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For all $\epsilon$ we have that $f(n)\le \epsilon n$ where n is a natural number. What can we say about the growth of $f(n)$? Clearly $f(n)=O(n)$, can we say anything sharper?

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If you mean for all $\epsilon>0$, then $f(n)\le 0$ for each $n$. –  David Mitra May 6 '12 at 0:18
    
Gee, $f(n)$ is small. Stunted, even. I guess that is not nice to say. Size challenged, that sounds good. –  André Nicolas May 6 '12 at 0:22
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For reference, the difference between what you've written and saying that $f(n)\in o(n)$ is that the latter just says that for each $\epsilon$ there's some $n_0$ s.t. $f(n)\leq\epsilon n$ for all $n\gt n_0$ (small caveat: for positive $f$); your condition says that this we could take $n_0=1$ for all $\epsilon$, and that as David notes, that implies (again, assuming non-negativity of $f$) that $f$ is identically zero. –  Steven Stadnicki May 6 '12 at 0:23
    
@Steven: I think the OP is looking for $f(n) \in o(n)$, so you may want to add it as an answer. –  TMM May 6 '12 at 0:29
    
If f is a real-valued function then all we can say is that f(n) = 0 for all n, and nothing else. If f is a natural-number-values function, then simply f(n) = 0 for all n. –  Adam Rubinson May 6 '12 at 0:30
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