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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{|q_n|^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{|q_n|^2}{|n-m|^{1+\alpha}} + \sum\limits_{\substack{m\in\mathbb{Z} \\ m\neq n}} \frac{|q_m|^2}{|n-m|^{1+\alpha}} \right ]\\ &= \sum\limits_{n\in\mathbb{Z}}\left [|q_n|^2 \sum\limits_{\substack{m\in\mathbb{Z} \\ m\neq n}}\frac{1}{|n-m|^{1+\alpha}} + \sum\limits_{\substack{m\in\mathbb{Z} \\ m\neq n}}|q_m|^2 \cdot \sum\limits_{\substack{m\in\mathbb{Z} \\ m\neq n}}\frac{1}{|n-m|^{1+\alpha}}\right ]\\ &= \zeta(1+\alpha) \left( \sum\limits_{n\in\mathbb{Z}}|q_n|^2 + \sum\limits_{n\in\mathbb{Z}}\sum\limits_{\substack{m\in\mathbb{Z} \\ m\neq n}}|q_m|^2 \right )\\ &= \zeta(1+\alpha)\left ( \mathcal{N} + \sum_{n\in\mathbb{Z}}\mathcal{N}\right ) \end{align}
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Mon, 15 May 2023 18:45 GMT