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I'm trying to determine a function $f(x)$ that is $O(f)$ but not $o(f)$ and also not $\Omega(f)$. Note the $f$ used in the asymptotic notation is not the same as $f(x)$.

Originally I thought of $f(x)=\log(x), O(x)$ but I am not convinced that $o(x)$ is invalid for this function.

Previously I thought it was, because I could always come up with some constant $c$ that would bring the function $x$ below $f(x)$. However, I could say the same for $o(2^x)$ because surely there is some infinitesimally small constant that I can find that will put $2^x$ below $f(x)$ at a given value $x$. Any advice in this matter?

Yuval Filmus
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drshmoo
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    Please finish improving the presentation of your question. All the math should be done with LaTeX. Furthermore, you are using only two function names and choose to call both f, whatever the font. This is apparently not needed. Plese change your notation, as it is very doubtful there is a good justification for it. – babou Mar 26 '15 at 20:28
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    Is your question: find functions $f,g$ such that $f=O(g)$, $f \neq o(g)$, $f \neq \Omega(g)$? – Yuval Filmus Mar 26 '15 at 20:57
  • If the question is what @YuvalFilmus says, we also have duplicates around. – Raphael Mar 27 '15 at 06:11

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Your example doesn't work, since in fact $\log x = o(x)$. However, the function $\phi$ below is $O(x)$ but not $o(x)$ or $\Omega(x)$: $$ \phi(x) = \begin{cases} x & \text{if $\lfloor x \rfloor$ is even}, \\ 1 & \text{if $\lfloor x \rfloor$ is odd}. \end{cases} $$

Yuval Filmus
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