2

Let $f: [a,\infty) \rightarrow \mathbb R$ be a twice differentiable function. If $\lim\limits_{x\to \infty}f(x)=f(a)$ then $\exists x \in (a,\infty)$ such $f''(x)=0$

So i was think to use the same idea here
But there is a easy and direct way to proof ? using for example Taylor's theorem with Lagrange remains. I have to prove $h*\lim\limits_{x\to \infty}f'(x)=0$ when i tried to use lagrange remains.

Thanks for any hint.

Community
  • 1
Ricardo Gomes
  • 439

2 Answers2

2

Suppose $f''(x) \neq 0$ for all $x$. Then $f'$ must be injective (otherwise the mean value theorem shows that $f''(\xi)=0$ at some point). Since $f'$ is continuous, it must be increasing or decreasing. Since we can deal with $f$ or $-f$, we might as well assume that $f'$ is increasing. In particular, this means that $f''(x) >0$ for all $x$, and so $f$ is convex.

Since $f$ is convex, the set $f^{-1}((-\infty,f(a)])$ is convex and contains $[a,\infty)$, so $f(x) \le f(a)$ for all $x$. Since $f$ is not constant (otherwise $f''(x) = 0$ everywhere), $f$ has a global minimum at some point $\hat{x} \in (a,\infty)$, and we have $f(y)>f(\hat{x})$ for some $y > \hat{x}$. Now let $z >y$, then $y = (1-t)\hat{x}+t z$ for some $t \in (0,1)$ and $f(y) \le (1-t)f(\hat{x})+t f(z)$, since $f$ is convex. Solving for $t$ and rearranging gives $f(z)= \left( { z-\hat{x} \over y-\hat{x} } \right) (f(y)-f(\hat{x})) +f(\hat{x})$, and so $\lim_{z \to \infty} f(z) = \infty$, a contradiction.

Hence we must have $f''(x) =0$ at some point.

copper.hat
  • 178,207
  • 1
    Or, using your proof that $f'$ is injective, you could note that $f'(x)<0$ for all $x$ since if, for any $x$, it held that $f'(x)\geq 0$, then it would follow that for any $y>x$ and all $z>y$ that $f'(z)>f'(y)>0$ implying that $\lim_{x\rightarrow\infty}f(x)$ diverges. However, this would mean that $f$ is decreasing, hence $f(a)>f(x)$ for any $x>a$. – Milo Brandt Jan 10 '15 at 03:14
  • @Meelo: Indeed, and you could finish noting that we must have $\int_a^\infty f't) dt = 0$, and so we must have $f'(t) = 0$ everywhere. However, I have a particular liking for proofs involving convexity (and compactness, for that matter), hence I took a longer route :-). – copper.hat Jan 10 '15 at 03:30
  • @copper.hat Thank you very much for this answer. i will wait if anyone will answer with a direct proof without using $f''(x) \neq 0$. It's not that your answer is wrong.. One more thing.. if is not bother you, can you send me a link or explain the why $f^{-1}((-\infty,f(a)])$ is convex and contains $[a,∞)$ – Ricardo Gomes Jan 10 '15 at 21:40
  • (The following is in the particular context of the proof above.) Since I take $f''(x)>0$, the function $f$ is convex, hence the level sets are convex. Since $f(x) \le f(a)$, we have $f(x) \in (-\infty, f(a)]$, hence $f^{-1}((-\infty, f(a)])$ must contain the whole domain (that is, $[a, \infty)$). – copper.hat Jan 10 '15 at 21:44
  • Note that the reason that $L=f^{-1}((-\infty,f(a)])$ contains $[a, \infty)$ needs some elaboration. It is clear that $a \in L$. Suppose there is some point $t$ such that $f(t) > f(a)$. Then this would contradict $f$ being convex (since $f(x) \to f(a)$), and so $f(x) \le f(a)$ for all $x \ge a$. My reasoning above is backwards (or I have forgotten my reasoning!). – copper.hat Jan 22 '15 at 01:24
0

Let's argue in the following manner. If $f''(x)$ does not vanish then it must maintain a constant sign (because of intermediate value property satisfied by any derivative). Without loss of generality we can assume $f''(x) > 0$ for all $x \geq a$ (otherwise we can apply argument to $-f(x)$).

This leads to the conclusion that $f'(x)$ is strictly increasing in $[a, \infty)$. If $f'(x) = 0$ at some point $c$ in this interval $[a, \infty)$ then $f'(x) > 0$ for all $x > c$. This would mean that $f(x)$ is strictly increasing in $[c, \infty)$. Let $d > c$ and $\epsilon = f'(d) > 0$. Since $\lim_{x \to \infty}f(x) = f(a)$, it follows that there is number $M > d$ such that $|f(M) - f(a)| < \epsilon / 2$ and $|f(M + 1) - f(a)| < \epsilon / 2$. Combining these two we get $|f(M + 1) - f(M)| < \epsilon$. By Mean Value Theorem we have $|f'(\xi)| < \epsilon$ for some $d < M < \xi < M + 1$. Since $f'$ is positive here we have $f'(\xi) < \epsilon = f'(d)$. Since $\xi > d$ this is contrary to the fact that $f'(x)$ is increasing.

It follows that $f'(x)$ does not vanish in $[a, \infty)$. Note that the same logic in the previous paragraph also proves that $f'(x) $ can't be positive in $[a, \infty)$ (the contradiction in previous paragraph was based on $f'(d) > 0$). Hence it follows that $f'(x) < 0$ for all $x \in [a, \infty)$. Thus $f(x)$ is decreasing and for $b > a$ we have $f(b) < f(a)$. And for $x > b$ we have $f(x) < f(b) < f(a)$. Letting $x \to \infty$ we see that $\lim_{x \to \infty}f(x) \leq f(b) < f(a)$. This contradiction shows that $f''(x)$ must vanish somewhere.

Paramanand Singh
  • 92,128