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On the whole space $\mathbb{R}^d$ consider the weight $\omega(x)=\sqrt{1+|x|}$. Under which conditions on $k,q$ is the embedding $$ L^p(\mathbb{R}^d,\omega(x)dx)\subset\subset (W^{k,q}(\mathbb{R}^d,\omega(x)dx))^{\prime} $$ compact ($p$ is given)?

I am writing $(W^{k,q}(\mathbb{R}^d,\Omega(x)dx))^{\prime}$ because I am not too sure about the notation $(W^{k,q})'=W^{-k,q'}$ for such weighted spaces...

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Note that your embedding is just the adjoint of the embedding from $W^{k,q}$ to $L^{p'}$. Hence, if you know that this embedding is compact, you get the compactness of your embedding. –  gerw May 10 '13 at 9:35
Thank you. My next question is then: Assume $X\subset\subset Y$ is compact (say both are separable reflexive Banach spaces). Is it automatically true that $X'\subset Y'$ is compact too? I thought this also required density of $X$ in $Y$? –  leo monsaingeon May 10 '13 at 22:08

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