INSTITUTE CALIFORNIA OF TECHNOLOGY
Selected Topics in Computer Science and Economics CS/EC/101b
K. Mani Chandy, John Ledyard Winter 2005 1.
Homework Set #1 Solutions
Issued: 7 Jan 05 Due: 17 Jan 05
2. (a) Let x H and y H be two generic tuples in H. The objective is to prove that for any a 1, we have that z = a x + (1 a) y is a tuple in H. By denition of H this is true i g(z) d. Due to that g is convex by hypothesis, we have that g(z) = g(a x + (1 a) y) a g(x) + (1 a) g(y) Since a 1 then a g(x) + (1 a) g(y) max{g(x), g(y)} d Combining 1 and 2 we can conclude that g(z) d. This ends the proof. (b) Suppose by contradiction that H is not closed. Then there exists y dom(g) such that g(y) > d and a sequence of tuples x1 , x2 , . . . , xn such that xi H limi xi = y Since the function g is convex by hypothesis, then g is also continuous in all domain. Recollect the continuous function theorem for sequences stating: If an is a sequence of real numbers, and if limn an = L, then the limn f (an ) = f (L) if f is a function that is continuous at L and dened at all an . It holds: lim g(xn ) = g( lim xn )
n
(1)
(2)
n
(3)
Since xi H then limn g(xn ) d. However g(limn xn ) = g(y) > d. Hence, a contradiction has been obtained and consequenly H must be closed. 3. (a) We start showing that the following statement holds: If for all j we have gj (x) bj and gj (x) qj for some xed x, then gj (x) a bj + (1 a) qj for any a 1. Let x = x0 a tuple which veries the hypothesis of the statement. Due to the previous exercise we know that Hj = {x : gj (x) bj } is a convex set. Therefore can we nd v, w Hj such that x0 = a v + (1 a) w for some a < 1, v, w Hj . Due to the assumption that gj is convex, we have that gj (a v + (1 a) w) a gj (v) + (1 a) gj (w) By hypothesis of the statement gj (v) bj and gj (w) qj , thus, we have that a gj (v) + (1 a) gj (w) a bj + (1 a) qj (5) (4)
This ends the proof of the statement. It can now be easily concluded that for any a < 1 and for any pair b, q of vectors z(a b + (1 a) q) max{z(b), z(q)} a z(b) + (1 a) z(q) (6)
The rst part of inequality 6 follows because every feasible solution for z(b) and z(q) is also a feasible solution for z(a b + (1 a) q) due to the statement. Since a < 1 it is easy to see that also the second part of the inequality holds. Hence, we have proven that z is a concave function. (b) z is in general nonlinear. In order to convince yourself, consider the function z(b), where b = (b1 , b2 ) is two dimensional vector dened as: z(b) = max x2 subje...