Let $\Omega \subset R^n$ be a bounded region with Lipschitz boundary. Is the trace operator $T: W^{1,1}(\Omega)\rightarrow L^1(\partial \Omega)$ surjective? If not, what is the image?
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2 Answers
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The trace operator $T:W^{1,1}(\Omega)\to L^1(\partial\Omega)$ is surjective.
This is a classical result of Gagliardo [1], see Theorem 18.13 in [2].
In other words for every function $g\in L^1(\partial\Omega)$ there is a function $Eg\in W^{1,1}(\Omega)$ such that $T(Eg)=g$. However, as was proved by Peetre [3], there is no bounded linear extension operator $E:L^1(\partial\Omega)\to W^{1,1}(\Omega)$. That is the extension operator $E$ is non-linear.
The original proof of Peetre is difficult to understand, but there are more direct (still difficult) proofs, see e.g. [4].
[1] E. Gagliardo, Caratterizzazioni delle tracce sulla frontiera relative ad alcune classi di funzioni in n variabili. Rend. Sem. Mat. Univ. Padova 27 1957 284–305.
[2] G. Leoni, A first course in Sobolev spaces. Second edition. Graduate Studies in Mathematics, 181. American Mathematical Society, Providence, RI, 2017.
[3] J. Peetre, A counterexample connected with Gagliardo's trace theorem. Special issue dedicated to Władysław Orlicz on the occasion of his seventy-fifth birthday. Comment. Math. Special Issue 2 (1979), 277–282.
[4] A. Pełczyński, M. Wojciechowski, Sobolev spaces in several variables in $L^1$-type norms are not isomorphic to Banach lattices. Ark. Mat. 40 (2002), 363–382.
answered May 6, 2018 at 18:39
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It is surjective for a $C^1$ boundary, see Demengel & Demengel, section 3.3. Following the proof will probably answer your question.
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1$\begingroup$ You could correct the link to the book by Demengel & Demengel, because it does not work. $\endgroup$ Commented May 7, 2018 at 1:05
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