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Could somebody help me to answer the following question?

Let $f:R_+ \rightarrow R_+$ be a nonindentically zero, nondecreasing, continuous, concave function with $f(0)=0$. Do we have that for any $s,t \in [0,1]$,

$$f(x)f(stx)\leq f(sx)f(tx), \quad \forall x \geq 0.$$

or equivalently, do we have that for any $t\in (0,1)$, $\frac{f(x)}{f(tx)}$ is nonincreasing on $x >0$.

Thanks!

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5 Answers 5

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The answer is no.

Mikael de la Salle has given a counterexample. In a more general setting, the necessary and sufficient condition for the inequality to hold is $g = \log \circ f \circ \exp$ is concave. The inequality can be rewritten into $g(y) + g(a + b + u) \le g(a + y) + g(b + y)$, where $y = \log x$, $a = \log s \le 0$ and $b = \log t \le 0$. This is precisely the necessary and sufficient condition for concavity of $\log \circ f \circ \exp$, which is certainly not the same as concavity of $f$.

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The answer is no.

For a counter-example, consider the piecewise affine function defined by $f(x)=x$ for $0 \leq x \leq 1$ and $f(x)=(x+1)/2$ for $x>1$. Take $x=4$, $s=t=1/2$. Then $f(x) f(stx)= 2.5$ and $f(sx)f(tx)=2.25$.

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I just got the proof by myself.

Without loss of generality, we assume $f$ is continuously differentiable. Otherwise, one can use the smoothing technique.

Notice that $$\left(\frac{f(x)}{f(tx)}\right)' = \frac{f'(x)f(tx)-tf(x)f'(tx)}{f^2(tx)}.$$ We only need to show $$\frac{f'(x)}{f(x)} \leq \frac{tf'(tx)}{f(tx)},$$ which is just $$[(\log\circ f)(x)]'\leq [(\log \circ f)(tx)]'.$$ Thus, we only need to show the composition $\log\circ f$ is concave. This is true since $f$ is concave and positive for $x>0$. Thus we complete the proof.

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  • $\begingroup$ I just found that this proof is wrong since $[(\log\circ f)(tx)]'$ is different from $(\log\circ f)'(tx)$. So, anyone can help? Thanks! $\endgroup$
    – user11870
    Commented May 31, 2011 at 12:43
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That $\log \circ f$ is concave follows from concavity of $\log$ and $f$ because $f$ is non-decreasing. I do not see how you could put positivity to use here.

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  • $\begingroup$ When you divide the inequality by f(x) or f(tx), it comes in handy. $\endgroup$
    – Michael Renardy
    Commented May 29, 2011 at 20:42
  • $\begingroup$ I meant that positivity of $f$ does not help in determining whether or not $log \circ f$ is concave. My answer should've been a reply to wmmiao's answer of course but I lack the permission to reply. $\endgroup$
    – anonymous
    Commented May 30, 2011 at 9:47
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    $\begingroup$ The positivity of $f$ is just for the domain of the $\log$ function. $\endgroup$
    – user11870
    Commented May 30, 2011 at 14:26
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We want $$f(x)f(stx)\le f(sx)f(tx)$$ for concave non-decreasing functions with $f(0) = 0$. Since we require this for every $x$ and $f$ we can assume $x = 1$, because our claim is invariant w.r.t scaling in the argument. That means $$f(1)f(st)\le f(s)f(t)$$ Now we're looking at a function $f \colon [0,1] \to \mathbb R^+$ that is continuous, thus bounded; without loss of generality again $f(1) = 1$ because our claim is invariant w.r.t scaling again (let's leave the case $f(1) = 0$ aside for a second). So we're looking at $$f(st)\le f(s)f(t)$$ for a concave non-decreasing function $f \colon [0,1] \to [0,1]$ with $f(1) = 1$.

Maybe this gets us any further?

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  • $\begingroup$ The case $f(1) = 0$ is actually irrelevant because then we'd have $f = 0$. $\endgroup$
    – anonymous
    Commented Jun 1, 2011 at 10:36

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