The ball formulation of the Kaplansky density theorem in nonselfadjoint algebras

Question

Let $H$ be a separable Hilbert space. Let $A \subseteq B(H)$ be a finitely generated unital algebra. Let $M$ be the strong operator topology closure of $A.$ Let $B_A$ be the closed ball in $A$ and $B_M$ be the closed ball in $M.$ Is $B_M$ the strong operator topology closure of $B_A?$ (More likely, is there a counterexample?)

Answer 1

W. R. Wogen, Some counterexamples in nonselfadjoint operator algebras, Ann. Math. 126 (1987), 415-427, constructs an operator $T$ for which the WOT and weak* closures of the algebra generated by $T$ and $I$ are distinct. Since the WOT closure equals the SOT closure (and WOT and weak* agree on bounded sets), this gives a counterexample to your version.