Why is the relative trace of Sobolev norms finite?

Question

I am reading the 2009 Paper on Effective Equidistribution by Einsiedler, Margulis and Venkatesh (EMV). I do not understand Section 5.3 on the proof of (3.10). They want to prove that the relative trace of certain Sobolev norms is finite. I believe the question can equivalently formulated for Sobolev norms on open subsets of $\mathbb{R}^n$.

Let $U\subset \mathbb{R}^n$ be a bounded open subset and let $f\in C_c^{\infty}(U)$. We define the degree $d$-Sobolev norm of $f$ as (following the notation of the paper) $$\mathcal{S}_d(f) = \sqrt{\sum_{||\alpha||_1 \leq d}||D_{\alpha}f||} , $$ where $\alpha \in \mathbb{N}_0^n$
We denote by $\mathcal{H}_0^d(U)$ the completion of $C_c^{\infty}(U)$ with respect to the Sobolev norm. A standard fact is that for $f \in C_c^{\infty}(U)$ \begin{equation} ||f||_{\infty} \ll \mathcal{S}_d(f) \end{equation} for $d > \tfrac{n}{2}$. EMV now claim that if $d' > d > 0$ are integers so that $d'-d > \tfrac{n}{2}$ then the relative trace (defined below) $\mathrm{tr}(S_d^2,S_{d'}^{2})$ is finite. In section 5.3. EMV proceed with the proof of this claim as follows. For $x \in U$ consider the linear form on $C_c^{\infty}(X)$ given by $$L_x : f \mapsto (D_{\alpha}f)(x)$$ for some $\alpha \in \mathbb{N}_0^{d}$ so that $||\alpha||_1 \leq d$. Then they claim that using the inequality that $||f||_{\infty} \ll S_d(f)$ it follows that $$\mathrm{tr}(|L_x|^2,S_{d'}^2)$$ is finite. This is exactly the point I do not understand. Because if $(f_n)_{n \in \mathbb{}N}$ is an orthonormal basis of $\mathcal{H}_{0}^{d'}(U)$ then we have that $$\mathrm{tr}(|L_x|^2,S_{d'}^2) = \sum_{n \geq 1} ||f_n||_{L_x}^2.$$ But then the fact that $||f||_{\infty} \ll S_d(f)$ only gives that each of the terms of the infinite sum is $\ll 1$. But then the infinite sum is not finite. So I don't understand how they conclude that $\mathrm{tr}(|L_x|^2,S_{d'}^2)$ is finite. I also didn't find a proof of this claim on my own.

Assuming that $\mathrm{tr}(|L_x|^2,S_{d'}^2)$ is finite is easily follows that $\mathrm{tr}(S_d, S_{d'})$ is finite by just integrating over $x$ using here that $U$ is bounded.

DEFINITION OF THE RELATIVE TRACE:

Here we summarize content form Appendix A of the 2002 paper by Bernstein and Reznikov called Sobolev norms of automorphic functionals. First consider a finite dimensional complex vector space $V$ and denote by $V^{+}$ the Hermitian dual consisting of anti-linear maps $f: V \to \mathbb{C}$. Let $\langle \cdot, \cdot \rangle_A$ and $\langle \cdot , \cdot \rangle_B$ be two positive definite Hermitian inner products on $V$.

We denote by $A_{+}$ the map $$A_{+} : V \longrightarrow V^{+}, \quad\quad v \longmapsto \langle v, \cdot \rangle_A,$$ where $B_{+}$ is analogously defined. As $A$ and $B$ are positive definite, $A_{+}$ and $B_{+}$ are isomorphism of finite dimensional vector spaces. Then we define the relative trace of $A$ and $B$ as $$\mathrm{tr}(A,B) = \mathrm{tr}(B_{+}^{-1}A_{+}).$$ If now $V$ is a second-countable infinite-dimensional vector space then we define the relative trace of $A$ and $B$ as $$\mathrm{tr}(A,B) = \sup_{W \subset V} \mathrm{tr}(A_W,B_W),$$ where the supremum is taken over all finite dimensional subspaces $W \subset V$ and $A_W$ respectively $B_W$ denotes the restriction of $A$ respectively $B$ onto $W$. It can be showed that if $(v_n)_{n \in \mathbb{N} }$ is a orthonormal basis of $B$ then $$\mathrm{tr}(A,B) = \sum_{n \geq 1} ||v_n||_A^2.$$

Answer 1

Following a hint by M. Einsiedler I am now able to present the following solution to my question.

Proposition: Let $d > d' > 0$ be integers so that $$d - d' > \frac{n}{2}.$$ Then the relative trace $\mathrm{tr}(\mathcal{S}_{d'}^2,\mathcal{S}_{d}^2)$ on the Hilbert space $\mathcal{H}^{d}_0(U)$ is finite.

Proof: We consider for a fixed $x \in U$ and $\alpha \in \mathbb{N}_0^d$ with $||\alpha||_1 \leq d'$ the map $$L_x : \mathcal{H}^{d}_0(U) \to \mathbb{C}, \quad\quad f \mapsto (D_{\alpha}f)(x),$$ where we note that this map is well defined by the Sobolev Embedding Theorem. We note that $L_x$ is not the zero map as $\mathcal{H}^{d}_0(U)$ contains $C_c(X)$ thus we observe that $\ker(L_x)$ is a closed proper subspace of $\mathcal{H}^{d}_0(U)$. Choosing some vector $g \in \ker(L_x)^{\perp}$ with $L_x(g) = 1$ we conclude that we have an orthogonal direct sum $$\mathcal{H}^{d}_0(U) = \ker(L_x) \oplus \mathbb{C}g,$$ where each function $f \in \mathcal{H}^{d}_0(U)$ has the decomposition $$f = f-L_x(f)g + L_x(f)g.$$ Finally we chose an orthonormal basis $f_1, f_2, \ldots$ of $\ker(L_x)$ with respect to $\mathcal{S}_{d}^2$ so that $g,f_1,f_2, \ldots$ is an orthogonal basis of $\mathcal{H}^{d}_0(U)$. Using that $L_x(f_n) = |L_x(f_n)|^2 = 0$ for all $n \geq 1$, we conclude \begin{align*} \mathrm{tr}({|L_x|^2},{\mathcal{S}_{d}^2}) &= \frac{\langle g,g \rangle_{L_x}}{\langle g,g \rangle_{\mathcal{H}^{d}_0(U)}} + \sum_{n \geq 1}\frac{\langle f_n, f_n \rangle_{L_x}}{\langle f_n,f_n \rangle_{\mathcal{H}^{d}_0(U)}} \\ &= \frac{\langle g,g \rangle_{L_x}}{\langle g,g \rangle_{\mathcal{H}^{d}_0(U)}} \\ &= \frac{|(D_{\alpha}g)(x)|^2}{\mathcal{S}_{d}^2} \ll 1, \end{align*} where used the above inequality $||f||_{\infty} \ll \mathcal{S}_d$. Integrating now over $x$, using that $X$ is a probability space, and summing over all $||\alpha||_1 \leq d'$ we conclude that the relative trace $\mathrm{tr}(\mathcal{S}_{d'},\mathcal{S}_{d'})$ on the Hilbert space $\mathcal{H}^{d}_0(U)$ is finite.