# Thread: convex function

1. ## convex function

Let $\displaystyle f:[a,+\infty)\to\mathbb{R}$ be twice differentiable. If $\displaystyle \displaystyle\lim_{x\to\+\infty}{f(x)}=f(a)$ then there exists $\displaystyle x\in (a,+\infty)$ such that $\displaystyle f''(x)=0.$

This is what I have so far:
Suppose f is twice differentiable. Well, there are three cases:
(1) $\displaystyle \exists x_0, x_1$ such that $\displaystyle f''(x_0)< 0$ and $\displaystyle f''(x_1)> 0$; or
(2) $\displaystyle f''(x)\geq 0, \forall x$; or
(3) $\displaystyle f''(x)\leq 0, \forall x$.

In the first case, the conclusion follows immediately from Darboux's theorem. The second case implies that $\displaystyle f$ is convex. And I'm stuck here. One idea is to try to prove that $\displaystyle \exists x_0, x_1$ such that $\displaystyle f''(x_0)< 0$ and $\displaystyle f''(x_1)> 0$ and use Darboux's theorem again (assuming $\displaystyle f$ is not constant). But I don't know how to do it. I don't know if this (attempted) solution seems too complicated. Maybe there is an easier way. I'm open to suggestions.

2. Originally Posted by jefferson_lc
Let $\displaystyle f:[a,+\infty)\to\mathbb{R}$ be twice differentiable. If $\displaystyle \displaystyle\lim_{x\to\+\infty}{f(x)}=f(a)$ then there exists $\displaystyle x\in (a,+\infty)$ such that $\displaystyle f''(x)=0.$

This is what I have so far:
Suppose f is twice differentiable. Well, there are three cases:
(1) $\displaystyle \exists x_0, x_1$ such that $\displaystyle f''(x_0)< 0$ and $\displaystyle f''(x_1)> 0$; or
(2) $\displaystyle f''(x)\geq 0, \forall x$; or
(3) $\displaystyle f''(x)\leq 0, \forall x$.

In the first case, the conclusion follows immediately from Darboux's theorem. The second case implies that $\displaystyle f$ is convex. And I'm stuck here. One idea is to try to prove that $\displaystyle \exists x_0, x_1$ such that $\displaystyle f''(x_0)< 0$ and $\displaystyle f''(x_1)> 0$ and use Darboux's theorem again (assuming $\displaystyle f$ is not constant). But I don't know how to do it. I don't know if this (attempted) solution seems too complicated. Maybe there is an easier way. I'm open to suggestions.

Look at the function $\displaystyle f'(x)$ . It can't be that $\displaystyle f'(x)>0\,\,or\,\,f'(x)<0\,\forall x\in(b,\infty)\,,\,\,b\geq a$

(can you see why?), so there are points $\displaystyle x_1\neq x_2\,\,s.t.\,\,f'(x_1)=f'(x_2)$ .

Now use the mean value theorem for derivatives...

Tonio

3. By sticking to your reasoning:

(1) Case solved
(2) Assume that $\displaystyle \forall x \geq a, f^\prime^\prime(x) > 0$
use the fact that a convex function always lies above its tangent(s) and show that there is at least one tangent with a strictly positive slope. Hence $\displaystyle f$ will tend to $\displaystyle + \infty$ on the right which is in contradiction with the initial assumption on$\displaystyle f$. Hence $\displaystyle \exists y / f^\prime^\prime (y) = 0$
(3) Assume that $\displaystyle \forall x \geq a,f^\prime^\prime(x) < 0$
use the fact that a concave function always lies below its tangent(s) and show that there is at least one tangent with a strictly negative slope. Hence $\displaystyle f$ will tend to $\displaystyle - \infty$ on the right which is in contradiction with the initial assumption on $\displaystyle f$. Hence $\displaystyle \exists y / f^\prime^\prime (y) = 0$

For (2)-(3) (existence of such tangent):
Since $\displaystyle f(a) = f(+\infty)$, $\displaystyle \exists x_0 / f^\prime(x_0) = 0$. So if we take $\displaystyle x_1 > x_0$, the strict positivity (resp. negativity) of $\displaystyle f^\prime^\prime$ means $\displaystyle f^\prime$ is strictly increasing (resp. strictly decreasing). Hence $\displaystyle f^\prime(x_1) > 0$ (resp. $\displaystyle f^\prime(x_1) < 0$). We've found a tangent with a strictly positive (resp. strictly negative) slope.

4. Originally Posted by tonio
Look at the function $\displaystyle f'(x)$ . It can't be that $\displaystyle f'(x)>0\,\,or\,\,f'(x)<0\,\forall x\in(b,\infty)\,,\,\,b\geq a$

(can you see why?)
I see what you're saying, but how do I prove this?

5. Originally Posted by jefferson_lc
I see what you're saying, but how do I prove this?

Say $\displaystyle f'(x)<0$ for all $\displaystyle x\geq b>a\Longrightarrow f(x)$ is monotone descending in $\displaystyle (b,\infty)\Longrightarrow$ as

$\displaystyle f(x)\xrightarrow [x\to\infty]{}f(a)$, it must be that $\displaystyle f(x_0)>f(a)$ for some $\displaystyle x_0\in(a,b]\Longrightarrow$

$\displaystyle \Longrightarrow$ the first derivative function of f(x) must be positive in some interval to the right of a and

negative in some other interval...

Tonio