### 数学代写|数论作业代写number theory代考|On Trace Class Operators

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## 数学代写|数论作业代写number theory代考|On Trace Class Operators

A classical result in an elementary linear algebra course states that, for an arbitrary $n \times$ $n$ matrix $B=\left(B_{j, k}\right){1 \leq j, k \leq n}$ of complex numbers with eigenvalues $\left{\beta{k}\right}_{k=1}^{n}$ (counting algebraic multiplicities), one has
$$\operatorname{tr}(B)=\sum_{j=1}^{n} B_{j, j}=\sum_{k=1}^{n} \beta_{k}$$
This result generalizes to an important class of compact operators in an infinitedimensional, separable Hilbert space $(\mathcal{H},(\cdot, \cdot) \mathcal{H})$, the so-called trace class operators. Specifically, a bounded linear operator $T: \mathcal{H} \rightarrow \mathcal{H}$ is trace class if, for some (and hence for all) orthonormal basis $\left{e_{j}\right}_{j \in \mathbb{N}}$ of $\mathcal{H}$, the sum
$$\sum_{j \in \mathbb{N}}\left(e_{j},\left(T^{*} T\right)^{1 / 2} e_{j}\right)_{\mathcal{H}}$$
is finite; see [56] [Sect. 3.6] for an in-depth discussion of this topic. In this case, the trace of $T$ is defined to be

$$\operatorname{tr}(T)=\sum_{j \in \mathbb{N}}\left(e_{j}, T e_{j}\right){\mathcal{H}}$$ this finite-valued sum is absolutely convergent and independent of the choice of orthonormal basis $\left{e{j}\right}_{j \in \mathbb{N}}$. In addition, if $\left{\tau_{k}\right}_{k \in J}$ (with $J \subseteq \mathbb{N}$ an appropriate index set) represent the eigenvalues of $T$, counting algebraic multiplicities, then by Lidskii’s theorem (see, e.g., [55] [Chap. 3], [56] [Sect. 3.12])
$$\operatorname{tr}(T)=\sum_{j \in \mathbb{N}}\left(e_{j}, T e_{j}\right){\mathcal{H}}=\sum{k \in J} \tau_{k} .$$
For the purpose of this note, we discuss a particular set of trace class integral operators. Let $[a, b] \subset \mathbb{R}$ be a compact interval and $K(\cdot, \cdot):[a, b] \times[a, b] \rightarrow \mathbb{C}$ be a continuous function. Define $T: L^{2}((a, b) ; d x) \rightarrow L^{2}((a, b) ; d x)$ by
$$(T f)(x)=\int_{a}^{b} d y K(x, y) f(y), \quad f \in L^{2}((a, b) ; d x)$$
and assume that $T \geq 0$ (implying $K(x, x) \geq 0, x \in[a, b]$ ). Then by Mercer’s Theorem (see, e.g.., [18] [Proposition 5.6.9], [56] [Theorem 3.11.9]), $T$ is trace class and
$$\operatorname{tr}(T)=\int_{a}^{b} d x K(x, x) \in[0, \infty)$$

## 数学代写|数论作业代写number theory代考|Some Generalizations

In our final section we probe some (z-dependent) extensions of Theorem $1 .$
More precisely, for $x, y \in[0,1], n \in \mathbb{N}$, we will be considering
$$\begin{array}{r} \left(\left(-\Delta_{D}\right)^{n}-z I\right)^{-1}(x, y)=K_{n}(z ; x, y)=2 \sum_{k \in \mathbb{N}} \frac{\sin (k \pi x) \sin (k \pi y)}{\left[(k \pi)^{2 n}-z\right]} \ z \in \mathbb{C} \backslash\left{(k \pi)^{2 n}\right}_{k \in \mathbb{N}} \end{array}$$
and
\begin{aligned} \left(-\Delta_{D}-z I\right)^{-n}(x, y) &=\widetilde{K}{n}(z ; x, y)=2 \sum{k \in \mathbb{N}} \frac{\sin (k \pi x) \sin (k \pi y)}{\left[(k \pi)^{2}-z\right]^{n}} \ &=\frac{1}{(n-1) !} \frac{d^{n-1}}{d z^{n-1}} K_{1}(z ; x, y), \quad z \in \mathbb{C} \backslash\left{(k \pi)^{2}\right}_{k \in \mathbb{N}} \end{aligned}
noting that,
\begin{aligned} &\left(-\Delta_{D}\right)^{-n}(x, y)=K_{n}(x, y)=\left.K_{n}(z ; x, y)\right|{z=0}=\left.\tilde{K}{n}(z ; x, y)\right|{z=0} \ &K{1}(z ; x, y)-\widetilde{K}{1}(z ; x, y), \quad z \in \mathbb{C} \backslash\left{(k \pi)^{2}\right}{k \in \mathbb{N}^{*}} \end{aligned}

We start with the case of $\widetilde{K}{n}(z ; \cdot, \cdot), n \in \mathbb{N}$, and recall the trace formula, \begin{aligned} &\sum{k \in \mathbb{N}}\left[(k \pi)^{2}-z\right]^{-\mu} \ &=\frac{\pi^{1 / 2}}{\left(2 z^{1 / 2}\right)^{\mu-(1 / 2)} \Gamma(\mu)} \int_{0}^{\infty} d x\left[e^{\pi x}-1\right]^{-1} x^{\mu-(1 / 2)} I_{\mu-(1 / 2)}\left(z^{1 / 2} x\right) . \ &\Re(\mu)>1 / 2,\left|\Re\left(z^{1 / 2}\right)\right|<\pi, \end{aligned}
ohtained as an elementary consequence of [27] [No. 6.6247] or [68] [Fq. (9) on p. 386] (originally due to Kapteyn [37]), where (cf. [46] [No. 10.27.6])
$$I_{v}(\zeta)=e^{\mp i(\pi / 2) v} J_{v}(\pm i \zeta), \quad \Re(v) \geq 0,-\pi \leq \pm \operatorname{Arg}(\zeta) \leq \pi / 2$$
where $\operatorname{Arg}(\cdot)$ represents the single-valued principal value of the argument function on $\mathbb{C} \backslash(-\infty, 0]$. Here $J_{v}(\cdot)$ (resp., $\left.I_{v}(\cdot)\right)$ denotes the Bessel function (resp., modified Bessel function) of order $v$ (cf., e.g.., [1] [Chap. 9], [46] [Chap. 10]).

## 数学代写|数论作业代写number theory代考|Notation

Let $\Omega \subset \mathbb{R}^{2}$ be a piecewise smooth bounded open domain (we will actually only work with convex polygonal domains), with boundary $\partial \Omega=\overline{\Gamma_{1} \sqcup \Gamma_{2}}$, where $\Gamma_{1}, \Gamma_{2}$ are two disjoint open subsets of $\partial \Omega$. We consider the eigenvalue problem

where $v$ is the outer unit normal along $\partial \Omega$ (defined almost everywhere).
Let $\left{\mu_{i}(\Omega, \mathfrak{d} \mathfrak{n}), i \geq 1\right}$ (resp. $\operatorname{sp}(\Omega, \mathfrak{d} \mathfrak{n})$ ) denote the eigenvalues (resp. the spectrum) of problem (1.1). We always list the eigenvalues in non-decreasing order, with multiplicities, starting with the index 1 . We simply write $\mu_{i}$, and skip mentioning the domain $\Omega$, or the boundary condition $\mathrm{on}$, whenever the context is clear. Examples of eigenvalue problems with mixed boundary conditions appear in Sects. 2 and $3 .$
Let $\mathcal{E}(\mu)$ denote the eigenspace associated with the eigenvalue $\mu$.
Define the min-index $\kappa(\mu)$ of the eigenvalue $\mu$ as
$$\kappa(\mu)-\min \left{m \mid \mu-\mu_{m}\right}$$

## 数学代写|数论作业代写number theory代考|On Trace Class Operators

tr⁡(乙)=∑j=1n乙j,j=∑ķ=1nbķ

∑j∈ñ(和j,(吨∗吨)1/2和j)H

tr⁡(吨)=∑j∈ñ(和j,吨和j)H这个有限值和是绝对收敛的，并且与正交基的选择无关\left{e{j}\right}_{j \in \mathbb{N}}\left{e{j}\right}_{j \in \mathbb{N}}. 此外，如果\left{\tau_{k}\right}_{k \in J}\left{\tau_{k}\right}_{k \in J}（和Ĵ⊆ñ一个适当的索引集）表示的特征值吨，计算代数重数，然后通过 Lidskii 定理（参见例如 [55] [Chap. 3]、[56] [Sect. 3.12]）

tr⁡(吨)=∑j∈ñ(和j,吨和j)H=∑ķ∈Ĵτķ.

(吨F)(X)=∫一个bd是ķ(X,是)F(是),F∈大号2((一个,b);dX)

tr⁡(吨)=∫一个bdXķ(X,X)∈[0,∞)

## 数学代写|数论作业代写number theory代考|Some Generalizations

\begin{array}{r} \left(\left(-\Delta_{D}\right)^{n}-z I\right)^{-1}(x, y)=K_{n}(z ; x, y)=2 \sum_{k \in \mathbb{N}} \frac{\sin (k \pi x) \sin (k \pi y)}{\left[(k \pi)^{ 2 n}-z\right]} \ z \in \mathbb{C} \反斜杠\left{(k \pi)^{2 n}\right}_{k \in \mathbb{N}} \end{大批}\begin{array}{r} \left(\left(-\Delta_{D}\right)^{n}-z I\right)^{-1}(x, y)=K_{n}(z ; x, y)=2 \sum_{k \in \mathbb{N}} \frac{\sin (k \pi x) \sin (k \pi y)}{\left[(k \pi)^{ 2 n}-z\right]} \ z \in \mathbb{C} \反斜杠\left{(k \pi)^{2 n}\right}_{k \in \mathbb{N}} \end{大批}

\begin{对齐} \left(-\Delta_{D}-z I\right)^{-n}(x, y) &=\widetilde{K}{n}(z ; x, y)=2 \ sum{k \in \mathbb{N}} \frac{\sin (k \pi x) \sin (k \pi y)}{\left[(k \pi)^{2}-z\right]^ {n}} \ &=\frac{1}{(n-1) !} \frac{d^{n-1}}{d z^{n-1}} K_{1}(z ; x, y ), \quad z \in \mathbb{C} \backslash\left{(k \pi)^{2}\right}_{k \in \mathbb{N}} \end{对齐}\begin{对齐} \left(-\Delta_{D}-z I\right)^{-n}(x, y) &=\widetilde{K}{n}(z ; x, y)=2 \ sum{k \in \mathbb{N}} \frac{\sin (k \pi x) \sin (k \pi y)}{\left[(k \pi)^{2}-z\right]^ {n}} \ &=\frac{1}{(n-1) !} \frac{d^{n-1}}{d z^{n-1}} K_{1}(z ; x, y ), \quad z \in \mathbb{C} \backslash\left{(k \pi)^{2}\right}_{k \in \mathbb{N}} \end{对齐}

\begin{aligned} &\left(-\Delta_{D}\right)^{-n}(x, y)=K_{n}(x, y)=\left.K_{n}(z ; x , y)\right|{z=0}=\left.\tilde{K}{n}(z ; x, y)\right|{z=0} \ &K{1}(z ; x, y) -\widetilde{K}{1}(z ; x, y), \quad z \in \mathbb{C} \反斜杠\left{(k \pi)^{2}\right}{k \in \mathbb {N}^{*}} \end{对齐}\begin{aligned} &\left(-\Delta_{D}\right)^{-n}(x, y)=K_{n}(x, y)=\left.K_{n}(z ; x , y)\right|{z=0}=\left.\tilde{K}{n}(z ; x, y)\right|{z=0} \ &K{1}(z ; x, y) -\widetilde{K}{1}(z ; x, y), \quad z \in \mathbb{C} \反斜杠\left{(k \pi)^{2}\right}{k \in \mathbb {N}^{*}} \end{对齐}

∑ķ∈ñ[(ķ圆周率)2−和]−μ =圆周率1/2(2和1/2)μ−(1/2)Γ(μ)∫0∞dX[和圆周率X−1]−1Xμ−(1/2)一世μ−(1/2)(和1/2X). ℜ(μ)>1/2,|ℜ(和1/2)|<圆周率,

## 数学代写|数论作业代写number theory代考|Notation

\kappa(\mu)-\min \left{m \mid \mu-\mu_{m}\right}

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