I am trying to recalculate 'Stability of Matter' Paper from Lieb.

I have a problem at the last step of the proof at page 3, where Lieb calculates the minimizer. I will explain my ideas and my problem so far.

I have the following functional to be minimized on $\rho\in L^1(\mathbb R^d)$. Here $\lambda$ is a Lagrange multiplier and $\rho\geq 0$.

$h(\rho) = \frac{1}{C_d}\left(\int_{\mathbb R^d} dx \rho(x)^\frac{d}{d-2}\right)^\frac{d-2}{d} - Z \int_{\mathbb R^d} dx \frac{\rho (x)}{|x|} +\lambda\left( \int_{\mathbb R^d} dx \rho(x) -1\right)$

My idea is the following:

$\frac{\text{d}}{\text{d}t}h(\rho_m +t \eta)|_{t=0} =0$, where $\rho_m$ is the minimizer and $\eta\in C_0^\infty(\mathbb R^d)$ arbitrary.

Then I obtain the following:

$0 = \int dx \,\eta(x) \left(\frac{1}{C_d}\|\rho_m\|_\frac{d}{d-2}^\frac{-2}{d-2}\rho_m(x)^\frac{2}{d-2} - \frac{Z}{|x|} + \lambda\right)$

Then by the fundamental lemma of variations it follows that:

$\frac{1}{C_d}\|\rho_m\|_\frac{d}{d-2}^\frac{-2}{d-2} \rho_m(x)^\frac{2}{d-2} - \frac{Z}{|x|} + \lambda = 0$

Now I use that $\rho_m\geq 0$ and therefore $\frac{Z}{|x|} - \lambda\geq 0$. Since one also sees that $\lambda$ has to be bigger or equal 0 as otherwise $\rho_m$ wouldn't be in $L^1$. I conclude that $\rho_m$ has the following form:

$\rho_m (x) =\begin{cases} C_d^\frac{d-2}{2} \|\rho_m\|_\frac{d}{d-2} \left(\frac{Z}{|x|} - \lambda \right)^\frac{d-2}{2} & \text{ if} |x|\leq \frac{Z}{\lambda} \\ 0 & \text{ if} |x|> \frac{Z}{\lambda} \end{cases}$

Now comes the point of my concerns: This $\rho_m$ has compact support and therefore cannot really satisfy $0 = \int dx \,\eta(x) \left(\frac{1}{C_d}\|\rho_m\|_\frac{d}{d-2}^\frac{-2}{d-2}\rho_m(x)^\frac{2}{d-2} - \frac{Z}{|x|} + \lambda\right)$. Is there a reason why this $\rho_m$ is however the correct minimizer?

Best wishes :)


You are minimizing under the constraint that $\rho \ge 0$. Hence your variation $\rho_m + t \eta$ might not be admissible. (Say if $\eta \ne 0$ and $\vert t\vert$ is large enough.)

The trick is to consider either any $\eta$ with $t \ge 0$, wich will give you the inequality $$ \frac{1}{C_d} \Vert \rho_m \Vert_{d/(d - 2)}^{-2/(d - 2)} \rho_m (x)^{2/(d - 2)} - \frac{Z}{\vert x \vert} + \lambda \ge 0. $$ or consider $\eta \in L^{\infty} (\mathbb{R}^d) \cap L^1 (\mathbb{R}^d)$ such that $$ \inf \{\rho_m (x) : \eta (x) \ne 0\} > 0. $$ Then, there exists $\tau > 0$ such that $u + t \eta \ge 0$ for each $t \in (-\tau, \tau)$. One concludes therefrom that if $\rho_m (x) > 0$, then $$ \frac{1}{C_d} \Vert \rho_m \Vert_{d/(d - 2)}^{-2/(d - 2)} \rho_m (x)^{2/(d - 2)} - \frac{Z}{\vert x \vert} + \lambda = 0. $$ This brings you to the desired solution, and the function satisfies the necessary conditions of extremum even if its support is compact.

Since the regularity of $\rho_m$ is not known a priori, it is important to take nonsmooth variations $\eta$.

  • $\begingroup$ Thanks a lot for your Answer! Do I not need a essinf instead of the normal inf? And I don't really see why it is not sufficient to perturbate with smooth functions. I need the fundamentallemma of variation, which need smooth functions. The derivative is still the correct step right? I just consider smalls ts? $\endgroup$ – Peter Feb 26 '14 at 14:33
  • 1
    $\begingroup$ If you consider functions in Lebesgue spaces as equivalence classes, an essinf would indeed be better defined. Instead of the fundamental lemma of calculus of variations you can just differentiate the integral with standard tools such as Lebesgue's dominated convergence theorem. $\endgroup$ – Jean Van Schaftingen Feb 26 '14 at 16:46
  • $\begingroup$ Thanks, is it so sufficient to consider $\eta \text{ as a }C_0^\infty$ functions or why should $\eta \in L^1 \cup L^\infty$? Furthermore I know that if $\rho_m(x) > 0 $ then I know how it looks like but I still don't know where $\rho_m(x)$ is greater than zero. $\endgroup$ – Peter Feb 26 '14 at 17:01
  • $\begingroup$ If $\rho_m$ is too rough, then smooth $\eta$ will not do it. Imagine for instance that $\rho$ is a characteristic function of a Cantor set of positive measure $C$. Then it is not possible to have $\eta$ which is supported in $C$ (since $C$ has empty interior). $\endgroup$ – Jean Van Schaftingen Feb 26 '14 at 18:12
  • $\begingroup$ This should work, as just taking the constant function 1, which is smooth. $\endgroup$ – Peter Feb 26 '14 at 21:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.