# A Proof for Generalized Rolle's Theorem

Can anyone please critique my proof?

# Problem

Generalized Rolle's Theorem Let $$f(x)$$ be differentiable over $$(-\infty,+\infty)$$, and $$\lim\limits_{x \to -\infty}f(x)=\lim\limits_{x \to +\infty}f(x)=l$$. Prove there exists $$\xi \in (-\infty,+\infty)$$ such that $$f'(\xi)=0.$$

# Proof

Consider proving by contradiction. If the conclusion is not true, then $$\forall x \in \mathbb{R}:f'(x)\neq 0$$. Thus, by Darboux's Theorem, $$f'(x)$$ can not change its sign, in another word, $$f'(x)$$ is either always positive or always negative.

Without loss of generality, we assume $$f'(x)>0$$, then $$f(x)$$ is strictly increasing. It's obvious that $$\inf_{x \in \mathbb{R}} f(x)=\lim_{x \to -\infty}f(x),~~~\sup_{x \in \mathbb{R}} f(x)=\lim_{x \to +\infty}f(x).$$ Therefore $$l=\inf_{x \in \mathbb{R}} f(x) which is absurd. The proof is completed.

• The proof is OK. – Will M. Dec 18 '18 at 6:29
• You could use Rolle's theorem directly. Define $h(x)=f\left(\frac{1}{a-x}+\frac{1}{b-x}\right)$ and also assert that $h(a)=h(b)=l$. Then all the conditions of Rolle's theorem are met, so there exists $x^*$ such that $h'(x^*)=0$. Using the chain rule this implies that $f'\left(x^{**}\right)=0$, where $x^{**}=\frac{1}{a-x^*}+\frac{1}{b-x^*}$ completing the proof. – Alex S Dec 18 '18 at 7:03
• Its fine but the usual proof for Rolle is simpler. – Paramanand Singh Dec 20 '18 at 8:08

The proof is solid. However, I do have some comments (as I frequently do):

1. You say, "It is obvious that...". While I do agree that the next conclusion is fairly obvious, at the level of an introductory real analysis course, there are some steps that many students would not be able to fill in. I understand omitting steps for brevity, but I would recommend being clear in your own head about how you would go about expanding the proof here, and if this proof is being submitted for an assessment, how you would go about convincing the marker that you do know what you're talking about.

2. The line $$l=\inf_{x \in \mathbb{R}} f(x) could be improved to say $$l=\inf_{x \in \mathbb{R}} f(x)\le f(-1) < f(0)< f(1) \le \sup_{x \in \mathbb{R}} f(x)=l.$$ In order to make the original line make sense, you would have to follow it with a "for all $$x \in \mathbb{R}$$", which would make the $$\sup$$ and $$\inf$$ poorly-formed. You could consider replacing the $$x$$ with a different pronumeral $$y$$, or (as I did) a specific number like $$0$$. I also included specific numbers to either side of $$0$$ to help neatly demonstrate why the inequality is strict.

Thanks for @Theo Bendit's critique. I'm willing to give a reply, but it seems to be a bit long, so please allow me to initiate a new post.

## Part 1

For the first point, I admit I omitted some details in the former post indeed. However, it's easy to add a proof for complement.

For any fixed $$x_0 \in \mathbb{R}$$, according to the monotonity of $$f(x)$$, we have $$\forall t>0 :f(x_0)

Let $$t \to +\infty$$ and take the limits of both sides. We have $$f(x_0)\leq \lim_{t \to +\infty}f(x_0+t)=l,$$ which shows that $$\forall x_0 \in \mathbb{R}:f(x_0)\leq l$$. Therefore $$l$$ is an upper bound of $$f(x).$$ Moreover, since $$\lim\limits_{x \to +\infty}f(x)=l$$, then $$\forall \varepsilon>0,\exists X>0, \forall x>X:|f(x)-l|<\varepsilon.$$ Thus, $$\exists x' \in (X,+\infty):f(x')>l-\varepsilon.$$ Now, we can claim that, $$l$$ is surely the supremum of $$f(x)$$. Likewise, we may prove $$l$$ is also the infimum of $$f(x)$$.

## Part 2

As for the second point, I can agree that $$\inf_{x \in \mathbb{R}} f(x)\leq f(x) \leq \sup_{x \in \mathbb{R}} f(x),\tag1$$ if we merely notice the property of the bound. But, taking the monotonicity into account, we can improve the conclusion, namely, cancel the equal mark.

Actually, the equality within $$(1)$$ can never hold at all. That's because, if $$\exists x''\in \mathbb{R}:f(x'')=\sup\limits_{x \in \mathbb{R}}f(x),$$ then $$\forall \xi>0: f(x''+\xi)>f(x'')=\sup\limits_{x \in \mathbb{R}}f(x)$$, which contradicts.