**Background:**
Fix a linear algebraic group $G$ over an algebraically closed field $k$ of arbitrary characteristic and let $B \subseteq G$ be a Borel subgroup with unipotent radical $N$. Let $\Delta^+$ denote the positive roots in a root system of a torus of $G$. Then we have the hyperalgebra $\bar U(N)$ of $N$ which is generated as a $k$-algebra by the divided-power elements $E_\beta^{(n)}$ for $\beta \in \Delta^+$ and $n \geq 0$. (I would also be happy to just consider the characteristic 0 case, where $\bar U(N)$ is just the enveloping algebra of Lie($N$) and is also generated by the non-divided power elements $E_\beta^n$).

Fix an ordering $\beta_1, \ldots, \beta_k$ of $\Delta^+$. Then the elements $E_{\beta_1}^{(n_1)} \cdots E_{\beta_k}^{(n_k)}$, for $n_1, \ldots, n_k \geq 0$, form a vector space basis of $\bar U(N)$. This gives a natural vector space grading on $\bar U(N)$ that I will denote by $\bar U(N)_n$ for $n \geq 0$. It is definitely not the case that $\bar U(N)_n$ is a multiplicative grading on $\bar U(N)$, as is easy to see. The following is, however, a well-known fact (by the PBW theorem): $$\bar U(N)_n \cdot \bar U(N)_m \subseteq \bar U(N)_{ \leq m+n } \,\,\, .$$ This gives an *upper* bound on the degree of a product of elements.

**Question:**
I am wondering if anyone has considered the question of a *lower* bound on the degree of a product of elements in $\bar U(N)$. That is, I would like to see a fact on the following kind: $$\bar U(N)_n \cdot \bar U(N)_m \subseteq \bar U(N)_{ \geq f(m,n) } \,\,\, ,$$ where $f(m,n)$ is some reasonable function of $m$ and $n$ that hopefully gives a sharp bound.

(Note that whereas the PBW theorem gives an upper bound on degrees for any hyperalgebra/enveloping algebra, I expect that this lower bound, if it exists, will very much depend on the fact that $N$ is a unipotent algebraic group.)