# Dual of Sobolev space $W^{1,p}(U)$ for $U$ an arbitrary subset of $\mathbb R^n$

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!

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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

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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.)

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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