A$ for all $z$ such that $|{\rm Arg}(z)|\le\alpha$.   

We now show that ${\bf D}\in D_C^{W,r}(f;0)$.  To see this, we have to pick
a neighborhood ${\bf U}$ of ${\bf D}$ in $\rr^{2\times 2}$ and construct 
a compact disc  $V_{{\bf{U}}}$ in $\comp$, centered at $0$,
and a sequence
$\{f_j\}$ in $C^1_{{\bf U}}(V_{{\bf{U}}}\cap C;\comp)$  that converges
uniformly to $f$ on $V_{{\bf{U}}}\cap C$ and is such that 
\beq\label{althoneigh}
Df_j(z)\in{\bf U}\,\,\,{\rm for\,\,\, all\,\,\,}j\in\nn\,,\,\,z\in
V_{{\bf{U}}}\cap C\,. 
\eeq
It turns out that in our example  we can actually  choose the sequence
$f_j$ independently of ${\bf U}$.

Actually, we will construct $f_j:C\to\comp$ such that $f_j\to f$
uniformly, and each $f_j$ is holomorphic on a neighborhood of $C$ and
satisfies $Df_j(p)\in{\bf D}$ for all $p\in C$.  To construct the
$f_j$, we pick a sequence $\{q_j\}$ of points of $C$ converging to $0$
and such that ${\rm{Re}}(q_j)>0$, and define $f_j(z)=f(z+q_j)$. Then
each $f_j$ is holomorphic on a neighborhood of $C$, since
$q_j+C\subseteq C'$.  Moreover, the $f_j$ clearly converge to $f$
uniformly on $C$, since $f$ is uniformly continuous on $\Omega\cup\{0\}$.
Finally, if $z\in C$ then $Df_j(z)=L_{\bar{x}_j}$, where
$\bar{x}_j=\frac{df_j}{dz}(z)=\frac{df}{dz}(z_j+q_j)$, so
$p_j=(1+i)(z+q_j)^i$, and then $a\le|p_j|\le b$, since
$|{\rm{}Arg}(z_j+q_j)|\le\alpha$. Therefore $p_j\in A$, so
$Df_j(z_j)\in{\bf {}W}$.  This completes the proof that
${\bf{}D}\in{}D_C^{W,r}(f;0)$.

We now show that ${\bf D}\notin D^{W,e}_C(f;0)$.  Assume that
${\bf{D}}\in D_C^{W,e}(f;0)$, and let $\{{\bf D}_j\}_{j=1}^\infty$ be a
decreasing sequence of compact neighborhoods of ${\bf D}$ in
$L(X,Y)$, such that ${\bf D}=\cap_j{\bf D}_j$. For each $j$ we can
find a $\delta_j>0$ such that $\delta_j<2^{-j}$, and a sequence
$\{f_{j,k}\}_{k=1}^\infty$ of $C^1$ maps
$f_{j,k}:\delta_j\disc\to\comp$ such that $f_{j,k}\to f$
uniformly on $C\cap\delta_j\disc$ and $Df_{j,k}(z)\in{\bf D}_j$ for
all $z\in\delta_j\disc$.  Since $f(0)=0$, we may subtract the
constant $f_{j,k}(0)$ from each $f_{j,k}$ and assume that
$f_{j,k}(0)=0$.
   

Let $\psi_{j,k}(z)=\frac{f_{j,k}(\delta_j z)}{\delta_j}$, for $z\in\disc$.
A simple application of the Chain Rule shows that $D\psi_{j,k}(z)=
Df_{j,k}(\delta_jz)$, so $D\psi_{j,k}(z)\in{\bf D}_j$  for all
$z\in{\bf D}$. 
This implies in particular that the differentials  $D\psi_{j,k}(z)$,
for all $j,k\in\nn$,  $z\in\disc$, belong to a fixed compact subset
of $\rr^{2\times 2}$. Since $\psi_{j,k}(0)=0$, we see that the set 
$S=\{\psi_{j,k}:j,k\in\nn\}$ is bounded and equicontinuous, 
hence relatively compact in $C^0(\disc,\comp)$.   

Let $K_\alpha=\{\rho e^{i\theta}:0\le\rho\le
1\,,\,|\theta|\le\alpha\}$.  Then, for each given $j$,
$$
\psi_{j,k}(z)\to \delta_j^{-1}f(\delta_jz)=(\delta_j)^iz^{1+i}
$$
uniformly on $K_\alpha$, as $k\to\infty$.  Because of the
compactness of $S$, we may pass to a subsequence, if necessary, and
assume that $\psi_{j,k}$ converges uniformly on $K_\alpha$ as
$k\to\infty$, to a limit $\psi_{j,\infty}\in C^0(\disc,\comp)$ that
obviously satisfies $\psi_{j,\infty}(z)=(\delta_j)^iz^{1+i}$ for
$z\in K_\alpha$.

Now let $\Delta:\rr^{2\times 2}\to\rr$ be the map that assigns to each
$2\times 2$ matrix $M=\{m_{ij}\}_{1\le i,j\le 2}$ the number
$|m_{11}-m_{22}|+|m_{12}+m_{21}|$.  Then $\Delta(M)=0$ iff
$M\in{\bf{M}}$.  Let $\Delta_j=\sup\{\Delta(M): M\in{\bf D}_j\}$.
Since $\cap_j{\bf D}_j={\bf D}$, $\Delta_j$ must go to zero as
$j\to\infty$.  Since the functions $\psi_{j,\infty}$ belong to the
closure of the relatively compact set $S$ we may assume, by passing to
a subsequence, that $\psi_{j,\infty}$ converges uniformly on $\disc$
to a limit function $\psi$.

For $j\in\nn$, $k\in\nninf$, write $\psi_{j,k}=u_{j,k}+iv_{j,k}$, 
with $u_{j,k}$, $v_{j,k}$
real-valued.  If $\varphi$ is an arbitrary real-valued function
on $\disc$, of class $C^\infty$ and compactly supported in the 
interior of $\disc$,  and $j,k\in\nn$, then 
$$
\int_{\disc}\biggl(\frac{\partial u_{j,k}}{\partial x}
-\frac{\partial v_{j,k}}{\partial y}\biggr)\varphi=
\int_{\disc}\biggl(v_{j,k}\frac{\partial\varphi}{\partial y}-
u_{j,k}\frac{\partial\varphi}{\partial x}\biggr)\,.
$$
Since the matrix $D\psi_{j,k}(z)$ belongs to ${\bf D}_j$ for
each $z\in\disc$, the estimate 
$\Big|\frac{\partial u_{j,k}}{\partial x}
-\frac{\partial v_{j,k}}{\partial y}\Big|\le\Delta_j$ holds pointwise.  So
\beq\label{taragui}
\Biggl|\int_{\disc}\biggl(v_{j,k}\frac{\partial\varphi}{\partial y}-
u_{j,k}\frac{\partial\varphi}{\partial x}\biggr)\Biggr|\le\Delta_j
|\!|\varphi|\!|_{sup}\,. 
\eeq
Since $\psi_{j,k}\to\psi_{j,\infty}$ uniformly as $k\to\infty$,
(\ref{taragui})  
also  holds for $k=\infty$. If we then write $\psi=u+iv$, and let
$j\to\infty$, we find that 
\beq\label{taragua}
\int_{\disc}\biggl(v\frac{\partial\varphi}{\partial y}-
u\frac{\partial\varphi}{\partial x}\biggr)=0\le\Delta_j\,. 
\eeq
So the continuous function $\psi=u+iv$ satisfies the Cauchy-Riemann
equation $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$
in the sense of  
distributions.  A similar argument shows that the other Cauchy-Riemann   
equation,
$\frac{\partial{u}}{\partial{y}}=-\frac{\partial{v}}{\partial{x}}$,
also holds  in the distributional sense. It then follows by well known
regularity  results  (e.g. that every distributional solution of the
Laplace equation is in fact smooth and a classical solution) that
$\psi$ is a holomorphic  function on ${\bf D}$.   

On $K_\alpha$, $\psi_{j,\infty}(z)$ equals $\delta_j^iz^{i+1}$.
Then $\delta_j^i=\psi_{j,\infty}(1)$ converges to $\psi(1)$,
so $\psi(z)=\psi(1)z^{i+1}$. Since $\delta_j^i=e^{i\,{\rm Ln}\,\delta_j}$,
and ${\rm Ln}\,\delta_j$  is real, we see that $|\delta_j|^i=1$.  Therefore
$|\psi(1)|=1$. So $\psi(1)\neq 0$. Then  
$$
z^{i+1}=\frac{\psi(z)}{\psi(1)}\,\,\,{\rm for}\,\,\,z\in K_{\alpha}\,,
$$ 
showing that the function $z\to z^{i+1}$  is the restriction to 
$K_{\alpha}$ of a holomorphic function on the full disc.  This is of course 
a contradiction. (For example, the derivative $(1+i)z^i$ has absolute value
$\sqrt{2}e^{-\theta}$ on the line ${\rm Arg}(z)=\theta$, so this derivative
has no limit as $z\to 0$.)  We have therefore proved that
${\bf D}\in D^{W,r}_C(f;0)$ but ${\bf D}\notin D^{W,e}_C(f;0)$. 




Let $S$ be the closed right half-plane $\{z\in\mathbb{C}:{\rm
Re}\,z\ge 0\}$.  Consider the function $S\ni z\mapsto
f(z)=z^i\in\mathbb{C}$, where $z^i$ is defined using the principal
branch of the logarithm, so that $z^i=e^{-\theta+i\,{\rm ln}\,r}$, if
$z=r\,e^{i\theta}$ and $-\frac{\pi}{2}
\le\theta\le\frac{\pi}{2}$.  If we let $f_\eps(z)=f(\eps+z)$ for 
$\eps>0$, $z\in S$, then $f_\eps$ is a holomorphic map from
$S$ to $\mathbb{C}$. The derivative $f'_\eps$ is given by
$f'_\eps(z)=i\,(\eps+z)^{i-1}$, i.e., $f'_\eps(z)=i\,z^{-1}f_\eps(z)$. 


*************************************************************************

\begin{lemma}\label{jointcontlemma}
Let $n,m\in\zz_+$. Let $\{(\alpha_j,h_j)\}_{j=1}^\infty$ be a sequence
of members of the product space 
\mbox{$S=\czer{[0,1]}{\rr^n}\times L^1([0,1],\rr^{m\times n})$} 
that converges in $S$ to a limit
$(\alpha_\infty,h_\infty)$. Assume that the sequence $\{\alpha_j\}$ 
is uniformly Lipschitz (that is, there exists a constant $r\in\rr$ such that
$\|\alpha_j(t)-\alpha_j(s)\|\le r|t-s|$ for all $j\in\nn$ and all
$t,s\in[0,1]$).  Then 
$$
h_j*\alpha_j\to h_\infty*\alpha_\infty\quad{\rm in}
\quad\czer{[0,1]}{\rr^{m}}\quad{\rm as}\quad j\to\infty\,.
$$
\end{lemma}
\begin{proof}
Let $r$ be a Lipschitz constant for the $\alpha_j$.
For $j\in\nn\cup\{\infty\}$, let 
$\beta_j=h_j*\alpha_j$.


Fix a positive number $\eps$.  Let $g:[0,1]\to\rr^{m\times n}$ be a 
function of class $C^1$ such that 
$$
\|g-h_{\infty}\|_{_{L^1}}\le\frac{\eps}{4r}\,.
$$
Then, if $t\in[0,1]$,
\begin{eqnarray*}
\beta_j(t)-\beta_\infty(t)&=&\int_0^th_j(s)\cdot\dot\alpha_j(s)\,ds\\
&\,&\,\,-\int_0^th_\infty(s)\cdot\dot\alpha_\infty(s)\,ds\\
&=&\,\,T_1+T_2+T_3+T_4\,,
\end{eqnarray*}\
where 
\begin{eqnarray*}
T_1&=&\int_0^t(h_j(s)-h_\infty(s))\cdot\dot\alpha_j(s)\,ds\,,\\
T_2&=&\int_0^t(h_\infty(s)-g(s))\cdot\dot\alpha_j(s)\,ds\,,\\
T_3&=&\int_0^tg(s)\cdot(\dot\alpha_j(s)-\dot\alpha_\infty(s))\,ds\,,\\
T_4&=&\int_0^t(g(s)-h_\infty(s))\cdot\dot\alpha_\infty(s)\,ds\,.
\end{eqnarray*}
Then 
\begin{eqnarray*}
\|T_1\|&\le& r\|h_j-h_\infty\|_{_{L^1}}\,,\\
\|T_2\|&\le&\frac{\eps}{4}\,,\\
\|T_4\|&\le&\frac{\eps}{4}\,.
\end{eqnarray*}
We can estimate $T_3$ using integration by parts:
\begin{eqnarray*}
T_3&=&\int_0^tg(s)\cdot(\dot\alpha_j(s)-\dot\alpha_\infty(s))\,ds\,,\\
&=&g(t)\cdot(\alpha_j(t)-\alpha_\infty(t))\\
&\,\,&\,\,-\int_0^t\dot g(s)\cdot(\alpha_j(s)-\alpha_\infty(s))\,ds\,,
\end{eqnarray*}
so 
$$
\|T_3\|\le\Big(\|g\|_{sup}+\|\dot g\|_{sup}\Big)\,\|\alpha_j-
\alpha_\infty\|_{sup}\,.
$$
Since $\|\alpha_j-\alpha_\infty\|_{sup}\to 0$ and 
$\|h_j-h_\infty\|_{_{L^1}}\to 0$ as $j\to\infty$, there exists a $j_0$ 
such that  $\|T_1\|+\|T_3\|\le\frac{\eps}{2}$ whenever 
$j\ge j_0$. Then $\|\beta_j-\beta_\infty\|_{sup}\le\eps$ whenever 
$j\ge j_0$. Since $\eps$ was an arbitrary positive number, we have 
shown that $\beta_j\to\beta_\infty$ uniformly,  and our proof 
is complete. 
\end{proof}

**************************************************************************

\begin{lemma}\label{lemmonpoly}
Let $X$ be a normed linear space. Then
\ben
\item[(1)] Every convex polyhedron in $X$ is compact.
\item[(2)] Every polyhedral cone  in $X$ is closed.
\item[(3)] If $C$ is a polyhedral cone $F$, and $F$ is a finite 
\mbox{subset}
of $X$ such that $C$ is the convex cone 
\mbox{generated} by $F$, then there exists
a positive constant $\kappa$ such that 
\mbox{every} $v\in C$ can be expressed
as a linear \mbox{combination}
\begin{equation}\label{orlando-aa}
v=\sum_{f\in F}\alpha_ff\,,\quad 0\le\alpha_f\le\kappa\|v\|\quad{\rm for\,\,\,all}
\quad f\in F\,.
\end{equation}
\item[(4)] If $K$ is a convex polyhedron, and $x\in K$, then 
\ben
\item[(4.a)] $T_xK$ is a polyhedral cone,
\item[(4.b)] there exists a $\rho\in\rr$ such that $\rho>0$ and
\mbox{$x+v\in K$} whenever 
$v\in T_xK$ and $\|v\|\le\rho$.
\een
\item[(5)] If $K$ is a convex polyhedron, then 
$\{T_xK:x\in K\}$ is a finite set.
\een
\end{lemma}
\begin{proof}
Statement (1) is trivial. We prove (2) and (3) together. Let $C$ be
the convex cone generated by a finite subset $F$ of $X$. It is clear
that $C$ is the set of all sums
\begin{equation}\label{orlando-a}
\sum_{f\in F}\alpha_ff\,,\quad\alpha_f\ge 0\quad{\rm for\,\,\,all}
\,\,\,f\in F\,.
\end{equation}
(Indeed, the set of all sums (\ref{orlando-a}) is obviosuly a convex cone
containing $F$, and all the sums (\ref{orlando-a}) clearly belong
to every convex cone containing $F$.)

Let $\mathcal{H}$ be the set
of all subsets $H$ of $F$ having the property that there exists a 
family $\{\alpha_h\}_{h\in H}$ of strictly positive numbers 
such that $\sum_{h\in H}\alpha_hh=0$.  It is easy to see that
$$
\Big(\,H_1\in\mathcal{H}\wedge H_2\in\mathcal{H}\,\Big)\Longrightarrow
H_1\cup H_2\in\mathcal{H}\,.
$$
Therefore, if we let
$$
H_{max}=\bigcup_{H\in\mathcal{H}}H\,,
$$
then $H_{max}\in\mathcal{H}$. Therefore there exist numbers $\bar\alpha_h$, for 
$h\in H_{max}$,  such that 
\begin{equation}\label{orlando-aaa}
\sum_{h\in H_{max}}\bar\alpha_hh=0\,,\quad\bar\alpha_h>0\quad
{\rm for\,\,\,every \,\,\,}h\in H_{max}\,.
\end{equation}
Let $D$ be the convex cone generated by $H_{max}$. If $v\in H_{max}$, and
(\ref{orlando-aaa}) holds, then 
$$
-v=\sum_{h\in H_{max},h\neq v}\frac{\bar\alpha_h}{\bar\alpha_v}\,h\,,
$$
so $-v\in D$. The set $D'=\{v\in D:-v\in D\}$ is obviously a convex cone, 
and we have shown that $H_{max}\subseteq D'$. Therefore $D\subseteq D'$, so 
$D=D'$. Hence $D$ is a linear subspace of $X$. 

Write $G=F\backslash H_{max}$, and let $\Gamma$ be the convex hull of $G$. 
Then $\Gamma$ is  compact, and 
\begin{equation}\label{orlando}
\Gamma\cap D=\emptyset\,.
\end{equation}
({\bem Proof.} If $x\in\Gamma\cap D$, then,
first of all, $x=\sum_{g\in G}\beta_gg$, with $\beta_g\ge 0$ and 
$\sum_{g\in G}\beta_g=1$. In addition, $x\in D$, so $-x\in D$, and then 
$-x=\sum_{h\in H_{max}}\gamma_hh$, with $\gamma_h\ge 0$. Therefore
$$
\sum_{g\in G}\beta_gg+\sum_{h\in H_{max}}\gamma_hh=0\,,
$$
and the coefficients $\beta_g$, $\gamma_h$ are nonnegative. After 
we eliminate those terms whose coefficients vanish, we are left with an 
identity 
$$
\sum_{g\in G'}\beta_gg+\sum_{h\in H'}\gamma_hh=0\,,
$$
where the coefficients $\beta_g$, $\gamma_h$ are positive, and
$G'$, $H'$ are subsets of $G$, $H_{max}$, respectively.  Then
$G'\cup H'\in\mathcal{H}$, so $G'\cup H'\subseteq H_{max}$, 
and then $G'\subseteq H_{max}$, implying that $G'=\emptyset$, because
$G'\subseteq G$ and $G\cap H_{max}=\emptyset$. On the other hand,
$G'$ must be nonempty, because $\sum_{g\in G}\beta_g=1$. So we have reached
a contradiction, and (\ref{orlando}) is proved.)

Let $\mathcal{E}$ be a maximal linearly independent subset of $H_{max}$, 
so  $\mathcal{E}$ is a basis of $D$.  Let 
$$
A=\min\{\,\bar\alpha_h:h\in H_{max}\,\}\,,\quad 
B=\max\{\,\bar\alpha_h:h\in H_{max}\,\}\,.
$$ 
Let $V$ be the set of all vectors $v\in D$ such that
$v=\sum_{e\in\mathcal{E}}\mu_e(v)e$ for some 
coefficients $\mu_e(v)$ such that $|\mu_e(v)|\le A$ for all 
$e\in\mathcal{E}$.  Then $V$ is a neighborhood of $0$ in $D$, so there exists
a positive number $R$ such that 
$(v\in D\wedge\|v\|\le R)\Longrightarrow v\in V$.

If $v\in V$, then 
\begin{eqnarray*}
v&=&0+v\\
%&=&\sum_{e\in\mathcal{E}}\lambda_ee\\
&=&\sum_{h\in H_{max}}\bar\alpha_hh+\sum_{e\in\mathcal{E}}\mu_e(v)e\\
&=&\sum_{h\in H_{max}}\beta_h(v)h\,,
\end{eqnarray*}
where the coefficients $\beta_h(v)$ are given by
$$
\beta_h(v)=\left\{\begin{array}{lcl}
\bar\alpha_h&{\rm if}&h\in H_{max}\backslash\mathcal{E}\,,\\
\bar\alpha_h+\mu_h(v)&{\rm if}&h\in\mathcal{E}\,,
\end{array}\right.
$$
so that $0\le\beta_h(v)\le A+B$ for all $h\in H_{max}$. 

If $v\in D$, then 
\begin{eqnarray*}
v&=&\frac{\|v\|}{R}\sum_{h\in H_{max}}\beta_h\Big(\frac{Rv}{\|v\|}\Big)h\\
&=&\sum_{h\in H_{max}}\lambda_h(v)h\,,
\end{eqnarray*}
where
$$
\lambda_h(v)h=\frac{\|v\|}{R}\beta_h\Big(\frac{Rv}{\|v\|}\Big)\,,
$$
so 
\begin{equation}\label{orlando-uvu-a}
0\le\lambda_h(v)\le \frac{A+B}{R}\,\|v\|\quad{\rm for\,\,\,all}\quad
h\in H_{max}\,. 
\end{equation}

If $G=\emptyset$, then $F=H_{max}$, $C=D$, so $C$ is closed,
and the conclusion of (3) is satisfied with 
$$
\kappa=\frac{A+B}{R}\,.
$$

Now suppose that $G\neq\emptyset$, so $\Gamma\neq\emptyset$ as well.
If $v\in C$, we can write
$$
v=\sum_{g\in G}\gamma_{g}g+\sum_{h\in H_{max}}\theta_{h}h\,,\quad
\gamma_{g}\ge 0\,,\,\,\theta_{h}\ge 0\,.
$$
Then
\begin{equation}\label{orlando-uvu}
v=ry+z\,,\quad r\ge 0\,,\quad y\in\Gamma\,,\quad z\in D\,.
\end{equation}
(Indeed, we can take
$$
r=\sum_{g\in G}\gamma_{g}\,,\quad z=\sum_{h\in H_{max}}\theta_{h}h\,.
$$ 
If $r>0$, we let 
$$
y=\sum_{g\in G}\frac{\gamma_{g}}{r}g\,.
$$
If $r=0$, we recall that $\Gamma$ is assumed to be nonempty, and
pick $y$ to be an arbitrary member of $\Gamma$.)

We now show that 
\ben
\item[(\#)] {\bem there exists a constant $\kappa_0$ such that 
$r\le\kappa_0\|v\|$ whenever $v\in C$  and $r,y,z$ are such that
(\ref{orlando-uvu}) holds.}
\een
Suppose statement (\#) was false. Then there would exist sequences
$\{v_j\}_{j=1}^\infty$, $\{y_j\}_{j=1}^\infty$, $\{z_j\}_{j=1}^\infty$, 
$\{r_j\}_{j=1}^\infty$, of members of the sets $C$, $\Gamma$, $D$, $[0,\infty\,[\,$, 
respectively, such that $v_j=r_jy_j+z_j$, and $r_j>j\|v_j\|$. 
%Then $v_j\neq 0$, because $v_j=0$ would imply $r_jy_j+z_j=0$, so
%$y_j=-\frac{z_j}{r_j}$, and then $y_j\in\Gamma\cap D$, contradicting 
%(\ref{orlando}).  

Since the set $\Gamma$ is compact we may assume, after passing to a subsequence,
that $y=\lim_{j\to\infty}y_j$ exists and belongs to $\Gamma$. Then
$$
\frac{z_j}{r_j}=\frac{v_j}{r_j}-y_j\,,
$$
and
$$
\lim_{j\to\infty}\frac{v_j}{r_j}=0\,,
$$
because $r_j>j\|v_j\|$. Therefore 
$$
\lim_{j\to\infty}\frac{z_j}{r_j}=-y\,.
$$
But $D$ is closed, and the vectors $\frac{z_j}{r_j}$ belong to $D$. Therefore
$-y\in D$, and then $y\in D$, because $D$ is a linear subspace.  Since $y\in\Gamma$, 
we have contradicted (\ref{orlando}).  Hence  (\#) is true.

Now, if $v\in C$, and $\kappa_0$ is as in (\#), we can write
\begin{eqnarray*}
v&=&ry+z\\
&=&r\sum_{g\in G}\sigma_gg+\sum_{h\in H_{max}}\lambda_h(z)h\\
&=&\sum_{f\in F}\alpha_ff\,,
\end{eqnarray*}
where $r\in\rr$, $0\le r\le\kappa_0\|v\|$, $z\in D$, $\sigma_g\ge 0$ for
$g\in G$, $\sum_{g\in G}\sigma_g=1$, and 
$$
\alpha_f=\left\{\begin{array}{lcl}
\lambda_f(z)&{\rm if}&f\in H_{max}\,,\\
r\sigma_f&{\rm if}&f\in G\,.\end{array}\right.
$$
Then
$$\|z\|=\|v-ry\|\le\|v\|+r\|y\|\le(1+\kappa_0\|y\|)\|v\|
\le(1+\kappa_0\kappa_1)\|v\|\,,
$$
where
$$
\kappa_1=\max\{\|w\|:w\in\Gamma\}\,.
$$
Therefore
$$
0\le\lambda_h(z)\le\frac{A+B}{R}(1+\kappa_0\kappa_1)\|v\|
\,\,\,{\rm for\,\,\,all}\,\,\,h\in H_{max}\,.
$$
Also 
$$
0\le r\sigma_g\le\kappa_0\|v\|\,\,\,{\rm for\,\,\,all}\,\,\,g\in G\,,
$$
because $0\le\sigma_g\le 1$ and $0\le r\le\kappa_0\|v\|$. If we then let
$$
\kappa=\max\Big(\kappa_0,\frac{A+B}{R}(1+\kappa_0\kappa_1)\Big)\,,
$$
we see that 
$$
0\le \alpha_f\le\kappa\|v\|\,\,\,{\rm for\,\,\,all}\,\,\,f\in F\,.
$$
This proves (3). To prove (2), we take a sequence $\{v_j\}_{j\in\nn}$
of points of $C$ that converges to a limit $v\in X$, and show that
$x\in C$. We write $v_j=r_jy_j+z_j$, with $0\le r_j\le\kappa_0\|v_j\|$,
$y_j\in\Gamma$, $z_j\in D$. Since $v_j\to v$, the sequence $\{r_j\}_{j\in\nn}$ 
is bounded. So we may assume, after passing to a subsequence, that the limit 
$r=\lim_{j\to\infty}r_j$ exists and is a nonnegative real number, and
the limit $y=\lim_{j\to\infty}y_j$ exists and belongs to $\Gamma$.
Let $z=rv-y$. Then 
$$
\lim_{j\to\infty}z_j=\lim_{j\to\infty}(r_jv_j-y_j)=rv-y=z\,.
$$
Therefore $z\in D$, because $z_j\in D$ and $D$ is closed. 
Since $v=ry+z$, $y\in\Gamma$, $z\in D$, $K\subseteq C$, and $D\subseteq C$, 
it follows that $v\in C$, and (2) is proved. 

To prove (4), we let $K$ be the convex hull of a finite set $F$, pick
$x\in K$, and let $C^x$ be the convex cone generated by the vectors
$f-x$, $f\in F$. Then $C^x$ is a polyhedral cone, so $C^x$ is closed by (2).
By definition, $f-x\in T_xK$ whenever $f\in F$, so $C^x\subseteq T_xK$.
On the other hand, if $y\in K$, then $y=\sum_{f\in F}\lambda_ff$ with
$\lambda_f\ge 0$ and $\sum_{f\in F}\lambda_f=1$. But then 
$y-x=\sum_{f\in F}\lambda_f(f-x)$, so $y-x\in C^x$. Therefore 
$C^x$ contains all the vectors $y-x$, $y\in K$ and, since $C^x$ is a closed
convex cone,  $T_xK\subseteq C^x$.  Hence $C^x=T_xK$.  
So $T_xK$ is a polyhedral cone, and (4.a) is proved.  

It then follows from (3) that there exists a constant $\kappa^x$ such that
every $v\in C^x$ has an expression $v=\sum_{f\in F}\alpha_f(f-x)$,
with $0\le\alpha_f\le\kappa^x\|v\|$.  Then, if $v\in C^x$, 
\begin{equation}\label{orlando-uvu-mo}
x+v=\Big(1-\sum_{f\in F}\alpha_f\Big)x+\sum_{f\in F}\alpha_ff\,.
\end{equation} 
Choose $\rho$ such that $\rho\kappa^x N\le 1$, where $N$ is the number of 
members of $F$. Then if $\|v\|\le\rho$ Formula 
(\ref{orlando-uvu-mo}) expresses $x+v$ 
as a convex combination of $x$ and the members of $F$, showing that 
$x+v\in K$. This proves (4.b).



We now turn to the proof of (5). First, we define a polyhedral cone 
${\bf C}(F;H)$, for each nonempty 
finite subset $F$ of $X$ and each nonempty subset  
$H$ of $F$, by letting 
${\bf C}(F;H)$ be the convex cone generated by the vectors
$f-h$, $f\in F$, $h\in H$. Given $F$, the set 
$$
\mathcal{C}(F)=\Big\{\,{\bf C}(F;H):\emptyset\neq H\subseteq F\,\Big\}\,,
$$
is a finite set of polyhedral cones.  We will prove (5) by showing that:
\ben
\item[(\#\#)] {\bem If $F$ is a finite subset of $X$ and $K$ is the convex hull
of $F$, then $T_xK\in\mathcal{C}(F)$ for every $x\in K$.}
\een
To prove (\#\#), we fix $F$, let $K$ be the convex hull of $F$, pick
$x\in K$, and show that $T_xK\in\mathcal{C}(F)$.

Let $F^x=\{f-x:f\in F\}$. As in the proof of (2) and (3), we define
$\mathcal{H}^x$ to be the class of all subsets $H$ of $F^x$ such that
$0=\sum_{h\in H}\alpha_hh$ for some family of numbers $\alpha_h$ such
that $\alpha_h>0$ for all $h\in H$, and we let $H_{max}^x$ be the union of 
all the members of $\mathcal{H}^x$.  Then 
\begin{equation}\label{orlando-uvu-babaa-oo-vava}
H_{max}^x\neq\emptyset\,.
\end{equation}
({\bem Proof.} Since $x\in K$,  $x$ is a convex combination of members of $F$, and
after eliminating the vanishing coefficients we may assume that
$x=\sum_{\ell\in L}\beta_\ell\ell$ for some subset $L$ of $F$, with
coefficients $\beta_\ell$ such that $\beta_\ell>0$ and $\sum_{\ell\in
L}\beta_\ell=1$. Then $\sum_{\ell\in L}\beta_\ell(\ell-x)=0$, so the
set $H=\{\ell-x:\ell\in L\}$ satisfies $\sum_{h\in H}\alpha_hh=0$ and
$\alpha_h>0$ for all $h\in H$, if
we let $\alpha_h=\beta_{h+x}$ for $h\in H$.  Therefore
$H\in\mathcal{H}^x$, and the fact that $\sum_{\ell\in L}\beta_\ell=1$
implies that $L\neq\emptyset$, because an indexed sum over the empty
set is equal to zero. Hence $H\neq\emptyset$.  So
(\ref{orlando-uvu-babaa-oo-vava}) holds.)

We let 
$$
\tilde H_{max}^x=\{f\in F:f-x\in H_{max}^x\}\,,
$$
so $\tilde H_{max}^x\subseteq F$. 
Then (\ref{orlando-uvu-babaa-oo-vava}) implies that 
\begin{equation}\label{orlando-uvu-babaa-oo-vavai}
\tilde H_{max}^x\neq\emptyset\,,
\end{equation}

We will show that
\begin{equation}\label{orlando-uvu-babaa}
T_xK={\bf C}(F;\tilde H_{max}^x)\,.
\end{equation}
Since $\tilde H_{max}^x$ is a nonempty subset of $F$, this will imply that
$T_xK\in\mathcal{C}(F)$, and (\#\#) will be proved.

As in the proof of (2) and (3), the set 
$H_{max}^x$ belongs to $\mathcal{H}^x$. Therefore there is a family
$\{\alpha_h^x\}_{h\in H_{max}^x}$ of numbers such that $\alpha_h^x>0$
whenever $h\in H_{max}^x$, 
and $\sum_{h\in H_{max}^x}\alpha_h^xh=0$.  Write
$\beta_h^x=\alpha_{h-x}^x$ for $h\in \tilde H_{max}^x$.  Then
\begin{eqnarray*}
0&=&\sum_{h\in H_{max}^x}\alpha_h^xh\\
&=&\sum_{h\in\tilde H_{max}^x}\beta_h^x(h-x)\\
&=&\sum_{h\in\tilde H_{max}^x}\beta_h^xh-B^xx\,,
\end{eqnarray*}
where
$$
B^x=\sum_{h\in\tilde H_{max}^x}\beta_h^x\,.
$$
It then follows that 
\begin{equation}\label{orlando-uvu-babaa-inc-a}
x=\sum_{h\in\tilde H_{max}^x}\gamma_h^xh\,,
\end{equation}
where the numbers $\gamma_h^x=\frac{\beta_h^x}{B^x}$ satisfiy
$\gamma_h^x>0$ and $\sum_{h\in\tilde H_{max}^x}\gamma_h^x=1$.

If $y\in K$, then 
$$
y=\sum_{f\in F}\lambda_ff\,,\quad\lambda_f\ge 0\quad{\rm for\,\,\,
all}\quad f\in F\,,\,\,\sum_{f\in F}\lambda_f=1\,.
$$
Therefore
\begin{eqnarray*}
y-x&=&\sum_{f\in F}\lambda_f(f-x)\\
&=&\sum_{f\in F}\lambda_f\Big(f-\sum_{h\in\tilde H_{max}^x}\gamma_h^xh\Big)\\
&=&\sum_{f\in F}\lambda_f\Biggl(\Big(\sum_{h\in\tilde H_{max}^x}\gamma_h^x
\Big)f-\sum_{h\in\tilde H_{max}^x}\gamma_h^xh\Biggr)\\ 
&=&\sum_{f\in F}\lambda_f\Biggl(\sum_{h\in\tilde H_{max}^x}\gamma_h^x
(f-h)\Biggr)\\ 
&=&\sum_{f\in F}\,\,\sum_{h\in\tilde H_{max}^x}\lambda_f\gamma_h^x(f-h)\,,
\end{eqnarray*}
showing that $y-x\in{\bf C}(F;\tilde H_{max}^x)$. Since this is true 
for arbitrary $y\in K$, and ${\bf C}(F;\tilde H_{max}^x)$ is a closed convex 
cone, the definition of $T_xK$ implies the inclusion
\begin{equation}\label{orlando-uvu-babaa-inc}
T_xK\subseteq{\bf C}(F;\tilde H_{max}^x)\,.
\end{equation}

To prove the opposite inclusion, we pick $f\in F$, $h\in \tilde H_{max}^x$.
Then (\ref{orlando-uvu-babaa-inc-a}) implies 
\begin{eqnarray*}
-\gamma_h^xh&=&\Biggl(\sum_{\ell\in\tilde H_{max}^x\backslash\{h\}}
\gamma_\ell^x\ell\Biggr)-x\\
&=&\Biggl(\sum_{\ell\in\tilde H_{max}^x\backslash\{h\}}\gamma_\ell^x\ell\Biggr)-
\Biggl(\sum_{\ell\in\tilde H_{max}^x}\gamma_\ell^x\Biggr)x\\
&=&\sum_{\ell\in\tilde H_{max}^x\backslash\{h\}}\gamma_\ell^x(\ell-x)
-\gamma_h^xx\,,
\end{eqnarray*}
so that
\begin{equation}\label{orlando-uvu-babaa-inc-aa}
x-h=\sum_{\ell\in\tilde H_{max}^x\backslash\{h\}}\delta_{\ell,h}^x(\ell-x)\,,
\end{equation}
where 
$$
\delta_{\ell,h}^x=\frac{\gamma_\ell^x}{\gamma_h^x}\,.
$$
Then 
\begin{eqnarray*}
f-h&=&f-x+x-h\\
&=&f-x+\sum_{\ell\in\tilde H_{max}^x\backslash\{h\}}\delta_{\ell,h}^x(\ell-x)\,,
\end{eqnarray*}
showing that $f-h$ belongs to the convex cone generated by the vectors 
$y-x$, $y\in K$, i.e. to $T_xK$. Since 
$f$ was an arbitrary member of $F$, and 
$h$ was an arbitrary member of $\tilde H_{max}^x$, we have in fact shown
that 
\begin{equation}\label{orlando-uvu-babaa-inc-op}
{\bf C}(F;\tilde H_{max}^x)\subseteq T_xK\,.
\end{equation}
Then (\ref{orlando-uvu-babaa-inc}) and (\ref{orlando-uvu-babaa-inc-op})
establish (\ref{orlando-uvu-babaa}), thereby proving (\#\#),
and completing the proof of (5).
\end{proof}
