# Are the norms $\sum_{n\in\mathbb{Z}}(1+n^{2s})|\hat u_n|^2$ and $\sum_{n\in\mathbb{Z}}(1+n^2)^s|\hat u_n|^2$ equivalent?

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?

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*}
• 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. – ManUtdBloke Oct 28 '19 at 6:19
• @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) – StarBug Oct 28 '19 at 10:51