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Let $f(x)$ a continuous function on $\Bbb{R}$.
Prove/disprove: If $\lim\limits_{n\to\infty} f(n)=\infty$, then $\lim\limits_{n\to\infty}f(f(n))=\infty,$ where the limits are taken over $n \in \mathbb N$.

I have an hunch this statement isn't always true, but couldn't find a proper function to show this.

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    $\begingroup$ It should be intuitively obvious at least. $\endgroup$ Aug 12, 2014 at 20:07
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    $\begingroup$ Where is the $n$ in your expression $f(x)$? $\endgroup$
    – DanZimm
    Aug 12, 2014 at 20:12
  • $\begingroup$ Edited again. Sorry for the confusion $\endgroup$ Aug 12, 2014 at 20:13
  • $\begingroup$ Expanding on what mathematician said, you know that if n goes to infinify, then f(n) goes to infinity. So in the second expression, the argument of the outer f is going to infinity. And we know when the argument of f goes to infinity, then f goes to infinity. $\endgroup$
    – NicNic8
    Aug 12, 2014 at 20:20

4 Answers 4

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Consider the function $$ f(x):=x\cos(2\pi x)+\frac{1}{4}. $$ Then $f$ is continuous in $\mathbb R$ such that $f(n)=n+\frac{1}{4}$ for $n\in\mathbb N$, which tends to $\infty$ as $n\to\infty$. But $f(f(n))=(n+\frac{1}{4})\cos(2\pi n+\frac{\pi}{2})+\frac{1}{4}=\frac{1}{4}$ for $n\in\mathbb N$.

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    $\begingroup$ Nice! How did you think about it? $\endgroup$ Aug 12, 2014 at 20:25
  • $\begingroup$ Pick a $c<1$, make a function with $f(n)=n+c$ and $f(n+c)=c$ (or $f(n+c)$ is some other constant, it doesn't matter). $\endgroup$
    – Ian
    Aug 13, 2014 at 4:05
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Let $M$ be given. There exists $N$ with the property that $x > N$ implies $f(x) > M$. There exists $K$ with the property that $x > K$ implies $f(x) > N$. Thus $$x > K \implies f(x) > N \implies f(f(x)) > M.$$

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  • $\begingroup$ Hey, first of all thank you for replying. I edited my question. $\endgroup$ Aug 12, 2014 at 20:11
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    $\begingroup$ With the altered question, this is no longer a correct answer. The new version only requires that the limit be carried out on integer $x$ rather than real $x$. $\endgroup$ Aug 12, 2014 at 20:35
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Counterexample:

$ f(x)= \begin{cases} \dfrac{1}{2}+x-2x(x-\lfloor{x}\rfloor) & \text{$0\leq{x}-\lfloor{x}\rfloor<\dfrac{1}{2}$} \\ \dfrac{1}{2}+2x(x-\lfloor{x}\rfloor-\dfrac{1}{2}) & \text{$\dfrac{1}{2}\leq{x}-\lfloor{x}\rfloor<1$} \end{cases} $


For $n\in\mathbb{N}$:

  • $f(n)=n+\dfrac{1}{2}\implies\lim\limits_{n\to\infty}f(n)=\infty$

  • $f(f(n))=\dfrac{1}{2}\implies\lim\limits_{n\to\infty}f(f(n))\neq\infty$


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When you say $f(n) \to \infty$ as $n \to \infty$, the usage of symbol $n$ usually means that $n$ is a positive integer (you have mentioned this explicitly also). Thus when the argument of $f$ tends to $\infty$ through positive integers then $f$ tends to $\infty$. In case of $f(f(n)) = f(t)$ the argument of $f$ is $t = f(n)$ and this itself may not be an integer. So even though the argument $t \to \infty$ it is not guaranteed to be an integer and hence $f(t)$ may not tend to $\infty$. If for all $n \in \mathbb{N}$ we ensure that $f(n) \in \mathbb{N}$ then $f(f(n)) \to \infty$ as $n \to \infty$.

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