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The action-functional $\Phi(q)$ is bounded above subject to the particle number constraint $\sum\limits_{n\in\mathbb{Z}}(q_n)^2 = \mathcal{N}$, and $\omega,\alpha \in\mathbb{R} > 0$
    \begin{align}
       \Phi(q) &= \sum\limits_{n \in \mathbb{Z}}\sum\limits_{\substack{m\in\mathbb{Z} \\ m\neq n}}\frac{|q_n - q_m|^2}{|n-m|^{1+\alpha}}- \left ( \frac{\omega}{2}\sum\limits_{n\in\mathbb{Z}}q_n^2  + \sum\limits_{n\in\mathbb{Z}}(q_n)^4\right)\\
           &\leq \label{eq:4.7} \mathcal{N} \left (8\zeta(1+\alpha)- \frac{\omega}{2} \right ) + \frac{1}{4}\mathcal{N}^2
    \end{align}

Note: $\Phi(0)=0$. So if we can show the function is positive then negative (resp. negative then positive) then we have shown existence of a minimum/maximum.

$$\Phi(q)\leq\mathcal{N}\left (8\zeta(1+\alpha) - \frac{\omega}{2} + \frac{\mathcal{N}}{4}\right )$$

Since $\mathcal{N}\geq 0$, then the above is negative only when $$8\zeta(1+\alpha) - \frac{\omega}{2} + \frac{\mathcal{N}}{4} < 0$$
Then, 
\begin{align}
16\zeta(1+\alpha) - 2\omega < \mathcal{N}
\end{align}
Therefore, 
$$\Phi(q) < 0 \iff 16\zeta(1+\alpha) - 2\omega < \mathcal{N}$$

Tue, 16 May 2023 16:06 GMT