Nondegenerate star-representations of a Banach star-algebra

Let $A$ be a complex Banach $\ast$-algebra without a required unit (Banach star-algebra without a required unit) and let $K$ be a complex Hilbert space (Hilbert space). A star-representation of $A$ on $K$ is a bounded complex-linear map $\sigma :A\to \mathcal{B}(K)$ (A bounded linear operator between normed spaces, Bounded Hilbert operators form a C star algebra) satisfying $\sigma (ab)=\sigma (a)\sigma (b),\sigma (a^{\ast })=\sigma (a{)}^{\ast }(a,b\in A),$ where the adjoint is the Hilbert adjoint on $\mathcal{B}(K)$. It is nondegenerate if the closed linear span of $\{\sigma (a)\xi :a\in A,\xi \in K\}$ (Linear combination of a finite list, and the span $\mathrm{span}(S)$ as the smallest linear subspace containing $S$) is all of $K$. When $A$ is a C*-algebra, a bounded star-representation of $A$ is exactly a bounded star-homomorphism $A\to \mathcal{B}(K)$ in the sense of C star algebra.

• Equivalence with the common-kernel condition. Assume Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)) for the Hilbert-space decomposition and adjoint suppliers in this paragraph. Nondegeneracy is equivalent to: every $\xi \in K$ with $\sigma (a)\xi =0$ for all $a\in A$ satisfies $\xi =0$. Indeed, let $S$ be the closed linear span of $\{\sigma (a)\xi \}$. By Orthogonal decomposition by a closed subspace one has $K=S\oplus S^{\perp }$, so $S=K$ exactly when $S^{\perp }=\{0\}$; and a vector $\eta$ lies in $S^{\perp }$ exactly when $⟨\sigma (a)\xi ,\eta ⟩=0$ for all $a\in A$ and $\xi \in K$. Since $⟨\sigma (a)\xi ,\eta ⟩=⟨\xi ,\sigma (a{)}^{\ast }\eta ⟩=⟨\xi ,\sigma (a^{\ast })\eta ⟩$ (Orthogonality and the orthogonal complement, Bounded Hilbert operators form a C star algebra), and since $\eta \perp K$ forces $\eta =0$, this holds exactly when $\sigma (a^{\ast })\eta =0$ for all $a$, i.e. when $\sigma (a)\eta =0$ for all $a$.
• Continuity is part of the definition. A star-representation is required to be bounded; no automatic-continuity statement is asserted here. For $A=L^1(G)$ the contractive bound is proved, not assumed, by the recovery lemma used on this page.