Fractional compactness on bounded collar domains

Let $U\in H^{1\mathrm{/}2}({\mathbb{R}}^n)$, put $T={\iota }_{1\mathrm{/}2}U$, and choose its unique density $g$ from [F3], so $⟨\xi {⟩}^{1\mathrm{/}2}\mathcal{F}T=u_g$ and $\mathrm{\parallel }g{\mathrm{\parallel }}_2=\mathrm{\parallel }U{\mathrm{\parallel }}_{H^{1\mathrm{/}2}}$. By [F9], $\mathcal{F}T=u_q$, where $q=⟨\xi {⟩}^{-1\mathrm{/}2}g\in L^2$ and $\mathrm{\parallel }q{\mathrm{\parallel }}_2\le \mathrm{\parallel }g{\mathrm{\parallel }}_2$. The $s=0$ case of [F3] therefore supplies $V\in H^0$ with ${\iota }_{0}V=T$. By [F8], $f=I_{0}V\in L^2$ satisfies $T=u_f$. The same fact gives $u_{{\mathcal{F}}_{2}f}=\mathcal{F}T=u_q$. Restricting this equality to compact tests by [F10] and applying [F7] to the locally integrable difference gives ${\mathcal{F}}_{2}f=q$ almost everywhere. Thus $⟨\xi {⟩}^{1\mathrm{/}2}{\mathcal{F}}_{2}f=g$ almost everywhere and $\mathrm{\parallel }f{\mathrm{\parallel }}_2=\mathrm{\parallel }q{\mathrm{\parallel }}_2\le \mathrm{\parallel }U{\mathrm{\parallel }}_{H^{1\mathrm{/}2}}$, with $\mathrm{\parallel }⟨\xi {⟩}^{1\mathrm{/}2}{\mathcal{F}}_{2}f{\mathrm{\parallel }}_2=\mathrm{\parallel }U{\mathrm{\parallel }}_{H^{1\mathrm{/}2}}$. This identification is linear and injective because the distribution embeddings and $I_0$ are linear and injective.

For $f\in L^2(G)$ and $v\in H_0^2(G)$, [F6] gives $\mathrm{\mid }{\int }_{G}f\bar{v}\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }v{\mathrm{\parallel }}_2\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }v{\mathrm{\parallel }}_{W^{2,2}}$. Thus the displayed pairing is a well-defined continuous conjugate-linear functional of $v$, and its operator norm is at most $\mathrm{\parallel }f{\mathrm{\parallel }}_2$. It is linear in $f$.

Under [F10], the distributional restriction of $T=u_f$ to $G$ acts on every test in $C_c^{\mathrm{\infty }}(G)$ by integration against $f{\mathrm{\mid }}_G$. Two whole-space extensions of the same restriction class therefore give $L^2(G)$ functions whose difference pairs to zero with all these tests; [F7] makes them equal almost everywhere. This defines the claimed map $H^{1\mathrm{/}2}(G)\to L^2(G)$. If its value is zero, its restricted distribution is zero, so its restriction class is zero, proving injectivity. For every whole-space extension $U$ of a class $u$, step 1.1 gives $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2(G)}\le \mathrm{\parallel }U{\mathrm{\parallel }}_{H^{1\mathrm{/}2}({\mathbb{R}}^n)}$; taking the infimum over extensions proves $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2(G)}\le \mathrm{\parallel }u{\mathrm{\parallel }}_{H^{1\mathrm{/}2}(G)}$. Hence this inclusion is well-defined, linear and continuous in exactly the restriction-quotient convention of the Statement.

Let $B$ be bounded in $H^{1\mathrm{/}2}(G)$ and identify $\mathcal{F}=E(B)$ with its $L^2$ classes by step 1.1. By [F2], ${\sup }_{U\in \mathcal{F}}\mathrm{\parallel }U{\mathrm{\parallel }}_{H^{1\mathrm{/}2}}<\mathrm{\infty }$, hence ${\sup }_{U\in \mathcal{F}}\mathrm{\parallel }U{\mathrm{\parallel }}_2<\mathrm{\infty }$. The distributions of these extensions are supported in one fixed compact set $K$. By [F10] they pair to zero with all compact tests in ${\mathbb{R}}^n\setminus K$, and [F7] makes their $L^2$ representatives vanish almost everywhere there. Thus their $L^2$ tails vanish uniformly by [F5]. Cauchy–Schwarz on $K$ gives $\mathrm{\parallel }U{\mathrm{\parallel }}_1\le \mathrm{\mid }K{\mathrm{\mid }}^{1\mathrm{/}2}\mathrm{\parallel }U{\mathrm{\parallel }}_2$, so each member also belongs to $L^1$.

For any $V\in L^1\cap L^2$, the two Fourier agreement identities in [F8] give $u_{\widehat{V}}=u_{{\mathcal{F}}_{2}V}$; by [F10] and [F7], $\widehat{V}={\mathcal{F}}_{2}V$ almost everywhere. If $U\in \mathcal{F}$, both $U$ and ${\tau }_{h}U$ belong to $L^1\cap L^2$, since substituting $x-h$ for $x$ preserves both norms. Applying this agreement and [F4], and then the Plancherel identity [F3], gives $\mathrm{\parallel }{\tau }_{h}U-U{\mathrm{\parallel }}_2^2=\int \mathrm{\mid }{e}^{-2\pi ih\cdot \xi }-1{\mathrm{\mid }}^2\mathrm{\mid }{\mathcal{F}}_{2}U(\xi ){\mathrm{\mid }}^{2}d\xi$. The elementary estimate $\mathrm{\mid }{e}^{-it}-1{\mathrm{\mid }}^2\le \min \{t^2,4\}\le 2\mathrm{\mid }t\mathrm{\mid }$ gives $\mathrm{\mid }{e}^{-2\pi ih\cdot \xi }-1{\mathrm{\mid }}^2\le 4\pi \mathrm{\mid }h\mathrm{\mid }⟨\xi ⟩$. By step 1.1, $\mathrm{\parallel }{\tau }_{h}U-U{\mathrm{\parallel }}_2^2\le 4\pi \mathrm{\mid }h\mathrm{\mid }\mathrm{\parallel }U{\mathrm{\parallel }}_{H^{1\mathrm{/}2}}^2$, uniformly for $U\in \mathcal{F}$.

The boundedness, tail and translation hypotheses of [F5] now hold for $\mathcal{F}$, so its closure is compact in $L^2({\mathbb{R}}^n)$. Restriction to $G$ is an $L^2$ contraction, and by [F2] and step 2.1 it restricts $Eu$ to the same $L^2(G)$ class as $u$ for every $u\in B$. Thus $B$ lies in the compact image of the closure of $\mathcal{F}$ under restriction, and its closure in $L^2(G)$ is compact. This proves compactness of the first inclusion.

If the functional in step 1.2 is zero, it vanishes on $C_c^{\mathrm{\infty }}(G)$ by [F6]. Complex conjugation preserves the test space, so ${\int }_{G}f\phi =0$ for every test function $\phi$. Since $f\in L^2(G)$ is locally integrable by Cauchy–Schwarz on compact subsets, [F7] yields $f=0$ almost everywhere. Hence the second map is injective; together with steps 5.1 and 1.2 this proves the stated triple. Conjugating the scalar value of each functional gives the stated continuous injective conjugate-linear map into the scalar-linear dual, with the same norm bound.