Semisimple Banach algebra ($J(A)=\{0\}$)

A Banach algebra is semisimple when its Jacobson radical is zero. This does not assert that every quasinilpotent element is zero. For example, the algebra of bounded operators on a Banach space of dimension greater than one is semisimple, but contains nonzero square-zero rank-one operators. This notion also does not require the finite-dimensional direct-sum decomposition associated with a semisimple algebra.

An algebra homomorphism onto a semisimple Banach algebra is continuous. To see the spectral mechanism, suppose $a_n\to 0$ and $T(a_n)\to b$. For any $c\in B$, choose $u,v\in A$ with $T(u)=b$ and $T(v)=c$, and set $d=cb$, $d_n=T(v{a}_n)$. Apply the spectral-radius three-circle inequality to $T((1-z)v{a}_n+zvu)$. On the outer circle its norm tends uniformly to $\Vert d\Vert$; on the circle of radius $1/R$ its spectral radius is bounded by $(1+1/R)\Vert v{a}_n\Vert +\Vert vu\Vert /R$. Hence $r(d{)}^2\le \Vert d\Vert \Vert vu\Vert /R$ for every $R>1$, so $r(cb)=0$. The unit criterion for the Jacobson radical gives $b\in J(B)=0$, and the closed graph theorem finishes the proof.