Banach Algebra Isomorphism

Question

Let $X=l^1$ with coordinate multiplication - it is commutative Banach algebra without unit. The Gelfand transformation is defined as $\widehat{x}(e_n)=x_n$ for $x \in l^1$. I would like to prove, that this algebra is isomorphic to $C^*$-algebra. Any ideas ? Thanks.

Answer 1

It is definitely not, even as a Banach space. Commutative C*-algebras are isomorphic to $C_0(X)$ for some locally compact Hausdorff space, which is infinite when the algebra is infinite-dimensional. In such case, you will find a subalgebra isomorphic to $c_0$. The space $c_0$ does not embed isomorphically into $\ell_1$ because the latter space is weakly sequentially complete.

Actually, it is easy to see that $\ell_1$ does not satisfy the C*-condition under the original norm. Take $x=e_1 + e_2$. Then $xx^* = x$ but $\|xx^*\|=2\neq \|x\|^2$. (In other words, observe that non-zero self-adjoint projections in C*-algebras have norm 1.)