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  "body": "\n## Definition\n\nLet $A$ be a commutative complex Banach algebra with unit, and let $\\Delta(A)$ be its [[character space|M7JVFJ3MMX]], a compact Hausdorff space.\nFor $a \\in A$, let $\\hat a$ be its [[Gelfand transform|VWNSHKJT92]].\nA *boundary* for $A$ is a subset $F \\subseteq \\Delta(A)$ such that, for every $a \\in A$,\n$$\\max_{\\varphi \\in \\Delta(A)} \\abs{\\hat a(\\varphi)} = \\sup_{\\varphi \\in F} \\abs{\\hat a(\\varphi)}.$$\nThe *Shilov boundary* $\\partial A$ is the smallest closed boundary for $A$: a closed boundary contained in every closed boundary.\n\n## Remarks\n\nThe existence of a smallest closed boundary is a theorem of Shilov: the intersection of all closed boundaries is again a boundary.\n\nExample: the disc algebra $A(\\mathbb{D})$ of continuous functions on the closed unit disc $\\overline{\\mathbb{D}}$ that are holomorphic in the open disc, with the sup norm.\nIts character space is $\\overline{\\mathbb{D}}$, through the evaluations at points.\nBy the maximum modulus principle, the unit circle $\\mathbb{T}$ is a closed boundary.\nNo proper closed subset $F \\subsetneq \\mathbb{T}$ is a boundary.\nIndeed, let $\\zeta \\in \\mathbb{T} \\setminus F$; the function $f(z) = (1 + \\overline{\\zeta} z) / 2$ satisfies $\\abs{f} \u003c 1$ on $\\overline{\\mathbb{D}} \\setminus \\set{\\zeta}$ and $f(\\zeta) = 1$.\nSo its supremum on the compact set $F$ is less than $1$.\nHence $\\partial A(\\mathbb{D}) = \\mathbb{T}$.\n\nNon-example: for the disc algebra, the circle $\\set{\\abs{z} = 1/2}$ is closed but not a boundary, since $f(z) = z$ has maximum modulus $1/2$ there and $1$ on $\\overline{\\mathbb{D}}$.\n\nFor $A = C(X)$ with $X$ compact Hausdorff, the Shilov boundary is all of $X$.\nIndeed, if $F \\subsetneq X$ is closed and $x \\notin F$, Urysohn's lemma gives $f \\in C(X)$ with $f(x) = 1$ and $f = 0$ on $F$, so $F$ is not a boundary.\n",
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