Results 1 to 5 of 5

Math Help - convex function

  1. #1
    Newbie
    Joined
    Jan 2009
    Posts
    17

    convex function

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Quote Originally Posted by jefferson_lc View Post
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2011
    Posts
    1
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jan 2009
    Posts
    17
    Quote Originally Posted by tonio View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Quote Originally Posted by jefferson_lc View Post
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove A Function Is Convex?
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: May 1st 2011, 01:48 PM
  2. Convex function
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: January 23rd 2011, 06:27 PM
  3. Convex function
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: September 21st 2010, 11:07 AM
  4. Concave / Convex function
    Posted in the Calculus Forum
    Replies: 5
    Last Post: June 17th 2009, 12:11 PM
  5. convex function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 30th 2008, 06:21 PM

Search Tags


/mathhelpforum @mathhelpforum