Limits of variants of Independent Set?
Question 1: For a function $f(n)$ lets say that $f$ is sub-polynomial if $\forall \epsilon &gt; 0, f(n) = O(n^\epsilon)$. The Exponential Time Hypothesis (ETH) states that 3-SAT needs $2^{\epsilon n}$ time for some $\epsilon &gt; 0$. Let's postulate a much weaker version, ``Weak Exponential Time Hypothesis'' (WETH): There exists an $\epsilon &gt; 0$ such that $3$-SAT requires $2^{\epsilon n^\epsilon}$ time. 

Claim: If $f$-bounded Independent Set is NP-hard for any sub-polynomial function $f$ then the WETH fails.

Proof: Supose $f$-bounded Independent Set is NP-hard under Turing-reductions for some any sub-polynomial function $f$. Then there exists a constant $c$ and an $O(n^c)$ time algorithm for $3$-SAT with $n$-variables that makes at most $O(n^c)$ oracle calls to $f$-bounded Independent Set with at most $O(n^c)$ vertices. 

From this algorithm we make an algorithm for $3$-SAT where the oracle calls to $f$-bounded Independent Set are replaced by the $O(|V(G)|^{k+2})$ time brute force algorithm. The running time of this algorithm is upper bounded by
$O((n^c) \cdot (n^c)^{f(n^c)+2}) = 2^{O(\log n \cdot f(n^c))}$. For any $\epsilon &gt; 0$, pick $\delta = \epsilon / 2c$. Since $f$ is sub-polynomial we have that $f(n^c) = O((n^c)^\delta) = O(n^{\epsilon / 2})$. Therefore the running time of the algorithm is upper bounded by $2^{O(\log n \cdot n^{\epsilon / 2})} = 2^{O(n^{\epsilon})}$, contradicting the WETH.



Question 2: Let's say that $f$ computably tends to infinity if there exists a computable function $g : \mathbb{N} \rightarrow \mathbb{N}$ such that for every $c$ and every $n \geq g(c)$, $f(n) \geq c$.

Claim: If $f$-bounded Independent Set is in P for any function $f$ that computably tends to infinity, then FPT = W[1].

Proof: Assuming the existence of a polynomial time algorithm for $f$-bounded Independent Set we will give an FPT algorithm for Independent Set, proving FPT=W[1]. 

Given as input $G$ and $k$, set $n = |V(G)|$ and compute $g(k)$. Note that computing $g(k)$ takes time that is only a function of $k$. If $g(k) \leq n$ run the polynomial time algorithm for $f$-bounded Independent Set, this will output the correct answer because $k \leq f(n)$. If $g(k) &gt; n$ then run the brute force $n^{k+2}$ algorithm, this takes time $g(k)^{k+2}$. In total the running time is bounded by $O(n^c + h(k))$ for some constant $c$ independent of $k$ and some function $h$.



Remark: Note that nothing here is specific to Independent Set. The first argument works for any problem with a $O(n^{k^c})$ time algorithm for some $c$, while the second argument works for any decidable problem.