Assume that $U$ is an open set in the complex plane $\mathbb{C}$ and $A$ is a real $2\times 2$ matrix.

We define

$$\mathcal{S}_{A}=\{f:U\to \mathbb{C}\mid Df.A=A.Df \}$$ where $Df$ is the $2\times 2$ Jacobian matrix of smooth function $f:U\to \mathbb{R}^{2}\simeq \mathbb{C}$.

Every function with this property is called an $A$-holomorphic function.

**Example:** When $J=\begin{pmatrix} 0&-1\\1&0\end{pmatrix}$, $\mathcal{S}_{J}$ is the space of (usual) holomorphic functions.

Assume that this vector space $\mathcal{S}_{A}$ is closed under "uniform convergence on compact subsets of $U$". That is: Assume that $f_{n}$ is a sequence in $\mathcal{S}_{A}$ and $f_{n}$ converges to $f$, uniformly on compact subsets of $U$. Then $f$ must be in $\mathcal{S}_{A}$.

Does this imply that $A$ is in the form $$A=\begin{pmatrix} a&-b\\b&a \end{pmatrix}$$

The question is motivated by the fact that the uniform limit of a sequence of holomorphic functions is a holomorphic function.

andargument in $C(A)$ and the Legendre remainder term has arguments in $C(A)$ (similar to my example above with the dual numbers replaced by the algebra $C(A)$). I've always wondered in which cases (other than over the complex numbers) the remainder term converges to 0 :-) Sorry, I could not be of much help. $\endgroup$ – M.G. Jul 11 '16 at 12:226more comments