= Solution
A <representation of a Banach algebra> is a complex-linear multiplicative map $\pi:A\to\operatorname{End}_{\mathbb C}(X)$. A nonzero irreducible action of a <unital algebra> automatically has $\pi(1)=I$: the image of the idempotent $\pi(1)$ is a nonzero invariant subspace and hence is all of $X$. An <algebraically irreducible representation of a Banach algebra> is a nonzero action if it has no nonzero proper <invariant submodule>, whether <closed> or not. A <normed representation of a Banach algebra> has $\pi(a)$ bounded for each $a$ on a preassigned <normed vector space> $X$; <continuity> of $\pi$ as an operator-norm-valued map is a further conclusion. This must not be confused with topological irreducibility, which excludes only <closed> <invariant subspaces>. Unitizing the algebra extends a nonunital action by $\pi(a+\lambda1)=\pi(a)+\lambda I$, so it is enough to discuss the <unital> case.
For $\xi\ne0$, irreducibility gives $A\xi=X$. The annihilator $L_\xi=\{a:\pi(a)\xi=0\}$ is a <maximal left ideal>: an intermediate <left ideal> gives an intermediate <submodule> of $X$. It is <closed> by the Neumann-series argument for <maximal left ideals>. Consequently $A/L_\xi$ transports a complete quotient <norm> to $X$, and left multiplication is contractive in this <norm>. Every <algebraically irreducible representation of a Banach algebra> can therefore be realized continuously on a <Banach space>. This alone does not prove <continuity> for its preassigned <norm>.
The kernel of $\pi$ is the intersection of the <closed> <left ideals> $L_\xi$ and is a <closed> <primitive ideal>. The <Jacobson radical> is the intersection of all <primitive ideals>, equivalently the intersection of <maximal left ideals>. Thus these quotient representations detect semisimplicity. Equivalent <algebra representations> are related by a bijective <module homomorphism>.
The algebra of <module endomorphisms> is exactly $\mathbb C I$. Indeed, if a <module endomorphism> $D$ sends a fixed cyclic <vector> $\xi$ to $a\xi$, then $D(b\xi)=ba\xi$. It is induced by bounded right multiplication by $a$ on $A/L_\xi$, so is bounded for the <quotient norm on an irreducible Banach module>. Every nonzero <module endomorphism> is injective and surjective by irreducibility, and its inverse is bounded by the same argument. This <commutant of an operator algebra> is a <closed> division <subalgebra> of the <bounded operators> on this Banach module; the <Gelfand-Mazur theorem> makes it $\mathbb C I$.
The <Jacobson density theorem> now says that for finitely many <linearly independent> $\xi_1,\ldots,\xi_n$ and arbitrary $\eta_1,\ldots,\eta_n$, one element of $A$ sends all $\xi_j$ to $\eta_j$. Here is the induction behind it. The one-vector assertion is $A\xi_1=X$. Assuming the first $n-1$ coordinates can be prescribed, set $L=\{a:a\xi_j=0\ (j<n)\}$. Its image $L\xi_n$ is an <invariant subspace>. If it is $X$, it adjusts the last image without changing the first ones. If it is zero, the rule $(a\xi_1,\ldots,a\xi_{n-1})\mapsto a\xi_n$ is a well-defined module map $X^{n-1}\to X$. Each coordinate map is scalar by the <module endomorphism> result, forcing $\xi_n$ to be a <linear combination> of the earlier <vectors>, a contradiction. This proves the assertion.
We can now give Johnson's <continuity> argument with its crucial construction. If $X$ is finite dimensional, the <closed> kernel has finite codimension and $A/\ker\pi$ is finite dimensional, so $\pi$ is <continuous>. Suppose instead that $X$ is infinite dimensional. Let
$$
Y=\{\xi\in X:a\mapsto\pi(a)\xi\text{ is continuous}\}.
$$
This is an <invariant submodule>, because the orbit map of $\pi(b)\xi$ is the orbit map of $\xi$ composed with the bounded map $a\mapsto ab$. Thus either $Y=X$ or $Y=0$.
To rule out $Y=0$, choose <linearly independent> unit <vectors> $\xi_0,\xi_1,\ldots$. There are elements $a_n\in A$ such that, with $c_n=a_n\cdots a_1$,
$$
\pi(c_m)\xi_n=0\quad(m>n),\qquad
\{\pi(c_n)\xi_j:j\ge n\}\text{ is linearly independent}.
$$
For completeness, construct them in the canonical quotient-norm Banach module. At stage $n$ the current tail <vectors> are independent. In the <closed> left annihilator of its first <vector>, density permits arbitrary images of any finite list of the remaining <vectors>. The condition that finitely many such images are independent is <open> in this Banach subspace and dense: perturb any element by $t$ times an element prescribing independent images; a suitable determinant is a nonzero <polynomial> in $t$, with only finitely many exceptional values. The <Baire category theorem> intersects these <open> dense conditions for all initial finite tail lists. The resulting $a_n$ kills the first current <vector> and retains independence of the entire remaining tail. This is the <countable annihilation sequence in an irreducible Banach module>.
The <gliding-hump continuity principle> supplies the other ingredient: if $E,F$ are <Banach spaces>, $S:E\to F$ is linear, $T_j:E\to E$ and $U_n:F\to F_n$ are bounded, and $U_nST_1\cdots T_m$ is <continuous> whenever $m>n$, then $U_nST_1\cdots T_n$ is <continuous> eventually. The product form of this principle is recorded in https://msp.org/pjm/1993/159-1/pjm-v159-n1-p08-p.pdf[Proposition 1.3 of Thomas's paper]. Here is the contradiction mechanism. Rescale so $\|T_j\|,\|U_n\|\le1$ and put $P_m=T_1\cdots T_m$. If infinitely many diagonal maps are discontinuous, choose increasing $n_j$ and successively tiny $x_j$ so
$$
\|U_{n_j}SP_{n_j}x_j\|>j+1+\sum_{i<j}\|U_{n_j}SP_{n_i}x_i\|.
$$
Also choose $\|x_j\|\le2^{-j}/(1+\max_{i<j}\|U_{n_i}SP_{n_i+1}\|)$. The convergent sum $x=\sum_jP_{n_j}x_j$ has, after its $j$th term, a tail factoring through $P_{n_j+1}$. Applying that <continuous> composite bounds the tail contribution by one. The display forces $\|U_{n_j}Sx\|>j$, contradicting $\|U_{n_j}Sx\|\le\|Sx\|$. Only finite decompositions and the <continuous> tail composites are used; no unjustified interchange of a discontinuous map with an infinite sum occurs.
Apply this principle with $E=A$, $F=\mathcal B(\overline X)$, where $\overline X$ is the completion in the original <norm>, and $S(a)$ the bounded extension of $\pi(a)$. Set $T_j(a)=aa_j$ and $U_n(R)=R\xi_n$. Then
$$
U_nST_1\cdots T_m(a)=\pi(a)\pi(a_m\cdots a_1)\xi_n.
$$
For $m>n$ this is zero, hence <continuous>. For $m=n$ it is the orbit map of a nonzero <vector>, hence discontinuous if $Y=0$. This contradicts the principle. Therefore $Y=X$.
Finally, apply the <Uniform boundedness principle> on the <Banach space> $A$ to the <continuous> maps $a\mapsto\pi(a)\xi$ indexed by $\|\xi\|\le1$. They are pointwise bounded because each individual $\pi(a)$ is bounded in the original <norm>. It gives
$$
\boxed{\|\pi(a)\|\le C\|a\|\quad(a\in A).}
$$
This proves <Johnson's continuity theorem for irreducible normed representations>, including when the original <normed vector space> is incomplete. The quotient <norm>, countable density construction, and gliding hump explain the substantive steps rather than assuming the desired <continuity> in advance.
Back to article page
