# Derivation of the Trace Formula: Diagonal Estimates

• Fritz Gesztesy
• Marcus Waurick
Chapter
Part of the Lecture Notes in Mathematics book series (LNM, volume 2157)

## Abstract

This chapter is a direct continuation of the preceding one and computes the actual trace of the operator $$\chi _{\varLambda }B_{L}(z) = z\chi _{\varLambda }\mathop{\mathrm{tr}}\nolimits _{N}\big(\left (L^{{\ast}}L + z\right )^{-1} -\left (LL^{{\ast}} + z\right )^{-1}\big)$$ for odd space dimensions n. An application of Montel’s theorem plays a decisive role in this trace computation, in addition it should be noted that the case n = 3 is more subtle than $$n\geqslant 5$$ and requires special attention to follow in Chap.

## Keywords

Diagonal Estimates Montel Trace Computable Local Boundedness Integral Kernel
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

In this chapter, we shall compute the trace of $$\chi _{\varLambda }B_{L}(z)$$, $$\varLambda> 0$$, $$z \in \varrho (-L^{{\ast}}L) \cap \varrho (-LL^{{\ast}}) \cap \mathbb{C}_{\mathrm{Re}>-1}$$, with B L given by (). After stating the next lemma (needed to be able to apply Lemma  and Proposition  to the sum in ()) we will outline the strategy of the proof.

We note that for the application of Lemma  to the first summand in (), one needs to establish continuous differentiability of the integral kernel of (). In this context we emphasize the different regularity of the kernels of () for n = 3 and $$n\geqslant 5$$, necessitating modifications for the case n = 3 due to the lack of differentiability of ().

### Lemma 8.1 ([22, Lemma 4, p. 224])

Let $$n \in \mathbb{N}_{\geqslant 3}$$ odd, $$L = \mathcal{Q}+\varPhi$$ be given by () , and let $$z \in \varrho \left (-LL^{{\ast}}\right ) \cap \varrho \left (-L^{{\ast}}L\right )$$, with Re (z) > −1. Denote the integral kernels of the following operators
$$\displaystyle\begin{array}{rcl} J_{L}^{j}(z)& =& \mathop{\mathrm{tr}}\nolimits _{ 2^{\hat{n}}d}\big(L\left (L^{{\ast}}L + z\right )^{-1}\gamma _{ j,n}\big) -\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(L^{{\ast}}\left (LL^{{\ast}} + z\right )^{-1}\gamma _{ j,n}\big), {}\\ A_{L}(z)& =& \mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\big[\varPhi,L^{{\ast}}\left (LL^{{\ast}} + z\right )^{-1}\big]\big) -\mathop{\mathrm{tr}}\nolimits _{ 2^{\hat{n}}d}\big(\big[\varPhi,L\left (L^{{\ast}}L + z\right )^{-1}\big]\big), {}\\ \end{array}$$
by GJ,j,z, j ∈{ 1,…,n}, and GA,z, respectively. Then GA,z is continuous and satisfies GA,z(x,y) → 0 if y → x for all $$x \in \mathbb{R}^{n}$$. If $$n\geqslant 5$$, GJ,j,z is continuously differentiable on $$\mathbb{R}^{n} \times \mathbb{R}^{n}.$$

### Proof

Appealing to Lemma , one recalls with R1+z, $$\mathcal{Q}$$, and C given by (), (), and (), respectively,
$$\displaystyle\begin{array}{rcl} J_{L}^{j}(z)& =& 2\mathop{\mathrm{tr}}\nolimits _{ 2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\left (R_{1+z}C\right )^{n-2}R_{ 1+z}\big) + 2\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\varPhi \left (R_{1+z}C\right )^{n-1}R_{ 1+z}\big) {}\\ & & +\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\big(\left (L^{{\ast}}L + z\right )^{-1} + \left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n}\big) {}\\ & & +\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\varPhi \big(\left (L^{{\ast}}L + z\right )^{-1} + \left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n}\big),\quad j \in \{ 1,\ldots,n\}, {}\\ \end{array}$$
and
$$\displaystyle\begin{array}{rcl} A_{L}(z)& =& \mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\big[\varPhi,\varPhi \big(2\left (R_{1+z}C\right )^{n}R_{ 1+z} +\big (\left (L^{{\ast}}L + z\right )^{-1} -\left (LL^{{\ast}} + z\right )^{-1}\big) {}\\ & & \qquad \quad \times \left (CR_{1+z}\right )^{n+1}\big)\big]\big) {}\\ & & \quad -\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\big[\varPhi,\mathcal{Q}\big(2\left (R_{1+z}C\right )^{n-1}R_{ 1+z} +\big (\left (L^{{\ast}}L + z\right )^{-1} + \left (LL^{{\ast}} + z\right )^{-1}\big) {}\\ & & \ \ \ \ \qquad \qquad \times \left (CR_{1+z}\right )^{n}\big)\big]\big). {}\\ \end{array}$$
By Proposition  (one recalls that Φ is admissible and hence $$\varPhi \in C_{b}^{\infty }\big(\mathbb{R}^{n}; \mathbb{C}^{d\times d}\big)$$ by Definition  (i)), one gets for all $$\ell\in \mathbb{R}$$,
$$\displaystyle{Q\gamma _{j,n}\left (R_{1+z}C\right )^{n-2}R_{ 1+z} \in \mathcal{B}\big(H^{\ell}(\mathbb{R}^{n})^{2^{\hat{n}}d },H^{\ell+2(n-2)+2-1}(\mathbb{R}^{n})^{2^{\hat{n}}d }\big).}$$
For $$n\geqslant 5$$, one obtains from $$(2(n - 2) + 2 - 1) = n - 3> 0,$$ the continuity of GJ, j, z by Corollary . Moreover, since $$\left (2(n - 2) + 2 - 1\right ) - n - 1 = n - 4> 0,$$ Corollary  also implies continuous differentiability of GJ, j, z. Similar arguments ensure the continuity of the integral kernel of A L (z) (for $$n\geqslant 3$$). Moreover, for $$n\geqslant 3$$, the integral kernel of A L (z) vanishes on the diagonal by Proposition . □

Next, we outline the idea for computing the trace of $$\chi _{\varLambda }B_{L}(z)$$. By Theorem  and Theorem , we know that the limit $$\lim _{\varLambda \rightarrow \infty }\mathop{\mathrm{tr}}\nolimits (\chi _{\varLambda }B_{L}(0))$$ exists. However, in order to derive the explicit formula asserted in Theorem  also for z in a neighborhood of 0, some work is required. As it will turn out, for z with large real part—at least for a sequence $$\{\varLambda _{k}\}_{k\in \mathbb{N}}$$—we can show that an expression similar to the one in Theorem  is valid.

For achieving the existence of the limit (without using sequences) for z in a neighborhood of 0, we intend to employ Montel ’s theorem. One recalls that for an open set $$U \subseteq \mathbb{C}$$, a set $$\mathcal{G}\subseteq \mathbb{C}^{U}\,:=\,\{\,f\,\vert \,f: U \rightarrow \mathbb{C}\}$$ is called locally bounded , if for all compact $$\varOmega \subset U$$,
$$\displaystyle{ \sup _{f\in \mathcal{G}}\sup _{z\in \varOmega }\vert \,f(z)\vert <\infty. }$$
(8.1)

### Theorem 8.2 (Montel ’s Theorem, see, e.g., [35, p. 146–154])

Let $$U \subseteq \mathbb{C}$$ open, $$\{f_{\varLambda }\}_{\varLambda \in \mathbb{N}}$$ a locally bounded family of analytic functions on U. Then there exists a subsequence $$\{f_{\varLambda _{k}}\}_{k\in \mathbb{N}}$$ and an analytic function g on U such that $$f_{\varLambda _{k}} \rightarrow g$$ as $$k \rightarrow \infty$$ in the compact open topology (i.e., for any compact set $$\varOmega \subset U$$ , the sequence $$\{f_{\varLambda _{k}}\vert _{\varOmega }\}_{k\in \mathbb{N}}$$ converges uniformly to g| Ω ).

For our particular application of Montel ’s theorem, we need to show that the family of analytic functions
$$\displaystyle{\{z\mapsto \mathop{\mathrm{tr}}\nolimits (\chi _{\varLambda }B_{L}(z))\}_{\varLambda }}$$
constitutes a locally bounded family. Thus, one needs to show that for all compact $$\varOmega \subset \mathbb{C}_{\mathrm{Re}>-1} \cap \varrho (-L^{{\ast}}L) \cap \varrho (-LL^{{\ast}})$$,
$$\displaystyle{ \sup _{\varLambda>0}\sup _{z\in \varOmega }\vert \mathop{\mathrm{tr}}\nolimits (\chi _{\varLambda }B_{L}(z))\vert <\infty. }$$
(8.2)
For this assertion, it is crucial that some integral kernel s involved in the computation of the trace vanish on the diagonal , see, for instance, Proposition . We note that generally, the expression
$$\displaystyle{ \sup _{\varLambda>0}\sup _{z\in \varOmega }\mathop{\mathrm{tr}}\nolimits (\vert (\chi _{\varLambda }B_{L}(z))\vert ), }$$
(8.3)
cannot be finite, as the example constructed in Appendix B demonstrates. In order to prove (8.2), we actually show for all $$\varOmega \subset \mathbb{C}_{\mathrm{Re}>-1} \cap \varrho (-L^{{\ast}}L) \cap \varrho (-LL^{{\ast}})$$ compact,
$$\displaystyle{ \sup _{\varLambda>0}\sup _{z\in \varOmega }\vert z\mathop{\mathrm{tr}}\nolimits (\chi _{\varLambda }B_{L}(z))\vert <\infty, }$$
(8.4)
and then appeal to the fact that condition (8.4) together with Theorem  implies (8.2), as the next result confirms:

### Lemma 8.3

Assume that $$\{\phi _{k}\}_{k\in \mathbb{N}}$$ is a sequence of analytic (scalar-valued) functions on $$B_{\mathbb{C}}(0,1)$$ . Assume that $$\{z\mapsto z\phi _{k}(z)\}_{k\in \mathbb{N}}$$ is locally bounded on B(0,1). Then $$\{\phi _{k}\}_{k\in \mathbb{N}}$$ is locally bounded on B(0,1).

### Proof

Assume that $$\{\phi _{k}\}_{k\in \mathbb{N}}$$ is not locally bounded on B(0, 1). Then there exists a subsequence $$\{\phi _{k_{\ell}}\}_{\ell\in \mathbb{N}}$$ and a corresponding sequence of complex numbers $$\{z_{k_{\ell}}\}_{\ell\in \mathbb{N}}$$ with the property that $$z_{k_{\ell}} \rightarrow 0$$ and $$\vert \phi _{k_{\ell}}(z_{k_{\ell}})\vert \rightarrow \infty$$ as $$\ell\rightarrow \infty$$. Since
$$\displaystyle{\{\psi _{\ell}\}_{\ell\in \mathbb{N}}\,:=\,\{z\mapsto z\phi _{k_{\ell}}(z)\}_{\ell\in \mathbb{N}}}$$
is locally bounded on B(0, 1) there exists an accumulation point ψ in the compact open topology of analytic functions $$\mathcal{H}(B(0,1))$$ on B(0, 1) by Montel ’s theorem. Without loss of generality, one can assume that ψ  → ψ in $$\mathcal{H}(B(0,1))$$ as $$\ell\rightarrow \infty$$. By construction, one has ψ (0) = 0 and for some r > 0,
$$\displaystyle\begin{array}{rcl} \left \vert \frac{1} {z}\psi _{\ell}(z) -\psi '(0)\right \vert & \leqslant & \left \vert \frac{1} {z}\left (\psi _{\ell}(z) -\psi _{\ell}(0)\right ) -\psi _{\ell}'(0)\right \vert + \vert \psi _{\ell}'(0) -\psi '(0)\vert {}\\ &\leqslant & \sup _{z\in B(0,r)}\vert (\psi '_{\ell}(z) -\psi '_{\ell}(0))\vert + \vert \psi _{\ell}'(0) -\psi '(0)\vert {}\\ \end{array}$$
for all z ∈ B(0, r)∖{0}. Since ψ  → ψ′ uniformly on compacts, it follows that
$$\displaystyle{\limsup _{\ell\rightarrow \infty }\sup _{z\in B(0,r)\setminus \{0\}}\left \vert \frac{1} {z}\psi _{\ell}(z)\right \vert <\infty.}$$
However, for sufficiently large, one concludes
$$\displaystyle{\sup _{z\in B(0,r)\setminus \{0\}}\left \vert \frac{1} {z}\psi _{\ell}(z)\right \vert \geqslant \left \vert \frac{1} {z_{k_{\ell}}}\psi _{\ell}(z_{k_{\ell}})\right \vert = \vert \phi _{k_{\ell}}(z_{k_{\ell}})\vert \mathop{\longrightarrow }\limits_{\ell\rightarrow \infty }\infty,}$$

### Remark 8.4

It turns out that the analyticity hypothesis in Lemma 8.3 is crucial. Indeed, for every $$n \in \mathbb{N}$$, there exists a $$C^{\infty }$$-function $$\psi _{n}: [0,1) \rightarrow [0,\infty )$$ with the properties,
$$\displaystyle{\psi _{n}\vert _{(0,1/(2n))} = 0,\quad 0\leqslant \psi _{n}(x)\leqslant \psi _{n}\left ( \frac{1} {n}\right ) = n,\quad \psi _{n}\vert _{(2/n,1)} = 0.}$$
Then ψ n (0) = 0 and $$0\leqslant x\psi _{n}(x)\leqslant (2/n)n = 2$$. Considering $$\phi _{n}(x + \mathrm{i}y)\,:=\,\psi _{n}(\vert x + \mathrm{i}y\vert )$$ for $$x,y \in \mathbb{R}$$, x + iy ∈ B(0, 1), $$n \in \mathbb{N}$$, one gets that ϕ n is real differentiable and the assumptions of Lemma 8.3, except for analyticity, are all satisfied. In addition, ϕ n (0) = 0, however, $$\phi _{n}\left (1/n\right ) = n \rightarrow \infty$$ as $$n \rightarrow \infty$$. Thus, $$\{\phi _{n}\}_{n\in \mathbb{N}}$$ is not locally bounded on B(0, 1). ◇
The next aim of this chapter is to establish Theorem 8.7, that is, an important step for obtaining (8.2). The terms to be discussed in Theorem 8.7 split up into a leading order term and the rest. The first term will be studied in Lemma 8.5 and the second one in Lemma 8.6. The strategy of proof in these lemmas is the same. It rests on the following observation: Let $$U \subseteq \mathbb{C}$$ open, $$U \ni z\mapsto T(z) \in \mathcal{B}\big(L^{2}(\mathbb{R}^{n})\big)$$. Assume that for all z ∈ U we have $$T(z) \in \mathcal{B}_{1}\big(L^{2}(\mathbb{R}^{n})\big)$$ and that $$z\mapsto \mathop{\mathrm{tr}}\nolimits (\vert T(z)\vert )$$ is locally bounded . Then
$$\displaystyle{\{z\mapsto \mathop{\mathrm{tr}}\nolimits (\chi _{\varLambda }T(z))\}_{\varLambda>0}}$$
is locally bounded as well. Indeed, the assertion follows from the boundedness of the family $$\{\chi _{\varLambda }\}_{\varLambda>0}$$ as bounded linear (multiplication) operators in $$\mathcal{B}(L^{2}(\mathbb{R}^{n}))$$ and the ideal property of the trace class . In the situations to be considered in the following, the trace class property for T(z) will be shown with the help of the results of Chap.

### Lemma 8.5

Let $$L = \mathcal{Q}+\varPhi$$ be given by () and for $$z \in \mathbb{C}$$ with Re (z) > −1 let R1+z be given by () and C as in () , $$n \in \mathbb{N}_{>1}$$ odd. For j ∈{ 1,…,n}, let $$\gamma _{j,n} \in \mathbb{C}^{2^{\hat{n}}\times 2^{\hat{n}} }$$ as in Remark 6.1 and $$\chi _{\varLambda }$$ as in () , $$\varLambda> 0$$. For $$z \in \mathbb{C}_{\mathrm{Re}>-1}$$ consider
$$\displaystyle{\psi _{\varLambda }(z)\,:=\,\chi _{\varLambda }\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\left [\mathcal{Q},\varPhi \left (CR_{1+z}\right )^{n}\right ]\big)}$$
and
$$\displaystyle{\tilde{\psi }_{\varLambda }(z)\,:=\,\chi _{\varLambda }\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\left [\mathcal{Q},\mathcal{Q}\left (CR_{1+z}\right )^{n}\right ]\big).}$$
Then for all $$z \in \mathbb{C}_{\mathrm{Re}>-1}$$, the operators $$\psi _{\varLambda }(z)$$, $$\tilde{\psi }_{\varLambda }(z)$$ are trace class and the families
$$\displaystyle{\{z\mapsto \mathop{\mathrm{tr}}\nolimits _{L^{2}(\mathbb{R}^{n})}(\psi _{\varLambda }(z))\}_{\varLambda>0}\,\text{ and }\,\big\{z\mapsto \mathop{\mathrm{tr}}\nolimits _{L^{2}(\mathbb{R}^{n})}\big(\tilde{\psi }_{\varLambda }(z)\big)\big\}_{\varLambda>0}}$$
are locally bounded (cf. (8.1) ).

### Proof

First we deal with $$\psi _{\varLambda }(z)$$. One computes,
$$\displaystyle{\psi _{\varLambda }(z) =\chi _{\varLambda }\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\left [\mathcal{Q},\varPhi \left (CR_{1+z}\right )^{n}\right ]\big) =\chi _{\varLambda }\mathop{\mathrm{tr}}\nolimits _{ 2^{\hat{n}}d}\big(\mathcal{Q}\varPhi \left (CR_{1+z}\right )^{n} -\varPhi \left (CR_{ 1+z}\right )^{n}\mathcal{Q}\big).}$$
Before we discuss the latter operator, we note that
$$\displaystyle\begin{array}{rcl} \mathcal{Q}\varPhi \left (CR_{1+z}\right )^{n}& =& \varPhi \mathcal{Q}\left (CR_{ 1+z}\right )^{n} + [\mathcal{Q},\varPhi ]\left (CR_{ 1+z}\right )^{n} {}\\ & =& \varPhi \bigg(\sum _{j=1}^{n}(CR_{ 1+z})^{j-1}[\mathcal{Q},C]R_{ 1+z}(CR_{1+z})^{n-j} + (CR_{ 1+z})^{n}\mathcal{Q}\bigg) {}\\ & & +[\mathcal{Q},\varPhi ]\left (CR_{1+z}\right )^{n}, {}\\ \end{array}$$
where the latter equality follows via an induction argument. Hence,
$$\displaystyle\begin{array}{rcl} & & \psi _{\varLambda }(z) =\chi _{\varLambda }\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\bigg(\varPhi \bigg(\sum _{j=1}^{n}(CR_{ 1+z})^{j-1}[\mathcal{Q},C]R_{ 1+z}(CR_{1+z})^{n-j}\bigg) \\ & & \phantom{vvvvvva}\qquad \qquad \ \ + [\mathcal{Q},\varPhi ](CR_{1+z})^{n}\bigg). {}\end{array}$$
(8.5)
Next, with the results of Chap. , we will deduce that the operator family
$$\displaystyle{ z\mapsto \bigg(\varPhi \bigg(\sum _{j=1}^{n}(CR_{ 1+z})^{j-1}[\mathcal{Q},C]R_{ 1+z}(CR_{1+z})^{n-j}\bigg) + [\mathcal{Q},\varPhi ]\left (CR_{ 1+z}\right )^{n}\bigg) }$$
(8.6)
is trace class , which—together with the estimates in Lemma 4.5—establishes the assertion for $$\psi _{\varLambda }$$: Indeed, the only difference between (8.5) and (8.6) is the prefactor $$\chi _{\varLambda }$$. So we get the assertion with the help of the reasoning prior to Lemma 8.5. In order to observe that each summand in (8.6) is trace class , we proceed as follows. Recall $$n = 2\hat{n} + 1$$ and let $$j \in \{ 1,\ldots,\hat{n}\}$$ (the case $$n - j \in \{ 1,\ldots,\hat{n}\}$$ can be dealt with similarly). Then, by the admissibility of Φ (see Hypothesis 6.11), one infers that $$[\mathcal{Q},C]$$ is a multiplication operator with
$$\displaystyle{\vert [\mathcal{Q},C](x)\vert \leqslant \kappa (1 + \vert x\vert )^{-1-\varepsilon },\quad x \in \mathbb{R}^{n}.}$$
Hence, as $$1+\varepsilon> 3/2$$ by Definition , Theorem  applies and guarantees that
$$\displaystyle{(CR_{1+z})^{j-1}[\mathcal{Q},C]R_{ 1+z}(CR_{1+z})^{\hat{n}-j}}$$
is Hilbert–Schmidt . Using Theorem , one deduces that $$(CR_{1+z})^{\hat{n}+1}$$ is also Hilbert–Schmidt and thus
$$\displaystyle{(CR_{1+z})^{j-1}[\mathcal{Q},C]R_{ 1+z}(CR_{1+z})^{\hat{n}-j}(CR_{ 1+z})^{\hat{n}+1}}$$
is trace class , by Theorem .
For $$\tilde{\psi }_{\varLambda }$$ one proceeds similarly. First one notes that
$$\displaystyle{ \tilde{\psi }_{\varLambda }(z) =\chi _{\varLambda }\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\bigg(\mathcal{Q}\bigg(\sum _{j=1}^{n}(CR_{ 1+z})^{j-1}[\mathcal{Q},C]R_{ 1+z}(CR_{1+z})^{n-j}\bigg)\bigg). }$$
(8.7)
Applying Theorems 4.7, 4.9, and 4.2, one infers the assertion for $$\tilde{\psi }_{\varLambda }$$. However, one has to use the respective assertions, where some of the resolvent s of the Laplacian is replaced by $$\mathcal{Q}$$ times the resolvent s. Indeed, in the sum in (8.7), the term for j = 1 yields
$$\displaystyle\begin{array}{rcl} & & \mathcal{Q}[\mathcal{Q},C]R_{1+z}(CR_{1+z})^{n-1} = [\mathcal{Q},[\mathcal{Q},C]]R_{ 1+z}(CR_{1+z})^{n-1} {}\\ & & \phantom{\qquad \qquad vvvvvvvvvvvvvvvvv} + [\mathcal{Q},C]\mathcal{Q}R_{1+z}(CR_{1+z})^{n-1} {}\\ & & \quad = [\mathcal{Q},[\mathcal{Q},C]]R_{1+z}(CR_{1+z})^{n-1} + [\mathcal{Q},C]R_{ 1+z}[\mathcal{Q},C]R_{1+z}(CR_{1+z})^{n-2} {}\\ & & \qquad + [\mathcal{Q},C]R_{1+z}C\mathcal{Q}R_{1+z}(CR_{1+z})^{n-2}, {}\\ \end{array}$$
and for j′ ∈ { 2, , n} one obtains
$$\displaystyle\begin{array}{rcl} & & \mathcal{Q}CR_{1+z}(CR_{1+z})^{j'-2}[\mathcal{Q},C]R_{ 1+z}(CR_{1+z})^{n-j'} {}\\ & & \quad = [\mathcal{Q},C]R_{1+z}(CR_{1+z})^{j'-2}[\mathcal{Q},C]R_{ 1+z}(CR_{1+z})^{n-j'} {}\\ & & \qquad + C\mathcal{Q}R_{1+z}(CR_{1+z})^{j'-2}[\mathcal{Q},C]R_{ 1+z}(CR_{1+z})^{n-j'}. {}\\ \end{array}$$
□
The next lemma is the reason, why we have to invoke Lemma 8.3 in our argument. The crucial point is that we can use the Neumann series expressions for the resolvent s $$(L^{{\ast}}L + z)^{-1}$$ and $$(LL^{{\ast}} + z)^{-1}$$ only for z with large real part. But for z in the vicinity of 0, we do not have such a representation. Using again the ideal property for trace class operators, we can, however, bound $$z(L^{{\ast}}L + z)^{-1}$$ for small z in the $$\mathcal{B}(L^{2}(\mathbb{R}^{n}))$$-norm. Introducing the sector $$\varSigma _{z_{0},\vartheta } \subset \mathbb{C}$$ by
$$\displaystyle{ \varSigma _{z_{0},\vartheta }\,:=\,\{z \in \mathbb{C}\,\vert \,\mathrm{Re}(z)> z_{0},\vert \arg (\mu )\vert <\vartheta \}, }$$
(8.8)
for some $$z_{0} \in \mathbb{R}$$ and $$\vartheta \in (0,\pi /2)$$, the result reads as follows:

### Lemma 8.6

Let $$L = \mathcal{Q}+\varPhi$$ be given by () and for $$z \in \mathbb{C}$$ with Re (z) > −1, let R1+z be given by () and C as in () , $$\chi _{\varLambda }$$ as in () , $$\varLambda> 0$$. For j ∈{ 1,…,n}, let $$\gamma _{j,n} \in \mathbb{C}^{2^{\hat{n}}\times 2^{\hat{n}} }$$ as in Remark  6.1 . For $$z \in \mathbb{C}_{\mathrm{Re}>-1} \cap \varrho (-L^{{\ast}}L) \cap \varrho (-LL^{{\ast}})$$ consider
$$\displaystyle{\eta _{\varLambda }(z)\,:=\,\chi _{\varLambda }\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\big[\mathcal{Q},\varPhi \big(\left (L^{{\ast}}L + z\right )^{-1} -\left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n+1}\big]\big)}$$
and
$$\displaystyle{\tilde{\eta }_{\varLambda }(z)\,:=\,\chi _{\varLambda }\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\big[\mathcal{Q},\big(\mathcal{Q}\big(\left (L^{{\ast}}L + z\right )^{-1} -\left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n+1}\big)\big]\big).}$$
Then for all $$z \in \mathbb{C}_{\mathrm{Re}>-1} \cap \varrho (-L^{{\ast}}L) \cap \varrho (-LL^{{\ast}})$$, the operators $$\eta _{\varLambda }(z)$$, $$\tilde{\eta }_{\varLambda }(z)$$ are trace class . There exists δ ∈ (−1,0), $$\vartheta \in (0,\pi /2)$$ such that the families
$$\displaystyle{\{\varSigma _{\delta,\vartheta } \cup \mathbb{C}_{\mathrm{Re}>0} \ni z\mapsto z\mathop{\mathrm{tr}}\nolimits _{L^{2}(\mathbb{R}^{n})}(\eta _{\varLambda }(z))\}_{\varLambda>0}}$$
and
$$\displaystyle{\left \{\varSigma _{\delta,\vartheta } \cup \mathbb{C}_{\mathrm{Re}>0} \ni z\mapsto z\mathop{\mathrm{tr}}\nolimits _{L^{2}(\mathbb{R}^{n})}\big(\tilde{\eta }_{\varLambda }(z)\big)\right \}_{\varLambda>0}}$$
are locally bounded (cf. (8.1) ).

### Proof

By the Fredholm property of L there exist δ ∈ (−1, 0) and $$\vartheta \in (0,\pi /2)$$ such that
$$\displaystyle{\varSigma _{\delta,\vartheta }\setminus \{0\} \ni z\mapsto z\left (L^{{\ast}}L + z\right )^{-1}\text{ and }\varSigma _{\delta,\vartheta }\setminus \{0\} \ni z\mapsto z\left (LL^{{\ast}} + z\right )^{-1}}$$
have analytic extensions to $$\varSigma _{\delta,\vartheta }$$. Let $$\varOmega \subset \varSigma _{\delta,\vartheta } \cup \mathbb{C}_{\mathrm{Re}>0}$$ be compact. One notes that
$$\displaystyle{\varOmega \ni z\mapsto z\left (L^{{\ast}}L + z\right )^{-1}\,\text{ and }\,\varOmega \ni z\mapsto z\left (LL^{{\ast}} + z\right )^{-1}}$$
define bounded families of bounded linear operators from $$L^{2}(\mathbb{R}^{n})^{2^{\hat{n}}d }$$ to $$H^{2}(\mathbb{R}^{n})^{2^{\hat{n}}d }$$. Indeed, by Proposition , one infers that $$\phi \mapsto \|(L^{{\ast}}L + 1)\phi \|$$ and $$\phi \mapsto \|(LL^{{\ast}} + 1)\phi \|$$ define equivalent norms on $$H^{2}(\mathbb{R}^{n})^{2^{\hat{n}}d }$$. Hence, for $$\phi \in L^{2}(\mathbb{R}^{n})^{2^{\hat{n}}d }$$ and z ∈ Ω∖{0} one computes
$$\displaystyle\begin{array}{rcl} \big\|(L^{{\ast}}L + 1)z(L^{{\ast}}L + z)^{-1}\phi \big\|& =& \vert z\vert \big\|(L^{{\ast}}L + z + 1 - z)(L^{{\ast}}L + z)^{-1}\phi \big\| {}\\ & \leqslant & \vert z\vert \|\phi \| + \vert z\vert \vert (1 - z)\vert \frac{1} {\vert z\vert }\|\phi \|. {}\\ \end{array}$$
Next, consider
$$\displaystyle\begin{array}{rcl} \eta _{\varLambda }(z)& =& \chi _{\varLambda }\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\big[\mathcal{Q},\varPhi \big(\left (L^{{\ast}}L + z\right )^{-1} -\left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n+1}\big]\big) {}\\ & =& \chi _{\varLambda }\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\mathcal{Q}\varPhi \big(\left (L^{{\ast}}L + z\right )^{-1} -\left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n+1} {}\\ & & \qquad \qquad \quad -\varPhi \big (\left (L^{{\ast}}L + z\right )^{-1} -\left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n+1}\mathcal{Q}\big). {}\\ \end{array}$$
For the first summand one observes that
$$\displaystyle\begin{array}{rcl} & & \mathcal{Q}\varPhi \big(\left (L^{{\ast}}L + z\right )^{-1} -\left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n+1} {}\\ & & \quad = C\big(\left (L^{{\ast}}L + z\right )^{-1} -\left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n+1} {}\\ & & \qquad +\varPhi \mathcal{Q}\big(\left (L^{{\ast}}L + z\right )^{-1} -\left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n+1}. {}\\ \end{array}$$
Employing our observation at the beginning of the proof and Theorem , one realizes that
$$\displaystyle{\varOmega \ni z\mapsto \mathcal{Q}z\big(\left (L^{{\ast}}L + z\right )^{-1} -\left (LL^{{\ast}} + z\right )^{-1}\big)}$$
defines a bounded family of bounded linear operators in $$L^{2}(\mathbb{R}^{n})^{2^{\hat{n}}d }$$. Thus, since $$\varOmega \ni z\mapsto \left (CR_{1+z}\right )^{n+1}$$ is a family of trace class operators,
$$\displaystyle\begin{array}{rcl} & & \varOmega \ni z\mapsto z\chi _{\varLambda }\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\mathcal{Q}\varPhi \big(\left (L^{{\ast}}L + z\right )^{-1} -\left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n+1}\big) {}\\ & & \phantom{vvvvvvvvv} =\mathop{ \mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(z\chi _{\varLambda }\big(\mathcal{Q}\varPhi \big(\left (L^{{\ast}}L + z\right )^{-1} -\left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n+1}\big)\big) {}\\ \end{array}$$
is uniformly bounded in $$\mathcal{B}_{1}$$, with bound independently of $$\varLambda> 0$$, upon appealing to the ideal property of trace class operators.
The second summand requires the observation that
$$\displaystyle{\left (CR_{1+z}\right )^{n+1}\mathcal{Q} = \left (CR_{ 1+z}\right )^{n}\left (CR_{ 1+z}\right )\mathcal{Q} = \left (CR_{1+z}\right )^{n}\left (C\mathcal{Q}R_{ 1+z}\right )}$$
defines a bounded family of trace class operators for z ∈ Ω, proving the assertion for $$\eta _{\varLambda }$$.
The corresponding assertion for $$\tilde{\eta }_{\varLambda }$$ is conceptually the same. In fact, it follows from the observation that
$$\displaystyle\begin{array}{rcl} & & \varOmega \ni z\mapsto \mathcal{Q}\mathcal{Q}z\big((L^{{\ast}}L + z)^{-1} - (LL^{{\ast}} + z)^{-1}\big) {}\\ & & \phantom{vvvvvvvvv} =\varDelta I_{2^{\hat{n}}d}z\big((L^{{\ast}}L + z)^{-1} - (LL^{{\ast}} + z)^{-1}\big) {}\\ \end{array}$$
is a bounded family of bounded linear operators by our preliminary observation that $$\varOmega \ni z \rightarrow z(L^{{\ast}}L + z)^{-1}$$ and $$\varOmega \ni z \rightarrow z(LL^{{\ast}} + z)^{-1}$$ define uniformly bounded operator families from $$L^{2}(\mathbb{R}^{n})^{2^{\hat{n}}d }$$ to $$H^{2}(\mathbb{R}^{n})^{2^{\hat{n}}d }$$, as well as using again the fact that
$$\displaystyle{\varOmega \ni z\mapsto \left (CR_{1+z}\right )^{n+1}\mathcal{Q}\,\text{ and }\,\varOmega \ni z\mapsto \left (CR_{ 1+z}\right )^{n+1}}$$
constitute bounded families of trace class operators. □

Lemmas 8.5 and 8.6 can be summarized as follows.

### Theorem 8.7

Let $$n \in \mathbb{N}_{\geqslant 3}$$ odd, let $$L = \mathcal{Q}+\varPhi$$ be given by () and for $$z \in \mathbb{C}$$ with Re (z) > −1 let R1+z be given by () and C as in () . For j ∈{ 1,…,n}, let $$\gamma _{j,n} \in \mathbb{C}^{2^{\hat{n}}\times 2^{\hat{n}} }$$ as in Remark 6.1 . For $$z \in \mathbb{C}_{\mathrm{Re}>-1} \cap \varrho (-LL^{{\ast}}) \cap \varrho (L^{{\ast}}L)$$, introduce
$$\displaystyle{\phi _{\varLambda }(z)\,:=\,\chi _{\varLambda }\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\big[\mathcal{Q},\varPhi \big((L^{{\ast}}L + z)^{-1} + (LL^{{\ast}} + z)^{-1}\big)(CR_{ 1+z})^{n}\big]\big)}$$
and
$$\displaystyle{\tilde{\phi }_{\varLambda }(z)\,:=\,\chi _{\varLambda }\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\big[\mathcal{Q},\big(\mathcal{Q}\big((L^{{\ast}}L + z)^{-1} + (LL^{{\ast}} + z)^{-1}\big)(CR_{ 1+z})^{n}\big)\big]\big).}$$
Then for all $$z \in \mathbb{C}_{\mathrm{Re}>-1} \cap \varrho (-LL^{{\ast}}) \cap \varrho (L^{{\ast}}L)$$, the operators $$\phi _{\varLambda }(z)$$, $$\tilde{\phi }_{\varLambda }(z)$$ are trace class . There exists δ ∈ (−1,0), $$\vartheta \in (0,\pi /2)$$ such that the families
$$\displaystyle{\{\varSigma _{\delta,\vartheta } \cup \mathbb{C}_{\mathrm{Re}>0} \ni z\mapsto \mathop{\mathrm{tr}}\nolimits _{L^{2}(\mathbb{R}^{n})}(z\phi _{\varLambda }(z))\}_{\varLambda>0}}$$
and
$$\displaystyle{\big\{\varSigma _{\delta,\vartheta } \cup \mathbb{C}_{\mathrm{Re}>0} \ni z\mapsto \mathop{\mathrm{tr}}\nolimits _{L^{2}(\mathbb{R}^{n})}\big(z\tilde{\phi }_{\varLambda }(z)\big)\big\}_{\varLambda>0}}$$
are locally bounded (cf. (8.1) ).

### Proof

One recalls from Eqs. () and () the expressions
$$\displaystyle\begin{array}{rcl} \left (L^{{\ast}}L + z\right )^{-1}& =& I + \left (L^{{\ast}}L + z\right )^{-1}CR_{ 1+z}, {}\\ \left (LL^{{\ast}} + z\right )^{-1}& =& I -\left (LL^{{\ast}} + z\right )^{-1}CR_{ 1+z}. {}\\ \end{array}$$
Hence, one gets
$$\displaystyle\begin{array}{rcl} \phi _{\varLambda }(z)& =& \chi _{\varLambda }\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(2\left [\mathcal{Q},\varPhi \left (CR_{1+z}\right )^{n}\right ]\big) {}\\ & & +\chi _{\varLambda }\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\big[\mathcal{Q},\varPhi \big(\left (L^{{\ast}}L + z\right )^{-1} -\left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n+1}\big]\big) {}\\ & =& 2\psi _{\varLambda }(z) +\eta _{\varLambda }(z), {}\\ \end{array}$$
and
$$\displaystyle\begin{array}{rcl} \tilde{\phi }_{\varLambda }(z)& =& \chi _{\varLambda }\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(2\left [\mathcal{Q},\mathcal{Q}\left (CR_{1+z}\right )^{n}\right ]\big) {}\\ & & +\chi _{\varLambda }\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\big[\mathcal{Q},\mathcal{Q}\big(\left (L^{{\ast}}L + z\right )^{-1} -\left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n+1}\big]\big) {}\\ & =& 2\tilde{\psi }_{\varLambda }(z) +\tilde{\eta } _{\varLambda }(z), {}\\ \end{array}$$
with the functions introduced in Lemmas 8.5 and 8.6. Thus, the assertion on the local boundedness follows from these two lemmas. □

The forthcoming statements are used for showing that for computing the trace the only term that matters is discussed in Proposition 8.13. We recall that by Remark , one can compute the trace of $$\chi _{\varLambda }B_{L}(z)$$ as the integral over the diagonal of its integral kernel . So the estimates on the diagonal derived in Chap.  will be used in the following. We shall elaborate on this idea further after having stated the next two auxiliary results. Both these results serve to show that some integral kernel s actually vanish on the diagonal .

### Lemma 8.8

Let $$n \in \mathbb{N}_{\geqslant 3}$$ be odd, $$z \in \mathbb{C}$$ with Re (z) > −1. Let R1+z, $$\mathcal{Q}$$, C, and $$\gamma _{j,n} \in \mathbb{C}^{2^{\hat{n}}\times 2^{\hat{n}} }$$, j ∈{ 1,…,n}, be given by () , () , () and as in Remark 6.1 , respectively. Let $$\varPhi: \mathbb{R}^{n} \rightarrow \mathbb{C}^{d\times d}$$ be admissible (see Definition 6.11 ). Then for all j ∈{ 1,…,n},
$$\displaystyle{ \mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\left (R_{1+z}C\right )^{n-2}R_{ 1+z}\big) = -\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\left (R_{1+z}\varPhi \mathcal{Q}\right )^{n-2}R_{ 1+z}\big). }$$
(8.9)

### Proof

One has
$$\displaystyle\begin{array}{rcl} \mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\left (\gamma _{j,n}\mathcal{Q}R_{1+z}CR_{1+z}\right )& =& \mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\left (\gamma _{j,n}\mathcal{Q}R_{1+z}\left (\mathcal{Q}\varPhi -\varPhi \mathcal{Q}\right )R_{1+z}\right ), {}\\ & =& \mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\left (\gamma _{j,n}R_{1+z}\mathcal{Q}\mathcal{Q}\varPhi R_{1+z}\right ) {}\\ & & -\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\left (\gamma _{j,n}\mathcal{Q}R_{1+z}\varPhi \mathcal{Q}R_{1+z}\right ) {}\\ & =& -\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\left (\gamma _{j,n}\mathcal{Q}R_{1+z}\varPhi \mathcal{Q}R_{1+z}\right ), {}\\ \end{array}$$
using Proposition A.8 to deduce that $$\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\left (\gamma _{j,n}R_{1+z}\mathcal{Q}\mathcal{Q}\varPhi R_{1+z}\right ) = 0$$. In order to proceed to the proof of (8.9), we now show the following: For all odd k ∈ { 3, , n} and  ∈ { 0, , k − 2} one has
$$\displaystyle\begin{array}{rcl} \begin{array}{rl} &\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\left (R_{1+z}C\right )^{k-2}R_{1+z}\big) \\ &\quad = \left (-1\right )^{\ell}\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\left (R_{1+z}\varPhi \mathcal{Q}\right )^{\ell}\left (R_{1+z}C\right )^{k-2-\ell}R_{1+z}\big).\end{array} & &{}\end{array}$$
(8.10)
In the beginning of the proof we have dealt with the case k = 3. One notes that Eq. (8.10) always holds for  = 0. Next, we assume that k ∈ { 5, , n} is odd, such that equality (8.10) holds for some  ∈ { 0, , k − 3}. Then one computes
$$\displaystyle\begin{array}{rcl} & & \mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\left (R_{1+z}C\right )^{k-2}R_{ 1+z}\big) {}\\ & & \quad = (-1)^{\ell}\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\left (R_{1+z}\varPhi \mathcal{Q}\right )^{\ell}\left (R_{1+z}C\right )^{k-2-\ell}R_{ 1+z}\big) {}\\ & & \quad = \left (-1\right )^{\ell}\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\left (R_{1+z}\varPhi \mathcal{Q}\right )^{\ell}R_{1+z}C\left (R_{1+z}C\right )^{k-2-\ell-1}R_{ 1+z}\big) {}\\ & & \quad = \left (-1\right )^{\ell}\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\left (R_{1+z}\varPhi \mathcal{Q}\right )^{\ell}R_{1+z}\left (\mathcal{Q}\varPhi -\varPhi \mathcal{Q}\right )\left (R_{1+z}C\right )^{k-2-\left (\ell+1\right )}R_{ 1+z}\big) {}\\ & & \quad = \left (-1\right )^{\ell}\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\left (R_{1+z}\varPhi \mathcal{Q}\right )^{\ell}\mathcal{Q}R_{1+z}\varPhi \left (R_{1+z}C\right )^{k-2-\left (\ell+1\right )}R_{ 1+z}\big) {}\\ & & \qquad + \left (-1\right )^{\ell+1}\mathop{ \mathrm{tr}}\nolimits _{ 2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\left (R_{1+z}\varPhi \mathcal{Q}\right )^{\ell+1}\left (R_{ 1+z}C\right )^{k-2-\left (\ell+1\right )}R_{ 1+z}\big). {}\\ \end{array}$$
By Corollary A.9, the first term on the right-hand side cancels, proving equation ( 8.10). Putting $$\ell= k - 2$$ in ( 8.10) implies the assertion. □

The following result is needed for Lemma 8.10, however, it is also of independent interest. Indeed, we will have occasion to use it rather frequently, when we discuss the case of three spatial dimensions specifically. Lemma 8.9 should be regarded as a regularization method, while preserving self-adjoint ness properties of the (L2-realization) of the underlying operators:

### Lemma 8.9

Let $$\varepsilon> 0$$, $$n \in \mathbb{N}$$, and $$T \in \mathcal{B}\big(H^{-(n/2)-\varepsilon }(\mathbb{R}^{n}),H^{(n/2)+\varepsilon }(\mathbb{R}^{n})\big)$$. Recalling equation () , we consider
$$\displaystyle{t: \mathbb{R}^{n} \times \mathbb{R}^{n} \ni (x,y)\mapsto \left \langle \delta _{\{ x\}},T\delta _{\{y\}}\right \rangle.}$$
For μ > 0 we denote $$T_{\mu }\,:=\,\left (1-\mu \varDelta \right )^{-1}T\left (1-\mu \varDelta \right )^{-1}$$ and tμ correspondingly. Then, for all $$(x,y) \in \mathbb{R}^{n} \times \mathbb{R}^{n}$$,
$$\displaystyle{t_{\mu }(x,y)\mathop{\longrightarrow }\limits_{\mu \downarrow 0}t(x,y).}$$

### Proof

It suffices to observe that for all $$s \in \mathbb{R}$$, $$\left (1-\mu \varDelta \right )^{-1}\mathop{\longrightarrow }\limits_{\mu \downarrow 0}I$$ strongly in $$H^{s}(\mathbb{R}^{n})$$ (see ()). □

In order to proceed to prove the trace theorem, we need to investigate the asymptotic behavior of the integral kernel of J L j (z) given by () on the diagonal . By Proposition  together with Lemma , we can use Gauss’ divergence theorem for computing the integral over the diagonal (see also ()). Thus, in the expression for the trace of $$\chi _{\varLambda }B_{L}(z)$$ we will use Gauss’ theorem for the ball centered at 0 with radius $$\varLambda$$. Having applied the divergence theorem, we integrate over spheres of radius $$\varLambda$$. The volume element of the surface measure grows with $$\varLambda ^{n-1}$$, so any term decaying faster than that will not contribute to the limit $$\varLambda \rightarrow \infty$$ in (). Consequently, any estimate of integral kernel s (or differences of such) to follow with the behavior of $$\vert x\vert ^{n-1+\gamma }$$ for some γ > 0 on the diagonal , can be neglected in the limit $$\varLambda \rightarrow \infty$$, when computing the expression $$\lim _{\varLambda \rightarrow \infty }\mathop{\mathrm{tr}}\nolimits (\chi _{\varLambda }B_{L}(z))$$.

### Lemma 8.10

Let $$n \in \mathbb{N}$$ odd, j ∈{ 1,…,n}, $$z \in \mathbb{C}$$, Re (z) > −1 and R1+z be given by () as well as $$\mathcal{Q}$$, C and γj,n given by () , () and as in Remark 6.1 . Then for $$n\geqslant 3$$, the integral kernel h2,j of
$$\displaystyle{2\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\varPhi \left (R_{1+z}C\right )^{n-1}R_{ 1+z}\big)}$$
satisfies,
$$\displaystyle{h_{2,j}(x,x) = h_{3,j}(x,x) + g_{0,j}(x,x),}$$
where h3,j is the integral kernel of $$2\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\left (\gamma _{j,n}\varPhi C^{n-1}R_{1+z}^{n}\right )$$ and g0,j satisfies for some κ > 0,
$$\displaystyle{\left \vert g_{0,j}(x,x)\right \vert \leqslant \kappa (1 + \vert x\vert )^{1-n-\varepsilon },\quad x \in \mathbb{R}^{n},}$$
where $$\varepsilon> 1/2$$ is given as in Definition 6.11 . In addition, if $$n\geqslant 5$$ and $$z \in \mathbb{R}$$, then the integral kernel h1,j of
$$\displaystyle{\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\left (R_{1+z}C\right )^{n-2}R_{ 1+z}\big)}$$
vanishes on the diagonal .

### Proof

We discuss h1, j first and consider the operator
$$\displaystyle{B_{n}\,:=\,\left (\varPhi \mathcal{Q}R_{1+z}\right )^{n-3}\varPhi =\varPhi \left (\mathcal{Q}R_{ 1+z}\varPhi \right )^{n-3},}$$
which is self-adjoint for all real z > −1. Indeed, this follows from the self-adjoint ness of Φ and the skew-self-adjoint ness of $$\mathcal{Q}R_{1+z}$$. For μ > 0 define $$B_{n,\mu }\,:=\,(1-\mu \varDelta )^{-1}B_{n}(1-\mu \varDelta )^{-1}$$. Then the integral kernel bn, μ of Bn, μ is continuous. Moreover, for all real z > −1, the operator Bn, μ is self-adjoint , by the self-adjoint ness of B n and so bn, μ is real and satisfies bn, μ(x, y) = bn, μ(y, x) for all $$x,y \in \mathbb{R}^{n}$$. By Lemma 8.8 one recalls
$$\displaystyle\begin{array}{rcl} & & \mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\left (R_{1+z}C\right )^{n-2}R_{ 1+z}\big) {}\\ & & \quad = -\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\left (R_{1+z}\varPhi \mathcal{Q}\right )^{n-2}R_{ 1+z}\big) {}\\ & & \quad = -\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}R_{1+z}\left (\varPhi \mathcal{Q}R_{1+z}\right )^{n-2}\big) {}\\ & & \quad = -\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}R_{1+z}\left (\varPhi \mathcal{Q}R_{1+z}\right )^{n-3}\varPhi \mathcal{Q}R_{ 1+z}\big) {}\\ & & \quad = -\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\left (\gamma _{j,n}\mathcal{Q}R_{1+z}B_{n}\mathcal{Q}R_{1+z}\right ). {}\\ \end{array}$$
By Fubini’s theorem and the symmetry of Bn, μ, one has for all j, k ∈ { 1, , n} and $$x \in \mathbb{R}^{n}$$, z > −1, μ > 0,
$$\displaystyle\begin{array}{rcl} \varPsi _{x,\mu }(\,j,k)& \,:=\,& \int _{\mathbb{R}^{n}\times \mathbb{R}^{n}}(\partial _{j}r_{1+z})(x_{1} - x)b_{n,\mu }(x_{1},x_{2})(\partial _{k}r_{1+z})(x_{2} - x)\,\mathrm{d}^{n}x_{ 1}\mathrm{d}^{n}x_{ 2} {}\\ & =& \int _{\mathbb{R}^{n}\times \mathbb{R}^{n}}(\partial _{j}r_{1+z})(x_{1} - x)b_{n,\mu }(x_{2},x_{1})(\partial _{k}r_{1+z})(x_{2} - x)\,\mathrm{d}^{n}x_{ 1}\mathrm{d}^{n}x_{ 2} {}\\ & =& \int _{\mathbb{R}^{n}\times \mathbb{R}^{n}}(\partial _{k}r_{1+z})(x_{1} - x)b_{n,\mu }(x_{1},x_{2})(\partial _{j}r_{1+z})(x_{2} - x)\,\mathrm{d}^{n}x_{ 1}\mathrm{d}^{n}x_{ 2} {}\\ & =& \varPsi _{x,\mu }(k,j). {}\\ \end{array}$$
By Lemma 8.9 one has for all $$x,y \in \mathbb{R}^{n}$$,
$$\displaystyle\begin{array}{rcl} h_{1,j}(x,y)& =& -\lim _{\mu \downarrow 0}\mathop{ \mathrm{tr}}\nolimits _{2^{\hat{n}}d}\bigg(\sum _{i_{2},i_{3}=1}^{n}\gamma _{ j,n}\gamma _{i_{2},n}\gamma _{i_{3},n}\partial _{i_{2}}\int _{\mathbb{R}^{n}\times \mathbb{R}^{n}}r_{1+z}(x - x_{1})b_{n,\mu }(x_{1},x_{2}) {}\\ & & \phantom{vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv} \times (\partial _{i_{3}}r_{1+z})(x_{2} - y)\,\mathrm{d}^{n}x_{ 1}\mathrm{d}^{n}x_{ 2}\bigg) {}\\ & =& \lim _{\mu \downarrow 0}\mathop{ \mathrm{tr}}\nolimits _{2^{\hat{n}}d}\bigg(\sum _{i_{2},i_{3}=1}^{n}\gamma _{ j,n}\gamma _{i_{2},n}\gamma _{i_{3},n}\int _{\mathbb{R}^{n}\times \mathbb{R}^{n}}(\partial _{i_{2}}r_{1+z})(x_{1} - x)b_{n,\mu }(x_{1},x_{2}) {}\\ & & \phantom{vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv} \times (\partial _{i_{3}}r_{1+z})(x_{2} - y)\,\mathrm{d}^{n}x_{ 1}\mathrm{d}^{n}x_{ 2}\bigg). {}\\ \end{array}$$
Thus, it follows from Corollary A.9 that
$$\displaystyle{h_{1,j}(x,x) =\lim _{\mu \downarrow 0}\mathop{ \mathrm{tr}}\nolimits _{2^{\hat{n}}d}\bigg(\sum _{i_{2},i_{3}=1}^{n}\gamma _{ j,n}\gamma _{i_{2},n}\gamma _{i_{3},n}\varPsi _{x,\mu }(i_{2},i_{3})\bigg) = 0,\quad x \in \mathbb{R}^{n}.}$$
The assertion about h2, j is a direct consequence of Remark  and the asymptotic conditions imposed on Φ. □

For the estimate on the diagonal of the integral kernel s of the operators under consideration in the next theorem we need to choose the real part of z large. In fact, we use the Neumann series expression for the resolvent s $$(L^{{\ast}}L + z)^{-1}$$ and $$(LL^{{\ast}} + z)^{-1}$$ and Remark , both of which making the choice of large Re(z) necessary. We shall also have an a priori bound on the argument of z, recalling the definition (8.8) of the sector $$\varSigma _{z_{0},\vartheta } \subset \mathbb{C}$$.

### Theorem 8.11

Let $$L = \mathcal{Q}+\varPhi$$ be given by () , and for $$z \in \mathbb{C}_{\mathrm{Re}>-1}$$, let R1+z be given by () and C as in () . For j ∈{ 1,…,n}, let $$\gamma _{j,n} \in \mathbb{C}^{2^{\hat{n}}\times 2^{\hat{n}} }$$ (cf. Remark  6.1 ), and $$\vartheta \in (0,\pi /2)$$. Then there exists z0 > 0, such that for all $$z \in \varSigma _{z_{0},\vartheta }$$ (see (8.8) ), the integral kernels g1,j and g2,j of the operators
$$\displaystyle{\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\varPhi \big(\left (L^{{\ast}}L + z\right )^{-1} + \left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n}\big)}$$
and
$$\displaystyle{\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\big(\left (L^{{\ast}}L + z\right )^{-1} + \left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n}\big),}$$
respectively, satisfy for some κ > 0,
$$\displaystyle{\big[\vert g_{1,j}(x,x)\vert + \vert g_{2,j}(x,x)\vert \big]\leqslant \kappa (1 + \vert x\vert )^{-n},\quad x \in \mathbb{R}^{n}.}$$

### Proof

We choose z0 such that $$\sqrt{z_{ 0}}> 2n$$ (one recalls Remark ) and that for $$M\,:=\,\sup _{x\in \mathbb{R}^{n}}\|\varPhi (x)\| \vee \|\left (\mathcal{Q}\varPhi \right )(x)\|$$ one has $$2M[z_{0}\cos (\vartheta )]^{-1/2}\leqslant 1/2$$. We treat g1, j first. Let $$z \in \varSigma _{z_{0},\vartheta }$$, then,
$$\displaystyle\begin{array}{rcl} & & \gamma _{j,n}\varPhi \big(\left (L^{{\ast}}L + z\right )^{-1} + \left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n} {}\\ & & \quad =\gamma _{j,n}\varPhi 2\left (R_{1+z}C\right )^{n}\sum _{ k=0}^{\infty }\left (R_{ 1+z}C\right )^{2k}R_{ 1+z} {}\\ & & \quad =\sum _{ k=0}^{\infty }\gamma _{ j,n}\varPhi 2\left (R_{1+z}C\right )^{n}\left (R_{ 1+z}C\right )^{2k}R_{ 1+z}. {}\\ \end{array}$$
For $$x \in \mathbb{R}^{n}$$ one infers (recalling δ{x} in ()),
$$\displaystyle\begin{array}{rcl} g_{1,j}(x,x)& =& \bigg\langle \delta _{\{x\}},\sum _{k=0}^{\infty }\gamma _{ j,n}\varPhi 2\left (R_{1+z}C\right )^{2k}\left (R_{ 1+z}C\right )^{n}R_{ 1+z}\delta _{\{x\}}\bigg\rangle {}\\ & =& \sum _{k=0}^{\infty }\big\langle \delta _{ \{x\}},\gamma _{j,n}\varPhi 2\left (R_{1+z}C\right )^{2k}\left (R_{ 1+z}C\right )^{n}R_{ 1+z}\delta _{\{x\}}\big\rangle. {}\\ \end{array}$$
Hence, by Lemma  together with Remark , there exists c > 0 such that for all $$x \in \mathbb{R}^{n}$$,
$$\displaystyle{\big\vert \big\langle \delta _{\{x\}},\gamma _{j,n}\varPhi 2\left (R_{1+z}C\right )^{2k}\left (R_{ 1+z}C\right )^{n}R_{ 1+z}\delta _{\{x\}}\big\rangle \big\vert \leqslant c\left ( \frac{2M} {\sqrt{1 + z_{0}}}\right )^{2k}\left ( \frac{1} {1 + \left \vert x\right \vert }\right )^{n}.}$$
Since $$2M[1 + z_{0}]^{-1/2}\leqslant 1/2$$, one concludes that
$$\displaystyle\begin{array}{rcl} \left \vert g_{1,j}(x,x)\right \vert & \leqslant & \sum _{k=0}^{\infty }\left \vert \langle \delta _{\{ x\}},\gamma _{j,n}\varPhi 2\left (R_{1+z}C\right )^{2k}\left (R_{ 1+z}C\right )^{n}R_{ 1+z}\delta _{\{x\}}\rangle \right \vert {}\\ &\leqslant & \sum _{k=0}^{\infty }c\left ( \frac{2M} {\sqrt{1 + z_{0}}}\right )^{2k}\left ( \frac{1} {1 + \left \vert x\right \vert }\right )^{n}\leqslant c\left ( \frac{1} {1 + \left \vert x\right \vert }\right )^{n}. {}\\ \end{array}$$
The analogous reasoning applies to g2, j. □

We conclude the results on estimates of certain integral kernel s on the diagonal with the following corollary, which, roughly speaking, says that the diagonal of the integral kernel s involved is determined by the integral kernel of the operator to be discussed in Proposition 8.13.

### Corollary 8.12

For $$z \in \mathbb{C}$$, Re (z) > −1, denote R1+z as in () , let $$\varPhi: \mathbb{R}^{n} \rightarrow \mathbb{C}^{d\times d}$$ be admissible (see Definition 6.11 ), and $$L = \mathcal{Q}+\varPhi$$ as in () , $$\vartheta \in (0,\pi /2)$$. In addition, denote JLj(z) for $$z \in \varrho \left (-L^{{\ast}}L\right ) \cap \varrho \left (-LL^{{\ast}}\right )$$ as in () for all j ∈{ 1,…,n}, and C as in () . Moreover, let $$\gamma _{j,n} \in \mathbb{C}^{2^{\hat{n}}\times 2^{\hat{n}} }$$, j ∈{ 1,…,n} (cf. Remark 6.1 ).
1. (i)
Let $$n \in \mathbb{N}_{\geqslant 5}$$, j ∈{ 1,…,n}. Then there exists z0 > 0, such that if $$z \in \varSigma _{z_{0},\vartheta }$$ (see (8.8) ), and h and g denote the integral kernel of $$2\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\left (\gamma _{j,n}\varPhi C^{n-1}R_{1+z}^{n}\right )$$ and JLj(z), respectively, then for some κ > 0,
$$\displaystyle{\left \vert h(x,x) - g(x,x)\right \vert \leqslant \kappa (1 + \vert x\vert )^{1-n-\varepsilon },\quad x \in \mathbb{R}^{n},}$$
where $$\varepsilon> 1/2$$ is given as in Definition 6.11 .

2. (ii)

The assertion of part (i) also holds for n = 3, if, in the above statement, J L j (z) is replaced by $$J_{L}^{j}(z) - 2\mathop{\mathrm{tr}}\nolimits _{2d}\left (\gamma _{j,3}\mathcal{Q}R_{1+z}CR_{1+z}\right )$$ .

### Proof

One recalls from Lemma ,
$$\displaystyle\begin{array}{rcl} J_{L}^{j}(z)& =& 2\mathop{\mathrm{tr}}\nolimits _{ 2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\left (R_{1+z}C\right )^{n-2}R_{ 1+z}\big) + 2\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\varPhi \left (R_{1+z}C\right )^{n-1}R_{ 1+z}\big) {}\\ & & +\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\big(\left (L^{{\ast}}L + z\right )^{-1} + \left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n}\big) {}\\ & & +\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\varPhi \big(\left (L^{{\ast}}L + z\right )^{-1} + \left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n}\big). {}\\ \end{array}$$
With the help of Theorem 8.11 one deduces that the integral kernel s of the last two terms may be estimated by $$\kappa (1 + \vert x\vert )^{-n}$$ on the diagonal . The integral kernel of the first term on the right-hand side vanishes on the diagonal , which is asserted in Lemma 8.10. Hence, it remains to inspect the second term of the right-hand side. Thus, the assertion follows from Lemma 8.10. □

Having identified the integral kernel g j of $$2\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\left (\gamma _{j,n}\varPhi C^{n-1}R_{1+z}^{n}\right )$$ to be the only term determining the trace of $$\chi _{\varLambda }B_{L}(z)$$ as $$\varLambda \rightarrow \infty$$, we shall compute the integral over the diagonal of g j :

### Proposition 8.13

Let $$n \in \mathbb{N}_{\geqslant 3}$$ odd, C as in () , $$z \in \mathbb{C}$$, Re (z) > −1, with R1+z given by () , $$\varPhi: \mathbb{R}^{n} \rightarrow \mathbb{C}^{d\times d}$$ be admissible (see Definition 6.11 ), $$\gamma _{j,n} \in \mathbb{C}^{2^{\hat{n}}\times 2^{\hat{n}} }$$, j ∈{ 1,…,n}, as in Remark 6.1 . Then for j ∈{ 1,…,n}, the integral kernel gj of
$$\displaystyle{2\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\left (\gamma _{j,n}\varPhi C^{n-1}R_{ 1+z}^{n}\right )}$$
satisfies,
$$\displaystyle\begin{array}{rcl} g_{j}(x,x)& =& (1 + z)^{-n/2}\left ( \frac{\mathrm{i}} {8\pi }\right )^{(n-1)/2} \frac{1} {\left [(n - 1)/2\right ]!} {}\\ & & \times \sum _{i_{1},\ldots,i_{n-1}=1}^{n}\varepsilon _{ ji_{1}\ldots i_{n-1}}\mathop{ \mathrm{tr}}\nolimits \big(\varPhi (x)(\partial _{i_{1}}\varPhi )(x)\ldots (\partial _{i_{n-1}}\varPhi )(x)\big),\quad x \in \mathbb{R}^{n}, {}\\ \end{array}$$
where $$\varepsilon _{ji_{1}\ldots i_{n-1}}$$ denotes the $$\varepsilon$$-symbol as in Proposition A.8 .

### Proof

We recall that $$n = 2\hat{n} + 1$$. With the help of Proposition A.8, g j is given by
$$\displaystyle\begin{array}{rcl} & & (x,y)\mapsto 2(2\mathrm{i})^{\hat{n}}\sum _{ i_{1}\ldots i_{n-1}=1}^{n}\varepsilon _{ ji_{1}\ldots i_{n-1}}\mathop{ \mathrm{tr}}\nolimits \big(\varPhi (x)(\partial _{i_{1}}\varPhi )(x)\ldots (\partial _{i_{n-1}}\varPhi )(x)\big) {}\\ & & \qquad \quad \;\; \times \int _{\left (\mathbb{R}^{n}\right )^{n-1}}r_{1+z}(x - x_{1})r_{1+z}(x_{1} - x_{2})\cdots r_{1+z}(x_{n-1} - y)\,\mathrm{d}^{n}x_{ 1}\cdots \mathrm{d}^{n}x_{ n-1}. {}\\ \end{array}$$
Hence, by substitution in the integral expression and putting x = y, one obtains
$$\displaystyle\begin{array}{rcl} g_{j}(x,x)& =& 2(2\mathrm{i})^{\hat{n}}\sum _{ i_{1}\ldots i_{n-1}=1}^{n}\varepsilon _{ ji_{1}\ldots i_{n-1}}\mathop{ \mathrm{tr}}\nolimits \big(\varPhi (x)(\partial _{i_{1}}\varPhi )(x)\ldots (\partial _{i_{n-1}}\varPhi )(x)\big) {}\\ & & \times \int _{\left (\mathbb{R}^{n}\right )^{n-1}}r_{1+z}(x_{1})r_{1+z}(x_{1} - x_{2})\cdots r_{1+z}(x_{n-1})\,\mathrm{d}^{n}x_{ 1}\cdots \mathrm{d}^{n}x_{ n-1}. {}\\ \end{array}$$
The last integral can be computed with the help of the Fourier transform and polar coordinates, as was done in Proposition . In fact, one gets (see also [57, 3.252.2]),
$$\displaystyle\begin{array}{rcl} & & \int _{\left (\mathbb{R}^{n}\right )^{n-1}}r_{1+z}(x_{1})r_{1+z}(x_{1} - x_{2})\cdots r_{1+z}(x_{n-1})\,\mathrm{d}^{n}x_{ 1}\cdots \mathrm{d}^{n}x_{ n-1} {}\\ & & \ \ = \left (2\pi \right )^{-n} \frac{2\pi ^{n/2}} {\varGamma \left (n/2\right )}\int _{0}^{\infty }r^{n-1} \frac{1} {\left (r^{2} + 1 + z\right )^{n}}\,\mathrm{d}r {}\\ & & \ \ = \left (2\pi \right )^{-n} \frac{2\pi ^{n/2}} {\varGamma \left (n/2\right )}(1 + z)^{-n/2} \frac{2^{-n}\sqrt{\pi }\varGamma \left (n/2\right )} {\left [(n - 1)/2\right ]!} {}\\ & & \ \ = \frac{1} {2^{2n-1}} \frac{1} {\pi ^{(n-1)/2}} \frac{1} {\left [(n - 1)/2\right ]!}(1 + z)^{-n/2}, {}\\ \end{array}$$
and notes that
$$\displaystyle\begin{array}{rcl} & & 2(2\mathrm{i})^{(n-1)/2} \frac{1} {2^{2n-1}} \frac{1} {\pi ^{(n-1)/2}} \frac{1} {\left [(n - 1)/2\right ]!} {}\\ & & \quad = \left (\frac{\mathrm{i}} {\pi } \right )^{(n-1)/2} \frac{1} {2^{(4n-4-n+1)/2}} \frac{1} {\left [(n - 1)/2\right ]!} {}\\ & & \quad = \left (\frac{\mathrm{i}} {\pi } \right )^{(n-1)/2} \frac{1} {2^{(3n-3)/2}} \frac{1} {\left [(n - 1)/2\right ]!} {}\\ & & \quad = \left ( \frac{\mathrm{i}} {8\pi }\right )^{(n-1)/2} \frac{1} {\left [(n - 1)/2\right ]!}. {}\\ \end{array}$$
□

Finally, we are ready to prove the (trace) Theorem , for $$n\geqslant 5$$, that is, we consider the operator $$L = \mathcal{Q}+\varPhi$$ with an admissible potential Φ, such that Φ is smooth and attains values in the self-adjoint , unitary d × d-matrices. In addition, we recall that the first derivatives of Φ behave like | x | −1 for large x, whereas higher-order derivatives decay at least with the behavior $$\vert x\vert ^{-1-\varepsilon }$$ for large x and some $$\varepsilon> 1/2$$. We note that we already established the Fredholm property of L in Theorem . We outline the proof of Theorem , for $$n\geqslant 5$$, as follows. The results in Chap.  yield the applicability of Theorem . More precisely, the operator $$\chi _{\varLambda }B_{L}(z)$$ is trace class with trace computable as the integral over the diagonal of the integral kernel of $$\chi _{\varLambda }B_{L}(z)$$. With Proposition  we will deduce that only the term involving J L j (z), being analysed in Lemma , matters for the computation of the index. Next, we will show that $$\{z\mapsto \mathop{\mathrm{tr}}\nolimits (\chi _{\varLambda }B_{L}(z))\}_{\varLambda>0}$$ is locally bounded using Lemma  (in particular ()). The local boundedness result is then obtained via Gauss’ divergence theorem and Lemma 8.10 as well as Theorem 8.7. Having proved local boundedness , we will use Montel ’s theorem for deducing that at least for a sequence $$\{\varLambda _{k}\}_{k\in \mathbb{N}}$$ the limit $$f\,:=\,\lim _{k\rightarrow \infty }\mathop{\mathrm{tr}}\nolimits (\chi _{\varLambda _{k}}B_{L}(\cdot ))$$ exists in the compact open topology , that is, the topology of uniform convergence on compacts. With the results from Corollary 8.12 and Proposition 8.13, choosing Re(z) sufficiently large, we get an explicit expression for f. The explicit expression for f, by the principle of analytic continuation, carries over to z in a neighborhood of 0. As we know, by Theorem , that the limit $$\lim _{\varLambda \rightarrow \infty }\mathop{\mathrm{tr}}\nolimits (\chi _{\varLambda }B_{L}(0))$$ exists and coincides with the index of L, we can then deduce that not only for the sequence $$\{\varLambda _{k}\}_{k\in \mathbb{N}}$$ but, in fact, the limit $$\lim _{\varLambda \rightarrow \infty }\mathop{\mathrm{tr}}\nolimits (\chi _{\varLambda }B_{L}(\cdot ))$$ exists in the compact open topology and coincides with f given in (). The detailed arguments read as follows.

### Proof (of Theorem  7.1 for $$n\geqslant 5$$)

By Theorem , $$\chi _{\varLambda }B_{L}(z)$$ is trace class for every $$\varLambda> 0$$. Moreover, by Remark , $$\mathop{\mathrm{tr}}\nolimits (\chi _{\varLambda }B_{L}(z))$$ can be computed as the integral over the diagonal of the respective integral kernel . Hence, by Proposition , Eq. (), recalling also Remark , one obtains
$$\displaystyle\begin{array}{rcl} 2\mathop{\mathrm{tr}}\nolimits (\chi _{\varLambda }B_{L}(z))& =& 2\int _{B(0,\varLambda )}\left \langle \delta _{\{x\}},B_{L}(z)\delta _{\{x\}}\right \rangle _{H^{-(n/2)-\varepsilon },H^{(n/2)+\varepsilon }}\,\mathrm{d}^{n}x \\ & =& \int _{B(0,\varLambda )}\bigg\langle \delta _{\{x\}},\bigg(\sum _{j=1}^{n}\big[\partial _{ j},J_{L}^{j}(z)\big] + A_{ L}(z)\bigg)\delta _{\{x\}}\bigg\rangle \,\mathrm{d}^{n}x \\ & =& \int _{B(0,\varLambda )}\bigg\langle \delta _{\{x\}},\sum _{j=1}^{n}\big[\partial _{ j},J_{L}^{j}(z)\big]\delta _{\{ x\}}\bigg\rangle \,\mathrm{d}^{n}x, {}\end{array}$$
(8.11)
where we used Lemma 8.1 to deduce that $$\left \langle \delta _{\{x\}},A_{L}(z)\delta _{\{x\}}\right \rangle = 0$$ for all $$x \in \mathbb{R}^{n}$$. Next, we prove that $$\{z\mapsto \mathop{\mathrm{tr}}\nolimits (\chi _{\varLambda }B_{L}(z))\}_{\varLambda>0}$$ is locally bounded . One recalls from Lemma ,
$$\displaystyle\begin{array}{rcl} J_{L}^{j}(z)& =& 2\mathop{\mathrm{tr}}\nolimits _{ 2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\left (R_{1+z}C\right )^{n-2}R_{ 1+z}\big) + 2\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\varPhi \left (R_{1+z}C\right )^{n-1}R_{ 1+z}\big) {}\\ & & +\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\big(\left (L^{{\ast}}L + z\right )^{-1} + \left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n}\big) {}\\ & & +\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\varPhi \big(\left (L^{{\ast}}L + z\right )^{-1} + \left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n}\big). {}\\ \end{array}$$
Hence,
$$\displaystyle\begin{array}{rcl} & & \sum _{j=1}^{n}\big[\partial _{ j},J_{L}^{j}(z)\big] =\sum _{ j=1}^{n}\big[\partial _{ j},\big(2\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}(R_{1+z}C)^{n-2}R_{ 1+z}\big) \\ & & \phantom{vvvvvvvvvvvvvvvvvvvv} + 2\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\varPhi (R_{1+z}C)^{n-1}R_{ 1+z}\big)\big)\big] \\ & & \quad +\mathop{ \mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\big[\mathcal{Q},\mathcal{Q}\big(\left (L^{{\ast}}L + z\right )^{-1} + \left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n}\big]\big) \\ & & \quad +\mathop{ \mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\big[\mathcal{Q},\varPhi \big(\left (L^{{\ast}}L + z\right )^{-1} + \left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n}\big]\big). {}\end{array}$$
(8.12)
Denoting by h j the integral kernel of $$2\text{tr}_{2^{\hat{n}}d}\left (\gamma _{j,n}\varPhi C^{n-1}R_{1+z}^{n}\right )$$, one observes that for some constant κ > 0, $$\vert h_{j}(x,x)\vert \leqslant \kappa (1 + \vert x\vert )^{1-n}$$, $$x \in \mathbb{R}^{n}$$. Hence, for any $$\varLambda> 0$$, invoking Lemma , Gauss’ theorem implies that
$$\displaystyle\begin{array}{rcl} & & \bigg\vert \int _{B(0,\varLambda )}\bigg\langle \delta _{\{x\}},\sum _{j=1}^{n}\left [\partial _{ j},2\text{tr}_{2^{\hat{n}}d}\left (\gamma _{j,n}\varPhi C^{n-1}R_{ 1+z}^{n}\right )\right ]\delta _{\{ x\}}\bigg\rangle \,\mathrm{d}^{n}x\bigg\vert \\ & &\ \ \ =\bigg \vert \int _{B(0,\varLambda )}\sum _{j=1}^{n}(\partial _{ j}h_{j})(x,x)\,\mathrm{d}^{n}x\bigg\vert \\ & &\ \ \ =\bigg \vert \int _{\varLambda S^{n-1}}\sum _{j=1}^{n}h_{ j}(x,x)\frac{x_{j}} {\varLambda } \,\mathrm{d}^{n-1}\sigma (x)\bigg\vert \\ & &\ \ \ \leqslant \int _{\varLambda S^{n-1}}\sum _{j=1}^{n}\left \vert h_{ j}(x,x)\right \vert \,\mathrm{d}^{n-1}\sigma (x) \\ & & \ \ \ \leqslant n\kappa (1+\varLambda )^{1-n}\varLambda ^{n-1}\omega _{ n-1}, {}\end{array}$$
(8.13)
(with ωn−1 being the (n − 1)-dimensional volume of the unit sphere $$S^{n-1} \subset \mathbb{R}^{n}$$, see ()). The latter is uniformly bounded with respect to $$\varLambda> 0$$.
Using Lemma 8.10, the definition of g0, j in that lemma as well as Gauss’ theorem, one arrives at
$$\displaystyle\begin{array}{rcl} & & \bigg\vert \int _{B(0,\varLambda )}\bigg\langle \delta _{\{x\}},\sum _{j=1}^{n}\big[\partial _{ j},2\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}\left (R_{1+z}C\right )^{n-2}R_{ 1+z}\big)\big]\delta _{\{x\}}\bigg\rangle \,\mathrm{d}^{n}x \\ & & \qquad +\int _{B(0,\varLambda )}\bigg\langle \delta _{\{x\}},\sum _{j=1}^{n}\big[\partial _{ j},\big(2\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\gamma _{j,n}\varPhi \left (R_{1+z}C\right )^{n-1}R_{ 1+z}\big)\big]\delta _{\{x\}}\bigg\rangle \,\mathrm{d}^{n}x \\ & & \qquad -\int _{B(0,\varLambda )}\bigg\langle \delta _{\{x\}},\sum _{j=1}^{n}\big[\partial _{ j},2\text{tr}_{2^{\hat{n}}d}\left (\gamma _{j,n}\varPhi C^{n-1}R_{ 1+z}^{n}\right )\big]\delta _{\{ x\}}\bigg\rangle \,\mathrm{d}^{n}x\bigg\vert \\ & &\ \ \ =\bigg \vert \int _{B(0,\varLambda )}\sum _{j=1}^{n}(\partial _{ j}g_{0,j})(x,x)\,\mathrm{d}^{n}x\bigg\vert \\ & &\ \ \ \leqslant \bigg\vert \int _{\varLambda S^{n-1}}\sum _{j=1}^{n}g_{ 0,j}(x,x)\frac{x_{j}} {\varLambda } \,\mathrm{d}^{n-1}\sigma (x)\bigg\vert \\ & &\ \ \ \leqslant \int _{\varLambda S^{n-1}}\sum _{j=1}^{n}\left \vert g_{ 0,j}(x,x)\right \vert \,\mathrm{d}^{n-1}\sigma (x) \\ & & \ \ \ \leqslant \int _{\varLambda S^{n-1}}n\kappa (1 + \vert x\vert )^{1-n-\varepsilon }\,\mathrm{d}^{n-1}\sigma (x) \\ & & \ \ \ \leqslant n\kappa (1+\varLambda )^{1-n-\varepsilon }\omega _{ n-1}\varLambda ^{n-1}\mathop{\longrightarrow }\limits_{\varLambda \rightarrow \infty }0. {}\end{array}$$
(8.14)
Next, Theorem 8.7 implies that
$$\displaystyle\begin{array}{rcl} & & \Big\{z\mapsto z\mathop{\mathrm{tr}}\nolimits \Big(\chi _{\varLambda }\Big(\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\big[\mathcal{Q},\mathcal{Q}\big((L^{{\ast}}L + z)^{-1} + (LL^{{\ast}} + z)^{-1}\big)(CR_{ 1+z})^{n}\big]\big) \\ & & \quad +\mathop{ \mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\big[\mathcal{Q},\varPhi \big((L^{{\ast}}L + z)^{-1} + (LL^{{\ast}} + z)^{-1}\big)(CR_{ 1+z})^{n}\big]\big)\Big)\Big)\Big\}_{\varLambda>0}{}\end{array}$$
(8.15)
is bounded on any compact neighborhood of 0 intersected with $$B(0,\delta ) \cup (\varrho (-LL^{{\ast}}) \cap \varrho (-L^{{\ast}}L))$$ for some δ > 0. Hence, summarizing Eqs. (8.11) and (8.12), we get for $$z \in \mathbb{C}_{\mathrm{Re}>-1} \cap \varrho (-L^{{\ast}}L) \cap \varrho (-LL^{{\ast}})$$:
$$\displaystyle\begin{array}{rcl} & & \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!z2\mathop{\mathrm{tr}}\nolimits (\chi _{\varLambda }B_{L}(z)) = z\int _{B(0,\varLambda )}\bigg\langle \delta _{\{x\}},\sum _{j=1}^{n}\big[\partial _{ j},J_{L}^{j}(z)\big]\delta _{\{ x\}}\bigg\rangle \,\mathrm{d}^{n}x {}\\ & =& z\int _{B(0,\varLambda )}\bigg\langle \delta _{\{x\}},\sum _{j=1}^{n}\big[\partial _{ j},2\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\mathcal{Q}(R_{1+z}C)^{n-2}R_{ 1+z}\big)\big]\delta _{\{x\}}\bigg\rangle \,\mathrm{d}^{n}x, {}\\ & & +z\int _{B(0,\varLambda )}\bigg\langle \delta _{\{x\}},\sum _{j=1}^{n}\big[\partial _{ j},2\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\gamma _{j,n}\varPhi (R_{1+z}C)^{n-1}R_{ 1+z}\big)\big]\delta _{\{x\}}\bigg\rangle \,\mathrm{d}^{n}x {}\\ & & +z\int _{B(0,\varLambda )}\!\!\!\bigg\langle \delta _{\{x\}},\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\big[\mathcal{Q},\mathcal{Q}\big(\left (L^{{\ast}}L + z\right )^{-1} {}\\ & & \qquad \qquad \qquad + \left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n}\big]\big)\delta _{\{ x\}}\bigg\rangle \,\mathrm{d}^{n}x {}\\ & & \ +z\int _{B(0,\varLambda )}\!\!\!\bigg\langle \delta _{\{x\}},\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\big[\mathcal{Q},\varPhi \big(\left (L^{{\ast}}L + z\right )^{-1} {}\\ & & \ \qquad \qquad \qquad + \left (LL^{{\ast}} + z\right )^{-1}\big)\left (CR_{ 1+z}\right )^{n}\big]\big)\delta _{\{ x\}}\bigg\rangle \,\mathrm{d}^{n}x {}\\ & \ \ =& z\int _{B(0,\varLambda )}\sum _{j=1}^{n}(\partial _{ j}g_{0,_{j}})(x,x)\,\mathrm{d}^{n}x + z\int _{ B(0,\varLambda )}\sum _{j=1}^{n}(\partial _{ j}h_{j})(x,x)\,\mathrm{d}^{n}x + {}\\ & & \ +z\mathop{\mathrm{tr}}\nolimits \Big(\chi _{\varLambda }\Big(\mathop{\mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\big[\mathcal{Q},\mathcal{Q}\big((L^{{\ast}}L + z)^{-1} + (LL^{{\ast}} + z)^{-1}\big)(CR_{ 1+z})^{n}\big]\big) {}\\ & & \ \quad \qquad +\mathop{ \mathrm{tr}}\nolimits _{2^{\hat{n}}d}\big(\big[\mathcal{Q},\varPhi \big((L^{{\ast}}L + z)^{-1} + (LL^{{\ast}} + z)^{-1}\big)(CR_{ 1+z})^{n}\big]\big)\Big)\Big). {}\\ \end{array}$$
Thus, with the estimates (8.13) and (8.14) together with (8.15), one infers that
$$\displaystyle{\{z\mapsto 2z\mathop{\mathrm{tr}}\nolimits (\chi _{\varLambda }B_{L}(z))\}_{\varLambda>0}}$$
is locally bounded on $$B(0,\delta ) \cup \mathbb{C}_{\mathrm{Re}>0}$$ for some δ > 0. By Lemma 8.3 together with Theorem , one infers that
$$\displaystyle{\{z\mapsto 2\mathop{\mathrm{tr}}\nolimits (\chi _{\varLambda }B_{L}(z))\}_{\varLambda>0}}$$
is locally bounded on $$B(0,\delta ) \cup \mathbb{C}_{\mathrm{Re}>0}$$. By Montel ’s Theorem, there exists a sequence $$\{\varLambda _{k}\}_{k\in \mathbb{N}}$$ of positive reals tending to infinity such that
$$\displaystyle{\{z\mapsto 2\mathop{\mathrm{tr}}\nolimits (\chi _{\varLambda _{k}}B_{L}(z))\}_{k\in \mathbb{N}}}$$
converges in the compact open topology . We denote by f the respective limit. Then Lemma  implies that for $$k \in \mathbb{N}$$,
$$\displaystyle{2\mathop{\mathrm{tr}}\nolimits (\chi _{\varLambda _{k}}B_{L}(z)) =\int _{B(0,\varLambda _{k})}\sum _{j=1}^{n}(\partial _{ j}g_{j})(x,x)\,\mathrm{d}^{n}x.}$$
and so
$$\displaystyle{f(z) =\lim _{k\rightarrow \infty }\int _{B(0,\varLambda _{k})}\text{div}\,\mathbb{G}_{J,z}(x)\,\mathrm{d}^{n}x.}$$
Here we denote $$\mathbb{G}_{J,z}\,:=\,\{x\mapsto g_{j}(x,x)\}_{j\in \{1,\ldots,n\}}$$, with g j being the integral kernel of J L j (z) for j ∈ { 1, , n}. Next, let $$\vartheta \in (0,\pi /2)$$ and choose z0 > 0 as in Corollary 8.12 (i). Let $$z \in \varSigma _{z_{0},\vartheta }$$, see (8.8). Recalling that h j is the integral kernel of $$2\text{tr}_{2^{\hat{n}}d}\left (\gamma _{j,n}\varPhi C^{n-1}R_{1+z}^{n}\right )$$, we define $$\mathbb{H}_{z}\,:=\,\{x\mapsto h_{j}(x,x)\}_{j\in \{1,\ldots,n\}}$$. Due to Corollary 8.12, one can find κ > 0 such that for $$k \in \mathbb{N}$$,
$$\displaystyle\begin{array}{rcl} & & \left \vert \int _{\varLambda _{k}S^{n-1}}\left ((\mathbb{G}_{J,z} - \mathbb{H}_{z})(x), \frac{x} {\varLambda _{k}}\right )_{\mathbb{R}^{n}}\,\mathrm{d}^{n-1}\sigma (x)\right \vert {}\\ & &\quad \leqslant \int _{\varLambda _{k}S^{n-1}}\|(\mathbb{G}_{J,z} - \mathbb{H}_{z})(x)\|_{\mathbb{R}^{n}}\,\mathrm{d}^{n-1}\sigma (x) {}\\ & & \quad \leqslant \kappa \int _{\varLambda _{k}S^{n-1}}(1 + \vert x\vert )^{1-n-\varepsilon }\,\mathrm{d}^{n-1}\sigma (x) {}\\ & & \quad =\kappa \varLambda _{ k}^{n-1}\omega _{ n-1}(1 +\varLambda _{k})^{1-n-\varepsilon }. {}\\ \end{array}$$
Consequently,
$$\displaystyle{\lim _{k\rightarrow \infty }\int _{\varLambda _{k}S^{n-1}}\left ((\mathbb{G}_{J,z} - \mathbb{H}_{z})(x), \frac{x} {\varLambda _{k}}\right )_{\mathbb{R}^{n}}\,\mathrm{d}^{n-1}\sigma (x) = 0.}$$
Hence, with the help of Gauss’ theorem,
$$\displaystyle\begin{array}{rcl} f(z)& =& \lim _{k\rightarrow \infty }\int _{B(0,\varLambda _{k})}\sum _{j=1}^{n}(\partial _{ j}g_{j})(x,x)\,\mathrm{d}^{n}x =\int _{ \mathbb{R}^{n}}\text{div}\,\mathbb{G}_{J,z}(x)\,\,\mathrm{d}^{n}x \\ & =& \lim _{k\rightarrow \infty }\int _{B(0,\varLambda _{k})}\text{div}\,\mathbb{G}_{J,z}(x)\,\mathrm{d}^{n}x =\lim _{ k\rightarrow \infty }\int _{\varLambda _{k}S^{n-1}}\left (\mathbb{G}_{J,z}(x), \frac{x} {\varLambda _{k}}\right )_{\mathbb{R}^{n}}\,\mathrm{d}^{n-1}\sigma (x) \\ & =& \lim _{k\rightarrow \infty }\int _{\varLambda _{k}S^{n-1}}\left (\mathbb{H}_{z}(x), \frac{x} {\varLambda _{k}}\right )_{\mathbb{R}^{n}}\,\mathrm{d}^{n-1}\sigma (x) \\ & =& \left ( \frac{\mathrm{i}} {8\pi }\right )^{(n-1)/2} \frac{1} {\left [(n - 1)/2\right ]!}(1 + z)^{-n/2}\lim _{ k\rightarrow \infty }\int _{\varLambda _{k}S^{n-1}} \\ & & \quad \times \sum _{j=1}^{n}\bigg(\sum _{ i_{1},\ldots,i_{n-1}=1}^{n}\varepsilon _{ ji_{1}\ldots i_{n-1}}\mathop{ \mathrm{tr}}\nolimits \big(\varPhi (x)(\partial _{i_{1}}\varPhi )(x)\ldots (\partial _{i_{n-1}}\varPhi )(x)\big)\bigg) \\ & & \qquad \quad \;\;\; \times \left (\frac{x_{j}} {\varLambda _{k}} \right )\,\mathrm{d}^{n-1}\sigma (x), {}\end{array}$$
(8.16)
where, for the last integral, we used Proposition 8.13. By Theorem  one has $$f(0) = 2\mathop{\mathrm{ind}}\nolimits (L)$$. In particular, any sequence $$\{\varLambda _{k}\}_{k\in \mathbb{N}}$$ of positive reals converging to infinity contains a subsequence $$\{\varLambda _{k_{\ell}}\}_{\ell}$$ such that for that particular subsequence the limit
$$\displaystyle\begin{array}{rcl} & & \lim _{\ell\rightarrow \infty }\int _{\varLambda _{k_{ \ell}}S^{n-1}}\sum _{j=1}^{n}\bigg(\sum _{ i_{1},\ldots,i_{n-1}=1}^{n}\varepsilon _{ ji_{1}\ldots i_{n-1}}\mathop{ \mathrm{tr}}\nolimits \big(\varPhi (x)(\partial _{i_{1}}\varPhi )(x)\ldots (\partial _{i_{n-1}}\varPhi )(x)\big)\bigg) {}\\ & & \qquad \qquad \qquad \qquad \times \left (\frac{x_{j}} {\varLambda _{k_{\ell}}} \right )\,\mathrm{d}^{n-1}\sigma (x) {}\\ \end{array}$$
exists and equals
$$\displaystyle{ \frac{2\mathop{\mathrm{ind}}\nolimits (L)\left [(n - 1)/2\right ]!} {\left [\mathrm{i}/(8\pi )\right ]^{(n-1)/2}}. }$$
(8.17)
Hence, the limit
$$\displaystyle\begin{array}{rcl} \begin{array}{rl} &\lim _{\varLambda \rightarrow \infty }\int _{\varLambda S^{n-1}}\sum _{j=1}^{n}\bigg(\sum _{i_{1},\ldots,i_{n-1}=1}^{n}\varepsilon _{ji_{1}\ldots i_{n-1}}\mathop{ \mathrm{tr}}\nolimits \left (\varPhi (x)\left (\partial _{i_{1}}\varPhi \right )(x)\ldots \left (\partial _{i_{n-1}}\varPhi \right )(x)\right )\bigg) \\ &\qquad \qquad \qquad \qquad \qquad \times \left (\frac{x_{j}} {\varLambda } \right )\,\mathrm{d}^{n-1}\sigma (x)\end{array} & &{}\end{array}$$
(8.18)
exists and equals the number in (8.17). On the other hand, for $$z \in \varSigma _{z_{0},\vartheta }$$, (see again Corollary 8.12) the family
$$\displaystyle{\{z\mapsto \mathop{\mathrm{tr}}\nolimits (\chi _{\varLambda }B_{L}(z))\}_{\varLambda>0}}$$
converges for $$\varLambda \rightarrow \infty$$ on the domain $$\varSigma _{z_{0},\vartheta }$$ if and only if the limit in (8.18) exists. Indeed, this follows from the explicit expression for the limit in (8.16). Therefore,
$$\displaystyle{\{z\mapsto \mathop{\mathrm{tr}}\nolimits (\chi _{\varLambda }B_{L}(z))\}_{\varLambda>0}}$$
converges in the compact open topology on $$\varSigma _{z_{0},\vartheta }$$. By the local boundedness of the latter family on the domain $$B(0,\delta ) \cup \mathbb{C}_{\mathrm{Re}>0}$$, the principle of analytic continuation implies that the latter family actually converges on the domain $$B(0,\delta ) \cup \mathbb{C}_{\mathrm{Re}>0}$$ in the compact open topology . In particular,
$$\displaystyle\begin{array}{rcl} & & 2f(z)(1 + z)^{n/2} \frac{[(n - 1)/2]!} {[\mathrm{i}/(8\pi )]^{(n-1)/2}} {}\\ & & \quad =\lim _{\varLambda \rightarrow \infty }\int _{\varLambda S^{n-1}}\sum _{j=1}^{n}\bigg(\sum _{ i_{1},\ldots,i_{n-1}=1}^{n}\varepsilon _{ ji_{1}\ldots i_{n-1}}\mathop{ \mathrm{tr}}\nolimits \big(\varPhi (x)(\partial _{i_{1}}\varPhi )(x)\ldots (\partial _{i_{n-1}}\varPhi )(x)\big)\bigg) {}\\ & & \qquad \qquad \qquad \qquad \qquad \times \bigg (\frac{x_{j}} {\varLambda } \bigg)\,\mathrm{d}^{n-1}\sigma (x). {}\\ \end{array}$$
□

## References

1. 22.
C. Callias, Axial anomalies and index theorems on open spaces. Commun. Math. Phys. 62, 213–234 (1978)
2. 35.
J.B. Conway, Functions of One Complex Variable I, 2nd edn., 7th corr. printing (Springer, New York, 1995)Google Scholar
3. 57.
I.S. Gradshteyn, I.M. Ryzhik, Table of Integrals, Series, and Products, corrected and enlarged edition, prepared by A. Jeffrey (Academic Press, San Diego, 1980)

© Springer International Publishing Switzerland 2016

## Authors and Affiliations

• Fritz Gesztesy
• 1
• Marcus Waurick
• 2
1. 1.Dept of MathematicsUniversity of MissouriColumbiaUSA
2. 2.Institut für AnalysisTU DresdenDresdenGermany