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midterm_2005Ans

# midterm_2005Ans - Fall 2005 ARE211 Midterm Exam Answer key...

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Unformatted text preview: Fall 2005 ARE211 Midterm Exam - Answer key Problem 1 : Local vs Global Conditions . (15 points) a) Let f : R n ! R and suppose that there exists x 2 R n and an open set U containing x such that f ( x ) f ( x ), for all x 2 U . Do not assume that f is a di erentiable function. i) Identify a condition on f that is su cient to ensure that f ( x ) < f ( x ), for all x 2 R n . Prove that the condition is su cient. The condition is that f is a strictly convex function. One de nition of strict convexity is that the epigraph of the function, is a strictly convex set. (The epigraph of a function, denoted epi ( f ) , is the \stu above the graph," i.e., epi ( f ) = f ( x ; y ) 2 R n ¢ R : y f ( x ) g . To prove that strict convexity is su cient for the property speci ed in i) assume that there exists x 2 R n and an open set U con- taining x such that f ( x ) f ( x ) , for all x 2 U , and y 2 R n , such that f ( y ) f ( x ) . We will show that f is not strictly convex, by showing that the line segment joining x ; f ( x ) ¡ and y ; f ( y ) ¡ is not contained in the interior of epi ( f ) . For 2 (0 ; 1) , let ( ) = x + (1 ) y ; f ( x ) + (1 ) f ( y ) ¡ . Since f ( y ) f ( x ) , it follows that. for all 2 (0 ; 1) , n +1 ( ) = f ( x ) + (1 ) f ( y ) ¡ f ( x ) . But since U is an open set, there exists > , such that for < < , ( 1 ( ) ; ::: n ( )) 2 U . By assumption, then, for < < , f ( x ) f ( 1 ( ) ; ::: n ( )) and hence ( ) does not belong to epi ( f ) . ii) Identify a second condition on f that is su cient to ensure that f ( x ) < f ( x ) for all x 2 R n and is a strictly weaker condition than the rst one you identi ed. Prove that the condition is su cient. The condition is that f is a strictly quasi-convex function. A necessary condition for strict quasi-convexity is that every lower contour set of f is a strictly convex set. To prove that strict quasi-convexity is su cient for the property speci ed in ii) assume that there exists x 2 R n and an open set U containing x such that f ( x ) f ( x ) , for all x 2 U , and y 2 R n , such that f ( y ) f ( x ) . Let LC ( x ) denote the lower contour set of f corresponding to f ( x ) . We will show that f is not strictly quasi-convex, by showing that the line segment joining x and y is not contained in the interior of LC ( x ) . By assumption, y 2 LC ( x ) . For 2 (0 ; 1) , let ( ) = x + (1 ) y . Since U is an open set, there exists > , such that for < < , ( ) 2 U . By assumption, then, for < < , f ( ( )) f ( x ) , and hence ( ) does not belong to the interior of LC ( x ) . Hence we have established that the line segment joining x and y is not contained in the interior of LC ( x ) . iii) Demonstrate with an example of a function from R 2 to R that the second condition is strictly weaker than the rst....
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midterm_2005Ans - Fall 2005 ARE211 Midterm Exam Answer key...

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