Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

Assume that F is a strictly increasing and continuous function , differentiable if needed, over [0,a] such that F(0)=0 and F(a)=1, a>0. Prove or disprove :

$\int_0^a (F(x)-4F^2(x)+3F^3(x))\, dx\leqslant 0$

I proved this for concave functions and also it is true for many non concave functions, like $x^k, k>1$ but I don't have a general proof nor a counter example.

Any other sufficient conditions for F for this to be true would be appreciated.

share|cite|improve this question
up vote 1 down vote accepted

To construct a counterexample, think about the map $y \mapsto y(1-4y+3y^2)$. This possesses roots at $y = 0,1/3,1$, and is positive on $(0,1/3)$ and negative on $(1/3,1)$. The problem can arise when $F$ spends time in $(0,1/3)$, contributing positive mass, or in $(1-\delta,1)$, contributing little negative mass. With this in mind, how about any function $F$ defined in such a way that $F(x) = 1/6 + x/1000$ for $x \in (\delta, 1-\delta)$, for $\delta > 0$ small (taking $a = 1$ for simplicity).

I checked this recipe for $\delta = 1/100$ and got that $$\int_{\delta}^{1-\delta} F(x) (1 - 4 F(x) + 3 F(x)^2) dx = 0.0680139$$, which means that no matter what values $F$ takes on the remainder of $[0,1]$, you'll still have an overall positive integral.

EDIT (Addendum): A sufficient condition for the inequality to hold: Write $G(y) = y(1-4y+3y^2)$. Then $G$ has a maximum value $z^* > 0$ at some point in $(0,1/3)$. The condition will be of the form $F|_{(\delta, a-\delta)} \in (1/3 + \epsilon, 1 - \epsilon)$, where $\epsilon, \delta$ will be determined in terms of $z^*$. We have, $$\int_0^a G \circ F(x) dx \leq z^* \delta + (a - 2 \delta) \max\{G(1/3 + \epsilon), G(1 - \epsilon)\} $$ You can now specify $\epsilon, \delta$ so that the right hand side is dominated by zero.

share|cite|improve this answer
To clarify, I'm saying that any $F$ satisfying the parameters of the problem which takes on the above form on the interval $(1/100, 99/100)$ will break the inequality. You can fill in $[0,1/100]$ and $[99/100]$ however you want to. – A Blumenthal Mar 5 '13 at 19:50
Good idea but it is not clear to me why (1/100,99/100) has to dominate (99/100,1). Derivative may be too high at 1. Will check your example by filling in (99\100,1). – Emre Per Mar 5 '13 at 20:11
The global minimum of $y(1-4y+3y^2)$ is about -.3 and so the contribution from the endpoints can be no worse than $-.3 * 2/100 = -.006$. – A Blumenthal Mar 5 '13 at 20:14
Moreover, this problem doesn't have anything to do with the derivative of $F$. Which derivative are you talking about? – A Blumenthal Mar 5 '13 at 20:15
Now, I see. If the calculation is correct, this shpuld be okay. Thank you for your work. Any idea on possible sufficient condition for F for the inequality to be true? – Emre Per Mar 5 '13 at 20:24

This doesn't have to be true. If $1>c-4c^2+3c^3>0$ for $0<c<1$, for example $c=1/4$, the integral of the constant function $F(x)=c-4c^2+3c^3$ from $0$ to $1$ is greater than $0$. Now just approximate $f(x)=c$ by a differentiable function that satisfies your criteria. One example (hope I didn't make a mistake !) $$ f(x)= \begin{cases} c\sin(\pi/2\cdot x/\epsilon),\quad 0\leq x <\epsilon\\ c,\quad \epsilon\leq x\leq 1-\epsilon\\ 1-(1-c)\cos(\pi/2\cdot (x-(1-\epsilon))/\epsilon),\quad 1-\epsilon<x\leq 1 \end{cases} $$ You can choose $\epsilon$ as small as you want, so the integral is as close to $c-4c^2+3c^3$ as you want.

share|cite|improve this answer

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.