Let g ( x ) : R → R + be a measurable function. Suppose { x : g ( x ) &gt; 0 } = ( a , b ). If g ( x ) − g ( x − α ) ≥ g ( y ) − g ( y − α ) , x &lt; y , α ≥ 0Can we say g ( . ) is concave? How can show this?

Question

Let   g  (  x  )  :      R    →            R        +   be a measurable function. Suppose   {  x  :  g  (  x  )  &amp;gt;  0  }  =  (  a  ,  b  ). If  g  (  x  )  −  g  (  x  −  α  )  ≥  g  (  y  )  −  g  (  y  −  α  )  ,    x  &amp;lt;  y  ,  α  ≥  0Can we say   g  (  .  ) is concave? How can show this?

getrdone07tl · Accepted Answer

Choosing   α  =  y  −  x shows that your condition implies Jensen&#039;s inequality  g      (                  x        +        y            2        )    ≤            g      (      x      )      +      g      (      y      )        2  for all   x  ,  y - this is known as Jensen convexity. Iterating this we get   t-convexity  g      (    (    1    −    t    )    x    +    t    y    )    ≤  (  1  −  t  )  g  (  x  )  +  t  g  (  y  )for all rational   t  ∈  [  0  ,  1  ]. (Maybe just convince yourself that this is true for dyadic rationals, which is a little easier.)If   g is additionally continuous, then the set  T  =  {  t  ∈  [  0  ,  1  ]  ∣  (  ∀  x  ,  y  )    g      (    (    1    −    t    )    x    +    t    y    )    ≤  (  1  −  t  )  g  (  x  )  +  t  g  (  y  )  }can be written as an intersection of preimages of closed sets by continuous functions, so it is closed and thus must be all of   [  0  ,  1  ].

Let g(x):R->R^+ be a measurable function. Suppose {x:g(x)>0}=(a,b). If (x)−g(x−α)>=g(y)−g(y−α),x<y,α>=0 Can we say g(.) is concave? How can show this?

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