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.

enter image description here

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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Jochen
    Sep 26 at 9:34

1 Answer 1


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.