= Spectral-radius three-circle inequality
{title2=$M_r(f)\leq M_{r_0}(f)^{1-\theta}M_{r_1}(f)^\theta$}
For a holomorphic <Banach algebra>-valued function, put $M_r(f)=\sup_{|z|=r}r(f(z))$. Apply the <Hadamard three-circle theorem> to the holomorphic powers $f^{2^n}$. The continuous functions $\|f(z)^{2^n}\|^{1/2^n}$ decrease to the <spectral radius>. <Supremum convergence for decreasing continuous functions> passes the boundary suprema to the limit, proving the inequality with the same logarithmic interpolation parameter. In particular, at the geometric-mean radius, the square of the middle spectral radius is at most the product of the two boundary suprema.
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