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I will consider Sobolev spaces with $p=2$, only, so that they are Hilbert spaces. Hence the Sobolev inner product identifies each Sobolev space with its dual. In other words, I have an isomorphism $H^m\to (H^m)^\ast$ given by $x\mapsto \langle x,\cdot\rangle_m$

Now, if $\sigma>0$, then I have an embedding $H^{m+\sigma}\hookrightarrow H^m$. Under the above isomorphism, how can I describe the image of $H^{m+\sigma}$ inside $(H^m)^\ast$? In particular, is there a $\tau>0$ such that $H^{m+\sigma}$ it identified with $(H^{m-\tau})^\ast$?

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    $\begingroup$ I think $H^m$ is more common notation than $W_m$ for this case. $\endgroup$
    – Nate Eldredge
    May 19, 2010 at 22:45
  • $\begingroup$ Maybe I'm misunderstanding your last sentence, but since there is an injection $W_{m} \to W_{m-\tau}$, the map $W^*_{m-\tau} \to W^*_m$ is a surjection for any $\tau \geq 0$. Notice also that $W_{m+\sigma}$ is dense in $W_m$ (so similarly for its image in the dual). $\endgroup$
    – Joel Fine
    May 19, 2010 at 23:36
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    $\begingroup$ No, I think the embedding $W_m\to W_{m-\tau}$ gives rise to another embedding $(W_{m-\tau})^\ast\to W_m$ on the dual spaces, precisely because $W_m\to W_{m-\tau}$ is dense. $\endgroup$
    – euklid345
    May 19, 2010 at 23:50
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    $\begingroup$ The question is: what is the embedding of $W_{m-\tau}^*$ inside $W_m^*$? As Harald indicated below, there is an isomorphism of $W_{m-\tau}^*$ with $W_m^*$, but presumably that's not what you want. So ask yourself what is the inclusion you are using, and the answer should be clear. $\endgroup$
    – Willie Wong
    May 20, 2010 at 8:54

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I am going to stick with the standard terminology $H^m$ here. Taking Fourier transforms one finds that $$\langle u,v\rangle_m=\int\hat u(\xi)\bar{\hat v}(\xi)(1+|\xi|^2)^m\,d\xi$$ (give or take the odd multiplicative constant), where $H^m$ consists precisely of those $u\in L^2$ for which $\langle u,u\rangle<\infty$. This works even for $m<0$, if you allow distributions whose Fourier transforms are functions. Everything follows from this, including the fact that $H^{-m}$ acts as the dual of $H^m$ simply by the distribution $u$ acting on the function $v$, which corresponds to the integral $$\langle u,v\rangle=\int \hat u(\xi)\bar{\hat v}(\xi)\,d\xi=\int \hat u(\xi)(1+|\xi|^2)^{-m/2}\cdot\hat{\bar v}(\xi)(1+|\xi|^2)^{m/2}\,d\xi$$ where I have split up the integrand into a product of two $L^2$ functions.

For this reason, it seems more natural to identify $H^{-m}$ with the dual of $H^m$ than to identify $H^m$ with its own dual. However, you can go ahead and identify any Sobolev space with the dual of any other just by inserting a suitable power of $1+|\xi|^2$ in the integral defining the pairing between the two.

Rather than coming straight out and answering your question, I'll leave it to you to ponder the consequences of the above. In particular, note that you we embed and identify you have to keep careful track of what space you have identified with whose dual, or you will be endlessly befuddled.

Addendum: To spell out a more direct answer to your question, $\langle\cdot,\cdot\rangle_m$ can identify $H^{m+\sigma}$ with the dual of $H^{m-\sigma}$, since we can write $$\langle u,v\rangle_m=\int \hat u(\xi)(1+|\xi|^2)^{(m-\sigma)/2}\cdot\bar{\hat v}(\xi)(1+|\xi|^2)^{(m+\sigma)/2}\,d\xi$$ where I have split the integrand into a product of two $L^2$ functions.

Edit: Changed a couple $\hat{\bar v}$ into $\bar{\hat v}$.

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  • $\begingroup$ Yes, I am endlessly befuddled. That's why I asked the question. :) I am aware that if I use the inner product $\langle x,y\rangle_0$, I end up identifying $(H^\sigma)^\ast$ with $H^{-\sigma}$. That's the case $m=0$ of my question. What about $m>0$? $\endgroup$
    – euklid345
    May 20, 2010 at 1:17
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    $\begingroup$ Why was Harald's answer voted down? To make what he is saying more explicit: let $k(x)$ be some smooth non-vanishing function, then the function space defined by the inner product $\langle f,g\rangle = \int \bar{f} \cdot g~ |k|^2dx$ has an isomorphism with $L^2$ just by sending $f\to kf$. So doing this on the Fourier side you can establish an identification that sends all $H^s \to H^{s+\sigma}$. Noting that the whole diagram (shifting $s$, and taking-the-dual-and-identifying-w.r.t.-a-norm) commutes what you seek should be clear. $\endgroup$
    – Willie Wong
    May 20, 2010 at 8:39
  • $\begingroup$ It might have been voted down because I did not give a detailed and complete answer. Anyhow, since the OP admits still to be befuddled, I added a bit of detail that hopefully makes it clear how to apply my general answer to his specific question. $\endgroup$
    – Harald Hanche-Olsen
    May 20, 2010 at 15:38

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