Two non-isometrically isomorphic spaces with the same duals. So that a such dual space could not be a W*-algebra under any multiplication and involution.

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c_{0} and c are both closed subspaces of l^{∞}. In addition for each x = (x_{n}) ∈ l^{1}, ρ_{x} : c_{0} ⟶ C given by (y_{n}) ↦ ∑_{n=1}^{∞}x_{n}y_{n} is a bounded linear functional on c_{0} with the norm ∥ ρ_{x} ∥=∥ x ∥. Clearly c_{0}^{#} is isometrically isomorphic to l^{1}. Also for each x = (x_{n}) ∈ l^{1}, η_{x} : c ⟶ C given by (y_{n}) ↦ x_{1}lim_{n}x_{n} + ∑_{n=1}^{∞}x_{n}y_{n} is a bounded linear functional on c with the norm ∥ η_{x} ∥=∥ x ∥. Obviously c^{#} is isometrically isomorphic to l^{1}. But by BA25.DVI the closed unit ball of c_{0} has no extreme point while the closed unit ball c contains at least (1, 1, 1, ⋯) as an extreme point (since if 1 = tx_{n} + (1 − t)y_{n} with |x_{n}| ≤ 1 and |y_{n}| ≤ 1, then 1 = tRex_{n} + (1 − t)Rey_{n} for all n, so that Rex_{n} = Rey_{n} = 1 and hence x_{n} = y_{n} = 1 for each n). Thus c_{0} and c_{1} are not isometrically isomorphic.

Now by [Sak1, Corollary 1.13.3], l^{1} can not be a W^{∗}-algebra.

Re.

[Sak1] S. Sakai, C^{∗}-algebras and W^{∗}-algebras, Springer-Verlag, 1971.