MathB.in - Share Mathematics with LaTeX and Markdown

$\mathcal{N} = \sum_{n \in \mathbb{Z}}|q_n|^2$

\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  + \frac{1}{4}\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}

The upper-bound must obtain its minimum value in  $I = (0,2\omega - 16\zeta(1+\alpha))$ because of both $\mathcal{N}\geq 0$ and being negative in that interval. Let $\mathcal{N}^{*}$ denote the minimum of the upper-bound.
 \begin{align*}
     \frac{d}{d\mathcal{N}} \left [ \mathcal{N} \left ( 8\zeta(1+\alpha) - \frac{\omega}{2}\right ) + \frac{\mathcal{N}^2}{4}\right ] &=0\\
     \implies \mathcal{N^{*}} &= \omega -16\zeta(1+\alpha)
 \end{align*}
 For $\mathcal{N}\in \mathbb{R}\setminus I$, the upper-bound is positive and monotonically increasing as the quadratic term dominates. Hence, the global minimum is achieved for $\mathcal{N^*}=\omega -16\zeta(1+\alpha)$.

Does the above imply $\Phi$ has a critical point? If so, can we find it?

Wed, 17 May 2023 00:42 GMT