= Solution
A <C-star algebra> is a complex <Banach algebra> with a conjugate-linear <involution> satisfying $(ab)^*=b^*a^*$, $(a^*)^*=a$ and $\|a^*a\|=\|a\|^2$. It is unital when it has an identity for multiplication. For $B(H)$ use composition and the Hilbert adjoint. Completeness in the <operator norm> is given; composition is submultiplicative and the adjoint has the required algebraic properties. For a unit vector $\xi$,
$$
\|T\xi\|^2=\langle T^*T\xi,\xi\rangle\leq\|T^*T\|.
$$
Taking the supremum and using the opposite bound $\|T^*T\|\leq\|T^*\|\|T\|=\|T\|^2$ proves the <C-star identity>. Its identity is $I_H$, proving \b[$B(H)$ is a unital <C-star algebra>].
<Compact operators> form a <norm>-closed linear space and are stable under composition with bounded operators. They are also adjoint stable. To see this directly, choose a finite $\varepsilon$-net of the compact image of the <unit ball> under $T$, and let $P$ project orthogonally onto the span of its centers. Then $\|(I-P)T\|\leq\varepsilon$. The <finite-rank operator> $T^*P$ approximates $T^*$ with the same error, so $T^*$ is compact by <norm> closure. Thus \b[$\mathcal K(H)$ is a <C-star algebra>], with the inherited <norm> and operations.
If it has a multiplicative identity $E$, apply $ER=R$ to every <rank-one operator> with arbitrary prescribed range vector $\xi$. Evaluating that equation on a vector sent to $\xi$ gives $E\xi=\xi$. Hence $E=I_H$. This identity is compact exactly when $H$ is finite dimensional: finite dimension gives compactness of the <unit ball>, while in infinite dimension an infinite <orthonormal sequence> has pairwise distance $\sqrt2$ and no convergent subsequence. Therefore \b[$\mathcal K(H)$ is unital exactly in finite dimension], with the zero-algebra convention understood if $H=0$.
Back to article page
