Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

What is the dual space of $ H¹(\Omega) = W^{1,2}(\Omega) $? What is the dual space of $ W^{m,p}(\Omega) $? I know for example that the dual space of $ L^{p}(\Omega) $ for $ 1 \le p < \infty $ is $ L^{q}(\Omega) $ where $ \dfrac{1}{p} + \dfrac{1}{q} = 1 $.

share|cite|improve this question
up vote 6 down vote accepted

The dual space of any topological vector space $X$ is the space of all bounded linear forms on $X$. Hence, the dual space of $W^{m,p}(\Omega)$ is the space of all linear $F:W^{m,p}(\Omega)\to \mathbb{R}$ such that $|F(v)|\leq C\|v\|_{W^{m,p}}$ for all $v\in W^{m,p}(\Omega)$.

If you are looking for explicit characterizations:

  • In the Hilbert space setting $p=2$, you can of course identify $H^m(\Omega)$ with its dual using the Riesz representation theorem (see also this question).
  • For $p=2$, you can also construct Sobolev spaces $H^s$ for any real $s$ using Fourier transforms. Then you can show that the dual of $H^s$ is $H^{-s}$.
  • If $p>n$ for $\Omega\subset\mathbb{R}^n$, then $W^{1,p}(\Omega)$ embeds densely into the continuous functions, and hence its dual space contains the space of regular Borel measures (such as the Dirac delta).

For details, I recommend Adams, R.A. and Fournier, J.J.F, Sobolev Spaces, 2nd ed., Academic Press, 2003.

share|cite|improve this answer
It's not a good idea to think that the Riesz-Fischer (conjugate...) identification of every Hilbert space with its dual behaves well with respect to maps among Hilbert spaces and adjoints thereof. It is rarely compatible... In particular, it is wildly incompatible with the inclusion $H^s\rightarrow H^{-s}$ and such. – paul garrett Jul 17 '13 at 0:51

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.