Thread Rating:
• 0 Votes - 0 Average
• 1
• 2
• 3
• 4
• 5
 If $\alpha = f(0-)=f(+\infty),\;f\,''$exists on $\mathbb{R}$, then $0\in f\,''((0,\infty)).$
04-05-2017, 12:22 AM (This post was last modified: 04-05-2017 11:02 AM by elim.)
Post: #1
 elim Moderator Posts: 580 Joined: Feb 2010 Reputation: 0
If $\alpha = f(0-)=f(+\infty),\;f\,''$exists on $\mathbb{R}$, then $0\in f\,''((0,\infty)).$
Let $f:R\rightarrow R$ a twice differentiable function, with
$\lim_{x\rightarrow 0}f(x)= \lim_{x\rightarrow +\infty }f(x)=\alpha \in\mathbb{R}$,
Prove that $0\in f\,''((0,\infty)).$ src
04-05-2017, 12:39 AM (This post was last modified: 04-05-2017 11:00 AM by elim.)
Post: #2
 musicbug Posts: 1 Joined: Apr 2017 Reputation: 0
A possibly related problem
Ask a (possibly) related question: If $f'(a)f'(b)< 0$ for some $[a,b]\subset\mathbb{R}$
are we sure that $0\in f'((a,b))$?
04-05-2017, 10:57 AM
Post: #3
 elim Moderator Posts: 580 Joined: Feb 2010 Reputation: 0
If $\alpha = f(0-)=f(+\infty),\;f\,''$ exists on $\mathbb{R}$, then $0\in f\,''((0,\infty))$
If $f$ is a constant, we are done. Assume, WLOG, that
$\alpha < f(\beta) = \max f((0,\infty)),\; f'(\beta) = 0.$ By the
mean-value theorem, there is some $\gamma > \beta$ such that
$u = f'(\gamma)< 0$ thus $[u,0]\subset f'([\beta,\gamma])$ (continuity of $f'$)
If $f'(x)\le v = u/2\;(x \ge \gamma)$, then $f\to -\infty<\alpha$
as $x\to\infty$ and so there must be some $\eta > \gamma$ such that
$f'(\eta) > v$ hence $[u,v]\subset f'([\gamma,\eta])$. Since $v\in [u,0]$,
$f'(a)=v=f'(b)$ for some $a\in (\beta,\gamma),\; b\in(\gamma,\eta).$
So $f''(c )={\small\dfrac{f'(b)-f'(a)}{b-a}} = 0$ for some $c\in (a,b).\;\square$
04-05-2017, 12:30 PM
Post: #4
 elim Moderator Posts: 580 Joined: Feb 2010 Reputation: 0
RE: A possibly related problem
(04-05-2017 12:39 AM)musicbug Wrote:  Ask a (possibly) related question: If $f'(a)f'(b)< 0$ for some $[a,b]\subset\mathbb{R}$
are we sure that $0\in f'((a,b))$?
Certainly. See here
 « Next Oldest | Next Newest »

Forum Jump: