= Solution
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 $r_0\leq|z|\leq r_1$, then for $r_0<r<r_1$ and $\theta=\log(r/r_0)/\log(r_1/r_0)$,
$$
\sup_{|z|=r}\|g(z)\|\leq\left(\sup_{|z|=r_0}\|g(z)\|\right)^{1-\theta}\left(\sup_{|z|=r_1}\|g(z)\|\right)^\theta.
$$
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 functions
$$
q_n(z)=\|f(z)^{2^n}\|^{1/2^n}
$$
decrease, because submultiplicativity gives $q_{n+1}\leq q_n$. 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 estimate
$$
M_r\leq M_{r_0}^{1-\theta}M_{r_1}^{\theta}.
$$
In particular, take $r_0=R^{-1}$, $r=1$, $r_1=R$, so $\theta=1/2$. Since $r(f(1))\leq M_1$,
$$
\boxed{r(f(1))^2\leq\left(\sup_{|z|=R}r(f(z))\right)\left(\sup_{|z|=R^{-1}}r(f(z))\right).}
$$
This proves the requested inequality without assuming continuity of the spectral radius on the circles.
For the final <spectral-radius Liouville theorem>, suppose $r(f(z))\leq C$ 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 gives
$$
M_a(r)\leq M_a(\delta)^{\log(R/r)/\log(R/\delta)}C^{\log(r/\delta)/\log(R/\delta)}.
$$
If $M_a(\delta)=0$, this forces $M_a(r)=0$. Otherwise let $R\to\infty$ to get $M_a(r)\leq M_a(\delta)$. Upper semicontinuity, proved in 2(a), gives $M_a(\delta)\leq r(f(a))+\varepsilon$ for all sufficiently small $\delta$. Thus every $z$ satisfies $r(f(z))\leq r(f(a))$. The center $a$ is arbitrary, so reversing the two points gives equality. Hence \b[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.
Back to article page
