Let f ( x ) be an increasing, strictly concave function with f ( 0 ) = 0. Show that given x &lt; y, f ( y + ε ) − f ( x + ε ) &lt; f ( y ) − f ( x ), where ε is a small, positive number.

Question

Let   f  (  x  ) be an increasing, strictly concave function with   f  (  0  )  =  0. Show that given   x  &amp;lt;  y,   f  (  y  +  ε  )  −  f  (  x  +  ε  )  &amp;lt;  f  (  y  )  −  f  (  x  ), where   ε is a small, positive number.

partatjar6t9 · Accepted Answer

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          &amp;gt;          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.

evitagimm9h · Accepted Answer

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          )          )                            &amp;gt;        (        1        −        α        )        f        (        y        +        ϵ        )        +        α        f        (        x        +        ϵ        )                            f        (        x        +        ϵ        )                    =                    f                  (          (          1          −          α          )          (          x          )          +          α          y          )                            &amp;gt;        (        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  )  &amp;gt;  f  (  y  +  ϵ  )  −  f  (  x  +  ϵ  )

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.

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