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Let $\Omega$ be a bounded open set in $\mathbb{R}^n$ and $f \in W^{1,p} (\partial \Omega)$. Can $f$ be extended to a function $u \in W^{1,p}(\Omega)$ such that $u|_{\partial \Omega}=f$ and

$$\lVert u\rVert_{W^{1,p}(\Omega)}\leq C\lVert f\rVert_{W^{1,p}(\partial \Omega)}?$$

What are the minimal assumptions that guarantee such continuous extensions?

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    $\begingroup$ Welcome to MathOverflow! How do you even define $W^{1,p}(\partial \Omega)$ without any regularity assumptions on $\partial \Omega$? $\endgroup$
    – Jochen Glueck
    Commented Oct 3, 2019 at 15:47
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    $\begingroup$ Thanks Jochen. We may assume that $\partial \Omega$ is Lipschitz or even $C^1$. $\endgroup$
    – MathLearner
    Commented Oct 3, 2019 at 18:30

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Theorem. If $\Omega\subset\mathbb{R}^n$, $n\geq 2$, is a bounded and smooth domain, then there is a bounded extension operator $$ E:W^{1,p}(\partial\Omega)\to W^{1,q}\cap C^\infty(\Omega), \quad \text{where $1<p<\infty$ and $q=\frac{np}{n-1}$.} $$

Note that $q>p$ so $W^{1,q}(\Omega)\subset W^{1,p}(\Omega)$. Fore more details and link to a proof see: https://mathoverflow.net/a/322635/121665

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  • $\begingroup$ I don't see how the corollary follows from the stated theorem: in:mathoverflow.net/a/322635/121665. I would appreciate if you could elaborate on this. Thanks. $\endgroup$
    – MathLearner
    Commented Oct 4, 2019 at 19:48
  • $\begingroup$ @MathLearner First note that the dimension of the boundary is $n-1$ so you need to replace $n$ by $n-1$ in the theorem and then you get the exponent $q$. Secondly to get boundary of $\Omega$ from $\mathbb{R}^{n-1}$ you use partition of unity and diffeomrohisms to flatten the boundary. This is a standard technique that you find in any textbook on Sobolev spaces, but there is not enough place to explain it here in details. $\endgroup$
    – Piotr Hajlasz
    Commented Oct 5, 2019 at 4:13
  • $\begingroup$ Thank you @Piotr Hajlasz $\endgroup$
    – MathLearner
    Commented Oct 5, 2019 at 15:01

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