Let $\Omega$ be a domain in $\mathbb{R}^3$, does there exist a vector-valued Sobolev-type space, or maybe space in other sense, $V$, satisfying the following:
1) $S:=\lbrace v\in V:\|v\|_{L^\infty(\Omega)}\leq 1\rbrace$ is bounded in $V$;
2) $V$ has one degree of regularity, or maybe weaker, so that the divergence can be defined, for example, $\textrm{div}v\in L^2(\Omega)$;
3) The trace of $v$ can be suitbly defined on $\partial\Omega$.
4) $C_0^\infty(\Omega)$ is dense in $V$.
5) $V$ is reflexive.



1 Answer 1


I think that such a space will not exists. In particular your requirements (1) and (2) contradicts in the following sense: (1) says basically, that the norm in $V$ is weaker than the $L^\infty(\Omega)$-norm, whereas (2) requires that you can control the divergence by the norm in $V$. Imho, this is not possible.

  • $\begingroup$ In general, the $L^\infty$ norm can be controlled by the Sobolev norm within the right parameters, but the converse cannot be true. Sobolev functions have some nice properties of the derivatives, but $L^\infty$ (even continuous, H$\"o$lder continuous) can have pathologically bad derivatives. $\endgroup$ Apr 24, 2013 at 7:37

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