Let f:R_(>=0)->R_(>=0), and g:R_(>=0)->R_(>=0) be two functions. Suppose that f(0)>g(0)=0, and f is strictly increasing and concave, while g is strictly increasing and strictly convex (so f′(x)>0,f′′(x)<=0,g′(x)>0, and g′′(x)>0). Suppose that for some x>0 we have that f(x)−g(x)>0. Is it true that then for every y in [0,x] it must be that f(y)−g(y)>=0?

Mark Rosales

Mark Rosales

Answered question

2022-11-07

Let f : R 0 R 0 , and g : R 0 R 0 be two functions. Suppose that f ( 0 ) > g ( 0 ) = 0, and f is strictly increasing and concave, while g is strictly increasing and strictly convex (so f ( x ) > 0 , f ( x ) 0 , g ( x ) > 0, and g ( x ) > 0 )). Suppose that for some x > 0 we have that f ( x ) g ( x ) > 0. Is it true that then for every y [ 0 , x ] it must be that f ( y ) g ( y ) 0?

Answer & Explanation

gortepap6yb

gortepap6yb

Beginner2022-11-08Added 19 answers

h ( x ) = f ( x ) g ( x ) is strictly concave. If h ( 0 ) 0 and h ( x ) 0 for some x > 0 then
h ( y ) > x y x h ( 0 ) + y x h ( x ) 0
for y ( 0 , x ).
The monotony of f and g is not needed for this conclusion.

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