> [!theoremb] Theorem > > Let $d \in \nat$, $s \in \real$, $k \in \nat$, $H^s(\real^d)$ be the [[Fractional Sobolev Space|Sobolev space]] and $C^k_0(\real^d)$ be the $k$-times [[Space of Continuously Differentiable Functions|continuously differentiable]] functions that [[Vanishes at Infinity|vanish at infinity]], then the following are equivalent: > 1. $s > k + d/2$. > 2. There exists a [[Linear Transformation|linear map]] $I: H^s \to C^k$ such that $If = f$ [[Almost Everywhere|a.e.]] for all $f \in H^s$. > 3. There exists a [[Bounded Linear Map|bounded linear map]] $I: H^s \to C_0^k$ such that $If = f$ a.e. for all $f \in H^s$. > > Moreover, if $\gamma = s - k - d/2 \in (0, 1)$, then the inclusion $I: H^s \to C_0^{k, \gamma}$ into the [[Hölder Space|Hölder space]] is also bounded. > > *Proof*. $(1) \Rightarrow (2)$: If $s > d/2$, then $I$ exists and is bounded by applying the uniform bound and density of the [[Schwartz Space|Schwartz space]]. > > $(2) \Rightarrow (3)$: Suppose that $I$ exists. Let $\seq{(f_n, If_n)} \subset \Gamma(I)$ be a [[Cauchy Sequence|Cauchy sequence]] in the graph of $I$. By [[Complete Metric Space|completeness]], there exists $f \in H^s$ and $g \in C_0^k$ such that $f_n \to f$ in $H^s$ and $If_n \to g$ in $C_0^k$. By $L^2$ and uniform limit, $f = g$ a.e., hence $If = g$. By the [[Closed Graph Theorem]], $I$ is bounded. > > $(3) \Rightarrow (1)$: By composition, $\partial^\alpha \delta \in (H^s)^* = H^{-s}$ for all $\alpha \in \nat_0^d$ with $\abs{\alpha} \le k$. Therefore $\xi^\alpha \in L^2(\mu_{-s})$, $M_k \in L^2(\mu_{-s})$, and $M_{k - s} \in L^2$, so $s > k + d/2$. > > If $\gamma = s - k - d/e \in (0, 1)$, then the Hölder bound yields the desired result. # Uniform and Hölder Bound > [!theorem] > > Suppose that $s > k + d/2$, then for any $f \in H^k$ and $\alpha \in \nat_0^d$, > $ > \norm{D^\alpha f}_u \le \normn{\wh{D^\alpha f}}_{L^1} \le (2\pi)^\abs{\alpha}\norm{M_{k - s}}_{L^2} \cdot \norm{f}_{H^s} > $ > with $D^\alpha f \in C_0$. > > *Proof*. By the [[Cauchy-Schwarz Inequality]], > $ > \begin{align*} > \int_{\real^d}\abs{\partial^\alpha f} &= \int_{\real^d}|M_\alpha \wh{f}| \le \int_{\real^d}M_k|\wh f| = \int_{\real^d}M_{k - s} \cdot M_s|\wh f| \\ > &\le \norm{M_{k - s}}_{L^2} \cdot \normn{M_s\wh f}_{L^2} = \norm{M_{k - s}}_{L^2} \cdot \norm{f}_{\ch^s} > \end{align*} > $ > where $D^\alpha f \in C_0$ comes from the Riemann-Lebesgue lemma. > [!theorem] > > Suppose that $\gamma = s - d/2 \in (0, 1)$, then > $ > \norm{\delta_x - \delta_y}_{\ch^{-s}} \le \abs{x - y}^\gamma \sqrt{\sigma(\partial B(0, 1))\braks{\frac{(2\pi)^2}{2 - 2\gamma} + \frac{2}{\gamma}}} > $ > for any $x, y \in \real^d$. > > *Proof*. Firstly, $\wh \delta_x(\xi) = e^{-2\pi i \angles{x, \xi}}$, so > $ > \norm{\delta_x - \delta_y}_{\ch^{-s}}^2 = \int_{\real^d}M_{-2s}(\xi)\braks{e^{-e\pi i \angles{x, \xi}} - e^{-2\pi i \angles{y, \xi}}}^2 d\xi > $ > By the [[Mean Value Theorem|mean value theorem]], > $ > \abs{e^{-e\pi i \angles{x, \xi}} - e^{-2\pi i \angles{y, \xi}}} \le \abs{2\pi \xi}\abs{x - y} > $ > Let $r = 1/\abs{x - y}$, then > $ > \begin{align*} > &\int_{\bracs{\abs{\xi} \le r}}M_{-2s}(\xi)\braks{e^{-e\pi i \angles{x, \xi}} - e^{-2\pi i \angles{y, \xi}}}^2 d\xi \\ > &\le (2\pi)^2 \int_{\bracs{\abs{\xi} \le r}}\frac{\abs{\xi}^2\abs{x - y}^2}{(1 + \abs{\xi}^2)^s}d\xi \\ > &\le (2\pi)^2 \abs{x - y}^2\int_{\bracs{\abs{\xi} \le r}}\abs{\xi}^{2 - 2s}d\xi > \end{align*} > $ > where using [[Polar Integration]], > $ > \begin{align*} > \int_{\bracs{\abs{\xi} \le r}}\abs{\xi}^{2 - 2s}d\xi &= \sigma(\partial B(0, 1))\int_0^r t^{d + 1 - 2s}dt \\ > &= \sigma(\partial B(0, 1))\int_0^r t^{1 - 2\gamma}dt \\ > &\le \sigma(\partial B(0, 1))\frac{r^{2 - 2\gamma}}{2 - 2\gamma} > \end{align*} > $ > so > $ > \int_{\bracs{\abs{\xi} \le r}}M_{-2s}(\xi)\braks{e^{-e\pi i \angles{x, \xi}} - e^{-2\pi i \angles{y, \xi}}}^2 d\xi \le (2\pi)^2\sigma(\partial B(0, 1))\frac{\abs{x - y}^{2\gamma}}{2 - 2\gamma} > $ > On the other hand, > $ > \begin{align*} > &\int_{\bracs{\abs{\xi} > r}}M_{-2s}(\xi)\braks{e^{-e\pi i \angles{x, \xi}} - e^{-2\pi i \angles{y, \xi}}}^2 d\xi \\ > &\le 4\int_{\bracs{\abs{\xi} > r}}M_{-2s} = 4\int_{\bracs{\abs{\xi} > r}}\abs{\xi}^{2s}d\xi \\ > &\le 4\sigma(\partial B(0, 1))\int_r^\infty t^{d - 1 - 2s}dt\\ > &\le 4\sigma(\partial B(0, 1))\int_r^\infty t^{-2\gamma - 1}dt = 4\sigma(\partial B(0, 1))\frac{2\abs{x - y}^{2\gamma}}{\gamma} > \end{align*} > $ > Combining the two bounds yields > $ > \norm{\delta_x - \delta_y}_{\ch^{-s}}^2 \le \abs{x - y}^{2\gamma}{\sigma(\partial B(0, 1))\braks{\frac{(2\pi)^2}{2 - 2\gamma} + \frac{2}{\gamma}}} > $