I want to express the following statement about a function $f(n)$: there exists $f_\Omega\in\Omega(h(n))$ such that $f\in O(g(f_\Omega(n))$. What's the correct notation for this? Is it $f\in O(g(\Omega(h(n))))$? Or does that notation mean that $f\in O(g(f_\Omega(n))$ for all $f_\Omega\in\Omega(h(n))$?
For example, take $f(n)=2^{-\alpha n}(n^4+n^2+\log(n))$ for $\alpha>0$. Such a function has that there exists $f_\Omega\in\Omega(n)$ (take $f_\Omega(n)=\alpha n/2$ for example) such that $f\in O(2^{-f_\Omega(n)})$. I want a nice succinct way to say this that stresses the important part (the exponential) and without mentioning the constant $\alpha$ as it is subject to change, like, $2^{-5n}(n^4+n^2+\log(n))\in O(2^{-\Omega(n)})$. But I'm not sure if this is "there exists $f_\Omega$" or "for all $f_\Omega$" and if it's the latter then what's the right way of writing this?
All Landau symbols are "there exists" by definition. Therefore $f(n)\in g(\Omega(h(n)))$ means simply "there exist $K_1$ such that $g(K_1h(n))\leq f(n)$ for all $n\geq n_0$.
Similarly, $f(n)\in\mathcal O(g(\Omega(h(n))))$ means "there exist $K_{0,1}$ and a function $e$ such that $f(n)\leq K_0 e(n)$ and $g(K_1h(n))\leq e(n)$ for all $n\geq n_0$.
However, I would certainly avoid such cryptic notations and write it out in words: "$f(n)\in\mathcal O(e(n))$ for some $e(n)\in g(\Omega(h(n))$."
Anyways, isn't combining of $\mathcal O$ and $\Omega$ a bit strange, since one is upper bound and one is lower bound?