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hw21solutions - = f y there is nothing to prove as that...

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Math 521 - Advanced Calculus I Instructor: J. Metcalfe Due: March 29, 2010 Assignment 21 1. Let I R be an interval and let f : I R be increasing on I . If c is not an endpoint of I , show that the jump j f ( c ) of f at c is given by inf { f ( y ) - f ( x ) : x < c < y, x, y I } . This follows from the formula on p. 150 of the text for increasing functions. j f ( c ) = inf { f ( x ) : x > c } - sup { f ( x ) : x < c } = inf { f ( y ) : y > c } + inf {- f ( x ) : x < c } = inf { f ( y ) - f ( x ) : y > c > x } . 2. Let f : [0 , 1] R be a continuous function that does not take on any of its values twice and with f (0) < f (1). Show that f is strictly increasing on [0 , 1]. Suppose not. That is, suppose there are x, y with 0 x < y 1 with f ( x ) f ( y ). If
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Unformatted text preview: ) = f ( y ), there is nothing to prove as that value is attained twice. Thus, assume f ( x ) > f ( y ). If f (0) < f ( x ), then ﬁx k ∈ R so that f (0) < k < f ( x ) and f ( y ) < k < f ( x ). Then by the intermediate value theorem, there exist points p between 0 and x and q between x and y where f ( p ) = f ( q ) = k . Thus f takes on a value twice. Similarly if f (0) ≥ f ( x ), then f (1) > f (0) ≥ f ( x ) > f ( y ) Thus we can choose k so that f ( y ) < k < f ( x ) and f ( y ) < k < f (1) and argue similarly. 1...
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