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I have seen the Sobolev norms on the interval $[0,2\pi]$ defined as both $$ ||u||_{H^s}^2 = \sum_{n\in\mathbb{Z}}(1+n^{2s})|\hat u_n|^2, $$ and $$ ||u||_{H^s}^2 = \sum_{n\in\mathbb{Z}}(1+n^2)^s|\hat u_n|^2. $$ So are both of these norms somehow equivalent?

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Yes. Recall that for any $a,b\geq 0$ we have \begin{align*} &0\leq s \leq 1: && (a+b)^s\leq (a^s+b^s)\leq 2^{1-s}(a+b)^s,\\ &1< s < \infty:&& 2^{1-s}(a+b)^s\leq (a^s+b^s)\leq (a+b)^s. \end{align*} Consequently, \begin{align*} &0\leq s \leq 1: && \sum_{n\in\mathbb{Z}} (1+n^{2})^s|\,\hat{u}_n|^2 \leq \sum_{n\in\mathbb{Z}} (1+n^{2s})|\,\hat{u}_n|^2\leq 2^{1-s}\sum_{n\in\mathbb{Z}} (1+n^{2})^s|\,\hat{u}_n|^2,\\ &1< s < \infty:&& 2^{1-s}\sum_{n\in\mathbb{Z}} (1+n^{2})^s|\,\hat{u}_n|^2 \leq \sum_{n\in\mathbb{Z}} (1+n^{2s})|\,\hat{u}_n|^2\leq \sum_{n\in\mathbb{Z}} (1+n^{2})^s|\,\hat{u}_n|^2. \end{align*}

Addition: I am trying to think of think of a good reference for the general inequalities above. They are for sure established in the book Inequalities by Hardy, Polya and Littlewood, but one can also show them as an excercise.

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    $\begingroup$ Are you sure the negative case is straightforward? Suppose $s=-1/2$, then as $n$ gets large the coefficient in the first series $(1+n^{-1}) \to 1$ whereas the coefficient of the second series $(1+n)^{-1} \to 0$. This is drastically different behaviour than the case of positive $s$ for which the coefficients in both terms go towards infinity as $n$ gets large. $\endgroup$ – eurocoder Oct 28 at 6:19
  • $\begingroup$ @eurocoder: You are right, the case $s<0$ is different. In this case, the first norm doesn't even make sense (due to $n^{2s}$ not being defined for $n=0$). That was a careless comment from my side (now removed) $\endgroup$ – StarBug Oct 28 at 10:51

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