I'm wondering if there is a "big-$O$" version of the graph counting lemma in the following sense. The ordinary graph counting lemma says that for a fixed graph $H$ with $h$ edges, any graph $G$ on $n$ with $o(n^h)$ copies of $H$ as subgraphs has $o(n^2)$ edges. (For the little-$o$ notation to make sense, we should think of $G$ as part of a family of graphs whose size $n$ is taken to infinity.)
In my case, I have a graph $H$ that has $5$ vertices and I'm able to show that my family of graphs $G$ has $\le n^3$ copies of $H$. Since $n^3 = o(n^5)$, the graph counting lemma says that the number of edges in $G$ must be $o(n^2)$. But since the number of copies of $H$ is quite a lot smaller than one would expect, I would really like to be able to get $O(n^c)$ where $c<2$ in some nontrivial way. Is this possible?
The reason I think this might be possible is that if we let $H$ be a path of length $2$ (so $h=3$) and we require that there are $\le n^2$ of them, then we find that the graph $G$ has $O(n^{3/2})$ edges. Is it possible that there is a more general phenomenon going on here?