$X$ Banach space. $\mathcal{L} \in B(X) $ is solvable Lie Algebra. Then for some n,
$\mathcal{L} \supset \mathcal{L}^{(1)}=[\mathcal{L},\mathcal{L}] \supset \mathcal{L}^{(2)}=[\mathcal{L}^{(1)},\mathcal{L}^{(1)}] \supset \mathcal{L}^3=[\mathcal{L}^{(2)},\mathcal{L}^{(2)}] \supset ... \supset \mathcal{L}^{(n+1)}=[\mathcal{L}^{(n)},\mathcal{L}^{(n)}] = \left\{ 0 \right\}$. Then $\mathcal{L}^{(n)}$ commutative and nilpotent.
I want show that $Z(\mathcal{L}^{(n)}) \subset Z(\mathcal{L})$.
Does $\mathcal{L}^{(n)}$ cartan subalgebra of $\mathcal{L}$?