10
$\begingroup$

Let $f:\mathbb{R} \to \mathbb{R}$ be a fuction such that $\displaystyle \lim_{x\to \infty} (f(x+1)-f(x))=1$. Is it true then that $\displaystyle \lim_{x\to \infty} \frac{f(x)}{x}=1$?

I think it is and here is how I went about it. Let $\varepsilon>0$. Then there is some $\delta_\varepsilon>0$ such that $$1-\varepsilon < f(x)-f(x-1)<1+\varepsilon, \forall x>\delta_\varepsilon.$$ Thus, we may write the following inequalities for an $x> \delta_\epsilon$:
$$1-\varepsilon < f(x)-f(x-1)<1+\varepsilon\\ 1-\varepsilon < f(x-1)-f(x-2)<1+\varepsilon\\ \vdots\\ 1-\varepsilon < f(\delta_\varepsilon+1)-f(\delta_\varepsilon)<1+\varepsilon$$
and after we sum these up we get that $$(x-\delta_\epsilon)(1-\epsilon)<f(x)-f(\delta_\varepsilon)<(x-\delta_\varepsilon)(1+\varepsilon), \forall x>\delta_\varepsilon.$$
This implies that $$\frac{f(\delta_\varepsilon)+(x-\delta_\varepsilon)(1-\varepsilon)}{x}<\frac{f(x)}{x}<\frac{f(\delta_\varepsilon)+(x-\delta_\varepsilon)(1+\varepsilon)}{x}, \forall x>\delta_\varepsilon.$$
If we take $\displaystyle\limsup_{x\to\infty}$, we get that $1-\varepsilon < \displaystyle\limsup_{x\to\infty} \frac{f(x)}{x}< 1+\varepsilon$, $\forall \varepsilon > 0$, so $\displaystyle\limsup_{x\to\infty} \frac{f(x)}{x}=1$. In the same way we get that $\displaystyle\liminf_{x\to\infty} \frac{f(x)}{x}=1$, so $\displaystyle\lim_{x\to\infty} \frac{f(x)}{x}=1$ as desired.

Is this proof correct? I am a bit unsure that I am taking that $\limsup$ correctly, even though I can't see why it could be wrong.

$\endgroup$
23
  • 6
    $\begingroup$ There should be another condition also that $f$ is bounded on every bounded interval. $\endgroup$
    – Paramanand Singh
    Commented Sep 20, 2021 at 10:39
  • 2
    $\begingroup$ The problem is that your argument works only when $x=\delta_{\epsilon} +n$ where $n$ is a positive integer. For it to work for all $x>\delta_{\epsilon} $ you will need $f$ to be bounded in $[\delta_{\epsilon}, \delta_{\epsilon} +1]$. I hope you understand the flaw. $\endgroup$
    – Paramanand Singh
    Commented Sep 20, 2021 at 11:10
  • 2
    $\begingroup$ @ParamanandSingh Ah, thanks, it does make sense, I felt that the way I tried to mimick what I would do for a sequence may be flawed. Now I understand, thank you very much! $\endgroup$
    – TheZone
    Commented Sep 20, 2021 at 11:14
  • 1
    $\begingroup$ @ArcticChar: things happen so fast here and I am limited in my speed due to mobile device. :) $\endgroup$
    – Paramanand Singh
    Commented Sep 20, 2021 at 11:25
  • 1
    $\begingroup$ @ParamanandSingh: 1) For me as a reader of SE, it would be quite confusing: Following a [duplicate] link that deviates in non-minor details. The answer to that other question cannot be applied to this one. 2) It's still preferable to have a proper answer (instead of having to dig into every comment quere). Afterall, the two proofs (one confirming, one counter-example) are quite different. 3) It's nice that the TO found help and can proceed; but IIUC, SE is also about sharing information with "passive" readers. Otherwise, all except the questions could be private. $\endgroup$ Commented Sep 20, 2021 at 11:32

3 Answers 3

15
$\begingroup$

Take the function $$f(x)=\begin{cases}\frac{1}{[x]} + \lfloor x\rfloor;& x\notin \mathbb Z\\ \lfloor x\rfloor ; &x\in\mathbb Z\end{cases}$$

where $[x]$ denotes the fractional part of $x$, i.e. $[x]=x-\lfloor x\rfloor $.

Then the function satisfies your condition, since $f(x+1)-f(x)=1$ is true for all $x$. However, the limit $$\lim_{x\to\infty}\frac{f(x)}{x}$$ does not exist, because the value of $f$ on every interval $[n, n+1]$ is unbounded.

$\endgroup$
3
  • $\begingroup$ thanks, so I guess that my proof fails here in this case: $\displaystyle \limsup_{x\to \infty}\frac{f(\delta_\epsilon)+(x-\delta_\epsilon)(1-\epsilon)}{x}$ (I don't really see any other place where it might). I thought that as $x\to \infty$ $f(\delta_\epsilon)$ would stay constant. $\endgroup$
    – TheZone
    Commented Sep 20, 2021 at 11:01
  • $\begingroup$ +1 for the fine counterexample. $\endgroup$
    – Paramanand Singh
    Commented Sep 20, 2021 at 11:29
  • $\begingroup$ @TheZone I think that's where your error is, yeah. $\endgroup$
    – 5xum
    Commented Sep 20, 2021 at 11:34
4
$\begingroup$

In your proof you are implicitly assuming that $x$ is such that $x-\delta_\epsilon=n$ is an integer $n$, so you have some constraint on the specific value of $x$ as $x\rightarrow \infty$. In the example given by 5xum this means that the term $\frac{1}{[x]}=\frac{1}{[\delta_\epsilon]}$ which you assume to be fixed. If you want to allow general $x$ you would have to modify your inequality to $$\frac{f(x-\lfloor x-\delta_\epsilon\rfloor)+\lfloor x-\delta_\epsilon\rfloor(1-\epsilon)}{x}<\frac{f(x)}{x}<\frac{f(x-\lfloor x-\delta_\epsilon\rfloor)+\lfloor x-\delta_\epsilon\rfloor(1+\epsilon)}{x}, \forall x>\delta_\epsilon.$$ Now you have to assume that $f(x-\lfloor x-\delta_\epsilon \rfloor)$ is bounded $\forall x>\delta_\epsilon$ in which case your proof would be correct. This however is not the case in 5xum's example.

$\endgroup$
2
  • $\begingroup$ thanks, now it is all really clear. $\endgroup$
    – TheZone
    Commented Sep 20, 2021 at 12:14
  • $\begingroup$ :P Wasn't sure it was already clear as you just assumed "I don't really see any other place where it might". $\endgroup$
    – Diger
    Commented Sep 20, 2021 at 12:23
2
$\begingroup$

Let $ f(x)= -\sin \left( \lfloor x\rfloor\cdot \dfrac{\pi}{2} \right) \cdot \cos\left( \lfloor x\rfloor\cdot \dfrac{\pi}{2} \right). $ Then \begin{align} f(x+1)-f(x) =& \sin \left( \lfloor x+1\rfloor\cdot \dfrac{\pi}{2} \right) \cdot \cos\left( \lfloor x+1\rfloor\cdot \dfrac{\pi}{2} \right) - \sin \left( \lfloor x\rfloor\cdot \dfrac{\pi}{2} \right) \cdot \cos\left( \lfloor x\rfloor\cdot \dfrac{\pi}{2} \right) \\ =&+\sin\left((\lfloor x+1\rfloor - \lfloor x\rfloor)\cdot \dfrac{\pi}{2}\right) \\ =& +\sin\left( 1\cdot \frac{\pi}{2} \right) \\ =& 1 \end{align} and by $-1\leq f(x)=- \sin \left( \lfloor x\rfloor\cdot \dfrac{\pi}{2} \right) \cdot \cos\left( \lfloor x\rfloor\cdot \dfrac{\pi}{2} \right)\leq +1$ we have $$ \lim_{x\to 0}\frac{f(x)}{x}=0. $$

$\endgroup$
1
  • $\begingroup$ Your third equality should be $-\sin(\pi \lfloor x \rfloor)=0$ not $1$. Btw: $\sin(x)\cos(x)=\sin(2x)/2$ so actually $f(x)=0$ for all $x$. $\endgroup$
    – Diger
    Commented Sep 20, 2021 at 15:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .