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Unformatted text preview: ARE211, Fall 2007 CHARFUNCT: TUE, NOV 6, 2007 PRINTED: DECEMBER 14, 2007 (LEC# 20) Contents 5. Characteristics of Functions. 1 5.1. Surjective, Injective and Bijective functions 1 5.2. Homotheticity 3 5.3. Homogeneity and Euler’s theorem 4 5.4. Monotonic functions 5 5.5. Concave and quasiconcave functions; Definiteness, Hessians and Bordered Hessians. 5 5. Characteristics of Functions. 5.1. Surjective, Injective and Bijective functions Definition: A function f : X → A is onto or surjective if every point in the range is reached by f starting from some point in the domain, i.e., if ∀ a ∈ A , ∃ x ∈ X such that f ( x ) = a . Example: consider f : R → R defined by f ( x ) = 5 for all x . This function is not surjective because there are points in the range (i.e., R ) that aren’t reachable by f (i.e., any point in R except 5 is unreachable). Definition: A function f : X → A is 11 or injective if for every point in the range, there is at most one point in the domain that gets mapped by f to that point. More precisely, f is injective if for all x,x prime such that x negationslash = x prime , f ( x ) negationslash = f ( x prime ) . 1 2 CHARFUNCT: TUE, NOV 6, 2007 PRINTED: DECEMBER 14, 2007 (LEC# 20) Example: consider f : R → R defined by f ( x ) = 5 for all x . This function is not injective because there is a point in the range (i.e., 5) that gets mapped to by f , starting from a large number of points in the domain. Definition: A function f : X → A is bijective if it is surjective and injective. Definition: A function f : X → A is invertible if there exists a function f 1 : A → X such that for all a ∈ A , f ( f 1 ( a )) = a and for all x ∈ X , f 1 ( f ( x )) = x . (We’ll show below that you need both of these conditions to capture the notion of invertibility.) If such a function exists, it is called the inverse of f. Theorem: A function f : X → A is invertible if and only if it is bijective. • You need f to be surjective (onto): otherwise there would be points in the range that aren’t mapped to by f from any point in the domain: E.g., take f : [0 , 1] → R defined by f ( x ) = 5: the image of [0 , 1] under f is { 5 } Now consider any point a negationslash∈ { 5 } . Since f doesn’t map anything to a , there cannot exist a function f 1 such that f ( f 1 ( a ) ) = a . • You need f to be injective (11): otherwise the inverse couldn’t be a function. E.g., take f : [0 , 1] → [01 / 4] defined by f ( x ) = x x 2 and take a = 3 / 16; this point is reached from both 1 / 4 and 3 / 4, i.e., f (1 / 4) = f (3 / 4) = 3 / 16. Now apply the definition to both 1 / 4 and 3 / 4: it is required that for all x ∈ X , f 1 ( f ( x )) = x . In particular, it is required that f 1 ( f (1 / 4)) = f 1 (3 / 16) = 1 / 4. But it is also required that f 1 ( f (3 / 4)) = f 1 (3 / 16) = 3 / 4. Thus f is required to map the same point to two different places, and hence can’t be a function....
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This note was uploaded on 08/01/2008 for the course ARE 211 taught by Professor Simon during the Fall '07 term at Berkeley.
 Fall '07
 Simon

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