Let $v\in L^1_\text{loc}(\mathbb{R}^3)$ and $f \in H^1(\mathbb{R}^3)$ such that \begin{equation} \int f^2 v_+ = \int f^2 v_- = +\infty. \end{equation} Here, $v_- = \max(0,-f)$, $v_+ = \max(0,f)$, i.e., the negative and positive parts of $v=v_+ - v_-$, respectively.

Question: Does $g\in H^1(\mathbb{R}^3)$ exist, such that \begin{equation} \int g^2 v_+ < \infty, \quad \int g^2 v_- = +\infty \quad ? \end{equation}

Some thoughts:

Let $S_\pm$ be the supports of $v_\pm$, respectively.

One can easily find $g\in L^2$ such that the last equation holds, simply multiply $f$ with the characteristic function of $S_-$. The intuitive approach is then by some smoothing of this function by a mollifier, or using a bump function to force the support of $g$ away from $S_+$.

However, the supports of $S_\pm$ can be quite complicated: for example, fat Cantor-like sets. Thus, a bump function technique or a mollifier may "accidentally" fill out any of $S_\pm$.

My motivation:

The problem comes from my original research on the mathematical foundations of Density Functional Theory (DFT) in physics and chemistry. Here, $f^2$ is proportional to the probability density of finding an electron at a space point, and $v$ is the potential energy field of the environmentn. $\int f^2 v$ is the total potential energy for the system's state. The original $f$ gives a meaningless "$\infty-\infty$" result, but for certain reasons, we are out of the woods if there is some $other$ density $g$ with the prescribed property.


Removed claim that $S_\pm$ must be unbounded. This does not follow from the stated assumptions.

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    $\begingroup$ I don't see why the supports of $v_{\pm}$ must be unbounded. If $f$ is not locally bounded, the integrals of $f^2v_{\pm}$ may be infinite even on a bounded set. Are you implicitly assuming that $f$ is "nice"? $\endgroup$ – user31373 Sep 8 '12 at 3:34
  • $\begingroup$ I agree with you. It was an error to claim that the supports must be unbounded. I have improved the question accordingly. $\endgroup$ – Jas Ter Sep 9 '12 at 9:21
  • $\begingroup$ Note that actually $g = v_{-} \in H^1(\mathbb{R}^3)$. You don't need to mollify. $\endgroup$ – Hans Engler Sep 9 '12 at 14:21
  • $\begingroup$ @Hans but $v_-$ is not known to be in $H^1$. (By the way, there is a typo in the definition of $v_-, v_+$ - instead of $f$ they should have $v$). $\endgroup$ – user31373 Sep 10 '12 at 5:15

The problem is, of course, that $v_+$ and $v_-$ can be closely related. Consider a cube $Q$ with sidelength $2a$ and the weight $v_Q$ such that $v_Q=1$ on its top half, $-1$ on its bottom half, and $0$ everywhere else. Then, applying the inequality $-|F(x)|^2+|F(x+a)|^2\ge -(|F(x)|+|F(x+a|)(\int_x^{x+a}|F'|)\le -\delta(|F(x)|^2+|F(x+a|^2)-\delta^{-1}a\int_{x}^{x+a}|F'|^2$ along each vertical line and integrating over the bottom half, we get $$ \int |g|^2 v_Q\ge -\delta \int_Q |g|^2-\delta^{-1}a^2\int_Q|\nabla g|^2. $$ Choosing $\delta=a$, we get $$ \int |g|^2 v_Q\ge - a\left[\int_Q |g|^2+\int_Q|\nabla g|^2\right]. $$ while $\int|v|=8a^3$. Now just consider a family of disjoint unit cubes $P_i$ tiled with cubes $Q_{ij}$ of sidelength $2a_j\to 0$ and put $v=\sum U_i v_{Q_{ij}}$ with $U_j>0$. To make sure that there exists $f\in H^1$ with $\int f^2v_+=\int f^2 v_-=+\infty$, it suffices to demand $U_j\to\infty$ (then you can build your $f$ as $\sum_i c_i f_i$ where $f_i$ is a unit size bump supported on $P_j$ and $\sum_i c_i^2<+\infty$, $\sum_i U_i c_i^2=+\infty$). On the other hand, to prevent the infinite disbalance between $\int g^2 v_\pm$, it suffices to ask that $U_ja_j$ is a bounded sequence. These two conditions are perfectly compatible, so, alas, you cannot do what you want without some extra assumptions.

  • $\begingroup$ Thanks! A very interesting construction, I learned a lot. $\endgroup$ – Jas Ter Sep 12 '12 at 15:07

When $v \in L^1_{loc}(\mathbb{R}^3)$, then $v$ is the density of an absolutely continuous signed measure $\mu$. Take a Hahn decomposition of $\mathbb{R}^3$ in two measurable sets, so that $\mathbb{R}^3$ is disjoint union of say $P$ and $N$, $P$ is positive for $\mu$ and $N$ is negative for $\mu$, defined as $\mu$ is non-negative on measurable subsets of $P$ and non-positive on measurable subsets of $N$. This gives a Jordan decomposition in measures $\mu^+$ and $\mu^-$, concentrated on $P$ and $N$ respective. Now $\mu^+$ on a measurable set $A$ is $\int_A v_+$ and $\mu_-$ on a measurable set $A$ is $\int_A v_-$. When $f \in H(\mathbb{R}^3) = L^2(R^3)$ has $\int f^2 v_+ = \infty $ and $ \int f^2 v_- = \infty$, then $f \,1_N \in L^2(\mathbb{R}^3)$, $\int f^2\,1_N v_- = \infty $ and $\int (f \, 1_N)^2 v_+ = \int f^2 \, 1_N v_+ = 0$. The not-shown measures in the integrals is Lebesgue measure on $\mathbb{R}^3$.

  • $\begingroup$ The problem is, $H^1(\mathbb R^3)$ is not $L^2(\mathbb R^3)$. It it the Sobolev space of functions with square-integrable (distributional) gradient. $\endgroup$ – user31373 Sep 10 '12 at 14:54

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