text:
\begin{document}

\title{Surface Riesz transforms and spectral property of elastic Neumann--Poincar\'e operators on less smooth domains in three dimensions\thanks{\footnotesize This work was
supported by NRF grants No. 2016R1A2B4011304 and 2017R1A4A1014735.}}

\author{Hyeonbae Kang\thanks{Department of Mathematics and Institute of Applied Mathematics, Inha University, Incheon
22212, S. Korea (hbkang@inha.ac.kr, k.goe.dai@gmail.com).} \and
Daisuke Kawagoe\footnotemark[2] }

\date{}
\maketitle

\begin{abstract}
It is proved in \cite{AKM18} that the Neumann--Poincar\'e operator for the Lam\'e system of linear elasticity is polynomially compact and, as a consequence, that its spectrum consists of three non-empty sequences of eigenvalues accumulating to certain numbers determined by Lam\'e parameters, if the boundary of the domain where the operator is defined is $C^\infty$-smooth. We extend this result to less smooth boundaries, namely, $C^{1,\Ga}$-smooth boundaries for some $\Ga>0$. The results are obtained by proving certain identities for surface Riesz transforms, which are singular integral operators of non-convolution type, defined by the matrix tensor on a given surface.
\end{abstract}

\noindent{\footnotesize {\bf AMS subject classifications}. 42B20 (primary), 35P05 (secondary)}

\noindent{\footnotesize {\bf Key words}. Neumann--Poincar\'e operator, Lam\'e system, polynomial compactness, spectrum, surface Riesz transform, composition of singular integral operators}

\section{Introduction}

The purpose of this paper is to prove certain identities for surface Riesz transforms on the boundary of a bounded domain in $\Rbb^3$, where the boundary is assumed to be $C^{1,\Ga}$ for some $\Ga>0$. We then use such identities to show that the elastic Neumann--Poincar\'e operator (the Neumann--Poincar\'e operator for the Lam\'e system of linear elasticity, abbreviated by eNP operator) on the boundary is polynomially compact. As a consequence, we show that the spectrum of the eNP operator consists of three non-empty sequences of eigenvalues accumulating to certain numbers determined by Lam\'e parameters.

Let $G(u)=(g_{ij})_{i,j=1,2}$ be a positive-definite symmetric matrix valued function on $\Rbb^2$ such that $G(u)=I$ (the identity matrix) for $u$ outside a compact set. We assume that $G$ is $C^\Ga$-smooth for some $\Ga>0$. In fact, $G$ is a metric tensor corresponding to a $C^{1,\Ga}$-smooth boundary $\p\GO$ of a certain bounded domain $\GO$ in $\Rbb^3$ (see (\ref{eq:g}) and (\ref{eq:G}) in section \ref{sec:Riesz}). Let
\beq \label{Luuv}
L(u, u-v) = \la u-v, G(u) (u-v) \ra^{-3/2}.
\eeq
The surface Riesz transform is defined by
\beq\label{Rgdef}
R_j^g[f](u) =\frac{1}{2\pi} \text{p.v.} \int_{\Rbb^2} {(u_j-v_j)}L(u, u-v) f(v) dv, \quad j=1,2.
\eeq
Here, p.v. stands for the Cauchy principal value and $u_j$ is the $j$-th component of the point $u$. The operator $R_j^g$ is a singular integral operator of non-convolution type and bounded on $L^2(\Rbb^2)$ (or $H^{-1/2}(\Rbb^2)$) (see, for example, \cite{Stein-book-70}). The Sobolev space $H^{-1/2}(\Rbb^2)$ is of particular interest in this paper because of its relation to the spectral theory of the eNP operator.

In this paper we prove the following theorem, for presentation of which we fix notation: $A \equiv B$ for two operators $A$ and $B$ bounded on $H^s(\Rbb^2)$ ($s = 0$ or $-1/2$) means that $A - B$ is compact on $H^s(U)$ for any bounded open set $U$.

\begin{theorem}\label{thm:Riesz}
Let $R_j^g$, $j=1,2$, be surface Riesz transforms defined by the metric tensor $G$. Suppose that $G$ is $C^\Ga$-smooth for some $\Ga>0$. Then, following identities hold:
\beq\label{commute}
{R_1^g} {R_2^g}-{R_2^g} {R_1^g} \equiv 0
\eeq
and
\beq\label{squaresum}
\det(G)(g_{11} ({R_1^g})^2 + 2g_{12} R_1^g R_2^g +g_{22} ({R_2^g})^2) \equiv -I.
\eeq
\end{theorem}

It is worth mentioning that, if the surface is flat or $G$ is the identity matrix, then surface Riesz transforms are usual Riesz transforms, i.e.,
$$
R_j^g [f](u) = R_j [f](u) = \frac{1}{2\pi} \text{p.v.} \int_{\Rbb^2} \frac{u_j-v_j}{|u-v|^3} f(v) \, dv, \quad j=1,2,
$$
which are singular integral operators of convolution type, and identities (\ref{commute}) and (\ref{squaresum}) are reduced to the following ones
\beq \label{RieszId}
R_1 R_2-R_2 R_1=0 \quad\mbox{and}\quad R_1^2 + R_2^2 =-I,
\eeq
which are also proved by taking the Fourier transform. See, for example, \cite{SW-book}.

The surface Riesz transform is closely related to the eNP operator in three dimensions like the Hilbert transform is related to it in two dimensions. In fact, we show the following theorems using Theorem \ref{thm:Riesz}.

\begin{theorem}[Polynomial compactness]\label{Polynomially compact}
Let $\GO$ be a bounded domain in $\Rbb^3$ with the $C^{1,\Ga}$-smooth boundary for some $\Ga>0$. Let $\BK$ be the eNP operator on $\p\GO$ corresponding to the pair of Lam\'e parameters $(\Gl, \Gm)$. Let $p_3(t)=t(t^2-k_0^2)$ where $k_0$ is given by
\beq\label{kzero}
k_0= \frac{\mu}{2(2\mu+\Gl)}.
\eeq
Then $p_3(\BK)$ is compact. Moreover, $\BK (\BK-k_0 \BI)$, $\BK (\BK + k_0 \BI)$ and $\BK^2 - k_0^2 \BI$ are \emph{not} compact.
\end{theorem}

\begin{theorem}[Spectral structure]\label{mainthm1}
Let $\GO$ and $\BK$ be as in Theorem \ref{Polynomially compact}. The spectrum of $\BK$ consists of three non-empty sequences of eigenvalues  which converge to $0$, $k_0$ and $-k_0$, respectively.
\end{theorem}

Theorem \ref{mainthm1} is a consequence of Theorem \ref{Polynomially compact} and the spectral mapping theorem which asserts that $p_3(\Gs(\BK))=\Gs(p_3(\BK))$, where $\Gs(\BK)$ denotes the spectrum of $\BK$ (see \cite{RS-book-80}). Theorem \ref{Polynomially compact} is proved in \cite{AKM18} under the assumption that $\p\GO$ is $C^\infty$-smooth. We describe below why this assumption was needed and how it is overcome in this paper, but we first make some motivational remarks.

We will be brief here and refer to \cite{AKM18} for more informative discussion on recent development on spectral theory of the Neumann--Poincar\'e operator (abbreviated by NP operator). The NP operator, sometimes called the double layer potential, is a boundary integral operator which naturally appears when solving classical boundary value problems for the Laplace operator using layer potentials. Its study goes back to C. Neumann \cite{Neumann-87} and Poincar\'e \cite{Poincare-AM-87} as the name of the operator suggests. The NP operator, which is not a self-adjoint operator on $L^2$ in general, can be realized as a self-adjoint operator by introducing a new inner product on the Sobolev space $H^{-1/2}$ \cite{KPS-ARMA-07}. If the boundary of the domain where the NP operator is defined is $C^{1,\Ga}$-smooth for some $\Ga>0$, then the NP operator is compact. So, its spectrum consists of eigenvalues converging to $0$.

However, its counterpart for the Lam\'e system, the eNP operator, is \emph{not} compact even if the boundary is smooth \cite{DKV-Duke-88}. Therefore, it was not clear how spectrum of eNP operator looked like. But, it is proved in \cite{AJKKY} that the eNP operator in two dimensions is polynomially compact if the domain is $C^{1,\Ga}$-smooth. More precisely, if we denote the eNP operator by $\BK$, then $\BK^2 - k_0^2 I$ is compact where $k_0$ the the number given by \eqnref{kzero}, and $\Gs(\BK)$ consists of two non-empty sequences of eigenvalues converging to $k_0$ and $-k_0$, respectively. The proof of this two-dimensional result cannot be extended to three dimensions since it uses the Hilbert transform, which relates the boundary values of harmonic functions with those of their conjugate harmonic functions.

In \cite{AKM18} the three-dimensional eNP operator is expressed in terms of surface Riesz transforms $R_j^g$, and identities \eqnref{commute} and \eqnref{squaresum} are proved when $\p\GO$ is $C^\infty$-smooth. In fact, the operator $R_j^g$ is realized as a classical $\psi$do (pseudo-differential operator) and its symbol is computed (see \eqnref{symbol}). Then, calculus of $\psi$do's immediately yields those two identities. For example, ${R_1^g} {R_2^g}-{R_2^g} {R_1^g}$ is a commutator of $\psi$do's and regularizing of order $1$. Theorem \ref{Polynomially compact} is proved using \eqnref{commute} and \eqnref{squaresum} when $\p\GO$ is $C^\infty$.

If $\p\GO$ is $C^{1,\Ga}$, then the metric tensor $G$ is merely $C^\Ga$, and so calculus of $\psi$do's may not be applied. In this paper we prove Theorem \ref{thm:Riesz} by directly dealing with compositions of singular integral operators of non-convolution type. We then prove Theorems \ref{Polynomially compact} and \ref{mainthm1} following the same argument as in \cite{AKM18}

The NP operators on $C^{1,\Ga}$ boundaries and Lipschitz boundaries exhibit drastically different spectrum. It is proved recently that if $\p\GO$ has corners, then the NP operator has continuous spectrum of the connected interval symmetric with respect to $0$ whose endpoints are determined by the angle of the corner \cite{PP-arXiv} (see also \cite{BZ-arXiv, HKL-17, KLY-17, PP-JAM-14}). In this regard Theorem \ref{mainthm1} and the corresponding result in two dimensions are quite interesting. If the domain has a corner, then it is expected that the eNP operator may have a continuous spectrum. However the continuous spectrum in two dimensions may not be a connected interval, since there are two accumulation points.

This paper is organized as follows. In section \ref{sec:Comm}, we introduce an approximation of compositions of surface Riesz transforms, and we prove Theorem \ref{thm:Riesz} by using the approximation. In section \ref{sec:Riesz} we review the relations between eNP operators and surface Riesz transforms, which was obtained in \cite{AKM18}, and describe how Theorem \ref{Polynomially compact} follows from Theorem \ref{thm:Riesz}.

\section{Surface Riesz transforms and proof of Theorem \ref{thm:Riesz}}\label{sec:Comm}

In what follows, we use the notation:
\beq\label{rj}
r_j(u, v) := v_j L(u, v), \quad j=1,2,
\eeq
where $L(u, v)$ is defined by \eqnref{Luuv}. Observe that
$$
R_i^g R_j^g [f](u) = \lim_{\Gd_1, \Gd_2 \downarrow 0} \dfrac{1}{4\pi^2} \int_{|u-v| > \Gd_1} r_i(u, u - v) \int_{|v-w| > \Gd_2} r_j(v, v - w) f(w)\,dw\,dv
$$
for a.e. $u$, where the limit exists either in the point-wise sense or $L^2$-sense. Define the operator $R_{ij}$ by
$$
R_{ij} [f](u) = \lim_{\Gd_1, \Gd_2 \downarrow 0} \dfrac{1}{4\pi^2} \int_{|u-v| > \Gd_1} r_i(u, u - v) \int_{|v-w| > \Gd_2} r_j(u, v - w) f(w)\,dw\,dv
$$
for a.e. $u$. We emphasize that the difference between $R_i^g R_j^g [f](u)$ and $R_{ij} [f](u)$ lies in the $r_j$ appeared in the formulas: the first one is $r_j(v, v - w)$ while the second one is $r_j(u, v - w)$.

The following proposition is the key ingredient in proving Theorem \ref{thm:Riesz}.

\begin{prop}\label{prop:CV}
If the metric tensor $G(u)$ is $C^\Ga$ for some $\Ga>0$, then
\beq\label{RiRjRij}
R_i^g R_j^g \equiv R_{ij}.
\eeq
for $i, j = 1, 2$.
\end{prop}

Let us prove Theorem \ref{thm:Riesz} first, and then give the proof of Proposition \ref{prop:CV} after that.

\medskip
\noindent{\sl Proof of Theorem \ref{thm:Riesz}}.
Thanks to \eqnref{RiRjRij}, it suffices to prove
\beq\label{commute2}
R_{12} - R_{21} = 0
\eeq
and
\beq\label{squaresum2}
\det(G)(g_{11} R_{11} + 2g_{12} R_{12} + g_{22} R_{22}) = -I.
\eeq

Note that
\begin{align*}
R_{ij}[f](u) &= \pv \dfrac{1}{4\pi^2} \int_{\Rbb^2} r_i(u, u - v)\, \pv \int_{\Rbb^2} r_j(u, v - w)f(w)\,dw\,dv \\
&= \dfrac{1}{2\pi} \int_{\Rbb^2} \Fcal[r_i(u, \cdot)](\xi) \Fcal[r_j(u, \cdot)](\xi) \Fcal[f](\xi)\,e^{\sqrt{-1} u \cdot \xi}\,d\xi
\end{align*}
for $i, j = 1, 2$,
where $\Fcal$ denotes the Fourier transform:
$$
\Fcal[f](\xi) := \dfrac{1}{2\pi} \int_{\Rbb^2} f(x) e^{-\sqrt{-1} x \cdot \xi}\,dx.
$$
Thus, \eqnref{commute2} follows immediately.

It is proved in \cite{AKM18} that
\beq \label{symbol}
\Fcal[r_i(u, \cdot)](\xi) = \dfrac{-\sqrt{-1}}{\det(G(u))^{1/2}} \dfrac{\sum_j g^{ij}(u) \xi_j}{\sqrt{\sum_{i, j} g^{ij}(u) \xi_i \xi_j}}, \quad i = 1, 2,
\eeq
where $(g^{ij})_{i, j = 1, 2}$ is the inverse metric tensor of $G$.
Thus, we have
\begin{align*}
&\det(G)(g_{11} R_{11} + 2g_{12} R_{12} + g_{22} R_{22})[f](u)\\
&= -\det(G(u)) \int_{\Rbb^2} \dfrac{1}{\sum_{i, j} g^{ij}(u) \xi_i \xi_j} \left[ \dfrac{g_{11}(u)}{\det{G(u)}} (g^{11}(u) \xi_1 + g^{12}(u) \xi_2)^2 \right.\\
& \quad + \dfrac{2g_{12}(u)}{\det{G(u)}}(g^{11}(u) \xi_1 + g^{12}(u) \xi_2)(g^{21}(u) \xi_1 + g^{22}(u) \xi_2)\\
& \quad +\left. \dfrac{g_{22}(u)}{\det{G(u)}} (g^{21}(u) \xi_1 + g^{22}(u) \xi_2)^2 \right] \Fcal[f](\xi) e^{\sqrt{-1} u \cdot \xi}\,d\xi\\
&= -\int_{\Rbb^2} \dfrac{1}{\sum_{i, j} g^{ij}(u) \xi_i \xi_j} \big[ \left( g_{11} (g^{11})^2 + 2g_{12} g^{11} g^{21} + g_{22} (g^{21})^2 \right) (u) \, \xi_1^2 \\
& \quad + 2 \left( g_{11} g^{11} g^{12} + g^{12} (g^{11}g^{22} + g^{12} g^{21} ) + g_{22} g^{21} g^{22} \right)(u) \, \xi_1 \xi_2\\
& \quad + \left( g_{11} (g^{12})^2 + 2g_{12} g^{12} g^{22} + g_{22} (g^{22})^2 \right)(u) \, \xi_2^2 \Big] \Fcal[f](\xi) e^{\sqrt{-1} u \cdot \xi}\,d\xi.
\end{align*}
Observing that $g^{11} = g_{22} / \det(G)$, $g^{12} = g^{21} = -g_{12} / \det(G)$ and $g^{22} = g_{11} / \det(G)$, we have
\begin{align*}
&g_{11} (g^{11})^2 + 2g_{12} g^{11} g^{21} + g_{22} (g^{21})^2\\
&= \det(G) (g^{22} (g^{11})^2 - 2(g^{12})^2 g^{11} + g^{11} (g^{21})^2)\\
&= \det(G) g^{11} (g^{22} g^{11} - (g^{12})^2)\\
&= g^{11}.
\end{align*}
Similarly one can show that
$$
g_{11} g^{11} g^{12} + g^{12} (g^{11}g^{22} + g^{12} g^{21} ) + g_{22} g^{21} g^{22}=  g^{12},
$$
and
$$
g_{11} (g^{12})^2 + 2g_{12} g^{12} g^{22} + g_{22} (g^{22})^2 = g^{22}.
$$
Thus, we have
\begin{align*}
&\det(G)(g_{11} R_{11} + 2g_{12} R_{12} + g_{22} R_{22})[f](u) \\
&= -\dfrac{1}{2\pi} \int_{\Rbb^2} \dfrac{g^{11}(u) \xi_1^2 + 2g^{12}(u) \xi_1 \xi_2 + g^{22}(u) \xi_2^2}{\sum_{i, j} g^{ij}(u) \xi_i \xi_j} \Fcal[f](\xi) e^{\sqrt{-1} u \cdot \xi}\,d\xi\\
&= -\dfrac{1}{2\pi} \int_{\Rbb^2} \Fcal[f](\xi) e^{\sqrt{-1} u \cdot \xi}\,d\xi\\
&= - f(u)
\end{align*}
by the Fourier inversion formula, which completes the proof. \qed

\medskip
\noindent{\sl Proof of Proposition \ref{prop:CV}}.
In this proof we use $R_k$ for the surface Riesz transform, dropping the superscript $g$ from the notation \eqnref{Rgdef}, for ease of notation.

Note that
\begin{align*}
& R_i R_j [f](u) - R_{ij} [f](u) \\
&= \lim_{\Gd_1, \Gd_2 \downarrow 0} \dfrac{1}{4\pi^2} \int_{|u - v| > \Gd_1} \int_{|v - w| > \Gd_2} r_i(u, u  - v) \left[ r_j(v, v - w) - r_j(u, v- w) \right] f(w)\,dw\,dv.
\end{align*}
By changing the order of integrations, we see that
\beq\label{orderchange}
R_i R_j [f](u) - R_{ij} [f](u)
= \lim_{\Gd_1, \Gd_2 \downarrow 0} \dfrac{1}{4\pi^2} \int_{\Rbb^2} k_{\Gd_1,\Gd_2}(u,w)  f(w)\,dw,
\eeq
where
\beq\label{defkonetwo}
k_{\Gd_1,\Gd_2}(u,w) :=
\int_{\{|u - v| > \Gd_1\} \cap \{|v - w| > \Gd_2\}} r_i(u, u  - v) \left[ r_j(v, v - w) - r_j(u, v- w) \right] \,dv .
\eeq
We will show that $k(u, w) := \lim_{\Gd_1, \Gd_2 \downarrow 0} k_{\Gd_1,\Gd_2}(u,w)$ exists and it is weakly singular, or more precisely, if $U$ is a bounded set in $\Rbb^2$, then
\beq\label{estkonetwo}
\left| k(u, w) \right| \le \dfrac{C}{|u - w|^{2 - \Gb}}, \quad u, w \in U
\eeq
for some constant $C$, where $\Gb=3\Ga / 4$. Then, \eqnref{RiRjRij} follows from \eqnref{estkonetwo} since a weakly singular integral operator is compact on $H^s(U)$. It is worth mentioning that the order of integrations and limits can be switched in \eqnref{orderchange} since the integral in \eqnref{orderchange} is absolutely convergent as is shown in the course of proving \eqnref{estkonetwo}.

Note that
\begin{align}
&\left| L(v, v - w) - L(u, v - w) \right| \nonumber \\
& = \left| \langle v - w, G(v)(v - w) \rangle^{3/2} - \langle v - w, G(u)(v - w) \rangle^{3/2} \right|
L(v, v - w) L(u, v - w) \nonumber \\
& \leq \left| \langle v - w, G(v)(v - w) \rangle^{3/4} - \langle v - w, G(u)(v - w) \rangle^{3/4} \right| \nonumber \\
& \quad \times L(v, v - w)^{1/2} L(u, v - w)^{1/2} \left( L(v, v - w)^{1/2} + L(u, v - w)^{1/2} \right). \label{1000}
\end{align}
Here, we invoke an inequality: for all $0< p < 1$,
$$
\left| | x |^p - | y|^p \right| \leq |x - y|^p, \quad x, y \in \Rbb.
$$
So, we have
\begin{align}
& \left| \langle v - w, G(v)(v - w) \rangle^{3/4} - \langle v - w, G(u)(v - w) \rangle^{3/4} \right| \nonumber \\
& \leq \left| \langle v - w, (G(v) - G(u))(v - w) \rangle \right|^{3/4} \nonumber  \\
& \leq \left( \sum_{i, j=1}^2 |v_i - w_i| |g_{ij}(v) - g_{ij}(u)| |v_j - w_j| \right)^{3/4}  \nonumber \\
& \leq C |u - v|^{\Gb} |v - w|^{3 / 2}, \label{1001}
\end{align}
where the last inequality holds since $G(u)$ is $C^\Ga$. Note that
there exist two positive constants $C_1$ and $C_2$ such that
$$
C_1 |v-w|^{-3} \le L(v, v - w) \le C_2 |v-w|^{-3}
$$
for all $v, w \in \Rbb^2$, and similar estimates are valid for $L(u, v - w)$ as well for all $u \in \Rbb^2$.  It then follows from \eqnref{1000} and \eqnref{1001} that
$$
\left| L(v, v - w) - L(u, v - w) \right| \leq C \dfrac{|u - v|^{\Gb}}{|v - w|^3}.
$$
Since $r_j(u, v) = v_j L(u, v)$, we also have
\beq\label{rjest}
\left| r_j(v, v - w) - r_j(u, v - w) \right| \leq C \dfrac{|u - v|^{\Gb}}{|v - w|^2}.
\eeq

We assume $|u - w| > \Gd$ for some $\Gd > 0$ and take $\Gd_1$ and $\Gd_2$ such that $2 \max \{ \Gd_1, \Gd_2 \} < \Gd$. We then decompose the domain of the integral in \eqnref{defkonetwo} into two disjoint subsets:
\beq \label{decom}
\{|u - v| > \Gd_1\} \cap \{|v - w| > \Gd_2\} = A \cup B,
\eeq
where
\begin{align*}
A &:= \{|u - v| > \Gd_1\} \cap \{ |u - v| \le |v - w| \},\\
B &:= \{|v - w| > \Gd_2\} \cap \{ |v - w| < |u - v| \}.
\end{align*}
Indeed, if there existed $v \in A \cap \{|v - w| \le \Gd_2\}$, then we would have
$$
|u - w| \le |u - v| + |v - w| \le 2|v - w| \le 2 \Gd_2 < \Gd,
$$
which contradicts the assumption that $|u - w| > \Gd$. Thus, we have $A = \{|u - v| > \Gd_1\} \cap \{|v - w| > \Gd_2\} \cap \{ |u - v| \le |v - w| \}$. In the same way, we see that $B = \{|u - v| > \Gd_1\} \cap \{|v - w| > \Gd_2\} \cap \{ |v - w| < |u - v| \}$. Thus (\ref{decom}) holds.

We write
$$
k_{\Gd_1,\Gd_2}(u,w) = I_A+I_B :=
\int_{A} + \int_B r_i(u, u  - v) \left[ r_j(v, v - w) - r_j(u, v- w) \right] \,dv .
$$

We first estimate $I_A$. Observe first that
$$
A = A_1 \cup A_2 := \{ \Gd_1 < |u - v| < |u - w| /2 \} \cup \{ |u - w| / 2 < |u - v| \le |v - w| \}.
$$
According to \eqnref{rjest} we have
\begin{align*}
|I_A| \leq \int_{A_1} + \int_{A_2} \dfrac{C}{|u - v|^{2 - \Gb} |v - w|^2}\,dv =: I_{A1} + I_{A2}.
\end{align*}
If $v \in A_1$, then
$$
|v - w| \ge |u-w| - |u-v| \ge |u-w|/2,
$$
and hence
$$
I_{A1} \le \dfrac{C}{|u - w|^2} \int_{\{ \Gd_1 < |u - v| < |u - w| /2\} } \dfrac{1}{|u - v|^{2 - \Gb}}\,dv \leq \dfrac{C}{|u - w|^{2 - \Gb}}.
$$
Here and afterwards, the constant $C$ appearing in the course of estimates may differ at each occurrence, and it is independent of $\Gd$, $\Gd_1$ and $\Gd_2$. We also have
$$
I_{A2} \le \int_{\{ |u - w| / 2 < |u - v| \}} \dfrac{C}{|u - v|^{4 - \Gb}}\,dv
\leq \dfrac{C}{|u - w|^{2 - \Gb}}.
$$
Thus we have
\beq\label{IA}
\left| I_A \right| \le \dfrac{C}{|u - w|^{2 - \Gb}}.
\eeq

We now deal with $I_B$. We decompose $B$ as
$$
B=B_1 \cup B_2 := \{ \Gd_2 < |v - w| \le |u - w|/2 \} \cup \{|u - w|/2 < |v - w| < |u - v| \},
$$
and write $I_B$ as
$$
I_B  = \int_{B_1} + \int_{B_2} r_i(u, u  - v) \left[ r_j(v, v - w) - r_j(u, v- w) \right] \,dv=: I_{B1} + I_{B2} .
$$

The integral $I_{B2}$ is easy to handle. Indeed, we have
\begin{align}
|I_{B2}| &\le \int_{\{|u - w|/2 < |v - w| < |u - v|\}} \dfrac{C}{|u - v|^{2-\Gb} |v - w|^2}\,dv \nonumber\\
& \le \int_{\{|u - w|/2 < |v - w|\}} \dfrac{C}{|v - w|^{4-\Gb}}\,dv \leq \dfrac{C}{|u - w|^{2-\Gb}} \label{IB2}.
\end{align}

The rest of the proof is devoted to estimating $I_{B1}$. We first observe that
\begin{align*}
& \int_{B_1} r_i(u, u - w) \left[ r_j(w, v - w) - r_j(u, v - w) \right] \,dv \\
& = r_i(u, u - w) \int_{B_1} \left[ r_j(w, v - w) - r_j(u, v - w) \right] \,dv = 0.
\end{align*}
Thus $I_{B1}$ can be written as
\begin{align*}
I_{B1} &= \int_{B_1} \big[ r_i(u, u - v) \left[ r_j(v, v - w) - r_j(u, v - w) \right]  \\
&\quad \quad - r_i(u, u - w) \left[ r_j(w, v - w) - r_j(u, v - w) \right] \big] \,dv .
\end{align*}
We then write the integrand as
\begin{align*}
& r_i(u, u - v) \big[ r_j(v, v - w) - r_j(u, v - w) \big]
 - r_i(u, u - w) \big[ r_j(w, v - w) - r_j(u, v - w) \big] \\
&= r_i(u, u - v) \big[ r_j(v, v - w) - r_j(w, v - w) \big] \\
& \qquad + (u_i-v_i) \big[ L(u,u-v) - L(u,u-w) \big] \big[ r_j(w, v - w) - r_j(u, v - w) \big] \\
& \qquad + (v_i-w_i) L(u,u-v) \big[ r_j(w, v - w) - r_j(u, v - w) \big].
\end{align*}
Thus we have
$$
I_{B1} = J_1 + J_2 + J_3,
$$
where $J_k$ ($k=1,2,3$) corresponds to the decomposition of the integral kernel above.

We have for $J_1$ that
\begin{align*}
|J_1| & \le \int_{B_1} |r_i(u, u - v)| \big| r_j(v, v - w) - r_j(w, v - w) \big| \,dv \\
& \le \int_{B_1} \frac{C}{|u-v|^2 |v-w|^{2-\Gb}} \,dv.
\end{align*}
Note that if $v \in B_1$, then
\beq\label{u-vu-w}
|u-v| \ge |u-w|/2.
\eeq
Thus we have
$$
|J_1| \le \frac{C}{|u-w|^2} \int_{B_1} \frac{1}{|v-w|^{2-\Gb}} \,dv \le \frac{C}{|u-w|^{2-\Gb}}.
$$

To estimate $J_2$, we observe in the same way as (\ref{1000}) that
\begin{align*}
&\left| L(u, u - v) - L(u, u - w) \right|\\
&\leq \left| \langle u - v, G(u) (u - v) \rangle^{3/4} - \langle u - w, G(u) (u - w) \rangle^{3/4} \right| \\
&\quad \times L(u, u - v)^{1/2} L(u, u - w)^{1/2} \left( L(u, u - v)^{1/2} + L(u, u - w)^{1/2} \right).
\end{align*}
One can see that
\begin{align*}
& \left| \langle u - v, G(u) (u - v) \rangle^{3/4} - \langle u - w, G(u) (u - w) \rangle^{3/4} \right| \\
& \le \left| \langle u - v, G(u) (u - v) \rangle - \langle u - w, G(u) (u - w) \rangle \right|^{3/4} \\
& \le \left| \langle w - v, G(u) (u - v) \rangle + \langle u - w, G(u) (w-v) \rangle \right|^{3/4} \\
& \le C |v-w|^{3/4} \left( |u-v|^{3/4} + |u-w|^{3/4} \right).
\end{align*}
It then follows that for $v \in B_1$,
\begin{align*}
&\left| L(u, u - v) - L(u, u - w) \right| \\
& \le  C |v-w|^{3/4} \left( |u-v|^{3/4} + |u-w|^{3/4} \right) \\
& \quad \times |u-v|^{-3/2}|u-w|^{-3/2} \left( |u-v|^{-3/2} + |u-w|^{-3/2} \right) \\
& \le  C |v-w|^{3/4} \left( |u-v|^{-3/4}|u-w|^{-3/2} + |u-v|^{-3/2}|u-w|^{-3/4} \right) \\
& \quad \times \left( |u-v|^{-3/2} + |u-w|^{-3/2} \right) \\
& \le C |v-w|^{3/4} |u-w|^{-15/4},
\end{align*}
where the last inequality follows from \eqnref{u-vu-w}. Then \eqnref{rjest} and the relation $|u - v| < |v - w|$ yield that
\begin{align*}
|J_2| \le \frac{C}{|u-w|^{15/4-\Gb}} \int_{B_1} \frac{1}{|v-w|^{1/4}} dv \le \frac{C}{|u-w|^{2-\Gb}}.
\end{align*}

Similarly, one can show that
\begin{align*}
|J_3| \le \frac{C}{|u-w|^{3-\Gb}} \int_{B_1} \frac{1}{|v-w|} dv \le \frac{C}{|u-w|^{2-\Gb}}.
\end{align*}
Thus we infer that
\beq \label{IB1}
|I_{B1}| \le \frac{C}{|u-w|^{2-\Gb}}.
\eeq

From (\ref{IA}), (\ref{IB2}) and (\ref{IB1}), we have
$$
\lim_{\Gd_1, \Gd_2 \downarrow 0} |k_{\Gd_1, \Gd_2}(u, w)| \leq \dfrac{C}{|u - w|^{2 - \Gb}}
$$
for $|u - w| > \Gd$, and since $\Gd > 0$ is arbitrary, this completes the proof.
\qed

\section{Polynomial compactness of the eNP operator}\label{sec:Riesz}

It is shown in \cite{AKM18} that the eNP operator can expressed in terms of surface Riesz transforms. In this section we review it and prove Theorem \ref{Polynomially compact} using Theorem \ref{thm:Riesz}.

Let $\GO$ be a bounded domain in $\Rbb^3$ whose boundary $\p\GO$ is $C^{1,\Ga}$-smooth for some $\Ga>0$.
Let $(\Gl, \Gm)$ be the Lam\'e parameters for $\GO$ satisfying the strong convexity condition: $\Gm > 0$ and $3\Gl + 2\Gm > 0$. The isotropic elasticity tensor $\Cbb = ( C_{ijkl} )_{i, j, k, l = 1}^3$ and the corresponding Lam\'e system $\Lcal_{\Gl,\Gm}$ are defined by
$$
C_{ijkl} := \Gl \, \Gd_{ij} \Gd_{kl} + \mu \, ( \Gd_{ik} \Gd_{jl} + \Gd_{il} \Gd_{jk} )
$$
and
$$
\Lcal_{\Gl,\Gm} u := \nabla \cdot \Cbb \hatna u = \Gm \Delta u + (\Gl + \Gm) \nabla \nabla \cdot u,
$$
where $\hatna$ denotes the symmetric gradient, namely,
$$
\hatna u := \frac{1}{2} \left( \nabla u + \nabla u^T \right) \quad (T \mbox{ for transpose}).
$$
The corresponding conormal derivative on $\p \GO$ is defined to be
$$
\p_\nu u := (\Cbb \hatna u) n = \Gl (\nabla \cdot u) n + 2\Gm (\hatna u) n \quad \mbox{on } \p \GO,
$$
where $n$ is the outward unit normal to $\p \GO$.

Let $\BGG(x) = \left( \Gamma_{ij}(x) \right)_{i, j = 1}^3$ be the Kelvin matrix of the fundamental solution to the Lam\'{e} operator $\Lcal_{\Gl, \mu}$, namely,
$$
  \Gamma_{ij}(x) =
    - \ds \frac{\Ga_1}{4 \pi} \frac{\Gd_{ij}}{|x|} - \frac{\Ga_2}{4 \pi} \ds \frac{x_i x_j}{|x|^3}, \quad x \neq 0,
$$
where
$$
  \Ga_1 = \frac{1}{2} \left( \frac{1}{\mu} + \frac{1}{2 \mu + \Gl} \right) \quad\mbox{and}\quad \Ga_2 = \frac{1}{2} \left( \frac{1}{\mu} - \frac{1}{2 \mu + \Gl} \right).
$$
The eNP operator is defined by
$$
\BK [f] (x) := \text{p.v.} \int_{\p \GO} \p_{\nu_x} {\bf \GG} (x-y) f(y) d \Gs(y) \quad \mbox{a.e. } x \in \p \GO.
$$
Here, the conormal derivative $\p_{\nu_x}\BGG (x-y)$ of the Kelvin matrix with respect to $x$-variables is defined by
$$
\p_{\nu_x}\BGG (x-y) b = \p_{\nu_x} (\BGG (x-y) b)
$$
for any constant vector $b$ (see \cite{Kup-book-65}).

Let
$$
\BK_1(x,y) = \frac{n_x (x-y)^T - (x-y) n_x^T}{2\pi |x-y|^{3}},
$$
where $n_x$ is the outward unit normal at $x$, and let
$$
\BT [f](x):= \text{p.v.} \int_{\p\GO} \BK_1(x,y) f(y) \, d \Gs(y), \quad x \in \p\GO.
$$
It is proved in \cite{AJKKY, AKM18} that
\beq\label{BKBT}
\BK \equiv k_0 \BT.
\eeq
Here \eqnref{BKBT} means that the difference $\BK - k_0 \BT$ is compact on $H^{-1/2}(\p \GO)^3$. We emphasize that $\BT$ is a singular integral operator and bounded on $H^{-1/2}(\p\GO)^3$ as well as on $L^2(\p\GO)^3$ (see \cite{CMM-AM-82}).

Denoting $n_x=(n_1(x), n_2(x), n_3(x))^T$, we have
$$
\BK_1(x,y) = \frac{1}{2\pi |x-y|^{3}}
\begin{bmatrix}
0 & K_{12} (x, y) & K_{13} (x, y) \\
- K_{12} (x, y) & 0 & K_{23} (x, y) \\
- K_{13} (x, y) & - K_{23} (x, y) & 0
\end{bmatrix},
$$
where
\begin{align*}
K_{12} (x, y) &= n_1(x)(x_2-y_2) - n_2(x) (x_1-y_1), \\
K_{13} (x, y) &= n_1(x)(x_3-y_3) - n_3(x) (x_1-y_1), \\
K_{23} (x, y) &= n_2(x)(x_3-y_3) - n_3(x) (x_2-y_2) .
\end{align*}
Let
$$
T_{ij}[f] (x) := \text{p.v.} \int_{\p\GO} \frac{K_{ij} (x, y)}{2\pi |x-y|^{3}} f(y) d\Gs(y),
$$
so that
$$
\BT =
\begin{bmatrix}
0 & T_{12} & T_{13} \\
- T_{12} & 0 & T_{23} \\
- T_{13} & - T_{23} & 0
\end{bmatrix}.
$$

Let $U$ be a coordinate chart in $\p\GO$ so that there is an open set $D$ in $\Rbb^2$ and a parametrization $\Phi: D \to U$, namely,
$$
x= \Phi(u)= (\Gvf_1(u), \Gvf_2(u), \Gvf_3(u)), \quad x \in U, \ \ u \in D.
$$
Then the metric tensor of the surface, denoted by $G(u)= (g_{ij}(u))_{i,j=1}^2$, is given by
\begin{align*}
dx_1^2+dx_2^2+dx_3^2
=g_{11} du_1^2 + 2 g_{12} du_1 du_2 + g_{22} du_2^2,
\end{align*}
where
\beq \label{eq:g}
g_{11}= |\p_1 \Phi|^2, \quad g_{12}=g_{21}=\p_1\Phi \cdot \p_2 \Phi, \quad g_{22}=|\p_2 \Phi|^2.
\eeq
Here and afterwards, $\p_j$ denotes the $j$-th partial derivative. In short, we have
\beq \label{eq:G}
G(u)= D\Phi(u)^T D\Phi(u),
\eeq
where $D\Phi$ is the $3 \times 2$ Jacobian matrix of $\Phi$. We then extend $G(u)$ to $\Rbb^2$ in such a way that $G(u)=I$ for $u$ outside a compact set. With this metric tensor, the surface Riesz transform is defined by \eqnref{Rgdef}.

Choose open sets $U_j$ ($j=1,2$) in $\p\GO$ so that $\ol{U_1} \subset U_2$ and $\ol{U_2} \subset U$. Let $\chi_j$ ($j=1,2$) be $C^{1, \Ga}$-smooth functions such that $\chi_1=1$ in $U_1$, $\mbox{supp} (\chi_1) \subset U_2$, $\chi_2=1$ in $U_2$, and $\mbox{supp} (\chi_2) \subset U$. We denote by $M_j$ the multiplication operator by $\chi_j$, i.e.,
$$
M_j [f](x) = \chi_j(x) f(x),
$$
and by $\widetilde{M}_j$ the multiplication operator by $\chi_j(\Phi(u))$ for $j=1,2$. Let $\Phi^*$ be a pull back operator, namely,
$$
\Phi^*[f](u): = f(\Phi(u)) |\p_1 \Phi \times \p_2 \Phi(u)|.
$$

For ease of notation, we set
\begin{align}
m_{11} &:= (g_{11} \p_2 \Gvf_3  - g_{12} \p_1 \Gvf_3), \label{m11} \\
m_{12} &:= (g_{21} \p_2 \Gvf_3  - g_{22} \p_1 \Gvf_3),\\
m_{21} &:= -(g_{11} \p_2 \Gvf_2  - g_{12} \p_1 \Gvf_2),\\
m_{22} &:= -(g_{21} \p_2 \Gvf_2  - g_{22} \p_1 \Gvf_2),\\
m_{31} &:= (g_{11} \p_2 \Gvf_1 - g_{12} \p_1 \Gvf_1),\\
m_{32} &:= (g_{21} \p_2 \Gvf_1 - g_{22} \p_1 \Gvf_1), \label{m32}
\end{align}
and denote by $M_{ij}$ the multiplication operator by $m_{ij}$. We emphasize that $m_{ij}$ are $C^\Ga$.

Let
\begin{align*}
X_{12} &:= \widetilde{M_2}(M_{11} {R_1^g} + M_{12}{R_2^g})\widetilde{M_1}, \\
X_{13} &:= \widetilde{M_2}(M_{21}{R_1^g}+ M_{22}{R_2^g})\widetilde{M_1} , \\
X_{23} &:= \widetilde{M_2}(M_{31}{R_1^g}+ M_{32}{R_2^g})\widetilde{M_1} ,
\end{align*}
and let
$$
\BR :=\begin{bmatrix}
	      0    &  X_{12} & X_{13}\\
	-X_{12} &      0    & X_{23}\\
	-X_{13} & -X_{23} &    0
	\end{bmatrix}.
$$
Then it is proved in \cite{AKM18} that the following relation holds:
$$
\Phi^* M_2 \BT M_1 \equiv \BR \Phi^*.
$$

Note that the crux of the matter in Theorem \ref{Polynomially compact} is that
\beq\label{p3BK}
p_3(\BK)=\BK(\BK^2-k_0^2 \BI)\equiv 0.
\eeq
In view of \eqnref{BKBT} this fact follows once we have
$$
\BT^3-\BT \equiv 0,
$$
which in turn follows from the following proposition:
\begin{prop} \label{prop:COMBR}
It holds that
\beq\label{BR3}
\BR^3 - \widetilde{M_1} \BR \equiv 0.
\eeq
\end{prop}
We refer to \cite[Section 5]{AKM18} for detailed argument to prove \eqnref{p3BK} from \eqnref{BR3}.
We now briefly show how Proposition \ref{prop:COMBR} is proved using Theorem \ref{thm:Riesz}.

\medskip
\noindent{\sl Proof of Proposition \ref{prop:COMBR}}. We first see that the following commutator relations hold:
\beq\label{3.1}
\widetilde{M_1} M_{ij} R_k^g \equiv R_k^g \widetilde{M_1} M_{ij}.
\eeq
Indeed, we have
\begin{align*}
& R_k^g \widetilde{M_1} M_{ij}[f](u) - \widetilde{M_1} M_{ij} R_k^g[f](u) \\
&= \dfrac{1}{2\pi} \int_{\Rbb^2} r_k(u, u - v) (\Gc_1(v) m_{ij}(v) - \Gc_1(u) m_{ij}(u)) f(v)\,dv,
\end{align*}
where $r_k(u, u - v)$ is defined by \eqnref{rj}. Since $m_{ij}$ is $C^{\Ga}$, we have
$$
\left| r_k(u, u - v) (\Gc_1(v) m_{ij}(v) - \Gc_1(u) m_{ij}(u)) \right| \le C|u-v|^{-2+\Ga}
$$
for some constant $C$. So \eqnref{3.1} follows, that is, $\widetilde{M_1} M_{ij} R_k^g - R_k^g \widetilde{M_1} M_{ij}$ is compact.

We then show that
\beq\label{Xij}
X_{12} X_{13} \equiv X_{13} X_{12}, \quad X_{12} X_{23} \equiv X_{23} X_{12}, \quad X_{13} X_{23} \equiv X_{23} X_{13}.
\eeq
In fact, we have
\begin{align*}
X_{12}X_{13} - X_{12}X_{13}
&= \widetilde{M_2}(M_{11} {R_1^g}+ M_{12} {R_2^g}) \widetilde{M_1}(M_{21} {R_1^g}+ M_{22} {R_2^g})\widetilde{M_1}\\
& \quad - \widetilde{M_2}(M_{21} {R_1^g}+ M_{22} {R_2^g})\widetilde{M_1}(M_{11} {R_1^g}+ M_{12} {R_2^g})\widetilde{M_1}.
\end{align*}
Here, we used the obvious identity: $\Gc_1 \Gc_2 = \Gc_1$. We then obtain using \eqnref{3.1} that
\begin{align*}
X_{12}X_{13} - X_{12}X_{13} \equiv (M_{12}M_{21}-M_{11}M_{22}) \widetilde{M_1}(R_2^g R_1^g - R_1^g R_2^g) \widetilde{M_1}.
\end{align*}
But, \eqnref{commute} implies that $\widetilde{M_1}(R_2^g R_1^g - R_1^g R_2^g) \widetilde{M_1}$ is compact. This proves the first identity in \eqnref{Xij}. The other identities there can be proved in the same way.

Cayley-Hamilton theorem and \eqnref{Xij} yield that
\beq\label{BR4}
\BR^3 + (X_{12}^2 + X_{13}^2 + X_{23}^2) \BR \equiv 0.
\eeq

One can show as before that
\begin{align*}
X_{12}^2 &= (\widetilde{M_2} M_{11} {R_1^g} \widetilde{M_1} + \widetilde{M_2} M_{12} {R_2^g} \widetilde{M_1})^2 \\
& \equiv \widetilde{M_1}(M_{11}^2 (R_1^g)^2 + 2 M_{11} M_{12} {R_1^g}{R_2^g} + M_{12}^2 (R_2^g)^2) \widetilde{M_1},
\end{align*}
likewise,
\begin{align*}
X_{13}^2 \equiv \widetilde{M_1}(M_{21}^2 (R_1^g)^2 + 2 M_{21} M_{22} {R_1^g}{R_2^g} + M_{22}^2 (R_2^g)^2) \widetilde{M_1},
\end{align*}
and
\begin{align*}
X_{23}^2 \equiv \widetilde{M_1}(M_{31}^2 (R_1^g)^2 + 2 M_{31} M_{32} {R_1^g}{R_2^g} + M_{32}^2 (R_2^g)^2)\widetilde{M_1}.
\end{align*}
Thus, we have
\begin{align*}
& X_{12}^2 + X_{13}^2 + X_{23}^2 \\
& \equiv \widetilde{M_1} \left[ \left( \sum_{i=1}^3 M_{i1}^2 \right) (R_1^g)^2 + 2 \left( \sum_{i=1}^3 M_{i1} M_{i2} \right) {R_1^g}{R_2^g} + \left( \sum_{i=1}^3 M_{i2}^2 \right) (R_2^g)^2 \right] \widetilde{M_1} .
\end{align*}
Then using the formulas \eqnref{m11}-\eqnref{m32} for $m_{ij}$ one can show that
\begin{align*}
X_{12}^2 + X_{13}^2 + X_{23}^2 &\equiv
\widetilde{M_1} \big[ (g_{22} g_{11}^2 - 2 g_{11} g_{12}^2 + g_{11} g_{12}^2) (R_1^g)^2\\
& \quad +2 (g_{22} g_{11} g_{21} - g_{11} g_{12} g_{22} - g_{21}^3 + g_{11} g_{12} g_{22}) {R_1^g}{R_2^g}\\
& \quad + (g_{22} g_{21}^2 - 2 g_{21}^2 g_{22} + g_{21} g_{22}^2) (R_2^g)^2 \big] \widetilde{M_1} \\
&= \widetilde{M_1} \det(G) (g_{11} (R_1^g)^2 + 2 g_{12} {R_1^g} {R_2^g} + g_{22} (R_2^g)^2 ) \widetilde{M_1}.
\end{align*}
It then follows from \eqnref{squaresum} that
$$
X_{12}^2 + X_{13}^2 + X_{23}^2 \equiv - \widetilde{M_1}.
$$
Now \eqnref{BR3} follows from \eqnref{BR4}. This completes the proof. \qed


TITLE: logarithmic integral function and asymptotic expansion
QUESTION [5 upvotes]: Show that Logarithmic integral function $$\int_2^x {1\over \log(t)} \, dt = Li(x)$$  has asymptotic expansion of the form $${x\over \log(x)}\cdot\sum_{j=0}^\infty a_j\cdot (\log(x))^{-j}.$$
I tried different stuff but I did not conclude to solve it. Any help for solving this?

REPLY [4 votes]: This is a very sketchy answer!
Observe, that
$$\text{Li(x)}=-\int_{\log(2)}^{\log(x)}\frac{e^{-y}}{y}=\text{Ei}(\log(x))-\text{Ei}(\log(2))$$
where $\text{Ei(x)}=\int_{x}^{\infty}\frac{e^{-q}}{q}$   the Expontential integral .
Doing intgration by parts N times we obtain
$$
\text{Ei(x)}=\frac{e^{-x}}{x}-\frac{e^{-x}}{x^2}+\frac{2!e^{-x}}{x^3}+....+(-1)^{N-1}\frac{e^{-x}(N-1)!}{x^n}+R(N)
$$
where $R(N)$ is the remaining integral.
We conclude that
$$
\text{Ei(x)}=\frac{e^{-x}}{x}\sum_{k=1}^{\infty}(-1)^{k+1}\frac{k!}{x^k}
$$
in an asymptotic sense.
Now plug in $\log(x)$ and $\log(2)$ and you are done.


TITLE: Are there infinitely many primes such that $4\notin\langle 6,8\rangle $?
QUESTION [1 upvotes]: Suppose $p\equiv 1\pmod{5}$, $\langle 6,8\rangle$ denotes the subgroup of $\mathbb{Z}_p^{*}$ generated by $6,8$. Is there any way to prove or disprove that:
There are infinitely many primes $p\equiv 1\pmod{5}$ such that $4\notin\langle 6,8\rangle$?

REPLY [3 votes]: Let $F=\mathbf{Q}(\sqrt[3]{6},\sqrt[3]{2},\zeta_{15})$. This is Galois with Galois group the semi direct product of $(\mathbf{Z}/15)^*$ by $(\mathbf{Z}/3)^2$. By Cebotarev, there are infinitely many primes whose a Frobenius is the element which is trivial in the quotient and is the order three element which fixes the first cube root and not the second. The fact that the Frobenius is trivial in the quotient implies that $p \equiv 1 \mod 15$, and the other property implies that $6$ is a cube module $p$ but $2$ (and thus $4$) is not. But now $4$ can’t be in the group generated by $6$ and $8$ since the latter are cubes. Hence there exist a positive density of such primes p.
The inverse is also true since if $p \not\equiv 1 \mod 3$ then $2$ and thus $4$ is already generated by $8$.


TITLE: Does unbounded curvature implies unbounded diameter?
QUESTION [1 upvotes]: Let $(M,g)$ be a Riemannian manifold. Assume that for each $C > 0$ there exists $p\in M$ and $X,Y\in T_pM$ unitary such that $K(X,Y) > C$. Does this implies that the diameter of $M$ is infinite?
I just have an intuition about it, related to the this figure:Gabriel's  Horn

REPLY [1 votes]: You could have a sequence of rapidly shrinking spheres all stitched to the same finite open section of a plane. Here, the manifold is not complete though.


TITLE: proof that f*d is an induced metric
QUESTION [0 upvotes]: Let X be a set and (M, d) be a metric space. Let f : X → M be an injective map. Then we define
f∗d := d ◦ (f × f): X × X →(o,infinity]
i.e.
f∗d(x, y) = d(f(x), f(y)) for x, y ∈ X .
I want to prove that f∗d is a distance function on X (“induced metric”). here's what ive tried so far ...
taking all of the above as fact ;
f is injective therefore f(x)=m1 and f(y)=m2  where m1 does not equal m2 and m1,m2 are elements of M.
=> fd(x,y)=d(m1,m2) which is defined already as a metric space X C(M,d) therefore fd is also a distance function.
...I feel like thats too easy and i must be missing something obvious ? any tips would be much appreciated but please dont answer the full question as i want to learn this on my own :)

REPLY [1 votes]: $f^*d(x,y)=d(m_1,m_2)$ which is defined already as a metric space ... therefore $f^*d$ is also a distance function.

This sentence doesn't make any sense.  To what is the “which” referring to?  The only possible antecedent in the sentence is the number $f^*d(x,y)=d(m_1,m_2)$.  A number isn't defined as a metric space.
If you're asserting that the facts that $d$ is a metric and $f^*d$ is defined in terms of $d$, make $f^*d$ a metric on $X$—well, that's exactly what you need to prove.
Go back to the definitions of metric:

You must show $f^*d(x,y) \geq 0$ for all $x,y \in X$, and if $f^*d(x,y) = 0$, then $x=y$.
You must show $f^*d(x,y) = f^*d(y,x)$ for all $x,y \in X$.
You must show the triangle inequality: For all $x$, $y$, $z\in X$,
$$
    f^*d(x,y) + f^*d(y,z) \geq f^*d(x,z)
$$

Each of those statements can be expressed in terms of $f$ and $d$. You'll know you're doing it right when you find that you need to use each of the requirements that must be satisfied for $d$ to be a metric, and the assumption that $f$ is injective.....


text:
\begin{document}

\begin{frontmatter}
\title{Minimax and minimax adaptive estimation in multiplicative regression: locally bayesian approach}
\runtitle{Locally bayesian approach}

\begin{aug}
\author{\fnms{M.} \snm{Chichignoud} \ead[label=e1]{chichign@cmi.univ-mrs.fr}}

\runauthor{M. Chichignoud}

\affiliation{Université Aix-Marseille 1}

\address{Université Aix-Marseille 1\\
  LATP, 39 rue Joliot Curie, 13453 Marseille cedex 13, FRANCE,\\
  \printead{e1}\\
\phantom{E-mail:\ }}
\end{aug}
\selectlanguage{english}
\begin{abstract}
The paper deals with the non-parametric estimation in the regression
with the multiplicative noise. Using the local polynomial fitting
and the bayesian approach, we construct the minimax on isotropic
Hölder class estimator. Next applying Lepski's method, we propose
the estimator which is optimally adaptive over the collection of
isotropic Hölder classes. To prove the optimality of the proposed
procedure we establish, in particular, the exponential inequality
for the deviation of locally bayesian estimator from the parameter
to be estimated. These theoretical results are illustrated by
simulation study.
\end{abstract}

\begin{keyword}[class=AMS]
\kwd{62G08} \kwd{62G20}
\end{keyword}

\begin{keyword}
\kwd{local bayesian fitting} \kwd{multiplicative regression}
\kwd{adaptive bandwidth selector} \kwd{Lepski's method}
\kwd{optimality criterion}
\end{keyword}

\end{frontmatter}

\section{Introduction}Let statistical experiment be generated by the couples of observations
$Y^{(n)}=(X_i,Y_i)_{i=1,...n},\:n\in\bN^{*}$ where $(X_i,Y_i)$
satisfies the equation
\begin{equation}
\label{model} Y_i=f(X_i)\times U_i,\quad i=1,\ldots, n.
\end{equation}
Here $f:[0,1]^d\rightarrow\mathbb{R}$ is unknown function and we are interested in estimating $f$ at a given point $y\in[0,1]^d$ from observation
$Y^{(n)}$.

The random variables (noise) $(U_i)_{i\in{1,\ldots,n}}$  are supposed to be independent and  uniformly distributed on $[0,1]$.

The design points $(X_i)_{i\in{1,...,n}}$ are deterministic and
without loss of generality we will assume that
$$
X_i\in \left\{1/n^{1/d},2/n^{1/d},\ldots, 1\right\}^{d},\:\:
i=1,\ldots,n.
$$
Along  the paper the unknown function $f$  is supposed to be smooth,
in particular, it belongs to the Hölder ball of functions
$\mathbb{H}_d({\beta},L,M)$ (see Definition \ref{def_holder} below).
Here ${\beta}>0$ is the smoothness of $f$, $M$ is the sum of upper bounds of
$f$ and its partial derivatives and $L>0$ is Lipschitz constant.

Moreover, we will consider only the functions $f$ separated away
from zero by some positive constant. Thus, from now on we will suppose
that there exists $0<A<M$ such that $f\in
\mathbb{H}_d({\beta},L,M,A)$, where
$$
 \mathbb{H}_d({\beta},L,M,A)=\left\{g\in\mathbb{H}_d({\beta},L,M): \inf_{x\in [0,1]^d}g(x)\geq A\right\}.
$$

\paragraph{Motivation.}
The theoretical interest to the multiplicative regression model (\ref{model}) with discontinuous noise
is dictated by the following fact.
The typical  approach to the study of the models with multiplicative
noise consists  in their transformation into the model with an additive noise
and in the application, after that, the  linear smoothing technique, based
on standard methods like kernel smoothing, local polynomials etc.
Let us illustrate the latter approach by the consideration of one of the
most popular non-parametric model namely multiplicative gaussian regression
\begin{equation}
\label{model-gauss} Y_i=\sigma(X_i) \xi_i,\quad i=1,\ldots, n.
\end{equation}
Here $\xi_i,\: i=1,\ldots, n $ are i.i.d. standard gaussian random variables and  the goal is to estimate
the variance $\sigma^{2}(\cdot)$.

Putting $Y^\prime_i=Y^2_i$ and $\eta_i=\xi_i^2-1$ one can transform the model (\ref{model-gauss}) into
the heteroscedastic additive regression:
$$
Y^\prime_i=\sigma^{2}(X_i)+ \sigma^{2}(X_i) \eta_i,\quad i=1,\ldots, n,
$$
where, obviously, $\bE\eta_i=0$. Applying any of the linear methods mentioned above to the estimation of $\sigma^{2}(\cdot)$ one can construct an
estimator whose estimation accuracy is given by $n^{-\frac{\beta}{2\beta+d}}$ and which is optimal in minimax sense (See Definition
\ref{def_minimax}). The latter result is proved under assumptions on $\sigma^{2}(\cdot)$ which are similar to the assumption imposed on the function
$f(\cdot)$. In particular, $\beta$ denotes the regularity of the function $\sigma^{2}(\cdot)$. The same result can be obtained  for any noise
variables $\xi_i$ with known, continuously  differentiable density, possessing
sufficiently many moments.
\par

The situation changes dramatically when one considers the noise with discontinuous distribution density. Although, the transformation of the original
multiplicative model to the additive one is still possible, in particular, the model (\ref{model}) can be rewritten as
$$
Y^\prime_i=f(X_i)+ f(X_i) \eta_i,\quad Y^\prime_i=2Y_i,\:\:\eta_i=2u_i-1,\quad i=1,\ldots, n,
$$
the linear methods are not optimal anymore. As it is proved in Theorem
\ref{th:lower_bound} the optimal accuracy is given by $n^{-\frac{\beta}{\beta+d}}$. To achieve this rate the non-linear estimation procedure,
based on locally bayesian approach, is proposed in Section \ref{sectionMinimax}.

\par

Another interesting feature is the selection from given family of estimators (see \cite{Barron_Birge_Massart99}, \cite{Goldenshluger_Lepski08}). Such
selections are used for construction of data-driven (adaptive) procedures. In this context, several approaches to the selection from the family of
linear estimators were recently proposed, see for instance \cite{Goldenshluger_Lepski08}, \cite{Goldenshluger_Lepski09},
\cite{Juditsky_Lepski_Tsybakov09} and the
references therein. However, these methods are heavily based on the linearity property. As we already mentioned
the locally bayesian estimators are non-linear and in Section \ref{sectionAdaptive} we propose the selection rule from this family. It requires, in
particular, to develop new non-asymptotical exponential inequalities, which may have an independent interest.

\medskip

Besides the theoretical interest, the multiplicative regression model is applied in various domains, in particular,
in the image processing, for example, in so-called {\it nonparametric frontier
model} (see \cite{Allon_Beenstock_Hackman_Passy_Shapiro}, \cite{Simar_Wilson00}) can be considered as the particular case of the model (\ref{model}).
Indeed, the reconstruction  of the regression function $f$ can be viewed as the estimation of a production set $\cP$. Indeed,  $Y_i\leq
f(X_i),\:\forall i$, and, therefore, the estimation of $f$ is reduced to finding the upper boundary of $\cP$. In this context, one can also cite
\cite{Korostelev_Tsybakov93} dealing with the estimation of function's support. It is worth to mention that although nonparametric estimation in the
latter models is studied, the problem of adaptive estimation was not considered in the literature.

config: arxiv


TITLE: limit of an expression?
QUESTION [0 upvotes]: Actually what I'm really looking for is the proof of the following limit:
$\frac{\left [\left ( \frac{k-i+1}{k}\right )^{j} - \left ( \frac{k-i}{k}\right )^{j}  \right ]}{\sum_{i=1}^{Q} \left ( \frac{k-i+1}{k}\right )^{j} - \left ( \frac{k-i}{k}\right )^{j}}$
as k is large enough, where $j$ is fixed and $1 \le i \le Q$ with $Q$ constant.
In the original proof, they just say the limit is $\frac{1}{Q}$, but no details are given.
Is there any indication please?
Thanks

REPLY [1 votes]: Hint
Divide numerator and denominator by $k$ and you then have $$A=(1-\frac {i-1}{k})^j-(1-\frac {i}{k})^j$$ and apply the binomial theorem.
For the very first terms, you should arrive to $$A=\frac{j}{k}+j\frac{-2 i j+2 i +j-1}{2
   k^2}+O\left(\left(\frac{1}{k}\right)^3\right)$$
I am sure that it was obvious to you that expanding the powers in the original expressionswill remove the $k^j$ which stays in the denominator.


TITLE: Reproduction of the explicit calculation of the Sunset diagram from the book Critical Properties of $\phi^4$ Theories (Kleinert)
QUESTION [3 upvotes]: The book starts with the equation
\begin{equation}
I(D) = \lambda^2 \int \frac{d^Dp_1}{(2\pi)^D}\frac{d^Dp_2}{(2\pi)^D} \frac{1}{\mathbf{p}_1^2 + m^2} \frac{1}{\mathbf{p}_2^2+m^2} \frac{1}{(\mathbf{p}_1+\mathbf{p}_2 + \mathbf{q})^2+m^2}\vphantom{\dfrac{a}{b}}
\tag{8.74}\label{8.74}
\end{equation}
then introduce the partial p
\begin{equation}
1 = \frac{1}{2D}\left( \frac{\partial p_1^\mu}{\partial p_1^\mu} + \frac{\partial p_2^\mu}{\partial p_2^\mu}\right)
\tag{8.75}\label{8.75}
\end{equation}
and get
\begin{equation}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!I(D) = - \frac{\lambda^2}{D-3} \int \frac{d^Dp_1}{(2\pi)^D}\frac{d^Dp_2}{(2\pi)^D} \frac{3m^2 + \mathbf{q}(\mathbf{p}_1 + \mathbf{p}_2 + \mathbf{q})}{(\mathbf{p}_1^2 + m^2) (\mathbf{p}_2^2+m^2)[(\mathbf{p}_1+\mathbf{p}_2 + \mathbf{q})^2+m^2]^2}
\tag{8.76}\label{8.76}
\end{equation}

REPLY [0 votes]: Starting with the integral,
\begin{equation}
I(d) = \int dq_1dq_2 \frac{1}{(q_1^2+m^2)(q_2^2+m^2)[(q_1+q_2+P)^2+m^2]},
\tag{1}\label{1}
\end{equation}
and inserting the partial q in (1), we have
\begin{equation}
I(d) = \frac{1}{2d}\int dq_1dq_2 \left(\frac{\partial q_1^\mu}{\partial q_1^\mu} + \frac{\partial q_2^\mu}{\partial q_2^\mu}\right)\frac{1}{(q_1^2+m^2)(q_2^2+m^2)[(q_1+q_2+P)^2+m^2]}.
\tag{2}\label{2}
\end{equation}
The application of the partial derivative is of the form,
$$\frac{\partial x}{\partial x}=1$$
$$\int 1\cdot f(x)dx = \int \frac{\partial x}{\partial x}f(x)dx$$
$$\int \left\{ \frac{\partial}{\partial x}\left[ xf(x)\right] - x\frac{\partial f(x)}{\partial x}\right\}dx$$
the surface term is equal to zero because the contour is infinite! So we have for our case,
\begin{equation}
\int f(x)dx = -\int x \frac{\partial f(x)}{\partial x}.
\tag{3}\label{3}
\end{equation}
Using (3) equation (2) becomes
\begin{equation}
I(d) = -\frac{1}{2d}\int dq_1dq_2 \left(q_1^\mu\frac{\partial}{\partial q_1^\mu} + q_2^\mu\frac{\partial}{\partial q_2^\mu}\right)\frac{1}{(q_1^2+m^2)(q_2^2+m^2)[(q_1+q_2+P)^2+m^2]}.
\tag{4}\label{4}
\end{equation}
Differentiating, we get
\begin{eqnarray}
I(d) &=& \frac{1}{d}\int dq_1dq_2 \Bigg\{\frac{q_1^2}{(q_1^2+m^2)^2(q_2^2+m^2)[(q_1+q_2+P)^2+m^2]}\\
&+& \frac{q_2^2}{(q_1^2+m^2)(q_2^2+m^2)^2[(q_1+q_2+P)^2+m^2]}\\
&+& \frac{(q_1+q_2)(q_1+q_2+P)}{(q_1^2+m^2)(q_2^2+m^2)[(q_1+q_2+P)^2+m^2]^2}\Bigg\}.
\tag{5}\label{5}
\end{eqnarray}
Inserting $m^2 - m^2$ in (5) and $P² - P²$ in the last integral of it
\begin{eqnarray}
I(d) &=& \frac{1}{d}\int dq_1dq_2 \Bigg\{\frac{q_1^2+m^2-m^2}{(q_1^2+m^2)^2(q_2^2+m^2)[(q_1+q_2+P)^2+m^2]}\\
&+& \frac{q_2^2 + m^2-m^2}{(q_1^2+m^2)(q_2^2+m^2)^2[(q_1+q_2+P)^2+m^2]}\\
&+& \frac{m^2-m^2 + (q_1+q_2+P^2-P^2)(q_1+q_2+P)}{(q_1^2+m^2)(q_2^2+m^2)[(q_1+q_2+P)^2+m^2]^2}\Bigg\}.
\tag{6}\label{6}
\end{eqnarray}
Using only algebra and rewriting in terms of I(d),
\begin{eqnarray}
I(d) &=& -\frac{1}{d-3}\int dq_1dq_2 \Bigg\{\frac{m^2}{(q_1^2+m^2)^2(q_2^2+m^2)[(q_1+q_2+P)^2+m^2]}\\
&+& \frac{m^2}{(q_1^2+m^2)(q_2^2+m^2)^2[(q_1+q_2+P)^2+m^2]}\\
&+& \frac{m^2+ P(q_1+q_2+P)}{(q_1^2+m^2)(q_2^2+m^2)[(q_1+q_2+P)^2+m^2]^2}\Bigg\}.
\tag{7}\label{7}
\end{eqnarray}
Using the following substitution, $q_1 = -q_2 - P - q$ on the first integral on the right hand side and $q_2 = -q_1 - P - q$ on the second integral on the right hand side of (7), it is easy to see that the three integrals proportions to $m²$ have the same solution.
Therefore, we can write (7) as
\begin{eqnarray}
\boxed{
I(d) = -\frac{1}{d-3}\int dq_1dq_2 \frac{3m^2+ P(q_1+q_2+P)}{(q_1^2+m^2)(q_2^2+m^2)[(q_1+q_2+P)^2+m^2]^2}}.
\tag{8}\label{8}
\end{eqnarray}


TITLE: Geometric Interpretation of the Basel Problem?
QUESTION [17 upvotes]: Does the Pi in the solution to the Basel problem have any geometric significance?  Every time I see Pi, I have to think of a circle.  I would love to see a nice intuitive picture connecting the Basel problem with geometrical figures.  Anyone here been lucky enough to stumble upon such a thing?

REPLY [7 votes]: 3Blue1Brown's recent take is very well crafted:

https://www.youtube.com/watch?v=d-o3eB9sfls


TITLE: Sizes of maximum matchings in a finite, simple, undirected graph
QUESTION [0 upvotes]: Let $G=(V,E)$ be a finite, simple, undirected graph. We say that a matching $M\subseteq E$ is a maximum matching if for all $e\in (E\setminus M)$ the set $M\cup\{e\}$ is not a matching any more.
Maximum matches can have different sizes.
Question. Is there a positive integer $k\in\mathbb{N}$ such that for all finite graphs $G=(V,E)$ and all maximum matchings $M_1, M_2\subseteq E$ we have $|M_1|-|M_2| \leq k$? If yes, what is the value of $k$?

REPLY [3 votes]: Let $m^-, m^+$ be the sizes of a smallest and largest maximal matching respectively.  The best possible inequality you can prove in general is that $m^- \geq m^+/2$.
To see that this inequality is true, let $M$ be a maximum matching and let $M'$ be any other maximal matching.  Then every edge of $M$ meets an edge of $M'$, else we could extend $M'$ using an edge from $M$.  But each edge of $M'$ meets at most $2$ edges of $M$, so $|M'| \geq |M|/2$.
To see that this inequality is best possible, it suffices construct a graph in which the above argument is tight.  So start with a matching $M$ with an even number of edges and add a second matching $M'$ such that every edge of $M'$ meets two edges of $M$ and no edge of $M$ meets more than one edge of $M'$.  Explicitly, this graph is a vertex-disjoint union of paths of length $3$.