3

$f$ is a continuous, strictly increasing function and $f(x+1) = f(x)+1$

We can know that $$\forall x \in \mathbb{R},\lim_{n\rightarrow + \infty} \frac{f^{n}(x)}{n}$$

exists, and its value doesn't rely on x

Is this proposition right?(Probably right)

And how can I prove it?

I have got to know that we can't write down the exact value of limitation probably, so we may try some indirect approaches(which make this question seem thorny)

Note:$f^{n}(x) =f( f^{n-1}(x))$

Dragon_math
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    I suggest that you write $f^n$ instead of $f^{(n)}$ because the latter notation is used for the $n$-th derivative. – Carsten S Mar 30 '23 at 16:55
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    That or $f^{\circ n}$ if you prefer (since you could read $f^n(x)$ as $(f(x))^n$ also). – Bruno B Mar 30 '23 at 17:17
  • Your $f(x)$ is essentially just $x+a;+$ some small perturbation that doesn't matter, and then obviously the limit is $a$. What I just said can (and should) be made rigorous. – Ivan Neretin Mar 30 '23 at 22:42
  • Ivan,I wonder whether we can find the exact value of 'a' to make the remaining(the small perturbation) not matter? – Dragon_math Mar 31 '23 at 02:15
  • @IvanNeretin That's wrong, for example take $f(x)=(x-\ell)^2+\ell$ when $\ell \le x \le \ell + 1$ for $\ell \in \mathbb Z$. Meaning you take the swoop of a quadratic connecting (0,0) and (1,1) in an increasing fashion, and then repeat that shape between (1,1) and (2,2) and so on. And you can add an arbitrary constant to that. It's simply the set of monotone increasing maps on a compact interval, to be more specific. – Snared Mar 31 '23 at 04:00
  • So NO your function is not $x+a+\text{some small pertubation}$, it is much more general than that. – Snared Mar 31 '23 at 04:01
  • That's what I said. The perturbation is *not* "small" in the sense of being $o(1)$ as $x\to\infty$ - no, that's simply impossible, since $f(x)-x$ is a periodic function. It is "small" in a different, yet still well-defined sense. – Ivan Neretin Mar 31 '23 at 06:59
  • were you able to figure out the answer to this question? – Snared Apr 29 '23 at 08:12

1 Answers1

-1

Since $f(x+1)=f(x)+1$, we have that $f(1)-f(0)=f(0)+1-f(0)=1$. Keep in mind here that $f$ is strictly increasing.

For $s \in \{-2, +2\}$, Define $\ell_s(x)=x+f(0)+s$. Notice that $$\ell_{+1}(0) = f(0)+2 > f(0)+1=f(1)$$

So for $x \in [0,1]$, $\ell_{+1}(x) > f(x)$. For integer $z$, since $\ell(z)$ is always larger than $f(z+1)$, being the maxima of $f$ over $[z,z+1]$ , we inductively have that $f < \ell_{+1}$. The natural logic applies to $\ell_{-1}$ to give

$$\ell_{-1}<f < \ell_{+1}$$

Do you see where this is going? We can now make computations using $\ell_s$ directly: $$\ell_s^{\circ n}(x)=\ell_s(x)+(n-1)\ell_s(0)$$

Immediately giving

$$\ell_{-1}(0)=\lim_{n\to\infty} \frac{\ell^{\circ n}_{-1}(x)}n \le \lim_{n\to\infty}\frac{f^{\circ n}(x)}n \le \lim_{n\to\infty} \frac{\ell^{\circ n}_{+1}(x)}n=\ell_{+1}(0)$$

Hence

$$\lim_{n\to\infty}\frac{f^{\circ n}(x)}n \text{ exists}$$

Snared
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  • It's wrong.We can notice that $l_{-1}(0)\neq 1$ and $l_{+1}(0)\neq 1$. – Dragon_math Mar 31 '23 at 05:10
  • I never claimed either of those to be 1 exactly. Literally the definition is right in front of you lol. Read it over again and write another comment – Snared Mar 31 '23 at 05:38
  • ah yeah I see what you mean, yup go ahead and fix that. Either way limit exists – Snared Mar 31 '23 at 05:39
  • I don't think it's easy to prove that the limit exists.Besides, it's also a task to figure out the value of it.The limitation is not easily determined by the $f$.It is the iteration that actually results in its complexity. – Dragon_math Mar 31 '23 at 06:27
  • what do you mean "I don't think it's easy to prove that the limit exists." my answer shows it ignoring the =1 and just that it's constant. It obviously exists because it's bounded by the l's – Snared Mar 31 '23 at 15:20
  • i updated the answer – Snared Mar 31 '23 at 15:21
  • No,it's wrong.That it's bounded doesn't mean the limit exists.For instance,$a_n = (-1)^n$ is bounded but the limit doesn't exist. – Dragon_math Apr 01 '23 at 05:09
  • And $l_{-1}(0) \neq l_{+1}(0)$ – Dragon_math Apr 01 '23 at 05:11
  • So you have to show that if the limit were to not exist it would diverge to $\pm \infty$, this limit would not be able to DNE by iterating across values. Now that you have it bounded you know it must exist. The reason why it can't be DNE inside a bounded range is due to the monotonicity of $f$ – Snared Apr 01 '23 at 18:37
  • But $\frac{f^{\circ n}(x)}{n} $is not monotonous – Dragon_math Apr 05 '23 at 11:45
  • @Dragon_math it is monotonous in $x$, right? Maybe not $n$ but we just need for $x$ – Snared Apr 05 '23 at 22:21
  • man this is such a tricky question. Did you get an ideal answer from your prof by any chance? Would be curious to see a better solution for this. It seems like the kind of problem where you two clever distinctions of the definitions to immediately bring the result forward.. – Snared Apr 05 '23 at 22:22
  • So what if $\frac{f^{\circ n}(x)}{n} = 1+0.1(-1)^n$?We can't make sure that this would not happen from that "f" is monotonous – Dragon_math Apr 07 '23 at 00:30
  • Why the final step is right?You need to explain it with details because I don't understand.(include the well-known truth that you've used) – Dragon_math Apr 07 '23 at 00:41
  • @Dragon_math nah this answer is bad you pointed out the flaw in this argument, so there's not much more to be done here. I should probably delete this answer – Snared Apr 07 '23 at 07:25