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Formulation:

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.

Edit:

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

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    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"?2012-09-08
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    I agree with you. It was an error to claim that the supports must be unbounded. I have improved the question accordingly.2012-09-09
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    Note that actually $g = v_{-} \in H^1(\mathbb{R}^3)$. You don't need to mollify.2012-09-09
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    @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 $v$).2012-09-10

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