Let $A$ be an associative algebra over a field. Then $A$ can be regarded as a Lie algebra via the Lie bracket defined by $[a,b]=abba$ for every $a,b\in A$. The algebra $A$ is called Lie locally nilpotent if it is locally nilpotent as a Lie algebra. Also, $A$ is said to be locally Lie nilpotent if every finitely generated associative subalgebra of $A$ is nilpotent as a Lie algebra. Clearly, if $A$ is locally Lie nilpotent then it is Lie locally nilpotent. Is the converse true?

$\begingroup$ Of course, my question is equivalent to the following. Suppose that a finitely generated associative algebra $A$ is locally nilpotent as a Lie algebra. Is $A$ necessary nilpotent as a Lie algebra? $\endgroup$ – Salvatore Siciliano Aug 5 '11 at 15:27

$\begingroup$ It maybe that the following paper: P. Etingof, J. Kim, X. Ma, On universal Lie nilpotent associative algebras, J. Algebra 321 (2009), N2, 697703 DOI: 10.1016/j.jalgebra.2008.09.042; arXiv:0805.1909, and/or references therein, are relevant. $\endgroup$ – Pasha Zusmanovich Aug 20 '11 at 16:12

$\begingroup$ Hi Pasha, thank you for pointing out this reference. I took a look at that paper, however a satisfactory answer to my question is not present there. $\endgroup$ – Salvatore Siciliano Aug 20 '11 at 16:59
I have to mention that for associative algebras over fields of characteristic not 3 the question has now a positive answer. This is a consequence of a very recent paper by Dias and Krasilnikov: https://arxiv.org/pdf/1709.05728.pdf