Prove it:Suppose f : R → R is a concave function and θ &gt; 1. Then θ k f ( x ) ≥ f ( θ k x )for all k = 1 , 2 , …

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Prove it:Suppose   f  :      R    →      R   is a concave function and   θ  &amp;gt;  1. Then      θ    k    f  (  x  )  ≥  f  (      θ    k    x  )for all   k  =  1  ,  2  ,  …

giosgi5 · Accepted Answer

What about   f  (  x  )  =  a  x  −  b for   a  ,  b  &amp;gt;  0? Then       θ    k    f  (  x  )  =      θ    k    a  x  −  b      θ    k    &amp;lt;      θ    k    a  x  −  b  =  f  (      θ    k    x  )Similarly   g  (  x  )  =  −      x    2    −  b for   b  &amp;gt;  0.Then inequality       θ    k    g  (  x  )  =  −      θ    k        x    2    −      θ    k    b  ≤  −      θ          2      k            x    2    −  b  =  g  (      θ    k    x  ) is equivalent to      x    2        θ    k    (      θ    k    −  1  )  ≤      θ    k    b  −  b , which doesn&#039;t hold for every   x.The problem in both cases was the value at   0. Indeed, let   f be concave. Then by concativity, for any   λ  ∈  (  0  ,  1  ),   x  ,  y  ∈      R   we have inequality:  f  (  λ  x  +  (  1  −  λ  )  y  )  ≥  λ  f  (  x  )  +  (  1  −  λ  )  f  (  y  )We want to prove   a  f  (  x  )  ≥  f  (  a  x  ) for   a  &amp;gt;  1  ,  x  ∈      R  . Taking   λ  =      1    a    ∈  (  0  ,  1  ),   x  =  a  t for some   t  ∈      R   and   y  =  0 we get:  a  f  (  t  )  ≥  f  (  a  t  )  +  (  a  −  1  )  f  (  0  )As we see, with   f  (  0  )  ≥  0 it holds, when   f  (  0  )  &amp;lt;  0 we have potential problems as showed in counterexamples above. I don&#039;t want to say that it does not hold for   f concave such that   f  (  0  )  &amp;lt;  0, because I do not have a proof (but I&#039;ll be glad to see one), but taking arbitrary concave   f and looking at       f    M    =  f  −  M, your inequality is equivalent to  a      f    M    (  t  )  ≥      f    M    (  a  t  )so that  a  f  (  t  )  −  a  M  ≥  f  (  t  )  −  Mand finally:      a    M    f  (  t  )  −  a  ≥      1    M    f  (  t  )  −  1Taking arbitrary   t  ∈      R    ,  a  &amp;gt;  1, we can take   M big enough so that it does not hold (since   a  &amp;gt;  1 and moreover, terms with       1    M   tend to   0). In other words, we showed that if we substract enough from arbitrary concave function, then your inequality does not hold.

Prove it: Suppose f:R->R is a concave function and theta>1. Then theta^kf(x)>=f(theta^kx) for all k=1,2,….

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