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Assume the compact space is nonempty. Write $M_n={\sup }_K{f}_n$. Each $M_n$ is finite and attained, and the sequence decreases to a number $L\ge 0$. Since $f\le f_n$, certainly ${\sup }_{K}f\le L$.

For any $t<L$, the sets $E_n=\{x\in K:f_n(x)\ge t\}$ are nonempty closed sets, and $E_{n+1}\subseteq E_n$. The finite intersection property and compactness give a point in their intersection. At that point $f\ge t$, so ${\sup }_{K}f\ge t$. Let $t\uparrow L$; together with the reverse bound this provesThis is supremum convergence for decreasing continuous functions. The limit need not be continuous, so the conclusion does not assert the uniform convergence of the Dini theorem. It is precisely the interchange of these decreasing suprema and pointwise limit needed in the next part.

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.