Given f: R->R, prove that concavity implies quasiconcavity but not vice versa. S

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Defintion of concavity: a function is concave if f( a*x + (1-a)*x' ) >= a*f(x) + (1-a)*f(x') (1) for all x, x', and for a belonging to (0,1) Definition of quasiconcavity: a function is quasi-concave if f (a*x + (1-a)*x' ) >= min( f(x) , f(x') ) (2) Now, let's first see the proof that a concave function is also quasi-concave. Given that the function is concave, then it satisfies equation (1). But then it's immediate that it's quasiconcave. To see this, without loss of generality, assume that f(x)>=f(x'). Then, a*f(x) + (1-a)*f(x') >= a*f(x') + (1-a)*f(x') = f(x') Therefore, by concavity of f, f( a*x + (1-a)*x' ) >= a*f(x) + (1-a)*f(x') >= f(x') But we assumed that f(x)>=f(x'), so min(f(x),f(x'))=f(x'), so we have that f( a*x + (1-a)*x' ) >= min(f(x),f(x')) which is definition of quasiconcavity. QED In order to see that the converse is not true, we must find a counterexample: a function that satisfies quasiconcavity but doesn't satisfy concavity. A possible counterexample if: f(x) = exp(x) (exp(x) means "e to the power of x"). It's easy to verify that this function is quasi-concave (actually, all monotonically increasing functions are quasi-concave). However, it's clear that exp(x) is a convex function. You can check this by looking at the second derivative of exp(x). Since the second derivative is also exp(x), which is greater than zero, then this function is convex. So this is an example of a quasiconcave function that is convex, providing a counterexample for the converse of the theorem above.

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