# Definitions of Sobolev Spaces - are they the same?

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?

• 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? – Vincent Apr 17 at 8:40
• Yes, thanks a lot. – Cy L Shih Apr 17 at 9:05
• $\mathcal{S}'$ 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) – Rhys Steele Apr 17 at 9:53
• Thanks Rhys, I have edited. – Cy L Shih Apr 17 at 10:16
• 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 $2$. – Rhys Steele Apr 17 at 11:56