I have read two definitions of Sobolev spaces.

Definition 1: We let $\lambda$ denote $\lambda^s(\xi)=(1+|\xi|^2)^\frac{s}{2}$ for $s \in \Bbb R$, $\xi \in \Bbb R^n$. We say that $u \in H^s$, if $u \in S'$ and $$||\lambda^s \hat{u} ||_2= (2 \pi)^{-n} \int (1 + |\xi|^2 )^s |\hat{u}(\xi)|^2 \, d \xi < \infty$$ under the identification of $L^2$ in $ S'$. $S'$ is the dual of the Schawrtz Space on $\Bbb R^n$, known as temperate distribution , with respect to the norm on Schawrtz space.

What differentiates this definition to the second definition:

Definition 2: On the Schwartz space, the Sobolev space $W^s$ is the completion of $S$ with respect to the $s$-norm. $$||u||_s^2 := (2 \pi)^{-n} \int (1 + |\xi|^2 )^s |\hat{u}(\xi)|^2 \, d \xi $$

Is that the first definition is working with topologicial dual on Schwartz space whilst latter is working directly with Schwartz space.

Are these the same? Why do we have a $H$ snd a $W$?

Sources: First definition is from Xavier's, Introudction to Pseudodifferential Manifolds, Second Definition is From Ebert's notes. I am quite confused, as both of these definitions do not appear in Sobolev spaces.

As suggested by user Rhys:

So we have a map $S \rightarrow S'$, given $$\phi \mapsto u_\phi= \left( \psi \mapsto \int \psi \bar{\phi} \right)$$ $$||\hat{u}_\phi||_2 = ||u_\phi||$$ by construction. It suffices to show $S$ in the $||\cdot||_2$ norm is

(i) Dense in $H^s$ (ii) and $H^s$ is complete.
Are these true?

  • $\begingroup$ You did not actually pose any question, but I guess what you want to know is why and how these two definition describe the same thing? $\endgroup$ – Vincent Apr 17 at 8:40
  • $\begingroup$ Yes, thanks a lot. $\endgroup$ – Cy L Shih Apr 17 at 9:05
  • $\begingroup$ $\mathcal{S}'$ is usually the topological dual space of $\mathcal{S}$. Are you sure you mean to write that $\mathcal{S}'$ is the space of seminorms on $\mathcal{S}$? (this is not even a vector space since seminorms must be non-negative) $\endgroup$ – Rhys Steele Apr 17 at 9:53
  • $\begingroup$ Thanks Rhys, I have edited. $\endgroup$ – Cy L Shih Apr 17 at 10:16
  • 1
    $\begingroup$ Here's a sketch of how to prove this. First check that definition $1$ gives you a complete space. Secondly, notice that you can consider $\mathcal{S}$ as a subset of $\mathcal{S}'$ by identifying it with its image under the map $\phi \mapsto \Phi$ where $\Phi(\psi) := \int \phi \psi dx$. Finally, check that under this identification $\mathcal{S}$ is a dense subset of $H^s$ (as defined by definition $1$). This gives that $H^s$ as in definition 1 is the completion that appears in definition $2$. $\endgroup$ – Rhys Steele Apr 17 at 11:56

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