# Inequality with integral

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.

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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.

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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.

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