Bounded Hilbert operators form a C star algebra

Composition in $\mathcal{B}(H)$ is bilinear and associative, and the operator norm is submultiplicative: $\mathrm{\parallel }ST\mathrm{\parallel }\le \mathrm{\parallel }S\mathrm{\parallel }\mathrm{\parallel }T\mathrm{\parallel }$.

The identity operator $I$ lies in $\mathcal{B}(H)$ and satisfies $IT=TI=T$, and $\mathrm{\parallel }I\mathrm{\parallel }=1$: since $H\mathrm{\ne }\{0\}$ every nonzero $x$ has $\mathrm{\parallel }x\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }=1$, so the unit-ball supremum defining $\mathrm{\parallel }I\mathrm{\parallel }$ equals $1$. Thus $\mathcal{B}(H)$ is a nonzero algebra whose unit $I$ has norm one.

The Hilbert adjoint is a map $\mathcal{B}(H)\to \mathcal{B}(H)$ which is conjugate-linear, involutive and isometric, satisfies $(ST{)}^{\ast }=T^{\ast }{S}^{\ast }$, and satisfies $\mathrm{\parallel }{T}^{\ast }T\mathrm{\parallel }=\mathrm{\parallel }T{\mathrm{\parallel }}^2$ for every $T$.

The space $H$ is Banach for its norm by [A1], so $\mathcal{B}(H)$ is complete for the operator norm by [A3]; together with the submultiplicativity, the identity of norm one and the nonvanishing just recorded, this makes $\mathcal{B}(H)$ a unital complex Banach algebra in the sense of [A5].

The space $\mathcal{B}(H)$, with the operator norm, composition, the identity and the Hilbert adjoint, fulfils every axiom of a unital complex C*-algebra listed in [A5], and the identity $\mathrm{\parallel }{T}^{\ast }T\mathrm{\parallel }=\mathrm{\parallel }T{\mathrm{\parallel }}^2$ holds.