convex function

**jefferson_lc** · Feb 16th 2011, 03:18 PM

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

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

In the first case, the conclusion follows immediately from Darboux's theorem. The second case implies that $f$ is convex. And I'm stuck here. One idea is to try to prove that $\exists x_0, x_1$ such that $f''(x_0)< 0$ and $f''(x_1)> 0$ and use Darboux's theorem again (assuming $f$ 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.

**tonio** · Feb 16th 2011, 06:52 PM

Originally Posted by jefferson_lc

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

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

In the first case, the conclusion follows immediately from Darboux's theorem. The second case implies that $f$ is convex. And I'm stuck here. One idea is to try to prove that $\exists x_0, x_1$ such that $f''(x_0)< 0$ and $f''(x_1)> 0$ and use Darboux's theorem again (assuming $f$ 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 $f'(x)$ . It can't be that $f'(x)>0\,\,or\,\,f'(x)<0\,\forall x\in(b,\infty)\,,\,\,b\geq a$

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

Now use the mean value theorem for derivatives...

Tonio

**Jasper** · Feb 16th 2011, 10:37 PM

By sticking to your reasoning:

(1) Case solved
(2) Assume that $\forall x \geq a, f^\prime^\prime(x) > 0$
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 $f$ will tend to $+ \infty$ on the right which is in contradiction with the initial assumption on $f$ . Hence $\exists y / f^\prime^\prime (y) = 0$
(3) Assume that $\forall x \geq a,f^\prime^\prime(x) < 0$
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 $f$ will tend to $- \infty$ on the right which is in contradiction with the initial assumption on $f$ . Hence $\exists y / f^\prime^\prime (y) = 0$

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

**jefferson_lc** · Feb 19th 2011, 02:55 PM

Originally Posted by tonio

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

(can you see why?)

I see what you're saying, but how do I prove this?

**tonio** · Feb 19th 2011, 06:35 PM

Originally Posted by jefferson_lc

I see what you're saying, but how do I prove this?

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

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

$\Longrightarrow$ 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

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