An Error in a Proof that all Banach algebras are symmetric In Folland's `A Course in Harmonic Analysis', a Banach $*$-algebra $A$ such that
$$ \phi(M^*) = \overline{\phi(M)} $$
holds for all characters $\phi$ is called Symmetric. I have a proof that all Banach $*$ algebras are symmetric, but I can't find an error anywhere in it.
If $M$ is an arbitrary element of a $B^*$ algebra, then $M = L + iK$ for some self-adjoint $L$ and $K$. Then $M^* = L - iK$, and we know $\sigma(L), \sigma(K) \subset \mathbf{R}$, because $\sigma(M^*) = \overline{\sigma(M)}$ (if $L(\lambda - A) = (\lambda - A)L = 1$, then $(\overline{\lambda} - A^*) L^* = L^* (\overline{\lambda} - A^*) = 1$). Thus if $\phi$ is any character, $\phi(L), \phi(R) \in \mathbf{R}$, and so
$$ \phi(M^*) = \phi(L - iK) = \phi(L) - i \phi(K) = \overline{\phi(L) + i \phi(K)} = \overline{\phi(L + iK)} = \overline{\phi(M)} $$
The proof is obviously false, because on the sequence space $L^1(\mathbf{Z})$ we have the involution $(a^*)_n = \overline{a_n}$, and for this involution, if we let $\phi(a) = \sum a_k z^k$ for some $z \in \mathbf{T}$, then $\phi(a)$ is definitely not real-valued if $a$ is self-adjoint ($a$ is a real valued sequence).
 A: The mistake is where you conclude that the spectrum of every self-adjoint element is real. All you really know is that the spectrum of a self-adjoint element is closed under complex conjugation. But, a subset of the complex plane can be closed under complex conjugation without being contained in the real line! 
To see this in action, consider the disc algebran $A_D$ consisting of those continuous complex-valued function on the closed unit disk $D$ whose restriction to the open disk are holomorphic. This has the structure of a unital Banach-$*$ algebra. The addition and multiplication are the usual pointwise operations, the norm is the supremum norm over the disk, and  the involution is given by $f^∗(z)=\overline{ f( \overline z)}$. The spectrum of an element $f \in A_D$ is just its range $f(D)$. Consider, say, the element $f(z) = z$. This is self-adjoint, since $\overline{\overline z} = z$, but its spectrum is the whole of $D$. 
By the way, this is also a good example for demonstrating that, in general Banach algebras, the spectrum of an element can depend on the ambient algebra in which you measure it. Restriction gives us a homomorphism $A_D \to C(S^1)$ to the continuous functions on the circle. Roughly by the maximum principle, the max of a function in $A_D$ is achieved on the boundary, so this homomorphism is an isometric embedding. Now we see the discrepancy. Given $f \in A_D$, its spectrum as an elemnent of $A_D$ is $f(D)$, while its spectrum in $C(S^1)$ is $f(S^1)$. 
