Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$f:\mathbf{R} \rightarrow \mathbf{R}$ 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.

share|improve this question
1  
The intuitive picture: 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, so it will pass below $y=l$ too, therefore $f'$ vanishes on $\mathbb R$. –  plm May 30 '12 at 11:33

3 Answers 3

up vote 1 down vote accepted

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 $f(x)=1-x^2$.

Graph of 1-x^2, with 2 tangents.

share|improve this answer

You know that $f\,'$ is a non-increasing function.

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

  2. 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 $x\in\Bbb R$.

share|improve this answer
    
Thanks - that makes much more sense now! –  Amy May 30 '12 at 11:52

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 $f'(y)\neq 0$?

share|improve this answer
    
Could you please tell me what really happen if $f'(y)\neq 0$? –  Une Femme Douce Jun 7 '12 at 13:20

Your Answer

 
discard

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.