If we have some function $f(x)$ that is real, strictly monotone increasing, and is $o(x)$, so $\lim_{x\rightarrow\infty}\frac{f(x)}{x}=0$, does this mean that we can find some polynomial $x^d$ with $0<d<1$ bounding $f$ from above?
I've tried thinking about this in terms of an asymptotic expansion of $f$ using different asymptotic scales but not made much progress.
If not, are there any further restrictions that would guarantee this?
The question, as I understand it, is whether there is any sublinear function that grows faster than sublinear polynomials. The answer is yes.
Take $f(x)=\frac{x}{\log x}$. Then $f(x)\in o(x)$ because $\frac{1}{\log x}\to 0$ (as $x\to \infty$), but $f(x)$ grows faster than any $x^d,0<d<1$ because:
$$ x^d=\frac{x}{x^{1-d}}$$
and powers grow faster than logarithms i.e. $x^{1-d}\in \Omega(\log x)$ so $\frac{x}{x^{1-d}}\in o\left(\frac{x}{\log x}\right)$.