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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: $$\lim_n \sup \frac{C_n}{2\sqrt{n}} \leq 1$$

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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

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