> [!theoremb] Theorem > > Let $d \in \nat$, $s \in \real$, $t < s$, and $H^s$ be the [[Fractional Sobolev Space|fractional Sobolev space]]. Let $\cb \subset H^s$ such that > 1. $M = \sup_{T \in \cb}\norm{T}_{\ch^{s}} < \infty$. > 2. There exists $K \subset \real^d$ [[Compactness|compact]] such that $\supp{T} \subset K$ for all $T \in \cb$. > > then $\cb$ is [[Totally Bounded|totally bounded]] in $H^t$. If $H^s_K = \bracs{T \in H^s: \supp{T} \subset K}$, then $\iota: H_K^s \to H^t$ is [[Compact Operator|compact]]. > > *Proof*. Let $\phi \in C_c^\infty(\real^d)$, then by assumption $(2)$, $\phi T = T$ and $\wh T = \wh \phi * \wh T$ for all $T \in \cb$. By the [[Arzelà-Ascoli Theorem]], $\bracsn{M_s\wh T: T \in \cb}$ is totally bounded with respect to the [[Uniform Norm|uniform norm]] on $B(0, R)$ for all $R > 0$. > > Let $\eps > 0$, then there exists $R > 0$ such that $(1 + R^2)^{t - s}M^2 \le \sqrt{\eps/2}$. Let $\seqf{T_j} \subset \cb$ such that for any $T \in \cb$, there exists $1 \le j \le n$ such that > $ > \normn{(M_s\wh T - M_s\wh T_j) \cdot \one_{B(0, 1)}}_{u}^2 < \frac{\sqrt{\eps/2}}{m(B(0, 1))R^d(1 + R^2)^{\max(t, 0)}} > $ > then by the Sobolev bound, > $ > \norm{T - T_j}_{\ch^s}^2 \le 2\sqrt{\eps/2}^2 = \eps > $ > so $\cb \subset \bigcup_{j = 1}^n B(T_j, \eps)$ with respect to $\ch^s$. # Distance Lemma > [!theorem] > > Let $\xi, \eta \in \real^d$, then > $ > (1 + \abs{\xi}^2)^{s}(1 + \abs{\eta}^2)^{-s} \le 2^{\abs{s}}(1 + \abs{\xi - \eta}^2)^{\abs{s}} > $ > *Proof*. Firstly, $\abs{\xi}^2 \le 2({\abs{\xi - \eta}^2 + \abs{\eta}^2})$, so > $ > (1 + \abs{\xi}^2) \le 2(1 +\abs{\xi - \eta}^2)(1 + \abs{\eta}^2) > $ > If $s \ge 0$, then > $ > (1 + \abs{\xi}^2)^s \le 2(1 +\abs{\xi - \eta}^2)^s(1 + \abs{\eta}^2)^{s} > $ > If $s < 0$, then > $ > (1 + \abs{\eta}^2)^{-s} \le 2(1 +\abs{\xi - \eta}^2)^\abs{s}(1 + \abs{\xi}^2)^{-s} > $ # Arzelà-Ascoli Bound > [!theorem] > > Let $\phi \in C_c^\infty(\real^d)$ and $T \in H^s$, then > $ > \normn{M_s \wh \phi * \wh T}_u \le 2^{\abs{s}}\norm{\phi}_{\ch^{\abs s}}\norm{T}_{\ch^s} > $ > and for any $R > 0$, there exists $C_{s, R, d, \phi} \ge 0$ such that > $ > \normn{D_j (M_s\wh \phi * \wh T)}_u \le C_{s, R, d, \phi}\norm{T}_{\ch^s} > $ > > *Proof*. Let $\xi \in \real^d$, then > $ > \begin{align*} > M_s\wh \phi * \wh T(\xi) &= M_s(\xi)\int_{\real^d}\wh \phi(\xi - \eta)\wh T(\eta)d\eta \\ > &= M_s(\xi)\int_{\real^d}M_{-s}(\eta)\wh \phi(\xi - \eta)M_s(\eta)\wh T(\eta)d\eta \\ > |M_s\wh \phi * \wh T(\xi)| &\le 2^{\abs s}\int_{\real^d}M_{\abs{s}}(\xi - \eta)|\wh \phi(\xi - \eta)|M_s|\wh T(\eta)|d\eta \\ > &\le 2^{\abs s}\normn{M_{\abs{s}}\wh \phi}_{L^2}\normn{M_s\wh T}_{L^2} = 2^{\abs s}\norm{\phi}_{\ch^{\abs{s}}}\norm{T}_{\ch^{s}} > \end{align*} > $ > Since $D_j(\wh \phi * \wh T) = (D_j \wh \phi) * \wh T$, > $ > \begin{align*} > \normn{M_s D_j( \wh \phi * \wh T)}_u &\le 2^{\abs s}\normn{M_{\abs s}D_j \wh \phi}_{L^2}\normn{M_s \wh T}_{L^2} \\ > &\le 2^{\abs s}\normn{M_{\abs s}D_j \wh \phi}_{L^2}\normn{T}_{\ch^s} > \end{align*} > $ > > Applying the product rule to the restriction of $M_s$ to $B(0, R)$ yields the desired result. # Sobolev Bound > [!theorem] > > For any $R > 0$, > $ > \begin{align*} > \norm{T}_{\ch^t}^2 &\le m(B(0, 1))R^d(1 + R^2)^{\max(t, 0)}\normn{\wh T \cdot \one_{B(0, R)}}_u^2 \\ > &+ (1 + R^2)^{t - s}\normn{\wh T}_{\ch^s}^2 > \end{align*} > $ > *Proof*. > $ > \begin{align*} > \norm{T}_{\ch^t}^2 &= \int_{\real^d}(1 + \abs{\xi}^2)^{t}|\wh T(\xi)|^2d\xi \\ > &= \int_{\bracs{\abs{\xi} \le R}}(1 + \abs{\xi}^2)^{t}|\wh T(\xi)|^2d\xi + \int_{\bracs{\abs{\xi} > R}}(1 + \abs{\xi}^2)^{t}|\wh T(\xi)|^2d\xi > \end{align*} > $ > where > $ > \begin{align*} > &\int_{\bracs{\abs{\xi} \le R}}(1 + \abs{\xi}^2)^{t}|\wh T(\xi)|^2d\xi\\ > &\le (1 + R^2)^{\max(t, 0)} \cdot \normn{\wh T \cdot \one_{B(0, R)}}_u^2 \cdot m(B(0, R))\\ > &= m(B(0, 1))R^d(1 + R^2)^{\max(t, 0)}\normn{\wh T \cdot \one_{B(0, R)}}_u^2 > \end{align*} > $ > and > $ > \begin{align*} > &\int_{\bracs{\abs{\xi} > R}}(1 + \abs{\xi}^2)^{t}|\wh T(\xi)|^2d\xi \\ > &= \int_{\bracs{\abs{\xi} > R}}(1 + \abs{\xi}^2)^{t - s}(1 + \abs{\xi}^2)^{s}|\wh T(\xi)|^2d\xi \\ > &\le (1 + R^2)^{t - s}\int_{\bracs{\abs{\xi} > R}}(1 + \abs{\xi}^2)^{s}|\wh T(\xi)|^2d\xi \\ > &= (1 + R^2)^{t - s}\normn{\wh T}_{\ch^s} > \end{align*} > $