# Proof of the Meyers–Serrin theorem (the "$H=W$" theorem)

I'm working on the proof of the "$$H=W$$" theorem in the book SOBOLEV SPACES by Adams and Fournier, and there is an argument that seems quite strange to me. Let me introduce some conventions first. $$H^{m,p}(\Omega)$$ is defined as the completion of $$\{u\in C^m(\Omega):\lVert u\rVert_{m,p}<\infty\}$$ w.r.t. the norm $$\lVert u\rVert_{m,p}=\left(\sum_{|\alpha|\leq m}\lVert D^\alpha u\rVert_p^p\right)^\frac{1}{p},$$ where $$D^\alpha$$ is understood in the weak sense, while $$W^{m,p}(\Omega)$$ is defined by weak derivatives in the usual way. As to completions, the authors mentioned in the preliminary chapter that every normed space $$X$$ is either a Banach space or a dense subset of a Banach space $$Y$$ called the completion of $$X$$ whose norm satisfies $$\lVert x\rVert_Y=\lVert x\rVert_X$$ for every $$x\in X$$. Finally, throughout our discussion, $$\Omega$$ denotes a nonempty open subset of the Euclidean space. Now let us see the proof to be understood.

The argument underlined with red is much annoying. I don't see any relevance between completeness of $$W^{m,p}(\Omega)$$ and the extension to the asserted isometric isomorphism. If the identity operator on $$S$$ means the inclusion of $$S$$ in $$W^{m,p}(\Omega)$$, how could I extend it to the isometric isomorphism? Before digging into the proof, I have reviewed the preliminary chapter, so I know the normed space $$S$$ is isometrically isomorphic to a dense subspace of a Banach space $$H^{m,p}(\Omega)$$? But this doesn't seem to be any helpful. How could I use this fact to build the extension. Help is much needed. Thank you for your precious time.

Update 1: I remember that if the codomain $$T$$ of a uniformly continuous map $$f:A\subseteq M\to T$$ is complete, then we can uniquely extend $$f$$ to the closure $$\bar{A}$$, preserving uniform continuity. But this seems to give nothing if I extend the inclusion of $$S$$ in $$W$$.

Update 2: Though it may be obvious to people who know about completions of normed spaces, I'd like to emphasize for once that every normed space $$X$$ is in fact isometrically isomorphic to a dense subspace of a Banach space and it is this Banach space that is defined as the completion of $$X$$. Then, by identifying isomorphic spaces, the authors conclude that every normed space is a dense subset of its completion, I guess.

• A problem with the completion of a normed spaces is that the definite article is misleading: It is not unique literally but only unique up to unique ismomorphisms (and this isomorohism is then even an isometry). The theorem $H=W$ should thus be formulated correctly as: The inclusion $i:C\hookrightarrow W$ is a completion. Sep 26 at 9:34

Let's start from $$S\subseteq W^{m,p}(\Omega)$$.
Consider the closure $$\bar{S}$$ of $$S$$ with respect to the topology on $$W^{m,p}(\Omega)$$ induced by its norm.

Claim 1: $$\bar{S}$$ is a linear subspace of $$W^{m,p}(\Omega)$$.
Proof. Check that $$\bar{S}$$ is closed under addition and scalar multiplication.

Claim 2: $$\bar{S}$$ is complete (hence it's a Banach space).
Proof. If you take a Cauchy sequence in $$\bar{S}$$, then it is Cauchy also in $$W^{m,p}(\Omega)$$, as the norm we are considering on $$\bar{S}$$ is the same as the one of $$W^{m,p}(\Omega)$$. Since $$W^{m,p}(\Omega)$$ is a Banach space, our Cauchy sequence needs to converge in $$W^{m,p}(\Omega)$$, but as $$\bar{S}$$ is closed, it must converge actually in $$\bar{S}$$.

Claim 3. $$\bar{S}$$ is the completion of $$S$$.
Proof. $$S$$ is clearly dense in $$\bar{S}$$, which is a Banach space by our second claim. Thus $$\bar{S}$$ is a completion of $$S$$, but it is also the unique one up to isometric isomorphism (because in general the completion of a normed vector spaces is unique up to isometric isomorphism). Thus we can identify $$H^{m,p}(\Omega)$$ with $$\bar{S}.$$

Hence $$H^{m,p}(\Omega)\subseteq W^{m,p}(\Omega)$$.
The step underlined in red is saying exactly this, with a different phrasing.

• Ha! I totally forgot uniqueness of completions! Now it looks like a nonsense to consider the property in Update 1. That's ridiculous. 2 days ago