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\begin{small}

  	      \vspace*{12pt}
  
          \begin{center}   
\begin{large}   
         {\bf Extra Proofs for: ``The Learning Curve in a Competitive
Industry'' }\\ 
  \end{large}   
 \end{center} 
         \vspace*{20pt}   
May 27, 1994  \\   
        \bigskip   
                       Emmanuel Petrakis,
Eric Rasmusen, and
                         Santanu Roy \\   
          
          \vspace*{24pt}
               
            \vspace*{ 36pt}
           \noindent 
\hspace*{20pt}	
Petrakis:  Departamento de Economia,
Universidad Carlos III de Madrid,
Calle Madrid 126 
28903 Getafe (Madrid) Spain.  
  Fax:  341-624-9875. Phone: 341-624-9652. 
Internet: Petrakis@eco.uc3m.es. \\
\hspace*{20pt}  Rasmusen: 	  Indiana University
School of Business,       
  10th Street  and Fee Lane,
  Bloomington, Indiana, U.S.A 47405-1701. 
  Phone: (812) 855-9219.  Fax: (812) 855-8679. Internet:
Erasmuse@indiana.edu.\\ 
 \hspace*{20pt} Roy: Econometric Institute,
Erasmus University, P.O.Box 1738, 3000 DR Rotterdam, The Netherlands.
Fax: 31-10-452-7746.  Phone: 31-10-408-1420.  Internet:
Santanu@wke.few.eur.nl. \\
 
    
 
  \newpage
    \noindent
            {\bf Extra Proofs for: ``The Learning Curve in a
Competitive Industry'' }\\ 
  
 One of the implications of assumption (A5) is that the minimum
one-period average cost with no learning, $p_m$, is attained at some
finite positive output level. This is part of a more general lemma:

LEMMA I.  Under assumptions (A1) - (A5), the functions
$[C(q,0)/q], [f(q,0)/q]$ and $[f(0,q)/q]$ attain their minima at
finite positive output levels. 

PROOF. First note that since $C(0,0) > 0 $, the average costs in the
statement of the lemma diverge to $+\infty$ as $q \rightarrow 0 $. 

     Let $m_1 = inf\{[f(q,0)/q]: q \geq 0 ]\}$. Suppose the infimum
is not attained at any finite $q$. Then, there exists $\{q_n \}
\rightarrow +\infty$ such that $ f(q_n ,0)/q_n \rightarrow m_1$ and,
further, for each $n$, 
   \begin{equation} \label{ee1}
              [f(q,0)/q] \geq [f(q_n ,0)/q_n ] \;\;for \;\;q \in
[0,q_n ] 
 \end{equation}
  There exists $N$ such that $q_n > K$ for all $n \geq N$. Then,
using (A5), for each $ n \geq N$, there exists $z < q_n , y < q_n$
such that $z + y = q_n$ and:
      $ f(q_n ,0) > f(z,0) + f(y,0)$, that is,
 $$
\begin{array}{l}
          f(q_n ,0)/q_n > [z/(z + y)][f(z,0)/z] + [y/(z +
y)][f(y,0)/y]\\
                   \geq [z/(z + y)][f(q_n ,0)/q_n ] + [y/(z +
y)][f(q_n ,0)/q_n ] 
                   = f(q_n ,0)/q_n,
 \end{array}
$$
  (using (\ref{ee1})), a contradiction.

     A similar method can be used to show that $ f(0,q)/q $ attains its
minimum at a finite positive output. Lastly, note that (A5) implies
that for any $q > K$, there exists $z,y \geq  0 , z + y \leq q$,
    $$
    C(0,0) + \delta C(q,0) > 2C(0,0) + \delta C(z,0) + \delta C(y,0)
    $$
 which implies
          $$
    C(q,0) > C(z,0) + C(y,0)
    $$
 Again, the same set of steps can be replicated to show that
$[C(q,0)/q]$ attains its minimum at a finite positive level. //

\bigskip


\bigskip
 PROPOSITION 2. Under assumptions (A1)-(A6), an equilibrium exists.
It is unique in prices, and it is socially optimal. 
    

 PROOF.  Section III of the text defines the social planner's
problem. Based on that definition, we can 
    define the social cost minimization problem for any $Q_1 \geq 0 ,
Q_2 \geq 0 $ to be produced by S-type firms as:
        $$
 (SCM1) \;\;\;\;\;\;Minimize \int_0^{n_S} �[C(q_1(i),0) + \delta
C(q_2(i),q_1(i))]di
   $$
 w.r.t. $n_S \geq  0 , q_t:[0,n_S] \rightarrow {\bf R}_+, t = 1,2,
q_t(.)$ integrable w.r.t. Lebesgue measure, subject to 
 the restrictions:
       $$
           \int_0^{n_S}    �q_t(i)di \geq  Q_t, t=1,2.                 
    $$
 Let �$ \psi (Q_1,Q_2)$ be the value of the minimization problem.

     Similarly, if $Q_E \geq 0 $ is the amount to be produced by
E-type firms in period 1, then the social cost minimization problem
is given by:
                $$
  (SCM2)   \;\;\;\;\;\;     Minimize  \int_0^{n_E}�C(q_E(i),0)di
   $$
 w.r.t. $n_E \geq 0 , q_E:[0,n_E] \rightarrow {\bf R}_+, q_E(.)$
integrable w.r.t.  Lebesgue measure, subject to the restriction:
            $$
            \int_0^{n_E}   q_E(i)di \geq  Q_E.                              
    $$
 Let $\psi_E(Q_E)$ be the value of the minimization problem.

     Lastly, if $Q_L \geq 0 $ is the amount to be produced by L-type
firms in period 2, 
 then the social cost minimization problem is given by:
            $$
 (SCM3)   \;\;\;\;\;\;     Minimize  \int_0^{n_L}�\delta C(q_L(i),0)di
    $$
 w.r.t. $n_L \geq  0 , q_L:[0,n_L]\rightarrow {\bf R}_+, q_L(.)$
integrable w.r.t.  Lebesgue measure, subject to the restriction:
      $$
            \int_0^{n_L}    �q_L(i)di \geq  Q_L                     (*)
   $$
 Let �$\psi_L(Q_L)$ be the value of this minimization problem.  

     First consider (SCM1). For $Q_1 = Q_2 = 0 $, the solution is
obviously $n_S = 0 $. Suppose $Q_2 = 0 $ and $Q_1 > 0 $. Let 
    \begin{equation} \label{ee2}
       m_1 = min\{[(C(q,0) + \delta C(0,q))/q]: q \geq   0 \}    
\end{equation}
    From Lemma I we have the existence of finite $q > 0 $ which
solves this minimization problem. Let $q(m_1) > 0 $ be any such
solution.  Consider the feasible set in (SCM1). One may without loss
of generality confine attention to the subset of the feasible set
where $q_2(i) = 0 $ a.e. and
           $$
          \int_0^{n_S}  �q_1(i)di = Q_1. 
     $$
 Let $(n_S, q_1(i),q_2(i))$ be any such feasible solution. 
 Then
       $$
     \begin{array}{ll}
     \int_0^{n_S} �[C(q_1(i),0) + \delta C(0,q_1(i))]di & \\
      & = \int_0^{n_S}�[(C(q_1(i),0) + \delta
C(0,q_1(i)))/q_1(i)]q_1(i)di\\
          & \geq    \int_0^{n_S} �m_1q_1(i)di \\
          &     =   m_1Q_1\\
    &  =    [C(q(m_1),0) + \delta C(0,q(m_1))]\hat{n}\\
\end{array}
 $$
 where $\hat{n} = Q_1/q(m_1)$.

 Thus, there exists a solution to SCM1 for $Q_1 >  0 , Q_2 =0 $ and �$
\psi (Q_1,0) = m_1Q_1$.

     Similarly, let $ m_2$ be defined by
\begin{equation} \label{ee3}
          m_2 = min\{[(C(0,0) + \delta C(q,0))/q]:q \geq  0 \}. 
\end{equation}
  Using Lemma I, there exists a finite positive solution to the
minimization problem in (\ref{ee3}).  Consider (SCM1) for the case
where $Q_1 = 0 , Q_2 > 0 $. One can show by similar arguments as
above, that there exists a solution to (SCM1) for this case and that
$\psi(0,Q_2) = m_2Q_2$.

     In fact, the same set of arguments will show that there exists
solution to (SCM2) and (SCM3) and that �$\psi_E(Q_E) = p_m Q_E$ and
$�\psi_L(Q_L) = \delta p_m Q_L.$

     Lastly, consider (SCM1) for the case where $Q_1 >  0 , Q_2 >  0.$
First note that $(n_S = 1, q_1(i) = Q_1, q_2(i) = Q_2)$ is a feasible
solution. Let $N $ be defined by
    $$
       N = [C(Q_1,0) + \delta C(Q_2,Q_1)]/[C(0,0)]
 $$
 Suppose there is a feasible solution $(n_S, q_1(i),q_2(i))$ where
$n_S > N$.  Then 
         $$
               \int_0^{n_S} �[C(q_1(i),0) + \delta C(q_2(i),q_1(i))]di
            \geq n_SC(0,0) > NC(0,0) = [C(Q_1,0) + \delta
C(Q_2,Q_1)].
 $$

We may, therefore, without loss of generality confine attention to
feasible points where $n_S \leq N$. Given assumption (A5), we may
confine attention to feasible solutions where $q_t(i) \leq K, i =
1,2$.  Lastly, it would be wasteful to introduce an S-type firm which
produces zero output in both periods. So, w.l.o.g. one can confine
attention to feasible points where $q_t(i) > 0 $ for some $t$, for
all $i �\in [0,n_S]. $

     Note that it is possible to extend any function $q_t(i)$ on
$[0,n_S]$ to an integrable function on $[0,N]$ by setting $q_t(i) =
0$ for $i �\in (n_S, N]$. 

     Let $I(q_1,q_2)= 1,$ if $q_t > 0 $ for some $t, q_t \leq K$ for
$t = 1,2 $ and 
               $I =  0  $ otherwise. 
 One can rewrite (SCM1) for  $Q_1,Q_2 >>  0 $ as
  $$
  (SCM1') \;\;\;\;\;  Minimize \int_0^N�G[q_1(i),C(q_2(i)]di
   $$
 w.r.t. $q_t:[0,N] x [0,K], q_t(.)$ integrable, 
 subject to
   $$
           \int_0^N�q_t(i)di \geq  Q_i, 
  $$
 where $G:{\bf R}^2_+ \rightarrow {\bf R}_+$ is given by
    $$
  G(q_1,q_2) = {C(q_1,0) + \delta C(q_2,q_1)}I(q_1,q_2).
 $$

Check that $G$ is a bounded function on ${\bf R}^2_+$ (bounded above
by $(1 + \delta )C(K,0))$. A direct appeal to Theorem 6.1 in Aumann
and Perles (1965) shows that there exists a solution to (SCM1').
This, in turn, implies that there exists a solution to (SCM1) for all
$Q_1,Q_2 \geq 0.$\footnote{R.J.  Aumann and M. Perles (1965) ``A
Variational Problem Arising in Economics.''  {\it Journal of
Mathematical Analysis and Applications}, 11: 488 - 503.}


     Let $�\lambda_1 $ and $\lambda_2$ be the Lagrangean multipliers
associated with the constraints (*) in (SCM1). Then at any optimal
solution $(n_S, q_1(i),q_2(i))$ the necessary conditions
 \begin{equation} \label{ee4}
 q_1(i) > 0 
  \; {\rm implies}\; C_q(q_1(i),0)+\delta
C_w(q_2(i),q_1(i)) = f_1 (q_1(i),q_2(i)) = \lambda_1(Q_1,Q_2),  
 \end{equation}
 and
 \begin{equation} \label{ee5}
 q_2(i) > 0 
  \; {\rm implies}\; C_q(q_2(i),q_1(i)) = f_2(q_1(i),q_2(i)) =
\lambda_2(Q_1,Q_2). 
 \end{equation}
 
     We claim that �$\psi (Q_1,Q_2)$ is a convex function on ${\bf
R}^2_+.$ Let us define the ``cost-possibility set'' of any firm $i
\in {\bf R}_+$ by
   $$
        f(i) = \{(q_1,q_2,-y) q_t \geq  0 , y \geq C(q_1,0) + \delta
C(q_2,q_1)\} \cup � \{(0,0,0)\}
 $$
 (no-entry is equivalent to (0,0,0))

$f(i)$ is identical for all $i$. Let $F$ be the ``cost-possibility
set'' of the social planner, i.e.  the set of all output-cost
combinations that are feasible for the social planner by using any
number of S-type firms and any distribution of output across such
firms.  Thus, $F$ is the integral of the set-valued correspondence
$f(i): {\bf R}_+ \rightarrow {\bf R}_+$ with respect to Lebesgue
measure.  A direct appeal to the Lyapunov-Richter theorem
\footnote{See L.1.3 in A. Mas-Colell (1985) {\it The Theory of
General Economic Equilibrium: A Differentiable Approach.} Cambridge
University Press, Cambridge, England.} shows that $F$ is a convex
set. By definition of �$\psi$ in (SCM1), $F$ is also the set of all
$ \{(Q_1,Q_2,-Y):\; Y \geq \psi(Q_1,Q_2), Q_t \geq 0 , t = 1,2\}$. It is
easily checked that convexity of $F$ implies that $\psi$ is a convex
function on ${\bf R}^2_+$.

     Next, we claim that $\psi$� is continuous on ${\bf R}^2_+$.
Continuity on ${\bf R}^2_{++}$ follows from its convexity. Continuity
on the border can be verified directly. For example, choose any
sequence $\{Q_1^m,Q_2^m\}\rightarrow (Q_1,Q_2)$ such that $Q_1 = 0 ,
Q_2 > 0 $. Let $q(m_2)$ be a solution to the minimization problem in
(\ref{ee2}).  Then
 $$
 m_2Q_2^m = \psi�(0,Q_2^m) \leq \psi�(Q_1^m,Q_2^m) \leq
\{C([(Q_1^mq(m_2))/Q_2^m],0) +\delta
C(q(m_2),
 $$
 and $[(Q_1^mq(m_2))/Q_2^m])\} \{Q_2^m/q(m_2)\} \rightarrow m_2Q_2 =
�\psi(0,Q_2)$ as $m \rightarrow +\infty.$
 

Since the left hand side of the inequality equals $ m_2Q_2^m
\rightarrow m_2Q_2 = \psi�(0,Q_2)$, we have that �$\psi(Q_1^m,Q_2^m)
\rightarrow �\psi(0,Q_2)$. Similar arguments can be used when the
limits $ (Q_1,Q_2)$ of the sequence $\{Q_1^m,Q_2^m\}$ are such that
$Q_2 = 0 , Q_1 > 0 $ and also when both $Q_t = 0 $.

     Next we want to establish that the partial derivatives of
$�\psi(Q_1,Q_2)$ exist on ${\bf R}^2_{++}$ . For any fixed $Q_2 = z >
0 $, let $g(Q_1) = �\psi(Q_1,z)$. We will show that $g$ is
differentiable on ${\bf R}_{++}$.

     Firstly, note that $g$ is a convex function and so its right and
left hand derivatives exist.  Furthermore, 
 \begin{equation} \label{ee6}
                    g'_+ \geq  g'_-.                    
\end{equation}
  Let $(n_S, q_1(i),q_2(i))$ be the optimal solution of (SCM1) at
$(Q_1,z)$.  For $\varepsilon >  0 $ small enough, consider the vector
$(Q_1 + \varepsilon ,z)$. Let $\hat{q_1}(i) = q_1(i) + \varepsilon
[q_1(i)/Q_1]$.  Note that $\hat{q_1}(i) =  0 $ if $q_1(i) =  0 $. It is easy
to check that $(n_S, \hat{q_1}(i),q_2(i))$ is a feasible way to produce
$(Q_1 + \varepsilon ,z)$.  Thus,
     $$
 \begin{array}{l}
 g(Q_1 + \varepsilon )  = \psi(Q_1 + \varepsilon ,z) 
               \leq  \int_0^{n_S}�[f(\hat{q_1}(i),q_2(i))]di \\
               = �\int_0^{n_S}[f([q_1(i) + \varepsilon
(q_1(i)/Q_1)],q_2(i))di
 \end{array}
  $$
 so that 
      $$
\begin{array}{ll}
  g'_+(Q_1) &= \;\;\stackrel{\textstyle lim}{\scriptstyle \varepsilon
\rightarrow 0 }\;\; \{[g(Q_1 + \varepsilon ) - g(Q_1)]/\varepsilon \}\\
          &   \leq \stackrel{\textstyle lim}{\scriptstyle \varepsilon
\rightarrow0 } \;\;(1/\varepsilon )\int_0^{n_S} \{f([q_1(i)+
\varepsilon (q_1(i)/Q_1)],q_2(i)) \\ - f(q_1(i),q_2(i))\}di. 
 \end{array}
      $$
              (Use the dominated convergence theorem and note that in
the last expression the term within curly brackets equals zero if $q_1(i) = 0
$.) 
          $$
    g'_+(Q_1) = �\stackrel{\textstyle lim}{\scriptstyle \varepsilon
\rightarrow0 }\;\; (1/\varepsilon )\int_0^{n_S}\{f([q_1(i)+
\varepsilon (q_1(i)/Q_1)],q_2(i)) - f(q_1(i),q_2(i)) \}I(i)di
   $$
 where $I(i)= 1$ if $q_1(i) >  0 $  and      $I(i)=  0 $  otherwise.
 
 Since $f(q_1,q_2)$ is differentiable and further using (\ref{ee4}),
we have from the above inequality 
 that:
  \begin{equation} \label{ee7}
 \begin{array}{l}
      g'_+(Q_1) \leq \int_0^{n_S}�f_1
(q_1(i),q_2(i))[q_1(i)/Q_1]I(i)di \\
           = \int_0^{n_S}\{[�\lambda_1(Q_1,z)]/Q_1\}�q_1(i)di \;\;
(using (\ref{ee4}))\\
          = �\lambda_1(Q_1,z).
  \end{array} \end{equation}
  By a similar argument one can show that:
   \begin{equation} \label{ee8}
     g'_- (Q_1) = \stackrel{\textstyle lim}{\scriptstyle \varepsilon
\rightarrow0 }\;\; [g(Q_1) - g(Q_1 - \varepsilon )](1/\varepsilon )
\geq �\lambda_1(Q_1,z).
  \end{equation}
  Thus (\ref{ee7}) and (\ref{ee8}) imply $$
     g'_+(Q_1) \leq \lambda_1 \leq g'_-(Q_1),
  $$ 
 which combined with (\ref{ee6}) yields
 $$
     g'_+(Q_1) = g'_-(Q_1) = �\lambda_1,
 $$
  that is, $g$ is differentiable at $Q_1$ and 
     \begin{equation} \label{ee9}
 g'(Q_1) = \psi_1(Q_1,Q_2) = \lambda_1(Q_1,Q_2).  
  \end{equation}
 Note that at $Q_2 = 0 , �\psi(Q_1,Q_2) = m_1Q_1$ so that
�$\psi_1(Q_1,0) = m_1$.  Similar reasoning shows that given $Q_1 \geq
0 $, the partial derivative of $\psi$ w.r.t. $Q_2$ exists on ${\bf
R}_{++}$ and is given by $$
          �\psi_2(Q_1,Q_2) = �\lambda_2(Q_1,Q_2)
$$
and, in particular, $�\psi_2(0,Q_2) = m_2$. 

     As the partial derivatives of $\psi$� exist at each point in
${\bf R}^2_{++}$ and �$\psi$ is convex, it follows that $\psi$ is
continuously differentiable on ${\bf R}^2_{++}$ (Section 42, Theorem
D and Section 44, Theorem E in Roberts and Varberg [1973]).\footnote{
Roberts, A. and D.F. Varberg (1973) {\it Convex Functions}, Academic
Press, New York.}
 

     We summarize the above discussion in the   Lemma II:

Lemma II: (a) There exists a solution to the (SCM1), (SCM2) and
(SCM3). 

(b) For any $Q_E \geq 0 , �\psi_E(Q_E) = p_m Q_E$ and in the solution
to (SCM2), $q_E(i) \in� \{q:[C(q,0)/q] = p_m \}$ 

(c) For any $Q_L \geq 0 , �\psi_L(Q_L) = \delta p_m Q_L$ and in the
solution to (SCM3), $q_L(i) \in� \{q:[C(q,0)/q] = p_m \}$. 

(d) For $Q_1 = 0 , �\psi(Q_1,0) = m_1Q_1$ where $ m_1 > 0 $ is
defined by (\ref{ee2}); For $Q_2 \geq 0 , �\psi(0,Q_2) = m_2Q_2$,
where $m_2$ is defined by (\ref{ee3}).

(e) $\psi$ � is continuous and convex on ${\bf R}^2_+$;

(f) For any $(Q_1,Q_2) \geq 0 $, there exists Lagrangean multipliers
$�\lambda_1(Q_1,Q_2), �\lambda_2(Q_1,Q_2)$, such that the solution
$(n_S, q_1(i),q_2(i))$ to (SCM1) is characterized by (\ref{ee4}) and
(\ref{ee5}).

(g) $\psi� $ is continuously differentiable on ${\bf R}^2_{++}$ and
its partial derivatives are given by �$\psi_t(Q_1,Q_2) =
�\lambda_t(Q_1,Q_2), t =1,2$; further, $�\psi_1(Q_1,0) = m_1$ and
$�\psi_2(0,Q_2) = m_2$.

  We now rewrite (SPP*) as the following problem (hereafter referred
to as SPP)
  $$
  \stackrel{\textstyle Maximize}{\scriptstyle Q_1, Q_2, Q_E, Q_L \geq
0 } W(Q_1,Q_2,Q_E,Q_L) $$
 where
$$
 W(Q_1,Q_2,Q_E,Q_L) = \{S(Q_1 + Q_E) + \delta S(Q_2 + Q_L) -
�\psi(Q_1,Q_2) - �\psi_E(Q_E) - �\psi_L(Q_L)\}, $$
  and $ S(y) =\int_0^y P(q)dq.$
   
Note that $S$ is a strictly concave function, continuous on ${\bf
R}_+$, continuously differentiable on ${\bf R}_{++}$ and $S'�(y) =
P(y)$. 

     Thus, using Lemma II, $W$ is continuous and strictly concave
on ${\bf R}^4_+$. Further, at any point $(Q_1 >  0 , Q_2 >  0 , Q_E \geq
0, Q_L \geq  0 )$, the partial derivatives of $W$ w.r.t all arguments
exist. In particular, 
   \begin{eqnarray} \label{ee10}
 \partial �W/ \partial �Q_1 = P(Q_1 + Q_E) - �\psi_1(Q_1,Q_2) = P(Q_1
+ Q_E) - �\lambda_1(Q_1,Q_2) \\
  \label{ee11} \partial �W/ \partial �Q_2 = P(Q_2 + Q_L) -
�\psi_2(Q_1,Q_2) = P(Q_2 + Q_L) - \lambda_2(Q_1,Q_2) \\ \label{ee12}
\partial �W/� \partial Q_E = P(Q_1 + Q_E) - �\psi_E'�(Q_E) = P(Q_1 +
Q_E) - p_m \\
 \label{ee13} \partial �W/ \partial �Q_L = \delta P(Q_2 + Q_L) -
�\psi_L'�(Q_L) = \delta [P(Q_2 + Q_L) - p_m ].  \end{eqnarray}

 Since $P(y)\rightarrow 0 $ as $y \rightarrow +\infty$ and $p_m > 0
$, there exists $Q_0 > 0 $, such that $\partial W/\partial Q_E < 0 $
for any $(Q_1,Q_2,Q_E,Q_L)$ where $Q_E > Q_0$ and $\partial
W/\partial �Q_L < 0 $ for $Q_L > Q_0$.  Note that
�$\lambda_2(Q_1,Q_2) = C_q(q_2(i),q_1(i)) \geq min\{C_q(q,x) : \; 0 \leq q
\leq K, 0 \leq w \leq K\}$. Using (A1) we can check that the minimum
in the previous expression is actually attained at some $(q'', w'')
�\in [0,K] \times [0,K]$.  Thus, for all $(Q_1,Q_2)$ such that $Q_2 > 0 $,
we have �$\psi_2(Q_1,Q_2) \geq C_q(q'', h'') = h$ where, using (A2),
we have $h > 0 $. So, there exists $Q^{\$} > 0 $ such that
$�\partial�W/�\partial�Q_2 < 0 $ for $Q_2 \geq Q^{\$}$ . Let $\hat{Q}
= max (Q^{\$}, Q_0)$. One can without loss of generality, rewrite SPP
as 
  $$
     \stackrel{\textstyle Maximize}{\scriptstyle Q_1 \geq 0 , Q_2,
Q_E, Q_L \in  [0,\hat{Q}]}\;\; W(Q_1,Q_2,Q_E,Q_L) $$

We claim there exists a solution to this maximization problem. Let
$W*$ be the supremum of the maximand (which can be $+\infty )$. Then,
there exists a sequence $\{Q_1^m, Q_2^m, Q_E^m, Q_L^m\}, m =1,2,...,$
where $W(Q_1^m, Q_2^m, Q_E^m, Q_L^m) \rightarrow W^*$. Suppose
$\{Q_1^m\}$ is bounded above. Then the sequence $\{Q_1^m, Q_2^m,
Q_E^m, Q_L^m\}$ is bounded above and has a convergent subsequence,
whose limit is the optimal solution, using continuity of $W$. 
 
     Now  suppose that $\{Q_1^m\}$ is not bounded above; abusing
notation somewhat, suppose that $\{Q_1^m\} \rightarrow +\infty $.
Choose $M$ such that $Q_1^M > 0 $. Then for $m \geq M$,
 $$
          �\psi_1(Q_1^m,Q_2^m)\geq �\psi_1(Q_1^M,Q_2^m) 
                         \geq inf\{�\psi_1(Q_1^M,x):\; 0 \leq x \leq K\}
                         =\Delta (say).
$$
      Let $f(x) = �\psi_1(Q_1^M,x)$. There exists a sequence $\{x_k\}
\in � [0,K]$, such that $f(x_k) \rightarrow \Delta$. We claim that
$\Delta >0$. To see this, consider a convergent subsequence of
$\{x_k\}$, e.g. $\{x_k'\} \rightarrow x*$.  If $x* > 0$, then from
continuous differentiability of $\psi$� (implying continuity of
partial derivatives) on $ {\bf R}^2_{++}$, we have that $f(x_n)
\rightarrow f(x*) = �\psi_1(Q_1^M,x*) = �\lambda_1(Q_1^M,x*) > 0 $.
Suppose $x* = 0$. Note that $�\psi_1(Q_1^M,0) = m_1$. For
$\varepsilon > 0 , �\psi_1(Q_1^M,\varepsilon ) \geq
[�\psi(Q_1^M,\varepsilon ) - �\psi(0,\varepsilon )]/Q_1^M$ (using
convexity of $\psi$�) so that taking the limit as
$\varepsilon\rightarrow 0 $ (using the continuity of $ �\psi $ on
${\bf R}^2_+$ and �the fact that $\psi(0,0) = 0 )$ yields: $$
           \stackrel{\textstyle lim}{\scriptstyle \varepsilon
\rightarrow0 }�\psi_1(Q_1^M,\varepsilon ) \geq �\psi(Q_1^M,0)/Q_1^M =
m_1 $$
  so that $\stackrel{\textstyle lim }{\scriptstyle k' \rightarrow
+\infty} f(x_k') = \Delta� > 0 $. 

     Since $Q_1^m \rightarrow +\infty$ , there exists $M' > 0 $
such that for $ m \geq M', P(Q_1^m) < \Delta�$. Thus, for $m \geq
max(M,M'), P(Q_E^m + Q_1^m) < �\psi_1(Q_1^m,Q_2^m)$ i.e. �$ \partial
W/\partial�Q_1 < 0 $ when evaluated at $m$ large enough.  This is a
contradiction. Thus, $\{Q_1^m\}$ must be bounded above. The proof of
existence is complete. 

     Note that since $W$ is strictly concave on ${\bf R}^4_+$, the
solution to (SPP) is unique. 

     It is easy to check that assumption (A6) implies $P(0) >
�\psi_E'�(0) = p_m$ and $\delta P(0) > �\psi_L'�(0) = \delta p_m$
which, in turn, is sufficient to assert that if $(Q_1,Q_2,Q_E,Q_L)$
solves (SPP) then $Q_1 + Q_E > 0 $ and $Q_2 + Q_L > 0 $. Further, it
is impossible that in the optimal solution $Q_t = 0 $ for either $t
=1$ or $t = 2$ or both. Suppose $Q_1 = 0 $ and $Q_2 > 0 $. Now,
consider the original form of the social planner's problem (SPP*).
Since $Q_1 + Q_E > 0 $, it must be true that $Q_E > 0 $. Then the
social planner can easily reduce total cost by letting the existing
$n_S$ firms of type $S$ (who currently produce zero in period 1)
produce a total amount $Q_1 = Q_E$ (setting $n_E = 0 $). The cost in
period 1 is unchanged and that in period 2 is reduced (using
assumption (A2)), a contradiction. Similarly, $ Q_1 > 0 , Q_2 = 0 $
is ruled out as this implies $Q_L > 0 $ and total cost can be reduced
by letting S-type firms (who currently produce nothing in period 2)
produce a total amount $Q_2 = Q_L$ (setting $n_L = 0 $). Lastly, if
both $Q_t$ are equal to zero, then $Q_E > 0 , Q_L > 0 $. Suppose we
let $\varepsilon$ be the measure of E-type firms that produce in
period 2--- i.e.  convert them to S-type firms, reduce the number of
L-type firms by $\varepsilon$ and transfer their output to these
$\varepsilon$ S-type firms. Then it is easy to see that total costs
are reduced, a contradiction.  To summarize:

\underline{ Lemma III:} There exists a solution to SPP and, hence,
(SPP*).  The solution to the social planner's problem is unique in
$(Q_1,Q_2,Q_E,Q_L)$. If $(Q_1^*,Q_2^*,Q_E^*,Q_L^*)$ is an optimal
solution in total output produced by different ``types'' of firms
then $Q_1^* > 0 $, $Q_2^* > 0 $. 

 
 
     Let us now write down the first order necessary conditions for
$(Q_1^*,Q_2^*,Q_E^*,Q_L^*)$ to be an optimal solution to (SPP)
 \begin{eqnarray}
 P(Q_1^* + Q_E^*) = �\psi_1(Q_1^*,Q_2^*) [ = �\lambda_1(Q_1^*,Q_2^*)]
\leq �\psi_E'�(Q_E^*) [ = p_m ] \label{ee14} \\
  P(Q_1^* + Q_E^*) = p_m \;\;if \;\;Q_E^* > 0 \label{ee15}\\
 \delta P(Q_2^* + Q_L^*) = �\psi_2(Q_1^*,Q_2^*) [ =
�\lambda_2(Q_1^*,Q_2^*)] \leq �\psi_L'�(Q_L^*) [ = \delta p_m ]
\label{ee16}\\
 P(Q_2^* + Q_L^*) = p_m \;\; if\;\; Q_L^* > 0 \label{ee17}
\end{eqnarray}
   Define $p_1^* = P(Q_1^* + Q_E^*), p_2^* = P(Q_2^* + Q_L^*)$.  Let
$(n_S^*,n_E^*,n_L^*,q_1^*(i),q_2^*(i),q_E^*(i),q_L^*(i))$ be the
solutions to (SCM1), (SCM2), and (SCM3) associated with
$(Q_1^*,Q_2^*), Q_E^*$ and $Q_L^*$ respectively. 

     We want to show that $[p_1^*, p_2^*, n_S^*, n_E^*, n_L^*,
(q_1^*(i),q_2^*(i),0 \leq i \leq n_S^*), (q_E^*(i), 0 \leq i \leq
n_E^*, (q_L^*(i), 0 \leq i \leq n_L^*)]$ constitutes an equilibrium.
Recall the conditions (i)-(xi) that define an equilibrium. By
definition of the prices, conditions (i) and (ii) in the definition
of equilibrium are satisfied. From (\ref{ee14}) and (\ref{ee15}) we
have that $p_1^* \leq p_m , p_2^* \leq p_m$ which implies conditions
(vii) and (viii) of the definition of equilibrium are satisfied. If
$n_E^* > 0 $, then it must be the case that $Q_E^* > 0 $ so that
(\ref{ee15}) implies $p_1^* = p_m$ and from Lemma II(b) we have that
$q_E^*(i)$ maximizes one period profit at price $p_m$ and the maximum
profit is equal to 0. Thus, conditions (iv) and (x) of the definition
of equilibrium are satisfied. Similarly (\ref{ee17}) and Lemma II(c)
imply that conditions (v) and (xi) are met. Since $n_S^*,Q_1^*,Q_2^*
>> 0 $, it just remains to show that conditions (iii) and (ix) are
satisfied (condition (vi) then holds automatically).  In other words,
we need to show that $(q_1^*(i),q_2^*(i))$ maximizes two period
discounted sum of profits at prices $(p_1^*,p_2^*)$ and that this
maximum is equal to 0.


     Consider (SCM1) at $(Q_1^*,Q_2^*) >> 0 $. From (\ref{ee14}) and
(\ref{ee16}) we have that $�\lambda_1(Q_1^*,Q_2^*) = p_1^*,
�\lambda_2(Q_1^*,Q_2^*) = p_2^*$. Then $(�\lambda_1 =
p_1^*,�\lambda_2 = p_2^*, q_1^*(i), q_2^*(i), n_S^*)$ minimizes the
Lagrangean function:
  $$
 L = �\int_0^{n_S} f(q_1(i),q_2(i))di + \lambda_1(Q_1-\int_0^{n_S}
q_1(i)di) + \lambda_2(Q_2 -\int_0^{n_L}�q_2(i)di) 
 $$ 
 with respect to $n_S \geq 0 , �\lambda_j \geq 0 , q_1:[0,n_S]
\rightarrow {\bf R}_+, q_2:[0,n_S]\rightarrow {\bf R}_+, q_t(.)$
integrable. 

     Then it must, in particular, be true that given $(�\lambda_1 =
p_1^*,�\lambda_2 = p_2^*)$, the vector 
 $(n_S^*,q_1^*(i),q_2^*(i))$ maximizes:
$$
    \int_0^{n_S}  �[p_1^*q_1(i) + p_2^*q_2(i) - f(q_1(i),q_2(i))]di
 $$ 
 with respect to $n_S \geq  0 , q_1:[0,n_S] \rightarrow {\bf R}_+,
q_2:[0,n_S] \rightarrow {\bf R}_+, q_t(.)$ integrable.  But this
implies that (almost everywhere)

(a) $(q_1^*(i),q_2^*(i))$ maximizes $[p_1^*q_1 + p_2^*q_2 - f(q_1,q_2)]$
with respect to $(q_1,q_2) \geq  0 $, and 

(b) $[p_1^*q_1(i) + p_2^*q_2(i) - f(q_1(i),q_2(i))] = 0 $.

Proof of (a) is obvious (for otherwise we could increase the maximand
by choosing a different value for $(q_1(i),q_2(i))$ on a positive
measure of firms. To see (b), suppose not. There are two
possibilities:

(1) $[p_1^*q_1(i) + p_2^*q_2(i) - f(q_1(i),q_2(i))] < 0 $ in which
case the maximand is increased by simply eliminating a positive
measure of such firms (reducing $n_S$ below $n_S^*$), a
contradiction;

(2) $[p_1^*q_1(i) + p_2^*q_2(i) - f(q_1(i),q_2(i))] >  0 $ in which case
the maximand can be increased to $+\infty$ by setting $n_S   = +\infty$ and
letting all $j \geq n_S^*$ produce the same output vector $(q_1(i),q_2(i))$,
a contradiction as $n_S^* < \infty$ .

This proves (b).  (a) and (b) imply that conditions (iii) and (ix) in
the definition of equilibrium hold. We have therefore proved that:

\underline{ Lemma IV:} Every solution to the social planner's problem
is implementable as a competitive equilibrium. In particular, let
$(n_S^*,n_E^*,n_L^*,q_1^*(i),q_2^*(i),q_E^*(i),q_L^*(i))$ be a
solution to (SPP*) with associated total output
$(Q_1^*,Q_2^*,Q_E^*,Q_L^*)$. Then, if $p_1^* = P(Q_1^* + Q_E^*),
p_2^* = P(Q_2^* + Q_L^*)$, then $[p_1^*, p_2^*, n_S^*, n_E^*, n_L^*,
(q_1^*(i),q_2^*(i),0 \leq i \leq n_S^*), (q_E^*(i), 0 \leq i \leq
n_E^*, (q_L^*(i), 0 \leq i \leq n_L^*)]$ is a competitive
equilibrium.  

Lemmas  III  and IV imply:

\underline{  Lemma V:} There exists an equilibrium. 

     Next, we show that every equilibrium is socially optimal. Let
$[p_1^{\$} , p_2^{\$} , n_S^{\$} ,$ $n_E^{\$} , n_L^{\$} ,
(\hat{q_1}^{\$} (i),q_2^{\$} (i), 0 \leq i \leq n_S), (q_E^{\$} (i), 0
\leq i \leq n_E^{\$} ), (q_L^{\$} (i), 0 \leq i \leq n_L^{\$} )]$ be
a equilibrium. Let $(Q_1^{\$}, Q_2^{\$} )$ be total output produced
by S-type firms in this equilibrium. From Lemma III, we have that
$Q_1^{\$} > 0 , Q_2^{\$} > 0 $.  Let $Q_E^{\$}$ and $Q_L^{\$}$ be the
total quantity produced by E and L type firms in their period of
stay. 

     Our first claim is that $(n_S^{\$} , \hat{q_1}^{\$} (i),q_2^{\$}
(i))$ is a socially cost minimizing way of producing $(Q_1^{\$}
,Q_2^{\$} )$ i.e. it solves (SCM1) given $(Q_1^{\$} ,Q_2^{\$} )$.
Suppose not.  Then there exists $(\hat{n},\hat{q_1}(i),\hat{q_2}(
i))$ which solves (SCM1) given $(Q_1^{\$} ,Q_2^{\$} )$ and 
        \begin{equation} \label{ee18}
        \psi(Q_1^{\$} ,Q_2^{\$} ) < \int_0^{n_S^\$} f(\hat{q_1}{\$}
(i),q_2^{\$} (i))di.  
   \end{equation}
  The sum of total profits of all S type firms in equilibrium is
zero. Therefore,
   $$
\begin{array}{l}
       0 = \int_0^{n_S^\$} [p_1^{\$} \hat{q_1}^{\$} (i) + \delta
p_2^{\$} q_2^{\$} (i) - f(\hat{q_1}^{\$} (i),q_2^{\$} (i))]di<
p_1^{\$} Q_1^{\$} + \delta p_2^{\$} Q_2^{\$} - \psi(Q_1^{\$} ,Q_2^{\$}
)\;\;(using \;\; (\ref{ee18}) )
                   \\
       = \int_0^{\hat{n}}�[p_1^{\$} \hat{q_1}(i) + \delta p_2^{\$}
\hat{q_2}( i) - f(\hat{q_1}(i),\hat{q_2}( i))]di\\
  \end{array}
 $$
 which implies, in turn, that there exists some i for which
$[p_1^{\$} \hat{q_1}(i) + \delta p_2^{\$} \hat{q_2}( i) -
f(\hat{q_1}(i),\hat{q_2}( i))] > 0$. But by definition of
equilibrium, the maximum possible discounted sum of profit at prices
$(p_1^{\$} ,p_2^{\$} )$ is 0. We have a contradiction. Hence,
  \begin{equation} \label{ee19}
            �(Q_1^{\$} ,Q_2^{\$} ) = \int_0^{n_S^\$} �f(\hat{q_1}^{\$}
(i),q_2^{\$} (i))di 
 \end{equation} and $(n_S^{\$} ,\hat{q_1}^{\$} (i),q_2^{\$} (i))$
does solve (SCM1) given $(Q_1^{\$} ,Q_2^{\$} )$.


 

     Next, suppose $n_E^{\$} > 0 $. Then from Proposition 1,
$p_1^{\$} = p_m$ and $q_E^{\$} (i) �\in \{q \geq 0 :[C(q,0)/q] = p_m
\}$ which means that total cost of production of $Q_E^{\$}$ is equal
to $p_m Q_E^{\$}$ which is equal to $�\psi_E(Q_E^{\$} )$, i.e.
$(n_E^{\$} ,q_E^{\$} (i))$ solves (SCM2) given $Q_E^{\$} $.
Similarly, one can show that if $n_L^{\$} > 0 $, then $(n_L^{\$}
,q_L^{\$} (i))$ solves (SCM3) given $Q_L^{\$}$ . 

     Therefore, in equilibrium, total output $(Q_1^{\$} ,Q_2^{\$}
,Q_E^{\$} ,Q_L^{\$} )$ produced by different types of firms are
produced in the socially cost minimizing way. Let the total social
welfare in equilibrium is equal to $W(Q_1^{\$} ,Q_2^{\$} ,Q_E^{\$}
,Q_L^{\$} )$, where the function $W$ is as defined before introducing
problem (SPP).  As noted earlier, $W(.)$ is strictly concave on ${\bf
R}^4_+$. The partial derivatives of $W$ exist at all
$(Q_1,Q_2,Q_E,Q_L) \geq 0 $, where $Q_1 > 0 , Q_2 > 0 $. 

     Suppose equilibrium is not socially optimal. Let
$(Q_1^*,Q_2^*,Q_E^*,Q_L^*)$ maximize social welfare.  Then,
  \begin{equation} \label{ee20}
    W(Q_1^{\$} ,Q_2^{\$} ,Q_E^{\$} ,Q_L^{\$} ) -
W(Q_1^*,Q_2^*,Q_E^*,Q_L^*) < 0. 
 \end{equation}
    From Lemma III, $Q_1^* > 0 , Q_2^* > 0 $. As noted above,
$Q_1^{\$} > 0 $, $Q_2^{\$} > 0 $. So strict concavity of $W$ implies
  \begin{equation} \label{ee21}
 \begin{array}{l}
W(Q_1^{\$},Q_2^{\$},Q_E^{\$},Q_L^{\$})-W(Q_1^*,Q_2^*,Q_E^*,Q_L^*)\\
\geq [\partial W(Q_1^{\$},Q_2^{\$},Q_E^{\$},Q_L^{\$})/ \partial Q_1]
[Q_1^{\$}-Q_1^*] \\ + [ \partial W(Q_1^{\$},
Q_2^{\$},Q_E^{\$},Q_L^{\$})/ \partial Q_2] [Q_2^{\$}-Q_2^*]\\
 +[ \partial  W(Q_1^{\$},Q_2^{\$},Q_E^{\$},Q_L^{\$})/ \partial Q_E]
[Q_E^{\$} - Q_E^*] \\
  +[ \partial W(Q_1^{\$},Q_2^{\$},Q_E^{\$},Q_L^{\$})/ \partial Q_L]
[Q_L^{\$}-Q_L^*]. \\
 \end{array}
\end{equation}
Note that:
    \begin{equation} \label{ee22}
  \partial W(Q_1^{\$} ,Q_2^{\$} ,Q_E^{\$} ,Q_L^{\$} )/ \partial Q_1 =
P(Q_1^{\$} + Q_E^{\$} ) - �\psi_1(Q_1^{\$} ,Q_2^{\$} ) 
     = p_1^{\$}  - �\lambda_1(Q_1^{\$} ,Q_2^{\$} ) \;\;\; 
     \end{equation}
    \begin{equation} \label{e0}
 = p_1^{\$} - f_1 (\hat{q_1}^{\$} (i),q_2^{\$} (i))\;\;(from\;\;
(\ref{ee4})). 
  \end{equation}
   As $Q_1^{\$} > 0 , Q_2^{\$} > 0 $, there exists positive measure
of $i$ such that $\hat{q_1}^{\$} (i) > 0 ,q_2^{\$} (i) > 0 $, so that
the first order conditions of profit maximization (condition (iii) in
definition of equilibrium) implies that right hand side of
(\ref{ee22}) is equal to zero, i.e. 
     \begin{equation} \label{ee23}
\partial  W(Q_1^{\$} ,Q_2^{\$} ,Q_E^{\$} ,Q_L^{\$} )/\partial �Q_1 =  0.             
\end{equation}
Similarly, one can show that
   \begin{equation} \label{ee24}
 \partial  W(Q_1^{\$} ,Q_2^{\$} ,Q_E^{\$} ,Q_L^{\$} )/ \partial Q_2 = 0.
\end{equation} 
 Next note that, from Proposition 1, $p_1^{\$} \leq p_m$ so that $
\partial W(Q_1^{\$} ,Q_2^{\$} ,Q_E^{\$} ,Q_L^{\$} )/ \partial Q_E =
p_1 - \psi_E�(Q_E^{\$} ) = p_1 - p_m \leq 0 $ and it is equal to zero
if $Q_E^{\$} > 0 $ (since $p_1^{\$} = p_m $). Thus,
    \begin{equation} \label{ee25}
      [ \partial W(Q_1^{\$} ,Q_2^{\$} ,Q_E^{\$} ,Q_L^{\$} )/ \partial
Q_E][Q_E^{\$} - Q_E^*] \geq 0.  \end{equation}
 Similarly, one can show that 
       \begin{equation} \label{ee26}
    [ \partial W(Q_1^{\$} ,Q_2^{\$} ,Q_E^{\$} ,Q_L^{\$} )/ \partial
Q_L][Q_L^{\$} - Q_L^*] \geq 0.  
  \end{equation}
   From (\ref{ee21}) , (\ref{ee23}) - (\ref{ee26}) we have 
        $ W(Q_1^{\$} ,Q_2^{\$} ,Q_E^{\$} ,Q_L^{\$} ) -
W(Q_1^*,Q_2^*,Q_E^*,Q_L^*) \geq 0 $ contradicting (\ref{ee20}) . The
proof is complete.  We have thus shown:
 
\underline{  Lemma VI:} Every competitive equilibrium is socially
optimal. 
 

Combining  Lemmas (III) - (VI)  yields Proposition 2. 


\bigskip  
  
PROPOSITION 3. Under assumptions (A1)-(A7), an equilibrium exists.
It is unique in prices, output, and number of firms, and it is
socially optimal.  


{\it Proof}.  It is sufficient to show that under (A7) there exists a
unique solution to (SCM1), (SCM2) and (SCM3). Proposition 2 then
implies the result. To see uniqueness in (SCM1) let $(n_S,
q_1(i),q_2(i)), (n',q_1'(i),q_2'(i))$ be any two optimal production
plans producing $(Q_1,Q_2) >> 0 $. Let $N = max (n_S, n')$. Suppose
$n_S < n'$.  Then extend $q_t(i)$ to the entire interval $[0,n']$ by
setting $q_t(i) = 0$ for $i > n_S$. Then let $\hat{q}_t(i) = (1/2)[q_t(i)
+ q_t'(i)], t =1,2$ be defined on $[0,n']$. It is easy to check that
this is a feasible production plan for $(Q_1,Q_2)$. Further, 
  $$
             \int_0^{n'}  �[f(\hat{q_1}(i),\hat{q_2}( i))]di
           <  \int_0^{n'} �(1/2)[f(q_1(i),q_2(i)) + f(q_1'(i),q_2'(i))]di
            =   �\psi (Q_1,Q_2),
$$
 a contradiction. 
 Uniqueness in (SCM2) and (SCM3) are similarly established. //

 
PROPOSITION 4. Under assumptions (A1)-(A7), the following is true in
equilibrium:
   \begin{itemize}
 \item[ (a)]
   Each of the staying firms behaves identically, and there exists a
positive measure of staying firms. There exist $q_1^*$ and $q_2^*$
such that $q_1^*(i)= q_1^*$ and $q_2^*(i)= q_2^*$ for all active
staying firms $i$. 
  \item[ (b)]
 If exiting firms exist, they produce at the initial minimum
efficient scale, which is less than the $q_1$ produced by the staying
firms.  If $n_E > 0 $, then $q_E(i) = q_m$ for all $i �\in [0,n_E]$,
where $q_m$ is the unique solution to minimization of $[C(q,0)/q]$
with respect to $q \geq 0 $, and $q_E < q_1^*$.  
 \item[ (c)] 
There exist no late-entering firms:  $n_L =  0 $.
     \end{itemize}

  
 
{\it Proof}.  The first part of (a) and (b) follow immediately from
strict concavity of the profit function for each type of firm. (Note
that since the total amount $(Q_1,Q_2)$ produced by all S-type firms
is always strictly positive, $(q_1^*,q_2^*) >> 0 $.)  The second part
of (b) results from Proposition 1, because the negative first-period
profits of the staying firms result from their high production for
the sake of learning.  It remains to show that $n_L = 0 $ for part
(c). 
 
     Suppose that $n_L > 0 $. Then from Lemma II, $n_E = 0 $ and $p_2
= p_m $. Under (A7), there exists a unique $q_m$ which minimizes
$[C(q,0)/q]$ over $q \geq 0 $. So, $q_L(i) = q_m$ and
     \begin{equation} \label{ee27}
     p_2 = p_m = C(q_m,0)/q_m = C_q(q_m,0).
  \end{equation} Furthermore,
     \begin{equation} \label{ee28}
     D(p_2) = D(p_m ) = n_Sq_2^* + n_Lq_m > n_Sq_2^*.   
 \end{equation}
From first order condition of profit maximization for firms which
produce in both periods we have that $C_q(q_2^*,q_1^*) = p_2 = p_m$
and, therefore (using (A7), (\ref{ee27})  and $q_1^* >  0 $)
       \begin{equation} \label{ee29}
       q_2^* \geq  q_m.                         
\end{equation}
Next we claim that the following inequality is true:
       \begin{equation} \label{ee30}
   C_w(q_2^*,q_1^*)q_1^* + C_q(q_2^*,q_1^*)q_2^* - C(q_2^*,q_1^*)
\geq 0.  \end{equation}
  By convexity of $C$ on ${\bf R}_+^2$,
  $$
     C(q_m,0) - C(q_2^*,q_1^*) \geq C_q(q_2^*,q_1^*)(q_m - q_2^*) +
C_w(q_2^*,q_1^*)(0 - q_1^*) 
 $$ 
 which implies that $ C_w(q_2^*,q_1^*)q_1^* +
C_q(q_2^*,q_1^*)q_2^* - C(q_2^*,q_1^*) \geq C_q(q_2^*,q_1^*)q_m - C(q_m,0) =
p_2q_m - C(q_m,0) = [p_2 - (C(q_m,0)/q_m)] =  0 $ (using (\ref{ee27}) ). 

From the first order conditions of profit maximization for S-type
firms and the fact that in equilibrium, the discounted sum of profits
is zero, we have: 
 $$
     C_q(q_1^*,0)q_1^* + \delta C_w(q_2^*,q_1^*)q_1^* + \delta
C_q(q_2^*,q_1^*)q_2^* - C(q_1^*,0) - \delta C(q_2^*,q_1^*) = 0.  $$
Using (\ref{ee30}) in the above equation we have:
  $$ 
 C_q(q_1^*,0)q_1^* - C(q_1^*,0) \leq 0 $$ which implies that 
        \begin{equation} \label{ee31}
  q_1^* \leq q_m                           
\end{equation}
so that, from (\ref{ee29}) , we have $ q_1^*  \leq q_2^*$. Thus,
        \begin{equation} \label{ee32}
       n_Sq_1^* \leq n_Sq_2^*.  
 \end{equation}
  From (\ref{ee28})  and (\ref{ee32})  we have
   $$
        D(p_2) > n_Sq_2^*  \geq n_Sq_1^* = D(p_1),  
$$
and so, $p_1 > p_2 = p_m$ , which violates condition (vii) of the 
definition of equilibrium.  
  //


\end{small}

 \end{document}
