MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $x >0$.


share|cite|improve this question
up vote 3 down vote accepted

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 $f$.

share|cite|improve this answer

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 $f(sx)f(tx)=2.25$.

share|cite|improve this answer

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.

share|cite|improve this answer
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! – user11870 May 31 '11 at 12:43

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.

share|cite|improve this answer
When you divide the inequality by f(x) or f(tx), it comes in handy. – Michael Renardy May 29 '11 at 20:42
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. – anonymous May 30 '11 at 9:47
The positivity of $f$ is just for the domain of the $\log$ function. – user11870 May 30 '11 at 14:26

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?

share|cite|improve this answer
The case $f(1) = 0$ is actually irrelevant because then we'd have $f = 0$. – anonymous Jun 1 '11 at 10:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.