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I am working through a proof which shows that the usual Sobolev norm $\|\cdot\|_{k,2}$ on $W^{k,2}(\Omega)$ is equivalent to the norm

$$ \|u\|_{H^m} := \| (1 + |\cdot|^2)^{\frac{k}{2}}\mathcal{F}u \|_2 $$

At one point we use the inequality

$$ 2^{-k} \leq \frac{1 + |\xi|^{2k}}{(1 + |\xi|^2)^k} \leq 2. $$

Since the fraction always is smaller than one, the right inequality is not a problem. But I can't find a way to prove the left inequality. Is there an elementary proof to this inequality?

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    $\begingroup$ $\|f\|^2_{W^{k,2}} = \sum_{|m|\le k} \|\partial_m f\|^2_{L^2} = \sum_{|m|\le k} \|\ |\xi|^m \hat{f}\|^2_{L^2} \sim\|\ (1+|\xi|)^{k} \hat{f}\|^2_{L^2}$ (where $\sim$ means the two norms are obviously equivalent) $\endgroup$ – reuns Sep 21 '16 at 17:28
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Yes. Distinguish the cases $|\xi|\leq 1$ and $|\xi|> 1$ and use crude bounds.

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  • $\begingroup$ Nice. Thanks a lot. $\endgroup$ – el_tenedor Sep 21 '16 at 17:34

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