What real distributions solve $f'=0$? [closed]

Question

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.

Answer 1

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.

Answer 2

$\newcommand\vpi\varphi\newcommand\R{\mathbb R}$Here is an elementary solution using only definitions.

Let $\Phi:=C^{\infty}_c(\R)$. Suppose that $f'=0$ for some $f\in\Phi^*$. Take any $\vpi\in\Phi$ such that $\int\vpi=0$ and let $\psi(x):=\int_{-\infty}^x\vpi$ for $x\in\R$. Then $\psi\in\Phi$ and $\psi'=\vpi$. So, $$f(\vpi)=f(\psi')=-f'(\psi)=0$$ for any $\vpi\in\Phi$ such that $\int\vpi=0$.

Now take any $\vpi\in\Phi$. Take also any $\vpi_0\in\Phi$ such that $\int\vpi_0=1$. Let $\vpi_1:=\vpi-(\int\vpi)\vpi_0$. Then $\int\vpi_1=0$ and hence $$0=f(\vpi_1)=f(\vpi-(\smallint\vpi)\vpi_0)=f(\vpi)-c_f\smallint\vpi,$$ where $c_f:=f(\vpi_0)$. So, $$f(\vpi)=c_f\smallint\vpi$$ for all $\vpi\in\Phi$, so that $f$ is identified with the constant $c_f$.

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Remark: Here $f$ does not even have to be assumed to be a distribution: it is enough that $f$ be any linear (not necessarily continuous) functional on $\Phi$, with $f'(\vpi)$ still defined as $-f(\vpi')$ for all $\vpi\in\Phi$.

Answer 3

Well, this is not exactly a research level question. But sice the answer is interesting in its own, I dare posting this answer. The following is a theorem:

Let $I\subset\mathbb R$ be an interval. Let $L=P(D)=a_0D^n+\cdots+a_n$ ($D=d/dx$) be a linear differential operator with constant coefficients. Then distributions $u\in{\mathcal D}'(I)$ that satisfy $Lu=0$ are exactly the ordinary functions satisfying $Lu=0$ in the classical sense.

The proof consists in proving that $\dim\ker_{{\mathcal D}'}L=n$. Since it contains the ordinary kernel, both coincide. To see the property above, we observe that $Lu=0$ means $\langle u,L^*\phi\rangle=0$ for every test function $\phi\in{\mathcal D}(I)$, where $L^*=P(-D)$. There remains to see that $L^*({\mathcal D}(I)$ has codimension $n$. Actually, if $f\in{\mathcal D}(I)$, the ODE $L^*\phi=f$, with $\phi\equiv0$ at the left end of $I$, is uniquely solvable. At the right end, $\phi$ is a solution of $L^*\phi=0$, thus belongs to an $n$-dimesional space $E_n$. Chosing a basis $\mathcal B$ of $E_n$, the coordinates of $\phi$ in $\mathcal B$ are $n$ independent, continuous linear forms. The range $L^*({\mathcal D}(I))$, being the intersection of their kernels, is of codimension $n$.

Your question is the special case $L=D$.

Answer 4

The other answers have already pointed out that there are no such distributions, but if we generalize the notion of a function even more, we can get such a thing. Particularly, among so-called improper priors you can see that $F(x)=x^{-2}\delta(1/x)$ possesses the desired property.