Are lacunary functions still lacunary over rings larger than $\mathbb{C}$?

Question

Take a lacunary function of your choice ex:

$$ f(z) = z + z^2 + z^4 + \cdots = \sum_{k=0}^\infty z^{2^n} $$

Obviously this cannot really be analytically or meromorphically continued outside the unit disk over $\mathbb{C}$.

But suppose now that $z \in \mathbb{C}^{2 \times 2}$ the set of $2\times 2$ complex matrices or some other bigger ring of your choice which contains a copy of $\mathbb{C}$

We could define some notion of distance on this space (I like the matrix $2$-norm). And now using our norm define open balls of the form $|z - z_0| \le k$ and now that we have open sets we can build sheaves and attempt to do a form of analytic continuation in this higher space.

So now the question is: does our function still resist continuation, or is it possible to extend this over the entire higher space?

Have people looked at problems like this before?

Answer 1

$f:M_k(\Bbb{C})\to M_k(\Bbb{C})$ is analytic on $\{ A\in M_k(\Bbb{C}),\sigma(A)<1\}$ and not beyond, where $\sigma(A)=\sup |\lambda_j(A)|$ is the spectral radius.

• If $\sigma(A)< 1$ then write the Jordan normal form $A=P (D+L)P^{-1}$ where $D$ is diagonal, $\|D\|_\infty=\sigma(A), DL=LD, L^k=0$ so that $A^n = P \sum_{m=0}^{k-1} {n\choose m} L^m D^{n-m} P^{-1}, \|A^n\|_\infty = O(n^k \sigma(A)^n)$ and hence the power series converges absolutely.

• If $\sigma(A)\ge 1$ then we are assuming that $f$ is analytic on a curve $0\to A$ so it will be analytic on a small disk around some $B$ such that $\sigma(B)=1$. In this disk there is some $C$ with $\sigma(C)=1$ and an eigenvalue $\exp(2i\pi s/2^t)$ with eigenvector $v$ so that $\lim_{r\to 1^-} \|f(r C) v\|=\infty$ contradicting that $f$ extends continuously to $C$.

No idea for more general infinite dimensional (Banach) $\Bbb{C}$-algebras.