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Let $q = \{q_n\}_{n\in\mathbb{Z}}$ be a real sequence and write $\mathcal{N} := \sum\limits_{i\in\mathbb{Z}}|q_i|^2$ \begin{align} \sum_{n\in\mathbb{Z}}\sum\limits_{\substack{m\in\mathbb{Z} \\ m\neq n}}\frac{|q_n - q_m|^2}{|n-m|^{1+\alpha}} &\leq \sum_{n\in\mathbb{Z}}\sum\limits_{\substack{m\in\mathbb{Z} \\ m\neq n}}\frac{2|q_n|^2 + 2|q_m|^2}{|n-m|^{1+\alpha}}\\ &= \sum\limits_{n\in\mathbb{Z}}\left [ \sum\limits_{\substack{m\in\mathbb{Z} \\ m\neq n}} \frac{2|q_n|^2}{|n-m|^{1+\alpha}} + \sum\limits_{\substack{m\in\mathbb{Z} \\ m\neq n}} \frac{2|q_m|^2}{|n-m|^{1+\alpha}} \right ]\\ &=\sum\limits_{n\in\mathbb{Z}}2|q_n|^2\sum\limits_{\substack{m\in\mathbb{Z}\\m\neq n}}\frac{1}{|n-m|^{1+\alpha}}+ \sum\limits_{n\in\mathbb{Z}}\sum\limits_{\substack{m\in\mathbb{Z} \\ m\neq n}} \frac{2|q_m|^2}{|n-m|^{1+\alpha}} \\ &= 4\mathcal{N}\zeta(1+\alpha)+\sum_{n\in\mathbb{Z}}4\mathcal{N}\zeta(1+\alpha) \end{align}
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Mon, 15 May 2023 19:56 GMT