> [!definition] > > Let $d \in \nat$, $s \in \real$, and define > $ > M_s: \real^d \to \real \quad \xi \mapsto (1 + \abs{\xi}^2)^{s/2} > $ > then $M_s \in C^\infty(\real^d)$ is [[Slowly Increasing Function|slowly increasing]]. If $\alpha \in \nat_0^d$ is a [[Multi-Index|multi-index]], denote $M_\alpha(\xi) = (2\pi i\xi)^\alpha$. Let $\cf: L^2(\real^d) \to L^2(\real^d)$ be the [[Fourier Transform|Fourier transform]], and define > $ > \Lambda^s: \cs' \to \cs' \quad T \mapsto \cf^{-1}(M_s\cf(T)) > $ > then $\Lambda^s$ is a [[Space of Toplinear Isomorphisms|toplinear isomorphism]] on the [[Tempered Distribution|tempered distributions]], with $(\Lambda^s)^{-1} = \Lambda^{-s}$. Define > $ > \ch^s = \bracs{T \in \cs': \Lambda^s T \in L^2} > $ > as the **fractional Sobolev space**, equipped with the [[Inner Product|inner product]] > $ > \angles{S, T}_{\ch^s} = \anglesn{\Lambda^sS, \Lambda^sT}_{L^2} = \anglesn{\Lambda^s\wh S, \Lambda^s \wh T}_{L^2} = \anglesn{\wh S, \wh T}_{L^2(\mu_s)} > $ > where $d\mu_s = M_s(x)dx$, then > 1. The Fourier transform $\cf: \ch^s \to L^2(\mu_s)$ is a [[Unitary Map|unitary]] isomorphism, making $\ch^s$ a [[Hilbert Space|Hilbert space]]. Moreover, $\cs$ is dense in $\ch^s$. > 2. For any $T \in \ch^s$, there exists $f \in \loci$ such that $\wh T = T_f$. Hence if $s \ge 0$, then $M_\alpha \wh T \in L^2$ for all $\alpha \in \nat_0^d$ with $\abs{\alpha} \le k$, and $T \in L^2$. > 3. If $s \in \nat$, then the mapping from the [[Sobolev Space|Sobolev space]] $I: H^s \to \ch^s$ defined by $f \mapsto T_f$ is a [[Space of Toplinear Isomorphisms|toplinear isomorphism]]. > > Hence we identify $H^s = \ch^s$ for $s \in \nat$, and define $H^s = \ch^s$ for $s \in \real$. Moreover, > > 4. $\Lambda_t: H^s \to H^{s - t}$ is a unitary isomorphism. > 5. For any $s \le t$, $\iota: H^t \to H^s$ is bounded. > 6. For any [[Multi-Index|multi-index]] $\alpha \in \nat_0^d$, $D^\alpha: H^s \to H^{s - \abs{\alpha}}$ is bounded. > > *Proof*. For $(1)$, it's sufficient to show that $\cf$ is surjective. To this end, let $f \in L^2(\mu_s)$, then $M_{s}f \in L^2$ is a tempered distribution. Since $M_{-s}$ is slowly increasing, $\cf^{-1}f = \cf^{-1}M_{-s}M_sf$ is also a tempered distribution. For density, it's sufficient to show that $\cs$ is dense in $L^2(\mu_s)$ after passing over the Fourier transform. Let $f \in L^2(\mu_s)$ and $\eps > 0$, then there exists $\phi \in \cs$ such that $\norm{\phi - M_sf}_{L^2} < \eps$. In which case, $M_{-s}\phi \in \cs$ with $\norm{M_{-s}\phi - f}_{L^2(\mu_s)} < \eps$. Therefore $\cs$ is dense in $L^2(\mu_s)$. > > For $(2)$, let $T \in \ch^s$, then $M_s\wh T \in L^2 \subset \loci$. Since multiplication by $M_{-s}$ is a map from $\loci$ to $\loci$, $\wh T = M_{-s}M_s\wh T \in \loci$ as well. > > For $(3)$, let $f \in H^s$, then expanding the polynomial terms and pulling the $2\pi$s out yields > $ > \begin{align*} > \angles{T_f, T_f}_{\ch^s} &= \int_{\real^d}(1 + \abs{\xi}^2)^{s/2}\wh f(\xi) \overline{\wh f(\xi)}d\xi \\ > &\le C_d\sum_{\abs{\alpha}\le s}\angles{D^\alpha f, D^\alpha f}_{L^2} \\ > \angles{f, f}_{H^s} &= \sum_{\abs{\alpha}\le s}\angles{D^\alpha f, D^\alpha f}_{L^2} \le C_d\angles{T_f, T_f}_{\ch^s} > \end{align*} > $ > so $f \mapsto T_f$ is continuous and admits a continuous inverse. On the other hand, for any $T \in \ch^s$, $\xi^\alpha T \in L^2$ for all $\abs{\alpha} \le s$ and > $ > \begin{align*} > \cf^{-1}D^\alpha \phi &= \td{\wh{D^\alpha \phi}} = (-1)^{\abs{\alpha}}M_\alpha \tilde{\wh{x}} > \end{align*} > $ > > Therefore for any $\phi \in \cd$, > $ > \begin{align*} > \angles{D^\alpha \phi, T}_{\cs} &= \anglesn{\cf^{-1}{D^\alpha \phi}, \wh T}_{\cs} = \anglesn{\td{\wh{D^\alpha \phi}}, \wh T}_{\cs} \\ > &= (-1)^{\alpha}\anglesn{M_\alpha \check \phi, \wh T}_{\cs} = (-1)^{\alpha}\anglesn{\check \phi, M_\alpha\wh T}_{\cs} \\ > & = (-1)^{\abs{\alpha}}\angles{\phi, \cf^{-1}(M_\alpha\cf(T))}_{\cs} > \end{align*} > $ > where $D^\alpha T = \cf^{-1}M_\alpha \cf(T) \in L^2$.