Let $\Omega\subset \mathbb{R}^d$ be an open bounded (possibly convex) domain. Suppose that $\Omega$ is divided up unto $N$ non-overlapping domains $(K_i)_{i=1}^N$ Suppose $f:\Omega\to \mathbb{R}$ is function that is a piecewise polynomial, meaning for all $K_i$ the restriction of $f$ to $K_i$ is a polynomial of degree at most $p$. Note that we don't require $f$ to be continuous. So for example if $p=1$ and the dimension of the problem, $d$, is 2 then $(f\big|_{K_i})(x,y) = a_{00} + a_{10} x+a_{01} y + a_{11} x y$.
Suppose that $(\Psi,\vec{\Phi})$ is the solution of the following mixed elliptic PDE
\begin{align*} 0& = \vec{\Phi} + \vec{\nabla} \Psi\ &\mathrm{for\ } x\in \Omega\\ f&= \vec{\nabla} \cdot \vec{\Phi}\, &\mathrm{for\ } x\in \Omega\\ 0&= \Psi\, &\mathrm{for\ } x\in \partial\Omega. \end{align*}
My question is whether or not we can prove that $\vec{\Phi}$ and $\Psi$ are also piecewise polynomials of degrees $p+1$ and $p+2$ respectively. It is true for $d=1$, are there any restrictions we need to unsure this is true for $d=2,3$?
Thanks
Followup,
It seems that the answer is "no" in general. If the domain $\Omega$ is a circle, there isn't a piecewise polynomial that even satisfies the boundary conditions. So assume that the the subdomains are all parallelograms for $d=2$ or parallelepipeds for $d=3$.