Confusion in GH Hardy's proof of $\lim_{x\to \infty} (f(x)+f'(x))=0\implies \lim_{x\to \infty} f(x)=0 \land \lim_{x\to \infty} f'(x)=0$

Question

The theorem to be proven is: $\lim_{x\to \infty} (f(x)+f'(x))=0\implies \lim_{x\to \infty} f(x)=0 \land \lim_{x\to \infty} f'(x)=0$.

Here's one Lemma that'll be used in the proof of the above theorem: If $\lim f(x)$ and $\lim f'(x)$ both exist then $\lim f'(x)=0\tag 1$ (here and henceforth in the post, $\lim$ would mean $\lim_{x\to \infty}$)

The outline of GH Hardy's proof is as follows:

Case 1: $f'$ takes positive values for all large $x$.

It means that there is some $M$ post which if $x$ lies then $f'(x)>0$. It follows that $f$ is strictly increasing on $(M,\infty)$. So there are only two possibilities for $\lim f(x)$ -either $\lim f(x)$ exists finitely or $\lim f(x)=+\infty$. Assuming the latter, it must be the case that $\lim f'(x)=-\infty$ (else the given hypothesis won't hold). But this is impossible in this case.

So $\lim f(x)$ must exist finitely, say the limit is $L$. Then to satisfy the given hypothesis $\lim f'(x)=-L$. By Lemma stated in $(1)$, this is possible only if $L=0$. So the theorem holds in this case.

Case 2: $f'$ takes negative values for all large $x$.

This case can be handled similar to the way Case 1 was handled above. The final conclusion from this case is "the theorem holds in this case too."

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Now comes the last case that confuses me: The exact wording of this case can be found here in the accepted answer. But I believe that the case is equivalent to the following case:

Case 3: $f'$ oscillates between positive and negative values when $x$ is large.

Now, here's what GH Hardy says (please treat $\phi$ as $f$ here):

"If $x$ has a large value corresponding to a maximum or a minimum of $\phi(x)$, then $\phi(x)+\phi'(x)$ is small and $\phi'(x)=0$, so that $\phi(x)$ is small. A fortiori the other values of $\phi(x)$ are small when $x$ is large." $\tag 2$

I understand the spirit of what's being said (minimum and maximum values of $f(x)$ tend to $0$ as $x\to \infty$ so $f(x)$ should also tend to $0$ as $x\to \infty$) but I don't understand why that's true. To be more specific, I have the following confusion(s):

1. In the event that some large $x$ corresponds to a maximum or minimum, then it is true that $f'(x)=0$ there but if this maximum/minimum is local then I don't see why "A fortiori" in $(2)$ is valid. The case when $(2)$ indeed works would be a very optimistic one -there are infinitely many large $x$ which correspond to global maximum/minimum so that "a fortiori all other values are small" works.

2. What happens if f doesn't have any extremum when $x$ is large?

In this case, it can be said (by Darboux's theorem) that $f'$ is $0$ at infinitely many large $x$. But such $x$ will correspond necessarily to an extremum, can not be said. This is not surprising as $h:x\mapsto x^3$ defined on $(-2,2)$ has no local extremum at $0$ even though $f'(0)=0$. So how does $(2)$ take care of this case (confusion 2)?

Please help me understand what I am missing. Thanks.

Answer 1

First, it is relevant that $\phi(x)+\phi'(x) \to 0$ in Hardy's proof. Also note that $\phi$ is continuous.

There are three cases, depending on the behaviour of $\phi'$ for large $x$.

(i) There is some $M$ such that $\phi'(x) \ge 0$ for all $x \ge M$,

(ii) There is some $M$ such that $\phi'(x) \le 0$ for all $x \ge M$ and

(iii) For all $M$ there exists $x, x^* \ge M$ such that $\phi'(x) >0$, $\phi'(x^*) <0$.

Cases (i), (ii) were dispatched above.

From Darboux's theorem we see that there in a strictly increasing sequence $x_n \to \infty$ such that $\phi'(x_n) = 0$. By assumption, we also have $\phi(x_n) \to 0$.

Let $\epsilon>0$ and find $M$ such that for $x \ge M$ we have $|\phi(x)+\phi'(x)| < \epsilon$. Choose the smallest $N$ such that $x_N \ge M$ and suppose $x \ge x_N$. Note that $|\phi(x_n)| < \epsilon $ for all $n \ge N$. In particular, note that $|\phi(x_N)| < \epsilon$.

I claim that $|\phi(x)| < \epsilon$ for all $x \ge x_N$. Suppose $|\phi(x)| \ge \epsilon$ for some $x$ and choose $N' > N$ such that $x< x_{N'}$. In particular, $|\phi|$ has a $\max$ at some $x^* \in (x_N, x_{N'})$ and so $\phi'(x^*) = 0$ and so $|\phi(x^*)| < \epsilon$, a contradiction.

Hence $\phi(x) \to 0$.

Answer 2

Here is a solution which has a "sequence" flavor instead of an "$\varepsilon$" flavor:

By hypothesis there exists an increasing sequence of points $x_n \geq 0$ for which $f'(x_n) = 0$ and such that $x_n \to \infty$. Because $f(x)$ is locally extremal at each such point, and by hypothesis $f'(x)$ is zero infinitely often for $x \geq 0$, then for all $x \geq x_n$ we have the bound (for each fixed $n$) $$ \lvert f(x) \rvert \leq \sup_{k \geq n}\,\lvert f(x_k) \rvert. $$ But $f(x) + f'(x) \to 0$ as $x \to \infty$, so in particular if we take the sequence $(x_n)$ which converges to $\infty$ as $n \to \infty$ we conclude that $$ f(x_n) = f(x_n) + f'(x_n) \to 0 $$ as $n \to \infty$. This means that $\sup_{k \geq n}\ \lvert f(x_k) \rvert \to 0$ as $n \to \infty$ as well, and therefore $$ \lim_{x \to \infty} \lvert f(x) \rvert \leq \lim_{n \to \infty} \sup_{k \geq n}\, \lvert f(x_k) \rvert = 0 $$ which completes the proof.