> [!theorem] > > Let $d \in \nat$ and $s \in \real$, then for any $T \in H^{-s}$ in the [[Fractional Sobolev Space|fractional Sobolev space]] and $\phi \in \cs$, > $ > \angles{\phi, T}_{\cs} = \int_{\real^d} \check \phi \wh T = \int_{\real^d} \wh \phi \check T > $ > and $\abs{\angles{\phi, T}_{\cs}} \le \norm{\phi}_{\ch^s} \cdot \norm{T}_{\ch^{-s}}$. By the [[Linear Extension Theorem]], $T$ admits a unique extension to $H^{s}$, denoted as > $ > T(g) = \angles{g, T}_H = \int_{\real^d} \check g \wh T = \int_{\real^d} \wh g \check T > $ > for any $g \in H^s$. The inclusion into the [[Topological Dual|dual]] $\iota: H^{-s} \to (H^{s})^*$ is a [[Unitary Map|unitary]] [[Space of Toplinear Isomorphisms|isomorphism]]. > > If $s > 0$, then $H^s \subset L^2 \subset H^{-s}$. > > *Proof*. By definition of the [[Fourier Transform|Fourier transform]], > $ > \angles{\phi, T}_{\cs} = \anglesn{\check \phi, \wh T}_{\cs} = \anglesn{\wh \phi, \check T}_{\cs} > $ > Since $\wh T, \check T \in \loci$, the above dual pairings admit integral representations, in which case > $ > \angles{\phi, T}_\cs = \int_{\real^d} M_{s}\check \phi \cdot M_{-s}\wh T > $ > so $\abs{\anglesn{\phi, T}_\cs} \le \norm{\phi}_{H^s} \cdot \norm{T}_{H^{-s}}$, and $\norm{T}_{(H^s)^*} \le \norm{T}_{H^{-s}}$. On the other hand, if $T \in (H^s)^*$, then $\check T \in L^2(\mu_s)^*$, and there exists $f \in L^2(\mu_s)$ such that > $ > \angles{\phi, T}_{\cs} = \int_{\real^d} M_s\wh \phi \cdot M_s f = \int_{\real^d}\wh \phi \cdot M_{2s}f > $ > In which case, $T = \cf^{-1}(M_{2s}f)$, and > $ > \norm{T}_{H^{-s}}^2 = \int_{\real^d}\abs{M_{2s}f}^2M_{-2s} = \int_{\real^d}M_{2s}\abs{f}^2 = \norm{f}_{L^2(\mu_s)}^2 = \norm{T}_{(H^s)^*} > $ > [!theorem] > > Let $d, k \in \nat$, $U \subset \real^d$ be open. Let $A = \bracsn{\alpha \in \nat_0^d: \abs{\alpha} \le d}$, and consider the space $L^2(U)^A$, equipped with the inner product > $ > \angles{S, T}_{L^2(U)^A} = \sum_{\alpha \in A}\angles{S_\alpha, T_\alpha}_{L^2(U)} > $ > Identify each $T \in L^2(U)^A$ with a linear functional in $H^k(U)$, such that > $ > \angles{f, T}_H = \sum_{\alpha \in A}\angles{D^\alpha f, T_\alpha}_{L^2(U)} > $ > and $\iota: L^2(U)^A \to H^k(U)^*$ be the inclusion map, then > > 1. $\iota$ is bounded with $\norm{\iota T}_{H^k(U)^*} \le \norm{T}_{L^2(U)^A}$ for all $T \in L^2(U)^A$. > 2. $\iota$ is surjective with $\norm{\phi}_{H^k(U)^*} = \inf\bracsn{\norm{T}_{L^2(U)^A}: \iota T = \phi}$ . > > *Proof*. Let $\phi \in H^k(U)^*$, then there exists $f \in H^k(U)$ such that $\angles{\cdot, \phi}_H = \angles{\cdot, f}_{H^k}$ and $\norm{\phi}_{H^k(U)^*} = \norm{f}_{H^k(U)}$. In which case, if $T \in L^2(U)^A$ with $T_\alpha = D^\alpha f$, then $\iota T = \phi$ with $\norm{\iota T}_{H^k(U)^*} = \norm{f}_{H^k(U)} = \norm{\phi}_{H^k(U)^*}$. > > Now let $S \in L^2(U)^A$ such that $\iota S = \phi$, then > $ > \norm{f}_{H^k(U)}^2 = \angles{f, f}_{H^k} = \angles{f, S}_{H} = \sum_{\alpha \in A}\angles{D^\alpha f, S_\alpha}_{L^2(U)} > $ > Since $\abs{D^\alpha f - S_\alpha} \ge 0$, > $ > \begin{align*} > 0 &\le \norm{T - S}_{L^2(U)^A} = \sum_{\alpha \in A}\int_{\real^d}(T_\alpha - S_\alpha)\overline{(T_\alpha - S_\alpha)} \\ > &= \norm{T_\alpha}_{L^2(U)^A} + \norm{S_\alpha}_{L^2(U)^A} - 2\angles{T, S}_{L^2(U)^A} \\ > \norm{T}_{L^2(U)^A} &\le \norm{S}_{L^2(U)^A} > \end{align*} > $