 Spectral-radius three-circle inequality

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 $\Vert f(z{)}^{2^n}{\Vert }^{1/2^n}$ 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.