# Proof about twice differentiable function.

I've been restudying Real Analysis (in one dimension) and got stuck on one problem. I thought that my proof was correct, but, when I checked the solution, the proof that I saw was way more complicated than mine, so I'm assuming mine was incorrect. With that being said, I would like to know why my solution fails to be correct (or if that's actually the case). The problem is the following:

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

Here is my proof: If $$f$$ is constant, we must have $$f(x)=f(a)$$. Hence, $$f''$$ vanishes everywhere, and we are done.

Let's then suppose that $$f$$ is not constant. If $$f'(a)$$ was both the global maximum and minimum of $$f'$$, then, for every $$x \in [a, + \infty)$$, we would have $$f'(a) \leq f'(x) \leq f'(a).$$ That is, $$f'$$ is constant, and $$f(x)=mx+b$$ for some $$m, b \in \mathbb{R}$$ with $$m \neq 0$$ (since we are assuming $$f$$ not to be constant). Therefore, $$\lim_{x \rightarrow \infty} f(x)=+ \infty$$ or $$\lim_{x \rightarrow \infty} f(x)=- \infty$$, a contradiction. Thus, $$f'(a)$$ cannot be both the global maximum and minimum of $$f'$$. In other words, $$f'$$ attains an extremum at some point $$x \in (a, + \infty)$$; but since $$f'$$ is differentiable and $$x$$ is an interior point of $$(a, + \infty)$$, we conclude that $$f''(x)=0,$$ as desired.

• If $f$ is not constant, then it does not necessarily have any global extrema, unless its domain if a closed interval. – Yiorgos S. Smyrlis Nov 28 '20 at 14:00
• Why do we assume that if $f$ is not constant that $f'(a)$ is both a global maximum and minimum? Surely this needs to be justified (or at least consider the case when this is not true)? Anyway what about the function $f:[\pi , \infty) \to \mathbb{R} , f(x):= \frac{sin(x)}{x}$? Hope this helps :) – THIG Nov 28 '20 at 14:01
• But doesn't the fact that $\lim_{x \rightarrow \infty} f(x)=f(a)$ implies that $f(x)$ has either a maximum or minimum? – Will199 Nov 28 '20 at 14:12

Lemma. (Theorem of Darboux) If $$f: [a,b]\to\mathbb R$$ is differentiable, and $$f'(a)<0, then there exists a $$c\in (a,b)$$, such that $$f'(c)$$.

If $$f''(x)\ne 0$$, for all $$x\in(a,\infty)$$, then according to the above result, either $$f''(x)>0$$, for all $$x\in(a,\infty)$$ or $$f''(x)<0$$, for all $$x\in(a,\infty)$$.

Assume that $$f''(x)>0$$, for all $$x\in(a,\infty)$$. (The case, $$f''(x)<0$$, for all $$x\in(a,\infty)$$ is treated similarly.) Then $$f'$$ is strictly increasing, and hence there are three possibilities:

I. $$f'(x)>0$$, for all $$x\in(a,\infty)$$.

II. $$f'(x)<0$$, for all $$x\in(a,\infty)$$.

III. There is a $$b>a$$, such that $$f'(x)<0$$, for all $$x\in(a,b)$$ and $$f'(x)>0$$, for all $$x\in(b,\infty)$$.

In the cases I and III, $$f$$ tends necessarily to infinity, as $$x\to\infty$$. Observe that in I, $$f'(x)>f'(a+1)>0$$, for $$x>a+1$$, and hence $$f(x)>(x-a-1)f'(a+1)+f(a+1)\to\infty$$. Similarly, in III, $$f'(x)>f'(b+1)>0$$, for $$x>b+1$$, and hence $$f(x)>(x-b-1)f'(b+1)+f(b+1)\to\infty$$.

In case II, $$f$$ is strictly decreasing, and hence $$\lim_{x\to\infty}f(x)\le f(a+1) So, all there cases lead to contradiction, and hence there exists a $$x\in(a,\infty)$$, where $$f''(x)=0$$.

• Thank you for the proof, Yiorgos, but do you mind verifying the veracity of my proof? – Will199 Nov 28 '20 at 17:53