TITLE: Reference for "distributional derivative being zero implies being constant"
QUESTION [5 upvotes]: I know that if a distribution (generalized function) has zero derivative, then it is a constant. I also know the proof. But I have a hard time finding a reference which contains a statement of this fact. Any thoughts? Thanks.
Update:
I indeed found it in "Théorie des distributions" by L. Schwartz. Section 2.4, Theorem 1.

REPLY [3 votes]: I may be aware that this is very bad timing, but I wanted to point out the fact that there is an easy proof also in $\mathbb{R}^N$. Source: lecture notes by Professor P. D'Ancona (in Italian). https://www1.mat.uniroma1.it/people/dancona/IstAnSup/dispense-esercizi/4-Distribuzioni-20191024.pdf
Proposition. Let $T \in \mathscr{D}^{\prime}\left(\mathbb{R}^{N}\right)$ such that $\nabla T=0 .$ Then there exists a constant $c$ such that $T=T_{c}$, where for $f\in L^1_{loc}$ I denote $T_f$ the distribution represented by $f$.
Proof. Let's define $f_{\varepsilon}:=\rho_{\varepsilon} * T$, the standard regularization of $T$. Then
$$
\nabla f_{\varepsilon}=\nabla\left(\rho_{\varepsilon} * T\right)=\rho_{\varepsilon} *(\nabla T)=0,
$$
thus $f_{\varepsilon}$ is a constant function $f_{\varepsilon}=c_{\varepsilon} .$ On the other hand, $T_{f_{\varepsilon}}=T_{c_{\varepsilon}}$ converge to $T$ in $\mathscr{D}^{\prime}(\Omega)$, hence the sequence of numbers $\left\{c_{\varepsilon}\right\}$ must be convergent, so there exists $c \in \mathbb{R}$ such that $T_{c_{\varepsilon}} \rightarrow T_{c}$.