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Thread: convex function

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by jefferson_lc View Post
    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
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  3. #3
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    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.
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  4. #4
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    Quote Originally Posted by tonio View Post
    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?
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  5. #5
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    Quote Originally Posted by jefferson_lc View Post
    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
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