Multiplicative linear functional on algebra of limit of polynomials Let $A$ be the space of all functions which are limit of polynomials over the unit ball $D$.
Then $A$ is a commutative Banach algebra. Then how do I show that $A$ has no non zero multiplicative linear functional. (As in my previous question I am not able to tackle it) Any help would be appreciated.
 A: It is not really necessary, but I think it is worthwile to not that $A$ is the algebra of uniformly continuous functions on $D$. Indeed, the uniform limit of a sequence of uniformly continuous functions is uniformly continuous. And polynomials are uniformly continuous on any bounded set. Conversely, every uniformly continuous function on $D$ can be uniquely extended to a (uniformly) continuous function on $\overline{D}$. By Stone-Weierstrass, you can approximate it uniformly by polynomials on $\overline{D}$, a fortiori on $D$.
So $A$ can actually be identified isometrically with $B=C(\overline{D})$.
This being said, every unital commutative Banach algebra has at least a character (=nonzero multiplicative bounded linear functional). This follows from Krull's theorem. 
In this case, just take $f\longmapsto f(0)$, for example.
Actually, the set of characters of $B$ (the spectrum of $B$) is exactly the set of all evaluations $f\longmapsto f(z)$ where $z$ runs over $\overline{D}$. Going back to $A$, this yields all the 
$$f\in A\longmapsto f(z)\qquad z\in D,$$ 
plus the 
$$f\in A\longmapsto \lim_{u\longrightarrow z}f(u)\qquad z\in \partial D=\overline{D}\setminus D.$$
So the spectrum of $A$ is indeed $\overline{D}$, and Gelfand tells you it is actually better to look at $B$ rather than $A$.
A: We can show that the non-zero linear functionals are characterized by $L_a\colon f\mapsto f(a)$, where $a\in D$. First, show that if $L$ is a linear multiplicative functional, then $L(P)=P(L(\mathbf 1))$, where $\mathbf 1$ is the function constant equal to $1$. As $L$ has to be continuous, we can extend the latest relation taking the uniform limit.
