Gelfand-Naimark theorem fails for Banach algebras?

Question

The Gelfand-Naimark theorem says that if $A$ is a commutative unital $C^*$-algebra, then $C(Spec(A))=A$, where $Spec(A)$ is the set of all characters on $A$.

Does the theorem fail for commutative unital Banach algebras?

To be clear, is there a counter-example to see that $C^*$ part is indeed important?

Answer 1

Hint.

Let $A=\ell^1(\mathbb Z)$ and for $z\in \mathbb T$, define $\phi_z: \ell^1(\mathbb Z) \to\mathbb C$ as $\phi_z(f)=\sum_n f(z)z^{-n}$ for all $f\in \ell^1(\mathbb Z)$.

Show that $\operatorname{Spec}(A)=\{\phi_z:z\in\mathbb T\}$ and then $C(\operatorname{Spec}(A)) = C(\mathbb T)$; the latter is not isomorphic to $A$.

Answer 2

Since $C(X)$ is always a $C^*$-algebra (where $X$ is a compact Hausdorff space), it suffices to give an example of a unital Banach algebra which is not a $C^*$-algebra. See for instance Why is $\ell^1(\mathbb{Z})$ not a $C^{*}$-algebra? and also Non-$C^{*}$ Banach algebras?.