The dual of the Sobolev space $W^{k,p}$

The dual of the Sobolev space if defined to be

$$(W^{k,p}(\Omega))' = W_0^{-k,p'}(\Omega)$$

where $\frac 1 p + \frac 1 {p'} = 1$.

Why makes this definition sense, especially why do we have $L^{p'}$-functions on right side that vanish on $\partial \Omega$?

• I don't think this is right. Under the usual identifications, the right side should have a $-k$. But the requirement to vanish on $\partial \Omega$ comes from the fact that integration by parts introduces a boundary term, and you want that term to vanish. – Nate Eldredge Jul 17 '13 at 0:38
• My answer below is based on the assumption that you meant $(W_0^{k,p}(\Omega))' = W^{-k,p'}(\Omega)$. – 40 votes Jul 21 '13 at 4:41

1 Answer

I'll use one-dimensional $\Omega=(-1,1)$, $p=2$, and $k=1$ for simplicity. The general idea is the same.

When $k\ge 1$, the derivative operator $\frac{d}{dx}$ maps $W^{k,2}$ onto $W^{k-1,2}$. This is such a natural property that we should expect it to hold for negative indices too. In particular, $W^{-1,2}$ should consist precisely of the distributional derivatives of $L^2$ functions.

What does it mean for distribution $T$ to be the derivative of $L^2$ function $g$? It means that $$T\varphi=-\int_{-1}^1 g\varphi'\tag1$$ for all test functions $\varphi\in C^\infty_c(\Omega)$. By Cauchy-Schwarz, $|T\varphi|\le \|g\|_{L^2} \|\varphi\|_{W^{1,2}}$. Therefore, the functional $T$ extends to the completion of $C^\infty_c(\Omega)$ in $W^{1,2}$ norm. Which is precisely $W^{1,2}_0(\Omega)$. We thus arrive at the identification of distributional derivatives of $L^2$ functions with the elements of $W^{1,2}_0(\Omega)'$.

Now that we see that vanishing on the boundary came from test functions having this property, it is natural to ask: why do we insist on test functions vanishing on the boundary? The answer is that we want to define distributions that live on $\Omega$, not somewhere else (in particular, not on $\partial \Omega$). Whatever "distribution on $\Omega$" means, it must be something we can identify by testing it against the functions supported in $\Omega$.

If we take the dual of $W^{1,2}(\Omega)$, we let in some distributions that vanish in $\Omega$, yet are not identically zero. For example, the functional $\varphi\to \varphi(1)$ belongs to $W^{1,2}(\Omega)'$. Clearly, this should not be an element of any function space on $\Omega$.

Remark. The surjectivity of differentiation comes in handy when one considers the Poisson equation with homogeneous boundary values: $\Delta:W_0^{1,2}(\Omega)\to W^{-1,2}(\Omega)$ turns out to be an isomorphism.

• OP put that zero subscript in the dual space, though I believe that was a typo, it makes me wonder is it making sense to properly phrasing the compactly supportedness for a distribution. – Shuhao Cao Jul 21 '13 at 4:39
• @ShuhaoCao Good point, I read what I expected to read, rather than what was written. No, the zero on the other side does not make sense. By making space larger, we also make its dual space larger, not smaller. So, the dual of $W^{1,2}$ contains $W^{-1,2}$, not the other say around. – 40 votes Jul 21 '13 at 4:41
• "we want to define distributions that live on $\Omega$, not somewhere else (in particular, not on $\partial \Omega$)". I asked this because I read papers discussing about negative $H(\mathrm{div})$'s and $H(\mathbf{curl})$'s tangential/normal components on boundary respectively. – Shuhao Cao Jul 21 '13 at 4:44
• @ShuhaoCao They can have boundary values in the sense of some limit, like ordinary function do. My point was that they should not be able to vanish everywhere in $\Omega$ yet be nonzero. – 40 votes Jul 21 '13 at 4:58
• @40votes, are you sure that a larger space has a larger dual? $L^2$ is a larger space that $H^1_0$, but the dual of $L^2$ is $L^2$, and $L^2$ is contained in the dual of $H^1_0$. – Ellya Aug 21 '15 at 8:29