TITLE: Decomposition of functionals on sobolev spaces
QUESTION [8 upvotes]: It is a well known theorem, that every signed measure can be split into its positive and negative parts (Hahn-Jordan-Decomposition). My question is, if something similar is possible for functionals on Sobolev spaces.
To be precise, let $\Omega \subset \mathbb{R}^n$ be some open domain and $\mu \in H^{-1}(\Omega) = H_0^1(\Omega)^*$. Are there $\mu^+, \mu^- \in H^{-1}(\Omega)$, which are positive in the sense that
$$\langle \mu^+, v \rangle \ge 0 \quad\text{for all } v \in H_0^1(\Omega), v \ge 0,$$
(and the same for $\mu^-$) and $\mu = \mu^+ - \mu^-$?
(By duality arguments, one obtains, that the set of all differences $\mu^+ - \mu^-$ of positive functionals $\mu^+, \mu^-$ is dense in $H^{-1}(\Omega)$.)
Edit: Found a very related question on MO: https://mathoverflow.net/questions/149151/is-any-order-bounded-continuous-linear-functionals-a-difference-of-positive-cont

REPLY [5 votes]: Just for reference, I post a different answer (which doesn't rely on the fact that positive distributions are measures).
Define $\Omega = (-1, 1)$, and
$$
f = \begin{cases} 0 & x \le 0, \\ |x|^{-1/3} & x > 0. \end{cases}
$$
Then, $f \in L^2(\Omega)$ and hence, $\mu$ defined by
$$
\langle \mu, v \rangle := \int_\Omega f \, v' \, \mathrm{d} x
$$
belongs to $H^{-1}(\Omega)$.
Now, assume that $\mu = \mu^+ - \mu^-$ for positive $\mu^+,\mu^- \in H^{-1}(\Omega)$. Let $v \in H_0^1(\Omega)$, $v \ge 0$ be given.
Then, for all $w \in H_0^1(\Omega)$ with $0 \le w \le v$, we have
$$
\langle \mu, w \rangle = 
\langle \mu^+, w \rangle -
\langle \mu^-, w \rangle
\le
\langle \mu^+, v \rangle -
\langle \mu^-, 0 \rangle
= \mathrm{const}.
$$
(In books about Banach lattices, this property is called "order bounded". This property is necessary and sufficient for the decomposition into a positive and a negative part.)
Now, we construct $v$ and $w$ which contradicts this boundedness. Let $v \ge 0$, $v \in H_0^1(\Omega)$ be such that $v = 1$ on $(-1/2, 1/2)$.
Define for $n \in \mathbb{N}$, $n \ge 2$,
$$
w = \begin{cases} 0 & |x| \ge 1/n,\\ 1+x\,n & x \in (-1/n, 0), \\ 1-x\,n & x \in (0,1/n).\end{cases}
$$
Then, $w \in H_0^1(\Omega)$ and $0 \le w \le v$.
However,
$$
\langle \mu, w \rangle = \int_0^{1/n} |x|^{-1/3} \, n \mathrm{d} x = \ldots = \frac{n^{1/3}}{2/3} \to \infty \quad\text{for } n \to \infty
$$
This is a contradiction. Hence, $\mu$ cannot be decomposed into a positive and a negative part.