Let $F[\mathbb{N},[0,1]]$ be the set of functions from $\mathbb{N}$ to $[0,1]$. Let $\alpha : \mathbb{N} \to [0,1]$ be any function in $F[\mathbb{N},[0,1]]$ and let $T : F[\mathbb{N},[0,1]] \to F[\mathbb{N},[0,1]] : \alpha \mapsto T[\alpha]$ be a higher-order function that maps the function $\alpha$ to the function $T[\alpha] : \mathbb{N} \to [0,1] : n \mapsto \alpha(n) / 2^{\sum_{j=1}^{n-1} \alpha(j)}$.
Let $\leq_\mathcal{O}$ be a total order on $F[\mathbb{N},[0,1]]$ induced by the big $\mathcal{O}$ notation, i.e., $f \leq_\mathcal{O} g \iff f \in \mathcal{O}(g)$. Let $A_\mathcal{O} := \{\hat{\alpha} | \forall \alpha \in F[\mathbb{N},[0,1]] : T[\hat{\alpha}] \geq_\mathcal{O} T[\alpha]\}$ be the set of maximal functions w.r.t. $\geq_\mathcal{O}$.
- Is it possible to find any element $\hat{\alpha} \in A_\mathcal{O}$? (In case $A_\mathcal{O} \neq \emptyset$.)
- If finding $\hat{\alpha}$ is infeasible, can we at least find the asymptotic behavior of $T[\hat{\alpha}]$? Concretely, how can one prove (disprove) that for any $\alpha \in F[\mathbb{N},[0,1]]$ it holds that $T[\alpha](n) \in \mathrm{o}(1/n)$?
My observation so far: If one restricts to "intuitive" functions $\alpha$, then the answer seems to lie between $\alpha(j) \sim 1/(j \ln(j))$ which yields $T[\alpha](n) \sim 1/(n\ln(n)^2)$ and $\alpha(i) \sim 1/(j\ln(j)^{1+\epsilon})$ which yields $T[\alpha](n) \sim 1/(n\ln(n)^{1+\epsilon})$. These follow from bounding the sum $\sum_{j=1}^{n-1} \alpha(j)$ by $\int_1^n \alpha(j) \mathrm{d}j$ (neglecting constant factors). However, this strategy fails when $\alpha$ is not integrable.