= Automatic continuity onto a semisimple Banach algebra
{title2=$T:A\twoheadrightarrow B\Longrightarrow T\text{ is continuous}$}
An <algebra homomorphism> onto a <semisimple Banach algebra> is <continuous>. To see the spectral mechanism, suppose $a_n\to0$ 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> $d=cb$, $d_n=T(va_n)$. Apply the <spectral-radius three-circle inequality> to $T((1-z)va_n+zvu)$. On the outer circle its <norm> tends uniformly to $\|d\|$; on the circle of radius $1/R$ its <spectral radius> is bounded by $(1+1/R)\|va_n\|+\|vu\|/R$. Hence $r(d)^2\le\|d\|\|vu\|/R$ 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.
Back to article page
