1
$\begingroup$

On page 270 of Evans' "Partial differential equations", the Extension theorem of the Sobolev functions states that

For any bounded domain $U$ with $C^1$ boundary and any bounded domain $V$ with $U\Subset V$, there exists a bounded linear operator $$ E: W^{1,p}(U)\to W^{1,p}(\mathbb{R}^n)$$ such that

  1. $Eu=u$ a.e. in $U$;
  2. $Eu$ has support in $V$;
  3. $\|Eu\|_{W^{1,p}}\leq C\|u\|_{W^{1,p}}$.

My question is, since the continuation is not unique, is the axiom of choice involved in the proof of the existence of the linear operator? Otherwise, how do I know which extension is the function $u$ mapped into?

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

The extension operator defined by Evans is fully deterministic. Essentially, he (cleverly) reflects the function along the boundary and then multiplies with a cut-off function to make sure that the support remains in $V$.

The proof relies on choosing a partition of unity of the boundary. Using a different partition will give a different extension operator, but as long as this is fixed there is no axiom of choice involved.

$\endgroup$
1
  • 2
    $\begingroup$ Also the construction of the reflection is by no means unique. $\endgroup$ Commented yesterday

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.