Let f(x) be an increasing, strictly concave function with f(0)=0. I have to show that given x<y, f(y+epsilon)−f(x+epsilon)<f(y)−f(x), where epsilon is a small, positive number.

inurbandojoa 2022-11-20 Answered
Let f ( x ) be an increasing, strictly concave function with f ( 0 ) = 0. Show that given x < y, f ( y + ε ) f ( x + ε ) < f ( y ) f ( x ), where ε is a small, positive number.
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Answers (2)

partatjar6t9
Answered 2022-11-21 Author has 8 answers
Your guess for a homotopy inverse is correct. A homotopy between the identity and ( x , t ) ( d ( s ( x ) ) , t ) is given by the expression
( ( x , t ) , u ) { ( x , t + u )  if  t + u 1 ( d ( s ( x ) ) , t + u 1 )  if  t + u > 1
where u [ 0 , 1 ] is the parameter of the homotopy. Geometrically this shifts a point in the mapping torus of d∘s towards the right a distance of u units.
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evitagimm9h
Answered 2022-11-22 Author has 5 answers
Let α = ϵ y x , which is between 0 and 1 for small positive ϵ
f ( y ) = f ( ( 1 α ) ( y + ϵ ) + α ( x + ϵ ) ) > ( 1 α ) f ( y + ϵ ) + α f ( x + ϵ ) f ( x + ϵ ) = f ( ( 1 α ) ( x ) + α y ) > ( 1 α ) f ( x ) + α f ( y ) .
Sum up these 2 inequalities, combine terms, and divide by 1 α, we get
f ( y ) f ( x ) > f ( y + ϵ ) f ( x + ϵ )
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