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

this question may be shameful, but nevertheless I can't help myself.

Let $U \subset \mathbb R^n$ be arbitrary, in particular not the whole of the space itself. I wonder about the dual of the space $W^{1,p}(U)$, for $p < \infty$.

For $U = \mathbb R^n$, we have $(W^{1,p})' = (W^{1,p'})$ with $p' = \frac{p}{p-1}$. How about different $U$?

For example, in case $U = B_1(0)$ being the closed $1$-Ball, it seems the dual is not a function space. Just recall that the trace is well-defined, linear and continuous on $W^{1,p}(U)$ and, with $S_1$ the boundary of $B_1(0)$ and $w \in L^p(S_1)$, we are given are continuous linear functional by

$ W^{1,p}(B_1(0)) \longrightarrow \mathbb C \, , f \mapsto \int_{S_1} w \cdot tr f dx $.

In fact, I wouldn't be surprised if the above example were somehow prototypical, but I have no clue how to proceed from this point. I regard this relevant, as these spaces are ubiquitous in analysis.

Thank you!

share|cite|improve this question

How about embedding $W^{1,p}$ in $L^p\times (L^p)^n$ and using Hahn-Banach and Riesz' representation theorem to get a nice characterization of elements in the dual

share|cite|improve this answer
Let $X$ be the (closed) image of the embedding into $Z = L^p \times (L^p)^n$ as described by you. Let $Y$ be the subspace of $Z'$ that vanishes on $X$. Then $X' \equiv Z'/Y$. We have $Z' = L^q \times (L^q)^n$. How do you employ the Hahn-Banach theorem? – shuhalo Dec 17 '10 at 1:54
With your notation, one can translate a functional $T$ in $W^{1,p}$ to one in $X$, say $S$, by the formula $Sx=Tw$ with $x$ the image of $w$ under the embedding. Using Hahn-Banach we extend this functional to $Z$ (call it $S'$), but by the Riesz representation theorem we find $u_0,...,u_n\in L^p$ such that $S'x=\int u_0x_0 + \sum_{k=1}^{n} \int u_kx_k$, because of the definition of $x$ we get $Tw=Sx=S'x=\int u_0w +\sum_{k=1}^{n} u_k \partial ^kw$ – Jose27 Dec 17 '10 at 19:58
+1, I like this approach. – Jonas Teuwen Jan 2 '11 at 15:44

I think you can find the answer yourself - I will just give you a hint: You are considering the wrong pairing between $W^{1,p}(U)$ and $W^{1,p'}(U)$. (The pairing you consider doesn't give you the desired isomorphism even in the $\mathbb R^n$ case.)

share|cite|improve this answer
You are right. $W^{1,p'} \subset (W^{1,p})'$, whilst for cases of sufficient regularity, as in case $p > n$ by Morrey's lemma, the dirac delta is already a continuous dual vector. However, I find it hard to give the space $(W^{1,p}(\mathbb R^n))'$ a 'nice' characterization. – shuhalo Dec 16 '10 at 16:04

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.