# Lieb: Stability of matter, problem with variational method

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 $\eta$.
• 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. Feb 26, 2014 at 17:01
• 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). Feb 26, 2014 at 18:12