such that p = hip). By the construction of 'h(-), this implies that p is an equilibrium. A graphical illustration is given below:
(d) Define g-. [0,4]2 [0,4]2 by gx(p) = - (px - l)(p1 - 2)(Pl - 3) - Pj + p2, gz(p) = - (p2 - l)(p2 - 2)(p2 - 3) + - p2. then g{0,0) = (6,6), £(4,4) = 6, - 6), g{-) is continuously differentiate and satisfies the GS property. Hence, by (b), it satisfies the SGS property. Moreover, g_1(0) 3 {(1,1),(2,25,(3,3)}.
(e) Suppose that g(p) = g(p') = 0. We shall now prove that there exists p
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