Sobolev spaces with negative exponent

Question

For $k\in \mathbb{N}$ and $p \geq 1,$ what is the motivation behind defining the Sobolev spaces with negative exponent $W^{-k,p}$ as the dual of $W_0^{k,p}$ and not as the dual of $W^{k,p}.$

Answer 1

When $1<p<\infty,$ there are two equivalent ways to define the negative Sobolev spaces.

• We define $W^{-k,p}(\Omega)$ as the dual space of $W^{k,p'}_0(\Omega),$ where $p' = \frac{p}{p-1}.$ This is equipped with the operator norm as usual.

• We define $W^{-k,p}(\Omega)$ to be the space of distributions that can be written as $$ f = \sum_{|\alpha| \leq k} D^{\alpha}f_{\alpha}, $$ where each $f_{\alpha} \in L^p(\Omega).$ The associated norm is defined as $\inf\{\sum_{|\alpha|\leq k} \lVert f_{\alpha}\rVert_{L^p(\Omega)}\},$ where the infimum is taken over all such representations.

The equivalence (up to isomorphism) can be established by noting if $f = \sum_{|\alpha| \leq k} D^{\alpha}f_{\alpha}$ and $u \in W^{k,p'}_0(\Omega),$ we have the duality pairing $$ \langle f, u \rangle = \sum_{|\alpha|=k} \langle D^{\alpha}f_{\alpha},u \rangle = \sum_{|\alpha|=k} (-1)^{|\alpha|}\int_{\Omega} f_{\alpha}D^{\alpha}u \,\mathrm{d} x.$$ Strictly speaking this holds for all $u \in \mathcal D(\Omega)$ by definition of the distributional derivative, and by density we can extend this to all $u \in W^{k,p}_0(\Omega).$ This in turn defines a bounded linear map $$ T : W^{-k,p}(\Omega) \longrightarrow W^{k,p'}_0(\Omega)^*, $$ and using the fact that $L^p(\Omega) \cong L^{p'}(\Omega)^*$ by a similar pairing we can argue $T$ is an isomorphism.

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Now the above provides several ways to motivate the duality definition.

• When we define $W^{-k,p}(\Omega)$ by duality, we want to be able to identify the space as the subspace of $\mathcal{D}'(\Omega).$ This can be done using the distributional pairing, as elements $f \in \mathcal{D}'(\Omega)$ such that the functional $\varphi \mapsto \langle \varphi, f \rangle$ extends uniquely from $\mathcal{D}'(\Omega)$ to a bounded linear operator on $W^{k,p'}_0(\Omega).$ For this it is necessary for the inclusion $\mathcal{D}'(\Omega) \hookrightarrow W^{k,p'}_0(\Omega)$ to have dense range.

• The second definition is arguably the more natural definition of negative order spaces, particularly when seeking the natural spaces that partial differential operators map into. The fact that it is the dual of $W^{k,p'}_0(\Omega)$ is simply a result of this fact.

• However in both cases, this boils down to ensuring we have the integration by parts formula $$ \int_{\Omega} f \,(\partial^{\alpha}g)\,\mathrm{d}x = (-1)^{|\alpha|} \int_{\Omega} (\partial^{\alpha}f )\, g \,\mathrm{d}x, $$ which classical holds for $f,g$ regular provided no boundary terms arise. This can be ensured by requiring either $f$ or $g$ to vanish suitably near the boundary, and this is the basis of defining the Schwarz distributions. Since distributions and $L^p$ functions don't have a well-defined notion of a trace, for the formal integration by parts formula to hold we need the Sobolev functions we test by to vanish suitably near the boundary.