Let f : R → R be a concave function. Let x 0 ∈ R and α &gt; 1. Show that for all x ∈ R , f ( x 0 + x ) + f ( x 0 − x ) ≥ f ( x 0 + α x ) + f ( x 0 − α x ) .

Question

Let   f  :      R    →      R   be a concave function. Let       x    0    ∈      R   and   α  &amp;gt;  1. Show that for all   x  ∈      R  ,  f  (      x    0    +  x  )  +  f  (      x    0    −  x  )  ≥  f  (      x    0    +  α  x  )  +  f  (      x    0    −  α  x  )  .

Jaydon Roth · Accepted Answer

By concavity, you have                    f        (        β                  x          ′                +        (        1        −        β        )                  y          ′                )                            ≥        β        f        (                  x          ′                )        +        (        1        −        β        )        f        (                  y          ′                )                            f        (        (        1        −        β        )                  x          ′                +        β                  y          ′                )                            ≥        (        1        −        β        )        f        (                  x          ′                )        +        β        f        (                  y          ′                )            for any   β  ∈  [  0  ,  1  ]. Now use       x    ′    =      x          0        −  α  x,       y    ′    =      x          0        +  α  x and   β  =            1      −      α              −      2      α        ∈  (  0  ,  1  ) in the sum of the two inequalities to get what you need.

Let f:R -> R be a concave function. Let x_0 in R and α>1. Show that for all x∈R, f(x_0+x)+f(x_0−x)>= f(x_0+αx)+f(x_0−αx).

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