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I am having a problem with the definition of the space $W^{-k,p}$. I use Adams's definition

$$ W^{-k,p} = \left\{T \in D'(\Omega) \ \middle| \sum \limits_{0 \leq |i| \leq k} (-1)^{|i|} \int_{\Omega} v_i D^{i} \phi \,dx \ \forall \phi \in D(\Omega), 0 \leq |i| \leq k\right\}, $$

and he says in his book Sobolev spaces that the dual space $(W_0^{k,q})'$ is isometrically isomorphic to $W^{-k,p}$ for all $1 \leq q < \infty$, where $q$ is the conjugate index of $p$. My question is if this is true also for $q = \infty$. I would really appreciate if you could tell me a source where this is proven/ claimed. This would really help me because I am trying to show that $W^{-1,1} \subset H^{-k}$ for some $k$. Do you have any other idea how I could show this embedding without using $W^{-1,1}$ as a dual space. Thank you very much in advance.

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$W^{-k,p}$ is the dual of $W_0^{k,q}$ if $p>1$. For $p=1$ this is not a natural definition. You should use the alternative definition that $W^{-1,1}$ is the set of all distributions of the form $f_0+\sum_m{\partial f_m\over\partial x_m}$, where $f_0,f_m\in L^1$. Using this definition for embeddings into $H^{-k}$ spaces presents no problem.

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  • $\begingroup$ Thank you very much for your quick reply! Unfortunately for me it is a problem. I don't think I understand the inclusion. I don't see that $H^{-k}$ is bigger than $W^{-1,1}$. Since $L^1$ is bigger than $L^2$, how can the distributional derivatives of elements that are in $L^1$ but not $L^2$ be in $H^{-k}$? I am a bit lost and would really appreciate if you could explain it more detailed. $\endgroup$
    – user292207
    Nov 24, 2015 at 17:21
  • $\begingroup$ $W^{k,1}$ embeds into $L^2$ if $k$ is large enough (Sobolev embedding). It follows that $W^{-1,1}$ embeds into $H^{-k-1}$. $\endgroup$ Nov 24, 2015 at 18:04

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