The real-valued version of Gelfand duality (ala Johnstone's "Stone Spaces") says every real Banach Algebra is equivalent to the algebra $C_b(Y,\mathbb R)$ of bounded real functions from some compact Hausdorff space $Y$.
Edit: No it doesn't. It only applies to real Banach Algebras with a compatible partial ordering.
It is well known that every complex Banach algebra is a real Banach algebra. Simply keep the same elements but forget about rescaling by complex scalars. Only allow rescaling by real scalars.
Denote by $C_b^{\mathbb R}(X,\mathbb C)$ this 'realisation' of the complex algebra $C_b(X,\mathbb C) $. The above says $C_b^{\mathbb R}(X,\mathbb C) \simeq C_b(Y,\mathbb R)$ for some space $Y$.
The space $Y$ is of course the set of $\mathbb R$-linear multiplictive functionals $C_b(X,\mathbb C) \to \mathbb R$. What do those functionals look like?
Suppose $C_b(X,\mathbb C) \to \mathbb R$ is one such functional. Letting $1: X \to \mathbb C$ be the function equal to $1$ everywhere consider the value of $\phi(i 1)$. We have $$-\phi(1) = \phi(-1) = \phi((i 1)^2)= \phi(i 1)^2 \ge 0$$
and so $$-\phi(1) \ge 0 \implies \phi(1) \le 0.$$ But also $$\phi(1)=\phi(1^2) = \phi(1)^2 \ge 0 $$ which implies $\phi(1) =0$. Thus for each $f$ we have $\phi(f) = \phi(1f) = \phi(1) \phi(f) =0$ and $\phi$ is trivial.
This suggests the space $Y$ is degenerate. But that would imply $C_b^{\mathbb R}(X,\mathbb C) \simeq \mathbb R$ which need not be true.
What is wrong with the above?
In particular, every complex Banach algebra is a real algebra.how does this follow from the first paragraph? – SBF May 30 '24 at 10:23