Let $A$ be a semisimple commutative Banach algebra with the maximal ideal space $X$. Further, assume that $A$ is regular i.e. for every closed set $E\subseteq X$ and $x\in X\setminus E$, there is some $a\in A$ such that $\widehat{a}|\_E\equiv 0$ and $\widehat{a}(x)=1$ (where $\widehat{a}$ is the Gelfand transform of $a$). 


I was wondering if one knows such a Banach algebra $A$ which is not closed under complex conjugate. In other words, there is some $a\in A$ such that for every other $b\in A$, 
$$\widehat{a}\neq \overline{\widehat{b}}.$$



Regular commutative Banach algebras which are not closed under complex conjugate
I'm writing in a hurry and away from various sources, but there are known examples of proper uniform algebras (i.e. not C(X)) which are regular. By Stone-Weierstrass, these can't be closed under conjugation.


The first examples are usually attributed to McKissick (1963) but I have heard some specialists say that the construction in that paper, erm, glosses over some details. See this [this talk](http://www.math.chalmers.se/Math/Grundutb/CTH/mve021/1112/conference/slides/feinstein.pdf)
for some background, and slide 20 of the talk for an explicit statement.


Actually, in these examples the algebra is R(X) (the uniform algebra generated by the rational functions whose poles lie outside X) and so probably one can show z-bar does not belong to the algebra by doing a suitable contour integral and using Cauchy's theorem; thus one can probably avoid use of Stone-Weierstrass.


