1
$\begingroup$

As you may know, there are two ways to define the Sobolev space $W^{k,p}$. One is to define $W^{k,p}$ as the set of all functions whose weak derivatives up to order $k$ are in the Lebesgue space $L^p$. The other is to define $W^{k,p}$ as the completion of a normed space of functions that can be continuously differentiated certain times. The first definition seems rigid and bears little variation in the literature; however, at least two versions of the second definition are conveyed among mathematicians. According to [Adams & Fournier, 2003], $W^{k,p}$ (actually, they use $H^{k,p}$ to distinguish this definition from the one via weak derivatives) is defined as the completion of $C^k$ in the norm $$\lVert u\rVert_{k,p}=\left(\sum_{|\alpha|\leq k}\lVert D^\alpha u\rVert_p^p\right)^\frac{1}{p}.$$ In a later chapter of their book, this definition is proved to be equivalent to the one via weak derivatives, a marvelous result, but recently I saw a definition still in the form of completions but with $C^k$ replaced by $C^\infty$. What's the difference between these two completions? Is it possible to complete two different spaces to the same space? Is the $C^\infty$ definition still equivalent to the one via weak derivatives? Any advice is appreciated. Thank you.

Update: After looking into the proof of the "$H=W$" theorem in [Adams & Fournier, 2003], I guess that these two completions are equivalent. Please see Theorem 3.17 if you have the same question.

$\endgroup$
1
  • 1
    $\begingroup$ I would rather say "there are several ways" instead of "there are two ways" ... you might not know other ways ... I would personally define $W^{k,p}$ as the space of distributions such that all the derivatives up to order $k$ in the sense of distributions are in $L^p$ (very close but formally different from weak derivatives). $\endgroup$
    – LL 3.14
    Sep 22 at 14:22

1 Answer 1

2
$\begingroup$

Yes, these are equivalent for $p<\infty$ (at least when working with functions defined on the whole space). The only result you need is the fact that $C^\infty_c(\Bbb R^d)$ is dense in $W^{k,p}(\Bbb R^d)$, and so any space in between (i.e. any space verifying $C^\infty_c(\Bbb R^d) \subset X \subset W^{k,p}(\Bbb R^d)$ is also dense in $W^{k,p}(\Bbb R^d)$.

Warning however about the case $p=\infty$ where $C^\infty_c(\Bbb R^d)$ is not dense in $W^{k,\infty}$ (the one defined with weak derivatives). Sometimes (see e.g. Triebel, Theory of functions spaces, or Maz'ya, Sobolev spaces) people write $\mathring{W}^{k,p}(\Bbb R^d)$ to denote the completion of $C^\infty_c(\Bbb R^d)$ with respect to the $W^{k,p}$ norm.

As an example, consider $L^\infty$. Then the completion of $C^\infty_c(\Bbb R^d)$ with respect to the $L^\infty$ norm is $C^0_0(\Bbb R^d)$, the space of continuous functions vanishing at infinity.

$\endgroup$

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.