set and hence we may use Tarski’s Fixed Point Theorem to conclude that there
exists a pair (x,y) such that B(x,y) = (rι(y), r2(x)). (x, y) is a pure strategy
Nash equilibrium.
Finally, by Lemma 4.1 we know that no equilibrium can be symmetric. ■
Proof. (Theorem 4.2) Define R, rι(y)∣∆U and r2(x)∣∆u as before. From
assumption B5, rι(y)∣∆U ∈ R and r2(x)∣∆u ∈ R and from assumption B10
they are continuous. Consider the mapping B : R → R such that B(x, y)=
(r1(y ), r2(x)). B is a continuous correspondence, given that both its compo-
nents are continuous, R is a compact set and hence we may use Brouwer’s
Fixed Point Theorem to conclude that there exists a pair (x∕, y∕) such that
B(x∕, y∕) = (r1 (y∕), r2 (x∕)). (x∕, y∕) is a pure strategy Nash equilibrium.
By Lemma 4.2 we know that no equilibrium can be symmetric. ■
8.4 Proofs of Section 6
To prove the theorem we first formulate a useful lemma.
Lemma 8.5 If A10, A2 and A4 hold, then there exist exactly one point d ∈ [0, c]
such that r1(d - ε) >d>r1(d + ε),ε > 0.
Proof. (Lemma 8.5) Consider ∆U . From A10 and Topkis’s Monotonicity The-
orem we know that reaction curve is decreasing in this area. The generalized
first order condition for a maximum to occur in (x, x) is (in the absence of dif-
ferentiability in this point) U1 (x, x) ≤ L1(x, x). Assumption A2 rules out this
possibility, thus no x ∈ (0, c) can be a best response to itself. Moreover, neither
(0, 0) nor (c, c) can be an equilibrium, since from A3 follows, that for player 1
it is always profitable to deviate from any of these points. Hence, the reaction
curve does not cross the 45o line, and there must exist a point, call it d ∈ [0, c]
such that r1(d - ε) >d>r1(d + ε).
Now, we prove uniqueness of this point. Consider r1(y) and r1(y) defined
as in Section 4. Denote W(y) = L(rι(y),y) and V(y) = U(rι(y),y). From the
Envelope Theorem, dwy(y) = L2(r1(y), y), and dVy(y) = U2(r1(y),y). Hence, if
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