Let f : R + → R + and g : R + → R + be two strictly concave, strictly increasing, twice differentiable functions, such that f ( x ) = O ( g ( x ) ) as x → ∞, i.e. there exists M &gt; 0 and x 0 such that f ( x ) ≤ M g ( x ) ∀ x ≥ x 0 .Is it true that f ′ ( x ) = O ( g ′ ( x ) ) as x → ∞?

Question

Let   f  :            R        +    →            R        +   and   g  :            R        +    →            R        +   be two strictly concave, strictly increasing, twice differentiable functions, such that   f  (  x  )  =  O  (  g  (  x  )  ) as   x  →  ∞, i.e. there exists   M  &amp;gt;  0 and       x    0   such that  f  (  x  )  ≤  M  g  (  x  )    ∀  x  ≥      x    0    .Is it true that       f    ′    (  x  )  =  O  (      g    ′    (  x  )  ) as   x  →  ∞?

Gilbert Petty · Accepted Answer

No. We first construct a function   h  :  [  −  1  ,  ∞  )  →      R   as follows: For integers   n  ∈  {  −  1  ,  0  ,  1  ,  2  ,  .  .  .  } we want   h to be:  h  (  n  )  =      1          (      n      +      2      )      !          ∀  n  ∈  {  −  1  ,  0  ,  1  ,  2  ,  .  .  .  }These values are positive and strictly decreasing. So, we can interpolate them to get the full function   h  (  x  ) that is differentiable, positive, and satisfies       h    ′    (  x  )  &amp;lt;  0 for all   x  ≥  −  1.Define   g  :  [  −  1  ,  ∞  )  →      R   by   g  (  x  )  =      ∫          −      1              x        h  (  t  )  d  t. Then       g    ′    (  x  )  =  h  (  x  )  &amp;gt;  0 and       g    ″    (  x  )  =      h    ′    (  x  )  &amp;lt;  0 for all   x  ≥  −  1. So   g is increasing, twice differentiable, and strictly concave.Now define   f  :  [  0  ,  ∞  )  →      R   by   f  :  [  0  ,  ∞  )  →      R  . Then  f  (  x  )  =  g  (  x  −  1  )  ≤  g  (  x  )    ∀  x  ∈  [  0  ,  ∞  )and so the conjecture       f    ′    (  x  )  =  O  (      g    ′    (  x  )  ) fails.

Let f:R^+->R^+ and g:R+->R+ be two strictly concave, strictly increasing, twice differentiable functions, such that f(x)=O(g(x)) as x->oo, i.e. there exists M>0 and x_0 such that f(x)<=Mg(x) forallx>=x0. Is it true that f′(x)=O(g′(x)) as x->oo?

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