This is the sequel to my previous question, inspired by Anixx's comment:
Holomorphicity for complex, split and dual functions
In it, Qiaochu Yuan helped establish what it means for these functions to be holomorphic. After some messing around, I've simplified the conditions (we can also define Wirtinger derivatives to do this, but this version looks nicer to me). Let $\kappa\in\{i,j,\varepsilon\}$ and $f(z)=u(z)+\kappa v(z)$ for a variable $z=x+\kappa y$. Then $f$ is holomorphic if
$$\kappa\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}$$
Now we want to test an important property, which is whether being holomorphic implies being infinitely differentiable. I imagine an inductive argument will suffice.
$$f\text{ is holomorphic}\implies\frac{df}{dz}\text{ is holomorphic}$$
To do so, I just differentiate the holomorphic condition with respect to $z$, since the holomorphic relation should be true everywhere on the domain, so it should hold true for them as a pair of functions. That makes me think that differentiating will retain equality.
$$\frac{d}{dz}\kappa\frac{\partial f}{\partial x}=\frac{d}{dz}\frac{\partial f}{\partial y}$$
Here comes the big assumption. I suppose the total $z$ derivative can be switched with the partial ones. Supposing that $u$ and $v$ are differentiable, their derivatives should be continuous. I would want to imagine this is justification, but I'm really not sure it is. Anyhow, supposing it is:
$$\kappa\frac{\partial}{\partial x}\frac{df}{dz}=\frac{\partial}{\partial y}\frac{df}{dz}$$
This would make $\frac{df}{dz}$ holomorphic, so some holomorphic $f$ is infinitely differentiable. Like last time though, something about the dual numbers is making me think this is invalid, and it's not just that $z$ depends on $x$ and $y$, so switching their total and partial derivatives feels dangerous.
For a holomorphic complex function $f=u+iv$, supposing $u$ and $v$ are twice differentiable gives that they would be harmonic conjugates:
$$\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}=0$$ $$\frac{\partial^{2}v}{\partial x^{2}}+\frac{\partial^{2}v}{\partial y^{2}}=0$$
For a holomorphic split function $f=u+jv$, supposing $u$ and $v$ are twice differentiable gives that they would be wave conjugates:
$$\frac{\partial^{2}u}{\partial x^{2}}=\frac{\partial^{2}u}{\partial y^{2}}$$ $$\frac{\partial^{2}v}{\partial x^{2}}=\frac{\partial^{2}v}{\partial y^{2}}$$
For a dual function, we get a simple system of P.D.E.s to solve, which are exactly the holomorphic conditions. Let $f=u+\varepsilon v$, then $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\text{ and }\frac{\partial u}{\partial y}=0$$
$$\frac{\partial u}{\partial y}=0\implies u(x,y)=u(x)$$ $$u'(x)=\frac{\partial v}{\partial y}\implies v(x,y)=yu'(x)+c$$
As such, an arbitrary holomorphic dual function should take the form $$f(z)=u(x)+\varepsilon yu'(x)+c$$
There's only one problem. Let $u\notin C^{n}(\mathbb{R})$. Then $f\notin C^{n-1}(\mathbb{E})$. This means we could construct a function which is holomorphic for a few derivatives, but fails after enough, a contradiction.
If I had to guess why, I'd suggest something to do with the second derivatives appearing in the complex and split case, when we got Laplace's equation and the wave equation, but not in the dual case. This may be important since Schwarz's theorem uses second partial derivatives. I was too blinded by variables last time to see the bigger picture, so I'm sure there's a simple fix I'm missing, I just don't know what.
Addendum: by expressing $f(x+\varepsilon y)$ as a Taylor series, we can find $$f(x+\varepsilon y)=f(x)+\varepsilon y f'(x)$$ This is the holomorphic condition from above (replacing the random $u$ with our known $f$), up to a constant, but the path to derive this follows from the assumption of infinite differentiability. This may be useful information, that an analytic real function applied to a dual number should satisfy the holomorphicity conditions.
