I mean specifically real-valued Schwartz distributions on the real line.  That is linear functionals  on $C^{\infty}\_c(\mathbb{R})$ continuous in the canonical LF topology.  My question is, what are all of these that have weak derivative 0? 
 
Right now the only examples I know are distributions corresponding to constant functions.   Are there non-constant locally integrable functions with distributional derivative 0?  Are there distributions on the real line with weak derivative 0 but not corresponding to any locally integrable function?


A locally integrable function with distributional derivative 0 must have no step discontinuity, and must have strong derivative 0 wherever it has a strong derivative.  But that is all I can prove about it now. 
Question https://mathoverflow.net/posts/415151 deals with a holomorphic version of this question.



What real distributions solve $f'=0$?
You can take Fourier transforms $\widehat{f'}(t)=it\widehat{f}(t)=0$ to conclude that $\widehat{f}$ is supported by $\{0\}$, so $\widehat{f}=\sum\_{j=0}^n c\_j \delta^{(j)}$, so $f$ is a polynomial and then $f=c$.


(If $f$ is locally integrable with distributional derivative in $L^1\_{\textrm{loc}}$, then $f$ is absolutely continuous, so can be recovered by integrating its derivative.)


Or you can do it by hand, in the more general version for arbitrary (not necessarily tempered) distributions. See Theorem 2.6 of my lecture notes [here.](http://www2.math.ou.edu/%7Ecremling/teaching/lecturenotes/ln-dist.pdf)


