text:
\begin{document}
\title[Riesz spaces with topologically full center]{Characterization of Riesz spaces with topologically full center}
\author{\c{S}afak Alpay}
\address{Mathematics Department, Middle East Technical University\\
06531 Ankara, Turkiye}
\email{safak@metu.edu.tr}
\author{Mehmet Orhon}
\address{Department of Mathematics and Statistics, University of New Hampshire\\
Durham, NH 03824, USA}
\email{mo@unh.edu}
\urladdr{}
\thanks{.}
\thanks{.}
\date{February 10, 2014}
\subjclass[2000]{Primary 47B38, 46B42; Secondary 47B60, 46H25}
\keywords{Riesz space, ideal center, orthomorphism, Arens extension}

\begin{abstract}
Let $E$ be a Riesz space and let $E^{\sim}$ denote its order dual. The
orthomorphisms $Orth(E)$ on $E,$ and the ideal center $Z(E)$ of $E,$ are
naturally embedded in $Orth(E^{\sim})$ and $Z(E^{\sim})$ respectively. We
construct two unital algebra and order continuous Riesz homomorphisms
\[
\gamma:((Orth(E))^{\sim})_{n}^{\sim}\rightarrow Orth(E^{\sim})\text{ }
\]
and
\[
m:Z(E)^{\prime\prime}\rightarrow Z(E^{\sim})
\]
that extend the above mentioned natural inclusions respectively. Then, the
range of $\gamma$ is an order ideal in $Orth(E^{\sim})$ if and only if $m$ is
surjective. Furthermore, $m$ is surjective if and only if $E$ has a
topologically full center. (That is, the $\sigma(E,E^{\sim})$-closure of
$Z(E)x$ contains the order ideal generated by $x$ for each $x\in E_{+}.$) As a
consequence, $E$ has a topologically full center $Z(E)$ if and only if
$Z(E^{\sim})=\pi\cdot Z(E)^{\prime\prime}$ for some idempotent $\pi\in
Z(E)^{\prime\prime}.$

\end{abstract}
\maketitle

\section{Introduction}

Let $E$ be a Banach lattice and let $Z(E)$ be its (ideal ) center. In general
$Z(E)$ is a subalgebra and sublattice of $Z(E^{\prime})$, the center of the
Banach dual $E^{\prime}$ of $E.$ It is possible to extend this embedding to a
contractive algebra and lattice homomorphism of $Z(E)^{\prime\prime}$ into
$Z(E^{\prime}).$ It is natural to ask when the homomorphism would be onto
$Z(E^{\prime})$. It is clear that $Z(E)$ has to be large, since $Z(E^{\prime
})$ is always large. It turns out that the concept of largeness best suiting
the center $Z(E)$ in this problem is that $Z(E)$ should be topologically full
in the sense of Wickstead \cite{W1}. Namely, for each $x\in E_{+},$ the
closure of $Z(E)x$ is the closed ideal generated by $x.$ Then it is shown that
the homomorphism is onto $Z(E^{\prime})$ if and only if $Z(E)$ is
topologically full \cite[Corollary 2]{O1}.

Our purpose in this $\ $paper is to consider the corresponding problem for a
Riesz space $E$ with point separating order dual $E^{\sim}$. In the case of
Riesz spaces however, there are, in general, two different algebras that act
on $E$ which should be considered. Namely, $Orth(E),$ the algebra of the
orthomorphisms on $E,$ and its subalgebra and (order) ideal $Z(E),$ the center
of $E.$ In the case of Banach lattices $Orth(E)=Z(E),$ therefore this problem
does not arise. Recall that an orthomorphism on $E$ is an order bounded
operator on $E$ that preserves bands, and $Z(E)$ is the ideal generated by the
identity operator on $E$ in the Riesz space $Orth(E).$ In fact $Orth(E)$ is an
$f$-algebra. The ideal center $Z(E)$, on the other hand, is a normed
$AM$-lattice where the identity operator is the order unit of the
$AM$-lattice. Similar to the Banach lattice case, $Orth(E)$ is embedded in
$Orth(E^{\sim})$ as an $f$-subalgebra and sublattice. Also as before, $Z(E)$
is embedded in $Z(E^{\sim})$ as a subalgebra and sublattice. We construct two
unital algebra and order continuous lattice homomorphisms
\[
\gamma:(Orth(E)^{\sim})_{n}^{\sim}\rightarrow Orth(E^{\sim})
\]
and
\[
m:Z(E)^{\prime\prime}\rightarrow Z(E^{\sim})
\]
that extend the two embeddings mentioned above respectively. Then the range of
$\gamma$ is an order ideal in $Orth(E^{\sim})$ if and only if $m$ is onto
$Z(E^{\sim})$ (Corollary \ref{c2}). Also, $m$ is onto $Z(E^{\sim})$ if and
only if $E$ has a topologically full center (Proposition \ref{p5}). That is
for each $x\in E_{+}$, the $\sigma(E,E^{\sim})$-closure of $Z(E)x$ in $E$
contains the ideal generated by $x.$ It follows that $E$ has a topologically
full center $Z(E)$ if and only if the ideal center of its order dual $E^{\sim
}$ is given as $Z(E^{\sim})=\pi\cdot Z(E)^{\prime\prime}$ for some idempotent
$\pi\in Z(E)^{\prime\prime}.$

We point out that the proofs of the above mentioned results differ from those
used in the Banach lattice case \cite{O1}. The method used in this paper owes
a lot to the work and results of Huijsmans and de Pagter \cite{HdP1} on the
bidual of an $f$-algebra.

In the Banach lattice case it is shown that if $Z(E)$ is topologically full
then it is maximal abelian \cite[Corollary 3]{O1}. Wickstead \cite{W2} showed
that the converse is not true. He constructed an interesting $AM$-lattice with
a center that is maximal abelian but is not topologically full. In the case of
Riesz spaces, even when $Z(E)$ is topologically full, it need not be maximal
abelian. In fact, when $Z(E)$ is topologically full, its commutant in the
order bounded operators is $Orth(E)$ (Corollary 4).

If $E$ is a Riesz space, by $E^{\sim}$ we will denote the Riesz space of order
bounded linear functionals on $E$. All Riesz spaces considered in this paper are assumed to have separating order duals. $E_{n}^{\sim}$ will
denote the order continuous linear functionals in $E^{\sim}$. We let
$L_{b}(E,F)$ denote the space of order bounded linear operators from the Riesz
space $E$ into the Riesz space $F.$ When $T:E\rightarrow F$ is an order
bounded operator between two Riesz spaces, the adjoint of $T$ carries
$F^{\sim}$ into $E^{\sim}$ and we will denote it by $T^{\prime}.$ The dual of
a normed space will be denoted by $E^{\prime}.$ In all terminology concerning
Riesz spaces we will adhere to the definitions in \cite{AB} and \cite{Z}.

Let us recall that for any associative algebra $A,$ a multiplication (called
the Arens multiplication) can be introduced in the second algebraic dual
$A^{\ast\ast}$ of $A$ \cite{Ar}. This is accomplished in three steps: given
$a,b\in A,$ $f\in A^{\ast}$ and $F,G\in A^{\ast\ast}$, one defines $f\cdot
a\in A^{\ast}$, $F\cdot f\in A^{\ast}$, and $F\cdot G\in A^{\ast\ast}$ by the
equations
\begin{align*}
(f\cdot a)(b)  &  =f(ab)\\
(F\cdot f)(a)  &  =F(f\cdot a)\\
(F\cdot G)(f)  &  =F(G\cdot f)
\end{align*}
For any Archimedean $f$-algebra $A,$ the space $(A^{\sim})_{n}^{\sim}$ is an
Archimedean $f$-algebra with respect to the Arens multiplication \cite{HdP1}.

We denote for any $F\in(A^{\sim})_{n}^{\sim}$ the mapping $f\rightarrow F\cdot
f$ by $\nu_{F}.$ The map $\nu_{F}$ is an orthomorphism on $A^{\sim}
$\cite{HdP1}. The mapping $\nu:(A^{\sim})_{n}^{\sim}\rightarrow Orth(A^{\sim
}),$ defined by $\nu(F)=\nu_{F}$ is an algebra and Riesz homomorphism for any
Archimedean $f$-algebra $A.$ Moreover $\nu$ is onto $Orth(A^{\sim})$ if and
only if $(A^{\sim})_{n}^{\sim}$ has a unit element. In that case, $\upsilon$
is injective by Theorem 5.2 in \cite{HdP1}. The main purpose of this $\ $paper
is to extend the latter result to an arbitrary Riesz space.

\section{The Arens Homomorphism}

Let $E$ be a Riesz space and consider the bilinear map
\begin{equation}
\ Orth(E)\,\times E\rightarrow E \tag{1}\label{eq1}
\end{equation}
defined by $(\pi,x)\rightarrow\pi(x)$ for each $\pi\in\ Orth(E)$
\thinspace\ and $x\in E.$ Related to (\ref{eq1}), we define the following
bilinear maps:
\begin{equation}
E\times E^{\sim}\rightarrow(\ Orth(E)\,)^{\sim}::(x,f)\rightarrow\psi
_{x,f}:\psi_{x,f}(\pi)=f(\pi x) \tag{2}\label{eq2}
\end{equation}
\begin{equation}
E^{\sim}\times(\ Orth(E)\,)^{\sim\sim}\rightarrow E^{\sim}::(f,F)\rightarrow
F\bullet f:F\bullet f(x)=F(\psi_{x,f}) \tag{3}\label{eq3}
\end{equation}
where $x\in E,$ $f\in E^{\sim},$ $\pi\in\ Orth(E)$\thinspace\ and
$F\in\ Orth(E)$\thinspace$)^{\sim\sim}.$ We call the map defined in
(\ref{eq3}) the\emph{ Arens extension} of the map in (\ref{eq1}).

For an Archimedean unital $f$-algebra $A$, we have $A^{\sim\sim}=(A^{\sim
})_{n}^{\sim}$ by Corollary 3.4 in \cite{HdP1}. Since $\ Orth(E)$
\thinspace\ is a unital $f$-algebra with point separating order dual, it is
Archimedean. This enables us to use the identification $(\ Orth(E)$
\thinspace$)^{\sim\sim}=((\ Orth(E)$\thinspace$)^{\sim})_{n}^{\sim}$
throughout this $\ $paper. It is straightforward to check that the Arens
product on the $f$-algebra $((\ Orth(E)$\thinspace$)^{\sim})_{n}^{\sim}$ is
compatible with the Arens extension defined in (\ref{eq3}), that is, $E^{\sim
}$ is a unital module over $((\ Orth(E)$\thinspace$)^{\sim})_{n}^{\sim}.$

We use (\ref{eq3}) to define a linear operator
\[
\gamma:((\ Orth(E)\,)^{\sim})_{n}^{\sim}\rightarrow L_{b}(E^{\sim})\text{ by
}\gamma(F)(f)=F\bullet f
\]
for each $F\in((\ Orth(E)$\thinspace$)^{\sim})_{n}^{\sim}$ and $f\in E^{\sim
}.$ We will call $\gamma$ the Arens homomorphism of the order bidual of
$\ Orth(E)$\thinspace$.$

\begin{proposition}
\label{p1} $\gamma$ is a unital algebra and order continuous Riesz
homomorphism such that $\gamma\left(  ((\ Orth(E)\,)^{\sim})_{n}^{\sim
}\right)  \subset Orth(E^{\sim}).$
\end{proposition}

\begin{proof}
It follows from the definition of $\gamma$ that \ $\gamma$ is a positive order
continuous unital algebra homomorphism. Also it is easily checked that
$\gamma(\pi)=\pi^{\prime}$ for each $\pi\in\ Orth(E)$\thinspace$.$ Let
$F\in((\ Orth(E)$\thinspace$)^{\sim})_{n}^{\sim}$ such that $|F|\leq\pi$ for
some $\pi\in\ Orth(E)$\thinspace$.$ Then $|\gamma(F)|\leq\gamma(|F|)\leq
\gamma(\pi)=\pi^{\prime}$ by the positivity of $\gamma.$ Since $\pi^{\prime
}\in Orth(E^{\sim}),$ we see that $\gamma(F)\in Orth(E^{\sim}).$ Then, since
$\gamma$ is order continuous and the ideal generated by $\ Orth(E)$
\thinspace\ is strongly order dense in $((\ Orth(E)$\thinspace$)^{\sim})_{n}^{\sim}
$,\ the range of $\gamma$ is contained in $Orth(E^{\sim}).$ That $\gamma$ is a
Riesz homomorphism follows from the fact that it is an algebra homomorphism by
Corollary 5.5 in \cite{HdP2}.
\end{proof}

If $A$ is an $f$-algebra then the range of the homomorphism $\nu:(A^{\sim
})_{n}^{\sim}\rightarrow Orth(A^{\sim})$ is contained in the range of the
Arens homomorphism $\gamma.$

\begin{proposition}
\label{p2} Let $A$ be an $f$-algebra with point separating order dual. The
range of the homomorphism $\nu:(A^{\sim})_{n}^{\sim}\rightarrow Orth(A^{\sim
})$ is contained in the range of the Arens homomorphism $\gamma.$
\end{proposition}

\begin{proof}
Let $P:A\rightarrow Orth(A)$ be the canonical embedding of $A$ into
$\ Orth(A)$\thinspace\ (i.e., $P(a)(b)=ab$ for all $a,b\in A$). It is well
known that $P(A)$ is a sublattice and an algebra ideal in $Orth(A).$ For each
$\mu\in(\ Orth(A)$\thinspace$)^{\sim}$, define $\widehat{\mu}\in A^{\ast}$ by
$\widehat{\mu}(a)=\mu(P(a)).$ Since positivity is preserved, it is clear that
$\widehat{\mu}\in A^{\sim}$ for each $\mu\in(\ Orth(A)$\thinspace$)^{\sim}.$
Given $a\in A,$ $f\in A^{\sim}$, we have $\psi_{a,f}\in(\ Orth(A)$
\thinspace$)^{\sim}.$ Then
\[
\widehat{\psi}_{a,f}(b)=\psi_{a,f}(P(b))=f(P(b)(a))=f(ab)=(f\cdot a)(b)
\]
for each $b\in A.$ That is $\widehat{\psi}_{a,f}=f\cdot a$ for all $a\in A,$
$f\in A^{\sim}.$ Now let $F\in(A^{\sim})_{n}^{\sim}.$ Define $\widehat{F}
\in((\ Orth(A)$\thinspace$)^{\sim})^{\ast}$ by $\widehat{F}(\mu)=F(\widehat
{\mu})$ for each $\mu\in(\ Orth(A)$\thinspace$)^{\sim}.$ Since $0\leq\mu$
implies $0\leq\widehat{\mu},$ we have $0\leq\widehat{F}$ whenever $0\leq F.$
That is $\widehat{F}\in((\ Orth(A)$\thinspace$)^{\sim})^{\sim}=((\ Orth(A)$
\thinspace$)^{\sim})_{n}^{\sim}.$ For each $F\in(A^{\sim})_{n}^{\sim},$ $f\in
A^{\sim}$ and $a\in A,$ we have
\begin{align*}
\nu_{F}(f)(a)  &  =F\cdot f(a)=F(f\cdot a)=F(\widehat{\psi}_{a,f})=\widehat
{F}(\psi_{a,f})\\
&  =\widehat{F}\bullet f(a)=\gamma(\widehat{F})(f)(a).
\end{align*}
Hence $\nu_{F}=\gamma(\widehat{F}).$
\end{proof}

Let us check the behavior of $\gamma$ in some specific examples.

\begin{example}
\label{ex1} Let $\omega$ denote all sequences and let $A=l^{1}$ the
$f$-subalgebra of $\omega$ consisting of absolutely summable sequences. Then
$A^{\sim}=l^{\infty}$ and $\ Orth(A)$\thinspace$=Orth(A^{\sim})=l^{\infty}.$
Since $(A^{\sim})_{n}^{\sim}=l^{1},$ $\nu$ is the inclusion map $l^{1}
\rightarrow l^{\infty}$ so that $\nu$ is one-to-one and not onto. On the other
hand $((\ Orth(A)$\thinspace$)^{\sim})_{n}^{\sim}=(l^{\infty})^{\prime\prime}$
and $\gamma$ is the band projection of $(l^{\infty})^{\prime\prime}$ onto
$l^{\infty}.$ Thus $\gamma$ is onto and not one-to-one.
\end{example}

\begin{example}
\label{ex2} Let $A=c_{0}$ be the $f$-subalgebra of $\omega$ consisting of the
sequences convergent to zero. Then $A^{\sim}=l^{1}$ and $\ Orth(A)$
\thinspace$=Orth(A^{\sim})=l^{\infty}.$ Since $(A^{\sim})_{n}^{\sim}
=l^{\infty},$ $\nu$ is the identity map on $l^{\infty}$ so that $\nu$ is
one-to-one and onto. On the other hand $((\ Orth(A)$\thinspace$)^{\sim}
)_{n}^{\sim}=(l^{\infty})^{\prime\prime}$ and $\gamma$ is the band projection
of $(l^{\infty})^{\prime\prime}$ onto $l^{\infty}.$ Thus $\gamma$ is onto and
not one-to-one.
\end{example}

\begin{example}
\label{ex3} Consider $C[0,1]$ with the product $\ast$ defined by $a\ast b=iab$
with $i(x)=x$ for all $x\in\lbrack0,1].$ Then $A=(C[0,1],\ast)$ is an
Archimedean $f$-algebra. As shown in \cite{HdP1}, $(A^{\sim})_{n}^{\sim}$ is
not semi-prime. Therefore $\upsilon$ is not one-to-one and not onto
\cite{HdP1}. On the other hand $\ Orth(A)$\thinspace$=Z(C[0,1])=C[0,1]$ and
$Orth(A^{\sim})=Z(C[0,1]^{\prime})=C[0,1]^{\prime\prime}.$ Since
$(C[0,1]^{\sim})_{n}^{\sim}=C[0,1]^{\prime\prime},$ $\gamma$ is the identity
map on $C[0,1]^{\prime\prime}.$ Therefore $\gamma$ is one-to-one and onto.
\end{example}

\section{Riesz spaces with topologically full center}

We start with the following definition that is due to Wickstead \cite{W1} in
the case of Banach lattices.

\begin{definition}
\label{df1} Suppose $E$ is a Riesz space. Then $E$ is said to have a
topologically full center if for each $x\in E_{+}$ the $\sigma(E,E^{\sim}
)$-closure of $Z(E)x$ contains the ideal generated by $x.$
\end{definition}

Banach lattices with topologically full center were initiated in \cite{W1}.
The class of Riesz spaces and the class Banach lattices with topologically
full center are quite large. For example, in a $\sigma$-Dedekind complete
Riesz space $E$ each positive element generates a projection band. Therefore
for each $x\in E_{+}$, $Z(E)x$ is an ideal and $Z(E)$ is topologically full.
Also Banach lattices with a quasi-interior point or with a topological
orthogonal system have topologically full center \cite{W2}. However not all
Riesz spaces have topologically full center.

\begin{example}
\label{ex4} (Zaanen \cite[p.664]{Z}) Let $E$ be the Riesz space of piecewise
affine continuous functions on $[0,1]$. Clearly the ideal generated by the
constant $1$ function equals $E.$ But, as shown by Zaanen, $Z(E)$ is trivial,
that is, it consists of the scalar multiples of the identity. Therefore $E$
does not have a topologically full center.
\end{example}

The first example of an AM-space that has trivial center was given in
\cite{G}. A thorough study of Banach lattices with trivial center was
undertaken in \cite{W3}. We refer the reader to \cite{W3} for further examples
of Banach lattices with trivial center as well as a careful treatment of the
following example of Goullet de Rugy mentioned above.

\begin{example}
\label{ex5}(\cite{G}) Let $K$ be a compact Hausdorff space with a point $p\in
K$ such that $\{p\}$ is not a $G_{\delta}$-set in $K$ (e.g., \cite[Example 4,
p.140]{S}). Let $C_{0}(K)$ denote the elements of $C(K)$ that vanish at $p.$
Let $H$ denote the positive unit ball of $C_{0}(K)^{\prime}$ with the relative
$\sigma(C_{0}(K)^{\prime},C_{0}(K))$-topology. Let $E=\{f\in C(H):f(r\mu
)=rf(\mu)$ for all $r\in\lbrack0,1]$ and for each $\mu\in H$ with $\Vert
\mu\Vert=1\}$. Then $E$ is an AM-space (without order unit). As a sublattice
of $l^{\infty}(H\smallsetminus\{0\}),$ one has $E^{d}=\{0\}.$ Therefore $Z(E)$
is embedded in $l^{\infty}(H\smallsetminus\{0\})$ \cite{W4}. Then one may
compute that $Z(E)$ consists of continuous bounded functions on
$H\smallsetminus\{0\}$ that are constant on the rays (i.e., $g(r\mu)=g(\mu)$
for all $r\in(0,1],$ for each $\mu\in H$ with $\left\Vert \mu\right\Vert =1$).
It follows by an argument in \cite[p.371]{G} that if there is a non-constant
function in $Z(E)$ then $\{0\}$ is a $G_{\delta}$-set in $H$. That in turn
implies that $\{p\}$ would be a $G_{\delta}$-set in $K.$ Therefore $Z(E)$ is
trivial and $E$ does not have a topologically full center.
\end{example}

$Z(E)$ is an Archimedean unital $f$-algebra with order unit. The order unit
norm induced on $Z(E)$ is an algebra and lattice norm. $\widehat{Z(E)}$, the
norm completion of $Z(E),$ is an AM-space and a partially ordered Banach
algebra where the order unit and the algebra unit coincide. Therefore by the
Stone algebra theorem $\widehat{Z(E)}\cong C(K)$ (isometric algebra and
lattice homomorphism) for some compact Hausdorff space $K.$ (Here $C(K)$
denotes the real valued continuous functions on $K.$) Then $Z(E)^{\prime
}=\widehat{Z(E)}^{\prime}=\widehat{Z(E)}^{\sim}=Z(E)^{\sim}$ and
$Z(E)^{\sim\sim}=Z(E)^{\prime\prime}.$ $Z(E)^{\prime\prime}$ is an AM-space
and with the Arens product, it is a partially ordered Banach algebra (an
Archimedean $f$-algebra with unit) where the order unit and the algebra unit
coincide. Therefore $Z(E)^{\prime\prime}\cong C(S)$ for some hyperstonian
space $S.$ That is the Arens product on $Z(E)^{\prime\prime}$ coincides with
the pointwise product on $C(S)$ \cite{Ar}.

Given the bilinear map
\begin{equation}
Z(E)\times E\rightarrow E \tag{4}\label{eq4}
\end{equation}
defined by $(T,x)\rightarrow Tx$ for each $T\in Z(E)$ and $x\in E,$ we define
the following bilinear maps:
\begin{equation}
E\times E^{\sim}\rightarrow Z(E)^{\prime}::(x,f)\rightarrow\mu_{x.f}
=\psi_{x,f}|_{Z(E)} \tag{5}\label{eq5}
\end{equation}
\begin{equation}
E^{\sim}\times Z(E)^{\prime\prime}\rightarrow E^{\sim}::(f,F)\rightarrow
F\circ f:F\circ f(x)=F(\mu_{x,f}) \tag{6}\label{eq6}
\end{equation}
where $x\in E,$ $f\in E^{\sim}$ and $F\in Z(E)^{\prime\prime}.$ The Arens
product on $Z(E)^{\prime\prime}$ is compatible with the bilinear map defined
in (\ref{eq6}). That is, $E^{\sim}$ is a unital module over $Z(E)^{\prime
\prime}.$ (\ref{eq6}) allows us to define a linear operator $m:Z(E)^{\prime
\prime}\rightarrow L_{b}(E^{\sim})$ where $m(F)(f)=F\circ f$ for all $f\in
E^{\sim}$ and $F\in Z(E)^{\prime\prime}.$ It is easily checked that
$m(T)=\gamma(T)$ whenever $T\in Z(E).$

We have the following analogue of Proposition \ref{p1} for the map $m$.

\begin{proposition}
\label{p3} $m$ is a unital algebra and order continuous lattice homomorphism
such that $m\left(  Z(E)^{\prime\prime}\right)  \subset Z(E^{\sim}).$
\end{proposition}

\begin{proof}
That $m$ is a positive order continuous algebra homomorphism is immediate from
the definition of $m$. That it is a lattice homomorphism follows as in
Proposition \ref{p1}.

For each $F\in Z(E)^{\prime\prime},$ there is a net $\{T_{\alpha}\}$ in $Z(E)$
such that $\parallel T_{\alpha}\parallel\leq\parallel F\parallel$ and
$T_{\alpha}\rightarrow F$ in $\sigma(Z(E)^{\prime\prime},Z(E)^{\prime}
)$-topology. Let $f\in E_{+}^{\sim}$ and $x\in E_{+}$. Then
\[
-||F||f(x)\leq F\circ f(x)=\underset{\alpha}{\lim}f(T_{\alpha}x)\leq
||F||f(x).
\]
So $F\in Z(E^{\sim}).$
\end{proof}

We will call the map $m:Z(E)^{\prime\prime}\rightarrow Z(E^{\sim})$ the Arens
homomorphism of the bidual of $Z(E)$ (into $Z(E^{\sim})$).

\begin{proposition}
\label{p4} Let $A$ be a unital $f$-algebra with point separating order dual.
Then the Arens homomorphism of the bidual of $Z(A)$ is onto $Z(A^{\sim}).$
\end{proposition}

\begin{proof}
Let $F\in Z(A^{\sim})$ with $0\leq F\leq1.$ Since $A$ is unital, Theorem 5.2
in \cite{HdP1} implies that the algebra and lattice homomorphism $\nu:(A^{\sim
})_{n}^{\sim}\rightarrow Orth(A^{\sim})$ of \cite{HdP1} is one-to-one and
onto. Therefore with a slight abuse of notation we will identify $(A^{\sim
})_{n}^{\sim}$ with $Orth(A^{\sim})$. In the duality $\langle(A^{\sim}
)_{n}^{\sim},A^{\sim}\rangle$, $A$ is dense in $(A^{\sim})_{n}^{\sim}$ with
respect to the weak topology. Since the locally solid convex topology
$|\sigma|((A^{\sim})_{n}^{\sim},A^{\sim})$ is compatible with the duality
$\langle(A^{\sim})_{n}^{\sim},A^{\sim}\rangle$ \cite[Theorem 11.13, p.170]
{AB}, there is a net $\{a_{\alpha}\}$ in $A$ such that $\{a_{\alpha}\}$
converges to $F$ in the $|\sigma|((A^{\sim})_{n}^{\sim},A^{\sim})$-topology.
Lattice operations are continuous in the locally solid convex $|\sigma
|((A^{\sim})_{n}^{\sim},A^{\sim})$-topology and $A$ is a sublattice of
$(A^{\sim})_{n}^{\sim}$ with $1\in A.$ Therefore we may suppose that $0\leq
a_{\alpha}\leq1.$ Hence, by the Alaoglu Theorem, there is $T\in Z(A)^{\prime
\prime}$ with $0\leq T\leq1$ such that a subnet $\{a_{\beta}\}$ converges to
$T$ in the $\sigma(Z(A)^{\prime\prime},Z(A)^{\prime})$-topology. Then, when
$f\in A^{\sim}$ and $b\in A,$
\[
T\circ f(b)=T(\mu_{b,f})=\underset{\beta}{\lim}\mu_{b,f}(a_{\beta}
)=\underset{\beta}{\lim}\psi_{b,f}(a_{\beta})=\underset{\beta}{\lim}
f(a_{\beta}b).
\]
On the other hand, since $\{a_{\beta}\}$ converges to $F$ in the
$|\sigma|((A^{\sim})_{n}^{\sim},A^{\sim})$-topology implies convergence also
in the $\sigma((A^{\sim})_{n}^{\sim},A^{\sim})$-topology, we have
\[
F\cdot f(b)=F(f\cdot b)=\underset{\beta}{\lim}f\cdot b(a_{\beta}
)=\underset{\beta}{\lim}f(a_{\beta}b).
\]
Therefore $m(T)=F.$
\end{proof}

Note that if $E$ is a Riesz space and if we set $A=Orth(E)$, then $Z(A)=Z(E).$
Also, since then $A$ is a unital $f$-algebra with point separating order dual,
Proposition \ref{p4} will be true for $A.$ In what follows we will use these
facts repeatedly.

\begin{corollary}
\label{c1} Let $E$ be a Riesz space. Let $A=Orth(E)$ and $m_{A}$ denote the
Arens homomorphism of the bidual of $Z(A)$ onto $Z(A^{\sim})$ (where
$Z(A)=Z(E)$). Then\newline(1) the following diagram is commutative
\[
\begin{array}
[c]{ccccc}
Z(E)^{\prime\prime} & \overset{m}{\rightarrow} & Z(E^{\sim}) & \overset
{i}{\rightarrow} & Orth(E^{\sim})\\
\shortparallel &  &  &  & \uparrow\gamma\\
Z(A)^{\prime\prime} & \overset{m_{A}}{\rightarrow} & Z(A^{\sim}) & \overset
{i}{\rightarrow} & (A^{\sim})_{n}^{\sim}
\end{array}
\]
where $i$ denotes the natural inclusion map;\newline(2) $\gamma(Z((A^{\sim
})_{n}^{\sim}))=m(Z(E)^{\prime\prime}).$
\end{corollary}

\begin{proof}
(1) Suppose $F\in Z(A)^{\prime\prime}$ with $0\leq F\leq1.$ Then, as in the
proof of Proposition \ref{p4}, let $0\leq a_{\alpha}\leq1$ be a net in
$Z(A)\subset A$ such that $\{a_{\alpha}\}$ converges to $F$ in the
$\sigma(Z(A)^{\prime\prime},Z(A)^{\prime})$-topology and also to $m_{A}
(F)\in(A^{\sim})_{n}^{\sim}$ in the $\sigma((A^{\sim})_{n}^{\sim},A^{\sim}
)$-topology. Since $(A^{\sim})_{n}^{\sim}=(Orth(E)^{\sim})_{n}^{\sim},$ for
each $x\in E$ and $f\in E^{\sim},$ we have
\begin{align*}
\gamma(m_{A}(F))(f)(x)  &  =m_{A}(F)\bullet f(x)=m_{A}(F)(\psi_{x,f}
)=\underset{\alpha}{\lim}\psi_{x,f}(a_{\alpha})\\
&  =\underset{\alpha}{\lim}\mu_{x,f}(a_{\alpha})=F(\mu_{x,f})=m(F)(f)(x).
\end{align*}

(2) As in proof of Proposition \ref{p4}, we may identify $(A^{\sim})_{n}
^{\sim}$ with $Orth(A^{\sim})$ via the homomorphism $\nu$. Then $Z(A^{\sim
})=Z((A^{\sim})_{n}^{\sim}).$ Since, by Proposition \ref{p4}, $m_{A}$ is onto
$Z(A^{\sim}),$ from part (1), we have $\gamma(Z((A^{\sim})_{n}^{\sim
})=m(Z(E)^{\prime\prime}).$
\end{proof}

\begin{corollary}
\label{c2} Let $E$ be a Riesz space. Then $\gamma((Orth(E)^{\sim})_{n}^{\sim
})$ is an order ideal in $Orth(E^{\sim})$ if and only if $m(Z(E)^{\prime
\prime})=Z(E^{\sim}).$
\end{corollary}

\begin{proof}
Let $A=Orth(E).$ Suppose $\gamma((A^{\sim})_{n}^{\sim})$ is an order ideal in
$Orth(E^{\sim}).$ Since $\gamma$ is an order continuous lattice homomorphism,
the kernel $Ker(\gamma)$ is a band in $(A^{\sim})_{n}^{\sim}.$ Let $(1-e)$ be
the band projection of $(A^{\sim})_{n}^{\sim}$ onto $Ker(\gamma).$ Then
$e\cdot(A^{\sim})_{n}^{\sim}$ is algebra and lattice isomorphic to the order
ideal $\gamma((A^{\sim})_{n}^{\sim})$ with $\gamma(e)=1.$ Let $T\in Z(E^{\sim
})$ with $0\leq T\leq1=\gamma(e).$ So there is $F\in(A^{\sim})_{n}^{\sim}$
such that $\gamma(F)=T.$ Moreover, since $\gamma$ is an algebra and lattice
homomorphism, we may choose $F$ so that $0\leq F\leq e\leq1.$ That is we may
suppose that $F\in Z((A^{\sim})_{n}^{\sim}).$ Then Corollary \ref{c1} part (2)
implies there is $G\in Z(E)^{\prime\prime}$ such that $m(G)=\gamma(F)=T.$

Conversely, suppose $m(Z(E)^{\prime\prime})=Z(E^{\sim}).$ Suppose $0\leq
T\leq\gamma(F)$ for some $0\leq F\in(A^{\sim})_{n}^{\sim}$ and for some $T\in
Orth(E^{\sim}).$ Since $Orth(E^{\sim})$ is Dedekind complete, there is
$\widehat{T}\in Z(Orth(E^{\sim}))=Z(E^{\sim})$ such that $0\leq\widehat{T}
\leq1$ and $T=\widehat{T}\gamma(F).$ By Corollary \ref{c1} part (2), there is
$G\in Z((A^{\sim})_{n}^{\sim})$ such that $\gamma(G)=\widehat{T}$ . Then
$T=\gamma(G)\gamma(F)=\gamma(G\cdot F)$ where $G\cdot F\in(A^{\sim})_{n}
^{\sim}.$
\end{proof}

Corollary \ref{c2} gives the first part of the result stated in the abstract.
We will complete the result by showing that the only Riesz spaces that satisfy
Corollary \ref{c2} are necessarily those with a topologically full center. For
Banach lattices a proof of this was given in \cite{O1}. Initially we will give
the proof of the sufficiency.

\begin{proposition}
\label{p5} Let $E$ be a Riesz space. Then $E$ has a topologically full center
if and only if the Arens homomorphism $m:Z(E)^{\prime\prime}\rightarrow
Z(E^{\sim})$ is surjective. Then there exits an idempotent $\pi\in
Z(E)^{\prime\prime}$ such that $Z(E^{\sim})=\pi\cdot Z(E)^{\prime\prime}$ and
$Ker(m)=(1-\pi)\cdot Z(E)^{\prime\prime}.$
\end{proposition}

\begin{proof}
(Sufficiency) Suppose $m$ is surjective. To show that $E$ has topologically
full center it is sufficient to show that each $\sigma(E,E^{\sim})$-closed
$Z(E)$ submodule of $E$ is an ideal. This is equivalent to showing that each
$\sigma(E^{\sim},E)$-closed $Z(E)$ submodule of $E^{\sim}$ is an ideal. Let
$M$ be a $\sigma(E^{\sim},E)$-closed $Z(E)$ submodule of $E^{\sim}$ and let
$T\in Z(E)^{\prime\prime}.$ There is a net $\{T_{\alpha}\}$ in $Z(E)$ that
converges to $T$ in the $\sigma(Z(E)^{\prime\prime},Z(E)^{\prime})$-topology.
For each $x\in E$ and $f\in M$ we have
\[
T_{\alpha}\circ f(x)=\mu_{x,f}(T_{\alpha})\rightarrow T(\mu_{x,f})=T\circ
f(x).
\]
Hence $M$ is a $Z(E)^{^{\prime\prime}}$-submodule of $E^{\sim}.$ Since
$Z(E^{\sim})=m(Z(E)^{\prime\prime}),$ $M$ is a $Z(E^{\sim})$-submodule.
$E^{\sim}$ is Dedekind complete. It is well known that a subspace of $E^{\sim
}$ is an ideal in $E^{\sim}$ if and only if it is a $Z(E^{\sim})$-submodule of
$E^{\sim}$ (e.g., \cite{W2}). Hence $M$ is an ideal in $E^{\sim}.$ Since $m$
is order continuous (Proposition \ref{p3}), $Ker(m)$ is a band in
$Z(E)^{\prime\prime}.$ Hence there exists a band projection $\pi\in
Z(E)^{\prime\prime}$ such that $Ker(m)=(1-\pi)\cdot Z(E)^{\prime\prime}$ and
$Z(E^{\sim})=\pi\cdot Z(E)^{\prime\prime}.$
\end{proof}

The proof of the converse requires some preparatory results.

Given a Riesz space $E$, let $A$ be a unital subalgebra of $Z(E).$ Let $A^{o}$
denote the polar of $A$ in $Z(E)^{\prime}$ and let $A^{oo}$ denote the polar
of $A^{o}$ in $Z(E)^{\prime\prime}.$ By standard duality theory, we have that
$A^{\prime}=Z(E)^{\prime}/A^{0}$ and $A^{\prime\prime}=A^{00}\subset
Z(E)^{\prime\prime}.$ Since $A$ is a normed algebra, $A^{\prime\prime}$ is a
Banach algebra with the Arens product \cite{Ar}. In fact $A^{\prime\prime}$ is
a subalgebra of $Z(E)^{\prime\prime}$ when $Z(E)^{\prime\prime}$ has its Arens product.

\begin{lemma}
\label{L1} Let $A$ be a unital subalgebra of $Z(E).$ Then $A^{00}$ is a
subalgebra of $Z(E)^{\prime\prime}$ and the algebra product on $A^{00}$ is
identical with the Arens product on $A^{\prime\prime}$ under the canonical
isomorphism of $A^{00}$ with $A^{\prime\prime}.$
\end{lemma}

\begin{proof}
Let $f\in A^{0}$ and $a\in A.$ It is easily checked that $f\cdot a\in A^{0}.$
Let $F,G\in A^{00}.$ There is a net $\{a_{\alpha}\}$ in $A$ that converges to
$F$ in the $\sigma(Z(E)^{\prime\prime},Z(E)^{\prime})$-topology. Let $f\in
A^{0}.$ Then
\[
F\cdot G(f)=F(G\cdot f)=\underset{\alpha}{\lim}G\cdot f(a_{\alpha}
)=\underset{\alpha}{\lim}G(f\cdot a_{\alpha})=0
\]
and $F\cdot G\in A^{00}.$

On the other hand, denote the isomorphism of $A^{\prime\prime}$ onto $A^{00}$
by $F\rightarrow\widehat{F}.$ Given $f\in A^{\prime}$, let $\widehat{f}\in
Z(E)^{\prime}$ denote any extension of $f$ on $Z(E).$ When $F,G\in
A^{\prime\prime},$ let $\{a_{\alpha}\}$ and $\{b_{\beta}\}$ be nets in $A$
that converge to $\widehat{F}$ and $\widehat{G}$ respectively in the
$\sigma(Z(E)^{\prime\prime},Z(E)^{\prime})$-topology. Then, evidently, the
respective nets also converge to $F$ and $G$ respectively in the
$\sigma(A^{\prime\prime},A^{\prime})$-topology. For any $f\in A^{\prime},$ we
have
\[
F\cdot G(f)=\underset{\alpha}{\lim}\underset{\beta}{\lim}f(a_{\alpha}b_{\beta
})=\underset{\alpha}{\lim}\underset{\beta}{\lim}\widehat{f}(a_{\alpha}
b_{\beta})=\widehat{F}\cdot\widehat{G}(\widehat{f}).
\]

\end{proof}

\begin{lemma}
\label{L2} Let $I$ be a closed algebra ideal in $Z(E)$ and consider $A=Z(E)/I$
with the quotient norm. Then $A$ is a normed algebra and with the Arens
product $A^{\prime\prime}$ may be identified with the subalgebra of
$Z(E)^{\prime\prime}$ given by
\[
(I^{00})^{d}=\{F\in Z(E)^{\prime\prime}:|F|\wedge|G|=0\text{ for all }G\in
I^{00}\}.
\]

\begin{proof}
Let $\widehat{Z(E)}=C(K^{\prime})$ for some compact Hausdorff space
$K^{\prime}$ and let $\overline{I}$ denote the closure of $I$ in $C(K^{\prime
}).$ There is a closed subset $K$ of $K^{\prime}$ such that $\overline
{I}=\{a\in C(K^{\prime}):a(K)=\{0\}\}$ and $C(K^{\prime})/\overline{I}=C(K).$
Furthermore $A$ is a subalgebra of $C(K)$. In fact $C(K)$ is the completion of
$A.$ Since $\overline{I}$ is an order ideal in $\widehat{Z(E)},$ $A^{\prime
}\cong I^{0}=(\overline{I})^{0}$ is a band in $Z(E)^{\prime}.$ Then
$A^{\prime\prime}\cong Z(E)^{\prime\prime}/I^{00}=(I^{00})^{d},$ since
$I^{00}$ is a band in $Z(E)^{\prime\prime}.$ It remains to check that the
Arens product of $A^{\prime\prime}$ is identical with the product on the
subalgebra $(I^{00})^{d}.$ Let $F,G\in Z(E)^{\prime\prime}$ and $\widehat
{F},\widehat{G}\in A^{\prime\prime}$ such that $\widehat{F}=F|_{I^{0}},$
$\widehat{G}=G|_{I^{0}}.$ Let $\{a_{\alpha}\}$ and $\{b_{\beta}\}$ be nets in
$Z(E)$ that converge to $F$ and $G$ respectively in the $\sigma(Z(E)^{\prime
\prime},Z(E)^{\prime}))$-topology. Let $[a]=a+I$ for each $a\in Z(E).$ Then,
it follows that, $\{[a_{\alpha}]\}$ and $\{[b_{\beta}]\}$ converge to
$\widehat{F}$ and $\widehat{G}$ respectively in the $\sigma(A^{\prime\prime
},A^{\prime})$-topology. Then, for $f\in I^{0},$ we have
\[
\widehat{F}\cdot\widehat{G}(f)=\underset{\alpha}{\lim}\underset{\beta}{\lim
}f([a_{\alpha}][b_{\beta}])=\underset{\alpha}{\lim}\underset{\beta}{\lim
}f(a_{\alpha}b_{\beta})=F\cdot G(f).
\]

\end{proof}
\end{lemma}

Let $J$ be an ideal in $E^{\sim}.$ For any $F\in Z(E)^{\prime\prime},$
$m(F)|_{J}\in Z(J).$ Each operator in $Z(J)$ has a unique extension to an
operator in the ideal center of the band generated by $J.$ Since bands in
$E^{\sim}$ are projection bands, we have $Z(J)=Z(E^{\sim})|_{J}.$

Suppose $J$ is an ideal in $E^{\sim}$ that separates the points of $E.$ Let
$A$ be a unital subalgebra of $Z(E).$

\begin{definition}
\label{df2} $E$ is called cyclic with respect to $A$ for the dual pair
$\langle E,J\rangle$ if there is $u\in E_{+}$ such that $(Au)^{0}=\{0\}$ in
$J.$

\begin{lemma}
\label{L3} Let $E$ be a Riesz space and $J$ be an ideal in $E^{\sim}$ that
separates the points of $E.$ Suppose $E$ is cyclic with respect to a unital
$f$-subalgebra $A$ of $Z(E)$ for the dual pair $\langle E,J\rangle.$ Then
$m(A^{\prime\prime})|_{J}=Z(J).$
\end{lemma}
\end{definition}

\begin{proof}
The norm completion $\widehat{A}$ of $A$ is a unital closed subalgebra of
$\widehat{Z(E)}.$ Therefore $\widehat{A}=C(K)$ for some compact Hausdorff
space $K$. Furthermore $A^{\sim}=A^{\prime}=C(K)^{\prime}$ and $A^{\prime
\prime}=C(K)^{\prime\prime}=C(S)$ for some hyperstonian space $S.$ (We again
mention that the usual multiplication on $C(S)$ is the Arens extension of the
product on $C(K)$ as shown in \cite{Ar}. Also the usual $C(S)$ -module
structure of $C(K)^{\prime}$ via its ideal center is the Arens homomorphism of
the bidual of $C(K)$ onto the ideal center of $C(K)^{\prime}$ (e.g.,
Proposition \ref{p4}).) In the rest of the proof, by Lemma \ref{L1}, we
consider $A^{\prime\prime}$ as a subalgebra of $Z(E)^{\prime\prime}.$

Let $u\in E_{+}$ be a cyclic vector and $f\in J_{+}.$ Then $\mu_{u,f}\in
Z(E)_{+}^{\prime}$ and $\mu_{u,f}|_{A}=\widehat{\mu}_{u,f}\in C(K)_{+}
^{\prime}.$ Let $P_{f}$ be the band projection of $C(K)^{\prime}$ onto the
band $B(\widehat{\mu}_{u,f})$ generated by $\widehat{\mu}_{u,f}.$ By the
Lebesgue Decomposition Theorem and the Radon-Nikodym Theorem, we have
\[
B(\widehat{\mu}_{u,f})=P_{f}\cdot C(K)^{\prime}=L^{1}(\widehat{\mu}
_{u,f})=\{\mu\in C(K)^{\prime}:|\mu|<<\widehat{\mu}_{u,f}\}.
\]
The first equality above follows by Proposition \ref{p4}.

Suppose $e\circ f=0$ for some idempotent $e\in A^{\prime\prime}=C(S).$ Let
$\{a_{\alpha}\}$ be a net in $A$ that converges to $e$ in the $\sigma
(A^{\prime\prime},A^{\prime})$-topology. Then for each $a\in A,$
\begin{align*}
e\cdot\widehat{\mu}_{u,f}(a)  &  =e(\widehat{\mu}_{u,f}\cdot a)=\underset
{\alpha}{\lim}\widehat{\mu}_{u,f}\cdot a(a_{\alpha})=\underset{\alpha}{\lim
}\widehat{\mu}_{u,f}(aa_{\alpha})=\underset{\alpha}{\lim}\mu_{u,f}(a_{\alpha
}a)\\
&  =\underset{\alpha}{\lim}f(a_{\alpha}au)=\underset{\alpha}{\lim}\mu
_{au,f}(a_{\alpha})=e(\mu_{au,f})=e\circ f(au)=\widehat{\mu}_{u,e\circ
f}(a)=0.
\end{align*}

Hence $0\leq e\leq1-P_{f}$. Conversely, since $u$ is a cyclic vector,
$\widehat{\mu}_{u,(1-P_{f})\circ f}=(1-P_{f})\cdot\widehat{\mu}_{u,f}=0$
implies that $(1-P_{f})\circ f=0.$ Therefore $1-P_{f}=\sup\{e\in C(S):e\circ
f=0$ and $e=e^{2}\}.$ That is, $a\circ f=0$ for some $a\in A^{\prime\prime}$
if and only if $P_{f}\cdot a=0.$

Let $T\in Z(E^{\sim})$ with $0\leq T\leq1.$ Let $f\in J_{+}.$ Then
$0\leq\widehat{\mu}_{u,Tf}\leq\widehat{\mu}_{u,f}$ in $A^{\prime}.$ Therefore,
by the Radon-Nikodym Theorem, there is $a_{f}\in L^{\infty}(\widehat{\mu
}_{u,f})\subset C(S)=A^{\prime\prime}$ such that $\widehat{\mu}_{u,Tf}
=a_{f}\cdot\widehat{\mu}_{u,f}=\widehat{\mu}_{u,a_{f}\circ f}$. Since $u$ is a
cyclic vector it follows that $Tf=a_{f}\circ f.$ Suppose $g\in J$ such that
$0\leq g\leq f.$ $E^{\sim}$ is Dedekind complete, there is $G\in Z(E^{\sim})$
with $0\leq G\leq1$ such that $g=Gf.$ Therefore
\[
m(a_{g})(g)=Tg=T(Gf)=G(Tf)=Gm(a_{f})(f)=m(a_{f})(Gf)=m(a_{f})(g).
\]
That is $(a_{g}-a_{f})\circ g=0.$ Therefore $(a_{g}-a_{f})\cdot P_{g}=0.$ Now
suppose $f,g\in J_{+}$ and $h=f\vee g.$ Then $(a_{f}-a_{h})\cdot P_{f}
=(a_{g}-a_{h})\cdot P_{g}=0.$ Hence $a_{g}\cdot P_{f}=a_{f}\cdot P_{g}$ for
all $f,g\in J_{+}.$ Since $S$ is Stonian and $0\leq a_{f}\leq P_{f}\leq1$ for
all $f\in J_{+},$ there is a unique $a\in A^{\prime\prime}$ such that
$P_{f}\cdot a=a_{f}$ and $(1-\sup\{P_{f}:f\in J_{+}\})\cdot a=0.$ Then
\[
Tf=a_{f}\circ f=(a\cdot P_{f})\circ f=a\circ(P_{f}\circ f)=a\circ f
\]
for all $f\in J_{+}.$
\end{proof}

Now we are ready to complete the proof of Proposition \ref{p5}.

\begin{proof}
(Necessity) Suppose $x\in E_{+}.$ Let $I(x)$ denote the ideal generated by $x$
in $E$ and let $Ann(x)=\{T\in Z(E):Tx=0\}.$ $Ann(x)$ is a closed order and
algebra ideal in $Z(E).$ Consider the map from $Z(E)$ into $Z(I(x))$ defined
by $T\rightarrow T|_{I(x)}.$ Clearly the map is a norm reducing positive
algebra homomorphism. Since the kernel of this map is equal to $Ann(x),$ the
map is also a lattice homomorphism. It induces a norm reducing lattice and
algebra homomorphism of $Z(E)/Ann(x)$ into $Z(I(x))$ where
$T+Ann(x)=[T]\rightarrow T|_{I(x)}.$ The induced map is an isometry. For
example, if $T\in Z(E)_{+}$ with norm $||T|_{I(x)}||$ in $Z(I(x)),$ then
$(T\wedge||T|_{I(x)}||1)|_{I(x)}=T|_{I(x)}$ and $||[T]||\leq||T|_{I(x)}||.$
Hence let $A=Z(E)|_{I(x)}$ be the normed unital $f$-subalgebra of $Z(I(x)).$
Then since $A\cong Z(E)/Ann(x)$ (isometric, lattice and algebra homomorphism),
we may think of $A^{\prime\prime}$ as a subalgebra of $Z(E)^{\prime\prime}$
(c.f., Lemma \ref{L2}). Equivalently, we may think of $A^{\prime\prime}$ as a
subalgebra of $Z(I(x))^{\prime\prime}$ (c.f., Lemma \ref{L1}).

Let $J=\{f|_{I(x)}:f\in E^{\sim}\}.$ Clearly $J$ is an ideal in $I(x)^{\sim}$
that separates the points of $I(x).$ Let $P_{x}\in Z(E^{\sim})$ denote the
band projection of $E^{\sim}$ onto $(I(x)^{0})^{d}.$ The map $f|_{I(x)}
\rightarrow P_{x}(f)$ is a lattice homomorphism of $J$ onto the band
$(I(x)^{0})^{d}.$ Let $m_{x}$ be the Arens homomorphism of $Z(I(x))^{\prime
\prime}$ into $Z(I(x)^{\sim})$ and let $m$ be the same for $Z(E)^{\prime
\prime}$ into $Z(E^{\sim}).$ We claim that on elements of $A^{\prime\prime},$
$m_{x}$ restricted to $J$ agrees with $m$ restricted to $(I(x)^{0})^{d}.$
Namely, let $F\in A^{\prime\prime},$ $f\in E^{\sim}$ and $y\in I(x).$ Choose a
net $\{a_{\alpha}\}$ in $A$ that converges to $F$ in the $\sigma
(A^{\prime\prime},A^{\prime})$-topology. Then
\begin{align*}
m_{x}(F)(f|_{I(x)})(y)  &  =F\circ f|_{I(x)}(y)=F(\mu_{y,f|_{I(x)}}
)=F([\mu_{y,f|_{I(x)}}])\\
&  =\underset{\alpha}{\lim}[\mu_{y,f|_{I(x)}}](a_{\alpha})=\underset{\alpha
}{\lim}\mu_{y,f|_{I(x)}}(a_{\alpha})=\underset{\alpha}{\lim}f(a_{\alpha}y)
\end{align*}
where $[\mu_{y,f|_{I(x)}}]=\mu_{y,f|_{I(x)}}+A^{0}\in A^{\prime}
=Z(I(x))^{\prime}/A^{0}.$ On the other hand
\begin{align*}
m(F)(P_{x}(f))(y)  &  =P_{x}(m(F)(f))(y)=m(F)(f)(y)=F\circ f(y)\\
&  =F(\mu_{y,f})=\underset{\alpha}{\lim}\mu_{y,f}(a_{\alpha})=\underset
{\alpha}{\lim}f(a_{\alpha}y).
\end{align*}
In the last string of equalities, the first equality follows because
$Z(E^{\sim})$ is commutative. The second equality follows because the range of
$1-P_{x}$ is $I(x)^{0}$ in $E^{\sim}.$ Finally the fifth equality follows
because $\mu_{y,f}\in Ann(x)^{0}=A^{\prime}$ in $Z(E)^{\prime}.$ Hence, the
claim is verified. In what follows we will keep the notation that we
established in this initial part of the proof.

Suppose that $E$ has topologically full center. This means that $I(x)$ is
cyclic with respect to the unital $f$-subalgebra $A$ of $Z(I(x))$ for the
duality $\langle I(x),J\rangle.$ Given $T\in Z(E^{\sim})$ with $0\leq T\leq1,$
let $\widehat{T}=T|_{(I(x)^{0})^{d}}\in Z(J)$ where $\widehat{T}
(f|_{I(x)})(y)=T(P_{x}(f))(y)$ for each $f\in E^{\sim}$ and $y\in I(x).$ Then
by Lemma \ref{L3} there is $a_{x}\in A^{\prime\prime}$ with $0\leq a_{x}\leq1$
such that $m_{x}(a_{x})=\widehat{T}$ on $J.$ Then we have that
\[
m(a_{x})(P_{x}(f))(y)=m_{x}(a_{x})(f|_{I(x)})(y)=\widehat{T}(f|_{I(x)}
)(y)=T(P_{x}(f))(y)
\]
for all $f\in E^{\sim}$ and $y\in I(x).$ Since the range of $1-P_{x}$ is
$I(x)^{0}$ in $E^{\sim},$ we have that
\[
m(a_{x})(f)(y)=T(f)(y)
\]
for all $f\in E^{\sim}$ and $y\in I(x).$

Let $1-e_{x}=\sup\{e\in Z(E)^{\prime\prime}:e=e^{2},$ $m(e)(f)(x)=0$ for all
$f\in E^{\sim}\}.$ Since $m$ is order continuous, we have $m(1-e_{x})(f)(x)=0$
for all $f\in E^{\sim}.$ Hence $F\circ f(x)=0$ for all $f\in E^{\sim}$ for
some $F\in Z(E)^{\prime\prime}$ if and only if $e_{x}\cdot F=0$. ( Note that
$\{\mu_{x,f}:f\in E^{\sim}\}$ is an ideal in $Z(E)^{\prime}$ and $1-e_{x}$ is
the band projection of $Z(E)^{\prime\prime}$ onto the band in $Z(E)^{\prime
\prime}$ that annihilates this ideal.) Now repeating the argument in the proof
of Lemma \ref{L3}, we find a unique $a\in Z(E)^{\prime\prime}$ such that
$0\leq a\leq1$, and $e_{x}\cdot a=a_{x}$ for each $x\in E_{+}.$ Then, for each
$f\in E^{\sim}$ and $x\in E_{+},$ we have
\[
T(f)(x)=m(a_{x})(f)(x)=m(a\cdot e_{x})(f)(x)=m(e_{x})m(a)(f)(x)=m(a)(f)(x).
\]
Therefore $m(a)=T.$
\end{proof}

We will conclude this paper by stating some immediate consequences of
Proposition \ref{p5}.

\begin{corollary}
\label{c3} Let $A$ be an $f$-algebra with point separating order dual such
that $(A^{\sim})_{n}^{\sim}$ has a unit. Then $A$ has topologically full center.
\end{corollary}

\begin{proof}
By Theorem 5.2 \cite{HdP1}, the homomorphism $\upsilon$ of $(A^{\sim}
)_{n}^{\sim}$ is onto $Orth(A^{\sim}).$ Then Proposition \ref{p2} implies that
the Arens homomorphism $\gamma:(Orth(A)^{\sim})_{n}^{\sim}\rightarrow
Orth(A^{\sim})$ is also onto. Hence, by Corollary \ref{c2}, $m$ is onto
$Z(A^{\sim}).$ Therefore, by Proposition \ref{p5}, $A$ has topologically full center.
\end{proof}

\begin{remark}
Characterizations of $f$-algebras $A$ such that $(A^{\sim})_{n}^{\sim}$ has
unit are given in \cite{HdP1},\cite{BJ},\cite{J}. Related to Corollary
\ref{c3} , we mention that we do not know any examples of semi-prime
$f$-algebras that do not have topologically full center.
\end{remark}

\begin{corollary}
\label{c4} Let $E$ be a Riesz space with topologically full center. Suppose
$T$ is an order bounded operator on $E.$ Then $T$ commutes with $Z(E)$ if and
only if $T$ is in $Orth(E).$
\end{corollary}

\begin{proof}
$T$ is order bounded implies $T^{\prime}:E^{\sim}\rightarrow E^{\sim}.$ If $T$
commutes with $Z(E)$, then $T^{\prime}$ commutes with $m(Z(E)^{\prime\prime
}).$ Since $Z(E)$ is topologically full, $m(Z(E)^{\prime\prime})=Z(E^{\sim}).$
Therefore $T^{\prime}$ commutes with $Z(E^{\sim}).$ That is, $T^{\prime}$
commutes with the band projections on $E^{\sim}.$ Since $E^{\sim}$ is Dedekind
complete, each band in $E^{\sim}$ is a projection band. So $T^{\prime}$ is
band preserving and therefore $T^{\prime}\in Orth(E^{\sim}).$ By a result in
\cite[Theorem 3.3]{W5}, $T\in Orth(E).$
\end{proof}

In view of Examples \ref{ex4} and \ref{ex5}, the corollary may fail if $Z(E)$
is not topologically full. On the other hand, the result may be true even when
$Z(E)$ is not topologically full. The example constructed by Wickstead in
\cite{W2} shows this. We refer the reader to the introduction for more
detailed information. Corollary \ref{c4} shows that $Orth(E)$ is maximal
abelian when $Z(E)$ is topologically full. On the other hand, Wickstead's
example in \cite{W2} shows that the converse is not true.

\begin{corollary}
\label{c5} Let $E$ be a Riesz space with topologically full center. Then the
following are equivalent:\newline(1) $E^{\sim\sim}=(E^{\sim})_{n}^{\sim}
.$\newline(2) $m$ is continuous when its domain has the $\sigma(Z(E)^{\prime
\prime},Z(E)^{\prime})$-topology and its range has the $\sigma(E^{\sim
},E^{\sim\sim})$-operator topology.
\end{corollary}

We leave the straightforward proof of the corollary to the interested reader.

Before stating our final corollary, we want to discuss its content and fix
some notation. Let $K$ be a hyperstonian space. That is $K$ is a Stonian
compact Hausdorff space and $C(K)$ is a dual Banach space. Let $C(K)_{\ast}$
denote the predual of $C(K).$ Recall that $C(K)_{\ast}=C(K)_{n}^{\prime}$, the
order continuous linear functionals on $C(K).$ Hence $C(K)_{\ast}$ is a band
in the Dedekind complete Banach lattice $C(K)^{\prime}.$ Since $Z(C(K)^{\prime
})=C(K)^{\prime\prime},$ there is an idempotent $p\in C(K)^{\prime\prime}$
such that $p$ is the band projection on $C(K)^{\prime}$ with range
$C(K)_{\ast}$. That is
\[
p\cdot C(K)^{\prime}=C(K)_{n}^{\prime}=C(K)_{\ast}.
\]
Let $E$ be a Riesz space. Its order dual $E^{\sim}$ is a Dedekind complete
Riesz space. Therefore $E^{\sim}$ has a topologically full center $Z(E^{\sim
}).$ Furthermore $Z(E^{\sim})$ is itself Dedekind complete as a Banach
lattice. In fact, it is familiar that $Z(E^{\sim})=C(K)$ for some hyperstonian
space $K$.(This will become clear in the proof of the corollary.) Let
$m:Z(E^{\sim})^{\prime\prime}\rightarrow Z(E^{\sim\sim})$ be the Arens
homomorphism of the bidual of $Z(E^{\sim})$. Since $Z(E^{\sim})$ is
topologically full, we have $m(Z(E^{\sim})^{\prime\prime})=Z(E^{\sim\sim})$
and $Ker(m)=(1-\pi)\cdot Z(E^{\sim})^{\prime\prime}$ for some idempotent
$\pi\in Z(E^{\sim})^{\prime\prime}=C(K)^{\prime\prime}$ (Proposition
\ref{p5}). We will show that $p\circ E^{\sim\sim}=(E^{\sim})_{n}^{\sim}$.

\begin{corollary}
\label{c6} Let $E$ be a Riesz space with point separating order dual $E^{\sim
}.$ Let $m$ be the Arens homomorphism of the bidual of $Z(E^{\sim})$ in
$Z(E^{\sim\sim}).$ Then

\begin{enumerate}
\item $Z(E^{\sim})$ is topologically full and $Z(E^{\sim})=C(K)$ for some
hyperstonian space $K.$

\item There is an idempotent $\pi\in C(K)^{\prime\prime}$ such that
\[
Z(E^{\sim\sim})=\pi\cdot C(K)^{\prime\prime}\text{\thinspace\thinspace
and\thinspace}\,Ker(m)=(1-\pi)\cdot C(K)^{\prime\prime}.
\]

\item There is an idempotent $p\in C(K)^{\prime\prime}$ with $p\leq\pi$ such
that
\[
p\cdot C(K)^{\prime}=C(K)_{\ast}=C(K)_{n}^{\prime}\text{ \thinspace and
\thinspace}p\circ E^{\sim\sim}=(E^{\sim})_{n}^{\sim}.
\]

\item $E^{\sim\sim}=(E^{\sim})_{n}^{\sim}$ if and only if $p=\pi.$
\end{enumerate}
\end{corollary}

\begin{proof}
1. Since $E^{\sim}$ is Dedekind complete, $Z(E^{\sim})=C(K)$ is topologically
full and $K$ is a Stonian compact Hausdorff space. (It is well known that $K$
is hyperstonian. We include a proof for the sake of completeness.) To show
that $K$ is hyperstonian, it is sufficient to see that the order continuous
linear functionals on $C(K)$ separate the points of $C(K)$ \cite{S}. Consider
$E\subset(E^{\sim})_{n}^{\sim}\subset E^{\sim\sim}.$ Take positive elements
$x\in E,$ $f\in E^{\sim}$ and $a_{\tau},a\in C(K)$ such that $\{a_{\tau}\}$ is
an increasing net with $\sup a_{\tau}=a$ in $C(K).$ Then, since $E^{\sim}$ is
Dedekind complete, $\sup a_{\tau}f=af$ in $E^{\sim}.$ Therefore $a_{\tau
}f(x)\uparrow af(x),$ since $x$ is an order continuous linear functional on
$E^{\sim}.$ Consider $\mu_{f,x}\in C(K)^{\prime}$ in the definition process of
$m,$ we have $\mu_{f,x}(b)=x(bf)=bf(x)$ for all $b\in C(K).$ Hence it follows
that $\mu_{f,x}\in C(K)_{n}^{\prime}$ for all $x\in E$ and $f\in E^{\sim}.$
Also it is clear that these linear functionals separate the points of the
center $Z(E^{\sim})=C(K).$ Therefore $K$ is hyperstonian and $C(K)_{\ast
}=C(K)_{n}^{\prime}$ is the predual of $C(K).$

2. The existence of $\pi$ is clear from Proposition \ref{p5}. An equivalent
means of defining $\pi\in C(K)^{\prime\prime}$ is by observing that $\pi$ is
the supremum of the band projections on $C(K)^{\prime}$ obtained by
considering the bands generated by each linear functional of the form
$\mu_{f,x^{\prime\prime}}\in C(K)^{\prime}$ when $f\in E^{\sim}$ and
$x^{\prime\prime}\in E^{\sim\sim}.$

3. Let $p\in C(K)^{\prime\prime}=Z(C(K)^{\prime})$ be the band projection onto
the band $C(K)_{n}^{\prime}=C(K)_{\ast}.$ An equivalent means of defining $p$
would be to observe that $p$ is the supremum of the band projections on
$C(K)^{\prime}$ obtained by considering the bands generated by each linear
functional of the form $\mu_{f,x}\in C(K)^{\prime}$ when $f\in E^{\sim}$ and
$x\in E\subset E^{\sim\sim}.$ Hence we have $p\leq\pi.$ It remains to show
that $p\circ E^{\sim\sim}=(E^{\sim})_{n}^{\sim}.$ Note that for each $f\in
E^{\sim}$ and each $x^{\prime\prime}\in E^{\sim\sim},$ we have $\mu_{f,p\circ
x^{\prime\prime}}=p\cdot\mu_{f,x^{\prime\prime}}$.( Namely, let $\{a_{\alpha
}\}$ be a net in $C(K)$ that converges to $p$ in $\sigma(C(K)^{\prime\prime
},C(K)^{\prime})$-topology. Then
\[
\mu_{f,p\circ x^{\prime\prime}}(a)=p\circ x^{\prime\prime}(af)=p(\mu
_{af,x^{\prime\prime}})=\underset{\alpha}{\lim}\mu_{af,x^{\prime\prime}
}\left(  a_{\alpha}\right)  =\underset{\alpha}{\lim}x^{\prime\prime}
(a_{\alpha}af)
\]
and
\[
p\cdot\mu_{f,x^{\prime\prime}}(a)=p(\mu_{f,x^{\prime\prime}}\cdot
a)=\underset{\alpha}{\lim}\mu_{f,x^{\prime\prime}}\cdot a(a_{\alpha
})=\underset{\alpha}{\lim}\mu_{f,x^{\prime\prime}}(aa_{\alpha})=\underset
{\alpha}{\lim}x^{\prime\prime}(a_{\alpha}af)
\]
for each $a\in C(K)$. Here the second set of displayed equalities follow from
the definition of the Arens product on the bidual of $C(K)$ when $C(K)$ is
considered as a unital $f$-algebra \cite{HdP1}.) But $p\cdot\mu_{f,x^{\prime
\prime}}\in C(K)_{\ast}=C(K)_{n}^{\prime}$ for each $f\in E^{\sim}.$ By
reversing the process we used in part (1), it follows that $p\circ
x^{\prime\prime}\in(E^{\sim})_{n}^{\sim}$. Conversely if $x^{\prime\prime}
\in(E^{\sim})_{n}^{\sim},$ the process we used in part (1) shows that
$\mu_{f,x^{\prime\prime}}\in C(K)_{\ast}$. Therefore
\[
\mu_{f,x^{\prime\prime}}=p\cdot\mu_{f,x^{\prime\prime}}=\mu_{f,p\circ
x^{\prime\prime}}
\]
for each $f\in E^{\sim}.$ That is, $p\circ x^{\prime\prime}=x^{\prime\prime}$
for all $x^{\prime\prime}\in(E^{\sim})_{n}^{\sim}.$ So $p\circ E^{\sim\sim
}=(E^{\sim})_{n}^{\sim}.$

Now (4) is clear from parts (2) and (3).
\end{proof}

\begin{remark}
The article \cite{HdP1} has initiated considerable research on Arens product
on the biduals of lattice ordered algebras, we include a partial list
\cite{BH}, \cite{Gr}, \cite{H}, \cite{Sd1}, \cite{Sd2}.
\end{remark}

config: arxiv


TITLE: Find $\lim_{x \to 0} (\frac {\sin x}{x})^{\frac{1}{\sin^2x}}$
QUESTION [2 upvotes]: Is there any other way to compute the problem $$\lim_{x \to 0} (\frac {\sin x}{x})^{\frac{1}{\sin^2(x)}}$$
I try to use L'Hospital rule but it's too complicated.

REPLY [1 votes]: $$\lim_{x \to 0} \left(\frac {\sin x}{x}\right)^{\frac{1}{\sin^2x}}$$
$$=\left(\lim_{x \to 0} \left(1+\dfrac{\sin x-x}x\right)^{\dfrac x{\sin x-x}}\right)^{\lim_{x \to 0} \dfrac{\sin x-x}{x^3}\cdot\left(\lim_{x \to 0}\dfrac x{\sin x}\right)^2}$$
Now by Are all limits solvable without L'Hôpital Rule or Series Expansion, $$\lim_{x \to 0}\dfrac{\sin x-x}{x^3}=\dfrac16$$
$\implies \lim_{x \to 0}\dfrac{\sin x-x}x=0$
Consequently, $$\lim_{x \to 0}\left(1+\dfrac{\sin x-x}x\right)^{\dfrac x{\sin x-x}}=e$$
Can you take it from here?


TITLE: Alternate methods for proving following equation
QUESTION [0 upvotes]: Is there an alternate method to prove the following?
If $2\cos A=x + \frac{1}{x}$, $2\cos B=y+\frac{1}{y}$ then show that $2\cos(A-B) = \frac{x}{y} + \frac{y}{x}$
I know that it can be done by converting the first two equations into quadratic equations and then calculating $\frac{x}{y}$ and $\frac{y}{x}$ and then adding them, but are there any simpler methods?

REPLY [0 votes]: If you stick to the reals, using @dxiv's remark, $x=\pm1,y=\pm1$ so that $A=k\pi,B=k'\pi$ and $A-B=k''\pi$, so that $2\cos(A-B)=\pm2=\dfrac xy+\dfrac yx$.


TITLE: Problem in solving a differential equation
QUESTION [0 upvotes]: This was a question on my exam that stumped me! I think I got lost in the maths. I really want to know how to solve this, because it is very simple.
Question:
A mass $m$ goes up an incline which is at angle $\theta$. It experiences linear drag $\vec{F_{\text{drag}}}=-~b\vec{v}$ as well as gravitational force. If it starts at the bottom with velocity $v$:
(a) How far up the slope does it go?
(b) What is the work done by drag, $W_{\text{drag}}$, and gravity $W_g$ on the mass $m$?
My attempt:
$$\sin\theta =\frac{-~mg\vec{y}}{F_g\vec{v}}$$
$$F_{\text{total}}=m \times \frac {\mathrm{d}v}{\mathrm{d}t}=-~bv-\frac {mg}{\sin\theta}$$
$$\frac {\mathrm{d}v}{\mathrm{d}t}=-\frac{bv}{m}-\frac {g}{\sin\theta}$$
Setting $F_{\text{drag}}$ equal to $F_{\text{gravity}}$ is terminal velocity.
$$F_{\text{drag(y)}}=F_g\vec{y}$$
$$-~bv\sin\theta=-~mg$$
$$V_{\text{term}}=\frac{mg}{b\sin\theta}$$
Plug this value into equation above:
$$\frac{-~bv}{m}=\frac{b\sin\theta}{mg} \times \frac{gv}{\sin\theta}$$
$$\frac {\mathrm{d}v}{\mathrm{d}t}=\left[-\frac{1}{V_{\text{term}}} \times \frac{gv}{\sin\theta}\right]-\frac {g}{\sin\theta}$$
$$\frac {\mathrm{d}v}{\mathrm{d}t}=g~\left[\frac{v}{\sin\theta~ {V_{\text{term}}}}-\frac {1}{\sin\theta}\right]$$
So I separate variables(This is where I'm super unsure about my math):
$$\left[\frac{\sin\theta~ {V_{\text{term}}}}{v}- {\sin\theta}\right]~\mathrm{d}v=g \ \mathrm{dt}$$
Integrating to get:
$${\sin\theta~ {V_{\text{term}}}} \times \ln(v)-\sin\theta~ v= gt$$
Then I realized that I was completely lost. Because I'm looking for height, not time or velocity! I wanted to perform a change in variables, but it didn't really work:
$$\frac {\mathrm{d}v}{\mathrm{d}t}=\frac {\mathrm{d}v}{\mathrm{d}x} \times \frac {\mathrm{d}x}{\mathrm{d}t}=\frac {\mathrm{d}v}{\mathrm{d}x} \times v$$
Then I ran out of time. And didn't even attempt B - which is even more complicated.
Could somebody show the right way to solve (a)?

REPLY [1 votes]: The component of the weight down the slope is $mg \sin \theta$ not $\frac{mg}{\sin \theta}$.
There is no terminal speed as the mass goes up the incline, stops, and then comes down again.
So you have your differential equation $\frac {\mathrm{d}v}{\mathrm{d}t}=-\frac{bv}{m}- g\sin\theta$ which you need to solve.
Rather than change variables I think it might be easier to stick to solving this equation.
If you are unsure about how to solve it plug solve[v'[t]=-a*v[t]-b] into WolframAlpha.
The constant of integration you can find by setting the intitial velocity at time $t=0$.
The time for maximum height $t_{\rm max}$ you can find by setting $v=0$.
You can then integrate your velocity equation to find the distance travelled along the slope before the velocity becomes zero.
The second part is perhaps easier than you think in that you can find the work done by the gravitational field on the mass by using the distance moved along the slope to find the vertical height achieved.
You can then do the same thing but assuming that there was no frictional force.
The difference is the work done on the mass by the frictional force.


TITLE: Counting the number of ways to partition N integers to N/3 triples
QUESTION [0 upvotes]: I am interested in counting the number of ways of partitioning N integers (N is a multiple of 3) to N/3 triples. I came up with this formula for which I am looking for a closed form or tight asymptotic upper-bound:
$C = \prod_{j=0}^{j=(N-3)/3}  \binom{N-3j}{3}
$
Trivial upper-bound is $N^N$. I am looking for tight upper-bound.

REPLY [2 votes]: Define $n=\frac N3$ as the number of triplets.  If you don't care about the order the triplets are selected, the number of ways is $$\frac {(3n)!}{(3!)^nn!}$$
You can get this by permuting the numbers in one of $(3n)!$ ways and taking the first three as the first triplet, the next three as the next triplet, and so on.  We don't care about the order of the three in each triplet, which gives us $n$ divisions by $3!$.  We also don't care about the order of the triplets, so we divide by $n!$  Applying Stirling's approximation to the large factorials we get
$$\frac {(3n)!}{(3!)^nn!}\approx \frac {(3n)^{3n}e^n\sqrt{2\pi3n}}{e^{3n}6^nn^n\sqrt{2\pi n}}=\sqrt 3\frac{n^{2n}27^n}{e^{2n}6^n}\approx \sqrt 3\ \cdot0.609^nn^{2n}$$


text:
\chapter{Cheeger constant and Gaussian isoperimetry}

While the previous chapter was devoted to Sobolev inequalities on Dirichlet spaces for which the semigroup satisfies ultracontractive estimates, the present chapter is devoted to situations where the Dirichlet form satisfies a Poincar\'e inequality or a log-Sobolev inequality.
Let $(X,\mu,\mathcal{E},\mathcal{F}=\mathbf{dom}(\mathcal{E}))$ be a symmetric Dirichlet space. Let $\{P_{t}\}_{t\in[0,\infty)}$ denote the Markovian semigroup associated with $(X,\mu,\mathcal{E},\mathcal{F})$. As usual, we assume that $P_t$ is conservative.

\section{Buser's type  inequality for the Cheeger constant of a Dirichlet space}
In the context of  a smooth compact  Riemannian manifold with Riemannian measure $\mu$, Cheeger  \cite{Ch}  introduced   the following isoperimetric constant
\[
h=\inf \frac{\mu (\partial A)}{\mu (A)},
\]
where the infimum runs over all open subsets $A$ with smooth boundary $\partial A$ such that $\mu(A)\le \frac{1}{2}$. Cheeger's constant can be used to bound from below the first non zero eigenvalue of the manifold. Indeed, it is proved in \cite{Ch} that
\[
\lambda_1 \ge \frac{h^2}{4}.
\]

Buser \cite{Bu} then proved that if the Riemannian Ricci curvature of the manifold is non-negative, then we actually have
\[
\lambda_1 \le C h^2
\]
where $C$ is a universal constant depending only on the dimension. Buser's inequality was reproved by Ledoux \cite{Le} using heat semigroup techniques. Under proper assumptions, by using the tools we introduced, Ledoux' technique can essentially reproduced in our general framework of Dirichlet spaces.

We assume in this section that $\mathcal{E}$ satisfies a spectral gap inequality and normalize the measure $\mu$ in such a way that $\mu(X)=1$.  For $\alpha \in (0,1]$, we define the $\alpha$-Cheeger's constant  of $X$ by
\[
h_\alpha=\inf \frac{\| \mathbf 1_E \|_{1,\alpha}}{\mu(E)}
\]
where the infimum runs over all measurable sets $E$ such that $\mu(E)\le \frac{1}{2}$ and $\mathbf 1_E \in \mathbf{B}^{1,\alpha}(X)$. We denote by $\lambda_1$ the spectral gap of $\mathcal{E}$.

\begin{theorem}
We have $h_\alpha \ge (1-e^{-1}) \lambda^\alpha_1$.
\end{theorem}

\begin{proof}
Let $A$ be a set with $P_\alpha (A) :=\| \mathbf 1_A \|_{1,\alpha}< +\infty$. As before, by symmetry and stochastic completeness of the semigroup, we have
\begin{align*}
\Vert \mathbf 1_A - P_t \mathbf 1_A \Vert_{L^1(X,\mu)} =& \int_A (1-P_t \mathbf 1_A) d\mu + \int_{A^c} P_t(\mathbf 1_A) d\mu\\
 =& \int_A (1-P_t \mathbf 1_A) d\mu + \int_A (P_t \mathbf 1_{A^c}) d\mu\\
 =& 2 \left( \mu(A)- \int_A P_t (\mathbf 1_A) d\mu \right)\\
 =& 2 \left( \mu(A)- \Vert P_\frac{t}{2} (\mathbf 1_A) \Vert_{L^2(X,\mu)}^2 \right).
\end{align*}
By definition, we have
\[
\| P_t \mathbf 1_A -\mathbf 1_A \|_{L^1(X,\mu)} \le t^\alpha P_\alpha(A).
\]
We deduce that
\[
\mu(A) \le \frac{1}{2} t^\alpha P_\alpha(A)+ \Vert P_\frac{t}{2} (\mathbf 1_A) \Vert_{L^2(X,\mu)}^2.
\]
Now, by the spectral theorem,
\[
\Vert P_\frac{t}{2} (\mathbf 1_A) \Vert_{L^2(X,\mu)}^2=\mu(A)^2 + \Vert P_\frac{t}{2} (\mathbf 1_A-\mu(A)) \Vert_{L^2(X,\mu)}^2 \le \mu(A)^2 +e^{-\lambda_1 t} \| \mathbf 1_A-\mu(A) \|_{L^2(X,\mu)}^2
\]
This yields
\[
\mu(A) \le \frac{1}{2} t^\alpha P_\alpha(A)+ \mu(A)^2 +e^{-\lambda_1 t} \| \mathbf 1_A-\mu(A) \|_{L^2(X,\mu)}^2.
\]
Equivalently, one obtains
\[
\frac{1}{2} t^\alpha P_\alpha(A) \ge \mu(A) (1 -\mu(A))(1-e^{-\lambda_1 t}).
\]
Therefore,
\[
h_\alpha \ge \sup_{t >0} \left( \frac{1-e^{-\lambda_1 t}}{t^\alpha} \right).
\]
\end{proof}

As already noted in \cite{BK}, let us observe that it is known that the Cheeger lower bound on $\lambda_1$ may be obtained under further assumptions on the Dirichlet space $(X,d,\mathcal{E})$. Indeed, assume  that $\mathcal{E}$ is strictly local with a carr\'e du champ $\Gamma$,  that Lipschitz functions are in the domain of $\mathcal{E}$ and that $\sqrt{\Gamma(f)}$ is an upper gradient in the sense that for any Lipchitz function $f$,
\[
\sqrt{\Gamma(f)}(x) =\lim \sup_{d(x,y)\to 0} \frac{ |f(x)-f(y)|}{d(x,y)}.
\]
In that case, if $A$ is a closed set of $X$, one defines its Minkowski exterior boundary measure by
\[
\mu_+(A)=\lim \inf_{\varepsilon \to 0} \frac{1}{\varepsilon} \left( \mu(A_\varepsilon) -\mu(A) \right),
\]
where $A_\varepsilon=\{ x \in X, d(x,A) <\varepsilon \}$. We can then define a Cheeger's constant of $X$ by
\[
h_+=\inf \frac{\mu_+(E)}{\mu(E)}
\]
where the infimum runs over all closed sets $E$ such that $\mu(E)\le \frac{1}{2}$. Then, according to Theorem 8.5.2 in \cite{BGL}, one has
\[
\lambda_1 \ge \frac{h^2_+}{4}.
\]

\section{Log-Sobolev and Gaussian isoperimetric inequalities}

We assume in this section that $\mathcal{E}$ satisfies a log-Sobolev inequality inequality and normalize the measure $\mu$ in such a way that $\mu(X)=1$.

We define the Gaussian isoperimetric constant of $X$ by
\[
k=\inf \frac{\| \mathbf{1}_E \|_{1,1/2}}{\mu(E)\sqrt{-\ln \mu(E)}}
\]
where the infimum runs over all sets $E$ such that $\mu(E)\le \frac{1}{2}$ and $\mathbf{1}_E \in \mathbf{B}^{1,1/2}(X)$. We denote by $\rho_0$ the log-Sobolev constant of $X$, that is the best constant in the inequality
\begin{equation}\label{log Sob}
\int f^2 \ln f^2 d\mu -\int f^2 d\mu \ln \int f^2 d\mu \le \frac{1}{\rho_0} \mathcal{E}(f,f).
\end{equation}

Following an argument of M. Ledoux \cite{Le}, one obtains:

\begin{theorem}

\[
\rho_0 \le C_{\text{ledoux}} k^2
\]
where $C_{\text{ledoux}}$ is a numerical constant.
\end{theorem}

\begin{proof}
Let $A$ be a measurable set such that $P(A):=\| \mathbf 1_A \|_{1,1/2} <+\infty$. By the same computations as before we have
\[
\mu(A) \le \frac{1}{2} \sqrt{t} P(A)+ \Vert P_\frac{t}{2} (\mathbf 1_A) \Vert_{L^2(X,\mu)}^2.
\]
Now we can use the hypercontractivity constant to bound $\Vert P_\frac{t}{2} (\mathbf 1_A) \Vert_2^2$. Indeed, from Gross' theorem it is well known that the logarithmic Sobolev inequality
\[
\int f^2 \ln f^2 d\mu -\int f^2 d\mu \ln \int f^2 d\mu \le \frac{1}{\rho_0} \mathcal{E}(f,f),
\]
is equivalent to hypercontractivity property
$$
\Vert P_t f \Vert_{L^q(X,\mu)} \leq \Vert f \Vert_{L^p(X,\mu)}
$$
 for all $f$ in $L^p(X,\mu)$ whenever $1<p<q<\infty$ and $e^{\rho_0 t}\geq \sqrt \frac{q-1}{p-1}$.

Therefore, with $p(t)= 1+e^{-2\rho_0 t}<2$, we get,
\begin{align*}
 \sqrt{t} P(A) \geq & 2 \left( \mu(A)- \mu(A)^\frac{2}{p(t)} \right)\\
 &\geq 2 \mu(A) \left(1- \mu(A)^\frac{1-e^{2-\rho_0 t}}{1+e^{-2\rho_0 t}} \right).
\end{align*}
By using then the computation page 956 in \cite{Le}, one deduces that if $A$ is a set which has a finite $P(A)$ and such that $0\leq \mu(A)\leq \frac{1}{2}$, then
$$
P( A)\geq \tilde{C} \sqrt{ \rho_0} \mu(A)\left(\ln \frac{1}{\mu(A)} \right)^\frac{1}{2},
$$
where $\tilde{C}$ is a numerical constant.
\end{proof}


text:
\section{Proofs of some basic properties of the MMSE and the free energy}

\subsection{Proof of Proposition~\ref{prop:mmse_dec}} \label{sec:proof_mmse_dec}
	Let $0 < \lambda_2 \leq \lambda_1$. Define $\Delta_1 = \lambda_1^{-1}$, $\Delta_2 = \lambda_2^{-1}$ and
	$$
	\begin{cases}
		\bbf{Y}_1 = \bbf{X} + \sqrt{\Delta_1} \bbf{Z}_1 \\
		\bbf{Y}_2 = \bbf{X} + \sqrt{\Delta_1} \bbf{Z}_1  + \sqrt{\Delta_2 - \Delta_1} \bbf{Z}_2 \,,
	\end{cases}
	$$
	where $\bbf{X} \sim P_X$ is independent from $\bbf{Z}_1,\bbf{Z}_2 \iid \cN(0,\Id_n)$. Now, by independence between $(\bbf{X},\bbf{Y}_1)$ and $\bbf{Z}_2$ we have
	\begin{align*}
		\MMSE(\lambda_1)
		&= \E \big\| \bbf{X} - \E [\bbf{X} | \bbf{Y}_1] \big\|^2
		= \E \big\| \bbf{X} - \E [\bbf{X} | \bbf{Y}_1, \bbf{Z}_2] \big\|^2
		= \E \big\| \bbf{X} - \E [\bbf{X} | \bbf{Y}_1, \bbf{Y}_2] \big\|^2
		\\
		&\leq \E \big\| \bbf{X} - \E [\bbf{X} | \bbf{Y}_2] \big\|^2
		= \MMSE(\lambda_2) \,.
	\end{align*}
	Next, notice that
	\begin{equation}\label{eq:upper_bound_mmse}
		\MMSE(\lambda_1)
		= \E \big\| \bbf{X} - \E [\bbf{X} | \bbf{Y}_1] \big\|^2
		\leq \E \big\| \bbf{X} - \E [\bbf{X}] \big\|^2 = \MMSE(0)\,.
	\end{equation}
	This shows that the $\MMSE$ is non-increasing on $\R_+$.
	It remains to prove the last point:
	$$
	0 \leq \MMSE(\lambda)
	= \E \| \bbf{X} - \E[\bbf{X}| \bbf{Y}]\|^2
	\leq \E \| \bbf{X} - \frac{1}{\sqrt{\lambda}} \bbf{Y}\|^2 = \frac{n}{\lambda} \xrightarrow[\lambda \to +\infty]{} 0 \,.
	$$

	\subsection{Proof of Proposition~\ref{prop:mmse_cont}} \label{sec:proof_mmse_cont}
	We start by proving that $\MMSE$ is continuous at $\lambda = 0$.
	Let $\lambda\geq0$ and consider $\bbf{Y},\bbf{X},\bbf{Z}$ as given by~\eqref{eq:inference}.

	By dominated convergence one has almost surely that
	$$
	\E[\bbf{X} | \bbf{Y}]
	=
	\frac{\int dP_X(\bbf{x}) \bbf{x} e^{-\frac{1}{2}\| \sqrt{\lambda} \bbf{x} - \bbf{Y}\|^2}}{\int dP_X(\bbf{x}) e^{-\frac{1}{2}\| \sqrt{\lambda} \bbf{x} - \bbf{Y}\|^2}}
	\xrightarrow[\lambda \to 0]{} \E[\bbf{X}] \,.
	$$
	Then by Fatou's Lemma we get
	$$
	\liminf_{\lambda \to 0} \MMSE(\lambda) \geq \E \Big[\liminf_{\lambda \to 0}
		\big\| \bbf{X} - \E[\bbf{X}| \bbf{Y}] \big\|^2
	\Big] = \E \big\| \bbf{X} - \E[\bbf{X}] \big\|^2 \,.
	$$

	Combining this with the bound $\MMSE(\lambda) \leq \E \| \bbf{X} - \E[\bbf{X}]\|^2$ gives $\MMSE(\lambda) \xrightarrow[\lambda \to 0]{} \E \| \bbf{X} -\E[\bbf{X}]\|^2$. This proves that the $\MMSE$ is continuous at $\lambda=0$.
	\\

	Let us now prove that the $\MMSE$ is continuous on $\R_+^*$. We need here a technical lemma:
	\begin{lemma}\label{lem:ui1}
		For all $\lambda > 0$, $p \geq 1$
		$$
		\E \| \bbf{X} - \langle \bbf{x} \rangle_{\lambda} \|^{2p} \leq \frac{2^p (2p !)}{\lambda ^{p} p!} n^{p+1} \,.
		$$
	\end{lemma}
	\begin{proof}
		We reproduce here the proof from~\cite{guo2011estimation}, Proposition~5. We start with the equality
		$$
		\sqrt{\lambda} \left(\bbf{X} - \langle \bbf{x} \rangle_{\lambda}\right)
		=
		\sqrt{\lambda} \bbf{X} - \E [ \sqrt{\lambda} \bbf{X} | \bbf{Y} ]
		=
		\bbf{Y} - \bbf{Z} - \E [ \bbf{Y} - \bbf{Z} | \bbf{Y} ]
		=
		\E [ \bbf{Z} | \bbf{Y} ] -\bbf{Z} \,.
		$$
		We have therefore
		\begin{align*}
			\E \| \bbf{X} - \langle \bbf{x} \rangle_{\lambda} \|^{2p}
			&= \frac{1}{\lambda ^{p}}
			\E \big\| \E [ \bbf{Z} | \bbf{Y} ] -\bbf{Z}\big\|^{2p}
			\leq
			\frac{2^{2p-1}}{\lambda ^{p}}
			\E \big[  \|\E [ \bbf{Z} | \bbf{Y} ]\|^{2p} +\|\bbf{Z}\|^{2p} \big]
			\leq
			\frac{2^{2p}}{\lambda ^{p}}
			\E \|\bbf{Z}\|^{2p} \,.
		\end{align*}
		It remains to bound
		$$
		\E \|\bbf{Z}\|^{2p}
		\leq n^p \E \left[\sum_{i=1}^n Z_i^{2p} \right]
		= n^{p+1} \frac{(2p)!}{2^p p!} \,.
		$$
	\end{proof}
	\\

	Let $\lambda_0 > 0$. The family of random variables $\big(\| \bbf{X} - \langle \bbf{x}\rangle_{\lambda}\|^2\big)_{\lambda \geq \lambda_0}$ is bounded in $L^2$ by Lemma~\ref{lem:ui1} and is therefore uniformly integrable. The function $\lambda \mapsto \| \bbf{X} - \langle \bbf{x}\rangle_{\lambda}\|^2$ is continuous on $[\lambda_0,+\infty)$, the uniform integrability ensures then that $\MMSE: \lambda \mapsto  \E\| \bbf{X} - \langle \bbf{x}\rangle_{\lambda}\|^2$ is continuous over $[\lambda_0,+\infty)$. This is valid for all $\lambda_0 >0$: we conclude that $\MMSE$ is continuous over $(0,+\infty)$.

	\subsection{Proof of the I-MMSE relation: Proposition~\ref{prop:i_mmse}} \label{sec:proof_i_mmse}
	\begin{align*}
		\MMSE(\lambda)
		= \E \| \bbf{X} - \langle \bbf{x} \rangle_{\lambda} \|^2
		= \E \| \bbf{X} \|^2 + \E \| \langle \bbf{x} \rangle_{\lambda} \|^2 - 2 \E \langle \bbf{x}^{\sT} \bbf{X} \rangle_{\lambda}
	\end{align*}
	Now, by the Nishimori property $\E \| \langle \bbf{x} \rangle_{\lambda} \|^2 = \E \big\langle (\bbf{x}^{(1)})^{\sT} \bbf{x}^{(2)} \big\rangle_{\lambda} = \E \langle \bbf{x}^{\sT} \bbf{X} \rangle_{\lambda}$. Thus
	\begin{equation}\label{eq:mmse_overlap}
		\MMSE(\lambda)
		= \E \| \bbf{X} \|^2 -  \E \langle \bbf{x}^{\sT} \bbf{X} \rangle_{\lambda} \,.
	\end{equation}
	By~\eqref{eq:mmse_overlap} and~\eqref{eq:f_i}, it suffices now to prove the second equality in~\eqref{eq:i_mmse}. This will follow from the lemmas below.

	\begin{lemma}\label{lem:free_energy_cont}
		The free energy $F$ is continuous at $\lambda = 0$.
	\end{lemma}
	\begin{proof} For all $\lambda \geq 0$,
		$$
		F(\lambda) = \E \log \int dP_X(\bbf{x}) e^{-\frac{1}{2} \| \bbf{Y}-\sqrt{\lambda} \bbf{x}\|^2 + \frac{1}{2} \|\bbf{Y}\|^2}
		= \E \log \int dP_X(\bbf{x}) e^{-\frac{1}{2} \| \sqrt{\lambda}\bbf{X}-\sqrt{\lambda} \bbf{x} + \bbf{Z}\|^2}
		+ \lambda \E \|\bbf{X}\|^2 + n \,.
		$$
		By dominated convergence $\int dP_X(\bbf{x}) e^{-\frac{1}{2} \| \sqrt{\lambda}\bbf{X}-\sqrt{\lambda} \bbf{x} + \bbf{Z}\|^2} \xrightarrow[\lambda \to 0]{} e^{-\frac{1}{2}\|\bbf{Z}\|^2}$. Jensen's inequality gives
		\begin{align*}
			\left|\log \int \!\!dP_X(\bbf{x}) e^{-\frac{1}{2} \| \sqrt{\lambda}\bbf{X}-\sqrt{\lambda} \bbf{x} + \bbf{Z}\|^2} \right|
			&=
			- \log \int \!\!dP_X(\bbf{x}) e^{-\frac{1}{2} \| \sqrt{\lambda}\bbf{X}-\sqrt{\lambda} \bbf{x} + \bbf{Z}\|^2}
			\\
			&\leq \frac{1}{2} \! \int \!\!dP_X(\bbf{x}) \| \sqrt{\lambda}\bbf{X}-\sqrt{\lambda} \bbf{x} + \bbf{Z}\|^2
			\leq \frac{3}{2}\big(\|\bbf{X}\|^2 + \E \|\bbf{X}\|^2 + \|\bbf{Z}\|^2 \big),
		\end{align*}
		for all $\lambda \in [0,1]$. One can thus apply the dominated convergence theorem again to obtain that $F$ is continuous at $\lambda=0$.
	\end{proof}
	\\

	\begin{lemma}\label{lem:der_f0}
		For all $\lambda \geq 0$,
		$$
		F(\lambda) - F(0) = \frac{1}{2} \int_{0}^{\lambda} \E \langle \bbf{x}^{\sT} \bbf{X} \rangle_{\gamma} d\gamma \,.
		$$
	\end{lemma}
	\begin{proof}
		Compute for $\lambda > 0$
		$$
		\frac{\partial}{\partial \lambda} \log \cZ(\lambda,\bbf{Y}) =
		\left\langle \frac{1}{2 \sqrt{\lambda}} \bbf{x}^{\sT} \bbf{Z} + \bbf{x}^{\sT} \bbf{X} - \frac{1}{2} \|\bbf{x}\|^2 \right\rangle_{\lambda} \,.
		$$
		Since $\E \|\bbf{X}\|^2 < \infty$, the right-hand side is integrable and one can apply Fubini's theorem to obtain
		$$
		F(\lambda_2) - F(\lambda_1) = \int_{\lambda_1}^{\lambda_2} \E
		\left\langle \frac{1}{2 \sqrt{\lambda}} \bbf{x}^{\sT} \bbf{Z} + \bbf{x}^{\sT} \bbf{X} - \frac{1}{2} \|\bbf{x}\|^2 \right\rangle_{\lambda} d\lambda\,.
		$$
		By Gaussian integration by parts, we have for all $i \in \{1, \dots, n \}$ and $\lambda >0$
		$$
		\E Z_i \langle x_i \rangle_{\lambda} = \E \frac{\partial}{\partial Z_i} \langle x_i \rangle_{\lambda}
		= \E \left[ \big\langle \sqrt{\lambda} x_i^2 \big\rangle_{\lambda} - \sqrt{\lambda} \big\langle x_i \big\rangle_{\lambda}^2 \right]
		= \sqrt{\lambda} \E \left[ \big\langle x_i^2 \big\rangle_{\lambda} - \big\langle x_i X_i \big\rangle_{\lambda} \right]
		\,,
		$$
		where the last equality comes from the Nishimori property (Proposition~\ref{prop:nishimori}). We have therefore
		$$
		F(\lambda_2) - F(\lambda_1) = \frac{1}{2} \int_{\lambda_1}^{\lambda_2} \E
		\left\langle  \bbf{x}^{\sT} \bbf{X} \right\rangle_{\lambda} d\lambda \,.
		$$
		By Lemma~\ref{lem:free_energy_cont}, $F$ is continuous at $0$ so we can take the limit $\lambda_1 \to 0$ to obtain the result.
	\end{proof}
	\\

	By Proposition~\ref{prop:mmse_cont}, the function $\lambda \mapsto \MMSE(\lambda)$ is continuous over $\R_+$. By~\eqref{eq:mmse_overlap} we deduce that $\lambda \mapsto \E \langle \bbf{x}^{\sT} \bbf{X} \rangle_{\lambda}$ is continuous over $\R_+$ and therefore Lemma~\ref{lem:der_f0} proves~\eqref{eq:i_mmse}.
	\\

	It remains only to show that $F$ is strictly convex when $P_X$ differs from a Dirac mass.
	We proceed by truncation. For $N \in \N$ and $x \in \R$ we write $x^{(N)} = x \, \1(-N \leq x \leq N)$. We extend this notation to vectors $\bbf{x} = (x_1,\dots, x_n) \in \R^n$ by $\bbf{x}^{(N)} = (x_1^{(N)}, \dots, x_n^{(N)})$.

	For $\bbf{X} \sim P_X$ we define $P_X^{(N)}$ as the distribution of $\bbf{X}^{(N)}$. $F^{(N)}$, $\MMSE^{(N)}$  and $\langle \cdot \rangle_{\lambda,N}$ will denote respectively the corresponding free energy, MMSE and posterior distribution.
	One can compute the second derivative (since $\bbf{X}^{(N)}$ is bounded, one can easily differentiate under the integral sign) and again, using Gaussian integration by parts and the Nishimori identity one obtains:
	\begin{equation}\label{eq:psi_seconde}
		F^{(N)\prime\prime}(\lambda) = \frac{1}{2} \E \left[
			\Tr \left(\left( \langle \bbf{x}\bbf{x}^{\sT} \rangle_{\lambda,N} - \langle \bbf{x} \rangle_{\lambda,N}\langle \bbf{x} \rangle_{\lambda,N}^{\sT}
			\right)^2\right)
	\right].

\end{equation}
By Cauchy-Schwarz inequality, we have for all positive, semi-definite matrix $\bbf{M} \in \R^{n \times n}$, $\Tr(\bbf{M})^2 \leq n \Tr(\bbf{M}^2)$. Hence
\begin{align*}
		F^{(N)\prime \prime}(\lambda)
		&\geq \frac{1}{2n} \E \left[
			\Tr \left( \langle \bbf{x}\bbf{x}^{\sT} \rangle_{\lambda,N} - \langle \bbf{x} \rangle_{\lambda,N}\langle \bbf{x} \rangle_{\lambda,N}^{\sT}
			\right)^2
	\right]
	\\
		&\geq \frac{1}{2n} \E \left[
			\Tr \left( \langle \bbf{x}\bbf{x}^{\sT} \rangle_{\lambda,N} - \langle \bbf{x} \rangle_{\lambda,N}\langle \bbf{x} \rangle_{\lambda,N}^{\sT}
			\right)
	\right]^2
		= \frac{1}{2n} \MMSE^{(N)}(\lambda)^2,
\end{align*}
	by Jensen's inequality.
	Let now $0<s<t$.

	By integrating~\eqref{eq:psi_seconde} we get
	\begin{equation}\label{eq:der_sec_int}
		F^{(N)\prime}(t) - F^{(N)\prime}(s) \geq \frac{1}{2n}
		\int_s^t
		\MMSE^{(N)}(\lambda)^2
		d\lambda \,.
	\end{equation}
	The sequence of convex functions $(F^{(N)})_N$ converges (by Proposition \ref{prop:free_wasserstein}) to $F$ which is differentiable. Proposition \ref{prop:deriv_convex} gives that the derivatives $(F^{(N)\prime})_N$ converge to $F'$ and therefore $\MMSE^{(N)}$ converges to $\MMSE$. Therefore, equation~\eqref{eq:der_sec_int} gives
	$$
		F'(t) - F'(s) \geq \frac{1}{2n}
		\int_s^t
		\MMSE(\lambda)^2
		d\lambda
		\geq \frac{1}{2n} (t-s) \MMSE(t)^2
		\,.
	$$
	If $P_0$ is not a Dirac measure, then the last term is strictly positive: this concludes the proof.

	\subsection{Pseudo-Lipschitz continuity of the free energy with respect to the Wasserstein distance}\label{sec:free_wasserstein}

	Let $P_{1}$ and $P_{2}$ be two probability distributions on $\R^n$, that admits a finite second moment. We denote by $W_2(P_1,P_2)$ the Wasserstein distance of order $2$ between $P_1$ and $P_2$.
	For $i = 1,2$ the free energy is defined as
	$$
	F_{P_i}(\lambda) = \E \log \int dP_i(\bbf{x}) \exp \Big(\sqrt{\lambda} \bbf{x}^{\sT} \bbf{Z} + \lambda \bbf{x}^{\sT} \bbf{X}- \frac{\lambda}{2} \|\bbf{x}\|^2 \Big) \,,
	$$
	where the expectation is with respect to $(\bbf{X},\bbf{Z}) \sim P_i \otimes \cN(0,\Id_n)$.

	\begin{proposition}\label{prop:free_wasserstein}
		For all $\lambda \geq 0$,
		$$
		\big| F_{P_1}(\lambda) - F_{P_2}(\lambda) \big| \leq \frac{\lambda}{2} \big(\sqrt{\E_{P_1}\|\bbf{X}\|^2} + \sqrt{\E_{P_2}\|\bbf{X}\|^2}\big)W_2(P_1,P_2) \,.
		$$
	\end{proposition}
A similar result was proved in~\cite{wu2012functional} but with a weaker bound for the $W_2$ distance.

	\begin{proof}
		Let $\epsilon > 0$. Let us fix a coupling $Q$ of $\bbf{X}_1 \sim P_1$ and $\bbf{X}_2 \sim P_2$ such that
		$$
		\big( \E \| \bbf{X}_1 - \bbf{X}_2 \|^2\big)^{1/2} \leq W_2(P_1,P_2) + \epsilon \,.
		$$
		Let us consider for $t \in [0,1]$ the observation model
		$$
		\begin{cases}
			\bbf{Y}^{(t)}_1 &=\  \sqrt{\lambda t} \bbf{X}_1  + \bbf{Z}_1 \,, \\
			\bbf{Y}^{(t)}_2 &=\  \sqrt{\lambda (1-t)} \bbf{X}_2 + \bbf{Z}_2 \,,
		\end{cases}
		$$
		where $\bbf{Z}_1,\bbf{Z}_2 \iid \cN(0,\Id_n)$ are independent from $(\bbf{X}_1,\bbf{X}_2) \sim Q$. Define
		$$
		f(t) = \E \log \int\! dQ(\bbf{x}_1,\bbf{x}_2) \exp \Big(
			\sqrt{\lambda t} \bbf{x}_1^{\sT} \bbf{Y}_1^{(t)} - \frac{\lambda t}{2} \|\bbf{x}_1\|^2
			+
			\sqrt{\lambda (1-t)} \bbf{x}_2^{\sT} \bbf{Y}_2^{(t)} - \frac{\lambda (1-t)}{2} \|\bbf{x}_2\|^2
		\Big).
		$$
		We have $f(0) = F_{P_2}(\lambda)$ and $f(1)=F_{P_1}(\lambda)$. By an easy extension of the I-MMSE relation~\eqref{eq:i_mmse} we have for all  $t \in [0,1]$:
		$$
		f'(t) = \frac{\lambda}{2}\E \Big\langle \bbf{X}_1^{\sT} \bbf{x}_1 - \bbf{X}_2^{\sT} \bbf{x}_2 \Big\rangle_{t} \,,
		$$
		where $\langle \cdot \rangle_t$ denotes the expectation with respect to $(\bbf{x}_1,\bbf{x}_2)$ sampled from the posterior distribution of $(\bbf{X}_1,\bbf{X}_2)$ given $\bbf{Y}_1^{(t)},\bbf{Y}_2^{(t)}$, independently of everything else.
		We have then
		\begin{align*}
			| \frac{2}{\lambda} f'(t) | & =
			\Big| \E \Big\langle \bbf{X}_1^{\sT} (\bbf{x}_1 - \bbf{x}_2) - (\bbf{X}_2 - \bbf{X}_1)^{\sT} \bbf{x}_2 \Big\rangle_{t} \Big|
			\\
			&\leq
			\Big( \E \|\bbf{X}_1\|^2 \E \big\langle \| \bbf{x}_1 - \bbf{x}_2\|^2\big\rangle_{t} \Big)^{1/2} +
				\Big( \E \big\langle\|\bbf{x}_2\|^2\big\rangle_{t}  \E  \|\bbf{X}_2 - \bbf{X}_1\|^2   \Big)^{1/2}
			\\
			&=
			\Big( \E \|\bbf{X}_1\|^2 \E \| \bbf{X}_1 - \bbf{X}_2\|^2 \Big)^{1/2} +
				\Big( \E \|\bbf{X}_2\|^2  \E  \|\bbf{X}_2 - \bbf{X}_1\|^2   \Big)^{1/2}
				\\
				& \leq
				\Big(
				\big(\E \|\bbf{X}_1\|^2 \big)^{1/2}
				+
				\big(\E \|\bbf{X}_2\|^2\big)^{1/2}
			\Big) (W_2(P_1,P_2) + \epsilon)\,,
		\end{align*}
		where we used successively the Cauchy-Schwarz inequality and the Nishimori property (Proposition~\ref{prop:nishimori}). We then let $\epsilon \to 0$ to obtain the result.
	\end{proof}

	\section{Convex analysis results}\label{sec:app_convex}

\begin{proposition}\label{prop:deriv_convex}
	Let $I \subset \R$ be an interval, and let $(f_n)_{n \geq 0}$ be a sequence of convex functions on $I$ that converges pointwise to a function $f$. Then for all $t \in I$ for which these inequalities have a sense
	$$
	f'(t^-) \leq \liminf_{n \to \infty} f_n'(t^-) \leq \limsup_{n \to \infty} f_n'(t^+) \leq f'(t^+).
	$$
\end{proposition}
\begin{proof}
	Let $t \in I$ and $h > 0$. By convexity
	$$
	f_n'(t^+) \leq \frac{f_n(t+h) - f_n(t)}{h} \xrightarrow[n \to \infty]{} \frac{f(t+h) - f(t)}{h} \xrightarrow[h \to 0]{} f'(t^+).
	$$
	The first inequality follows from the same arguments.
\end{proof}

	\subsection{Basic results on the monotone conjugate}

	\begin{definition}\label{def:monotone_conjugate}
		We define the \textit{monotone conjugate} (see \cite{rockafellar2015convex} p.110) of a non-decreasing convex function $f: \R_+ \to \R$ by:
		\begin{equation}\label{eq:def_monotone}
			f^*(x) = \sup_{y \geq 0}\, \{ xy - f(y) \}.
		\end{equation}
	\end{definition}

	The most fundamental result on the monotone  conjugate is the analog of the Fenchel-Moreau theorem:
\begin{proposition}[\cite{rockafellar2015convex}~Theorem~12.4]\label{prop:inverse}
	Let $f$ be a non-decreasing lower semi-continuous convex function on $\R_+$ such that $f(0)$ is finite. Then $f^*$ is another such function and $(f^*)^* = f$.
\end{proposition}

\begin{proposition}\label{prop:der_conjugate}
	Let $f$ be a non-decreasing lower semi-continuous convex function on $\R_+$ such that $f(0)$ is finite. Then for all $x,y \geq 0$:
	$$
	x \in \partial f^*(y)
	\ \Longleftrightarrow \
	f(x) + f^*(y) = xy
	\ \Longleftrightarrow \
	y \in \partial f(x).
	$$
\end{proposition}
\begin{proof}
	Let $x \in \partial f^*(y)$. We get that $(f^*)^*(x) = x y - f^*(y)$ and therefore that
	$f^*(y) = x y - f(x)$, by Proposition \ref{prop:inverse}. This gives that $x$ maximizes $s \mapsto s y - f(s)$ over $\R_+$ and thus $y \in \partial f(x)$.
	It remains to show that $y \in \partial f(x) \implies x \in \partial f^*(y)$. This follows from Proposition \ref{prop:inverse} and the implication $ x \in \partial f^*(y) \implies y \in \partial f(x)$ that we just showed.
\end{proof}

	\subsection{A supremum formula}
The goal of this section is to prove:
\begin{proposition}\label{prop:sup_inf}
	Let $f,g$ be two convex Lipschitz functions on $\R_+$.

	For $(q_1,q_2) \in \R_+$ we define $\varphi(q_1,q_2) = f(q_1) + g(q_2) - q_1 q_2$
	and $\psi(q_1,q_2) = q_1 q_2 - f^*(q_2) - g^*(q_1)$.
	Then the set $\Gamma = \big\{(q_1,q_2) \in \R_+^2 \, \big| \, q_1 \in \partial g(q_2), \ q_2 \in \partial f(q_1) \big\}$ is non-empty and:
	\begin{equation}\label{eq:sup_inf}
		\sup_{(q_1,q_2) \in \Gamma} \varphi(q_1,q_2) \
		= \sup_{q_1,q_2 \geq 0} \psi(q_1,q_2) \
		=\sup_{q_1 \geq 0} \inf_{q_2 \geq 0} \varphi(q_1,q_2),
	\end{equation}
	and the two first suprema above are achieved and precisely at the same couples $(q_1,q_2)$.
	\\

	If moreover $f$ and $g$ are both differentiable and strictly convex, then the same result holds for $\Gamma$ replaced by
	\begin{equation}\label{eq:simpli_gamma}
	\widetilde{\Gamma} = \big\{ (q_1,q_2) \in \R_+^2 \, \big| \, q_2 = f'(q_1) \ \text{and} \ q_1 = g'(q_2) \big\}.
	\end{equation}
\end{proposition}
\begin{proof}
	Let $L_f$ (resp.\ $L_g$) be the Lipschitz constant of $f$ (resp.\ $g$). For $x > L_f$, $f^*(x) = +\infty$ and (since $f^*$ is lower semi-continuous by Proposition \ref{prop:der_conjugate}) $f^*(x) \to +\infty$ as $x \to L_f$. Analogously, $g(x) \to +\infty$ as $x \to L_g$. The function $\psi$ is therefore continuous on $[0,L_g) \times [0,L_f)$ and goes to $-\infty$ on the border $\big(\{L_g\} \times [0,L_f]\big) \cup \big( [0,L_g] \times \{ L_f \} \big)$.

	The functions $\psi$ achieves therefore its maximum at some $(q_1,q_2) \in [0,L_g) \times [0,L_f)$. $(q_1,q_2)$ verifies then $q_2 \in \partial g^*(q_1)$ and $q_1 \in \partial f^*(q_2)$ which gives $(q_1,q_2) \in \Gamma$ by Proposition \ref{prop:der_conjugate}. The set $\Gamma$ is therefore non-empty and
	$$
	\sup_{q_1,q_2 \geq 0} \psi(q_1,q_2) \leq \sup_{(q_1,q_2)\in \Gamma} \varphi(q_1,q_2).
	$$
	By definition of the conjugates $f^*$ and $g^*$ we have for all $q_1,q_2 \geq 0$
	$$
	\begin{cases}
		f(q_1) + f^*(q_2) \geq q_1 q_2 \\
		g(q_2) + g^*(q_1) \geq q_1 q_2.
	\end{cases}
	$$
	We get that $\varphi(q_1,q_2) \geq \psi(q_1,q_2)$ with equality if and only if $(q_1,q_2) \in \Gamma$, by Proposition \ref{prop:der_conjugate}. This gives in particular that
	$$
	\sup_{q_1,q_2 \geq 0} \psi(q_1,q_2) \geq \sup_{(q_1,q_2)\in \Gamma} \varphi(q_1,q_2).
	$$
	Hence, both supremum are equal and are achieved over the same couples because we have seen that all couple $(q_1,q_2)$ that achieves the supremum of $\psi$ is in $\Gamma$.

	We consider now the second equality.
	Using the definition of the monotone conjugate \eqref{eq:def_monotone} and Proposition \ref{prop:inverse}:
	\begin{align*}
		\sup_{q_1 \geq 0} \inf_{q_2 \geq 0} \varphi(q_1,q_2)
		&= \sup_{q_1 \geq 0} \big\{ f(q_1) - g^*(q_1) \big\}
		= \sup_{q_1 \geq 0} \sup_{q_2 \geq 0} \big\{ q_1 q_2 - f^*(q_2) - g^*(q_1) \big\}.
	\end{align*}

	Let us now prove the second part of the Proposition: we now assume that $f$ and $g$ are differentiable, strictly convex.
	Let $(q_1,q_2) \in \Gamma$ be a couple that achieves the maximum of $\varphi$ over $\Gamma$. It suffices to show that $(q_1,q_2) \in \widetilde{\Gamma}$.
	If $q_1 > 0$ and $q_2 >0$, then this is trivial because $f$ and $g$ are differentiable.

	Suppose now that $q_1 = 0$ (the case $q_2 = 0$ follows by symmetry). Since $0 = q_1 \in \partial g(q_2)$ and $g$ is strictly increasing, we get that $q_2 = 0$,
	so that $\sup \psi = \varphi(q_1,q_2) = f(0) + g(0)$.
	Notice that for $f^*(f'(0)) = -f(0)$ and $g^*(g'(0)) = -g(0)$ so
	\begin{equation}\label{eq:lower_bound0}
		f(0) + g(0) = \sup \psi \geq \psi\big(g'(0),f'(0)\big) = f(0) + g(0) + f'(0)g'(0) \geq f(0) + g(0).
	\end{equation}
	We get that $(g'(0),f'(0))$ achieves the supremum of $\psi$, which implies that $(g'(0),f'(0)) \in \Gamma$. This gives that $g'(0) = 0 = f'(0)$ by strict convexity of $f$ and $g$.
	This proves that $(q_1,q_2) = (0,0) \in \widetilde{\Gamma}$.
\end{proof}

	\section{Differentiation of a supremum of functions} \label{sec:appendix_envelope}

We recall in this section two results about the differentiation of a supremum of functions from Milgrom and Segal~\cite{milgrom2002envelope}.
Let $X$ be a set of parameters and consider a function $f: X \times [0,1] \to \R$. Define, for $t \in [0,1]$
\begin{align*}
	V(t) &= \sup_{x \in X} f(x,t) \,,\\
	X^*(t) &= \big\{ x \in X \, \big| \, f(x,t) = V(t) \big\} \,.
\end{align*}

\begin{proposition}[Theorem~1 from~\cite{milgrom2002envelope}\,] \label{prop:envelope}
	Let $t \in [0,1]$ such that $X^*(t) \neq \emptyset$. Let $x^* \in X^*(t)$ and suppose that $f(x^*,\cdot)$ is differentiable at $t$, with derivative $f_t(x^*,t)$.
	\begin{itemize}
		\item  If $t > 0$ and if $V$ is left-hand differentiable at $t$, then $V'(t^-) \leq f_t(x^*,t)$.
		\item  If $t < 0$ and if $V$ is right-hand differentiable at $t$, then $V'(t^+) \geq f_t(x^*,t)$.
		\item  If $t \in (0,1)$ and if $V$ is differentiable at $t$, then $V'(t) = f_t(x^*,t)$.
	\end{itemize}
\end{proposition}

\begin{proposition}[Corollary~4 from~\cite{milgrom2002envelope}\,] \label{prop:envelope_compact}
	Suppose that $X$ is nonempty and compact. Suppose that for all $t\in [0,1]$, $f(\cdot,t)$ is continuous. Suppose also that $f$ admits a partial derivative $f_t$ with respect to $t$ that is continuous in $(x,t)$ over $X \times [0,1]$. Then
	\begin{itemize}
		\item $\displaystyle V'(t^+) = \max_{x^* \in X^*(t)} f_t(x^*,t)$ for all $t\in [0,1)$ and $\displaystyle V'(t^-) = \min_{x^* \in X^*(t)} f_t(x^*,t)$ for all $t\in (0,1]$.
		\item $V$ is differentiable at $t \in (0,1)$ is and only if \,$\displaystyle \Big\{ f_t(x^*,t) \, \Big| \, x^* \in X^*(t) \Big\}$
			is a singleton. In that case $V'(t) = f_t(x^*,t)$ for all $x^* \in X^*(t)$.
	\end{itemize}
\end{proposition}


TITLE: Find the minimum of an integral
QUESTION [1 upvotes]: Let M be the set of the Borel integrable functions $f:[0,\pi]\to \mathbb{R}$ such that
$$\int_0^\pi f(x) \sin(x) dx= \int_0^\pi f(x) \cos(x) dx=1.$$ Find $$\min_{f\in M} \int_0^\pi f^2(x) dx.$$ Thank you.

REPLY [0 votes]: Use Cauchy–Schwarz inequality:
1) $\left (\int_{0}^{\pi }f(x)sin(x)dx  \right )^2\leq \left (  \int_{0}^{\pi }f^{2}(x)dx\right )\left (  \int_{0}^{\pi }sin^{2}(x)dx\right )$
\begin{align}\int_{0}^{\pi }f^{2}(x)dx \geq \frac{1}{\int_{0}^{\pi }sin^{2}(x)dx}\end{align}
2)$\left (\int_{0}^{\pi }f(x)cos(x)dx  \right )^2\leq \left (  \int_{0}^{\pi }f^{2}(x)dx\right )\left (  \int_{0}^{\pi }cos^{2}(x)dx\right )$
\begin{align}\int_{0}^{\pi }f^{2}(x)dx \geq \frac{1}{\int_{0}^{\pi }cos^{2}(x)dx}\end{align}
Therefore:
\begin{align}\int_{0}^{\pi }f^{2}(x)dx \geq max\left ( \frac{1}{\int_{0}^{\pi }sin^{2}(x)dx},\frac{1}{\int_{0}^{\pi }cos^{2}(x)dx} \right ) \end{align}
or:
\begin{align}\int_{0}^{\pi }f^{2}(x)dx \geq max\left ( \frac{4}{2\pi-sin(2\pi)},\frac{4}{2\pi+sin(2\pi)} \right )= \frac{2}{\pi}\end{align}


TITLE: Fitting a line through a complex 2D shape, use of multi-dimensional projection?
QUESTION [0 upvotes]: I am working with a complex 2-dimensional shape (similar to this image), and would like to pass a single, continuous line through it to aid future analyses. I have distinct data points on each pixel of this object and need to keep them independent and ordered.
After a while of trying to fit a line to it with different compromises/unfolding, I remembered that in linear regression, you map data points to higher dimensions, turning data that cannot be fitted by a line, into data that can be fitted by a line.
Is it possible to do a similar thing in this case? Would it be reasonable for me to try to map this object into 3-4+ dimensional space to get a line through it/ make it linear?
Does the fact that there are several "end points" mean that it would not be possible, at all, to linearize my object?

Edit (to add details):
I have data points on an object surface that looks like the above linked image. They are in a complex 2-dimensional shape, and analyzing them effectively is not practical while in this shape.
I want to transfer my data onto a line. On this line, I would like to keep the order of my data, and relationships between points.
I would like to do this while losing as little data as possible.
An example of what I was thinking of:

Circle and 5 lines drawn in X,Y.
Cut the circle at one point and unfold it (circle + 5lines would turn into line + 5 lines).
Rotating my object in Z, and align my 5 lines to the 1 main line, as to have all lines in the same vertical plane as my object.
Collapse my points.

This method is very inoptimal, and I would be:
1) losing a lot of data from it
2) disrupting some of the data on my main line (or circle).
Another example:

"Blow up" my structure so as to have each line represented in two ways (center to edge, and edge back to center).
Collapse the structure onto a line by disconnecting the circle at all line points
What I would have is: Arc segment 1 > line1 up > line1 down > Arc segment 2 > line2 up > line2 down > Arc segment 3... etc

This method would make me lose the proximity relationship between the points on my arc.
I was wondering if mapping this structure onto higher dimensions before transforming it into a line would yield better results.

REPLY [0 votes]: I don't know if I understood your problem, but if you have a sequence of 2-D points
$$
(x_1,y_1), (x_2,y_2), \ldots, (x_n,y_n),
$$
then you can turn them into 3-D points like that:
$$
(x_1,y_1,h), (x_2,y_2,2h), \ldots, (x_n,y_n,nh),
$$
where $h$ is any non-zero quantity.
Could that help you?


TITLE: Relation $R$ on $V$ is given by $x+y$ is even
QUESTION [0 upvotes]: A relation $R$ on $V$ is given by $x+y$ is even. How can we show that if integers $x$ and $y$ are $R$-related then either $x$ and $y$ are both even or $x$ and $y$ are both odd? I've been looking through Google for information on how to answer relation questions but without any luck at all. I think I need to use equivalence relations but I am not entirely sure.

REPLY [3 votes]: If $x$ is even and $y$ is odd, then $x=2k$ for some integer $k$ and $y=2l+1$ for some integer $l$, so $x+3y=2k+3(2l+1)=2(k+3l+1)+1$ is odd.
If $x$ is odd and $y$ is even, then $x=2m+1$ for some integer $m$ and $y=2n$ for some integer $n$, so $x+3y=2m+1+3(2n)=2(m+3n)+1$ is odd.


TITLE: Derivation of the second total derivative of a function
QUESTION [1 upvotes]: Let there be a function $f$ of two variables $x$ and $y$ who are dependent on a single variable $t$. I tried to derive the second order total derivative of f, $\frac{df}{dt}$
Before asking this, I have checked the topic Intuition behind second total derivative
Here the poster is not asking how to derive it, they want to get the intuition behind it. Here, I have a problem deriving that six-term expression.
So, the first total derivative is
$\frac { \mathrm d f } { \mathrm d t } = \frac { \partial f } { \partial x } \frac { \mathrm d x } { \mathrm d t } + \frac { \partial f } { \partial y } \frac { \mathrm d y } { \mathrm d t } $
and the second total derivative is
$\frac{\mathrm d}{\mathrm d t }(\frac { \mathrm d f } { \mathrm d t }) =\frac{\mathrm d}{\mathrm d t }( \frac { \partial f } { \partial x } \frac { \mathrm d x } { \mathrm d t }) + \frac{\mathrm d}{\mathrm d t }(\frac { \partial f } { \partial y } \frac { \mathrm d y } { \mathrm d t } )$
Applying the product rule to two terms to the right of equal sign:
$\frac { \mathrm d ^ 2 f } { \mathrm d t ^ 2 }$=$(\frac{\mathrm d}{\mathrm d t } \frac { \partial f } { \partial x })\frac { \mathrm d x } { \mathrm d t }+\frac { \partial f } { \partial x }\frac{\mathrm d}{\mathrm d t }\frac { \mathrm d x } { \mathrm d t }+(\frac{\mathrm d}{\mathrm d t } \frac { \partial f } { \partial y })\frac { \mathrm d y } { \mathrm d t }+\frac { \partial f } { \partial y }\frac{\mathrm d}{\mathrm d t }\frac { \mathrm d y } { \mathrm d t }$
and then:
$\frac { \ d ^ 2 f } { \mathrm d t ^ 2 }=\frac { \partial ^2f } { \partial  x ^ 2 }\frac { \mathrm d x } { \mathrm d t }\frac { \mathrm d x } { \mathrm d t }+\frac { \partial f } { \partial x }\frac { \mathrm d^2 x } { \mathrm d t^2 }+\frac { \partial ^ 2 f } { \partial y ^ 2 }\frac { \mathrm d y } { \mathrm d t }\frac { \mathrm d y } { \mathrm d t }+\frac { \partial f } { \partial y }\frac { \mathrm d^2 y } { \mathrm d t^2 }$
Finally:
$\frac { \mathrm d ^ 2 f } { \mathrm d t ^ 2 }=\frac { \partial ^ 2 f } { \partial x ^ 2 } ( \frac { \mathrm d x } { \mathrm d t } ) ^ 2 +\frac { \partial f } { \partial x }\frac { \mathrm d^2 x } { \mathrm d t^2 }+\frac { \partial ^ 2 f } { \partial y ^ 2 } ( \frac { \mathrm d y } { \mathrm d t } ) ^ 2+\frac { \partial f } { \partial y }\frac { \mathrm d^2 y } { \mathrm d t^2 }$
However, the correct expression is
$\frac { \mathrm d ^ 2 f } { \mathrm d t ^ 2 } = \frac { \partial ^ 2 f } { \partial x ^ 2 } ( \frac { \mathrm d x } { \mathrm d t } ) ^ 2 + \frac { \partial ^ 2 f } { \partial y \, \partial x } \frac { \mathrm d y } { \mathrm d t } \frac { \mathrm d x } { \mathrm d t } + \frac { \partial f } { \partial x } \frac { \mathrm d ^ 2 x } { \mathrm d t ^ 2 } + \frac { \partial ^ 2 f } { \partial x \, \partial y } \frac { \mathrm d x } { \mathrm d t } \frac { \mathrm d y } { \mathrm d t } + \frac { \partial ^ 2 f } { \partial y ^ 2 }  ( \frac { \mathrm d y } { \mathrm d t }  ) ^ 2 + \frac { \partial f } { \partial y } \frac { \mathrm d ^ 2 y } { \mathrm d t ^ 2 }  $
I am missing $\frac { \partial ^ 2 f } { \partial y \, \partial x } \frac { \mathrm d y } { \mathrm d t } \frac { \mathrm d x } { \mathrm d t }$ and $\frac { \partial ^ 2 f } { \partial x \, \partial y } \frac { \mathrm d y } { \mathrm d t } \frac { \mathrm d x } { \mathrm d t }$
Where do these two terms come from?

REPLY [1 votes]: I'm confused about your comment ‘the first and third terms again need the product rule’; if you used the product rule on them, then they would again split into two terms each, and you'd have your six terms total.  They're not products, so they don't really need the product rule; however, they do need another rule that splits things into two terms each, which is the rule for the total derivative.
Specifically, just as $$ \frac { \mathrm d } { \mathrm d t } f = \frac { \partial } { \partial x } f \cdot \frac { \mathrm d x } { \mathrm d t } + \frac { \partial } { \partial x } f \cdot \frac { \mathrm d y } { \mathrm d t } $$ in the first total derivative, so $$ \frac { \mathrm d } { \mathrm d t } \biggl ( \frac { \partial f } { \partial x } \biggr ) = \frac { \partial } { \partial x } \biggl ( \frac { \partial f } { \partial x } \biggr ) \cdot \frac { \mathrm d x } { \mathrm d t } + \frac { \partial } { \partial y } \biggl ( \frac { \partial f } { \partial x } \biggr ) \cdot \frac { \mathrm d y } { \mathrm d t } $$ in the first of your four terms for the second total derivative, and similarly in your third term.  This is where you get the additional terms with the mixed second partial derivatives.
It's also worth noticing that (as long as $ f $ is twice differentiable as a function of $ x $ and $ y $) these two terms are equal, so they can now be combined into a single term.


TITLE: The set of points of a sequence cannot form any other thing than a subsequence?
QUESTION [2 upvotes]: Let $x_n$ be a sequence and let $X=\{x_1,x_2,\ldots,x_i,\ldots\}$ be the set of points of $x_n$. The intuition tells me that if I create any sequence of elements of $X$ (By which I mean taking $x_i$ in any order, not necessarily increasing the index), then it is indeed a subsequence of $x_n$, at least since certain $n$ onwards. I would like to know if this intuition is true and if it is, how is this supposed to be proved? Thanks.

REPLY [2 votes]: It is not true: consider the sequence
$$\langle x_2,x_1,x_4,x_3,x_6,x_5,\ldots\rangle\;,$$
formed by reversing the order of each pair $x_{2n}$ and $x_{2n-1}$. No matter where you start, no tail of this is a subsequence of the original sequence.
Added: The most that you can guarantee is that your subsequence has an infinite subsequence that is also a subsequence of the old sequence. This is because if your subsequence is $\langle x_{\sigma(n)}:n\in\Bbb Z^+\rangle$, then the sequence $\langle\sigma(n):n\in\Bbb Z^+\rangle$ always has an infinite strictly increasing subsequence.


TITLE: Finding period of $\sin\left(2\left(\tan^{-1}\left(\frac{\tan(x)}k\right)\right)\right)$
QUESTION [1 upvotes]: Could anyone kindly help me evaluate the period of the following function:
$$f(x)=\sin\left(2\left(\tan^{-1}\left(\frac{\tan(x)}k\right)\right)\right)$$
where $k$ is some positive real constant.
Had it been only $\sin2(\tan^{-1}(\tan(x)))$, I could have found its period but this constant $k$ is proving to be a challenge for me.
I guess, for certain values of $k$, f may not be periodic at all, can we find under what condition this function will be periodic?
Thanks for your time.
PS: I am in electrical engineering and I used to find the period of function ten years back, but now I have really forgotten the concepts and formulas and I don't have any book at my disposal either.

REPLY [1 votes]: As written the function
$$f(x):=\sin\left(2\arctan{\tan x\over k}\right)$$
is undefined at odd multiples of ${\pi\over2}$. But it turns out that these singularities are removable, and that $f$ is real analytic on all of ${\mathbb R}$ when $k>0$. Put
$$\arctan{\tan x\over k}=:\alpha\in\left]{-{\pi\over2}},{\pi\over2}\right[\ .$$
Then
$${\tan x\over k}=\tan\alpha$$
whenever $\tan x $ is defined, and we 0btain
$$f(x)=\sin(2\alpha)={2\tan\alpha\over1+\tan^2\alpha}={2k\tan x\over k^2+\tan^2 x}={k\sin(2x)\over k^2\cos^2 x+\sin^2 x}\ ,$$
whereby now the RHS makes sense for all $x\in{\mathbb R}$. This RHS is of period $\pi$ by inspection, and it is easy to see that no number of the form ${\pi\over n}$, $n\geq2$,  can be a period, whatever the value of $k>0$.