MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...

share|cite|improve this question
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.