Boundary spectral counting and Sobolev scale

Let $(U_j,{\varphi }_j{)}_{j=1}^J$ and $({\psi }_j{)}_{j=1}^J$ be the boundary charts and square partition supplied by [F1], with ${\varphi }_j(U_j)⋐(1\mathrm{/}4,3\mathrm{/}4{)}^n$. Define $\mathcal{G}f=(({\psi }_{j}f)\circ {\varphi }_j^{-1}{)}_{j=1}^J,\mathcal{L}(v_1,\dots ,v_J)={\sum }_{j=1}^J{\psi }_j(v_j\circ {\varphi }_j),$ where each chart component is extended by zero outside its chart image and each summand by zero off $U_j$. Since every ${\psi }_j$ is compactly supported in $U_j$, the zero extensions preserve smoothness for smooth inputs. Change of variables, the bounded chart Jacobians, and the product and chain rules show that $\mathcal{G}$ and $\mathcal{L}$ are bounded on $L^2$ and $H^1$. Direct substitution gives $\mathcal{L}\mathcal{G}f={\sum }_j{\psi }_j^{2}f=f$ almost everywhere. For $N\ge 1$, Fourier Parseval [F4] gives componentwise $\mathrm{\parallel }(I-P_N)\mathcal{G}f{\mathrm{\parallel }}_{\oplus L^2({\mathbb{T}}^n)}^2\le (2\pi N{)}^{-2}\mathrm{\parallel }\mathrm{\nabla }\mathcal{G}f{\mathrm{\parallel }}_{\oplus L^2({\mathbb{T}}^n)}^2\le C{N}^{-2}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^1(M)}^2\mathrm{.}$ For $N=0$, boundedness of $\mathcal{G}$ on $H^1$ gives the same estimate after increasing $C$. Therefore $\mathrm{\parallel }f-\mathcal{L}{P}_N\mathcal{G}f{\mathrm{\parallel }}_2\le C(1+N{)}^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^1(M)}\mathrm{.}$ The finite Fourier sums $\mathcal{L}{P}_N\mathcal{G}f$ are smooth functions on $M$ and converge to $f$ in $L^2$, so smooth functions are dense in $L^2(M)$.

The estimate in step 1.1 proves $H^1(M)\hookrightarrow L^2(M)$ is compact: the finite-rank maps $\mathcal{L}{P}_N\mathcal{G}$ approximate the identity uniformly on the $H^1$ unit ball, and their images are bounded subsets of finite-dimensional spaces, hence have finite nets of every positive radius. Thus the unit ball is totally bounded in the complete space $L^2(M)$.

Put $q(u,v)={\int }_M(u\bar{v}+{\mathrm{\nabla }}_{M}u\cdot \overline{{\mathrm{\nabla }}_{M}v})dS$ on $H^1(M)$. For $f\in L^2(M)$, [F2] gives a unique $u=\tilde{K}f\in H^1(M)$ with $q(u,v)=⟨f,v⟩$ for every $v\in H^1(M)$ and $\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}\le \mathrm{\parallel }f{\mathrm{\parallel }}_2$. Let $J:H^1(M)\hookrightarrow L^2(M)$ and $K=J\tilde{K}$. By step 2.1, $K$ is compact. Hermitian symmetry of $q$ gives $⟨Kf,g⟩=⟨f,Kg⟩$ and $⟨Kf,f⟩=q(\tilde{K}f,\tilde{K}f)\ge 0$. If $Kf=0$, the weak equation and smooth density from step 1.1 imply $⟨f,v⟩=0$ for every $v\in L^2(M)$, so $f=0$. Thus $K$ is compact, positive, self-adjoint and injective.

By [F3], $K$ has an orthonormal eigenbasis $e_j$ with positive eigenvalues ${\mu }_j\to 0$; order these so that ${\mu }_j$ is nonincreasing. Put ${\lambda }_j={\mu }_j^{-1}-1$. From $K{e}_j={\mu }_j{e}_j$ and the weak equation, ${\int }_M{\mathrm{\nabla }}_M{e}_j\cdot \overline{{\mathrm{\nabla }}_{M}v}dS={\lambda }_j⟨e_j,v⟩$ for every $v\in H^1(M)$. Taking $v=e_j$ gives ${\lambda }_j={\int }_M\mathrm{\mid }{\mathrm{\nabla }}_M{e}_j{\mathrm{\mid }}^{2}dS\ge 0$, and these are the eigenvectors of the stated weak Laplace–Beltrami operator. In the form inner product $q$, the vectors $(1+{\lambda }_j{)}^{-1\mathrm{/}2}{e}_j$ are orthonormal and complete: a vector $v$ orthogonal to every $e_j$ in $q$ satisfies $(1+{\lambda }_j)⟨v,e_j⟩=0$ for every $j$, hence vanishes in $L^2$. Form Parseval gives $q(v,v)={\sum }_j(1+{\lambda }_j)\mathrm{\mid }⟨v,e_j⟩{\mathrm{\mid }}^2$, proving the $s=1$ norm identity.

Let $E_{\mathrm{\Lambda }}=\mathrm{span}\{e_j:{\lambda }_j\le \mathrm{\Lambda }\}$. For $u\in E_{\mathrm{\Lambda }}$, step 4.1 gives $\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}^2\le (1+\mathrm{\Lambda })\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$. If $\dim E_{\mathrm{\Lambda }}$ exceeded the rank of $P_N\mathcal{G}{\mathrm{\mid }}_{E_{\mathrm{\Lambda }}}$, rank-nullity would give nonzero $u\in E_{\mathrm{\Lambda }}$ with $P_N\mathcal{G}u=0$. Since $\mathcal{L}\mathcal{G}=I$ and $\mathcal{L}$ is bounded, step 1.1 then gives $\mathrm{\parallel }u{\mathrm{\parallel }}_2\le C(1+N{)}^{-1}\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}\le C\sqrt{1+\mathrm{\Lambda }}(1+N{)}^{-1}\mathrm{\parallel }u{\mathrm{\parallel }}_2$. Choose $N\ge 2C\sqrt{1+\mathrm{\Lambda }}$ to contradict $u\mathrm{\ne }0$. Thus $N_M(\mathrm{\Lambda })=\dim E_{\mathrm{\Lambda }}\le \mathrm{rank}(P_N\mathcal{G})\le J(2N+1{)}^n\le C^{\mathrm{\prime }}(1+\mathrm{\Lambda }{)}^{n\mathrm{/}2}$.

For $p=n\mathrm{/}2+\tau$, group eigenvalues into $[2^k,2^{k+1})$. The bound from step 5.1 gives ${\sum }_j(1+{\lambda }_j{)}^{-p}\le C{\sum }_{k\ge 0}{2}^{-kp}{N}_M(2^{k+1})\le C^{\mathrm{\prime }}{\sum }_{k\ge 0}{2}^{-k(p-n\mathrm{/}2)}<\mathrm{\infty }$; the finitely many eigenvalues below $1$ contribute finitely. This proves the stated summability for every $\tau >0$. Defining $H_{\mathrm{\Delta }}^s(M)$ by the weighted coefficient norm in the Statement completes the spectral-scale claim.