Does there exist such an upper bound for $g(n)$?

Question

If a function $g(n)$ satisfies $g(n)<n^5$ where $n$ is a positive real, does that mean there exists a real number $m,m<5$ such that $g(n)<n^{m}$

This isn’t from any particular question I was just wondering if this was possible, because my logic was that $g(n)$ never touches the line $n^5$, therefore, there should be a line $n^{m}$ between lines $g(n)$ and $n^5$

Answer 1

Consider a function like $g(n)=n^5-10^{-10}$. There is no $m<5$ such that $g(n)<n^m$ for all $n$.

To see that, suppose that we had such an $m$. We can suppose $m>0$, since the claim is clear for $m≤0$. Then we'd have $$10^{-10}>n^5-n^m=n^m\left(n^{5-m}-1\right)$$

for large $n$. But $n^m\to \infty$ for large $n$ and $n^{5-m}-1>1$ since $5>m$, hence we have a contradiction.

Answer 2

No. For any $m<5$, with $g(n):=n^{(m+5)/2}$ you have

$$n^m<g(n)<n^5$$ so for any $m$ you will always find a $g$ that contradicts.

There is plenty of room between $n^m$ and $n^5$.

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By the way, assume that $m=4.9$ was a solution, and $g(n)<n^5\implies g(n)<n^{4.9}$. But then, as $5$ has nothing special, why not $g(n)<n^{4.9}\implies g(n)<n^{4.8}$, also causing $g(n)<n^5\implies g(n)<n^{4.8}$ ? And what would stop the exponent from decreasing ?