Decomposition of functionals on sobolev spaces

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

Solution 1:

No, such a decomposition is not possible. The reason is that

every distribution which is non-negative on non-negative functions is of the form $\varphi\mapsto \int \varphi \, d\mu$ for some (positive) Radon measure.

Hence, $\mu=\mu^+-\mu^-$ would be given by the difference of two Radon measures. But the elements of $H^{-1}$ can be more singular than measures. For example, take $n=1$, $\Omega=(-1,1)$, and $f(x)=|x|^{-1/3}$. Then $f\in L^2(\Omega)$, which implies that the distributional derivative $f'$, that is, the functional $v\mapsto -\int_\Omega f(x)v'(x)\,dx $, belongs to $H^{-1}(\Omega)$. If this functional were given by a signed measure $\mu$, that is, $$-\int_\Omega f(x)v'(x)\,dx = \int_\Omega v(x) d\mu \ \text{ for all }\ v\in H^1_0(\Omega)$$ then after integration by parts on the right we would obtain $f(x)=C+\mu((-1,x))$ a.e., which is impossible because $f$ is not a function of locally bounded variation.

It is instructive to try to decompose $f'$ formally: $$f'(x)=-\frac{1}{3}|x|^{-4/3}\operatorname{sign}x = \mu^+-\mu^-$$ where $$\mu_+=\frac{1}{3}|x|^{-4/3}\chi_{(-1,0)}\quad \text{and}\quad \frac{1}{3}|x|^{-4/3}\chi_{(0,1)}$$ The problem is, neither $\mu_+$ nor $\mu_-$ define bounded linear functionals on $H^1$. These are positive measures that are not locally finite, and therefore cannot be integrated even against nice bump functions. The fact that $f'$ can be (in a sense) integrated against nice functions is due to cancellation between $\mu^+$ and $\mu^-$. The positive and negative parts really need each other.

Solution 2:

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.