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Given a concave function $f(x)$ defined for $x \ge 0$, I am trying to understand whether $g(x) = f(x)- x f'(x)$ should be positive or not.

From what I am reading it seems that it should be positive, but I cannot understand why. Any help?

user3026610
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    Concavity is not affected by adding constants to $f$. Positivity of $f - xf'$ is. So no, the claimed inequality is false. –  Aug 12 '16 at 05:29
  • In the case of concave functions, is there any specific relation between $f$ and $xf'$ that I cannot see? – user3026610 Aug 12 '16 at 05:37

3 Answers3

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COUNTEREXAMPLE

$$f(x)=-x^2-1 \Rightarrow g(x)=x^2-1$$

$g(x)$ is not always positive nor is it always negative.

So your inequality is false.

S.C.B.
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Assume that $f$ has the second derivative and it is concave for $x\geq 0$.

If $f(0)>0$ then the property holds. Infact $g(0)=f(0)>0$ and for $x>0$, $g'(x)=f'(x)-f'(x)-xf''(x)=-xf''(x)\geq 0$. Hence $g$ is increasing and always positive for $x\geq 0$.

Otherwise you have counterexamples: for any concave function $h(x)$ take $f(x)=h(x)-h(0)-1$ then $f$ is still concave and $g(0)=-1<0$.

Robert Z
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  • Is the reverse true? I.e does $f(x)>xf’(x)$ imply that $f$ is concave (I think if we take the derivative of both sides and simplify we find that $f’’(x)<0 $ which implies $f$ is strictly concave?) – user106860 Feb 18 '20 at 22:08
  • @user106860 $f>g$ does not imply that $f'>g'$ – Robert Z Feb 18 '20 at 22:21
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The first order Taylor approximation of $f(a)$ is $f(a)+f'(a)(x-a)$. Now, this approximation clearly bounds it above, since $f(a)-af'(a)+f'(a)x$ is the tangent to $f(a)$, a concave function. Taking this together, we arrive in:

$$f(a) \le f(a)+f'(a)(x-a)$$ $$f(0) \le f(a)-af'(a) $$

Now, we can't assume that $f(0)$ would be $0$ or greater, so your inequality is not always true.

Eemil Wallin
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