Problem_Set_9_Solutions

# Problem_Set_9_Solutions - Problem Set 9 Solutions Economics...

This preview shows pages 1–3. Sign up to view the full content.

Problem Set 9 Solutions Economics 6170 TA: Christopher Handy December 5, 2008 Problem 1 (a) Since ¯ x C is a point of local maximum of F , there is some β > 0 such that for all x C satisfying d ( x, ¯ x ) < β , we have F ( x ) F x ). And since C is open, there is some γ > 0 such that for all x R n satisfying d ( x, ¯ x ) < γ , we have x C . Now, let r = 1 2 min { β,γ } , so that B B x, 2 r ) = { x R n | d ( x, ¯ x ) < 2 r } ⊂ C . Then for all x B , we have that F x ) F ( x ). (b) For any k ∈ { 1 ,...,n } , deﬁne a ( k ) = ¯ x - re k , b ( k ) = ¯ x + re k Now, a ( k ) B and b ( k ) B because d ( a ( k ) , ¯ x ) = d ( b ( k ) , ¯ x ) = d ( re k , 0) = r < 2 r . Since the open ball B is convex, we have that tb ( k ) + (1 - t ) a ( k ) B for all t I = [0 , 1]. Deﬁne f : I R by f ( t ) = F ( tb ( k ) + (1 - t ) a ( k ) ) for all t I Since tb ( k ) + (1 - t ) a ( k ) B C for all t I and F is deﬁned on C , the function f is well-deﬁned. Let A = (0 , 1). We showed in problem 1 of Problem Set 8 that f is continuous on [0 , 1], so it follows that f is continuous on A . Since F is continuously diﬀerentiable on the open set C , we can use the Chain Rule to diﬀerentiate f on A . For all t A , we have f 0 ( t ) = n X i =1 ± D i F ( tb ( k ) + (1 - t ) a ( k ) ) ∂t ( tb i ( k ) + (1 - t ) a i ( k ) ) ² = n X i =1 ³( b i ( k ) - a i ( k ) ) D i F ( tb ( k ) + (1 - t ) a ( k ) = ( b ( k ) - a ( k ) ) F ( tb ( k ) + (1 - t ) a ( k ) ) = (2 re k ) F ( ¯ x + (2 t - 1) re k ) = 2 rD k F ( ¯ x + (2 t - 1) re k ) 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Since F is continuously diﬀerentiable on C and tb ( k )+(1 - t ) a ( k ) = ¯ x +(2 t - 1) re k C whenever t A , we have that f 0 ( t ) is continuous on A . So f is continuously diﬀerentiable on A . (c) Since tb ( k ) + (1 - t ) a ( k ) = ¯ x + (2 t - 1) re k , we have f ( 1 2 ) = F x ). And since tb ( k ) + (1 - t ) a ( k ) B for all t I and F x ) F ( x ) for all x B , we have by the deﬁnition of f that f ( 1 2 ) f ( t ) for all t I , and thus for all t A . (d) Using the result in part (c), we have by the stated Theorem that f 0 ( 1 2 ) = 0. And from part (b), we have that f 0 ( 1 2 ) = 2 rD k F x ). Since r > 0, these two equations imply that D k F x ) = 0. Because k ∈ { 1 ,...,n } was chosen arbitrarily, we have F x ) = 0. This proves the Corollary. Problem 2 We will prove this by verifying a series of claims. First, we will show that the Hessian of f is negative deﬁnite in some neighborhood around ¯ x . Second, we will use Taylor’s Theorem to show that ¯ x is a point of strict local maximum of f . Finally, we will use the quasi-concavity of f on A to show that ¯ x is the unique point of global maximum of f on
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 10/03/2009 for the course ECON 6170 taught by Professor Mitra during the Fall '08 term at Cornell University (Engineering School).

### Page1 / 6

Problem_Set_9_Solutions - Problem Set 9 Solutions Economics...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online