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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.

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    $\begingroup$ Post your proof here and then you can use here as a reference. :) $\endgroup$
    – user223391
    Aug 14, 2015 at 22:04
  • $\begingroup$ I think that I studied a proof on one of Laurent Schwarz's books... $\endgroup$
    – Siminore
    Aug 14, 2015 at 22:07
  • $\begingroup$ @Siminore Thank you. I will try to locate it there. $\endgroup$
    – Uchiha
    Aug 14, 2015 at 22:12
  • $\begingroup$ @avid19 lol, only if my professor does not get mad at me $\endgroup$
    – Uchiha
    Aug 14, 2015 at 22:13

2 Answers 2

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Ok, you said you knew how to prove it. Others may not. And you may like this better than what you have (it's the second thing I always think of when this comes up, and I like it a lot better than the first thing I think of...)

Say $u$ is a distribution and $u'=0$. By definition $u(\phi')=0$ for any test function $\phi$. Hence $u(\phi)=0$ for any test function $\phi$ with $\int\phi=0$.

Fix $\psi_0$ with $\int\psi_0=1$, and let $c=u(\psi_0)$. Now for an arbitrary test function $\phi$, let $\alpha=\int\phi$. Then $u(\alpha\psi_0-\phi)=0$, which says $$u(\phi)=c\int\phi.$$Which is exactly what "$u=c$" means.


Detail: We used the following fact above: Given a test function $\phi$ on $\Bbb R$, there exists a test function $\psi$ with $\phi=\psi'$ if and only if $\int\phi=0$. In case this is not clear: First, if $\phi=\psi'$ then $\int\phi=\int\psi'=0$ because $\psi$ has compact support. Suppose on the other hand that $\int\phi=0$, and define $\psi(x)=\int_{-\infty}^x\phi$. Then $\psi'=\phi$ and hence $\psi$ is infinitely differentiable, while the fact that $\int\phi=0$ shows that $\psi$ has compact support.

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    $\begingroup$ Hi! I know that it seems to be late, but can I ask you a question? You wrote: "Hence $u(\phi)=0$ for any test function $\phi$ with $\int \phi =0$. Why do we need $\int \phi = 0$ condition? Thank you! $\endgroup$
    – Kerr
    Jan 28, 2016 at 23:47
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    $\begingroup$ @Jane If $\psi$ is a test function then $\int\psi'=0$, since $\psi$ has compact support. It follows that given a test function $\phi$, there exists a test function $\psi$ with $\phi=\psi'$ if and only if $\int\phi=0$. (I just proved one direction. For the other direction, assume $\int\phi=0$ and define $\psi(x)=\int_{-\infty}^x\phi$; note that $\psi$ has compact support precisely because $\int\phi=0$.) $\endgroup$ Jan 29, 2016 at 0:25
  • $\begingroup$ I see now! Thank you very much! $\endgroup$
    – Kerr
    Jan 29, 2016 at 0:29
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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 $T_{c_{\varepsilon}} \rightarrow T_{c}$.

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    $\begingroup$ Just to make this clear: $\mathscr D(\mathbb R^n)$ is the usual space of compactly supported smooth functions $C_{\text{c}}^\infty(\mathbb R^n)$, where the topology is given as usual (cf. Rudin functional analysis, Theorems 6.4, 6.5). $\endgroup$ Aug 9, 2021 at 16:48

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