This is a follow-up question to Decomposition of functionals on sobolev spaces.

Let $\Omega \subset \mathbb{R}^n$ be a bounded, open set and $\mu \in H^{-1}(\Omega) = H_0^1(\Omega)^*$. Moreover, let $\mu(v) \le C \, \|v\|_{L^\infty(\Omega)}$ for all $v \in H_0^1(\Omega) \cap C_0(\Omega)$. Then, we can extend $\mu$ uniquely to $C_0(\Omega)$ and by the Riesz representation theorem, it is a regular Borel measure with finite total variation. Moreover, we have the Hahn-Jordan decomposition $\mu = \mu^+ - \mu^-$.

Do we have $\mu^+ \in H^{-1}(\Omega)$ as well?

(This means: if we restrict $\mu^+$ to $H_0^1(\Omega) \cap C_0(\Omega)$, it is continuous w.r.t. the $H_0^1(\Omega)$-norm and we can extend it uniquely to a bounded functional on $H_0^1(\Omega)$.)

I already know the following:

  • It does not work, if we drop the assumption that $\mu$ is already a measure (see the linked question above). Moreover, the counterexamples in that question are not measures.

  • It does work in case $n = 1$, since every finite Borel measure is in $H^{-1}(\Omega)$ due to the continuous embedding $H_0^1(\Omega) \hookrightarrow C(\bar\Omega)$.

  • We cannot get a bound $\| \mu^+ \|_{H^{-1}} \le C \| \mu \|_{H^{-1}}$. This can be seen by considering $\mu_n(x) = \sin(n \, x)$ on $\Omega = (0,1)$. Note that the assertion of the question still holds.

  • It would be sufficient to show that $\mu$ is order bounded, see https://mathoverflow.net/questions/149151/is-any-order-bounded-continuous-linear-functionals-a-difference-of-positive-cont, but I was not successful. This would mean: for $v,w \in H_0^1(\Omega)$ we find $C > 0$ such that $$|\mu(y)| \le C \qquad \text{for all } y \in H_0^1(\Omega), v \le y \le w.$$ Then, the link would yield the desired splitting.

  • $\begingroup$ Since you have counterexamples to $ \| \mu^+ \|_{H^{-1}} \le C \| \mu \|_{H^{-1}}$, would not gluing them (with disjoint subsets) give a negative answer to the question? Make it so that $\sum \|\mu_n\|_{H^{-1}}^2$ converges but $\sum \|\mu_n^+\|_{H^{-1}}^2$ diverges (disjoint supports, so they are orthogonal). $\endgroup$
    – user147263
    Aug 21, 2015 at 0:18
  • $\begingroup$ @NormalHuman: The $H^{-1}$ norm is not local, thus measures with disjoint support may not be orthogonal. Moreover, it is not clear why this strategy would fail in 1 dimensions. I think I managed to construct a counterexample in dimension 2, but I need to check it again. $\endgroup$
    – gerw
    Aug 22, 2015 at 7:58

1 Answer 1


The following provides a counterexample.

$\newcommand\R{\mathbb{R}}\newcommand\norm[1]{\lVert#1\rVert}\newcommand\abs[1]{\lvert#1\rvert}\newcommand\d{\mathrm{d}}\newcommand\MM{\mathcal{M}}\newcommand{\floor}[1]{\lfloor #1 \rfloor}\newcommand{\dual}[2]{\langle #1 ,\, #2 \rangle}$Let $\Omega = B_1(0) \subset \R^2$ be the (open) unit ball. We denote by $\delta_{\hat r}$ the line measure which is located at the radius $\hat r \in (0,1)$ and with mass $2 \, \pi$ (i.e., line density $1/\hat r$).

By $(-\Delta_0)^{-1}$ we denote the solution mapping associated with the Laplace equation with homogeneous Dirichlet boundary condition on $\Omega$.

We have \begin{equation*} v_{\hat r}(x,y) := (-\Delta_0)^{-1}(\delta_{\hat r}) (x,y) = \begin{cases} \log(1 / \hat r), & \text{if } r \le \hat r , \\ \log(1 / r), & \text{if } r > \hat r . \end{cases} \end{equation*} Here and in the sequel, we use $r = \sqrt{x^2 + y^2}$. We find \begin{equation*} \frac{\partial}{\partial r} v_{\hat r}(x,y) = \begin{cases} 0 & \text{if } r \le \hat r \\ -1/r & \text{if } r > \hat r \end{cases} \end{equation*} and, thus, \begin{equation*} \norm{v_{\hat r}}_{H_0^1(\Omega)}^2 = \int_\Omega \abs{\nabla v}^2 \, \d (x,y) = \int_\Omega (\frac{\partial}{\partial r} v)^2 \, \d (x,y) = \int_{\hat r}^1 1/r \, \d r = \log(1/\hat r). \end{equation*}

Now, let $q \in (0,1)$ and a sequence $\{c_i\}_{i = 1}^\infty \subset \R^+$ be given. We set $r_i = q^i$.

We define a sequence $\{\mu_k\} \subset H^{-1}(\Omega) \cap \MM(\Omega) = (H_0^1(\Omega))^* \cap (C_0(\Omega))^*$ by \begin{equation*} \mu_k := \sum_{i = 1}^k c_i \, (\delta_{r_{2 \, i}} - \delta_{r_{2\,i-1}}) . \end{equation*} Since all line measures have mass $2 \, \pi$, the sequence $\{\mu_k\}$ is Cauchy in $\MM(\Omega)$ iff $\{c_i\}$ is summable.

In order to compute the $H^{-1}(\Omega)$-norm of $\mu_k$, we set \begin{equation*} v_k := (-\Delta_0)^{-1} \mu_k \end{equation*} and have \begin{equation*} \norm{v_k}_{H_0^1(\Omega)} = \norm{\mu_k}_{H^{-1}(\Omega)}. \end{equation*}

Since \begin{equation*} \frac{\partial}{\partial r} v_k(x,y) = \begin{cases} -c_i/r & \text{if } r_{2\,i} \le r \le r_{2\,i-1} \text{ with } i \in \{1,\ldots,k\}, \\ 0 & \text{else}, \end{cases} \end{equation*} we find for $n \le k$ \begin{align*} \norm{\mu_n - \mu_k}_{H^{-1}(\Omega)}^2 &= \norm{v_n - v_k}_{H_0^1(\Omega)}^2 \\ &= \int_\Omega \abs{\nabla (v_n - v_k)}^2 \, \d (x,y) \\ &= \int_\Omega (\frac{\partial}{\partial r} (v_n - v_k))^2 \, \d (x,y) \\ &= \sum_{i = n + 1}^k c_i^2 \, \int_{r_{2\,i}}^{r_{2\,i-1}} 1/r \, \d r \\ &= \sum_{i = n + 1}^k c_i^2 \, \log(r_{2\,i-1}/r_{2\,i}) \\ &= \log(1/q) \, \sum_{i = n + 1}^k c_i^2 \end{align*} Hence, the sequence $\{\mu_k\}$ is Cauchy in $H^{-1}(\Omega)$ iff $\{c_i\}$ is square-summable.

In case $\{c_i\}$ is summable, the limits of $\mu_k$ in $H^{-1}(\Omega) = H_0^1(\Omega)^*$ and $\MM(\Omega) = C_0(\Omega)^*$ coincide, since $C_0^\infty(\Omega)$ is a dense subspace of $H_0^1(\Omega)$ and of $C_0(\Omega)$.

Now, we choose $c_i = i^p$ for some $-3/2 < p < -1$. Then $c_i$ is summable and square summable.

The positive part of \begin{equation*} \mu = \sum_{i = 1}^\infty c_i \, (\delta_{r_{2\,i}} - \delta_{r_{2\,i-1}}) \end{equation*} is \begin{equation*} \mu^+ = \sum_{i = 1}^\infty c_i \, \delta_{r_{2\,i}}. \end{equation*} Note that $\mu$ belongs to $\MM(\Omega)$ and $H^{-1}(\Omega)$.

Let $\varphi \in C_0^\infty(\Omega)$ with $0 \le \varphi \le 1$ and $\varphi \equiv 1$ on $B_q(0)$ be given. For $0 < s < 1/2$, the function \begin{equation*} v(x,y) = \log \big(1/r\big)^s \, \varphi(x,y) \end{equation*} belongs to $H_0^1(\Omega)$.

For given $K > \log(1/q)^s$, we consider the truncation $v_K$ of $v$ at $K$ and have \begin{equation*} v_K(x,y) = \begin{cases} K & r \le \exp(-K^{1/s}), \\ v(x,y) & r > \exp(-K^{1/s}). \end{cases} \end{equation*} Moreover, $v_K \to v$ in $H_0^1(\Omega)$ as $K \to \infty$.

But \begin{equation*} \dual{\mu^+}{v_K} = 2\,\pi\,\sum_{i = 1}^\infty c_i \, v_K(r_i) \ge 2\,\pi\,\sum_{i = 1}^{n(K)} c_i \, \log \big(1/q^i\big)^s = 2\,\pi\, \log \big(1/q\big)^s \, \sum_{i = 1}^{n(K)} i^{p + s} , \end{equation*} where $n(K) = \floor{K^{1/s} / \log(1/q)}$. Note that $n(K) \to \infty$ as $K \to \infty$ and, hence, \begin{equation*} \dual{\mu^+}{v_K} \ge 2\,\pi\, \log \big(1/q\big)^s \, \sum_{i = 1}^{n(K)} i^{p + s} \to \infty \end{equation*} as $K \to \infty$ if $p + s \ge -1$. Note that for all $p \in (-3/2,-1)$, we can choose $s \in (0,1/2)$ such that $p + s \ge -1$.

This shows that $\mu^+ \in \MM(\Omega)$ is not bounded on $H_0^1(\Omega) \cap C_0(\Omega)$ w.r.t.\ the $H_0^1(\Omega)$-norm.

I would be glad if anyone who checked the counterexample would leave a comment.


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