Let $\mathfrak{g}$ be a finite dimensional simple Lie algebra over $\mathbb{C}$ and $\mathfrak{h}$ be its Cartan subalgebra (or maximal toral subalgebra). We know that the universal enveloping algebra $U$ of $\mathfrak{g}$ admits a filtration given by $U^0 \subset U^1 \subset......\subset U^n \subset....$ (where $U^0 = \mathbb{C}$ and $U^n$ is generated by all the products $x_1....x_k (x_i \in \mathfrak{g}$ and $k \leq n$ for $n \in \mathbb{N}$)) and consequently we have an associated graded algebra over $\mathbb{C}$ denoted by $grU = \bigoplus_{n \in Z_{+}} \dfrac{U^n}{U^{n-1}}$ (where $U^{-1}=(0)$ and $Z_{+}$ is the set of all non-negative integers).
Now $\mathfrak{g}$ acts on $U$ via $x.u=x.u-u.x$ which is the nothing but the action induced from the adjoint action of $\mathfrak{g}$ on itself. Under this action, $\mathfrak{h}$ acts diagonally on $U$ and thus we have a decomposition of $U$ given by $U= \bigoplus_{\beta \in Q} U_{\beta}$ where $Q$ $(\subset \mathfrak{h^{*}})$ is the root lattice of $\mathfrak{g}$ and $U_{\beta}=\{u \in U: h.u = \beta(h)u$ $\forall h \in \mathfrak{h}\}$. Note that for any $\beta \in Q$, $U_{\beta}$ acts on $U_0 (0\in Q)$ both as a left module as well as a right module.
Moreover note that $\mathfrak{h}$ preserves the above filtration of $U$ and hence acts diagonally also on $grU$. Consequently, we also have a decomposition of $grU$ given by $grU = \bigoplus_{\beta \in Q} (grU)_{\beta}$ where $(grU)_{\beta}=\{u \in grU: h.u = \beta(h)u$ $\forall h \in \mathfrak{h}\}$. Again observe that for any $\beta \in Q$, $(grU)_{\beta}$ acts on $(grU)_0(0 \in Q)$ both as a left module as well as a right module.
Now $\bf assume$ that for any $\beta \in Q$, $(grU)_{\beta}$ is a finitely generated $(grU)_0$-module. Then is it true that $U_{\beta}$ is a finitely generated $U_0$-module ?
I guess that the assertion is true as many properties that hold good in case of the associated graded algebra carry over to the original algebra, but I am not able to prove it rigorously.
Any help will be greatly appreciated.