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Let $q = \{q_n\}_{n\in\mathbb{Z}}$ be a real sequence and $\alpha > 0$. 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}}\sum\limits_{\substack{m\in\mathbb{Z}\\m\neq n}}\frac{2|q_n|^2}{|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}} \\
&=\sum\limits_{n\in\mathbb{Z}}\sum\limits_{\substack{m\in\mathbb{Z}\\m\neq n}}\frac{2|q_n|^2}{|n-m|^{1+\alpha}}+ \sum\limits_{m\in\mathbb{Z}}\sum\limits_{\substack{n\in\mathbb{Z} \\ m\neq n}}  \frac{2|q_m|^2}{|n-m|^{1+\alpha}} \\
&=\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_{m\in\mathbb{Z}} 2|q_m|^2 \sum\limits_{\substack{n\in\mathbb{Z} \\ m\neq n}} \frac{1}{|n-m|^{1+\alpha}} \\
&=2\mathcal{N}\cdot 2\zeta(1+\alpha) + 2\mathcal{N}\cdot 2\zeta(1+\alpha) \\
&=8\mathcal{N}\zeta(1+\alpha)
\end{align}

Mon, 15 May 2023 23:05 GMT