Let $A$ be a Banach algebra (say, complex and unital) and suppose that every (closed) commutative subalgebra of $A$ is finite dimensional.


**Question.** Does it follow that $A$ is finite dimensional?


**Remark.** Clearly, every element of $A$ is algebraic (i.e. annihilated by a polynomial) and thus has finite spectrum. If $A$ is semisimple, it therefore follows from a result of Kaplansky that $A$ is finite dimensional (Lemma 7 in "I. Kaplansky: Ring Isomorphisms of Banach Algebras (1954)". So the question is concerned with the non-semi-simple case.



Criterion for a Banach algebra to be finite dimensional
I think it's true by Dixon's theorem ([JLMS 1974](https://doi.org/10.1112/jlms/s2-8.2.325)) which says (on the third page) that any Banach algebra consisting only of algebraic elements is nilpotent-by-finite. Thanks to this, we may assume $A$ is nilpotent. We moreover assume $A$ is infinite-dimensional and will construct an infinite-dimensional commutative subalgebra. Take the largest $n$ such that the ideal $B := \overline{\rm span}\, A^n$ is infinite-dimensional. For $x\in B$, we denote its left and right multiplication operator on $B$ by $L\_x$ and $R\_x$. These have finite rank. Now, start with a nonzero $b\_1 \in B$ and recursively choose nonzero $b\_k \in \bigcap\_{i=1}^{k-1}(\ker L\_{b\_i} \cap \ker R\_{b\_i})$ that is linearly independent of $\{ b\_i : i < k\}$. Then, the algebra generated by $\{ b\_1,b\_2,\ldots\}$ is commutative and infinite dimensional.


