Independent Set (IS) is the $\mathsf{NP}$-complete decision problem
Input: graph $G$ with $v=|V(G)|$, integer $k$
Question: is there an independent set $S \subseteq V(G)$ with at least $k$ vertices?
Similarly, we can define $f(v)$-bounded IS as the promise version of IS, where the promise is that $k \le f(v)$. Note that the function parameter depends only on the number of vertices in the input graph.
It is straightforward to show that when $f(v) = v/c$ for some constant $c$, then $f(v)$-bounded IS remains $\mathsf{NP}$-hard. This is because IS can be reduced to $f(v)$-bounded IS by blowing up the number of vertices by at most a factor of $c$. Similarly, when $f(v)=v^{1/c}$ for some constant $c$, then $f(v)$-bounded IS is also still $\mathsf{NP}$-hard, by blowing up the number of vertices by a power of $c$.
On the other hand, when $k$ is a constant, then $k$-bounded IS can be decided in polynomial time by exhaustively checking each of the $\binom{v}{k} = \Theta(v^k)$ different $k$-subsets, in at most $O(v^{k+2})$ steps.
This suggests that there is a spectrum of hardness defined via functions restricting the largest independent set that is sought in any instance. This is, in fact, the case: Bodirsky and Grohe 2008 (preprint) and Chen, Thurley, Weyer 2008 (preprint) established such results non-constructively via versions of Ladner's theorem, conditional on $\mathsf{P} \ne \mathsf{NP}$.
Keeping in mind that asymptotic behaviour does not induce a total order on functions,
what is a "smallest" function $f(v)$ such that $f(v)$-bounded IS is $\mathsf{NP}$-hard?
In particular, is there some function $f(v)=o(v^{1/c})$ for every constant $c>0$ for which bounded IS is hard? (I'd be happy with an argument that is conditional on a common hypothesis such as ETH or $W[1] \ne \mathsf{FPT}$.)
Note that while $f(v) = v^{1/\lg v} = 2$ yields a polynomial upper bound on time, the somewhat faster growing function $f(v)=v^{1/\lg\lg v}$ yields a bound for exhaustive search that is superpolynomial but subexponential. It would therefore also be interesting to establish whether there is a function $f(v) = \omega(1)$ such that $f(v)$-bounded IS is still in $\mathsf{P}$.