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I have just encountered the following question: Let $C_n$ be a sequence of real numbers with the following three properties: 1) $C_n$ is subadditive, such that $$C_{m+n} \leq C_m +C_n$$

2) $C_n=O(\sqrt{n})$

3) $$C_{q+1} \leq 2 \sqrt{q}$$ for every prime power $q=p^m$

But I have no idea what's the use of the first condition in order to prove that: $$\lim_n \sup \frac{C_n}{2\sqrt{n}} \leq 1$$

Rob
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  • Perhaps you mean $\le1$? Because $C_n=0$ identically seems to be a counterexample. – anon Sep 11 '11 at 21:34
  • @anon: Would $C_n=0$ satisfy $C_n=O(n)$? Normally, a constant function is $O(1)$, not $O(n)$. – Arturo Magidin Sep 11 '11 at 21:37
  • @Arturo: $|0|\le1\cdot|n|$ for all $n\ge0$, so I believe it's $O(n)$. In analytic number theory there's lots of discussion of improving exponents in big-$O$ asymptotes, so just because $O(n)$ can be improved to $O(1)$ here doesn't mean $O(n)$ isn't true as well. There might be some convention that distinguishes an exponent of $0$ I'm not aware of, but that would seem to contradict the definition of $O$. What say you? – anon Sep 11 '11 at 22:08
  • @anon: I'm honestly asking, not correcting. I don't usually work with $O$ notation. Looking at the definition given in Apostol's "Intro to Analytic Number Theory", though, I see that if $f$ is $O(1)$, then it would also be $O(n)$. – Arturo Magidin Sep 11 '11 at 22:16
  • @Thijs: Maybe you have this question in mind? – t.b. Sep 11 '11 at 22:33
  • @Thijs: Ah, right. Not that it matters much, but it is interesting to note that the question was deleted by the OP himself -- and it was a list of four homework problems on which the OP worked hard according to a comment in the now deleted question and the question is now sold as "just encountered"... Anyway, the post I linked to should lead to a proof in a relatively straightforward way. Also, the $\limsup$ used to be $\leq 1$ instead of $= 1$ (the $\leq$ makes more sense). – t.b. Sep 11 '11 at 23:45

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