Shilov boundary

In functional analysis, the Shilov boundary is the smallest closed subset of the structure space of a commutative Banach algebra where an analog of the maximum modulus principle holds. It is named after its discoverer, Georgii Evgen'evich Shilov.

Precise definition and existence

Let ${\displaystyle {\mathcal {A}}}$ be a commutative Banach algebra and let ${\displaystyle \Delta {\mathcal {A}}}$ be its structure space equipped with the relative weak*-topology of the dual ${\displaystyle {\mathcal {A}}^{*}}$. A closed (in this topology) subset ${\displaystyle F}$ of ${\displaystyle \Delta {\mathcal {A}}}$ is called a boundary of ${\displaystyle {\mathcal {A}}}$ if ${\textstyle \max _{f\in \Delta {\mathcal {A}}}|f(x)|=\max _{f\in F}|f(x)|}$ for all ${\displaystyle x\in {\mathcal {A}}}$. The set ${\textstyle S=\bigcap \{F:F{\text{ is a boundary of }}{\mathcal {A}}\}}$ is called the Shilov boundary. It has been proved by Shilov^[1] that ${\displaystyle S}$ is a boundary of ${\displaystyle {\mathcal {A}}}$.

Thus one may also say that Shilov boundary is the unique set ${\displaystyle S\subset \Delta {\mathcal {A}}}$ which satisfies

1. ${\displaystyle S}$ is a boundary of ${\displaystyle {\mathcal {A}}}$, and
2. whenever ${\displaystyle F}$ is a boundary of ${\displaystyle {\mathcal {A}}}$, then ${\displaystyle S\subset F}$.

Examples

Let ${\displaystyle \mathbb {D} =\{z\in \mathbb {C}$ :|z|<1\}} be the open unit disc in the complex plane and let ${\displaystyle {\mathcal {A}}=H^{\infty }(\mathbb {D} )\cap {\mathcal {C}}({\bar {\mathbb {D} }})}$ be the disc algebra, i.e. the functions holomorphic in ${\displaystyle \mathbb {D} }$ and continuous in the closure of ${\displaystyle \mathbb {D} }$ with supremum norm and usual algebraic operations. Then ${\displaystyle \Delta {\mathcal {A}}={\bar {\mathbb {D} }}}$ and ${\displaystyle S=\{|z|=1\}}$.

References

• "Bergman-Shilov boundary", Encyclopedia of Mathematics, EMS Press, 2001 [1994]