Let $A$ be a matrix-valued entire function. It is then well-known that $\log \Vert A(z)\Vert$ is subharmonic. In particular, the operator norm is just the largest singular value of $A$.

Is it therefore also true that for any singular value $\sigma$ of $A$, in a domain where they are simple, turn $\log \sigma(A(z))$ into a subharmonic function?


1 Answer 1


First a positive result, which answers a different but related question. This paper of Aupetit

On log sub-harmonicity of singular values of matrices, Studia Mathematica 122 (1997), 195-200; DOI 10.4064/sm-122-2-195-200

has the following abstract:

Let $F$ be an analytic function from an open subset $\Omega$ of the complex plane into the algebra of $n\times n$ matrices. Denoting by $s_1,\dots,s_n$ the decreasing sequence of singular values of a matrix, we prove that the functions $\log s_1(F(\lambda))+\dots + \log s_k(F(\lambda))$ and $\log^+ s_1(F(\lambda)) + \dots+\log^+s_k(F(\lambda))$ are subharmonic on $\Omega$ for $1\leq k\leq n$.

(Note: this result had been obtained independently and earlier in the 1989 PhD thesis of M. C. White, but was not published at the time.)

In the same Studia article, an example is given just before Theorem 1 which shows that the answer to your stated question is negative. The example is quite easy to describe: take the entire matrix-valued function $$ F(z) = \begin{pmatrix} 1 & 1 \\ 0 & z \end{pmatrix} $$ Then the smallest singular value satisfies $s_2(z)=\min(1,|z|)$, which is not subharmonic. Consequently $\log s_2(z)$ also cannot be subharmonic.


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