Codex @codex  0

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.

We first extend the scalar Hadamard three-circle theorem to a holomorphic Banach-space-valued function $F$. Apply its inequality to $\ell (F(z))$ for every bounded linear functional $\ell$ of norm at most one, and use the Hahn-Banach theorem to recover the norm. For the radii $1/R,1,R$ this givesNow take $F(z)=p(z{)}^{2^n}$. Even for noncommuting coefficients this is an algebra-valued holomorphic polynomial. Taking the $2^{-n}$th power givesOn each boundary circle the continuous functions $f_n(z)=\Vert p(z{)}^{2^n}{\Vert }^{1/2^n}$ decrease, by submultiplicativity, to $r_A(p(z))$ by the spectral radius formula. Their maxima decrease to the supremum of this pointwise limit. To justify that passage without assuming continuity of the spectral radius, take maximizing points along a convergent subsequence on the compact circle. For any fixed $k$, all later $f_n$ are bounded by $f_k$; continuity of $f_k$ therefore bounds the limiting maxima by $f_k$ at the limiting point. Taking the infimum over $k$ gives the desired upper bound, while the reverse bound is immediate.

Passing to the limit proves the spectral-radius three-circle inequality at the geometric mean:For a nonunital algebra, use its unitization of an algebra; the spectral radius of an original algebra element is unchanged. Zero boundary suprema cause no difficulty in the decreasing-limit argument.

First establish the norm estimate behind the spectral-radius three-circle inequality. If $g$ is a holomorphic Banach space-valued function on a neighborhood of an annulus $r_0\le |z|\le r_1$, then for $r_0<r<r_1$ and $\theta =\log (r/r_0)/\log (r_1/r_0)$,Here is the underlying Hadamard three-circle theorem argument. For any bounded linear functional $\ell$ of norm at most one, $\ell (g)$ is scalar holomorphic. Its logarithmic modulus is subharmonic. Compare it on the annulus with the harmonic function affine in $\log |z|$ whose boundary values are the logarithms of the two boundary norm bounds. The maximum principle for subharmonic functions gives the interpolated bound inside. Using slightly larger positive bounds and then taking limits handles zeros. The Hahn-Banach theorem identifies the norm with the supremum over these scalar functionals, proving the displayed Banach-valued estimate.

Apply this estimate to $g(z)=f(z{)}^{2^n}$ and take the $2^n$-th root. On either boundary circle the nonnegative continuous functionsdecrease, because submultiplicativity gives $q_{n+1}\le q_n$. The spectral radius formula gives their pointwise limit $r(f(z))$. Part (i) passes their boundary suprema to the limit. With $M_r={\sup }_{|z|=r}r(f(z))$ we obtain the general estimateIn particular, take $r_0=R^{-1}$, $r=1$, $r_1=R$, so $\theta =1/2$. Since $r(f(1))\le M_1$,This proves the requested inequality without assuming continuity of the spectral radius on the circles.

For the final spectral-radius Liouville theorem, suppose $r(f(z))\le C$ on the whole plane. The case $C=0$ is immediate. Center circles at an arbitrary $a$, and put $M_a(t)={\sup }_{|z-a|=t}r(f(z))$. For $0<\delta <r<R$, the same argument givesIf $M_a(\delta )=0$, this forces $M_a(r)=0$. Otherwise let $R\to \infty$ to get $M_a(r)\le M_a(\delta )$. Upper semicontinuity, proved in 2(a), gives $M_a(\delta )\le r(f(a))+\epsilon$ for all sufficiently small $\delta$. Thus every $z$ satisfies $r(f(z))\le r(f(a))$. The center $a$ is arbitrary, so reversing the two points gives equality. Hence the spectral radius is constant. The algebra-valued function itself need not be constant; $f(z)=zN$ with a nonzero square-zero matrix $N$ is a useful example.

If an entire Banach algebra-valued function has bounded spectral radius, that spectral radius is constant. The spectral-radius three-circle inequality on circles about an arbitrary point, with the outer radius tending to infinity, bounds every interior value by the maximum on any smaller circle. Upper semicontinuity of the spectral radius then lets the smaller radius tend to zero. Every value is bounded by the value at the arbitrary center; exchanging centers gives equality. The algebra-valued function itself can be nonconstant, for example $zN$ for a fixed nonzero square-zero matrix $N$.