Function bounded below, with negative 2nd derivative. Show function is constant. $f:\mathbf{R} \rightarrow \mathbf{R}$ is twice differentiable. $f''(x) \leq 0$ $\forall x \in \mathbf{R}$. $f$ is also bounded below. Show $f$ is a constant function.
I've got to 
$$f(x+y)-f(x) \leq f(x)-f(x-y)\, \forall x \in \mathbf{R}, y>0$$ using Rolle's Theorem
and I think I'm making an argument about $f'(x)$ being a decreasing function, but I can't see how to get to $f$ constant.
 A: You know that $f\,'$ is a non-increasing function.


*

*Suppose that $f\,'(x_0)>0$ for some $x_0\in\Bbb R$; show that $\lim\limits_{x\to-\infty}f(x)=-\infty$.

*Suppose that $f\,'(x_0)<0$ for some $x_0\in\Bbb R$; show that $\lim\limits_{x\to\infty}f(x)=-\infty$.
Conclude that $f\,'(x)=0$ for all $x\in\Bbb R$.
A: We have for $x<y$ that 
$$f(x)-f(y)=\int_y^xf'(t)dt\leq \int_y^xf'(y)dt=f'(y)(x-y)$$
hence $\inf_t f(t)\leq f(x)\leq f(y)+f'(y)(x-y)$. What happens if $f'(y)\neq 0$?
A: I'll include a picture and post my comment as an answer. :)
The intuition:
Assume $f(x)>l$, $f'$ is nonincreasing so if it is not $0$ at some point $x_0$, take the tangent to the graph of $f$ there. Since this line has nonzero slope if will cut the lower bound line $y=l$ either on the left or on the right of $x_0$, and the graph of $f$ lies below that tangent (to make this rigorous use the fundamental theorem of calculus), so it will pass below $y=l$ too, therefore $f'$ vanishes on $\mathbb R$.
You can see this situation with the horizontal line $y=-1$ below, for $f(x)=1-x^2$.

