The maximal ideal space of $A$ is contained in the unit ball of $A^\ast$

Question

On page 281 of Rudin's book Functional Analysis, let $\Delta$ be the maximal ideal space of a commutative Banach algebra $A$, $K$ be the norm-closed unit ball of $A^*$, then $\Delta\subset K$ (by Theorem 10.7). Can someone give me more detail?

Answer 1

The maximal ideal space is in bijection with the set of characters on $A$; if $M$ is a maximal ideal, then the correspondence is given by $M \leftrightarrow \phi$, where the character $\phi: A \rightarrow \mathbb{C}$ is a nontrivial algebra homomorphism whose kernel is precisely $M$.

Because $\phi$ is multiplicative and nonzero, then one can prove that the operator norm of $\phi$ when regarded as a linear operator is at most $1$, so hence each $\phi \in K$. By passing through the above bijection, then it can be regarded that $\Delta \subset K$.

Answer 2

This is true after identification of closed maximal ideals with non-zero characters on $A$. More preciesly each maxiamal closed ideal in $A$ is a kernel of uniqely defined character. One can show that its norm is $1$. For details see this answer.

Answer 3

Here is a proof that every multiplicative linear functional has norm at most $1$.

First we have the following result:

For $a \in A$ (where $A$ is a unital commutative Banach algebra with unity $e$)

$$ \phi(a) \in sp(a) \;\;\forall\;\; \phi \in \chi(A) $$

where $\chi(A)$ is the set of all non-zero multiplicative linear functionals on $A$.

This is easily proved as $\phi(a - \phi(a)e) = 0$ and hence $a - \phi(a)e$ cannot be invertible.

Then for $\phi \in \chi(A)$, we have $|\phi(a)| \leq r(a) \leq ||a||$, where $r(a)$ is the spectral radius of $a$. Thus, $||\phi|| \leq 1$.

Now the set of maximal ideals of $A$ is in bijective correspondence with the set $\chi(A)$ and hence after an identification, you have what you need.