Show that functions in O(1) can't grow faster than their composition with themselves

Question

Let $f(n)$ be a function s.t $f(n)\geq 1 $ for every $n$.

I want to disprove that if $f(n) = \omega (f(f(n)))$ then it means that $f(n) = O(1)$.

I thougt of 2 approaches to show that this statement is incorrect:

1. Give an Example of a function that $f(n) = \omega (f(f(n)))$ for it,

for example: $f(n) = \sqrt n$

and then I'm supposed to get a contradiction, and show that $\sqrt n \neq O(1)$ but that's the step I'm stuck at, but I'm not sure how that contradiction is derived. I mean formally, because I understand it by intuition.

2. Start with the function that suffice $f(n) = O(1)$ and then show that it's impossible to get the condition under that function (It's sort of going from the end to the beginning)

what's the right approach and how do I get it right?

Answer 1

To complete the first approach and show that $\sqrt{n}\notin O(1)$, you just need to show that the statement "There exist constants $c$ and $n_0$ such that, for all $n\geq n_0$, $\sqrt{n}\leq c\cdot1$" is false. That is, you need to show that, for any $c$ and $n_0$, there is an $n\geq n_0$ such that $\sqrt{n}>c$. For example, you can take $n=n_0+c^2$ (or $\max\,\{n_0,c^2\}$, if you prefer).

I don't really understand your second approach. If $f(n)=O(1)$, it's perfectly possible that $f(n)\in \omega(f(f(n)))$: take any constant function, for example.

Answer 2

Regarding the second approach, suppose that $f(n) = O(1)$. Then for some $C \geq 1$, $f(n) \leq C$ for all $C$. On the other hand, since $f(n) \geq 1$, we have $f(f(n)) \geq 1$. Therefore $f(n) \leq Cf(f(n))$, implying that $f(n) = O(f(f(n)))$. In particular, it is not the case that $f(n) = \omega(f(f(n)))$.

Note, however, that this doesn't disprove the claim you were interested in. Let $P(f)$ be the property that $f(n) = \omega(f(f(n))$, and let $Q(f)$ be the property that $f(n) = O(1)$. Also, let $F$ be the set of functions $f$ such that $f(n) \geq 1$ for all $n$. The claim you want to disprove is $$ \forall f \in F, P(f) \Rightarrow Q(f). $$ The negation of this claim is $$ \exists f \in F, P(f) \text{ and not } Q(f). $$ Your second approach instead shows that $$ \forall f \in F, Q(f) \Rightarrow \lnot P(f). $$ This does not disprove the original claim. For example, it might be the case that $P(f)$ is always false, and then the original claim would actually be correct.