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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.

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I think it's true by Dixon's theorem (JLMS 1974) 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.

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