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Let $D\subset\mathbb{R}^d$ be an open domain and let consider the open cylinder $D\times (0,T)\subset\mathbb{R}^{d+1}$ where $T\in (0,+\infty)$ arbitrary. Let $H^{1}(D\times (0,T))$ be the Sobolev space defined as usual. It is known that: if we assume that $D$ satisfies the cone condition, then we have the continuous embeding $$H^{1}(D\times (0,T))\mapsto L^{2}(D\times \{\tau=t \})$$ for every $t\in (0,T)$ (see the book of Adams).

My question is: If $D$ is an unbounded domain with finite volume and it hasn't the cone condition,does the above embedding still true? and under which assypmtions? Thanks.

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    $\begingroup$ What do you mean by "the continuous embedding"? Shouldn't it be a well-defined version of $f \mapsto f(\cdot,t)$? This is never injective and hence, I would not call this an embedding. $\endgroup$
    – Jochen Wengenroth
    Commented Mar 1, 2018 at 12:21
  • $\begingroup$ It means that: $||f(.,t)||_{L^{2}(D)}\leq k ||f||_{H^{1}(D)}$ with $K$ is a constant $\endgroup$
    – Rym Touibi
    Commented Mar 1, 2018 at 13:31
  • $\begingroup$ Hence, for every $t\in (0,T)$, we get $x\mapsto f(x,t) \in L^{2}(D) ,$ what I need. $\endgroup$
    – Rym Touibi
    Commented Mar 1, 2018 at 13:57

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