Given a Commutative Banach Algebra say $\mathcal{A}$, call it Maximal Ideal Space $\mathcal{M}_{\mathcal{A}}$. Let $\mathcal{B}_{\lambda}$ be a set where every $\phi \in \mathcal{M}_{\mathcal{A}}$ obtain its maximum. We call $\mathcal{B}_{\lambda}$ a boundary of $\mathcal{A}$. My question is given a general Commutative Banach Algebra, is the intersection of two boundaries always a boundary?
I came across a term called Shilov Boundary and in the book called Commutative Normed Rings written by Gelfand (Chapter 11), I found the proof for the unique existence of minimal boundary, which is the Shilov Boundary. However, no explicit construction of Shilov Boundary is given.
Thanks for Martin's correction, I realize in some $C(X)$ (e.g: in the example given by him) the intersection of two boundary could be empty. We need to go into the function system (or function algebra) of $C(X)$, which is a closed subalgebra what separate points and contain constant functions. Let $\mathcal{U}$ be a function system in $C(X)$. In this case I can view boundaries of $\mathcal{U}$ as subsets of $X$. Given two boundaries $\mathcal{B}_{\lambda_1}$ and $\mathcal{B}_{\lambda_2}$, their intersection will not be empty (by definition of boundary and Urysohn's Lemma). Hence $\bigcap_{\lambda \in \Lambda}\,\mathcal{M}_{\lambda}$ will not be empty. If finite intersection of boundaries is also a boundary, then for eacah $f \in C(X)$, there will be a net indexed by $\mathcal{F} = \{\bigcap_{F \subseteq \Lambda}\,\vert\, \vert\,F\,\vert < \infty\}$ (say $\{x_F\}_{F \in \mathcal{F}}$) such that $f(x_F) = \|f\|_{\infty}$. This net will converge to a point in $\bigcap_{\lambda \in \Lambda}\,\mathcal{M}_{\lambda}$ and this also proves the unique existence of Shilov Boundary.
Note:
The use of net is originally from Commutative Normed Rings Chapter 11 and this question is inspired by Problem 2.27 from Banach Algebra Technique in Operator Theory written by Douglas
The answer is "no", in general.
Note that if you take all of $C(X)$, then the only boundary is $X$, precisely because of Urysohn. The Silov boundary is an interesting object when you consider proper subalgebras (or even subspaces) of $C(X)$; more properly, one can show that the Silov boundary exists for function systems (that is, subspaces of $C(X)$ that contain $1$ and the conjugates of its elements, and that separate points). Here are a few super basic examples (note that any closed set that contains the Silov boundary is a boundary):
On $C[0,1]$, let $\mathcal F=\operatorname{span}\{1,x\}$. Then the Silov boundary is $\{0,1\}$.
On $C(\mathbb T)$, let $\mathcal F=\operatorname{span}\{1,z,\bar z\}$. Then the only boundary of $\mathcal F$ is $\mathbb T$.
On $C(\overline{\mathbb D})$, let $\mathcal F=\operatorname{span}\{1,z,\bar z\}$. The the Silov boundary of $\mathcal F$ is $\mathbb T$.
The proof that I know of the existence of the Silov boundary for function systems inside $C(X)$ is not complicated, but not trivial either. Regarding your argument, you say that the intersection of boundaries is a boundary; this is trivially true in your case because the only boundary is $X$. In the case of a function system inside $C(X)$, the only proof I know that the intersection of boundaries is a boundary comes from first showing that a minimal boundary exists (i.e., the Silov boundary exists).
The requirement that the function system separates points is essential. Otherwise, and this answers your question, consider for instance $$ \mathcal F=\{f\in C[0,1]:\ f(t)=f(1-t),\ t\in[0,1/2]\}|. $$ This is a Banach algebra (a C$^*$-algebra, actually), but not a function system because it does not separate points. In this case both $[0,1/2]$ and $[1/2,1]$ are boundaries, but their intersection is clearly not a boundary. In fact, with a simple tweak one can get disjoint boundaries: for instance the subalgebra $\{f(3-t)=f(t)\}\subset C([0,1]\cup[2,3])$ has $[0,1]$ and $[2,3]$ as boundaries.