Recently, I've been reading Notes of Thierry Ramond about distributions theory and I arrived to the following result (Page 72),
Proposition 4.2.3: Let $\Omega \subset \mathbb R^p$ be an open connected set, and let $T \in D'(\Omega)$. \begin{equation*} \partial_j T = 0 \quad \text{for all} \ j \in \{1,\dotsc,p\} \Longleftrightarrow T=T_C \quad \text{for some constant} \ C \in \mathbb C \end{equation*}
The proof showed in the notes just discuses the case $\Omega=\mathbb R^p$ and it relies in
Proposition 4.2.2
The set of linear combinations of tensor products of $p$ functions in $C_0^\infty(\mathbb R) $ is dense in $C_0^\infty(\mathbb R^p)$
I've been trying to complete the proof. At first I tried extending a distribution of $D'(\Omega)$ to a distribution of $D'(\mathbb R^p)$ but now I know it seems not always possible (see extension of distributions).
Is it possible to prove this result using some kind of extension of the Proposition 4.2.2 to the case $D'(\Omega)$ instead of $D'(\mathbb R^p)$? I mean, Proposition 4.2.2 works because $\mathbb R^p= \mathbb R \times \mathbb R \times \cdots \times \mathbb R$, but it's not always possible to write $\Omega$ as a cartesian product.
EDIT: As I suspected and Christian Remling confirmed, we also need $\Omega$ connected.
EDIT-EDIT: Thanks to Daniel Fischer, I started thinking in considering in proving an alternative version of Proposition 4.2.2,
Proposition 4.2.2 (alternative)
Let $\Omega=\Omega_1 \times \cdots \times \Omega_p$ be an open hypercube of $\mathbb R^p$. Then, the set of linear combinations of tensor products of $p$ functions in $C_0^\infty(\Omega_1)\times \cdots \times C_0^\infty(\Omega_p)$ is dense in $C_0^\infty(\Omega)$.
After that, I think that since $\Omega$ open connected can be written as union of non-disjoint open hypercubes (is it true?), then, proving Proposition 4.2.3 locally in each hypercube and then using the fact that those hypercubes aren't disjoint would give the desired result (is this idea valid despite the fact the union is not necessarily countable?)
