I've struggled with the problem below and yet been able to solve it.
Let $f$ be an increasing, concave function of $x\geq x_1$, where $x,x_1\in\mathbb{R}$. Assume that $f(x)$ is constant for $x\geq x_2$, where $x_2\in\mathbb{R}$ and $x_1 < x_2$. Show that $f$ is strictly increasing for $x\in[x_1,x_2]$.
Could anyone help me out?
$f(x)=0$ for all $x$ gives a counter-example.
However if you assume that $f$ is not a constant on $(x_2-r,\infty)$ for any $r>0$ then the result is true. One way of proving this is to use the fact that the right hand derivative of $f$ is in decreasing and it is $0$ for $x \geq x_2$. If $f$ is not strictly increasing in $[x_1,x_2]$ then (it is constant in some subinterval and ) there exists $y <x_2$ with $f'(y+)=0$ and this makes $f'(t+)=0$ for all $t >y$. But then $f'(t)=0$ for all $t >y$ making $f$ constant on $(y, \infty)$.
