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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.