Modified and Extended Oligopolies - Nonlinear Oligopolies: Stability and Bifurcations - page 194

Lemma 4.1. The reaction function R k .Q;S k / defined above is a decreasing func-
tion of both variables Q and S k .

Proof. Let g. z k ;Q;S k / denote the left hand side of (4.97), then

@g
@ z k D f 0 .Q/ C 0 k . z k /<0;
@g
@Q D f 0 .Q/ C . z k C S k /f 00 .Q/ 0;
@g
@S k D f 0 .Q/<0;

so g is strictly decreasing in z k and S k , and is non-increasing in Q. Assume first
that Q .1/ <Q .2/ , and fix the value of S k . If with Q D Q .1/ the first case of (4.96)
occurs then assumption (B) implies that the same hold also for Q D Q .2/ ,soR k
remains 0, and if with Q D Q .1/ the second case occurs, then R k cannot increase
any further. If with Q D Q .1/ the third case of (4.96) occurs, then the monotonicity
of g in Q implies that with Q D Q .2/ the second case is impossible so the best
response with Q D Q .2/ is either zero or the solution of (4.97). In the first case R k
clearly decreases. It will also be shown that if the third case of (4.96) occurs with
both Q D Q .1/ and Q D Q .2/ , then still R k .Q .1/ ;S k / R k .Q .2/ ;S k /. Assume
this is not the case, then

0 D g.R k .Q .1/ ;S k /;Q .1/ ;S k />g.R k .Q .2/ ;S k /;Q .1/ ;S k /
g.R k .Q .2/ ;S k /;Q .2/ ;S k / D 0;

which is an obvious contradiction. Assume next that S .1/
k

<S .2/

k , and fix the value of
Q. If with S k D S .1 k the first case of (4.96) occurs, then f 0 <0implies that the same
holds for S k D S .2 k ,soR k remains 0, and if the second case occurs, then R k cannot
increase any further. If with S k D S .1/

the third case of (4.96) occurs, then f 0 <0

k

implies that with S k D S .2/

k the second case is impossible, so the best response is
either zero or the solution of (4.97). In the first case R k decreases. Assume finally
that with both S k D S .1/
k

and S k D S .2/

k the third case of (4.96) occurs. It can easily
be shown that in this case R k .Q;S .1 k />R k .Q;S .2 k /. Assume not, then

0 D g.R k .Q;S .1 k /;Q;S .1 k />g.R k .Q;S .1 k /;Q;S .2 k /
g.R k .Q;S .2 k /;Q;S .2 k / D 0;

which is again a contradiction.