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I'm new to Sobolev Space I'm not quite understanding the following:

Proposition: If $s$ and $s' \in \mathbb{R}, s < s'$ then we have $H^{s'}(\mathbb{R}^{n}) \subset H^{s}(\mathbb{R}^{n})$ where the inclusion is strict and where we denote the Sobolev space of order $s$ on $\mathbb{R}^{n}$ by

$$ H^{s}(\mathbb{R}^{n}):= \{u \in \mathcal{S}'(\mathbb{R}^{n}): (1+|\xi|^{2})^{\frac{s}{2}}\hat{u} \in L^{2}(\mathbb{R}^{n})\}$$

To show the inclusion is strict we use consider $u \in \mathcal{S}'(\mathbb{R}^{n})$ such that $ \hat{u}(\xi) = <\xi>^{-s-\frac{n}{2}- \epsilon}, \epsilon \in (0, s'-s)$

where $ <\xi>:=(1+|\xi|^{2})^{\frac{1}{2}}$

I don't understand how applying the above we obtain the following

$$ <\xi>^{s}\hat{u} \in L^{2}, <\xi>^{s'}\hat{u} \notin L^{2}$$

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With this choice of $u$ you have: $$\langle \xi \rangle^t \hat{u}= \langle \xi\rangle^{t-s-\frac{n}{2}-\epsilon}$$ so as by definition $\langle \xi\rangle^2=(1+|\xi|^2)$: $$\int_{\mathbb{R}^d}|\langle \xi \rangle^t \hat{u}(\xi)|^2 d \xi=\int_{\mathbb{R}^d}(1+|\xi|^2)^{t-s-\frac{n}{2}-\epsilon} d\xi$$ using polar coordinate, with $C_n$ the surface of the unit ball of dimension $n$ you obtain: $$\int_{r \in (0,+\infty)}(1+r^2)^{t-s-\frac{n}{2}-\epsilon} C_n r^{n-1}dr$$ as the integrand is continuous in $[0,+\infty)$ and: $$(1+r^2)^{t-s-\frac{n}{2}-\epsilon} r^{n-1} \sim_{r \to \infty} r^{2(t-s)-n-2\epsilon+n-1}=r^{2(t-s)-2\epsilon-1}$$ we obtain:

  • if $2(t-s)-2\epsilon-1<-1$ then integral is finite
  • if $2(t-s)-2\epsilon-1\geq-1$ then integral is infinite

and in your case:

  • $2(s-s)-2\epsilon-1<-1$
  • $2(s'-s)-2\epsilon-1=2((s'-s)-\epsilon)-1 <-1$
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