Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 7 / 6

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.

It suffices first to treat unital algebras and a unital homomorphism. A surjective homomorphism between unital algebras automatically preserves the identity. If $B=0$, continuity is immediate. In the nonunital case, extend to the forced unitizations by $T^{+}(a+\lambda 1)=T(a)+\lambda 1$. This is still surjective and the unitization of a semisimple algebra is semisimple: its radical projects to zero in its scalar quotient, and its remaining intersection with the original algebra is its original radical. Continuity of the extension implies continuity of the original map.

An algebra homomorphism decreases the spectral radius, even without any continuity assumption, because it takes an inverse of $\lambda 1-a$ to an inverse of $\lambda 1-T(a)$. Suppose $a_n\to 0$ in $A$ and $T(a_n)\to b$ in $B$. Fix any $c\in B$ and choose $u,v\in A$ with $T(u)=b$, $T(v)=c$. PutConsider the polynomial $q_n(z)=T((1-z)w_n+zw)=(1-z)d_n+zd$. Its value at one is exactly $d$. On the outer circle $|z|=R$,On the inner circle $|z|=1/R$, spectral-radius decrease givesApplying part (i) in $B$ and then letting $n\to \infty$, with $R$ fixed, givesThis holds for every $R>1$, so $r_B(cb)=0$ for every $c\in B$.

This forces $b$ into the Jacobson radical. To see the criterion directly, if $b$ were outside a maximal left ideal $L$, then $L+Bb=B$, so $1=l+cb$ for some $l\in L$, $c\in B$. The equality $r_B(cb)=0$ makes $1-cb$ invertible, but this element equals $l\in L$, impossible in a proper left ideal. Hence $b$ lies in every maximal left ideal. Since $B$ is semisimple, $b=0$.

We have shown that the separating space of $T$ is zero. If $a_n\to a$ and $T(a_n)\to y$, apply the result to $a_n-a$ to obtain $y=T(a)$. Its graph is closed, and the closed graph theorem now provesThe crucial estimate uses part (i) with an exactly fixed central value; it does not assume the false general assertion that the spectral radius is continuous under norm limits.