Is there always a Big Oh complexity strictly between any two others?

I'm learning about asymptotic analysis, and have seen some exotic looking complexities living between other common ones. For instance "log log n" is strictly between 1 and log n. It makes me wonder if one can always find complexities between any other two.

Specifically, for any functions f and g with O(f) ⊂ O(g) does there always exist an h such that O(f) ⊂ O(h) ⊂ O(g)?

This isn't homework or anything. I'm just curious if anyone knows.

Yes: take a function in the middle, for some suitable definition of middle. You have a wide choice.

If $O(f) \subset O(g)$ (where the inclusion is strict), then $g \in O(g) \setminus O(f)$ (because if $g \in O(f)$ and $f \in O(g)$ then $\Theta(f) = \Theta(g)$). Take the geometric mean: let $h = \sqrt{f \cdot g}$ (since we're talking about complexity here, I assume the functions are positive).

Then $f \in O(h)$ and $h \in O(g)$ (if this is not immediately obvious, prove it using the definition of $O$), i.e. $O(f) \subseteq O(h) \subseteq O(g)$. If $O(f) = O(h)$ then $g = f \in O(f)$, which is not the case since we assumed $g \notin O(f)$. It remains to prove that $O(h) \ne O(g)$, and we'll have $O(f) \subset O(h) \subset (g)$.

If $O(h) = O(g)$ then $g \in O(h)$, i.e. there exists $A$ and $C \gt 0$ such that $\forall x \ge A, g(x) \le C \, h(x) = C \sqrt{f(x) g(x)}$. Then $g(x) \le C^2 f(x)$ (take the square and divide by $g(x)$; again, I assume positive functions), thus $g \in O(f)$, which goes against our initial assumption. The hypothesis $O(h) = O(g)$ leads to a contradiction, which concludes the proof.

StackExchange Computer Science Q#18993, answer score: 10