Suppose $F(x)∈[0,1]$ and $x∈[0,X]$. Suppose $F(0)=0$ and $F(X)=1$. Suppose that $F(x)$ is continuous and strictly concave for $x∈[F(x^∗)x^∗,X]$. Suppose that $\frac{x^* F(x^*)}{X} \leq F(x^* F(x^*))$.
Show that:
$F(x^∗)^2<F(x^*F(x^∗))$
By using concavity I have shown that if $F(x)$ is strictly concave for $x\in[0,X]$, then the inequality is satisfied. This can be done applying concavity with $x_1=x^*$, $x_2=0$ and $\lambda=F(x^*)$. Given that result, I believe that the above one can be proven as well, but not sure how to do it. We cannot longer use the "trick" of setting $x_2=0$. Any nice alternative?
Please, answer if you can solve it or if you can give me any insight on how to attack the problem. It will be of enormous help!
It seems bad for me. Look at the picture below:
Now I'm going to explain what is illustrated here. First, points $O$ and $P$ on a coordinate plane stand for $(0, 0)$ and $(X, F(X))$, respectively. Then, $B = (x^*, 1)$, and $A = \left(x^*, F\left(x^*\right)\right)$. We assume that $A$ is located above $OP$ as on the pic.
Now $C$ is defined as the intersection of $OB$ and a horizontal line through $A$. One can see that $C$ is $\left(x^*F\left(x^*\right), F\left(x^*\right)\right)$. The condition that
$$\frac{x^*F\left(x^*\right)}{X}\leqslant F\left(x^*F\left(x^*\right)\right)$$
means that point $\left(x^*F\left(x^*\right), F\left(x^*F\left(x^*\right)\right)\right)$ is located somewhere above $OP$.
Let this point lie below $OA$. Then we can connect points $\left(x^*F\left(x^*\right), F\left(x^*F\left(x^*\right)\right)\right)$, $A$ and $P$ with a concave curve and continue it onto $\left[0, x^F\left(x^\right)\right) properly.
One can see that the intersection of $OA$ and the vertical line through $C$ has $y$-coordinate equal to
$$F\left(x^*\right)\cdot\frac{x^*F\left(x^*\right)}{x^*} = F\left(x^*\right)^2,$$
and we've just constructed a function where it exceeds $F\left(x^*F\left(x^*\right)\right)$.
Maybe I misunderstood the statement?