text:
\begin{document}
\title{ Singularity of  macroscopic variables near boundary for gases  with  cut-off inverse-power potential      }
\author[IC]{I-Kun Chen}
\address{Department of Mathematics, National Taiwan University, Taipei 10617, Taiwan }
\email{ikun.chen@gmail.com}
\author[CH]{Chun-Hsiung Hsia}
\address{Institute of Applied Mathematical Sciences, National Taiwan University, Taipei 10617, Taiwan }
\email{willhsia@math.ntu.edu.tw}
\date{\today}
\begin{abstract}
In this article, the boundary singularity for stationary solutions of the linearized Boltzmann equation with cut-off inverse power potential is analyzed. In particular, for cut-off hard-potential cases, we establish the asymptotic approximation for the gradient of the moments. Our analysis indicates the logarithmic singularity of the gradient of the moments.
\end{abstract}
 \maketitle
 \section{introduction}
 The Boltzmann equation is
 \begin{equation}\frac{\partial F}{\partial t}+\xi \cdot\frac{\partial F}{\partial t}=Q(F,F),\end{equation}
 where $F=F(t,x,\xi)$.   $Q$ above  is the collision operator only involves velocity as follows:
 \begin{align}\label{QFF} Q(F,F)&=\int \int_0^{2\pi}\int_0^{\frac{\pi}2}(F'F_*'-FF_*)B(|V|,\theta) d\theta d \epsilon d\xi_*,\end{align}
 where $V=\xi_*-\xi$ and $\alpha$ is a unit vector on a hemisphere  parametrized by $\theta$ and $\epsilon$ such that $\alpha\cdot V=|V|\cos\theta $ and
 \begin{align}
 &F=F(\xi),\ \ F_*=F(\xi_*),\ F'=F(\xi'),\ \ F_*'=F(\xi_*'),\\&
 \xi'=\xi+(\alpha\cdot V)\alpha,\\&
 \xi_*'=\xi_*-(\alpha\cdot V)\alpha.
 \end{align}
 The $B(|V|,\theta)\geq 0$ is called the cross-section. If we consider inverse power force between particles, i.e.,  $Force= \frac1{r^s}$,
 then the  cross-section is in the form
 \begin{equation}
B(|V|,\theta)=|V|^{\gamma}\beta(\theta),
 \end{equation}
 where  $\gamma=\frac{s-5}{s-1}$. The fact  $\beta\sim (\frac{\pi}2-\theta)^{-\frac{s+1}{s-1}}$ as $\theta\to \frac{\pi}2$, which is not integrable in $\theta$, makes us unable to separate \ref{QFF} into gain and lost parts.  To avoid this mathematical difficulty, it was Grad's idea,  \cite{Grad}, to consider the cross-section such that
 \begin{equation}
 B(|V|,\theta)\leq C|V|^{\gamma}\cos\theta\sin\theta.
 \end{equation}
 We will refer these cases as Grad's angular cut-off potential.
 In particular, in our research we will consider the cases that $B$ as a product of a function of $|V| $ and one of $\theta$, i.e.,
  \begin{equation}\label{IP}\begin{split}
 B(|V|,\theta)=|V|^{\gamma}\beta(\theta),\ \
\beta(\theta)\leq C\cos\theta\sin\theta.\end{split}
 \end{equation}
 To make distinctions, we will refer the cases above, \eqref{IP}, as  power-law potential with angular cut-off in this paper.
 We first non-dimensialize the equation so that the Maxwellian  we are interested becomes the standard one:
 \begin{align}w=\frac{1}{\sqrt{\pi}^3}e^{-{|\zeta|^2}}.\end{align}

  We  linearize the  equation around standard Maxwellian  so that
  \begin{align}F=w+w^{\frac12}f.\end{align}
  We have,

  \begin{equation}sh\frac{\partial f}{\partial t}+\zeta \cdot\frac{\partial f}{\partial t}=\frac1{\kappa}L(f),\end{equation}
  where $L(f)=2w^{-\frac12}Q(w,w^\frac12f).$ Under the assumption of Grad's angular cut-off, the linearized collision operator can be decomposed into a damping multiplicative operator $-\nu$ and a smoothing integral operator $K$: \begin{equation}L(\phi)(\zeta)=-\nu(\zeta)\phi(\zeta)+K(\phi)(\zeta).\label{decomposeL}\end{equation}
The the following properties of the linearized collision operator were studied by Grad \cite{Grad} and Caflisch \cite{Caflisch}.   The collision frequency satisfies the following estimate
  \begin{equation}\nu_0(1+|\zeta|)^\gamma\leq \nu(\zeta)\leq \nu_1(1+|
  \zeta|)^\gamma,
  \end{equation}
  where $0<\nu_0<\nu_1$ and $-2\leq\gamma\leq1$ is a parameter from interaction between particles.  $\gamma=1$ is called the hard sphere model; $\gamma=0$ is called the Maxwellian case. Positive $\gamma$'s corresponded to hard potential; negative $\gamma$'s correspond to soft-potential.

If we  consider power-law potential with angular cut-off, we have further properties :\begin{equation}\nu(\zeta)\mbox{ is a function of } |\zeta|.
 \end{equation}
 \begin{equation} \label{dKLp}\Vert \partial_{\zeta_i} K(f)\Vert_{L^p}\leq C\Vert f\Vert_{L^p}, \ p\geq 1.\end{equation}

  In this paper, we restrict our study to the cases $0<\gamma\leq1$.  We define
   \begin{equation}\Vert f(\zeta)\Vert_{L^\infty_a}=\sup_{\zeta}(1+|\zeta|)^a|f(\zeta)|.\end{equation}

   The integral operator  improves the decay:
   \begin{align}
&\label{L2Linfty}\Vert K(f)\Vert_{L^\infty_{\frac32-\gamma}} \leq C\Vert f\Vert_{L^2}, \\ &\Vert K(f)\Vert_{L^\infty_{2+a-\gamma}} \leq C\Vert f\Vert_{L_a^\infty}.\label{Linfty}\end{align}

  We consider the stationary equation:

    \begin{equation}
 \zeta_1\partial_xf(x,\zeta)=L(f)(\zeta),  \mbox{for}  \  0<x\leq l. \label{localf}
  \end{equation}
  The functional space we are considering is as follows:

  \begin{definition}\begin{align} L^*_\zeta(\mathbb R^3) &=\{ f,  \Vert f(\zeta)\Vert_*<\infty\},
  \end{align}where \begin{equation}  \Vert f(\zeta)\Vert_*=\Big (\int f^2(\zeta)\nu(\zeta)d\zeta\Big)^{\frac12}.\end{equation}
  Also, \begin{equation}|||f|||:=\sup_{0\leq x\leq l}||f||_*.\end{equation}
  \end{definition}
 We say $f\in L^\infty_x\big([0,l],L_\zeta^*(\mathbb{R}^3)\big)$ is a solution to \eqref{localf} if it satisfies the following integral equation:

  \begin{equation}\label{Integraleq1}
  f(x,\zeta)=\left\{\begin{array}{ll}e^{-\frac{\nu(\zeta)}{|\zeta_1|}x}f(0,\zeta)+\int_0^x\frac1{|\zeta_1|}e^{-\frac{\nu(\zeta)}{|\zeta_1|}(x-s)}K(f)(s,\zeta)ds, & \mbox{for}\  \zeta_1>0,\\ e^{-\frac{\nu(\zeta)}{|\zeta_1|}(l-x)}f(l,\zeta)+\int_x^l\frac1{|\zeta_1|}e^{-\frac{\nu(\zeta)}{|\zeta_1|}(s-x)}K(f)(s,\zeta)ds, & \mbox{for} \ \zeta_1<0.\end{array}\right.\end{equation}
 \begin{remark}
 The solution spaces for both Milne and Kramar's problems given in \cite{GP} are in $ L^\infty_x\big([0,l],L_\zeta^*(\mathbb{R}^3)\big)$ if  $x$ is restricted  to $[0,l].$
 \end{remark}

     The moments are defined as follows:
  \begin{definition}\label{moments}The $\alpha$ moment is defined as
  \begin{equation}
  \sigma_\alpha(x)=\int f(x,\zeta)\phi_\alpha (\zeta)d\zeta,
  \end{equation}
  where \begin{equation}\notag
\alpha=(\alpha_1,\alpha_2,\alpha_3),    \mbox{ $\alpha_i'$s are nonnegative  integers,  }\end{equation}and\begin{equation}
\phi_\alpha(\zeta)=\zeta^\alpha E^{\frac12}=\pi^{-\frac34}\zeta_1^{\alpha_1}\zeta_2^{\alpha_2}\zeta_3^{\alpha_3}e^{-\frac{|\zeta|^2}2}.\end{equation}
  \end{definition}
  We introduce a constant depending on $\alpha$:
  \begin{definition} Set
  \begin{align}A_\alpha&=(2\alpha_1)^{\frac{\alpha_1}2}(2\alpha_2)^{\frac{\alpha_2}2}(2\alpha_3)^{\frac{\alpha_3}2}e^{-\frac{\alpha_1+\alpha_2+\alpha_3}2},
  \end{align}where we follow the convention $0^0=1$.
  \label{defaalpha}
  \end{definition}
   The macroscopic variables are obtained through the moments. For example, $\sigma_{(0,0,0)}$ is the density, $\sigma_{(1,0,0)}$ is the  flow velocity in $x_1$ direction, and $\frac23(\sigma_{(2,0,0)}+\sigma_{(0,2,0)}+\sigma_{(0,0,2)})-\sigma_{(0,0,0)}$ is the temperature.
   The following inequality will be frequently used later :\begin{equation}|\phi_\alpha|\leq C A_\alpha e^{-\frac{|\zeta|^2}{4}}.\end{equation}

The Main Theorem is as follows \begin{theorem}\label{mainthm}Suppose $f\in L^\infty_x\big([0,l],L_\zeta^*(\mathbb{R}^3)\big)$ is a solution to \eqref{localf}  with power law potential with angular cut-off with $0<\gamma\leq1$ and
  $  \nabla f(0,\zeta)\in L^p_\zeta(\mathbb{R}^{3+})$ for $p>1$, $f(0,\zeta)\in  L^ \infty_\zeta(\mathbb{R}^{3+})$ , and $f(l,\zeta)\in  L^ \infty_\zeta(\mathbb{R}^{3-}) $.  Then,  for $x$ small enough,
 \begin{equation}\begin{split}|\partial_x\sigma_\alpha(x)|= &-\ln x\int\int \phi_\alpha(0,\zeta_2,\zeta_3)L(f)(0,0^+,\zeta_2,\zeta_3)d\zeta_2d\zeta_3\\&+O(A_\alpha \langle f \rangle'),\end{split}
\end{equation}  where
\begin{equation}
L(f)(0,0^+,\zeta_2,\zeta_3):=\lim_{\zeta_1\to 0^+}L(f)(0,\zeta_1,\zeta_2,\zeta_3)\end{equation}  and
\begin{equation}\langle f\rangle':=1+|||f|||+\Vert f(0,\zeta)\Vert_{L^ \infty_\zeta(\mathbb{R}^{3+})}+\Vert f(l,\zeta)\Vert_{L^ \infty_\zeta(\mathbb{R}^{3-})}+\Vert \nabla f(0,\zeta)\Vert_{L^ p_\zeta(\mathbb{R}^{3+})}.\end{equation}\end{theorem}

We first investigate the problem for Grad's angular cut-off potential.  We consider the the distribution function for $\zeta_1>0$. The case for $\zeta_1<0$ can be treated  similarly.  Differentiating \eqref{Integraleq1} for $\zeta_1>0$, we have

\begin{equation}\begin{split}\label{Dsigmap}
\frac{\partial}{\partial x}f(x,\zeta)=&-\frac{\nu(\zeta)}{|\zeta_1|}e^{-\frac{\nu(\zeta)}{|\zeta_1|}x}f(0,\zeta)+\frac{1}{|\zeta_1|}K(f)(x,\zeta)\\&-\int_0^x\frac{\nu(\zeta)}{|\zeta_1|^2}e^{-\frac{\nu(\zeta)}{|\zeta_1|}(x-s)}K(f)(s,\zeta)ds.\end{split} \end{equation}
We  observe that the first  term is nice in $\zeta$ when $x$ is away from zero and has a singularity at  $x=0$. On the other hand, the second and third terms have  factors $|\zeta_1|^{-1}$ and $|\zeta_1|^{-2}$ in $\zeta$, which cause a  difficulty in our analysis.  In order to overcome this difficulty, we reorganize  the equation \eqref{Dsigmap}:
\begin{equation}\label{dfdxf}\begin{split}
\frac{\partial}{\partial x}f(x,\zeta)=&-\frac{\nu(\zeta)}{|\zeta_1|}e^{-\frac{\nu(\zeta)}{|\zeta_1|}x}f(0,\zeta)+\frac{1}{|\zeta_1|}e^{-\frac{\nu(\zeta)}{|\zeta_1|}x}K(f)(x,\zeta)\\&+\int_0^x\frac{\nu(\zeta)}{|\zeta_1|^2}e^{-\frac{\nu(\zeta)}{|\zeta_1|}(x-s)}\big(K(f)(x,\zeta)-K(f)(s,\zeta)\big)ds\\&{=\frac{1}{|\zeta_1|}e^{-\frac{\nu(\zeta)}{|\zeta_1|}x}L(f)(0,\zeta)+\frac{1}{|\zeta_1|}e^{-\frac{\nu(\zeta)}{|\zeta_1|}x}(K(f)(x,\zeta)-K(f)(0,\zeta))}\\&{+\int_0^x\frac{\nu(\zeta)}{|\zeta_1|^2}e^{-\frac{\nu(\zeta)}{|\zeta_1|}(x-s)}\big(K(f)(x,\zeta)-K(f)(s,\zeta)\big)ds.}\end{split}
\end{equation}
The contribution from the second and third terms is uniformly bounded. Roughly speaking,  we organize the equation in such a way in order to use the "H{\"o}lder continuity" of $K(f)$ in $x$ to obtain  "differentiability" of $f$  in a very  week sense, which will be explained in detail in Section \ref{section2}. Also in Section \ref{section2}, we will deal with the contribution from $\zeta_1<0$.  In section \ref{section3}, with further assumption of invers-power potential with angular cut-off and regularity on boundary data, we can extract the
singularity from the contribution of the first term on the right hand side of  \eqref{dfdxf}, which concludes the Theorem \ref{mainthm}.

\section{Upper Bound Estimates \label{section2} }
As outlined in the introduction,  the goal of this section is to prove \begin{lemma}\label{lmGrad}Suppose $f\in L^\infty_x\big([0,l],L_\zeta^*(\mathbb{R}^3)\big)$ is a solution to \eqref{localf} with Grad's angular cut-off potential with $0<\gamma\leq 1$ and $f(0,\zeta)\in  L^ \infty_\zeta(\mathbb{R}^{3+})$  and $f(l,\zeta)\in  L^ \infty_\zeta(\mathbb{R}^{3-}) $.

  Then,
\begin{align}\label{lmGrad1}
& | \int_{\zeta_1>0}\phi_\alpha\frac{1}{|\zeta_1|}e^{-\frac{\nu(\zeta)}{|\zeta_1|}x}L(f)(0,\zeta)d\zeta| \leq C(| \ln x|+1)A_\alpha\langle f \rangle,\\&\label{lmGrad2}
| \int_{\zeta_1>0}\phi_\alpha\frac{1}{|\zeta_1|}e^{-\frac{\nu(\zeta)}{|\zeta_1|}x}(K(f)(x,\zeta)-K(f)(0,\zeta))d\zeta|\leq  CA_\alpha\langle f \rangle,\\&| \int_{\zeta_1>0}\phi_\alpha\int_0^x\frac{\nu(\zeta)}{|\zeta_1|^2}e^{-\frac{\nu(\zeta)}{|\zeta_1|}(x-s)}\big(K(f)(x,\zeta)-K(f)(s,\zeta)\big)dsd\zeta|\leq CA_\alpha\langle f \rangle \label{lmGrad3}
, \end{align}

where
\begin{equation}
\langle f \rangle:=|||f|||+\Vert f(0,\zeta)\Vert_{L^ \infty_\zeta(\mathbb{R}^{3+})}+\Vert f(l,\zeta)\Vert_{L^ \infty_\zeta(\mathbb{R}^{3-})}.
\end{equation}\end{lemma}

\begin{proof}

We observe that $f$ is in fact bounded for all $x$ and $\zeta$ if $f(0,\zeta)\in  L^ \infty_\zeta(\mathbb{R}^{3+})$  and $f(l,\zeta)\in  L^ \infty_\zeta(\mathbb{R}^{3-}) $. For  $\zeta_1>0$,
\begin{equation}\begin{split} |f(x,\zeta)|&\leq |f(0,\zeta)|+C\Vert f\Vert_*\int_0^x\frac1{|\zeta_1|}e^{-\frac{\nu(\zeta)}{|\zeta_1|}(x-s)}ds\\&\leq \Vert f(0,\zeta)\Vert_{L^ \infty_\zeta(\mathbb{R}^{3+})}+C\Vert f\Vert_*\nu_0^{-1}\int_0^{\frac{x}{|\zeta_1|}}e^{-z}dz\\&\leq C( \Vert f(0,\zeta)\Vert_{L^ \infty_\zeta(\mathbb{R}^{3+})}+\Vert f\Vert_*).\end{split}\end{equation}
 A similar  inequality holds for $\zeta_1<0$. Therefore,
\begin{equation} \Vert f \Vert_{L^\infty_{x,\zeta}} \leq C\langle f \rangle.\end{equation}

 We observe
 \begin{equation}\begin{split}
 &| \int_{\zeta_1>0}\phi_\alpha\frac{1}{|\zeta_1|}e^{-\frac{\nu(\zeta)}{|\zeta_1|}x}L(f)(0,\zeta)d\zeta| \\&\leq CA_\alpha\langle f \rangle|\int_{\zeta_1>0}e^{-\frac{|\zeta|^2}4}\frac{1}{|\zeta_1|}e^{-\frac{\nu(\zeta)}{|\zeta_1|}x}d\zeta|\leq CA_\alpha\langle f \rangle(|\ln x|+1),\end{split}
 \end{equation}
 which concludes \eqref{lmGrad1}

We will present the proof for \eqref{lmGrad3}.

Replacing $0$, $l$ by $s$, $x$ in the \eqref{Integraleq1}, we can derive
\begin{equation}\begin{split}\label{Kexpres}
&K(f)(x,\zeta)-K(f)(s,\zeta)=\int_{\zeta_{*1}>0}k(\zeta,\zeta_*)(e^{-\frac{\nu(\zeta_*)}{|\zeta_{*1}|}|x-s|}-1)f(s,\zeta_*)d\zeta_*\\&+\int_{\zeta_{*1}>0}k(\zeta,\zeta_*)\int^x_{s}\frac1{|\zeta_{*1}|}e^{-\frac{\nu(\zeta_*)}{|\zeta_{*1}|}|x-t|}K(f)(t,\zeta_*)dtd\zeta_*\\&+\int_{\zeta_{*1}<0}k(\zeta,\zeta_*)(1-e^{-\frac{\nu(\zeta_*)}{|\zeta_{*1}|}|x-s|})f(x,\zeta_*)d\zeta_*\\&+\int_{\zeta_{*1}<0}k(\zeta,\zeta_*)\int^x_{s}\frac1{|\zeta_{*1}|}e^{-\frac{\nu(\zeta_*)}{|\zeta_{*1}|}|s-t|}K(f)(t,\zeta_*)dtd\zeta_*\\&=: H_1+H_2+H_3+H_4.
\end{split}\end{equation}
The term $H_2$ and $H_4$ have the property to be proved later
\begin{equation}\label{H2Holder}
|H_2|\leq C||f||_*|x-s|^{\beta}, \ |H_4|\leq C||f||_*|x-s|^{\beta},
\end{equation}
where $0<\beta<\frac{\gamma}{2+\gamma}$.
We let \begin{equation}\begin{split} &\int_{\zeta_1>0}\int_0^x\phi_\alpha(\zeta)\frac{\nu(\zeta)}{|\zeta_1|^2}e^{-\frac{\nu(\zeta)}{|\zeta_1|}(x-s)}\big(K(f)(x,\zeta)-K(f)(s,\zeta)\big)dsd\zeta\\&=\int_{\zeta_1>0}\int_0^x\phi_\alpha(\zeta)\frac{\nu(\zeta)}{|\zeta_1|^2}e^{-\frac{\nu(\zeta)}{|\zeta_1|}(x-s)}\big(H_1+H_2+H_3+H_4)dsd\zeta \\&=:B_1+B_2+B_3+B_4.\end{split}\end{equation}Therefore, for $i=2$ and $4,$

\begin{equation}\begin{split}
&|B_i|\leq CA_{\alpha}||f||_*|\int_{\zeta_1>0}e^{-\frac{|\zeta|^2}4}\int_0^x\frac{\nu(\zeta)}{|\zeta_1|^2}e^{-\frac{\nu(\zeta)}{|\zeta_1|}(x-s)}|x-s|^\beta
 ds d\zeta|\\&\leq CA_\alpha ||f||_*|\int_{\zeta_1>0}e^{-\frac{|\zeta|^2}4}{\nu(\zeta)^{-\beta}}|\zeta_1|^{\beta-1}\int_0^{\frac{\nu(\zeta)x}{|\zeta_1|}}z^{\beta}e^{-z}
 dz d\zeta|\\&\leq CA_\alpha ||f||_*,\end{split}\end{equation}
where $z=\frac{\nu(\zeta)}{|\zeta_1|}(x-s)$.
Estimates for $B_1$ and $B_3$ are not so obvious and the analysis is more demanding. We shall present the case for $B_1$ only and the case for $B_3$ can be done similarly. We claim
\begin{equation}\begin{split}
&|B_1|=\\&|\int_{\zeta_1>0}\phi_\alpha\int_0^x\frac{\nu(\zeta)}{|\zeta_1|^2}e^{-\frac{\nu(\zeta)}{|\zeta_1|}(x-s)}
 \int_{\zeta_{*1}>0}k(\zeta,\zeta_*)(e^{-\frac{\nu(\zeta_*)}{|\zeta_{*1}|}|x-s|}-1)f(s,\zeta_*)d\zeta_*ds d\zeta|\\&\leq CA_\alpha \langle f\rangle.\end{split}\end{equation}

Change the order of integration, we have

\begin{equation}\begin{split}
&|B_1|=\\&|\int_0^x\int_{\zeta_{*1}>0}\Big(\int_{\zeta_1>0}k(\zeta,\zeta_*)\phi_\alpha\frac{\nu(\zeta)}{|\zeta_1|^2}e^{-\frac{\nu(\zeta)}{|\zeta_1|}(x-s)}
 d\zeta\Big)(e^{-\frac{\nu(\zeta_*)}{|\zeta_{*1}|}|x-s|}-1)f(s,\zeta_*)d\zeta_*ds|.\end{split}\end{equation}

 We observe, for $a\geq0$
 \begin{align}
&\Vert\phi_\alpha\frac{\nu(\zeta)}{|\zeta_1|^2}e^{-\frac{\nu(\zeta)}{|\zeta_1|}(x-s)}\Vert_{L_a^1} \leq C_aA_\alpha |x-s|^{-1}(1+\big|\ln|x|\big|),\\&\Vert\phi_\alpha\frac{\nu(\zeta)}{|\zeta_1|^2}e^{-\frac{\nu(\zeta)}{|\zeta_1|}(x-s)}\Vert_{L_a^\infty} \leq C_aA_\alpha |x-s|^{-2}\label{phiainf}.\end{align}
Interpolating the inequalities above, we obtain
\begin{equation}\begin{split}\label{interf}
&\Vert\phi_\alpha\frac{\nu(\zeta)}{|\zeta_1|^2}e^{-\frac{\nu(\zeta)}{|\zeta_1|}(x-s)}\Vert_{L_a^p} \leq C_aA_\alpha |x-s|^{-2+\frac1p}(1+\big|\ln|x|\big|)^{\frac1p},\end{split}\end{equation}
where $1\leq p \leq \infty.$ In particular,
\begin{equation}\label{L2phialpha}
\Vert\phi_\alpha\frac{\nu(\zeta)}{|\zeta_1|^2}e^{-\frac{\nu(\zeta)}{|\zeta_1|}(x-s)}\Vert_{L^2} \leq C A_{\alpha} |x-s|^{-\frac32}(1+\big|\ln|x|\big|)^{\frac12}.
\end{equation}

Let $h(\theta,\gamma,a)=(\frac32-\gamma)\theta+(2+a-\gamma)(1-\theta)=2+a(1-\theta)-\gamma-\frac12\theta$, where $0\leq\theta\leq1$. We have \begin{equation} \begin{split}\Vert K(f)\Vert_{L^\infty_h}=&\sup_{\zeta\in\mathbb{R}^3}\left(|f(\zeta)|(1+|\zeta|)^{\frac32-\gamma}\right)^\theta\left(|f(\zeta)|(1+|\zeta|)^{2+a-\gamma}\right)^{1-\theta}\\& \leq \Vert K(f)\Vert_{L^\infty_{\frac32-\gamma}}^\theta\Vert K(f)\Vert_{L^\infty_{2+a-\gamma}}^{1-\theta}\leq C\Vert f\Vert_{L^2}^\theta\Vert f\Vert_{L^\infty_a}^{1-\theta}.\label{interpoK}\end{split}
\end{equation}

 Combining \eqref{L2Linfty}, \eqref{Linfty}, \eqref{phiainf} ,  \eqref{L2phialpha}, and \eqref{interpoK}, we have
 \begin{equation} \Vert\int_{\zeta_1>0}\phi_\alpha\frac{\nu(\zeta)}{|\zeta_1|^2}e^{-\frac{\nu(\zeta)}{|\zeta_1|}(x-s)}d\zeta_1\Vert_{L^\infty_h}\leq C C_a^{1-\theta}A_\alpha|x-s|^{-2+\frac12\theta}(1+\big|\ln|x-s|\big|)^{\frac12\theta},\label{KfLinfh}
\end{equation} Applying \eqref{KfLinfh} above with fixed $0<\theta<1$ and  $a$ large enough,
we have
\begin{equation}\begin{split}\label{B1est}
&{|B_1|}\leq|\int_0^x\int_{\zeta_{*1}>0}(1+|\zeta_*|)^{-h(\theta,\gamma,a)}|x-s|^{-(2-\frac12\theta)}(1+\big|\ln|x-s|\big|)^{\frac12\theta}|\\&(e^{-\frac{\nu(\zeta_*)}{|\zeta_{*1}|}|x-s|}-1)|f(s,\zeta_*)|d\zeta_*ds|\\&\leq{ C A_\alpha}
|\int_0^x|x-s|^{-(2-\frac12\theta)}(1+\big|\ln|x-s|\big|)^{\frac12\theta}\\&\ \ \ \ \int_0^{|x-s|}\int_{\zeta_{*1}>0}\frac{\nu(\zeta_*)}{|\zeta_{*1}|}e^{-\frac{\nu(\zeta_*)}{|\zeta_{*1}|}t}(1+|\zeta_*|)^{-h(\theta,\gamma,a)}|f(s,\zeta_*)|d\zeta_*dtds|\\&\leq C A_\alpha\langle f\rangle
|\int_0^x|x-s|^{-(1-\frac12\theta)}(1+\big|\ln|x-s|\big|)^{1+\frac12\theta}ds| \leq CA_\alpha\langle f\rangle.\end{split}\end{equation}

 The proof for \eqref{lmGrad2} is similar and simpler.
  Replacing $s$ in \eqref{Kexpres} by 0 and denoting these terms as $H'_i$s, we write

\begin{equation} \begin{split}&\int_{\zeta_1>0} \phi_\alpha\frac{1}{\zeta_1}e^{-\frac{\nu(\zeta)}{|\zeta_1|}x}\left(K(f)(x,\zeta)-K(f)(0,\zeta)\right)d\zeta\\&=\int_{\zeta_1>0} \phi_\alpha\frac{1}{\zeta_1}e^{-\frac{\nu(\zeta)}{|\zeta_1|}x}(H_1'+H_2'+H_3'+H_4')d\zeta\\&=:B_1'+B_2'+B_3'+B_4'.\end{split}\end{equation}

Using the fact \begin{equation} |H_2'|\leq C|x|^\beta\Vert f\Vert_* ,\ \ |H_4'|\leq C|x|^\beta\Vert f\Vert_*,\end{equation}
we have
\begin{equation}\begin{split}|B_2'+B_4'|&\leq \Vert f\Vert_*\int_{\zeta_1>0} \phi_\alpha\frac{1}{\zeta_1}e^{-\frac{\nu(\zeta)}{|\zeta_1|}x}|x|^\beta d\zeta \\&\leq C\Vert f\Vert_*\int_{\zeta_1>0} \phi_\alpha\frac{1}{\nu(\zeta)|\zeta_1|^{1-\beta}} d\zeta\leq C\Vert f\Vert_*.\end{split} \end{equation}

The treatment for $B_1'$ and $B_3'$ is similar, and therefore we present the case for $B_1'$ only. Changing the order of integration, we have

\begin{equation} \begin{split}&B_1'=\int_{\zeta_1>0} \phi_\alpha\frac{1}{\zeta_1}e^{-\frac{\nu(\zeta)}{|\zeta_1|}x}\int_{\zeta_{*1}>0}k(\zeta,\zeta_*)(e^{-\frac{\nu(\zeta_*)}{|\zeta_{*1}|}x}-1)f(0,\zeta_*)d\zeta_*d\zeta\\&=\int_{\zeta_{*1}>0} (e^{-\frac{\nu(\zeta_*)}{|\zeta_{*1}|}x}-1)f(0,\zeta_*)\left(\int_{\zeta_{1}>0}k(\zeta,\zeta_*)\phi_\alpha\frac{1}{\zeta_1}e^{-\frac{\nu(\zeta)}{|\zeta_1|}x}d\zeta_*\right)d\zeta.\end{split}\end{equation}

Note that\begin{align}
&\Vert\phi_\alpha\frac{1}{\zeta_1}e^{-\frac{\nu(\zeta)}{|\zeta_1|}x}\Vert_{L^1}\leq CA_\alpha(1+\big|\ln x\big|), \\
&\Vert\phi_\alpha\frac{1}{\zeta_1}e^{-\frac{\nu(\zeta)}{|\zeta_1|}x}\Vert_{L^\infty_a}\leq C_aA_\alpha(\frac1x). \label{LinfBprime}
\end{align}
By interpolation, we have
\begin{equation}
\Vert\phi_\alpha\frac{1}{\zeta_1}e^{-\frac{\nu(\zeta)}{|\zeta_1|}x}\Vert_{L^2} \leq CA_\alpha(\frac1x)^{\frac12}(1+\big|\ln x\big|)^{\frac12}\label{L2Bprime}.
\end{equation}
Combining \eqref{L2Linfty}, \eqref{Linfty}, \eqref{LinfBprime}, \eqref{L2Bprime}, and \eqref{interpoK}, we have
\begin{equation}\Vert\int_{\zeta_1>0} k(\zeta,\zeta_*)\phi_\alpha\frac{1}{\zeta_1}e^{-\frac{\nu(\zeta)}{|\zeta_1|}x}d\zeta\Vert_{L^\infty_h}\leq
CC_a^{1-\theta}A_\alpha (\frac1x)^{1-\frac{\theta}2}(1+\big|\ln x\big|)^{\frac12\theta}. \end{equation}
Similar to \eqref{B1est}, for fixed $0<\theta<1$ and $a$ large enough, we have

\begin{equation} \begin{split}&|B_1'|\leq CA_\alpha\int_{\zeta_{*1}>0} (e^{-\frac{\nu(\zeta_*)}{|\zeta_{*1}|}x}-1)f(0,\zeta_*)(1+|\zeta_*|)^{-h(\theta,\gamma,a)}(\frac1x)^{1-\frac{\theta}2}(1+\big|\ln x\big|)^{\frac12\theta}d\zeta_*\\&\leq CA_\alpha (\frac1x)^{1-\frac{\theta}2}(1+\big|\ln x\big|)^{\frac12\theta}\int_0^x\int_{\zeta_{*1}>0}\frac{\nu(\zeta_*)}{|\zeta_{*1}|}e^{-\frac{\nu(\zeta_*)}{|\zeta_{*1}|}t}(1+|\zeta_*|)^{-h(\theta,\gamma,a)}|f(0,\zeta_*)|d\zeta_*dt\\&\leq CA_\alpha \langle f\rangle x^{\frac{\theta}2}(1+\big|\ln x\big|)^{1+\frac12\theta}\leq CA_\alpha \langle f\rangle .\end{split}\end{equation}

We still have to prove \eqref{H2Holder}.  We will present the proof for $H_2$ only and the one for $H_4$ is similar. We will first present the following lemma
\begin{lemma}If  $f\in L^2_*$ and $\theta\in(\frac{2}{2+\gamma},1)$, then \label{lemmatheta}
\begin{equation}
\int\frac1{|\zeta_{*1}|^{2-2\theta}\nu(\zeta_*)^{2\theta}}|K(f)|^2d\zeta_*\leq C\Vert f\Vert_*^2.
\end{equation}
\end{lemma}

The proof Lemma \ref{lemmatheta} follows the idea of the one of  Lemma  4.2  in \cite{GP}. We  will present the proof at the end of this section to make this paper self-contained.

With the Lemma \ref{lemmatheta} above, we have
 \begin{equation}\begin{split}|H_2|&=|\int^x_{s}\int_{\zeta_{*1}>0}k(\zeta,\zeta_*)\frac1{|\zeta_{*1}|}e^{-\frac{\nu(\zeta_*)}{|\zeta_{*1}|}|x-t|}K(f)(t,\zeta_*)d\zeta_*dt|\\&\leq  C(\int_s^x(\int_{\zeta_{*1}>0}|\frac1{|\zeta_{*1}|}e^{-\frac{\nu(\zeta_*)}{|\zeta_{*1}|}|x-t|}K(f)(t,\zeta_*)|^2d\zeta_*)^{\frac12}dt\\&
 \leq    C(\int_s^x\frac1{|x-t|^\theta}(\int_{\zeta_{*1}>0}|\frac1{|\zeta_{*1}|^{2-2\theta}\nu(\zeta_*)^{2\theta}}K(f)(t,\zeta_*)|^2d\zeta_*)^{\frac12}dt\\&\leq C|x-s|^{1-\theta}=C|x-s|^{\beta}. \end{split}\end{equation} \end{proof}

For $\zeta_1<0$, we can yield a similar lemma.  Together, we have

  \begin{lemma}\label{sigmapm}Suppose $f\in L^\infty_x\big([0,l],L_\zeta^*(\mathbb{R}^3)\big)$ is a solution to \eqref{localf} with Grad's angular cut-off potential with $0<\gamma\leq 1$ and $f(0,\zeta)\in  L^ \infty_\zeta(\mathbb{R}^{3+})$  and $f(l,\zeta)\in  L^ \infty_\zeta(\mathbb{R}^{3-}) $.
 Then,
  \begin{align}|\partial_x\sigma^+_\alpha(x)|&\leq C(\big|\ln|x|\big|+1),\label{sigmap}\\|\partial_x\sigma^-_\alpha(x)|&\leq C(\big|\ln|l-x|\big|+1),\label{sigmam}
\end{align}  where\begin{equation}\sigma^+_\alpha(x)=\int_{\zeta_1>0}\phi_\alpha f d\zeta,\ \ \sigma^-_\alpha(x)=\int_{\zeta_1<0}\phi_\alpha f d\zeta.\end{equation}\end{lemma}

 \begin{proof}[Proof for Lemma \ref{lemmatheta}]
 We observe
 \begin{equation}\begin{split}&\int|Kf|^2d\zeta=\int\left(\int k(\zeta,\zeta_*)f(\zeta*)d\zeta*\int k(\zeta,\zeta')f(\zeta')d\zeta'\right)d\zeta\\&\leq C\Vert f\Vert_{L^2}\int | f(\zeta_*)|\int k(\zeta,\zeta_*)(1+|\zeta|)^{-(\frac32-\gamma)} d\zeta d\zeta_*\\ &\leq C\Vert f\Vert_{L^2}\int| f(\zeta_*)|(1+|\zeta*|)^{-(\frac72-2\gamma)}d\zeta_*\leq C\Vert f\Vert_{L^2}\Vert f\Vert_{*}\leq C\Vert f\Vert_{*}^2.\end{split}\end{equation}
Together with \eqref{L2Linfty},  we know $|Kf|^2\in L^\infty\cap L^1$. Interpolating between these two inequalities, we have
\begin{equation}\Vert |Kf|^2 \Vert_{L^p}\leq C \Vert f\Vert_*^2 \mbox{ for }  1 \leq p\leq \infty .\end{equation}
Therefore, we now only need to prove for  some proper $0<\theta<1$ and  H{\"o}lder conjugate of $p$, $p'\in [1,\infty]$,
\begin{equation}
\int\frac1{|\zeta_{*1}|^{(2-2\theta)p'}|1+|\zeta_*|)|^{2\gamma\theta p'}}d\zeta_*<\infty,
\end{equation}
which yields the following condition:

\begin{equation}
(2-2\theta)p'<1;\ (2-2\theta+2\gamma\theta)p'>3.
\end{equation}
Such $p'$ exists if and only if
\begin{equation}
\theta<1,\ 3(2-2\theta)<2-2\theta+2\gamma\theta,
\end{equation}
which concludes Lemma \ref{lemmatheta}.
\end{proof}

\section{Asymptotic formula\label{section3}}In the precious section, we obtain an upper bound for $|\partial_x\sigma_\alpha|$, which diverges to infinity at boundary like a logarithmic function. Through the analysis, we also localize  the source of singularity,
which is the contribution from the first term on the right hand side in \eqref{dfdxf}. In this section, restricted   to the inverse-power potential with angular cut-off,  the goal is to further single out and factorize the singularity and form an asymptotic formula, i.e.,
\begin{lemma} \label{DsigmaS}Suppose $f(0,\zeta)\in L_*$ and $\nabla f(0,\zeta)\in L^p_\zeta(\mathbb{R}^{3+})$. Then, \begin{equation} \begin{split}\frac{\partial}{\partial x}\sigma_{\alpha1}^+:=&\int_{\zeta_1>0}\phi_{\alpha}
\frac{1}{|\zeta_1|}e^{-\frac{\nu(|\zeta|)}{|\zeta_1|}x}L(f)(0,\zeta)d\zeta\\&=-\ln x\int\int \phi_\alpha(0,\zeta_2,\zeta_3)L(f)(0,0^+,\zeta_2,\zeta_3)d\zeta_2d\zeta_3+O(A_\alpha \langle g \rangle'),\end{split}
\end{equation}  where
\begin{equation}
L(f)(0,0^+,\zeta_2,\zeta_3):=\lim_{\zeta_1\to 0^+}L(f)(0,\zeta_1,\zeta_2,\zeta_3).\end{equation}

\end{lemma}

   \begin{proof}

  If we change to spherical coordinates so that \[\zeta=(\rho\cos\theta, \rho\sin\theta\cos\phi,\rho\sin\theta\sin\phi),\] we have

  \begin{equation}\frac{\partial}{\partial x}\sigma_{\alpha1}^+=\int_0^\infty\int_0^{2\pi}\int_0^{\frac{\pi}{2}}\label{sigmapsh}
\frac{1}{\rho\cos\theta}e^{-\frac{\rho^2}2}e^{-\frac{\nu(\rho)}{\rho\cos\theta}x}\bar{F}(\rho,\theta,\phi)\rho^2\sin\theta d\theta d\phi d\rho,\end{equation}
  where
  \begin{equation}\bar{F}(\rho,\theta,\phi)=\pi^{-\frac34}\zeta_1^{\alpha_1}\zeta_2^{\alpha_2}\zeta_3^{\alpha_3}L(f)(0,\zeta)=:g(\zeta).\end{equation}
  Further letting  $z=\cos\theta$, we obtain
   \begin{equation}\frac{\partial}{\partial x}\sigma_{\alpha1}^+=\int_0^\infty\int_0^{2\pi}\left[\int_0^{1}\label{sigmapshz}
\frac{1}{z}e^{-\frac{\nu(\rho)}{\rho z}x}{F}(\rho, z,\phi) d z \right]e^{-\frac{\rho^2}2}\rho d\phi d\rho,\end{equation}
  where
  \begin{equation}F(\rho,z ,\phi)=\bar{F}(\rho,\cos^{-1}z,\phi).\end{equation}

  Here, we introduce a well-know special function, exponential integral,
  \begin{equation} E_1(x)=\int_0^1\frac1{z}e^{-\frac{x}{z}}dz.\end{equation}
  The $E_1(x)$ has the following  properties, \cite{handbook}:

\begin{equation}
E_{1}(x)=-\gamma-\ln x+\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^{k}}{k\cdot k!},\label{expandion}
\end{equation}

\begin{equation}
\frac{1}{2}e^{-x}\ln(1+\frac{2}{x})\leq E_{1}(x)\leq e^{-x}\ln(1+\frac{1}{x})\ \ \mbox{for}\ x>0.\label{e1far}
\end{equation}

From the properties above, we have
\begin{equation}
E_{1}(x)=-\ln x+O(1)\label{E1o1}\  \mbox{, for } 0<x\leq1,
\end{equation}

  Let \begin{equation} H(z,x)=-\int_z^1\frac{1}{u}e^{-\frac{\nu(\rho)}{\rho u}x}du.\end{equation}
  Notice that  $\frac{\partial}{\partial z} H(z,x)=\frac{1}{z}e^{-\frac{\nu(\rho)}{\rho z}x}$and $H(0,x)=E_1(\frac{\nu(\rho)}{\rho }x)$.
  Performing integration by parts for the inner most integral in \eqref{sigmapshz}, we obtain
  \begin{equation}\begin{split}&\int_0^1\left(\frac{\partial}{\partial z}H(z)\right)F(\rho,z,\phi)dz\\&=E_1(\frac{\nu(\rho)}{\rho }x)F(\rho,0,\phi)-\int_0^1H(z)\left(\frac{\partial}{\partial z}F(\rho,z,\phi)\right)dz\\.\end{split}\end{equation}
  The first  term on the right hand side above is the source of singularity and will be explained in detail later. We will prove the contribution from the second term above is finite.
  Let \begin{align}I:&=\int_0^\infty\int_0^{2\pi}E_1(\frac{\nu(\rho)}{\rho}x)F(\rho,0,\phi)e^{-\frac{\rho^2}2}\rho d\phi d\rho.\label{Isigular},\\II:&=\int_0^\infty\int_0^{2\pi}\int_0^1H(z)\left(\frac{\partial}{\partial z}F(\rho,z,\phi)\right)dz e^{-\frac{\rho^2}2}\rho d\phi d \rho.&\end{align}

  Notice that
  \begin{equation} |H(z,x)|\leq |\ln z|,\end{equation}

  \begin{equation}\frac{\partial}{\partial z} F(\rho, z,\phi)=\rho\frac{\partial}{\partial \zeta_1}g+\frac{\rho z}{\sqrt{1-z^2}}\cos\phi \frac{\partial}{\partial \zeta_2}g+\frac{\rho z}{\sqrt{1-z^2}}\sin\phi\frac{\partial}{\partial \zeta_3}g.\end{equation}

  We have \begin{equation} \begin{split}\left|II\right| &\leq \left|\int_0^\infty\int_0^{2\pi}\int_0^1H(z)\left(\frac{\partial}{\partial z}F(\rho,z,\phi)\right)dz e^{-\frac{\rho^2}2}\rho d\phi d \rho\right| \\&\leq \left|\int_0^\infty\int_0^{2\pi}\left(\int_0^1C(|\frac{\partial}{\partial \zeta_3}g|+|\frac{\partial}{\partial \zeta_2}g|)dz +\int_0^1|\ln z||\frac{\partial}{\partial \zeta_1}g| dz \right)e^{-\frac{\rho^2}2}\rho^2 d\phi d \rho\right|\\ &\leq CA_\alpha \int_{\zeta_1>0}e^{-\frac{|\zeta|^2}8}\left(|L(0,\zeta)|+|\frac{\partial}{\partial \zeta_2}L(0,\zeta)|+|\frac{\partial}{\partial \zeta_3}L(0,\zeta)|\right) d\zeta\\&+A_\alpha\left|\int_0^\infty\int_0^{2\pi}\left(\int_0^1|\ln z|^q dz \right)^{\frac1q}\left(\int_0^1|\frac{\partial}{\partial \zeta_1}L(0,\zeta)|^p dz \right)^{\frac1p}e^{-\frac{\rho^2}4}\rho^2 d\phi d \rho\right|,\end{split}\end{equation}

  where $p>1$ and $\frac1p+\frac1q=1$.

  Let the second term on the right hand side above be $II'$.
  \begin{equation}\begin{split} &|II'| \leq \left(\int_0^\infty\int_0^{2\pi}\left(\int_0^1|\ln z|^q dz \right)e^{\frac{q\rho^2}8}\rho^2 d\phi d \rho\right)^{\frac1q}\left(\int_0^\infty\int_0^{2\pi}\left(\int_0^1|\frac{\partial}{\partial \zeta_1}L(0,\zeta)|^p dz \right)e^{\frac{p\rho^2}8}\rho^2 d\phi d \rho\right)^{\frac1p}\\&\leq C_p A_\alpha\left( \int_{\zeta_1>0}|\frac{\partial}{\partial \zeta_1}L(0,\zeta)|^p e^{-\frac{p|\zeta|^2}8} d\zeta\right)^{\frac1p}\end{split} .\end{equation}
  Therefore,\begin{equation}\begin{split}|II|&\leq CA_\alpha \int_{\zeta_1>0}e^{-\frac{|\zeta|^2}8}\left(|L(0,\zeta)|+|\frac{\partial}{\partial \zeta_2}L(0,\zeta)|+|\frac{\partial}{\partial \zeta_3}L(0,\zeta)|\right) d\zeta\\&+ C_p A_\alpha\left( \int_{\zeta_1>0}|\frac{\partial}{\partial \zeta_1}L(0,\zeta)|^p e^{-\frac{p|\zeta|^2}8} d\zeta\right)^{\frac1p},\end{split}\end{equation}
  where $p>1$.

  Using the assumption and \eqref{dKLp}, we have
  \begin{equation}|II|\leq C A_\alpha(\Vert\nabla f(0,\zeta)\Vert_{L^p_\zeta(\mathbb{R}^{3+})}+\Vert  f(0,\zeta)\Vert_*)\end{equation}

  Finally, we are going to extract the singularity from  $I$.
   We let \begin{equation}\rho_0=\sup\{\rho|\frac{\nu(\rho)}\rho x>1\}.\end{equation}
 We devided the domain of integration of $I$ in \eqref{Isigular} into two, $0\leq\rho\leq \rho_0$ and $\rho_0\leq \rho$, and denoted the integral as $I_s$ and $I_l$ correspondently.
 Note that if $0\leq \rho<\rho_0$, then  \begin{equation}\frac{\nu(\rho)}{\rho}x> \frac{c_0}{c_1}.\end{equation} Applying \eqref{e1far}
, we have  \begin{equation}
 |I_s|\leq C A_\alpha \langle f \rangle.\end{equation}   Using the asymptotic formula \eqref{E1o1}, We obtain
   \begin{equation}\begin{split}I_l&=-\ln(x) \int_{\rho_0}^\infty\int_0^{2\pi}F(\rho,0,\phi)e^{-\frac{\rho^2}2}\rho d\phi d\rho+O(A_\alpha\langle f\rangle)\\&= -\ln x\int\int \phi_\alpha(0,\zeta_2,\zeta_3)L(f)(0,0^+,\zeta_2,\zeta_3)d\zeta_2d\zeta_3+O\left(A_\alpha \langle f \rangle(1+\rho_0^2|\ln x| )\right).\end{split}
   \end{equation}
   The remaining task is to estimate $\rho_0$. If we assume $x\leq\frac1{2c_1}$,  then\begin{equation}\label{nurho}
   1=\frac{\nu(\rho_0)}{\rho_0}x\leq c_1\frac{(1+\rho_0)^\gamma}{\rho_0}x\leq \frac{(1+\rho_0)^\gamma}{2\rho_0}.\end{equation}
   If $\gamma=1$, we see $\rho_0\leq 1.$
   Observe that, for $0\leq\gamma<0$,
   $ \rho$  grows faster then the
  $\frac12(1+\rho)^\gamma$ as $\rho\to \infty$. Therefore, \eqref{nurho} implies $\rho_0\leq m$ for some $m<\infty$.
  Therefore, we have
 \begin{equation}\label{nurho}
   1=\frac{\nu(\rho_0)}{\rho_0}x\leq c_1\frac{(1+m)^\gamma}{\rho_0}x.\end{equation}
   We have
   \begin{equation}
   |\rho_0^2\ln x|\leq Cx^2|\ln x| \leq C
      \end{equation}
      and conclude the lemma.\end{proof}

      Finally, combining \eqref{lmGrad2} and \eqref{lmGrad3} in Lemma \ref{lmGrad}, \eqref{sigmam} in Lemma \ref{sigmapm}, and Lemma \ref{DsigmaS}, we concludes Theorem \ref{mainthm}.


TITLE: Finding the value of function defined as the limit of multivariable function
QUESTION [2 upvotes]: Suppose we have a function
$f(x,y) = \frac{\cos{x}-\cos{y}}{2(x^2+y^2)}$
and we define
$F(x) := \lim_{y \to 0} f(x,y)$
How would I go about finding $F(0)$?
I am unclear whether to plug the value in first, and then take the limit, eg.
$F(0) = \lim_{y \to 0} f(0,y) = \frac{1}{4}$
or take the limit first, and then plug in the value.
$F(x) = \lim_{y \to 0} f(x,y) = \frac{\cos{x}-1}{2(x)^2} $
$F(0) = \frac{\cos{0}-1}{2(0)^2} $
In this case, the function is undefined.
I have attempted to compute the value in Maple, and I've gotten the following
f := (x, y) -> (cos(x) - cos(y))/(2*y^2 + 2*x^2)
F := x -> limit(f(x, y), y = 0)
F(0) = 1/4

This would suggest the first approach being correct, but I am still unsure whether this is actually true and why it is correct.

REPLY [1 votes]: Well, the first solution is definitely the most intuitive. It follows the definition
$$F(x)=\lim_{y\to0}f(x,y)\implies F(0)=\lim_{y\to0}f(0,y)$$
Taking the limit of $\frac{\cos{x}-1}{2x^2}$ implies nothing else than the fact that the function is discontinuous, as $F(0)=\frac{1}{4}$ and the limit $\lim_{x\to0}\frac{\cos{x}-1}{2x^2}=-\frac{1}{4}$.
Thus, $F$ is defined at $x=0$, although $\frac{\cos{x}-1}{2x^2}$ is not. This is due to the function being defined as a limit, not the result of the limit.


TITLE: Find the radius of convergence for $\sum_{n=1}^{\infty}\frac{2^n}{3^n+4^n}z^n$
QUESTION [0 upvotes]: $$\sum_{n=1}^{\infty}\frac{2^n}{3^n+4^n}z^n$$
What I've done is try to evaluate the expression sans $z^n$ with the root test.
$$\sqrt[n]{\frac{2^n}{3^n+4^n}}=\frac{2}{\sqrt[n]{3^n+4^n}}$$
But I'm stuck with the denominator, how does one evaluate such expressions?

REPLY [3 votes]: Hint: Note that $4^n+3^n=4^n\cdot(1+(3/4)^n)$.


TITLE: Boolean Logic using proofs
QUESTION [0 upvotes]: ABC' + C = AB + C
I understand this using venn diagrams and intuition. However, I am not able to derive the proof for getting from one side to the other. It's probably very simple step that I keep missing. Please enlighten me.

REPLY [1 votes]: $$AB \overline C  + C$$
Identity Law: $X • 1 = X$
$$AB \overline C  + 1 • C$$
Annulment Law: $X + 1 = 1$
$$AB \overline C  + (AB + 1) C$$
Distributive Law: $X • (Y + Z) = X Y + X Z$
$$AB \overline C  + ABC + C$$
Distributive Law: X Y + X Z = $X • (Y + Z)$
$$AB (\overline C  + C) + C$$
Complement Law: $X + \overline X = 1$
$$AB + C$$
$$AB \overline C  + C = AB + C$$


TITLE: If $(x_n)$ is a sequence such that $\sum_{k=1}^n\frac{{x^4_k}}n\to 0$, then $\sum_{k=1}^n\frac{{x_k}}n\to 0$
QUESTION [2 upvotes]: Suppose that $(x_n)$ is a sequence such that
$\lim_{n\to\infty}\sum_{k=1}^n\frac{{x^4_k}}n=0.$
How do I show that $\lim_{n\to\infty}\sum_{k=1}^n\frac{{x_k}}n=0$?

REPLY [5 votes]: By using the Cauchy-Schwarz inequality, $(u\cdot v)^2 \leq ||u||^2 ||v||^2$, on the vectors $u=(1,1,\ldots,1)$ and $v=(x_1,x_2,\ldots,x_n)$ we get
$$\left(\sum_{k=1}^n x_k\right)^2 \leq \sum_{k=1}^n x_k^2\sum_{k=1}^n 1 = n \sum_{k=1}^n x_k^2$$
Now applying the same inequality, but on the vectors $u=(1,1,\ldots,1)$ and $v=(x_1^2,x_2^2,\ldots,x_n^2)$ we get
$$\left(\sum_{k=1}^n x_k^2\right)^2 \leq \sum_{k=1}^n x_k^4\sum_{k=1}^n 1 = n\sum_{k=1}^n x_k^4$$
Putting the two inequalities above togeather we find
$$\left(\sum_{k=1}^n x_k\right)^4 \leq n^2 \left(\sum_{k=1}^n x_k^2\right)^2 \leq n^3\sum_{k=1}^nx_k^4$$
or
$$\left|\sum_{k=1}^n \frac{x_k}{n}\right| \leq \left(\sum_{k=1}^n\frac{x_k^4}{n}\right)^{1/4}$$
Since the right hand side vanishes in the limit $n\to \infty$ so does the left hand side.


TITLE: Show that if $(x_n)\rightarrow 2$ then $(1/x_n)\rightarrow 1/2$
QUESTION [2 upvotes]: Given a convergent sequence $(x_n)\rightarrow 2$, I am asked to prove $(1/x_n)\rightarrow 1/2$ without the use of the algebraic limit theorem.
Here is what I have tried: Let $\epsilon>0$ be arbitrary. We must prove that $|1/x_n-1/2|<\epsilon$. Observe that $$|1/x_n-1/2|=\frac{|x_n-2|}{2|x_n|}$$
and we can make $|x_n-2|$ as small as we like. However, I'm not sure how to choose $N$, and what to do with the $|x_n|$ in the denominator.

REPLY [1 votes]: Let $\varepsilon > 0$ be given.
As $\left( x_n \right)_{n \in \mathbb{N} }$ converges to $2$, so there exists a natural number $N_1$ such that $$ \left\lvert x_n - 2 \right\rvert < 1 \tag{1} $$
for all $n \in \mathbb{N}$ such that $n > N_1$.
Thus, for all $n \in \mathbb{N}$ such that $n > N_1$, we have
$$
\begin{align}
\left\lvert x_n \right\rvert &= \left\lvert x_n - 2  + 2  \right\rvert \\
&\geq \left\lvert \ \left\lvert x_n - 2 \right\rvert - \lvert 2 \rvert \ \right\rvert  \qquad \mbox{ [ using a form of the triangle inequality for real numbers ] } \\
&= \left\lvert \ \lvert 2 \rvert  - \left\lvert  x_n - 2  \right\rvert \ \right\rvert  \\
&\geq \lvert 2 \rvert - \left\lvert x_n - 2 \right\rvert \qquad \mbox{ [ note this step ] } \\
&= 2 - \left\lvert x_n - 2 \right\rvert  \\
&> 2 - 1 \qquad \mbox{ [ using (1) ] } \\
&= 1,
\end{align}
$$
and hence upon taking the reciprocals of the ends of the last chain of inequalities, we find that, for all $n \in \mathbb{N}$ such that $n > N_1$, we have
$$ 0 <  \frac{ 1 }{ \left\lvert x_n \right\rvert } < 1. \tag{2}  $$
$$
\begin{align}
 \left\lvert \frac{1}{x_n} - \frac{1}{2} \right\rvert &= \left\lvert \frac{ 2 - x_n  }{ 2 x_n  } \right\rvert \\
&= \frac{ \left\lvert  2 - x_n  \right\rvert  }{ \left\lvert  2 x_n \right\rvert   } \\
&= \frac{ \left\lvert  x_n - 2   \right\rvert  }{ 2 \left\lvert   x_n \right\rvert   } \\
&< \frac{ \left\lvert  x_n - 2   \right\rvert  }{ 2 }. \tag{3} \qquad \mbox{ [ using (2) above ] }
\end{align}
$$
Now as the sequence $\left( x_n \right)_{n \in \mathbb{N} }$ converges to $2$, so there exists a natural number $N_2$ such that
$$ \left\lvert x_n - 2 \right\rvert < 2\varepsilon \tag{4} $$
for all $n \in \mathbb{N}$ such that $n > N_2$.
Finally, let $N$ be any natural number such that $N \geq \max \left\{ \ N_1, N_2 \ \right\}$.
Then for all $n \in \mathbb{N}$ such that $n > N$, we have
$$
\begin{align}
 \left\lvert \frac{1}{x_n} - \frac{1}{2} \right\rvert &< \frac{ \left\lvert  x_n - 2   \right\rvert  }{ 2 } \qquad \mbox{ [ using (3) above; note that $N \geq N_1$ and so $n > N_1$ also ] } \\
&< \frac{ 2 \varepsilon }{ 2 } \qquad \mbox{ [ using (4) above; note that $N \geq N_2$ and so $n > N_2$ also ] } \\
&= \varepsilon.
\end{align}
$$
Thus we have shown that, corresponding to any real number $\varepsilon > 0$, there exists a natural number $N$ such that
$$  \left\lvert \frac{1}{x_n} - \frac{1}{2} \right\rvert < \varepsilon $$
for all $n \in \mathbb{N}$ such that $n > N$. Hence the sequence $\left( \frac{1}{x_n} \right)_{n \in \mathbb{N} }$ converges to $\frac{1}{2}$, whenever $\left( x_n \right)_{n \in \mathbb{N} }$ is any sequence of real numbers converging to $2$.
Hope this helps.


TITLE: Restrictions on Bell-type inequalities
QUESTION [1 upvotes]: While deriving and proving Bell-type inequalities of the form
$|E(a,b)-E(a,b')|+|E(a',b)+E(a',b')|\leq 2$
I know that the conditions on the operators $O_a$ and $O_b$ are that they must be bounded by $\pm 1$.
Joint operator $O_{ab}\equiv O_a O_b$ is consequently bounded by $\pm 1$.
However, is there any such bound on the correlation $E(a,b)$ given by operating by $O_{ab}$ on whatever state you're studying?  Does $E(a,b)$ necessarily NEED to be bounded by $\pm 1$ as per the definition?
(I know that it sometimes is as a result of, say, operating on the singlet state, but is this a consequence or a condition?)
Thanks in advance!

REPLY [0 votes]: The CHSH inequality was not proved for any state, but for the spin singlet or for the photon singlet, s.t. to your question
"is there any such bound on the correlation E(a,b) given by operating by O_{ab} on whatever state you're studying? Does E(a,b)  necessarily NEED to be bounded by ±1  as per the definition?
As I say, you cannot work with whatever state.
On other states there are other inequalities, you can follow in the  arXiv quant-ph  the works of Adan Cabello.
But WHY are you interested to break the limit +- 1 on E(a,b)? What is exactly the problem?
Good luck,
Sofia


TITLE: Find an equation for the ellipse with foci $(\pm 4,0)$ passing through $(-4,1.8)$
QUESTION [0 upvotes]: Find an equation for the conic that satisfies the given conditions:

Ellipse, Foci $(-4,0)$ and $(4,0)$, passes through $(-4,1.8)$.

I know how to do these questions with the vertices, but I'm kinda lost figuring this one out.

REPLY [0 votes]: Matteo (https://math.stackexchange.com/users/686644/matteo), Find an equation for the ellipse with foci $(\pm 4,0)$ passing through $(-4,1.8)$, URL (version: 2019-07-23): https://math.stackexchange.com/q/3301684


TITLE: The Origin of the Musical Isomorphisms
QUESTION [26 upvotes]: In Riemannian geometry, the "lowering indices" operator is denoted by $\flat:TM \to T^*M$ and the "raising indices" operator by $\sharp:T^*M \to TM$. These isomorphisms are sometimes referred to as musical isomorphisms, as stated on Wikipedia and in several other sources. Surely, the motivation for such terminology is clear. I would nevertheless like to know who decided to adpot these (rather amusing) notations, so here is a question:

What was the first paper / textbook that made use of the notations $\flat$ and $\sharp$?

and a possible follow-up question:

If such notations were not adopted widely after the first appearance, who popularized them?

REPLY [13 votes]: The musical isomorphisms already appear in 1971 on page 21 of BERGER M. et al, Le spectre d'une variete riemannienne, Lecture Notes in Math. 194, 1971, Springer, see http://ci.nii.ac.jp/naid/10003477917/
However, I am not convinced that this is the first place where they appear, or that Berger was the inventor.  One would have to trace German textbooks in Riemannian geometry from the 1960s or perhaps earlier.  Interesting question!


TITLE: The distance between the centers of two circles K1 and K2 is 2cm. Determine the surface area of the figure that belongs to both circles.
QUESTION [0 upvotes]: The distance between the centers of two circles K1 and K2 is 2cm. They intersect under a right angle, the radius of K1 is 1cm. Determine the surface area of the figure that belongs to both circles.
[
So I drew it like this. It is obvious that the radius of K2 is then the square root of 3. I have no idea how to get the surface area of this weird shape. Any tips?

REPLY [3 votes]: You can calculate the area of a circular cap as shown below. Two such circular caps produce your 'weird shape'.


TITLE: Understanding the proof of Yoneda Embedding
QUESTION [1 upvotes]: Definitions
Recently I was going through the proof Yoneda Embedding from Steven Roman's An Introduction to the Language of Category Theory. But before I go into the problem that I am having with the proof let me first state Roman's definition of an embedding (I do so because according to this there is "no satisfactory and generally accepted definition of embeddings that is applicable in all categories").

Let $\mathcal{C},\mathcal{D}$ be two categories and $F:\mathcal{C}\to\mathcal{D}$ be a functor. Then,
Definition 1. The restriction of $F$ to $\text{hom}_\mathcal{C}(A,B)$ as a local arrow part of $F$ and by the object part of $F$ we mean $F:\text{Obj}(\mathcal{C})\to\text{Obj}(\mathcal{D})$ where $\text{Obj}(\mathcal{C})$ and $\text{Obj}(\mathcal{D})$ denotes the set of objects of respectively $\mathcal{C}$ and $\mathcal{D}$.
Definition 2. $F$ is full if all of its local arrow parts are surjective.
Definition 3. $F$ is faithful if all of its local arrow parts are injective.
Definition 4. $F$ is fully faithful if all of its local arrow parts are bijective.
Definition 5. $F$ is an embedding of $\mathcal{C}$ in $\mathcal{D}$ if it is fully faithful and the object part of $F$ is injective.

Background
Now let me come directly to my problem. What I was going through is the following theorem (number unaltered),

Theorem 48
Let $\mathcal{C}$ be a category. The contravariant functor $y: \mathcal{C} \to \sf{Set}^\mathcal{C}$ defined by,
  $$y(A)=\text{hom}_\mathcal{C}(A,\cdot)$$and, $$y(h)=\{{h_X}^\rightarrow\}:\text{hom}_\mathcal{C}(A,\cdot)\to\text{hom}_\mathcal{C}(B,\cdot)$$for all $A\in\mathcal{C}$ and all $h\in\text{hom}_\mathcal{C}(B,A)$
  is a contravariant embedding of $\mathcal{C}$ into the functor
  category $\sf{Set}^\mathcal{C}$, called the Yoneda embedding of $\mathcal{C}$ in $\mathsf{Set}^\mathcal{C}$.

Where ${h_X}^\rightarrow:\text{hom}_\mathcal{C}(A,X)\to\text{hom}_\mathcal{C}(B,X)$ defined by, $${h_X}^\rightarrow(g)=g\circ h_X$$
In the proof he writes,

The object part of $y$ maps $A$ to $\text{hom}_\mathcal{C}(A,\cdot)$ and since $\text{hom}_\mathcal{C}(A,\cdot)$ and $\text{hom}_\mathcal{C}(B,\cdot)$ are distinct for distinct objects $A$ and $B$, the object part of $y$ is injective.
As to injectivity, if $y_{A,B}(h) = y_{A,B}(k)$ for $h,k: B \to A$, then
$$\{h_X^\to:X\in\mathcal{C}\}=\{k_X^\to:X\in\mathcal{C}\}$$
In particular, for the components associated with $X=A$, we can apply them to $1_A$ (the identity morphism from $A$ to $A$) to get
$$1_A ∘ h = 1_A ∘ k$$
and so $h = k$.

My Questions

How does it follow that $A\ne B$ implies that $\text{hom}_\mathcal{C}(A,\cdot)\ne\text{hom}_\mathcal{C}(B,\cdot)$?
If $\{h_X^\to:X\in\mathcal{C}\}=\{k_X^\to:X\in\mathcal{C}\}$ then all we can say is the following, $$1_A ∘ h_Y = 1_A ∘ k_Z$$for some $Y,Z\in\mathcal{C}$. Why can we remove the subscripts?

REPLY [3 votes]: Yes - you understood my comment on 1. Point 2, is similar. Namely: Suppose we are given $h\colon B \to A$. Then $h$ induces, for each $X$, a  function $$h_X\colon {\rm hom} (A, X) \to {\rm hom }( B, X), $$ defined by the rule $ h_X (g) = g \circ h $. [No subscript on the right.]
[ Continuing the proof of Yoneda - although you're not asking for it: ]
Suppose we  are given 'another' morphism $k\colon B \to A$ (same source and target as $h$).
That means that we  now have, for  each $X$, two functions
 $$h_X,k_X \colon {\rm hom} (A, X) \to {\rm hom }( B, X). $$
But what  happens, if for each $X$, $h_X$ and $k_X$ coincide? That is, what happens if, for each $X$,  $$h_X = k_X?$$
Then the  previous equality holds, in particular, with $X=A$. That is,  $$h_A = k_A.$$
That means, for every $g\in {\rm hom} (A, A)$, that
$$ h_A (g) = k_A(g),$$ or equivalently $g \circ h = g \circ k$. In particular, we can take $g= 1_A$. Therefore $1_A\circ h = 1_A\circ k$, and $h = k$ - and we are done.
To summarize: if $h_X=k_X$ for every $X$, then $h_A=k_A$, so that $h_A(1_A)=k_A(1_A)$, so that $1_A\circ h = 1_A\circ k$, so that $h=k$.
[This answer was a follow-up to a comment above - but was itself too long for a comment.]


TITLE: Number of questions to attempt to pass a multiple choice test
QUESTION [1 upvotes]: There are $100$ multiple choice questions in a test, with $4$ options each.  The probability of choosing a right answer for a question is $0.25$.  Each right answer gets $4$ marks and each wrong answer gets $-1$ mark.  The passing criteria is $50\%$. How many questions should a person attempt to pass the test, given that they are sure of getting at least $30$ questions right.

I tried:
To get $50\%$ (200 marks) in the test,
\begin{align*}
(100 - x)4 -x & = 200\\
400 - 5x & = 200\\
5x & = 200\\
x & = 40
\end{align*}
incorrect
Should the person attempt $60$ questions to pass the test?

REPLY [0 votes]: The formulation of the question is a little bit odd to me, since whatever the number of questions the person answers, there is a possibility of failing. Obviously, the person will get $120$ points from the questions they are sure of knowing the answer to, so they need $80$ additional points out of the $70$ other questions. They may answer $20$ questions and luckily get them all correct, or answer $70$ questions and get the all wrong, losing $70$ points. What we can do is to compute the probability of passing the test as a function of the number of questions answered randomly, and mazimize this probability
Assume that the person answers randomly $n$ questions, where $0\le n\le 70$ and $Y$ the number of correct answers to these questions. Then $Y$ has a binomial distribution of parameters $n$ and $1/4$. Their score will be
$$X=120 + 4Y + (-1)(n-Y) = 120-n + 5Y$$
so the probability $p(n)$ of passing the test is
$p(n)=P(X\ge 200) = P(120-n+5Y \ge 200) = P(Y \ge 16 + \frac{n}{5})= \sum_{\lceil 16+\frac{n}{5} \rceil}^n \binom{n}{k}\left(\frac{1}{4}\right)^k\left(\frac{1}{4}\right)^{n-k}$
Using a computer, we see that $p(n)$ is very small for every $n$ and it is maximum when $n=70$ with a whooping $p(70)=0.000824$ probability of passing the test. They'd better study than rely on luck to pass the exam.

We may also count the average number of points the person will make on the test by answering randomly $n$ questions. Using the notations above, we have $E(Y)=\frac{n}{4}$ so
$$E(X)=120-n+ 5E(Y) = 120 + \frac{n}{4}$$
We have $E(X)\ge 200 \iff n \ge 320$. This explains why the probability of passing the test is extremely low. Intuitively, each time the person answers $4$ questions randomly, they will lose $3$ points and gain $4$ so a net gain of only $1$ point.


TITLE: Friedmann vacuum solution linear rather than exponential?
QUESTION [0 upvotes]: Does the Friedmann vacuum equation have a linear solution rather than an exponential one?
Using natural units one can write Friedmann's equation for the vacuum as
$$
\begin{eqnarray}
\left(\frac{\dot a}{a}\right)^2 &=& \frac{8\pi G}{3}\rho_{vac}\\\tag{1}
&=& L^2 \left(\frac{\rho_0}{L^4}\right)
\end{eqnarray}
$$
where I define the Planck length $L=(8\pi G \hbar / 3 c^3)^{1/2}$, $\hbar = c = 1$, and $\rho_0$ is a dimenionless constant.
Now let us interpret the Planck length $L$ to be the size of the smallest volume of space that can be described by general relativity.
But the Weyl postulate, together with cosmological observations, also imply that space is expanding.
Therefore we must have
$$L = a(t) L_0\tag{2}$$
where $L_0$ is the Planck length measured at the reference time $t_0$ where $a(t_0)=1$.
Inserting Eq.(2) into Eq.(1) we find
$$\left(\frac{\dot a}{a}\right)^2 = L_0^2 \left(\frac{\rho_0}{a^2L_0^4}\right)\tag{3}$$
where the Friedmann equation (3) has been rescaled in terms of the Planck length $L_0$ measured at the reference time $t_0$.
Eq.(3) has a linear solution
$$a(t) = \frac{t}{t_0}.$$
The scaled mass density $\rho(t)$ of the vacuum is not constant but rather given by
$$\rho(t) = \frac{\rho_0}{a^2 L_0^4} = \frac{1}{t^2 L_0^2}.$$

REPLY [1 votes]: Depending on who you are asking, it's called

empty Milne universe (Milne)
coasting universe (Kolb)
$R_h = ct$ universe (Melia)
linear universe (in your case)

Everyone and her grandmother jumped on the bandwagon.


TITLE: Covariant form of Green's function for wave equation
QUESTION [1 upvotes]: In J. D. Jackson's "Classical Electrodynamics", page 614 in the 3rd edition, he states that you can write the Green's functions for the wave equation in covariant form using the fact that
\begin{align*}\delta(x-x')^2 &= \delta\left[\left(x_{0} - {x_{0}'}\right)^2-\left\lvert\vec x -\vec x'\right\rvert^2\right] \\
&= \delta\left[\left(x_{0}-x_{0}'-R\right)\left(x_{0}-x_{0}'+R\right)\right]
\\
&=\frac{1}{2R}\left[\delta\left(x_{0}-x_{0}'-R\right)+\delta(x_{0}-x_{0}'+R)\right]
\end{align*}
I'm not seeing that last step. I have the feeling he's using something like
$$ \delta(f(x)) = \sum_{r\in f^{-1}(0)}\frac{\delta(x-r)}{f'(r)}$$
But since there are multiple variables here I'm not sure how to use that in this case.

For reference, this is for the wave equation $\square f = (\frac{1}{c^{2}}\partial_{t}^{2}-\nabla^{2})f = \delta^{(4)}(x-x')$ and the two Green's functions are:

$D_{r}(x-x') = \frac{\theta(x_{0}-x_{0}')}{4\pi|\vec x - \vec x'|}\delta(x_{0}-x_{0}'-|\vec x - \vec x'|)$ or in easier notation, $\frac{\theta(t-t')}{4\pi R}\delta\left[t' - \left(t-\frac{R}{c}\right)\right]$ (retarded Green's function)

$D_{a}(x-x') = \frac{\theta(x_{0}'-x_{0})}{4\pi|\vec x - \vec x'|}\delta(x_{0}-x_{0}'+|\vec x - \vec x'|)\quad$ (advanced Green's function)

And in covariant form they end up and $D_{r} = \frac{\theta(x_{0}-x_{0}')}{2\pi}\delta[(x-x')^2]$ and $D_{a} = \frac{\theta(x_{0}'-x_{0})}{2\pi}\delta[(x-x')^2]$.

REPLY [1 votes]: The formula for Dirac delta function has an absolute value in it, so it differs from yours (Dirac delta function - Composition with a function), that's why you are not getting the third step. Just take $x_0-x_0'$ as a variable and use the correct formula.


text:
\section{Proof of \fref{thm:LMSPE}}
\label{app:LMSPE}

The proof proceeds in two steps detailed as follows.

\paragraph{Mean Matrix}

We first compute the mean matrix~$\bW_0$. Since \fref{eq:optimalestimator} is a quadratic form, we can take the derivative in~$\widetilde\bW_0^H$ and set it to zero, i.e.,
\begin{align*}
\frac{d}{d\widetilde\bW_0^H}\Ex{}{\left\| \widetilde\bW_0 + \sum_{m=1}^M \setT(y_m) \widetilde\bW_m-\bmx\bmx^{H}\right\|_F^2} \!= 0.
\end{align*}
Basic matrix calculus yields
\begin{align} \label{eq:optimalmean}
\widetilde{\bW}_0 =  \textstyle \bK_{\bmx} - \sum_{m=1}^M  \overline\setT(y_m) \widetilde\bW_m
\end{align}
with $\overline\setT(y_m)=\Ex{}{\setT(y_m)}$ and $\bK_{\bmx}=\Ex{}{\bmx\bmx^{H}}$.

\paragraph{Linear Estimation Matrix}

With \fref{eq:optimalmean} and the fact that~\fref{eq:optimalestimator} is a quadratic form in the matrices $\bW_m$, $m=1,\ldots,M$, we take the derivatives in~$\bW_m^H$ and setting them to zero:

\begin{align*}

 {\textstyle  \frac{d}{d\widetilde\bW_m^H}}\mathbb{E} \! \Bigg[\bigg\|\!\!\sum_{m=1}^M\! (\setT(y_m)\!-\!\overline\setT(y_m)) \widetilde\bW_m \!
 \!-\! ( \bmx\bmx^{H}\! \!\!-\! \bK_{\bmx}) \bigg\|_F^2\! \Bigg] \!\!= \!0.
\end{align*}
By interchanging the derivative with expectation and with basic manipulations, we obtain the following set of optimality conditions for $\bW_m$ for $m=1,\dots,M$:
\begin{align} \label{eq:conditions}
& \textstyle \sum_{m'=1}^M \widetilde\bW_{m'} \Ex{}{(\setT(y_m)-\overline\setT(y_m))(\setT(y_{m'})-\overline\setT(y_{m'}))} \notag  \\

& \qquad \quad = \Ex{}{(\setT(y_m)-\overline\setT(y_m))(\bmx\bmx^{H}-\bK_{\bmx})}\!.
\end{align}
In compact matrix form, the above condition reads
\begin{align} \label{eq:matrixcondition}
(\bT\kron\bI_{N\times N}) \underline\bW = \underline\bV,
\end{align}
where we used the following shortcuts:
\begin{align*}
\bT & = \Ex{}{(\setT(\bmy)-\overline\setT(\bmy))(\setT(\bmy)-\overline\setT(\bmy))^T} \\
\underline\bW & = [ \widetilde\bW_1^T, \ldots, \widetilde\bW_m^T, \ldots, \widetilde\bW_M^T ]^T \\
\bV_m \!& = \Ex{}{(\setT(y_m)\!-\!\overline\setT(y_m))(\bmx\bmx^{H}\!\!-\bK_{\bmx})}\!, m=1,\ldots,M \\
\underline\bV & = [ \bV_1^T, \ldots, \bV_m^T, \ldots, \bV_M^T ]^T.
\end{align*}

The condition in \fref{eq:matrixcondition} can be solved for the estimation matrices in $\underline\bW$ leading to
$\underline\bW = (\bT^{-1}\kron\bI_{N\times N})  \underline\bV$,
where we require the matrix $\bT$ to be full rank.

To obtain the linear spectral estimator matrix, we simplify as
\begin{align*}
\bD_\bmy & = \bK_{\bmx} + ((\setT(\bmy)-\overline\setT(\bmy))^T \kron \bI_{N\times N}) \underline\bW \\

& = \textstyle  \bK_{\bmx} + \sum_{m=1}^M t_m \bV_m, \notag
\end{align*}
where we define the vector $ \bmt = \bT^{-1}(\setT(\bmy)-\overline\setT(\bmy))$.

\section{Proof of \fref{thm:MSEofLMSPE}}
\label{app:MSEofLMSPE}

To compute the spectral MSE in \fref{eq:MSE}, we simplify
\begin{align*}
\textit{S-MSE} & =\textstyle  \Ex{}{\left\|  \sum_{m=1}^M t_m \bV_m-(\bmx\bmx^{H}-\bK_{\bmx})\right\|_F^2}\!.
\end{align*}
We expand this expression into four terms
\begin{align}
& \textstyle \Ex{}{ \left\|  \sum_{m=1}^M t_m \bV_m-(\bmx\bmx^{H}-\bK_{\bmx})\right\|_F^2} \notag \\
& = \textstyle  \Ex{}{\left\|  \sum_{m=1}^M t_m \bV_m\right\|_F^2} \label{eq:proofterm1} \\
& \textstyle \quad +  \Ex{}{\left\|  \bmx\bmx^{H}-\bK_{\bmx}\right\|_F^2} \notag  \\
& \textstyle \quad - \Ex{}{ \tr\!\left( (\bmx\bmx^{H}-\bK_{\bmx})^H\left(\sum_{m=1}^M t_m \bV_m\right) \right) }  \label{eq:proofterm2} \\
&  \textstyle \quad - \Ex{}{ \tr\!\left( \left(\sum_{m=1}^M t_m \bV_m\right)^H(\bmx\bmx^{H}-\bK_{\bmx})  \right) } \label{eq:proofterm3}
\end{align}
and simplify each expression individually. We start with~\fref{eq:proofterm1} and use the fact that
\begin{align*}
\textstyle \sum_{m=1}^M t_m \bV_m = ((\setT(\bmy)-\overline\setT(\bmy))^T \bT^{-1}\kron\bI_{N\times N})  \underline\bV
\end{align*}
and rewrite the quantity within expectation as follows:
\begin{align*}
&  \underline\bV^H ( \bT^{-1} (\setT(\bmy)-\overline\setT(\bmy)) \kron\bI_{N\times N}) \\
& \quad \times   ((\setT(\bmy)-\overline\setT(\bmy))^T \bT^{-1}\kron\bI_{N\times N})  \underline\bV \\
 & = \underline\bV^H (( \bT^{-1} (\setT(\bmy)-\overline\setT(\bmy)) \\
 & \quad \times  (\setT(\bmy)-\overline\setT(\bmy))^T \bT^{-1}) \kron\bI_{N\times N}) \underline\bV.
\end{align*}
We now evaluate the expectation which leads to
\begin{align*}
\textstyle \Ex{}{\left\|  \sum_{m=1}^M t_m \bV_m\right\|_F^2} = \tr\!\left( \underline\bV^H ( \bT^{-1} \kron\bI_{N\times N}) \underline\bV \right)
\end{align*}
or, equivalently, to
\begin{align*}
\textstyle \Ex{}{\left\|  \sum_{m=1}^M \!t_m \! \bV_m\right\|_F^2} \!\!=\displaystyle  \!\!\sum_{m=1}^M\! \sum_{m'=1}^M [ \bT^{-1}]_{m,m'} \tr\!\left( \bV^H_m \bV_{m'} \right)\!.
\end{align*}

We next will simplify \fref{eq:proofterm2}. Recall that
\begin{align*}
\textstyle t_m = \sum_{m'=1}^M [ \bT^{-1}]_{m,m'} (\setT(y_{m'})-\overline\setT(y_{m'})),
\end{align*}
which enables us to write \fref{eq:proofterm2} as

\begin{align}
& \textstyle \Ex{}{ \tr\!\left( (\bmx\bmx^{H}-\bK_{\bmx})^H\left(\sum_{m=1}^M t_m \bV_m\right) \right) } \notag \\

&\textstyle =  \sum_{m=1}^M \sum_{{m'}=1}^M [ \bT^{-1}]_{m,{m'}}  \notag\\
&\textstyle \quad \times \tr\!\left(  \Ex{}{(\bmx\bmx^{H}-\bK_{\bmx})^H(\setT(y_{m'})-\overline\setT(y_{m'}))} \bV_m \right) \notag  \\

& \textstyle=  \sum_{m=1}^M \sum_{{m'}=1}^M [ \bT^{-1}]_{m,{m'}}    \tr\!\left( \bV_{m'}^H \bV_m \right)\!. \label{eq:proofcombine2}
\end{align}
Seeing as \fref{eq:proofterm3} is the Hermitian conjugate of \fref{eq:proofterm2}, we have
\begin{align}
& \textstyle \Ex{}{ \tr\!\left( \left(\sum_{m=1}^M t_m \bV_m\right)^H(\bmx\bmx^{H}-\bK_{\bmx})  \right) }\notag \\

& \textstyle = \sum_{m=1}^M \sum_{{m'}=1}^M [ \bT^{-1}]^*_{m,{m'}} \tr\!\left(  \bV_m^H  \bV_{m'} \right)\!.  \label{eq:proofcombine3}
\end{align}

Combining  all these terms yield the spectral MSE
\begin{align}
  \textit{S-MSE} & =  \textstyle\Ex{}{ \left\|  \sum_{m=1}^M t_m \bV_m-(\bmx\bmx^{H}-\bK_{\bmx})\right\|_F^2} \notag \\

& \textstyle=  \bC_{\bmx\bmx^{H}}-  \tr\!\left( \underline\bV^H ( \bT^{-1} \kron\bI_{N\times N}) \underline\bV \right)\!. \notag
\end{align}
with $\bC_{\bmx\bmx^{H}}=\Ex{}{\left\|  \bmx\bmx^{H}-\bK_{\bmx}\right\|_F^2}$.

\section{Proof of \fref{cor:LMSPEapproxerror}}
\label{app:LMSPEapproxerror}

We bound the estimation error with the spectral MSE of the LSPE as follows. For a given instance, we have
\begin{align}
&  \| \hat\bmx\hat\bmx^H- \bmx\bmx^{H} \|_F^2   =   \| \hat\bmx\hat\bmx^H- \bD_\bmy +\bD_\bmy - \bmx\bmx^{H} \|_F^2 \notag  \\
 &  \quad  \overset{\text{(a)}}{\leq}   2  \| \hat\bmx\hat\bmx^H- \bD_\bmy\|_F^2 + 2\| \bD_\bmy - \bmx\bmx^{H} \|_F^2 \notag \\
 &  \quad \overset{\text{(b)}}{\leq} 4 \| \bD_\bmy - \bmx\bmx^{H} \|_F^2, \notag
\end{align}
where (a) follows from the squared triangle inequality and (b) because $\hat\bmx \hat\bmx^H$ is the best rank-1 approximation of $\bD_\bmy$. Averaging over all instances finally yields
\begin{align*}
 \Ex{}{ \| \hat\bmx\hat\bmx^H- \bmx\bmx^{H} \|_F^2 } \leq 4\,\textit{S-MSE}_\text{LSPE}.
\end{align*}

config: arxiv


TITLE: The subgroup of an algbraic group generated by two non-irreductible closed subsets need not to be closed
QUESTION [1 upvotes]: Actually I have difficulties with the Section 7 Exercise 10 of the book J. E. Humphrey "Linear algebraic groups".
This exercises is about what is mentionned in the title. We have to make a demonstration by example and the book alrady gave one:
Consider in $GL_2(\mathbb{C})$ the subgroup $H$ generated by the elements
$\left( \begin{array}{cc}
 1 & 0 \\
 0 & -1
\end{array}
\right)$
and
$\left( \begin{array}{cc}
 1 & 1 \\
 0 & -1
\end{array}
\right)$
In addition, I would like to find its Zariski closure $\overline{H}$ and I don't know how to.
Thank you in advance for any hint to solve this.
K. Y.

REPLY [2 votes]: First hint: Let $G$ be a subgroup of $GL_n(\mathbb{C})$, then if $G$ is both countable and Zariski closed, it is finite. Use this to deal with the exercise.
Second hint: Show that the kernel of the determinant map restricted to $H$ is of index $2$ in $H$ and explicitly compute its kernel. Deduce from this the connected component of the Zariski closure of $H$  and then $\overline{H}$.