Representation of a Banach algebra ($\pi :A\to {\mathrm{End}}_{\mathbb{C}}(X)$)

A representation of a complex Banach algebra is a complex-linear multiplicative action on a vector space. For a unital action require $\pi (1)=I$; a nonunital algebra can be handled by extending the action to its unitization. A nonzero action is algebraically irreducible when it has no nonzero proper invariant vector subspace, equivalently when it defines a simple module.

Algebraic irreducibility excludes every proper nonzero invariant vector subspace, including those that are not closed. It is stronger than topological irreducibility, which excludes only closed invariant subspaces. For a nonzero irreducible action, every nonzero vector is cyclic in the algebraic sense: $A\xi =X$.

For countably many linearly independent vectors in an infinite-dimensional irreducible complex Banach module, there are elements $a_n$ satisfying the displayed annihilation conditions while the tail images $\{\pi (a_n\cdots a_1){\xi }_j:j\ge n\}$ remain linearly independent. Use the quotient norm on an irreducible Banach module to make each finite annihilator a closed Banach subspace. The Jacobson density theorem makes each condition of independence of finitely many tail images open and dense there; the Baire category theorem intersects these countably many conditions. This is stronger than prescribing only finitely many images once.

For a unital irreducible action and $\xi \ne 0$, the annihilator $L_{\xi }=\{a:\pi (a)\xi =0\}$ is a maximal left ideal and is closed. Thus $A/L_{\xi }$ supplies the displayed complete norm on the module, with contractive action. Every module endomorphism is bounded for this quotient norm: its value at the cyclic vector is $a\xi$, so it is induced by bounded right multiplication by $a$ on the quotient. The Gelfand-Mazur theorem then makes the module's commutant consist of scalar operators. This constructs one compatible norm; it does not by itself establish continuity in a different preassigned norm.

A normed representation acts by individually bounded operators on a normed vector space. This condition alone does not assume that $a\mapsto \pi (a)$ is continuous in the operator norm, nor that the representation space is complete. A continuous representation has a common bound $\Vert \pi (a)\Vert \le C\Vert a\Vert$.

Every algebraically irreducible representation of a Banach algebra that is normed is continuous. The continuous orbit vectors form an invariant subspace. If all nonzero orbit maps were discontinuous, apply the countable annihilation sequence in an irreducible Banach module and the gliding-hump continuity principle to right multiplication on the algebra and evaluation of the operator-valued representation. Off-diagonal maps vanish, whereas diagonal maps are discontinuous, a contradiction. Once every orbit map is continuous, the Uniform boundedness principle on the Banach algebra gives the common operator-norm bound. Completing the original normed representation space supplies the Banach operator space needed in the argument.