TITLE: Closed form of a recursive equation
QUESTION [0 upvotes]: I have an assignment where I need to bring the following recursive equation in a closed form:
$$R(n) = \dfrac{2}{n} \cdot (R(0) +R(1) +. . .+R(n−1)) + c$$
I can use
$$\sum_{i=1}^{n} \dfrac{1}{i(i+1)} = \dfrac{n}{n+1}$$
from a previous assignment.
So far I came to $nR(n)=2(R(0) +R(1) + \cdots +R(n−1)) + cn$
and then to
$(n-1)R(n-2)=2(R(0) +R(1) + \cdots +R(n−1)) + c(n-1)$
But I have no idea how to continue and how to use the other equation.
Any help is appreciated.

REPLY [1 votes]: Sometimes it helps to just plot a few values:
$R(0) = 0$
$R(1) = \frac{2}{1} \cdot 0 + c = c$
$R(2) = \frac{2}{2} \cdot (0 + c) + c= 2c$
$R(3) = \frac{2}{3} \cdot (0 + c + 2c) + c = 3c$
$R(4) = \frac{2}{4} \cdot (0 + c + 2c + 3c) + c= 4c$
$R(5) = \frac{2}{5} \cdot (0 + c + 2c + 3c + 4c) + c= 5c$
Which strongly suggests that $R(n) = nc$. Let's prove it by induction:
Base case: $R(0) = 0 \cdot c = 0$, which is true. And likewise $R(1) = 1 \cdot c = c$, which is true.
Inductive step: For all $0<n$ up to this point, suppose $R(n) = c + \frac{2}{n} \cdot \sum_{i=0}^{n-1}R(i) = nc$. Let's show that $R(n+1) = (n+1) \cdot c$
Then we have:
$$R(n+1) = c + \frac{2}{n+1} \cdot \sum_{i=0}^{n} R(i) = c + \frac{2}{n+1} \sum_{i=0}^{n}ic \\= c + \frac{2}{n+1} \cdot \frac{1}{2}cn(n + 1) = (n+1) \cdot c$$
Which proves the result.


text:
\begin{document}
\setcounter{page}{1}

\title[Schr\"odinger equation on the sphere ]{Sharp Strichartz estimates for the Schr\"odinger equation on the sphere }

\author[D. Cardona]{Duv\'an Cardona}
\address{
  Duv\'an Cardona S\'anchez:
  \endgraf
  Department of Mathematics: Analysis, Logic and Discrete Mathematics
  \endgraf
  Ghent University, Belgium
  \endgraf
  {\it E-mail address} {\rm duvanc306@gmail.com, duvan.cardonasanchez@ugent.be}
  }

  \author[L. Esquivel]{Liliana Esquivel}
\address{
 Liliana Esquivel:
  \endgraf
 Gran Sasso Science Institute, CP 67100
  \endgraf
  L' Aquila, Italia.
  \endgraf
  {\it E-mail address} {\rm liliana.esquivel@gssi.it}
  }

\subjclass[2010]{Primary: 35Q40, { Secondary: 42B35, 42C10, 35K15}.}

\keywords{}

\thanks{The first author was supported by the FWO Odysseus 1 grant G.0H94.18N: Analysis and Partial Differential Equations of Professor Michael Ruzhansky.}

\begin{abstract}
In this contribution we investigate the  Schr\"ordinger equation associated to the Laplacian on the sphere in the form of sharp Strichartz estimates. We will provided   simple proofs for our main theorems  using purely  the  $L^2\rightarrow L^p$ spectral estimates for the operator norm of the spectral projections (associated to the spherical harmonics)  proved in \cite{KS}. A sharp index of regularity is established for the initial data in spheres of arbitrary dimension $d\geq 2$.

\end{abstract} \maketitle

\section{Introduction}

In this work we provide
sharp Strichartz estimates for the Schr\"odinger equation for the Laplacian on the sphere $\mathbb{S}^d,$
$\Delta_{\mathbb{S}^d},$ which is given by
\begin{equation}\label{PVI}
   \textnormal{(PVI):}
     \begin{cases}
      iu_{t}+\Delta_{\mathbb{S}^d}u=0,&\quad (t,x)\in \mathbb{T}\times \mathbb{S}^d,
      \\
       u(0,x)=f(x), &\quad x\in \mathbb{S}^d,
     \end{cases}
\end{equation} where $\mathbb{T}\cong [0,2\pi),$ $0\sim 2\pi,$ denotes the one dimensional torus. The fractal nature of the solutions for \eqref{PVI} has been discussed, e.g. in Taylor \cite{Taylor2003} and references therein. To illustrate our results let us recall, that \eqref{PVI} is an analogy of the standard Schr\"odinger equation on $\mathbb{R}^n,$
\begin{equation}\label{SEq}
iu_t(t,x)+\Delta_{\mathbb{R}^n} u(t,x)=0,
\end{equation} with initial data $u(0,\cdot)=f.$ The solution for \eqref{SEq}, satifies the following Strichartz estimate
\begin{equation}\label{KT'}
\Vert e^{it\Delta_{\mathbb{R}^n}}f\Vert_{L^q[(0,\infty), L_x^p(\mathbb{R}^d)]}\leq C\Vert f\Vert_{L^2(\mathbb{R}^n)},
\end{equation}
which holds, if and only if, $1/q=n/2(1/2-1/p),$ for $n=1$ and $2\leq p\leq \infty,$ $n=2$ and $2\leq p<\infty,$ and, $2\leq p<\frac{2n}{n-2}$ if $n\geq 3.$
This fundamental result proved by Keel and Tao \cite{KeelTao} is the source of inspiration for several extensions of \eqref{SEq} to general compact and non-compact manifolds.

The key point for obtaining \eqref{KT'}, is the following estimate
\begin{equation}\label{Analogue}
    \Vert e^{it\Delta_{\mathbb{R}^n}}\Vert_{L^1(\mathbb{R}^n)\rightarrow L^\infty(\mathbb{R}^n) }\lesssim_{n} C|t|^{-\frac{n}{2}},
\end{equation}which expresses the dispersive property of the linear Schr\"odinger equation on $\mathbb{R}^n.$ If $\mathbb{R}^n$ is replaced by a compact manifold $M,$ the analogue of the dispersion estimate \eqref{Analogue} fails globally in time, as the example of constant solutions shows (see Burq, Gerard and Tzvetkov \cite{BurqGerardT} for details). It is less obvious, but still true, that it also fails locally in time (see the discussion in the beginning of Section 3 of \cite{BurqGerardT}). For the case  of the $d$-dimensional torus, $M=\mathbb{T}^d,$ Bourgain \cite{Bourgain93,Bourgain94,Bourgain99} proposed a different approach to Strichartz estimates by using the Fourier series representation of the Schr\"odinger group,
\begin{equation*}
   u(t,x)\equiv e^{it\Delta_{\mathbb{T}^d}}f(x):=\sum_{k\in \mathbb{Z}^d} e^{i2\pi (t|k|^2+k\cdot x)}\widehat{f}(k).
\end{equation*} Notice that this representation shows that the solution is also periodic in time. By computing explicitly
\begin{equation}\label{Bourgain}
     \Vert u \Vert^4_{L^4(\mathbb{T}\times \mathbb{T}^d )}= \Vert u^2 \Vert^2_{L^2(\mathbb{T}\times \mathbb{T}^d )},
\end{equation}   Bourgain obtains similar Strichartz estimates as in the fundamental paper due to Keel and Tao \cite{KeelTao}.

In general, on a compact manifold without boundary $M,$ it was proved by Burq, Gerard and Tzvetkov \cite{BurqGerardT} the following estimate
\begin{align}\label{Burq}
\Vert e^{it\Delta_{M}}f\Vert_{L^q[[0,T], L_x^p(M)]}\lesssim_{T,p,q,s} \Vert f\Vert_{W^{s}(M)},\,\,s\geq \frac{1}{q},
\end{align}for $1/q=\textnormal{dim}(M)/2(1/2-1/p),$ $q\geq 2,$ and  $p<\infty.$ As in our case, Taylor in \cite{Taylor2003} considered the case of spheres $M=\mathbb{S}^d,$ where the author obtained sharp $L^p$-estimates on the solution operator at those times which are rational multiples of $\pi.$ This idea that the profile of linear dispersive equations depend on the algebraic properties of time have been further exploited in the papers of Kapitanski-Rodniaski \cite{KapitanskiRodniaski99}, Taylor  \cite{Taylor2003} and Rodnianski \cite{Rodnianski} where the fractal nature for the solution of the Schr\"odinger equation on spheres and torus were considered.

Our main goal is to study \eqref{PVI}, by using the $L^2\rightarrow L^p$ bounds for the spectral resolution of  the Laplacian $\Delta_{\mathbb{S}^d}.$ More precisely, we deal with the problem of establishing an analogy of \eqref{SEq} in the case of the Laplacian on the sphere $\mathbb{S}^d.$  Our main results are, the following Theorem \ref{duvantheorem}, and Theorem \ref{LiliTheorem}, where the Schr\"odinger equation with potential \eqref{PVI'}, (this means, associated to $-\Delta_{\mathbb{S}^d}+V(x,t),$ for a suitable potential $V(x,t)$) is considered.

\begin{theorem}\label{duvantheorem}
Let $d\geq 2$ and $p,q$ satisfy $2\leq p\leq \infty,$ and $2\leq q<\infty.$ Then the following Strichartz estimate
\begin{equation}\label{TheoremM}
     \Vert e^{it\Delta_{\mathbb{S}^d}}f(z) \Vert_{L^{p}_z[\mathbb{S}^{d},\,L^q_t(\mathbb{T})]}\lesssim_{p,q,s} \Vert f \Vert_{W^{s}(\mathbb{S}^d)},\footnote{Observe that under the symmetric compromise  $p=q,$ $\Vert u(t,z) \Vert_{L^{p}_z[\mathbb{S}^{d},\,L^q_t(\mathbb{T})]}=\Vert u(t,z) \Vert_{L^{p}(\mathbb{S}^{d}\times \mathbb{T})},$ and for $p=4,$ $  \Vert u \Vert^4_{ L^4(\mathbb{S}^d\times\mathbb{T}  )}= \Vert u^2 \Vert^2_{L^2(\mathbb{S}^d\times \mathbb{T})} .$ We use the notation $A\lesssim B$ to indicate $A\leq cB,$ where $c>0$ does not depend on fundamental quantities.}
\end{equation}holds true for $s\geq \varkappa_{p,q},$ where
$$   \varkappa_{p,q}: = \left\{
     \begin{array}{lr}
       \frac{d-1}{2}\left(\frac{1}{2}-\frac{1}{p}\right)+\left(\frac{1}{2}-\frac{1}{q}\right), &   2\leq p\leq \frac{2(d+1)}{d-1},\\
       d\left(\frac{1}{2}-\frac{1}{p}\right)-\frac{1}{q}, &   p>\frac{2(d+1)}{d-1}. \\

     \end{array}
   \right.  $$
 For $q=2,$ and $s<\varkappa_{p,2},$ the estimate in \eqref{TheoremM} does not hold for any non-trivial initial data $f\in L^2(\mathbb{S}^d)$. So, in this case, the regularity order $\varkappa_{p,2} $ is sharp in any dimension $d,$ in the sense that \eqref{TheoremM}  does not hold for all $s<\varkappa_{p,2} $.
\end{theorem}
By keeping the notation $\varkappa_{p,2}$ in Theorem \ref{duvantheorem}, we consider the case of Schr\"odinger equation associated with the Hamiltonian $$\mathbb{H}:=-\Delta_{\mathbb{S}^d}+V(x,t).$$ Indeed, we prove:
\begin{theorem}\label{LiliTheorem} Let us consider the following Schr\"odinger equation with potential
\begin{equation}\label{Lilicase}
   \textnormal{(PVI'):}
     \begin{cases}
      iu_{t}=(-\Delta_{\mathbb{S}^d}+V(x,t))u,&\quad (t,x)\in \mathbb{T}\times \mathbb{S}^d,
      \\
       u(0,x)=f(x), &\quad x\in \mathbb{S}^d
     \end{cases}
\end{equation} where    $\Vert V \Vert_{L^{q}_x(\mathbb{S}^d,L^\infty_t(\mathbb{T}))}$ is small enough. Let us assume that  $\frac{1}{q}+\frac{1}{p/2}=1.$ Then, there exists a unique solution  $u\in {C}(\mathbb{T}, W_x^{s}(\mathbb{S}^d))\cap L^p_x(\mathbb{S}^d, L^2_t(\mathbb{T}))$ for \eqref{Lilicase}, provided that $s\geq \varkappa_{p,2}$
\end{theorem}

This paper is organised as follows. In Section \ref{Preliminaries} we present some basic properties about the spectral decomposition of the Laplacian and we introduce some auxiliar (Triebel-Lizorkin) spaces, that will be useful in our analysis. In Section \ref{t1}  we prove our Theorem \ref{duvantheorem}. Finally, in Section  \ref{potential} we consider the Schr\"odinger equation with potential.
We finish this introduction by  discussing briefly our main results.
\begin{remark} For $M=\mathbb{S}^d,$ let us observe that \eqref{Burq} is in general a better estimate (in the case when $p>2(d+1)/(d-1)$) that the obtained in Theorem
\ref{duvantheorem} in the case, where, $$1/q=d/2(1/2-1/p),\footnote{By following the usual nomenclature,   such pair $(p,q)$ is called admissible.}$$ but, both estimates agree for $2\leq p\leq 2(d+1)/(d-1).$ However, the main point in Theorem \ref{duvantheorem} is the sharpness of the index $\varkappa_{p,2}$ in any dimension, and that the estimate that we obtain is in general independent of the known restriction $1/q=\textnormal{dim}(M)/2(1/2-1/p),$ between the dimension $d,$ and the pair $(p,q).$ Several sharp results on the sphere are know only for slow dimensions $d,$ (see e.g. Taylor \cite{Taylor2003}), or for $p=q=4$ (see Burq, Gerard and Tzvetkov \cite{BurqGerardT}), but the sharp Sobolev regularity order  $\varkappa_{p,2}$ is to our knowledge, a  new result in this context and also in the context of Theorem \ref{LiliTheorem} for the Hamiltonian $\mathbb{H}$.
\end{remark}
\begin{remark}We will discuss some implications of our results to general manifolds by following a similar analysis as in Ruzhansky and Turunen \cite[page 578]{RuzhanskyTurunenBook2010}.
Under the existence of a $C^\infty$-diffeomorphism $i:\mathbb{S}^d\rightarrow M,$ where $M$ is a Riemannian manifold, (these kind of diffeomorphisms are known for $d=3,$ with $M$ being a connected a simply connected manifold in view of the Perelman proof \cite{Perelman1,Perelman2,Perelman3,Perelman4} for the Poincar\'e conjecture \cite{Poincare} and in higher dimensions), if $X_{1},\cdots,X_{d}$ is an orthonormal basis for the tangent bundle $T\mathbb{S}^d,$ in such a way that
\begin{equation}
    \Delta_{\mathbb{S}^d}:=-\sum_{1\leq j\leq d}X_j^2,
\end{equation}let us denote by
\begin{equation*}
    \Delta_{M}:=-\sum_{1\leq j\leq d}Y_j^2,
\end{equation*}  the second order differential operator induced by $i.$ This means that, $Y_{j}:=i_{*}(X_{j}),$ is the vector field induced by the differential of $i,$ which is a diffeomorphism between tangent bundles, $i_{*}:T\mathbb{S}^d\rightarrow TM.$ Theorem \ref{duvantheorem} implies the estimate
\begin{equation}\label{TheoremMM}
     \Vert e^{it\Delta_M}u_0 \Vert_{L^{p}_z[M,\,L^q_t(\mathbb{T})]}\leq C\Vert u_0 \Vert_{W^{s}(M)},
\end{equation} for $s\geq \varkappa_{p,q},$ with $\varkappa_{p,2}$ being sharp, and also Theorem \ref{LiliTheorem}, implies the existence of a unique $u\in {C}(\mathbb{T}, W_x^{s}(M))\cap L^p_x(M, L^2_t(\mathbb{T}))$ satisfying the differential problem
\begin{equation}\label{Lilicase2}
   \textnormal{(PVI''):}
     \begin{cases}
      iu_{t}=(-\Delta_{M}+V(x,t))u=0,&\quad (t,x)\in \mathbb{T}\times M,
      \\
       u(0,x)=u_0(x), &\quad x\in M,
     \end{cases}
\end{equation} for a potential $V$ with  $\Vert V \Vert_{L^{q}_x(M,L^\infty_t(\mathbb{T}))}$ small enough.

\end{remark}

\section{preliminaries}\label{Preliminaries}
Now we present some basics about the spectral decomposition of the Laplacian on the sphere, for this we follow Schoen and Yau \cite{T}.
Let $d\geq 1,$  $d\in \mathbb{N}.$ Let $$\mathbb{S}^{d}:=\{x=(x_0,x_1,\cdots,x_d)\in \mathbb{R}^{d+1}:|x_0|^2+|x_1|^2+\cdots+|x_d|^2=1\},$$ be the $d$-dimensional unit sphere contained in $\mathbb{R}^{d+1}$ and $\mathcal{H}^{d}_n$ be the space of spherical harmonics of degree $n.$ There is a standard metric $g$ on $\mathbb{S}^{d},$ induced by the standard metric on $\mathbb{R}^{d+1},$ and conrrespondly, a Laplace operator $\Delta_\mathbb{S}^d$ which can be defined, initially on $C^\infty(\mathbb{S}^{d})$ as follows. For $f\in C^\infty(\mathbb{S}^d),$ let $\tilde{f}(x):=f(x/|x|),$ then
\begin{equation*}
 \Delta_{\mathbb{S}^d} f:=\Delta_{\mathbb{R}^{d+1}}\tilde{f}|_{\mathbb{S}^d},
\end{equation*}where, by following the usual nomenclature, $$\Delta_{\mathbb{R}^{d+1}}:=-\partial_{x_0}^2-\partial_{x_1}^2-\cdots- \partial_{x_d}^2$$ is the standard positive Laplacian on $\mathbb{R}^{d+1}.$ It can be shown that for $r=|x|,$ one can write
\begin{equation}
    \Delta_{\mathbb{R}^{d+1}}\tilde{f}(x)=r^{-2}(\Delta_{\mathbb{S}^d} f).
\end{equation}In polar coordinates $r=|x|,$ $\omega=x/r,$
\begin{equation*}
   \Delta_{\mathbb{S}^d}= r^2\Delta_{\mathbb{R}^{d+1}}-
    r^2\partial_{r}^2-dr\partial_{r}.
    \end{equation*}
It is well known, from the Fourier analysis associated to $\Delta_{\mathbb{S}^d},$ the spectral decomposition
\begin{equation*}
    L^{2}(\mathbb{S}^{d})=\bigoplus_{n=0}^\infty \mathcal{H}^{d}_n,\,\,\dim(\mathcal{H}^{d}_n)\sim n^{d-1},
\end{equation*} where $L^2(\mathbb{S}^d)\equiv L^2(\mathbb{S}^d,dx)$ is the usual $L^2$-space\footnote{As usually,  defined by those measurable functions on $\mathbb{S}^d,$ such that  $\Vert f\Vert_{L^2(\mathbb{S}^d)}:=\left(\,\int\limits_{\mathbb{S}^d}|f(\omega)|^2d\omega\right)^{\frac{1}{2}}<\infty.$ In general, on a measure space $(X,dx),$ we use $\Vert f\Vert_{L^p(X)}:=\left(\,\int\limits_{X}|f(x)|^pdx\right)^{\frac{1}{p}},$ for all $1\leq p<\infty.$} associated to the surface measure  on the sphere, $dx:=d\sigma_{\mathbb{S}^d}(x),$ and that, $\mathcal{H}^{d}_n$ is  the eigenspace of the Laplace-Beltrami operator $\Delta_{\mathbb{S}^d}$ associated  with the eigenvalue $\lambda_{n}=n(n+d-1).$ Let  $H^d_{n}$ be the spectral projection to the subspace $\mathcal{H}^{d}_n$ for every $n\in \mathbb{N}_{0}.$ Then, for every $f\in L^2(\mathbb{S}^{d})$ we can write
\begin{equation*}
    f=\sum_{n=0}^{\infty}H^d_{n}f,
\end{equation*} where the series converges in the $L^2(\mathbb{S}^d)$-norm. In particular, the spectral theorem for $\Delta_{\mathbb{S}^d}$ implies
\begin{equation*}
 m(\Delta_{\mathbb{S}^d})=   \sum_{n=0}^{\infty}m(\lambda_n)H^d_{n}f,
\end{equation*}for every continuous function $m$ on $\mathbb{R},$ where the domain of $m(\Delta_{\mathbb{S}^d})$ is given by
\begin{equation*}
    \textnormal{Dom}(m(\Delta_{\mathbb{S}^d})):=\{f\in  L^2(\mathbb{S}^{d}):\sum_{n=0}^\infty|m(\lambda_n)|^2\Vert H^d_{n}f\Vert^2_{L^2(\mathbb{S}^d)}<\infty \}.
\end{equation*}
By observing that, for  $m(\lambda):=e^{\pm it\lambda },$ $\textnormal{Dom}(m(\Delta_{\mathbb{S}^d}))=L^{2}(\mathbb{S}^d),$  the Schr\"odinger propagator $e^{it \Delta_{\mathbb{S}^d} }:L^2(\mathbb{S}^{d})\rightarrow L^2(\mathbb{S}^d)$ is given by
\begin{equation*}
   e^{it \Delta_{\mathbb{S}^d}}f:= \sum_{n=0}^{\infty}e^{i\lambda_n t}H^d_{n}f,
\end{equation*} and for $f\in C^\infty(\mathbb{S}^d)$ the solution to \eqref{PVI} is given by
\begin{equation*}\label{equsol}
    u(t,x)=\sum_{n=0}^{\infty}e^{i\lambda_n t}H^d_{n}f(x),\quad (t,x)\in \mathbb{T}\times \mathbb{S}^d.
\end{equation*}

In this paper we want to estimate the mixed norms $L^p_x(L^q_t)$ of  solutions to Schr\"odinger equations by using the following version of Triebel-Lizorkin associated to the family of spectral projections of the Laplacian $\Delta_{\mathbb{S}^d}$.
 \begin{definition}
Let us consider $0<p\leq \infty,$ $r\in\mathbb{R}$ and $0<q\leq \infty.$ The Triebel-Lizorkin space associated to the family of projections $H_{n}^d,$ $n\in  \mathbb{N}_0,$ and to the parameters $p,q$ and $r$ is defined by those complex functions $f$ satisfying
\begin{equation}\label{5}
\Vert f\Vert_{{F}^r_{p,q}(\mathbb{R}^n)}\equiv \Vert f\Vert_{{F}^r_{p,q}(\mathbb{R}^n,\,\,(H_{n}^d)_{n\in \mathbb{N}})}:=\left\Vert \left(\sum_{n=0}^\infty n^{rq}|H_n^d f|^{q}\right)^{\frac{1}{q}}\right\Vert_{L^p(\mathbb{S}^d)}<\infty.
\end{equation}
 \end{definition}
 The definition considered above differs from those arising with dyadic decompositions on compact homogeneous manifolds (see e.g. Nursultanov, Ruzhansky and Tikhonov \cite{NRT}). Indeed, when working with dyadic decompositions one recover the Littlewood-Paley theorem, about the equivalence of certain dyadic Triebel-Lizorkin spaces with the Lebesgue spaces $L^p(\mathbb{S}^d),$ which with the current definition we do not necessarily have. The following are natural embedding properties of such spaces.   Sobolev spaces $W^{s,p}$  in $L^p(\mathbb{S}^d)$-spaces and associated to $\Delta_{\mathbb{S}^d},$ can be defined by the norm
 \begin{equation}\label{Sobp}
 \Vert f\Vert_{W^{s,p}}\equiv \Vert \Delta^s_{\mathbb{S}^d} f\Vert_{L^p}:=\left(\,\int\limits_{\mathbb{S}^d}|\Delta^s_{\mathbb{S}^d} f(x)|^pdx \right)^{\frac{1}{p}}\footnote{ Here, $W^{s,p}:=\textnormal{Dom}(\Delta^s_{\mathbb{S}^d})$ is considered as the completion of $C^\infty(\mathbb{S}^d)$ with respect to the norm defined in  \eqref{Sobp}.}.
 \end{equation}
  We will use the notation $W^s:={W^{s,2}}$ for the Sobolev space modelled on $L^2(\mathbb{S}^d).$  Indeed, we  have
 \begin{equation*}
     \Vert u\Vert_{W^s(\mathbb{S}^d)}\asymp\left( \sum_{\ell=0}^\infty \ell^s\Vert H_\ell^d u\Vert_{L^2(\mathbb{S}^d)}  \right)^\frac{1}{2}\footnote{Where we use $\asymp$ because $\ell\sim\lambda_\ell:= [\ell(\ell+d-1)]^{\frac{1}{2}}$ for $\ell\rightarrow\infty.$} .
 \end{equation*}

 The following properties can be deduced  by following a similar analysis as in  \cite{NRT}.
 \begin{itemize}
\item[-] ${{F}}^{r+\varepsilon}_{p,q_1}\hookrightarrow {{F}}^{r}_{p,q_1}\hookrightarrow {{F}}^{r}_{p,q_2}\hookrightarrow {F}^{r}_{p,\infty},$  $\varepsilon>0,$ $0<p\leq \infty,$ $0<q_{1}\leq q_2\leq \infty.$
\item[-]  ${F}^{r+\varepsilon}_{p,q_1}\hookrightarrow {F}^{r}_{p,q_2}$, $\varepsilon>0,$ $0<p\leq \infty,$ $1\leq q_2<q_1<\infty.$
\item[-] ${F}^0_{2,2}=L^2$ and consequently, for every $s\in\mathbb{R},$ $W^{s}={F}^s_{2,2}.$
\end{itemize}
It is a very special fact, how at rational multiples of $\pi,$ the propagator $e^{it \Delta_{\mathbb{S}^d}}$ behaves on Sobolev spaces. Indeed, let us remark that (see Proposition 2.1 of Taylor \cite{Taylor2003}) for all $1<p<\infty,$ and all $s\in \mathbb{R},$
\begin{equation}
    e^{2\pi i(\frac{m}{2k})\Delta_{\mathbb{S}^d}}:W^{s,p}(\mathbb{S}^d)\rightarrow W^{s-(d-1)|\frac{1}{2}-\frac{1}{p}|,p}(\mathbb{S}^d),
\end{equation}extends to a bounded operator. For the sharpness of this estimate we refer the reader to  Taylor \cite[page 148]{Taylor2003}.

\section{Proof of  Theorem \ref{duvantheorem}}\label{t1}
In this section we will prove one of our main results.   We will see that information about the Strichtaz estimates for the Schr\"odinger propagator, is encoded in the $L^2\rightarrow L^p$-operator norm of the projections ${H}_n^d,$ and this will essential in order to give a short proof for  Theorem \ref{duvantheorem}.

The space $L^2_{\textnormal{fin}}(\mathbb{S}^{d})$ consists of those finite linear combinations of elements in the sub-spaces  $\mathcal{H}_n^d.$
\begin{lemma}
Let us consider $f\in {F}^0_{p,2} (\mathbb{S}^{d} )$, then for all $1\leq p\leq \infty$  we have
\begin{equation}\label{eql2lp}
\Vert u(t,z) \Vert_{L^p_z[\mathbb{S}^{d},\,L^2_t(\mathbb{T})]}=\sqrt{2\pi}\Vert f \Vert_{{F}^0_{p,2}(\mathbb{S}^{d})}.
\end{equation}
\end{lemma}
\begin{proof}
Let us consider $f\in {F}^0_{p,2}(\mathbb{S}^{d}).$ By a standard argument of density we only need to assume $f\in L^2_{\textnormal{fin}}(\mathbb{S}^{d}).$ The solution $u(t,z)$ for \eqref{PVI} is given by
\begin{equation}
u(t,z)=\sum_{n=0}^{\infty}e^{it \lambda_n}H_{n}^df(z).
\end{equation}
Let us note that the previous sums runs over a finite number of $n's$ because $f\in L^2_{\textnormal{fin}}(\mathbb{S}^{d}).$ So, we have,
\begin{align*}
\Vert u(t,z) \Vert_{L^2_t(\mathbb{T})}^2&=\int\limits_{0}^{2\pi}u(t,z)\overline{u(t,z)}dt=\int\limits_{0}^{2\pi} \sum_{n,n'}e^{it(\lambda_n-\lambda_{n'})}H_{n}^df(z)\overline{ H_{n'}^df(z)  }  dt .
\end{align*}Observe that if $n\neq n',$ then $$\lambda_n=n(n+d-1)\neq n'(n'+d-1)=\lambda_{n'},$$ and consequently, from the  $L^2$-orthogonality of the trigonometric polynomials we get,
\begin{align*}
\Vert u(t,z) \Vert_{L^2_t(\mathbb{T})}^2&=\int\limits_{0}^{2\pi}u(t,z)\overline{u(t,z)}dt
= \sum_{n,n'}\int\limits_{0}^{2\pi}e^{it(\lambda_n-\lambda_{n'})}H_{n}^df(z)\overline{ H_{n'}^df(z)  }  dt\\
&=\sum_{n=0}^{\infty}2\pi\,\cdot|H_{n}^df(z)|^2,
\end{align*} where we have changed summation with integration because $f\in L^2_{\textnormal{fin}}(\mathbb{S}^{d}).$
Thus, we conclude the following fact
\begin{equation}\label{identity1}
\Vert u(t,z) \Vert_{L^2_t(\mathbb{T})}=\left(\sum_{n=0}^\infty2\pi\,\cdot|H_{n}^df(z)|^2\right)^{\frac{1}{2}},\,\,\,  f\in L^2_{\textnormal{fin}}(\mathbb{S}^{d}).
\end{equation}
Finally,
\begin{equation}
\Vert u(t,z)\Vert_{L_2(\mathbb{S}^{d},L^2_t(\mathbb{T}) )}=\sqrt{2\pi}\Vert f\Vert_{{F}^0_{p,2}(\mathbb{S}^{d})},
\end{equation}as claimed.

\end{proof}
The following lemma, allows us to compare the $L^p_{z}(L^q_{t})$-mixed norms of the solution $u(t,z)$ and some Triebel-Lizorkin norms of the initial data.

\begin{lemma}\label{T1}
Let $p,q$ and $s_q$ be such that $0<p\leq \infty,$ $2\leq q<\infty$ and $s_{q}:=\frac{1}{2}-\frac{1}{q}.$ Then
\begin{equation}
C_p'\Vert f\Vert_{{F}^0_{p,2}}\leq \Vert u(t,z)\Vert_{L^p_{z}(\mathbb{S}^{d}, L^q_{t}(\mathbb{T}))} \leq C_{p,s}\Vert f\Vert_{{F}^s_{p,2}},
\end{equation}
holds true for every $s\geq s_{q}.$
\end{lemma}
\begin{proof}
Let us  consider, by a density argument, the initial data $f\in L^2_f(\mathbb{R}^n).$ In order to estimate the norm  $\Vert u(t,z) \Vert_{L^p_z[\mathbb{S}^{d},\,L^q_t(\mathbb{T})]}$ we can use the Wainger Sobolev embedding Theorem:
\begin{equation}
\left\Vert \sum_{n\in \mathbb{Z},n\neq 0}|n|^{-\alpha}\widehat{F}(n)e^{in t}\right\Vert_{L^q(\mathbb{T})}\leq C\Vert F\Vert_{L^r(\mathbb{T})},\,\,\alpha:=\frac{1}{r}-\frac{1}{q}.
\end{equation}
For $s_q:=\frac{1}{2}-\frac{1}{q}$  we have
\begin{align*}
\Vert u(t,z)\Vert_{L^q(\mathbb{T})} &=\left\Vert \sum_{n=0}^{\infty} e^{it\lambda_n}H_n^df(z)  \right\Vert_{L^q(\mathbb{T})} \\
&\leq C\left\Vert \sum_{n=0}^{\infty} n^{s_q}e^{it\lambda_n}H_n^df(z)  \right\Vert_{L^2(\mathbb{T})}\\
&=C\left\Vert \sum_{n=0}^{\infty} e^{it\lambda_n}H_n^d[\mathcal{L}^{s_q}f(z)]  \right\Vert_{L^2(\mathbb{T})}\\
&=C\left( \sum_{n=0}^{\infty} |H_n^d[\mathcal{L}^{s_q}f(z)|^2  \right)^{\frac{1}{2}}\\
&:=T'(\mathcal{L}^{s_q}f)(z),
\end{align*}where we have denoted
\begin{equation*}
 \mathcal{L}^{s_q}:=   \sum_{n=0}^{\infty}n^{s_q}H^d_{n}.
\end{equation*}
 So, we have
\begin{align*}
\Vert u(t,z) \Vert_{L^p_z[\mathbb{S}^{d},\,L^q_t(\mathbb{T})]}&\leq C\Vert T'(H^{s_q}f) \Vert_{L^p(\mathbb{S}^{d})}\\&= C\Vert \mathcal{L}^{s_q}f\Vert_{{F}^0_{p,2}(\mathbb{S}^{d})}\\
&=C\Vert f\Vert_{{F}^{s_q}_{p,2}(\mathbb{S}^{d})}.
\end{align*}
We end the proof by taking into account the embedding ${F}^s_{p,2}\hookrightarrow {F}^{s_q}_{p,2}$ for every $s\geq s_q,$ and the following estimate for all $2\leq q<\infty,$
\begin{align*}
\Vert f\Vert_{{F}^0_{p,2}}&\lesssim  \Vert T'f\Vert_{L^p}= C \Vert u(t,z) \Vert_{L^p_z[\mathbb{S}^{d},\,L^2_t(\mathbb{T})]}\lesssim \Vert u(t,z) \Vert_{L^p_z[\mathbb{S}^{d},\,L^q_t(\mathbb{T})]}.
\end{align*}
\end{proof}
The following lemma, about the $L^2\rightarrow L^p$-operator norm of the spectral projection for the Laplacian on the sphere, is a consequence of the main results in Kwon and Lee \cite{KS}.
\begin{lemma}\label{mainlemmata}
Let $n\in \mathbb{N},$ and let $2\leq p\leq \infty.$ Let us consider the orthogonal projection $H_{n}^d:L^2(\mathbb{S}^d)\rightarrow \mathcal{H}^d_n,$ $n\in \mathbb{N}.$ Then we have
\begin{equation}
    \Vert H_{n}^df \Vert_{L^p(\mathbb{S}^{d})}\leq Cn^{\varkappa_p}\Vert f\Vert_{L^2(\mathbb{S}^{d})},
\end{equation}where $$\varkappa_p:=\frac{d-1}{2}\left(\frac{1}{2}-\frac{1}{p}\right),\quad \textnormal{ if, }2\leq p\leq \frac{2(d+1)}{d-1},$$ and  $$ \varkappa_p: =d\left(\frac{1}{2}-\frac{1}{p}\right)-\frac{1}{2},\quad \textnormal{ if, } p>\frac{2(d+1)}{d-1}. $$  The exponent $\varkappa_p$ is sharp, in the sense that there is not $f\in L^2(\mathbb{S}^d),$ $f\neq 0,$ such that
\begin{equation}
    \Vert H_{n}^df \Vert_{L^p(\mathbb{S}^{d})}\leq C'n^{s}\Vert f\Vert_{L^2(\mathbb{S}^{d})},
\end{equation}
for all $s<\varkappa_p.$
\end{lemma}
Now, with the analysis developed above and by using Theorem \ref{mainlemmata} we will provide a short proof for the  main result of this section.
\begin{proof}[Proof of Theorem \ref{duvantheorem}]
We observe that from Lemma \ref{T1}, $$ C'_{p,s}\Vert f\Vert_{{F}^0_{p,2}}\leq \Vert u(t,z)\Vert_{L^p_{z}(\mathbb{S}^{d}, L^q_{t}(\mathbb{T}))} \leq C_{p,s}\Vert f\Vert_{{F}^s_{p,2}},  $$ so, we only need to estimate the ${F}^s_{p,2}$-norm of the initial data $f,$ by showing that $\Vert f\Vert_{{F}^s_{p,2}}\lesssim \Vert f\Vert_{W^{s}},$ for every $s\geq\varkappa_{p,q}. $ Moreover, by the embedding $W^{s}\hookrightarrow W^{\varkappa_{p,q}}$ for every $s\geq \varkappa_{p,q}$ we only need to check the previous estimate when $s=\varkappa_{p,q}.$ By the condition $2\leq p<\infty,$  together with  the Minkowski integral inequality and Theorem \ref{mainlemmata}, we  give
\begin{align*}
\Vert f\Vert_{{F}^{s_q}_{p,2}} &=\left\Vert \left( \sum_{n=0}^{\infty} {n}^{2s_q}|H_{n}^df(z)|^2  \right)^{\frac{1}{2}}\right\Vert_{L^p}\\&\leq \left( \sum_{n=0}^{\infty} {n}^{2s_q}\Vert H_{n}^df\Vert^2_{L^p}  \right)^{\frac{1}{2}}\\
&\lesssim \left( \sum_{n=0}^{\infty} {n}^{2(s_q+\varkappa_p)}\Vert H_{n}^df\Vert^2_{L^2}  \right)^{\frac{1}{2}}\\
&=\Vert f\Vert_{W^{\varkappa_{p,q}}},
\end{align*}
where we have used that $\varkappa_{p,q}=s_q+\varkappa_p.$
Let us note that the previous estimates are valid for $p=\infty.$ The sharpness of the result for $q=2$ can be proved by choosing $f=H_n^{d}g$ for some arbitrary, but non-trivial  function $g\in L^2(\mathbb{S}^d).$ In this case, $\varkappa_{p,q}=\varkappa_p$ because $s_q=0,$ and we have
\begin{align*}
    \Vert u(t,z) \Vert_{L^{p}_z[\mathbb{S}^{d},\,L^2_t(\mathbb{T})]}&=\Vert H_n^{d}g\Vert_{L^p}
    \leq n^{\varkappa_p}\Vert H_n^{d} g\Vert_{L^2}\\
    &=\Vert g\Vert_{W^{\varkappa_p}}.
\end{align*}The sharpness of this inequality is consequence of the sharpness for the exponent $\varkappa_p$ in Lemma \ref{mainlemmata}. Indeed,  an improvement of the previous estimate
would yield improved estimates for the spectral projection operator $H_n^{d}$, which is not
possible. Thus, we finish the proof.
\end{proof}

\section{The Schr\"odinger equation with potential.  }\label{potential}

In this section we study the problem,
\begin{equation}\label{PVI'}
   \textnormal{(PVI'):}
     \begin{cases}
      iu_{t}=(-\Delta_{\mathbb{S}^d}+V(x,t))u,&\quad (t,x)\in \mathbb{T}\times \mathbb{S}^d,
      \\
       u(0,x)=f(x), &\quad x\in \mathbb{S}^d
     \end{cases}
\end{equation}for a suitable potential $V.$
Our starting point is the following lemma.
\begin{lemma}
The Schr\"odinger propagator satisfies
\begin{equation} \label{aaaa1}
\left\Vert  \int\limits_0 ^t e^{i(t-\tau)\Delta_{\mathbb{S}^d}} \left( G(\cdot,\tau) \right)(x)d\tau\right\Vert_{L^p_x(\mathbb{S}^d,L^2_t(\mathbb{T}))}\leq  C  \Vert G \Vert_{L^{p'}_x(\mathbb{S}^d,L^2_t(\mathbb{T}))}
\end{equation}
\end{lemma}
\begin{proof}
Using the duality argument, we have
\begin{equation}
\begin{array}{l}
\displaystyle \left\Vert  \int\limits_0 ^t e^{i(t-\tau)\Delta_{\mathbb{S}^d}} \left( G(\cdot,\tau)\right)(x)d\tau\right\Vert_{L^p_x(\mathbb{S}^d,L^2_t(\mathbb{T}))}\\
\displaystyle =\sup \left\{ \left|\,  \int \limits_{\mathbb{S}^d} \int\limits_0^{2\pi}  \int\limits_0 ^t e^{i(t-\tau)\Delta_{\mathbb{S}^d}} \left( G(\cdot,\tau)\right)(x) d\tau h(x,t)dt \ dx\right|:\Vert h \Vert_{L^{p'}_x(\mathbb{S}^d,L^2_t(\mathbb{T}))}=1\right\}.
\end{array} \label{duality}
\end{equation}
Now, since $\Delta_{\mathbb{S}^d}$ is a positive operator, this implies $e^{it\Delta_{\mathbb{S}^d}}$ is unitary, this by various applications of Fubini's theorem  we have
$$
\begin{array}{l}
\displaystyle \int\limits_{\mathbb{S}^d} \int\limits_0^{2\pi}  \int\limits_0 ^t e^{i(t-\tau)\Delta_{\mathbb{S}^d}} \left( G(\cdot,\tau)\right)(x) d\tau h(x,t)dt \ dx\\
 \hspace{4cm}= \displaystyle \int\limits_0^{2\pi}  \int\limits_0 ^t  \int _{\mathbb{S}^d} e^{i(t-\tau)\Delta_{\mathbb{S}^d}} \left( G(x,\tau)\right)(x)h(x,t)dx\ d\tau \ dt\\
  \hspace{4cm}= \displaystyle \int\limits_{\mathbb{S}^d}  \left(\int\limits_0 ^t  e^{-i\tau\Delta_{\mathbb{S}^d} } G(x,\tau) d\tau\right) \left(  \int\limits_0^{2\pi}e^{it \Delta_{\mathbb{S}^d}} h(x,t)dt\right) dx,
\end{array}
$$
via Hölder's inequality we have
\begin{equation}
\begin{array}{l}
\displaystyle \left|\,  \int \limits_{\mathbb{S}^d} \int\limits_0^{2\pi}  \int\limits_0 ^t e^{i(t-\tau)\Delta_{\mathbb{S}^d}} \left( G(\cdot,\tau)\right)(x) d\tau h(x,t)dt \ dx\right|\\
\hspace{2cm}\displaystyle \leq \left\Vert  \int\limits_0 ^{2\pi} e^{-i\tau\Delta_{\mathbb{S}^d} } G(x,\tau) d\tau \right\Vert _{ L^2(\mathbb{S}^d)}\left\Vert  \int\limits_0 ^{2\pi} e^{it\Delta_{\mathbb{S}^d} } h(x,t) dt \right\Vert  _{ L^2(\mathbb{S}^d)}.
\end{array}\label{aaa}
\end{equation}
Since
$$\left\Vert  \int\limits_0 ^{2\pi} e^{-i\tau\Delta_{\mathbb{S}^d} } G(x,t) dt \right\Vert  _{ L^2(\mathbb{S}^d)} \leq C\Vert G \Vert_{L^{p'}_x(\mathbb{S}^d,L^2_t(\mathbb{T}))}$$
from \eqref{duality} and \eqref{aaa} we have \eqref{aaaa1}
\end{proof}
We define the Banach space
\begin{equation}
X:={C}(\mathbb{T}, W_x^{s}(\mathbb{S}^d))\cap L^p_x(\mathbb{S}^d, L^2_t(\mathbb{T}))
\end{equation}
via the norm
\begin{equation}
\Vert u\Vert_{X}:=\sup _{t\in \mathbb{T}}\Vert u(t,\cdot)\Vert_{L^2_z}+ \Vert u \Vert_{L^p_x(\mathbb{S}^d,L^2_t(\mathbb{T}))}.
\end{equation}

Now we prove the main result of this section.

\begin{proof}[Proof of Theorem \ref{LiliTheorem}] In view of the Duhamel principle, the solution for (PVI') can be written as
\begin{equation}
    u(t,z)=e^{it \Delta_{\mathbb{S}^d}  }f(x)-\int\limits_{0}^te^{i(t-\tau) \Delta_{\mathbb{S}^d}}(V(\cdot,\tau)u(\cdot))(x)d\tau. \
\end{equation}
We use the standard contraction mapping argument, we need to prove  the nonlinear map
\begin{equation}
\Phi w:=e^{it \Delta_{\mathbb{S}^d}  }f(x)-\int\limits_{0}^te^{i(t-\tau) \Delta_{\mathbb{S}^d}}(V(\cdot,\tau)w(\cdot))(x)d\tau.\label{operator}
\end{equation}
defines a contraction map on $X$.
From the unitary group properties and the estimation given in Theorem \ref{duvantheorem} there exists $C_0>1$ such that
\begin{equation}\label{eee1}
\Vert e^{it \Delta_{\mathbb{S}^d}  }f(x) \Vert_{X}\leq C_0 \Vert f\Vert_{W^{s}(\mathbb{S}^d)}.
\end{equation}
Now we  estimate the integral part in \eqref{operator}. Using \eqref{aaaa1} and the estimation in Theorem \ref{duvantheorem} for $q=2$, we also have
\begin{equation*}
\left\Vert  \int\limits_0 ^t e^{i(t-\tau)\Delta_{\mathbb{S}^d}} \left( V(\tau,\dot)w(\cdot) \right)(x)d\tau\right\Vert_{X}\leq  (C_0+C_0^2)  \Vert  Vw \Vert_{X},
\end{equation*}
Note, $\Vert  Vw \Vert_{L^{p'}_x(\mathbb{S}^d,L^2_t(\mathbb{T}))},\leq \Vert V \Vert_{L^{q}_x(\mathbb{S}^d,L^\infty_t(\mathbb{T}))}\Vert  w \Vert_{L^{p}_x(\mathbb{S}^d,L^2_t(\mathbb{T}))},$ as long as $\frac{1}{q}+\frac{2}{p}=1$, therefore
\begin{equation}
\left\Vert  \int\limits_0 ^t e^{i(t-\tau)\Delta_{\mathbb{S}^d}} \left( V(\tau,\dot)w(\cdot) \right)(x)d\tau\right\Vert_{X}\leq  (C_0+C_0^2)  \Vert V \Vert_{L^{q}_x(\mathbb{S}^d,L^\infty_t(\mathbb{T}))}\Vert  w \Vert_{X}.\label{eee2}
\end{equation}
Hence
\begin{equation}
    \Vert \Phi w \Vert\leq C_0 \Vert f\Vert_{W^{s}(\mathbb{S}^d)}+ (C_0+C_0^2)  \Vert V \Vert_{L^{q}_x(\mathbb{S}^d,L^\infty_t(\mathbb{T}))}\Vert  w \Vert_{X}.
\end{equation}
So, if we take  $\Vert V \Vert_{L^{q}_x(\mathbb{S}^d,L^\infty_t(\mathbb{T}))}$, being small enough, such that
$$(C_0+C_0^2)  \Vert V \Vert_{L^{q}_x(\mathbb{S}^d,L^\infty_t(\mathbb{T}))} \leq \frac{1}{2},$$
we observe, $\Phi$ maps $B=\{w\in X: \Vert w\Vert_X\leq 2(C_0+1)\Vert f\Vert_{W^s(\mathbb{S}^d)}\}$ into itself. Now, if $w, v\in B$, the same argument shows that
\begin{equation}
    \Vert \Phi(w)-\Phi(v) \Vert_{X}\leq \frac{1}{2}\Vert w-v\Vert_{X}
\end{equation}
Thus, by the contraction mapping principle, there exists a unique solution to \eqref{PVI'}. Thus, we finish the proof.

\end{proof}

\bibliographystyle{amsplain}

config: arxiv


TITLE: Average Number of Blocks in a Partition Containing a Large Block
QUESTION [11 upvotes]: Notation

Let $[n]$ denote the set of integers from 1 up to $n$,
inclusive.
Let $A_n$ denote the average number of blocks over all partitions of $[n]$.
Let $B_n$ denote the number of partitions of $[n]$ (the $n$th Bell number).

Question
For any given $n \geq 1$, how small can $m$ be (as a function of $n$) and still ensure
$$
A_{n-m} + 1 < A_n$$
(or, equivalently,
$$
\frac{B_{n+1-m}}{B_{n-m}} < \frac{B_{n+1}}{B_n}-1)?
$$
Remarks
The lefthand side of the inequality comes from fixing a particular block of size $m$ and considering all partitions of $[n]$ containing that fixed block. The idea is that if a partition contains a single large block, the expected number of blocks must fall. For my purposes, smaller values of $m$ are superior, but I would be interested to see anything.
The following reasoning led me to believe $m = \lceil n / A_n \rceil$ would be big enough. The average size of a block in a partition of $[n]$ is at most $\lceil n / A_n \rceil$. Having a single larger-than-average block should reduce the average number of blocks appearing in such a partition. Exploring with Maple shows $\lceil n / A_n \rceil$ is not quite big enough to establish the inequality, but $\lceil n / A_n \rceil + 1$ works for $n \leq 500$.
I keep mentioning averages because I'm hoping there is a quick proof of the kind I attempted in the preceding paragraph. If that vantage point is not helpful, there is the alternative statement of the inequality in terms of Bell numbers. This expression can possibly be analyzed asymptotically (though the Moser-Wyman expansion fails here), but I was rather hoping to avoid it. Still, any suggestions in this direction are welcome.

REPLY [5 votes]: We can think of a partition of $[n+1]$ as being formed by taking a parent partition of $[n]$ and either adding $n+1$ to an existing set or putting it in as a new set (so each parent with $k$ blocks has $k$ children with $k$ blocks and $1$ child with $k+1$ blocks; this is the same coupling used in Mike Spivey's answer to this question).  If we let $X_n$ denote the number of blocks in a random partition of $[n]$ (so $A_n=E(X_n)$), this gives
\begin{eqnarray*}
A_{n+1} &=& \frac{\sum_k (k^2+(k+1)) P(X_n=k)}{\sum_k (k+1) P(X_n=k)} \\
&=& \frac{E(X_n^2)+ A_n+1}{A_n+1} \\
&\geq& \frac{A_n^2+A_n+1}{A_n+1}  \textrm{ (since } E(X_n^2) \geq (E(X_n))^2 \textrm{)}\\
&=& A_n+\frac{1}{A_n+1}
\end{eqnarray*}
So inductively we have
$$A_{n}-A_{n-m} \geq \sum_{j=n-m}^{n-1} \frac{1}{A_j+1} \geq \frac{m}{A_{n-1}+1}$$
since $A_j$ is non-decreasing (as was shown in the linked question).  It follows we can take $m=\lceil A_{n-1}+1 \rceil$.

As was noted in the comments, this bound is far from tight.  If you're interested in asymptotic bounds, then one possibility may be to leave in the term I left out, writing
$$A_{n+1}=A_n+\frac{1}{A_n+1}+\frac{Var(X_n)}{A_n+1}.$$
Both $A_n$ and the variance seem to be well-studied -- Theorem 8.60 of Toufik Mansour's recent book (referencing Theorem VIII.41 in Flajolet and Sedgwick's Analytic Combinatorics) gives their values as $\frac{n}{\log n}(1+o(1))$ and $\frac{n}{\log^2 n} (1+o(1))$ respectively, which would give you $m=\log n$ asymptotically.


TITLE: MLE for lower bound of Uniform Distribution
QUESTION [1 upvotes]: Let $X_1$, $X_2$, . . . , $X_n$ be a random sample from a $Uniform(θ, 1)$ population, where $θ < 1$.
(a) Find the MLE $\widehat{\theta}$ of $θ$.
(b) Find constants c and d (possibly depending on n) such that, c+d$\widehat{\theta}$ is unbiased for $θ$.
The likelihood of $θ$ being greater than $X_1$ is zero, but what is the likelihood of it being lower $X_1$. Is it the $θ^{-n}$ (same as when we try to estimate the higher limit of the Uniform Distribution.
I understand that $θ$ should be the minimum of the sample and 1 should be the max of the sample (ideally), but how to derive it mathematically?

REPLY [1 votes]: The likelihood of a sample $\mathbf x=(x_1,\ldots,x_n)$ in $(0,1)$ is
$$
L(\theta\mid\mathbf x)=\prod_{k=1}^n\frac{\mathbf 1_{[\theta,1]}(x_k)}{1-\theta}=\frac{\mathbf 1_{\theta\leqslant\min\mathbf x}}{(1-\theta)^n}.
$$
Surely you can deduce which value $\hat\theta(\mathbf x)$ of $\theta$ maximizes this.


text:
\begin{document}

\begin{center}
   {\large \bf
   Modules, comodules and cotensor products \\
   over Frobenius algebras
   }
   \\
   \medskip
   {\large Lowell Abrams} \\
   \vspace{1pt}
   {\small
   Department of Mathematics \\
   Rutgers University \\
   New Brunswick, NJ 08903  \\
   \vspace{1pt}
   labrams@math.rutgers.edu
   }
\end{center}

\vspace{.5in}

\begin{abstract}
We characterize noncommutative Frobenius algebras $A$ in terms of the
existence of a coproduct which is a map of left $\Ae$-modules. We show
that the category of right (left) comodules over $A$, relative to this
coproduct, is isomorphic to the category of right (left) modules. This
isomorphism enables a reformulation of the cotensor product of
Eilenberg and Moore as a functor of modules rather than comodules.

We prove that the cotensor product $M \Box N$ of a right $A$-module
$M$ and a left $A$-module $N$ is isomorphic to the vector space of
homomorphisms from a particular left $\Ae$-module $D$ to $N \otimes
M$, viewed as a left $\Ae$-module. Some properties of $D$ are
described. Finally, we show that when $A$ is a symmetric algebra, the
cotensor product $M \Box N$ and its derived functors are given by the
Hochschild cohomology over $A$ of $N \otimes M$.

\vspace{1pt}
\noindent
Keywords: Frobenius algebra, comodule, cotensor product, Hochschild
cohomology
\end{abstract}

\section{Introduction}

Eilenberg and Moore originally introduced the cotensor product $M \Box
N$ and its derived functors Cotor$(M,N)$ on comodules $M, N$ as tools
for the calculation of the homology of the fiber space in a fibration
\cite{EilMoo66}. This paper investigates these functors in the context
where the coalgebra is a Frobenius algebra (defined in section
\ref{sec:FA}).

The Frobenius case is not far removed from that of Eilenberg and
Moore, whose coalgebra is the set of normalized singular chains in
some space $X$; in the presence of sufficient flatness, all the
relevant constructions yield exactly the same data upon passing to
homology [{\it ibid.}]. When the space $X$ under consideration is
compact and oriented, its homology is in fact a Frobenius algebra.

Nevertheless, our approach diverges from that of Eilenberg and Moore
in an important way. The results presented here rest on a new
characterization of Frobenius algebras as algebras possessing a
coassociative counital comultiplication $\delta \colon A \rightarrow A
\otimes A$ which is a map of regular bimodules. (This is formulated
slightly differently as theorem \ref{th:comult} below.)  This
comultiplication is decidedly different from the one used by Eilenberg
and Moore. The relationship between the two coproducts will be
discussed elsewhere.

The Frobenius algebra coproduct, and in particular the element
$\delta(1_A)$, has already begun to find its place in a variety of
contexts. In two dimensional topological quantum field theory, it
gives rise to the handle operator \cite{Abr96}. In quantum cohomology
it provides a generalization of the classical Euler class
\cite{Abr97}. It also plays an important role in the study of quantum
Yang-Baxter equations and, under certain conditions, serves as a
separability idempotent \cite{BeiFonSto97}. Here, we will consider
left submodules of $A \otimes A$ generated by $\delta(1_A)$ and $T
\circ \delta(1_A)$. These will be discussed more later in this
section.

The bimodule property of the Frobenius algebra coproduct implies
another important property of Frobenius algebras, appearing as theorem
\ref{th:modcomod}: The category of right modules over a Frobenius
algebra $A$ is isomorphic to the category of right comodules over
$A$. This result makes it possible to view Eilenberg and Moore's
functors on comodules as functors on modules. Now, using the Snake
Lemma, one can show that the cotensor product is left exact in both
variables. (This also follows from theorem \ref{th:cotensorhom}, of
course.) This suggests that the right module $M \Box N$ should be
expressible as a module of homomorphisms from some left module $D$ to
$N \otimes M$. In fact, this is the case, as stated in theorem
\ref{th:cotensorhom}. The concern is to develop a satisfactory
understanding of the module $D$.

Specifically, $D$ denotes the left $\Ae$-submodule of $A \otimes A$
generated by $T \circ \delta(1_A)$, where $T \colon A \otimes A
\rightarrow A \otimes A$ denotes the canonical involution. This is not
the same as the left $\Ae$-submodule $\delta(A)$ of $A \otimes A$
generated by $\delta(1_A)$. The latter module is a very natural object
to consider, since $\delta$ itself is a left $\Ae$-module map, but the
importance of $D$ in this context is somewhat surprising. Under
certain conditions, delineated in \ref{pr:symmdelta} and
\ref{cor:Ddelta}, $D$ and $\delta(A)$ are in fact the same up to a
canonical involution. But in other cases, such as the one presented
below as example 4.\ref{ex:1}, this is not so.

There are two important corollaries to the main results discussed
above. One (\ref{cor:cotorext} below) is that the right derived
functors of the cotensor product $M \Box N$, \ie Cotor$^*(M,N)$, are
in fact the modules Ext$^*(D,N \otimes M)$. The other
(\ref{cor:cotorHoch} below) is that when $A$ is a symmetric algebra,
the cotensor product $M \Box N$ and its derived functors are given by
the Hochschild cohomology over $A$ of $N \otimes M$.

The author extends heartfelt thanks to Chuck Weibel who, in addition
to being free with helpful advice, is a living index to
\cite{Wei}. Peter May also offered several useful comments and
suggestions.

\vspace{.25\baselineskip}

\noindent
{\bf Notation and Conventions}

All algebras $A$ considered here are assumed to be finite dimensional
as a vector space over their coefficient field $K$, and to possess a
multiplicative identity element $1_A$. We let $\mu \colon A \otimes A
\rightarrow A$ denote the multiplication map. The symbols $A^n$ will
always denote $A^{\otimes n}$, \ie the tensor product of $n$ copies of
$A$, and never the Cartesian product. For any object $X$, we will use
``X'' or ``$\cdot$'' to denote the identity map $X \rightarrow X$, and
the symbols $\cdot \otimes \cdot$ will be abbreviated ``$\cdot
\cdot$''.

\section{Noncommutative Frobenius Algebras}
   \label{sec:FA}

An algebra $A$ is defined to be a {\bf Frobenius algebra} if it
possesses a left $A$-module isomorphism $\lambdal \colon A \rightarrow
A^*$ with its vector space dual.  Here, $A$ is viewed as the left
regular module over itself, and $A^*$ is made a left $A$-module by the
action $(a \cdot \zeta)(b) \, := \, \zeta(ba)$ for any $a,b \in A$ and
$\zeta \in A^*$. It is easy to show that the existence of the
isomorphism of left modules implies the existence of an isomorphism
$\lambdar$ of right modules, where the right module structures are
defined analogously.

There are many equivalent definitions of Frobenius algebras; see
\cite{CurRei} for more information. For our purposes, the new
characterization of Frobenius algebras presented below is very useful.

\begin{thm}
   \label{th:comult}
An algebra $A$ is a Frobenius algebra if and only if it has a
coassociative counital comultiplication $\delta \colon A \rightarrow A
\otimes A$ which is a map of left $\Ae$-modules.
\end{thm}

Here, $\Ae$ denotes the ring $A \otimes A^{\mathrm{op}}$, and $A$ has
the left $\Ae$-action defined by $(b \otimes b') \cdot a:= bab'$.

In many respects, the proof of this result follows the proof of an
analogous result for the commutative case, found in \cite{Abr96}. For
the sake of space, we merely indicate how this proof differs from the
one given there.

\begin{prf}
Assume $A$ denotes a Frobenius algebra with left-module isomorphism
$\lambdal \colon A \rightarrow A^*$. Let $\mut \colon A \otimes A
\rightarrow A$ denote the composition $\mu \circ T$. Define the
comultiplication map $\deltal \colon A \rightarrow A \otimes A$ to be
the composition $(\lambdal^{-1} \otimes \lambdal^{-1}) \circ \mut^*
\circ \lambdal$. With the appropriate adjustments, the discussion in
\cite{Abr96} shows that the following diagram commutes:
\[
\begin{diagram}
   \node{A \otimes A} \arrow{e,t}{\mu}
                      \arrow{s,r}{\cdot \otimes \deltal}
       \node{A} \arrow{s,r}{\deltal} \\
   \node{A \otimes A \otimes A} \arrow{e,t}{\mu \otimes \cdot}
      \node{A \otimes A}
\end{diagram}
\]
In words, $\deltal$ is a map of left $A$-modules.

Using the right-module isomorphism $\lambdar \colon A \rightarrow
A^*$, it is an analogous exercise to define $\deltar$ and show that
this comultiplication map is a map of right modules. Let $\epsilon
\colon A \rightarrow K$ denote $\lambdar(1_A)$. Note that
$\lambdar(1_A) = \lambdal(1_A)$, and thus that $\epsilon$ serves as a
counit for both $\deltar$ and $\deltal$.

Now consider the following diagram:
\[
\begin{diagram}
   \node[2]{A} \arrow{se,t}{\deltar} \\
    \node{A^2} \arrow{se,b}{\deltar \otimes \deltal}
                        \arrow{e,t}{\deltar \otimes \cdot}
                        \arrow{ne,t}{\mu}
       \node{A^3}
                 \arrow{s,r}{\cdot \otimes A \otimes \deltal}
                  \arrow{e,b}{\cdot \otimes \mu}
         \node{A^2}
                   \arrow{s,r}{\deltal}
                   \arrow{se,t}{\cdot \cdot} \\
      \node[2]{A^4}
                        \arrow{e,b}{\cdot \otimes \mu \otimes A}
      \node{A^3}
                    \arrow{e,b}{\cdot \otimes \epsilon \otimes \cdot}
      \node{A^2}
\end{diagram}
\]
This diagram commutes because of the properties of $\deltar$,
$\deltal$ and $\epsilon$ mentioned just above. It follows that
$\deltar \circ \mu$ is the same as the composition of maps from the
far left down and along the bottom row to the lower right-hand
corner. A corresponding diagram shows that $\deltal \circ \mu$ is also
the same as that composition, \ie $\deltar \circ \mu = \deltal \circ
\mu$. Since $A$ has an identity element, we see that $\deltar =
\deltal$.  Define $\delta := \deltar = \deltal$. We have just shown
that this map $\delta \colon A \rightarrow A \otimes A$ is a map of
bimodules, \ie is an $\Ae$-module map, and has a counit.

The remainder of the proof follows as in \cite{Abr96}.
\end{prf}

Throughout the sequel, $\delta$ and $\epsilon$ will denote the
comultiplication and counit respectively. Let $\delta(A)$ denote the
image of $\delta$.

\begin{cor}
   \label{cor:delta}
The map $\delta$ is an injection of left $\Ae$-modules.
\end{cor}
\begin{prf}
By theorem \ref{th:comult}, $\delta$ is a map of left $\Ae$-modules. Since
$\delta$ has a counit, it is certainly injective.
\end{prf}

\section{Modules and Comodules}

We let $1_A \colon K \rightarrow A$ denote the map sending $1_K$ to
$1_A$. Since $X$ and $X \otimes K$ are canonically isomorphic, for any
map $f \colon X \rightarrow X$ we will abuse notation and write $f
\otimes 1_A \, \colon X \rightarrow X \otimes A$ instead of $f \otimes
1_A \, \colon X \otimes K \rightarrow X \otimes A$. When discussing
compositions of maps, the term ``switch'' will always refer to
reversing the order of noninteracting maps.

Suppose $M$ is a right $A$-module with structure map $m \colon M
\otimes A \rightarrow M$. Define the map $\comod_m \colon M
\rightarrow M \otimes A$ to be the composition:
\[
M    \myrightarrow{\cdot \otimes 1_A}
M \otimes A    \myrightarrow{\cdot \otimes \delta}
M \otimes A^2    \myrightarrow{m \otimes \cdot}
M \otimes A
\]

\begin{lemma}
   \label{lem:comodmod} The map $\comod_m$ endows $M$ with the
structure of a right $A$-comodule.
\end{lemma}
\begin{prf}
It is necessary to show that the following diagram commutes:
\begin{equation}
   \label{diag:comod}
\begin{diagram}
   \node{M} \arrow{e,t}{\comod_m} \arrow{s,r}{\comod_m}
      \node{M \otimes A} \arrow{s,r}{\cdot \otimes \delta} \\
   \node{M \otimes A} \arrow{e,t}{\comod_m}
      \node{M \otimes A^2}
\end{diagram}
\end{equation}
Expanding each of the occurrences of $\comod_m$ in accordance with the
definition of that map yields the outer edge of this diagram:
\[
\begin{diagram}
   \node{M}  \arrow{e,t}{\cdot \otimes 1_A}
             \arrow{s,r}{\cdot \otimes 1_A}
      \node{M \otimes A} \arrow{e,t}{\cdot \otimes \delta}
                         \arrow{s,r}{\cdot \otimes \delta}
      \node{M \otimes A^2} \arrow{e,t}{m \otimes \cdot}
                      \arrow{s,r}{\cdot \cdot \otimes \delta}
      \node{M \otimes A}
                      \arrow{s,r}{\cdot \otimes \delta}
   \\
   \node{M \otimes A}  \arrow{e,t}{\cdot \otimes \delta}
                       \arrow{s,r}{\cdot \otimes \delta}
      \node{M \otimes A^2}
                      \arrow{e,t}{\cdot \otimes \delta \otimes \cdot}
      \node{M \otimes A^3}
                      \arrow{e,t}{m \otimes \cdot \cdot}
      \node{M \otimes A^2}
   \\
   \node{M \otimes A^2}
         \arrow{e,t}{\cdot \cdot \otimes 1_A \otimes \cdot}
         \arrow{se,b}{m \otimes \cdot}
      \node{M \otimes A^3}
         \arrow{e,t}{\cdot \otimes A \otimes \delta \otimes \cdot}
         \arrow{n,r}{\cdot \otimes \mu \otimes \cdot}
      \node{M \otimes A^4}
         \arrow{n,r}{\cdot \otimes \mu \otimes A \otimes \cdot}
         \arrow{e,t}{m \otimes \cdot \cdot \cdot}
      \node{M \otimes A^3}
         \arrow{n,r}{m \otimes \cdot \cdot}
   \\
   \node[2]{M \otimes A}
            \arrow{e,b}{\cdot \otimes 1_A \otimes \cdot}
      \node{M \otimes A^2}
            \arrow{ne,b}{\cdot \otimes \delta \otimes \cdot}
\end{diagram}
\]
From left to right and top down, the squares inside this large diagram
commute for the following reasons: Vacuity, coassociativity of
$\delta$, switch, property of the multiplicative identity, $\delta$
being a module map, $m$ being a module map. The hexagon on the bottom
is commutative because it only involves a switch.

It follows that the outer edge forms a commutative square, \ie diagram
(\ref{diag:comod}) is commutative.
\end{prf}

Suppose now that $M$ is a right $A$-comodule, with comodule structure
map $\comod \colon M \rightarrow M \otimes A$. Define the map
$m_{\comod} \colon M \otimes A \rightarrow M$ to be the composition:
\[
M \otimes A  \myrightarrow{\comod \otimes \cdot}
M \otimes A^2    \myrightarrow{\cdot \otimes \mu}
M \otimes A   \myrightarrow{\cdot \otimes \epsilon}
M
\]

\begin{lemma}
   \label{lem:modcomod}
The map $m_{\comod}$ endows $M$ with the structure of a right
$A$-module.
\end{lemma}
\begin{prf}
It is necessary to show that the following diagram commutes:
\begin{equation}
   \label{diag:mod}
\begin{diagram}
   \node{M \otimes A^2}
         \arrow{e,t}{m_{\comod} \otimes \cdot}
         \arrow{s,r}{\cdot \otimes \mu}
      \node{M \otimes A}
         \arrow{s,r}{m_{\comod}}
   \\
   \node{M \otimes A}
         \arrow{e,t}{m_{\comod}}
      \node{M}
\end{diagram}
\end{equation}

Expanding each occurrence of $m_{\comod}$ in accordance with the
definition of that map yields the outer edge of the following diagram:

\[
\begin{diagram}
   \node{M \otimes A^2}
         \arrow{e,t}{\comod \otimes \cdot \cdot}
         \arrow{se,t}{\comod \otimes \cdot \cdot}
         \arrow[3]{s,r}{\cdot \otimes \mu}
      \node{M \otimes A^3}
         \arrow{e,t}{\cdot \otimes \mu \otimes \cdot}
         \arrow{se,t}{\comod \otimes \cdot \cdot \cdot}
      \node{M \otimes A^2}
         \arrow{e,t}{\cdot \otimes \epsilon \otimes \cdot}
         \arrow{se,t}{\comod \otimes \cdot \cdot}
      \node{M \otimes A}
         \arrow{se,t}{\comod \otimes \cdot}   \\
   \node[2]{M \otimes A^3}
         \arrow{e,t}{\cdot \otimes \delta \otimes  A \otimes \cdot}
         \arrow{se,t}{\cdot \otimes \mu \otimes \cdot}
         \arrow[2]{s,r}{\cdot \cdot \otimes \mu}
      \node{M \otimes A^4}
         \arrow{e,t}{\cdot \otimes A \otimes \mu \otimes \cdot}
      \node{M \otimes A^3}
         \arrow{e,t}{\cdot \cdot \otimes \epsilon \otimes \cdot}
      \node{M \otimes A^2}
         \arrow{s,r}{\cdot \otimes \mu}  \\
   \node[3]{M \otimes A^2}
         \arrow{ne,t}{\cdot \otimes \delta \otimes \cdot}
         \arrow[2]{e,t}{\cdot \otimes \mu}
      \node[2]{M \otimes A}
         \arrow{s,r}{\cdot \otimes \epsilon}\\
   \node{M \otimes A}
         \arrow{e,t}{\comod \otimes \cdot}
      \node{M \otimes A^2}
         \arrow[2]{e,t}{\cdot \otimes \mu}
      \node[2]{M \otimes A}
          \arrow{ne,t}{\cdot \cdot}
          \arrow{e,t}{\cdot \otimes \epsilon}
      \node{M}
\end{diagram}
\]
The subdiagrams of this diagram are commutative for the following
reasons: In the top row of squares, the leftmost square expresses the
comodule property of $\comod$. The other two squares simply involve
switches, as does the large square on the far left. The square in the
center (between the second and third rows of maps) uses the module
property of $\delta$. The square to its right uses the counit property
of $\epsilon$. The large pentagon on the bottom expresses the
associativity of $\mu$. The triangle in the lower right hand corner is
vacuous.

It follows that the outer edge forms a commutative square, \ie diagram
(\ref{diag:mod}) is commutative.
\end{prf}

Lemmas \ref{lem:comodmod} and \ref{lem:modcomod} show that there are
canonical maps between the category of modules over $A$ and the
category of comodules over $A$. In fact, these provide an isomorphism.

\begin{thm}
   \label{th:modcomod}
The category of right modules over a Frobenius algebra $A$ is
isomorphic to the category of right comodules over $A$.
\end{thm}
\begin{prf}
First we will show that the constructions $m \mapsto \comod_m$ and
$\comod \mapsto m_{\comod}$ are mutual inverses. Then we will show
that every module map is a comodule map for the corresponding comodule
structures, and vice-versa.

Suppose $m \colon M \otimes A \rightarrow M$ is a right module
structure map. Consider the following diagram:

\[
\begin{diagram}
   \node{M \otimes A}
         \arrow{e,t}{\cdot \otimes 1_A \otimes \cdot}
         \arrow{s,r}{\cdot \otimes \delta}
      \node{M \otimes A^2}
         \arrow{e,t}{\cdot \otimes \delta \otimes \cdot}
         \arrow{s,r}{\cdot \cdot \otimes \delta}
      \node{M \otimes A^3}
         \arrow{e,t}{m \otimes \cdot \cdot}
         \arrow{s,r}{\cdot \cdot \cdot \otimes \delta}
      \node{M \otimes A^2}
         \arrow{se,t}{\cdot \otimes \mu}
         \arrow{s,r}{\cdot \cdot \otimes \delta}
   \\
   \node{M \otimes A^2}
         \arrow{e,t}{\cdot \otimes 1_A \otimes \cdot \cdot}
         \arrow{s,r}{\cdot \cdot \cdot}
      \node{M \otimes A^3}
         \arrow{e,t}{\cdot \otimes \delta \otimes A \otimes \cdot}
         \arrow{sw,b}{\cdot \otimes \mu \otimes \cdot}
      \node{M \otimes A^4}
         \arrow{e,t}{m \otimes \cdot \cdot \cdot}
         \arrow{s,l}{\cdot \otimes A \otimes \mu \otimes \cdot}
      \node{M \otimes A^3}
         \arrow{s,l}{\cdot \otimes \mu \otimes \cdot}
      \node{M \otimes A}
         \arrow[2]{s,r}{\cdot \otimes \epsilon}
   \\
   \node{M \otimes A^2}
          \arrow[2]{e,b}{\cdot \otimes \delta \otimes \cdot}
      \node[2]{M \otimes A^3}
          \arrow{e,b}{m \otimes \cdot \cdot}
          \arrow{se,b}{\cdot \cdot \otimes \epsilon \otimes \epsilon}
      \node{M \otimes A^2}
          \arrow{ne,b}{\cdot \otimes \epsilon \otimes \cdot}
    \\
    \node[4]{M \otimes A}
          \arrow{e,t}{m}
      \node{M}
\end{diagram}
\]

\noindent
The composition of maps across the top and down the right is nothing
other than the definition of the map $m_{\comod_m} \colon M \otimes A
\rightarrow M$. Since the composition of maps down the left and across
the bottom is $m$ itself (by the counit property), the identity
$m_{\comod_m} \equiv m$ will follow if the diagram is
commutative. This is in fact the case, because the subdiagrams are
commutative for the following reasons: With the exception of those
that will now be mentioned explicitly, the subdiagrams are
commutative simply because they involve switches. The triangle on the
lower left uses the multiplicative unit property. The square to its
right expresses the module property of $\delta$. The square on the far
upper right is commutative because it is essentially the outer edge of
the following diagram:
\[
\begin{diagram}
   \addtolength{\dgARROWLENGTH}{-1.4em}
   \addtolength{\dgVERTPAD}{-1ex}
   \node[2]{A^2}
         \arrow{sw,t}{\cdot \otimes \delta}
         \arrow{s,r}{\mu}
         \arrow{se,t}{\mu}
   \\
   \node{A^3}
         \arrow{se,b}{\mu \otimes \cdot}
      \node{A}
         \arrow{s,r}{\delta}
      \node{A}
   \\
   \node[2]{A^2}
         \arrow{ne,b}{\epsilon \otimes \cdot}
\end{diagram}
\]
This latter diagram is commutative because the square on the left
expresses the module property of $\delta$, and the square on the right
express the counit property of $\epsilon$.

It follows that $m_{\comod_m} \equiv m$. Suppose, on the other hand,
that $\comod \colon M \rightarrow M \otimes A$ is a comodule
structure. We now show that $\comod_{m_{\comod}} \equiv
\comod$. Consider the following diagram:
\[
\begin{diagram}
   \node{M}
         \arrow{e,t}{\cdot \otimes 1_A}
         \arrow{se,b}{\comod}
      \node{M \otimes A}
         \arrow{e,t}{\cdot \otimes \delta}
         \arrow{se,b}{\comod}
      \node{M \otimes A^2}
         \arrow{e,t}{\comod \otimes \cdot \cdot}
      \node{M \otimes A^3}
         \arrow{e,t}{\cdot \otimes \mu \otimes A}
      \node{M \otimes A^2}
         \arrow{s,r}{\cdot \otimes \epsilon \otimes \cdot}
   \\
   \node[2]{M \otimes A}
         \arrow{e,b}{\cdot \otimes \cdot \otimes 1_A}
      \node{M \otimes A^2}
         \arrow{ne,b}{\cdot \otimes A \otimes \delta}
         \arrow{e,b}{\cdot \otimes \mu}
      \node{M \otimes A}
          \arrow{ne,b}{\cdot \otimes \delta}
          \arrow{e,b}{\cdot \cdot}
      \node{M \otimes A}
\end{diagram}
\]
From left to right, the subdiagrams are commutative for the following
reasons: Switch, switch, the module property of $\delta$, the counit
property of $\epsilon$. Because the composition of maps across the top
and down the right of this diagram is simply the definition of
$\comod_{m_{\comod}}$, and the composition of maps down the left and
across the bottom is just $\comod$ (by the unit property of $1_A$), we
see that $\comod_{m_{\comod}} \equiv \comod$.

Suppose that $M$ and $N$ are right $A$-modules with module structure
maps $m$ and $n$ respectively. In order to verify that a map $f \colon
M \rightarrow N$ of right modules is also a map of right comodules
(for the corresponding comodule structures), consider the following
diagram:
\[
\begin{diagram}
   \node{M}
      \arrow{e,t}{\cdot \otimes 1_A}
      \arrow{s,r}{f}
   \node{M \otimes A}
      \arrow{e,t}{\cdot \otimes \delta}
      \arrow{s,r}{f \otimes \cdot}
   \node{M \otimes A^2}
      \arrow{e,t}{m \otimes \cdot}
      \arrow{s,r}{f \otimes \cdot \cdot}
   \node{M \otimes A}
     \arrow{s,r}{f \otimes \cdot}
  \\
   \node{N}
      \arrow{e,t}{\cdot \otimes 1_B}
   \node{N \otimes A}
      \arrow{e,t}{\cdot \otimes \delta}
   \node{N \otimes A^2}
      \arrow{e,t}{n \otimes \cdot}
   \node{N \otimes A}
\end{diagram}
\]
Two of the subdiagrams simply involve switches. The third is
commutative because $f$ is a map of modules. Thus, the outer edges
form a commutative diagram as well. But this diagram asserts that $f$
is a map of comodules, where the comodule structure maps are
$\comod_m$ and $\comod_n$.

If $f \colon M \rightarrow N$ is assumed to be a map of right
comodules, where the comodule structure maps are $\comod$ and
$\comod'$, then by reasoning analogous to that of the previous
paragraph, the following diagram shows that $f$ is a map of right
modules:
\[
\begin{diagram}
   \node{M \otimes A}
      \arrow{e,t}{\comod \otimes \cdot}
      \arrow{s,r}{f \otimes \cdot}
   \node{M \otimes A^2}
      \arrow{e,t}{\cdot \otimes \mu}
      \arrow{s,r}{f \otimes \cdot \cdot }
   \node{M \otimes A}
      \arrow{e,t}{\cdot \otimes \epsilon}
      \arrow{s,r}{f \otimes \cdot}
   \node{M}
     \arrow{s,r}{f}
  \\
   \node{N \otimes A}
      \arrow{e,t}{\comod' \otimes \cdot}
   \node{N \otimes A^2}
      \arrow{e,t}{\cdot \otimes \mu}
   \node{N \otimes A}
      \arrow{e,t}{\cdot \otimes \epsilon}
   \node{N}
\end{diagram}
\]
This completes the proof.
\end{prf}

With appropriate changes, all the results and proofs in this section
apply to left modules and left comodules as well.

\section{Cotensor Product}
   \label{sect:cotensor}

Suppose that $M$ is a right $A$-module with module structure map $m$,
and that $N$ is a left $A$-module with module structure map
$n$. By theorem \ref{th:modcomod}, $M$ is a right comodule with
structure map $\comod_m$ and $N$ is a left comodule with structure map
$\comod_n$. Let $\phi$ denote the map
\[
   \phi := \ \comod_m \! \otimes N \, - \, M \otimes \! \comod_n
        \, \colon \, M \otimes N
        \longrightarrow
        M \otimes A \otimes N .
\]
The {\bf cotensor product} \cite{EilMoo66} $M \Box N$ of $M$ and $N$
is defined to be the kernel of $\phi$.

Viewing $A$ as both the right and left regular modules over itself
(\ie the module structure maps are both $\mu$), we can form $A \Box
A$. Note that $\comod_{\mu}$ is just the map $\delta$, by the module
property of $\delta$.

\begin{prop}
   \label{pr:ABoxA}
The cotensor product $A \Box A$ is exactly $\delta(A)$.
\end{prop}
\begin{prf}
By the definition of $\phi$, to show that $\delta(A) \subseteq A \Box
A$ it suffices to show that the two maps $(\comod_{\mu} \otimes A)
\circ \delta$ and $(A \otimes \comod_{\mu}) \circ \delta$ are the
same. But these two maps are just $(\delta \otimes A) \circ \delta$
and $(A \otimes \delta) \circ \delta$, respectively. These are the
same, by the coassociativity of $\delta$.

Now consider any element $x := \sum_i a_i \otimes b_i \, \in A \Box
A$. We have $(\delta \otimes A)x = (A \otimes \delta)x$, and thus
\[
x \, = \, (\epsilon \otimes A^{\otimes 2}) \circ (\delta \otimes A)x
\, = \, (\epsilon \otimes A^{\otimes 2}) \circ (A \otimes \delta)x
\, = \, \sum_i \epsilon(a_i)\delta(b_i) \, .
\]
It follows that $A \Box A \subseteq \delta(A)$.
\end{prf}

\begin{definition}
Let $D$ denote the left $\Ae$-submodule of $A \otimes A$ generated by
$T \circ \delta(1_A)$.  Note that $D$ and $\delta(A)$ (see corollary
\ref{cor:delta} above) are different objects.
\end{definition}

For any Frobenius algebra $A$, the map $\eta \colon A \otimes A
\rightarrow K$ defined by $\eta(a \otimes b) := \lambdal(1_A)(ab)$ is
a nondegenerate associative bilinear form \cite{CurRei}. If $\eta
\equiv \eta \circ T$, then $A$ is called a {\bf symmetric algebra}
[{\it ibid}].

\begin{prop}
   \label{pr:symmdelta} If $A$ is a symmetric algebra, then $D$ and
$\delta(A)$ are the same left $\Ae$-module.
\end{prop}
\begin{prf}
It suffices to show that if $A$ is a symmetric algebra, then
$\delta(1_A)$ is symmetric, \ie $T \circ \delta(1_A) =
\delta(1_A)$. Let $e_1, \ldots , e_n$ denote a basis for $A$, and let
$e_1^{\#}, \ldots , e_n^{\#}$ denote the dual basis of $A$ relative to
$\eta$, \ie the basis satisfying $\eta(e_i^{\#} \otimes e_j) =
\delta_{ij}$. The proof of proposition 5 in \cite{Abr96} (bearing in
mind the adjustments made in the proof of theorem \ref{th:comult} here
for noncommutativity) shows that $\delta(1_A) \ = \ \sum_j e_j \otimes
e_j^{\#}$.  Since, by assumption, we have $\eta(e_i^{\#} \otimes e_j)
= \eta(e_j \otimes e_i^{\#})$, a change of basis shows that
$\delta(1_A) \ = \ \sum_i e_i^{\#} \otimes e_i = T \circ \delta(1_A)$.
\end{prf}

\begin{cor}
   \label{cor:Ddelta}
If $A$ is commutative or semisimple or a group algebra then $D$ and
$\delta(A)$ are the same left $\Ae$-module.
\end{cor}
\begin{prf}
If $A$ is commutative then it is surely a symmetric algebra. Thus the
hypothesis of proposition \ref{pr:symmdelta} is automatically
satisfied.

By Wedderburn's first structure theorem, to prove the result in the
case when $A$ is semisimple it suffices to assume that $A$ is a matrix
ring. In that case, $A$ has a Frobenius algebra structure given by the
map $\lambdal(1_A)(a) := \mathrm{Tr}(a)$. It is an easy exercise to
show that this provides $A$ with the structure of a symmetric algebra.

In the case of a group algebra $A$ over group $G$, the Frobenius
algebra structure is given by the map which returns the coefficient of
the identity element. The coproduct then sends $1_A$ to $\sum_{g \in
G}g \otimes g^{-1}$, which is clearly symmetric.
\end{prf}

When $A$ is not a symmetric algebra, proposition \ref{pr:symmdelta}
does not necessarily apply.

\begin{example}
   \label{ex:1}
Let $A$ denote the exterior algebra on two generators, $x$ and
$y$. Then
\[
\delta(1_A) \, = \, 1_A \otimes xy \, + \, xy \otimes 1_A
          \, - \, x \otimes y \, + \, y \otimes x ,
\]
and $\delta(A)$ has the basis
\[
   \left\{ \delta(1_A),\ \
      x \otimes xy \, + \, xy \otimes x,\ \, \
      y \otimes xy \, + \, xy \otimes y,\ \, \
      xy \otimes xy \right\} \, ,
\]
whereas $D$ has the basis
\[
   \left\{ T \circ \delta(1_A),\ \
      x \otimes xy \, - \, xy \otimes x,\ \, \
      y \otimes xy \, - \, xy \otimes y,\ \, \
      xy \otimes xy \right\} \, .
\]
\end{example}

\vspace{3pt}

Given a right module $M$ and a left module $N$ as above, endow $N
\otimes M$ with the obvious left $\Ae$-module structure. Let $\Hom{D}$
denote the vector space of left $\Ae$-module maps.

\begin{thm}
   \label{th:cotensorhom}
There is a vector space isomorphism
\[
M \Box N \ \cong \ \Hom{D} \, .
\]
\end{thm}
\begin{prf}
Note first that an element $f \in \Hom{D}$ is determined by its value
on $T \circ \delta(1_A)$, the generator of $D$.

The following diagram is commutative, since $f$ is a map of modules:
\[
\begin{diagram}
   \addtolength{\dgARROWLENGTH}{-1.4em}
    \node{K}
         \arrow[2]{e,b}{\delta(1_A) \, \otimes \, T \circ \delta(1_A)}
      \node[2]{A^4}
         \arrow[3]{e,b}{(\cdot \cdot \otimes \mu) \circ T_{1432} \
                   - \ (\cdot \otimes \mu \otimes \cdot)}
         \arrow{s,r}{\cdot \cdot \otimes f}
      \node[3]{A^3}
         \arrow{s,r}{\cdot \otimes f}
   \\
   \node[3]{A^2 \otimes N \otimes M}
         \arrow[3]{e,t}{(\cdot \cdot \otimes m) \circ T_{1432} \
                   - \ (\cdot \otimes n \otimes \cdot)}
      \node[3]{A \otimes N \otimes M}
\end{diagram}
\]
By the comodule property of $\delta$, the composition of maps across
the top of the diagram is $0$. Since the composition of maps from the
upper left, down and across the bottom is $T_{132} \circ \phi \circ f
\left[T \circ \delta(1_A) \right] $, it follows that $f \left[ T \circ
\delta(1_A) \right] \in M \Box N$. Thus, there is a well defined
injective map $\sigma \colon \Hom{D} \rightarrow M \Box N$ sending $f
\mapsto f \left[ T \circ \delta(1_A) \right]$. Since each element $e
\in N \otimes M$ defines a unique $\Ae$-module map $\tau(e) \colon T
\circ \delta(1_A) \mapsto e$, restriction of $\tau$ to $M \Box N$
provides an inverse to $\sigma$.
\end{prf}

Allowing for abuse of notation, define the cotensor product functor
$\Box_A$ by $\Box_A \colon M \otimes N \mapsto M \Box N$, and let
$\mathrm{Cotor}^i_A(M,N)$ denote its right derived functors. Let
$H^i(A, -)$ denote the Hochschild cohomology functors.

\begin{cor}
   \label{cor:cotorext}
Over a Frobenius algebra $A$, the Cotor functor is given by
\[
\mathrm{Cotor}^*_A(M,N) \cong \mathrm{Ext}^*_{A^e}(D,N \otimes M) \, .
\]
\end{cor}
\begin{prf}
In light of theorem \ref{th:cotensorhom}, this is purely a matter of
definitions.
\end{prf}

\begin{cor}
   \label{cor:cotorHoch}
If $A$ is a symmetric algebra, then cotensor product and its derived
functors are Hochschild cohomology, \ie
\[
\mathrm{Cotor}^*_A(M,N) \cong H^*(A,N \otimes M)\, .
\]
\end{cor}
\begin{prf}
By corollary \ref{cor:Ddelta} we have $\delta(1_A) = T \circ
\delta(1_A)$ and thus $D = \delta(A)$. Since, by corollary
\ref{cor:delta}, $\delta$ is an injective map of left $\Ae$-modules
(determined by its value on $\delta(1_A)$), $D$ and $A$ are isomorphic
as $\Ae$-modules. It follows from theorem \ref{th:cotensorhom} that $M
\Box N \cong \Hom{A}$. But this is exactly $H^0(A,N \otimes M)$
\cite[pg. 301]{Wei}. Since $H^*(A,-) \cong \mathrm{Ext}^*_{A^e}(A,-)$
[{\it ibid.} pg. 303], this corollary follows from \ref{cor:cotorext}.
\end{prf}


TITLE: We throw a die 5 times probability question
QUESTION [0 upvotes]: We throw a die $5$ times, what is the probability of throwing $6$ three consecutive times but not four consecutive times?
My thought process:
We know that the probability of a die showing $6$ is $\frac{1}{6}$, that the die shows any other number is $\frac{5}{6}$.
{$666XX$} the probability for the sequence of throws on the left would be: $\frac{1}{6^3}\cdot \frac{5}{6}\cdot\frac{5}{6}$.
Then the probability for the sequence of throws {$X666X$}: $\frac{5}{6}\cdot\frac{1}{6^3}\cdot\frac{5}{6}$
And then with the last sequence we get: $\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{1}{6^3}$.
We can use the theorem that $P(A\cap B)= \{\}\implies P(A\cup B)= P(A)+P(B)$
If I sum them all together I get clearly a wrong answer.
The solution should be:
$$\frac{1 \cdot 1\cdot 1 \cdot 5 \cdot 6}{6^5}+\frac{5\cdot1\cdot1\cdot1\cdot5}{6^5}+\frac{6\cdot5\cdot1\cdot1\cdot1}{6^5}=\frac{85}{7776}=0.0109.$$
However, I do not understand how to get to this solution.

REPLY [4 votes]: In the sequence
$$X-X-6-6-6$$
the first occurrence can be any number, thus with probability 1.
The same happens in the sequence $6-6-6-X-X$ for the last occurrence.
the probability of these two sequences is
$$1\cdot\frac{5}{6}\cdot \frac{1}{6^3}$$
while for the second sequence you showed you have a probability of
$$\frac{5}{6}\cdot \frac{1}{6^3}\cdot\frac{5}{6}$$
thus the total probability is
$$2\cdot\frac{5}{6}\cdot \frac{1}{6^3}+\frac{5}{6}\cdot \frac{1}{6^3}\cdot\frac{5}{6}=\frac{85}{7776}$$


TITLE: Quick General Question regarding Interior Point Methods
QUESTION [1 upvotes]: So I noticed that equality constraints cannot be treated with interior barriers, which means we can only work with inequality constraints. I know that a trick sometimes used is to double the amount of inequality constraints by keeping $g_i(x) \le 0$ and generating new $-g_i^{new} \ge 0$. However I am guessing this will not work with interior barrier algorithms right?
This makes me wonder whether this also applies to the "basic interior point method?" for which i have included a photo of the algorithm (not sure if there is a more specific name for this provided algorithm)?
So to be more specific with my question, can i also use this algorithm using only equality constraints? If i can, does this mean we remove the 2nd column and 2nd row from the matrix in step b), as well as the 2nd entry in the left and right hand vector? Or do we simply set all inequalities $g_i(x) \le 0$ as being nothing (zero) and move along as if nothing were different?
I am trying to get this to work with my optimization problem
$$
\min_{\theta \in \mathbb{C}^{n} } Tr((\textbf{HH}^H)^{-1}) \; \; s.t. \; \; Re^2(\theta_i)+Im^2(\theta_i) -1 =0 \; \; \forall i \in \{1,\ldots,64\}
$$
Where $\textbf{H}$ is a linear function of $\theta$, and $n=64$
Super appreciate any help, I am a total noob to this stuff.

REPLY [1 votes]: Equality constraints are naturally supported by IPM. They simply appear "as is" in the KKT conditions and linearized in the linear system (augmented or KKT system) resulting from the Newton step.
This is actually true for all Newton-based methods (IPM, SQP, etc). The difference is how they handle inequality constraints. Modern IPM add slacks (which results in equality constraints only), then use a barrier term for the bounds (original variables + slacks). In modern formulations, all the iterates are interior wrt the primal bounds (original variables + slacks) and the dual bounds, not wrt the original constraints.
Edit: I'm guessing that $h$ in your screenshot are equality constraints.


text:
\begin{definition}[Definition:Isolated Point (Complex Analysis)]
Let $S \subseteq \C$ be a [[Definition:Subset|subset]] of the [[Definition:Real Number|set of real numbers]].
Let $z \in S$.
Then $z$ is an '''isolated point of $S$''' {{iff}} there exists a [[Definition:Neighborhood (Complex Analysis)|neighborhood of $z$]] in $\C$ which contains no points of $S$ except $z$:
:$\exists \epsilon \in \R_{>0}: \map {N_\epsilon} z \cap S = \set z$
\end{definition}

config: wiki