text:
\begin{document}

\title[Convolutions on  Fourier algebras]
{Convolutions on the Haagerup tensor products of Fourier algebras}
\author[M. Rostami and N. Spronk]
{Mehdi Rostami and Nico Spronk}

\begin{abstract}
We study the ranges of the maps of convolution $u\otimes v\mapsto u\ast v$
and a `twisted' convolution $u\otimes v\mapsto u\ast \check{v}$ ($\check{u}(s)=u(s^{-1})$)
and on the Haagerup tensor product of a Fourier algebra
of a compact group $\falg$ with itself.   We compare the results to result of factoring these
maps through projective and operator projective tensor products.  We notice
that $(\fal(G),\ast)$ is an operator algebra and observe an unexpected set of spectral synthesis.
\end{abstract}

\maketitle

\footnote{{\it Date}: \today.

2000 {\it Mathematics Subject Classification.} 46L07, 43A10, 43A30.
{\it Key words and phrases.} Fourier algebra, convolution, Haagerup tensor product

M. Rostami.
Faculty of Mathematical and Computer Science, Amirkabir University of Technology,
424 Hafez Avenue, 15914 Tehran, Iran. {\tt mross@aut.ac.ir}

N. Spronk.
Department of Pure Mathematics, University of Waterloo, 200 University Avenue West,
Waterloo, Ontario, N2L 3G1, Canada. {\tt nspronk@uwaterloo.ca}

M. Rostami  conducted this work whilst visiting University of Waterloo.
N. Spronk was partially supported by NSERC Grant 312515-2010.}

\section{Introduction}

Let $G$ be a compact group and $\fal(G)$ be its Fourier algebra, in the sense of \cite{eymard}.

In \cite{forrestss}, questions of the following nature were addressed:  what are the ranges
of convolution and `twisted' convolution, when applied to $\fal(G\times G)=\fal(G)\hat{\otimes}\fal(G)$
(operator projective tensor product).  The authors' motivation was two-fold.  First, these
particular maps played a fundamental role in the famous result of B.~Johnson (\cite{johnson})
that $\fal(G)$ is sometimes non-amenable, and the authors were interested in seeing how these
techniques related to the completely bounded theory of Fourier algebras.  This perspective
led the authors to the results of \cite{forrestss2}.
Secondly, it was observed is that `twisted' convolution
averages $\fal(G\times G)$ over left cosets of the diagonal subgroup $\Del=\{(s,s):s\in G\}$,
whereas convolution averages $\fal(G\times G)$ over orbits of the group action
$(r,(s,t))\mapsto (sr^{-1},rt):G\times(G\times G)\to G\times G$.  Thus the images may be
rightly regarded as Fourier algebras of certain homogeneous/orbit spaces of
$G\times G$ in the sense
of \cite{forrest}.  The homogeneous space $G\times G/\Del$ and the orbit space
$G\times G/\check{\Del}$ may be naturally identified with $G$.  Thus we define
$\Gam,\check{\Gam}:\fC(G\times G)\to\fC(G)$ by
\[
\Gam u(s)=\int_Gu(sr,r)\,dr\text{ and }\check{\Gam}u(s)=\int_Gu(sr,r^{-1})\,dr.
\]
It is easy to check that $\Gam(u\otimes v)=u\ast\check{v}$ and $\check{\Gam}(u\otimes v)
=u\ast v$.

In \cite{effrosr1}, it was shown that the Haagerup tensor product
$\fal(G)\otimes^h\fal(G)$ is a Banach algebra.
By \cite[Thm.\ 2]{tomiyama} this algebra has spectrum $G\times G$.
We shall note, below, that $\fal(G)\otimes^h\fal(G)$ is, in fact, semi-simple, and may thus be
regarded as an algebra of functions on $G\times G$.  Hence it is natural to ask whether
we discover anything new if we apply the maps $\Gam$ and $\check{\Gam}$ to
$\fal(G)\otimes^h\fal(G)$.  While we gain no new spaces, we learn interesting
comparisons between $\fal(G)\otimes^h\fal(G)$, $\fal(G\times G)$ and
$\fC(G)\otimes^h\fC(G)$.  See \S \ref{sec:comparison}, below.

\subsection{Some basic results}
We let for each $\pi$ in $\what{G}$,
$\trig_\pi=\spn\{s\mapsto\inprod{\pi(s)\xi}{\eta}:\xi,\eta\in\fH_\pi\}$ and
$\trig(G)=\bigoplus_{\pi\in\what{G}}\trig_\pi$.  This has linear dual space
$\trig(G)'=\prod_{\pi\in\what{G}}\fB(\fH_\pi)$ via dual pairing
\begin{equation}\label{eq:trigdual}
\langle u, T\rangle=\sum_{\pi\in\what{G}}d_\pi\tr(\hat{u}(\pi)T_\pi)
\end{equation}
where $\hat{u}(\pi)=\int_Gu(s)\pi(s^{-1})\,ds$.
If $T\in\trig(G)'$, we let $\check{T}$ be defined by
$\langle u,\check{T}\rangle=\langle \check{u},T\rangle$ in the duality (\ref{eq:trigdual}).
Here $\check{u}(s)=u(s^{-1})$.
The following is surely well-known, and recorded here for later convenience.

\begin{lemma}\label{lem:checkdual}
For $T\in\trig(G)'$, then for $\pi$ in $\what{G}$, $\check{T}_\pi={T_{\bar{\pi}}}^\trans$.
Here the conjugation and transpose are realised with respect to the same orthonormal basis
for $\fH_\pi$.
\end{lemma}

\begin{proof}
We first compute that
\[
\what{\check{u}}(\pi)=\int_G u(s)\pi(s)\,ds=\int_G u(s)\bar{\pi}(s^{-1})^\trans\,ds=\hat{u}(\bar{\pi})^\trans.
\]
Hence $\tr(\what{\check{u}}(\pi)T_\pi)=\tr(\hat{u}(\bar{\pi}){T_\pi}^\trans)$.  Thus in
(\ref{eq:trigdual}) we simply change $\pi$ to $\bar{\pi}$.
\end{proof}

We will identify the left regular representation up to quasi-equivalence as
\[
\lam=\bigoplus_{\pi\in\what{G}}\pi\quad\text{on}\quad\fH=\ell^2\text{-}\bigoplus_{\pi\in\what{G}}\fH_\pi.
\]
It is standard that $\lam(G)$ is weak*-dense in $\trig(G)'$ in terms of the duality
(\ref{eq:trigdual}).  The von Neumann algebra generated by $\lam(G)$ is thus the operator space
direct product
\[
\vn(G)=\ell^\infty\text{-}\bigoplus_{\pi\in\what{G}}\fB(\fH_\pi).
\]
Observe that this algebra has centre $\cent\vn(G)
=\ell^\infty\text{-}\bigoplus_{\pi\in\what{G}}\Cee I_\pi$.
The following is also surely well-known.  Again, we provide a proof for convenience of the reader.

\begin{proposition}\label{prop:expectation}
For each $\pi$ in $\what{G}$, we have that $A\mapsto\int_G\pi(s^{-1})A\pi(s)\,ds:
\fB(\fH_\pi)\to\Cee I_\pi$ is a tracial expectation; hence
$\int_G\pi(s^{-1})A\pi(s)\;ds=\frac{1}{d_\pi}\tr(A)I_\pi$.  Thus
$T\mapsto \int_G\lam(s^{-1})T\lam(s)\,ds:\vn(G)\to\cent\vn(G)$ is a tracial expectation
given by
\[
\int_G\lam(s^{-1})T\lam(s)\;ds=\bigoplus_{\pi\in\what{G}}\frac{1}{d_\pi}\tr(T_\pi)I_\pi.
\]
\end{proposition}

\begin{proof}
It is easy to see that $\int_G\pi(s^{-1})A\pi(s)\,ds$ commutes with each element of $\pi(G)$,
hence by Schur's lemma is a scalar operator, and preserves $I_\pi$.
Hence, $\int_G\pi(s^{-1})A\pi(s)\;ds=\omega(A)I_\pi$
for some functional $\omega$.  Likewise
\begin{align*}
\int_G\pi(s^{-1})A\pi(t)\pi(s)\,ds&=\int_G\pi(s^{-1})A\pi(ts)\,ds \\
&=\int_G\pi(s^{-1}t)A\pi(s)\,ds=\int_G\pi(s^{-1})\pi(t)A\pi(s)\,ds
\end{align*}
and as $\spn\pi(G)=\fB(\fH_\pi)$ we see that
$\int_G\pi(s^{-1})AB\pi(s)\,ds=\int_G\pi(s^{-1})BA\pi(s)\,ds$.  By uniqueness of the
normalised trace, $\omega=\frac{1}{d_\pi}\tr$.  The second result is immediate. \end{proof}

The Fourier algebra is the predual of $\vn(G)$ via the dual pairing (\ref{eq:trigdual}).
Hence we obtain complete isometric identification
\begin{equation}\label{eq:falg}
\fal(G)=\ell^1\text{-}\bigoplus_{\pi\in\what{G}}d_\pi\smat^1_{d_\pi}
\end{equation}
where $\smat_d^1$ denotes the $d\times d$-matrices with trace norm; i.e.\ for
$u$ in $\fal(G)$ we have
\[
\norm{u}_{\fal}=\sum_{\pi\in\what{G}}d_\pi\norm{\hat{u}(\pi)}_{\smat^1}
\]
where $\norm{\hat{u}(\pi)}_{\smat^1}$ is the trace-norm of the $d_\pi\times d_\pi$-matrix
$\hat{u}(\pi)$.

An {\it operator space structure} on a given complex linear space $\fX$ is an assignment
of norms, one on each space $\mat_n(\fX)$ for natural number $n$, which satisfies
certain compatibility conditions; see, for example M1 and M2 of \cite[p.\ 20]{effrosr}.
We shall not require these explicit axioms here.  Of importance to us, are the following facts. First, any
von Neuman algebra $\fV$, in particular $\vn(G)$, will have assigned to each $\mat_n(\fV)$ the
unique norm which realises it as a von Neumann algebra.  Maps
between operator spaces, $\Phi:\fX\to\fY$ are those maps whose matrix amplifications
$\Phi^{(n)}:\mat_n(\fX)\to\mat_n(\fY)$, $\Phi^{(n)}[X_{ij}]=[\Phi(X_{ij})]$ are
uniformly bounded:  $\norm{\Phi}_{cb}=\sup_n\norm{\Phi^{(n)}}<\infty$.  The space
$\fC\fB(\fX,\fY)$ of completely bounded maps is itself an operator space via the
isometric identifications $[\Phi_{ij}]\mapsto (X\mapsto[\Phi_{ij}(X)]):\mat_n(\fC\fB(\fX,\fY))\to
\fC\fB(\fX,\mat_n(\fY))$, where $\mat_m(\mat_n(\fX))=\mat_{mn}(\fX)$.  In particular
linear functionals are automatically completley bounded, and $\fX^*=\fC\fB(\fX,\Cee)$
inherits the operator space structure perscribed above.  If $\fV$ is a von Neumann algebra,
then its predual $\fV_*$ inherits the operator space structure from the inclusion $\fV_*\hookrightarrow
\fV^*$.  A map $\Phi:\fX\to\fY$ is called a complete isometry of each $\Phi^{(n)}$ is an isometry, and
a complete quotient if each $\Phi^{(n)}$ is a quotient map.  In the latter case we say $\fY$ is
a complete quotient space of $\fX$.  See \cite[I.3]{effrosr} for details.

Now if $\fA$ is any Banach space
of functions on $G$ for which $\trig(G)$ is dense in $\fA$, then $\fA^*$ may be realised
as a subspace of $\trig(G)'$.  Generally $\fA$ will be a subspace of $\fal(G)$
(these will arise form application of $\Gam$ and $\check{\Gam}$; see
\cite[Prop.\ 1.3]{forrestss}), whence
$\fV=\vn(G)$ will be frequently used as below.

Given an operator space $\fA$, a subspace $\fV$ of $\fA^*$ is {\it completely norming}
if for each $n$ and $[x_{ij}]$ in $\matr_n(\fA)$ we have
\[
\norm{[x_{ij}]}_{\matr_n(\fA)}=
\sup\left\{\norm{[f_{kl}(x_{ij})]}_{\matr_{mn}}:[f_{kl}]\in\matr_m(\fV),
\norm{[f_{kl}]}_{\matr_m(\fA^*)}\leq 1,m\in\En\right\}.
\]
If $\fV$ is any weak*-dense subspace of $\fA^*$, then it is completely norming.
Indeed, this follows from the fact that the embedding of $\fA$ into $\fA^{**}$ is
a complete isometry \cite[3.2.1]{effrosr}, and then that $\matr_n(\fV)$ is weak*-dense in
$\matr_n(\fA^*)$, where the latter is identified as $\fC\fB(\fA,\matr_n)
\cong(\fA\what{\otimes}\smat^1_n)^*$ \cite[(7.1.6)]{effrosr}.

\begin{lemma}\label{lem:opspace}
Let $\fX$ and $\fY$ be  operator spaces, $\Lam:\fX\to\fY$ be a bounded linear map
with dense range, and $\fV$ be a weak*-dense subspace of $\fY^*$.  Then
$\fA=\Lam(\fX)$, assigned the operator space structure in such a manner that it is regarded as a
complete quotient of $\fX$, admits $\fV$ as a weak*-dense subspace
of it dual $\fA^*$.  In particular, when $\fV$ is given the operator space structure by which
$\Lam^*:\fV\to\fX^*$ is a complete isometry, then $\fV$ is completley norming for $\fA$.
\end{lemma}

\begin{proof}
The density of $\fA$ in $\fY$ allows us to view $\fV$ as a subspace of $\fA^*$.
Furthermore, if $a$ in $\fA$ satisfies $f(a)=0$ for each $f$ in $\fV$, then $f(a)=0$ for
each $f$ in $\fY^*$, hence $a=0$.  Thus $\{f\in\fV:f|_\fA=0\}=\{0\}$,
and bipolar theorem allows us to conclude that $\fV$ is weak*-dense in $\fY^*$.
Moreover, $\Lam^*$ is a complete isometry exactly when $\Lam$ is a complete quotient map,
thanks to \cite[4.1.9]{effrosr}.  Hence we see that $\fV$ is completely norming, thanks
to remarks in the last paragraph, above.
\end{proof}

\subsection{The Haagerup tensor product of Fourier algebras}
Fix a Hilbert space $\fH$.  In \cite{smith,blechers} it is shown that
each weak*-weak* continuous completley bounded operator $\Phi$ on $\fB(\fH)$ --- we shall
write $\Phi\in\fC\fB^\sig(\fB(\fH))$ --- is given by
\begin{equation}\label{eq:ncb}
\Phi(T)= \sum_{i\in I}V_iTW_i
\end{equation}
where $\{V_i,W_i\}_{i\in I}$ is a family in $\fB(\fH)$ for which each of the series
$\sum_{i\in I}V_iV_i^*$ and
$\sum_{i\in I}W_i^*W_i$ is  weak*-convergent.  We shall write $\Phi=
\sum_{i\in I}V_i\otimes W_i$, accordingly.  Furthermore,
we have completely bounded norm
\[
\norm{\Phi}_{cb}=\min\left\{\norm{\sum_{i\in I}V_iV_i^*}^{1/2}
\norm{\sum_{i\in I}W_i^*W_i}^{1/2}:\Phi=\sum_{i\in I}V_i\otimes W_i\text{ as in (\ref{eq:ncb})}\right\}
\]
and operator composition
\begin{equation}\label{eq:cbcomposition}
\sum_{i\in I}V_i\otimes W_i\circ\sum_{i'\in I}V'_{i'}\otimes W'_{i'}
=\sum_{i\in I}\sum_{i'\in I}V_iV'_{i'}\otimes W'_{i'}W_i.
\end{equation}
Hence it is sensible to write
\[
\fC\fB^\sig(\fB(\fH))=\fB(\fH)\otimes^{w^*h}\fB(\fH)
\]
and call this space the {\it weak* Haagerup tensor product}.  This is known to be the
same as the extended Haagerup tensor product of \cite{effrosr1}; see,
for example, the treatment of \cite[\S 2]{spronk}.

Let $\fV\subseteq\fB(\fH)$ be a von Neumann algebra and $\fV'$ its commutant.  It is
shown in \cite{smith,blechers} that subspace of $\fV'$-bimodule maps in
$\fC\fB^\sig(\fB(\fH))$ are exactly those element which admit a representation as in
(\ref{eq:ncb}) with each $V_i,W_i$ in $\fV$.  The description of the norm, with minimum
taken over elements $V_i$ and $W_i$ form $\fV$, and the operator composition are maintained,
making this a closed subalgebra of $\fB(\fH)\otimes^{w^*h}\fB(\fH)$.  We denote this space by
$\fV\otimes^{w^*h}\fV$.  Let $\fV_*$ denote the predual of $\fV$.  We define for an
elementary tensor $u=v\otimes w$ in $\fV_*\otimes\fV_*$ and $\Phi=\sum_{i\in I}V_i\otimes W_i$
in $\fV\otimes^{w^*h}\fV$, the dual pairing
\[
\langle u,\Phi\rangle
=\sum_{i\in I}\langle v,V_i\rangle\langle w,W_i\rangle.
\]
and define the Haagerup norm on $\fV_*\otimes\fV_*$ by
\[
\norm{u}_h=\sup\{|\langle u,\Phi\rangle|:\Phi\in\fV\otimes^{w^*h}\fV,\norm{\Phi}_{cb}\leq 1\}.
\]
We then let $\fV_*\otimes^h\fV_*$
denote the completion of $\fV_*\otimes\fV_*$ with respect to this norm.
Then, as shown in \cite{blechers} the dual pairing above extends to a duality
\begin{equation}\label{eq:htpdual}
(\fV_*\otimes^h\fV_*)^*\cong\fV\otimes^{w^*h}\fV.
\end{equation}
Then $\fV_*\otimes^h\fV_*$
gains it operator space structure it gains as being a distinguished predual of
$\fV\otimes^{w^*h}\fV\subset\fC\fB^\sig(\fB(H))$, as in \cite[(3.2.2) \& (3.2.11)]{effrosr}.
[The more traditional method
of defining $\fV_*\otimes^h\fV_*$ is to assign $\fV_*$ the predual operator space structure
and use tensor formulas such as in \cite[II.9]{effrosr}.  This gives an equivalent but non-intuitive
description of $\fV_*\otimes^h\fV_*$.  We will make extensive use only of (\ref{eq:htpdual}), above.]

Let us return to the Fourier algebra $\fal(G)$ of a compact group.
Recall, as in the last section, that $\fal(G)$ is the predual of
the von Neumann algebra $\vn(G)\subset\fB(\fH)$ where
$\fH=\ell^2\text{-}\bigoplus_{\pi\in\what{G}}\fH_\pi$.
Then we define the Haagerup tensor product of $\fal(G)$ with itself in terms of the completely
isometric duality, as in (\ref{eq:htpdual}), above
\begin{equation}\label{eq:fahtpdual}
(\fal(G)\otimes^h\fal(G))^*\cong\vn(G)\otimes^{w^*h}\vn(G)\subset \fC\fB(\fB(\fH)).
\end{equation}

Let us note a result promised above.

\begin{proposition}
The Banach algebra $\fal(G)\otimes^h\fal(G)$ is semi-simple.
\end{proposition}

\begin{proof}
We observe that $\vn(G)\otimes\vn(G)$ is weak*-dense in $\vn(G)\otimes^{w^*h}\vn(G)\cong
(\fal(G)\otimes^h\fal(G))^*$, by \cite[Thms.\ 3.1 \& 3.2]{blechers}.  It is obvious that
$\spn\lam(G)\otimes\lam(G)$ is weak*-dense in $\vn(G)\otimes\vn(G)$,
where $\lam:G\to\vn(G)$ identifies $G$ with the spectrum of $\falg$.
Thus the bipolar theorem tells us that $(\spn\lam(G)\otimes\lam(G))_\perp=\{0\}$,
whence $\lam(G)\otimes\lam(G)$ separates points in $\fal(G)\otimes^h\fal(G)$.
\end{proof}

\section{Main results}

\subsection{`Twisted' convolution}\label{sec:twistedconvolution}
Our main method for dealing with understanding $\Gam$, as defined on either
of $\fal(G)\hat{\otimes}\fal(G)$, or on $\fal(G)\otimes^h\fal(G)$, is to study its
adjoint.  To this end consider $\Gam:\trig(G)\otimes\trig(G)\to\trig(G)$.  If
$u,v\in\trig(G)$ and $t\in G$ we have
\[
\langle \Gam(u\otimes v),\lam(t)\rangle=
\int_G u(s)v(t^{-1}s)\; ds=\int_G\langle u\otimes v,\lam(s)\otimes\lam(t^{-1})\lam(s)\rangle\,ds
\]
and hence we have that $\Gam^*(\lam(t))=\int_G\lam(s)\otimes\lam(t^{-1})\lam(s)\, ds$, in a
weak* sense.  By weak*-density of $\spn\lam(G)$ in $\trig(G)'$ we conclude that
for $T$ in $\trig(G)'$ we have
\begin{equation}\label{eq:gamadjoint}
\Gam^*(T)=\int_G\lam(s)\otimes\check{T}\lam(s)\, ds
\end{equation}
where the integral is understood in the weak* sense and $\check{T}$ is defined as in
Lemma \ref{lem:checkdual}.

We define $\fal_\Del(G)=\Gam(\fal(G)\hat{\otimes}\fal(G))$.  In \cite{forrestss} this was
regarded as a quotient space of $\fal(G\times G)\cong\fal(G)\hat{\otimes}\fal(G)$ and
assigned a norm accordingly.
We augment the concrete realisation of this norm, computed in
\cite[Thm.\ 2.2]{forrestss}, by specifying the operator space structure on
$\fdelg$ in a concrete manner.

\begin{theorem}\label{theo:opsponfdel}
The operator space structure on $\fdelg$, qua complete quotient of $\fal(G)\hat{\otimes}\fal(G)$
by $\Gam$, is given by the weighted operator space direct sum
\[
\fdelg=\ell^1\text{-}\bigoplus_{\pi\in\what{G}}d_\pi^{3/2}\smat^2_{d_\pi,r}
\]
where $\smat^2_{d,r}$ denotes the $d\times d$ matrices with
Hilbert-Schmidt norm and row operator space structure.
\end{theorem}

\begin{proof}
We recall that $(\fal(G)\hat{\otimes}\fal(G))^*\cong\vn(G)\bar{\otimes}\vn(G)$
by \cite{effrosr0} (see \cite[7.2.4]{effrosr}).
We consider the adjoint $\Gam^*:\vn(G)\to\vn(G)\bar{\otimes}\vn(G)$ with
$n$th amplification $\Gam^{*(n)}:\mat_n(\vn(G))\to\vn(G)\bar{\otimes}\mat_n(\vn(G))$.
We have from (\ref{eq:gamadjoint}) that
\[
\Gam^{*(n)}[T_{ij}]=\int_G \lam(t)\otimes[\check{T}_{ij}] \lam(t)^{(n)}\,ds
\]
and hence
\begin{align*}
\Gam^{*(n)}[T_{ij}]^*&\Gam^{*(n)}[T_{ij}]
=\int_G\int_G \lam(s^{-1}t)\otimes \lam(s^{-1})^{(n)}[\check{T}_{ij}]^*[\check{T}_{ij}]
\lam(t)^{(n)}\,ds\,dt \\
&=\int_G\int_G \lam(s^{-1})\otimes \lam(s^{-1})^{(n)}\lam(t^{-1})^{(n)}
\sum_{k=1}^n[\check{T}_{ki}^*\check{T}_{kj}]\lam(t)^{(n)}\,ds\,dt \\
&=\int_G\lam(s)\otimes \lam(s)^{(n)}\,ds\cdot
\sum_{k=1}^nI\otimes\int_G [\lam(t^{-1})\check{T}_{ki}^*\check{T}_{kj}\lam(t)]\,dt.
\end{align*}
We observe that
\[
P_n=\int_G\lam(s)\otimes \lam(s)^{(n)}\,ds
\]
is evidently a self-adjoint projection.  Furthermore, the Schur orthogonality relations
tell us that on
$\fH\otimes^2\fH^n=\ell^2\text{-}\bigoplus_{\pi',\pi\in\what{G}}\fH_{\pi'}\otimes^2\fH_\pi^n$
we have
\[
P_n=\bigoplus_{\pi\in\what{G}}\int_G \bar{\pi}(s)\otimes\pi(s)^{(n)}\,ds
\]
and hence this projection is non-zero only in each anti-diagonal component of \linebreak
$\vn(G)\bar{\otimes}\mat_n(\vn(G))=\ell^\infty\text{-}
\bigoplus_{\pi',\pi\in\what{G}\times\what{G}}\fB(\fH_{\pi'})
\bar{\otimes}\mat_n(\fB(\fH_\pi))$.  However, by Proposition \ref{prop:expectation} we see that
\[
\sum_{k=1}^nI\otimes\int_G [\lam(t^{-1})\check{T}_{ki}^*\check{T}_{kj}\lam(t)]\,dt
=\sum_{k=1}^n\bigoplus_{\pi\in\what{G}}\frac{1}{d_\pi}I\otimes
[\tr(\check{T}_{ki,\pi}^*\check{T}_{kj,\pi})I_\pi]
\]
Thus we see that
\begin{align*}
\norm{\Gam^{*(n)}[T_{ij}]}
&=\norm{\Gam^{*(n)}[T_{ij}]^*\Gam^{*(n)}[T_{ij}]}^{1/2} \\
&=\sup_{\pi\in\what{G}}\frac{1}{d_\pi^{1/2}}
\norm{\sum_{k=1}^n[\tr(\check{T}_{ki,\pi}^*\check{T}_{kj,\pi})]} ^{1/2} \\
&=\sup_{\pi\in\what{G}}\frac{1}{d_\pi^{1/2}}
\norm{\sum_{k=1}^n[\tr(T_{ki,\bar{\pi}}^*T_{kj,\bar{\pi}})]} ^{1/2}
\end{align*}
where we have appealed to Lemma \ref{lem:checkdual} in the last line.
According to \cite[(3.4.4)]{effrosr} we obtain completely isometric embedding
\[
\Gam^*(\vn(G))\subseteq\ell^\infty\text{-}\bigoplus_{\pi\in\what{G}}
\frac{1}{d_\pi^{1/2}}\smat^2_{d_\pi,c}
\]
Thus Lemma \ref{lem:opspace} and the duality of operator Hilbert spaces
\cite[(3.4.4)]{effrosr} provide the desired result.  We observe, moreover, that
our dual pairing (\ref{eq:trigdual}) is bilinear, which is why we need not concern ourselves
with conjugate spaces.
\end{proof}

The operator space structure of Theorem \ref{theo:opsponfdel} provides alternate explanation
for curious results discovered in the some prior articles.

\begin{corollary}
{\bf (i)} {\rm \cite[Prop.\ 2.5]{forrestss}} We have $\fdelg\hat{\otimes}\fdelg\cong\fal_\Del(G\times G)$.

{\bf (ii)} {\rm \cite[Prop.\ 4.3]{forrestss2}}  The map $u\mapsto\check{u}$ is a complete isometry
on $\fdelg$.
\end{corollary}

\begin{proof}
Consider the identifications
\begin{align*}
\left(\ell^1\text{-}\bigoplus_{\pi\in\what{G}}d_\pi\smat_{d_\pi,r}^2\right)\hat{\otimes}
\left(\ell^1\text{-}\bigoplus_{\pi'\in\what{G}}d_{\pi'}\smat_{d_{\pi'},r}^2\right)
&\cong \underset{\pi,\pi'\in\what{G}\times\what{G}}{\ell^1\text{-}\bigoplus}
d_\pi d_{\pi'}\smat_{d_\pi,r}^2
\hat{\otimes}\smat_{d_{\pi'},r}^2 \\
&\cong  \underset{\pi,\pi'\in\what{G}\times\what{G}}{\ell^1\text{-}\bigoplus}
d_\pi d_{\pi'}\smat_{d_\pi d_{\pi'},r}^2
\end{align*}
where we have appealed to the fact that the tensor product or row spaces is again a row space
(\cite[9.3.5]{effrosr}).  Since $\what{G\times G}\cong\what{G}\times
\what{G}$, (i) follows.

The map $u\mapsto \check{u}:\trig_\pi\to\trig_{\bar{\pi}}$,
which is an isometry from $d_\pi\smat_{d_\pi}^2$ to
$d_{\bar{\pi}}\smat_{d_{\bar{\pi}}}^2$ -- practically the transpose
-- is a complete isometry with row structure, thanks to \cite[3.4.2]{effrosr}.
Thus by the structure of the direct product,
we get (ii).
\end{proof}

The next result is a bit of a surprise.  It shows that $\fal(G)\otimes^h\fal(G)$ behaves exactly
as does $\fal(G)\hat{\otimes}\fal(G)$ with respect to $\Gam$.

\begin{theorem}\label{theo:gamonhaag}
We have that $\Gam(\fal(G)\otimes^h\fal(G))=\fdelg$.  Moreover,
if $\fdelg$ is given the operator space structure in Theorem \ref{theo:opsponfdel}, above,
then $\Gam:\fal(G)\otimes^h\fal(G)\to\fdelg$ is a complete quotient map.
\end{theorem}

\begin{proof}
Let $\fal^h_\Del(G)=\Gam(\fal(G)\otimes^h\fal(G))$, and assign it the operator space
structure which makes $\Gam$ a complete quotient map.  The completley contractive
inclusion $\fal(G)\hat{\otimes}\fal(G)\hookrightarrow\fal(G)\otimes^h\fal(G)$ gives, via
the fact that $\fdelg$ is a complete quotient of $\fal(G)\hat{\otimes}\fal(G)$, a
completely contractive inclusion $\fdelg\hookrightarrow\fal^h_\Del(G)$.
Since $\trig(G)$ is dense in both subspaces, $\fdelg$ is dense in $\fal^h_\Del(G)$,
so there the adjoint $\fal^h_\Del(G)^*\hookrightarrow\fdelg^*$ is completely contractive and injective.
We wish to see that this map is a complete isometry.  It suffices to appeal
to Lemma \ref{lem:opspace} and verify that on $\matr_n(\vn(G))$, we have
\begin{equation}\label{eq:dualineq}
\norm{[T_{ij}]}_{\matr_n(\fal^h_\Del(G)^*)}\leq\norm{[T_{ij}]}_{\matr_n(\fal_\Del(G)^*)}.
\end{equation}

We recall the duality relation (\ref{eq:fahtpdual}).
Hence for $[T_{ij}]$ in $\matr_n(\vn(G))$ we have by (\ref{eq:gamadjoint}) and the last
identification immediatley above
\begin{align*}
&\norm{[T_{ij}]}_{\matr_n(\fal^h_\Del(G)^*)}
=\norm{\Gam^{*(n)}[T_{ij}]}_{\fC\fB(\fB(\fH),\matr_n(\fB(\fH)))} \\
&\phantom{mm}
=\norm{A\mapsto \int_G [\lam(s)A\check{T}_{ij}\lam(s)]\,ds}_{\fC\fB(\fB(\fH),\matr_n(\fB(\fH)))} \\
&\phantom{mm}=\sup\left\{\left|\int_G \inprod{[\lam(s)A_{kl}\check{T}_{ij}\lam(s)]\xi}{\eta}\,ds\right|:
\begin{matrix} [A_{kl}]\in\mathrm{ball}(\matr_m(\fB(\fH))) \\
\xi,\eta\in\mathrm{ball}(\fH^{mn}),\, m\in\En \end{matrix}\right\}.
\end{align*}
We observe for operator matrix $[A_{kl}]$ and vectors $\xi,\eta$ as above that
\begin{align*}
&\left|\int_G \inprod{[\lam(s)A_{kl}\check{T}_{ij}\lam(s)]\xi}{\eta}\,ds\right| \\
&\phantom{mmm}\leq \int_G\norm{[A_{kl}\check{T}_{ij}\lam(s)]\xi}\norm{\lam(s^{-1})^{(nm)}\eta}\,ds \\
&\phantom{mmm}\leq \left( \int_G\norm{[A_{kl}\check{T}_{ij}\lam(s)]\xi}^2\right)^{1/2}
\left( \int_G\norm{\lam(s^{-1})^{(nm)}\eta}^2\,ds\right)^{1/2} \\
&\phantom{mmm}\leq  \left( \int_G
\inprod{[A_{kl}\check{T}_{ij}\lam(s)]^*[A_{kl}\check{T}_{ij}\lam(s)]\xi}{\xi}\right)^{1/2} \\
&\phantom{mmm}\leq
\inprod{\int_G \sum_{k=1}^n[\lam(s^{-1})\check{T}_{ki}^*\check{T}_{kj}\lam(s)]\xi}{\xi}^{1/2} \\
&\phantom{mmm}\leq
\norm {\int_G \sum_{k=1}^n[\lam(s^{-1})\check{T}_{ki}^*\check{T}_{kj}\lam(s)]}^{1/2}.
\end{align*}
However, calculations in the proof of Theorem \ref{theo:opsponfdel} reveal that the last
quantity is exactly
\[
\sup_{\pi\in\what{G}}\frac{1}{d_\pi^{1/2}}
\norm{\sum_{k=1}^n[\tr(T_{ki,\bar{\pi}}^*T_{kj,\bar{\pi}})]} ^{1/2}
=\norm{[T_{ij}]}_{\matr_n(\fal_\Del(G)^*)}.
\]
Thus (\ref{eq:dualineq}) is established.
\end{proof}

\subsection{Convolution}\label{sec:convolution}
We now wish to consider the map $\check{\Gam}$ on $\fal(G)\otimes^h\fal(G)$.
Consider $\check{\Gam}:\trig(G)\otimes\trig(G)\to\trig(G)$.  If
$u,v\in\trig(G)$ and $t\in G$ we have
\[
\langle \check{\Gam}(u\otimes v),\lam(t)\rangle=
\int_G u(s)v(s^{-1}t)\; ds=\int_G\langle u\otimes v,\lam(s)\otimes\lam(s^{-1})\lam(t)\rangle\,ds
\]
and hence we have that $\Gam^*(\lam(t))=\int_G\lam(s)\otimes\lam(s^{-1})\lam(t)\, ds$, in a
weak* sense.  By weak*-density of $\spn\lam(G)$ in $\trig(G)'$ we conclude that
for $T$ in $\trig(G)'$ we have
\begin{equation}\label{eq:cgamadjoint}
\check{\Gam}^*(T)=\int_G\lam(s)\otimes\lam(s^{-1})T\, ds
\end{equation}
where the integral is understood in the weak* sense.

\begin{theorem}\label{theo:cgamonhaag}
We have $\check{\Gam}(\fal(G)\otimes^h\fal(G))=\fal(G)$.  Morover,
$\check{\Gam}:\fal(G)\otimes^h\fal(G)\to\fal(G)$ is a complete quotient map.
\end{theorem}

\begin{proof}
Let $T\in\vn(G)$.  Then $\check{\Gam}^*(T)$, via the identification
(\ref{eq:fahtpdual}) and using the composition product of (\ref{eq:cbcomposition}),  factors as
\begin{gather*}
\check{\Gam}^*(T)=\int_G\lam(s)\otimes\lam(s^{-1})T\, ds
=(I\otimes T)\circ\int_G\lam(s)\otimes\lam(s^{-1})\, ds \\
\text{i.e.\ } \check{\Gam}^*(T)(A)=\left(\int_G \lam(s)A\lam(s^{-1})\,ds\right)T.
\end{gather*}
Hence if $[T_{ij}]\in\matr_n(\vn(G))$ then, since the map from Proposition \ref{prop:expectation}
is an expectation, hence completely contractive, we have
\begin{align*}
&\norm{\check{\Gam}^{*(n)}[T_{ij}]}_{\fC\fB(\fB(\fH),\matr_n(\fB(\fH)))} \\
&\phantom{mmm}\leq\norm{A\mapsto [AT_{ij}]}_{\fC\fB(\fB(\fH),\matr_n(\fB(\fH)))}
\norm{A\mapsto \int_G\lam(s)A\lam(s^{-1})\,ds}_{\fC\fB(\fB(\fH))} \\
&\phantom{mmm}\leq\norm{[T_{ij}]}_{\matr_n(\vn(G))}.
\end{align*}
Conversely, inspecting this operator at $A=I$ we observe
\[
\norm{\check{\Gam}^{*(n)}[T_{ij}]}_{\fC\fB(\fB(\fH),\matr_n(\fB(\fH)))}
\geq \norm{[T_{ij}]}_{\matr_n(\vn(G))}.
\]
Thus equality holds.

Since $\check{\Gam}(\fal(G)\otimes^h\fal(G))$ contains
$\trig(G)$ which is dense in $\fal(G)$, and since $\check{\Gam}^*$ is a complete isometry,
the desired results hold.
\end{proof}

We obtain an immediate corollary of Theorem  \ref{theo:cgamonhaag}, by
using the result \cite{blecher}
(see \cite[5.2.1]{blecherl}) that any algebra and operator space
$\fA$ whose product completely boundedly
factors through $\fA\otimes^h\fA$ is completely isomorphic
to an operator algebra, i.e., an algebra of operators acting on a Hilbert space.

\begin{corollary}
The convolution algebra $(\fal(G),\ast)=(\fal(G),\check{\Gam})$ is
completely isomorphic to an operator algebra.
\end{corollary}

In fact, by \cite[5.2.8]{blecherl}, we see that the representation $\rho:(\falg,\ast)\to\fB(\fK)$
can be set to satisfy $\norm{\rho}_{\mathrm{cb}}\leq 2$ and $\norm{\rho^{-1}}_{\mathrm{cb}}\leq 1$
(the latter on $\rho(\falg)$).  Since $(\falg,\ast)$ is a Segal algebra in $\mathrm{L}^1(G)$, it does not
admit a bounded approximate identity (\cite[Theo.\ 1.2]{burnham}), in particular it does not admit
a contractive approximate identity.  Hence we do not have enough information
to establish if $\rho$ can be made a complete isometry (as would follow from the
Blecher-Ruan-Sinclair theorem as stated in \cite[2.3.2]{blecherl}, for example).

This stands in mild contrast to  \cite[5.5.8]{blecherl}, where it is shown that
the `matrix' algebra $\smat^1_\infty$ is not an operator algebra.  Of course, the global
structure of $(\fal(G),\ast)$ is much different, since convolution on
$\trig_\pi\otimes\trig_\pi\cong d_\pi\smat^1_{d_\pi}\otimes d_\pi\smat^1_{d_\pi}$ is really
matrix multiplication times a scalar factor $\frac{1}{d_\pi}$ (see \cite[(27.20)]{hewittrII}); hence
we obtain a contraction $d_\pi\smat^1_{d_\pi}\otimes^hd_\pi\smat^1_{d_\pi}\to
d_\pi\smat^1_{d_\pi}$.

Let us close this section
with a remark on convolution applied to $\fal(G\times G)$.  The methods are very close
to those of \cite[Thm.\ 4.1]{forrestss}, but we make the extra effort to gain the operator space structure.

\begin{proposition}
We have a completely isometric identification
\[
\check{\Gam}(\fal(G\times G))=\ell^1\text{-}\bigoplus_{\pi\in\what{G}}d_\pi^2\smat^1_{d_\pi}
\]
where we regard this space as a complete quotient space of $\fal(G\times G)$ by $\check{\Gam}$.
\end{proposition}

We remark that the space above was denoted $\fgamg$ in \cite{johnson}.  In
\cite[Def.\ 2.6]{lees} $\fgamg$ is regarded as a Beurling-Fourier algebra,
and is given the same operator
space structure, though in terms of a certain weighted dual pairing with $\vn(G)$,
which is a different perspective than the one taken here.

\begin{proof}
As above, we need only determine the norm of $\check{\Gam}^{*(n)}[T_{ij}]$ in \linebreak
$\matr_n(\vn(G)\bar{\otimes}\vn(G))\cong\matr_n(\vn(G\times G))$, for
$[T_{ij}]$ in $\matr_n(\vn(G))$.  Using (\ref{eq:cgamadjoint}) and factoring in the product
in $\mat_n(\vn(G)\bar{\otimes}\vn(G))$ we have
\begin{align*}
\check{\Gam}^{*(n)}[T_{ij}]&=\left(\int_G \lam(s)\otimes\lam(s^{-1})\, ds\right)^{(n)}[I\otimes T_{ij}] \\
&=\bigoplus_{\pi\in\what{G}}\left(\int_G\pi(s)\otimes\pi(s^{-1})\,ds\right)^{(n)}[I\otimes T_{ij,\pi}].
\end{align*}
It is straightforwrd to calculate that each $\int_G\pi(s)\otimes\pi(s^{-1})\,ds=\frac{1}{d_\pi}U_\pi$
where $U_\pi$ is a unitary, in fact a permutation matrix.  Thus we obtain a completely
isometric embedding
\[
\check{\Gam}^*(\vn(G))
\subseteq\ell^\infty\text{-}\bigoplus_{\pi\in\what{G}}\frac{1}{d_\pi}\smat_{d_\pi}^\infty
\]
and the desired result follows.
\end{proof}

\section{Comparison of results}\label{sec:comparison}

We let $\fC(G)$ denote the space of continuous functions on $G$.  The
{\it Varopoulos algebra} is given by
\[
\mathrm{V}(G\times G)=\fC(G)\otimes^\gam\fC(G)=\fC(G)\hat{\otimes}\fC(G)
=\fC(G)\otimes^h\fC(G)
\]
where isomorphic equality of the spaces is provided by Grothendieck's inequality
\cite[14.5]{defantf}.  We shall take $\otimes^h$ to define our canonical
norm on $\mathrm{V}(G\times G)$.
Since the map $u\mapsto\check{u}$ is a complete isometry
on $\fC(G)$, we have that $\check{\Gam}(\mathrm{V}(G\times G))=
\Gam(\mathrm{V}(G\times G))$ completely isometrically.  We recall that
$\falg\hat{\otimes}\falg=\fal(G\times G)$ completely isometrically, by virtue of
the facts that $(\falg\hat{\otimes}\falg)^*=\vn(G)\bar{\otimes}\vn(G)$,
as indicated in the proof of Theorem \ref{theo:opsponfdel}, and the latter
space is $\vn(G\times G)$; while
$\falg\otimes^\gam\falg=\fal(G\times G)$ isomorphically only when
$G$ has an abelian subgroup of finite index \cite{losert}, and isometrically only when
$G$ is abelian.  The fact that $u\mapsto\check{u}$
is an isometry on $\fal(G)$ means that $\Gam(\falg\otimes^\gam\falg)=
\check{\Gam}(\falg\otimes^\gam\falg)$.

\begin{tabular}{|c|c|c|l|}\hline
algebra &   image under $\Gam$ & image under $\check{\Gam}$ &  references \\ \hline\hline
$\mathrm{V}(G\times G)$ & $\falg$ & $\falg$ &  \cite{spronkt} \\
$\falg\otimes^h\falg$ & $\fdelg$ & $\falg$ & \S\ref{sec:twistedconvolution}, \S\ref{sec:convolution} \\
$\fal(G\times G)$ & $\fdelg$ &  $\fgamg$ & \cite{forrestss} (\S\ref{sec:twistedconvolution},
\S\ref{sec:convolution}) \\
$\falg\otimes^\gam\falg$ & $\fgamg$ & $\fgamg$ & \cite{johnson}  \\ \hline
\end{tabular}

\noindent Notice that in each of the first three rows, $\Gam$ and $\check{\Gam}$ can
be regarded as a complete quotient map, as shown in \S\ref{sec:twistedconvolution} and
\S\ref{sec:convolution}, above.

Let us use Theorems  \ref{theo:gamonhaag} and \ref{theo:cgamonhaag}
to observe some further connections
between $\fal(G\times G)$ and $\fal(G)\otimes^h\fal(G)$, and also between
$\mathrm{V}(G\times G)$ and $\fal(G)\otimes^h\fal(G)$.
This addresses a question asked in \cite[p.\ 21]{daws}.
We use the same definitions as in \cite{spronkt,forrestss}.
The equivalence of (i), (ii) and (iii) below, is an immediate consequence of
 \cite[Theo.\ 1.4]{forrestss}.  The equivalence of (i'), (ii') and (iii')
is proved essentially as \cite[Thm.\ 3.1]{spronkt}; see
\cite[Lem.\ 2.3]{forrestss2}.  We refer the reader to those sources for the proof.

\begin{proposition}\label{theo:synthesis}
Let $\theta,\check{\theta}:G\times G\to G$ be given by
$\theta(s,t)=st^{-1}$ and $\check{\theta}(s,t)=st$.
Also, let $E\subset G$ be closed.  Then the following are equivalent:

{\bf (i)} $E$ is a set of spectral synthesis for $\fdelg$;

{\bf (ii)} $\theta^{-1}(E)$ is a set of spectral synthesis for $\fal(G\times G)$;

{\bf (iii)} $\theta^{-1}(E)$ is a set of spectral synthesis for $\fal(G)\otimes^h\fal(G)$.

\noindent Also, the following are equivalent:

{\bf (i')} $E$ is a set of spectral synthesis for $\fal(G)$;

{\bf (ii')} $\check{\theta}^{-1}(E)$ is a set of spectral synthesis for $\mathrm{V}(G\times G)$;

{\bf (iii')} $\check{\theta}^{-1}(E)$ is a set of spectral synthesis for $\fal(G)\otimes^h\fal(G)$.
\end{proposition}

It is well-known, see for example \cite{herz}, that point sets are spectral for
$\fal(G)$.  Since $\check{\theta}^{-1}(\{e\})=\{(s,s^{-1}):s\in G\}$, we gain the following.

\begin{corollary}
The anti-diagonal $\check{\Delta}=\{(s,s^{-1}):s\in G\}$ is a set of
spectral synthesis for $\fal(G)\otimes^h\fal(G)$.
\end{corollary}

 This stands in marked contrast to the case
for $\fal(G\times G)$:  $\check{\Delta}$ is a set of spectral synthesis for $\fal(G\times G)$
if and only if the connected component of the identity $G_e$ is abelian (\cite[Thm.\ 2.5]{forrestss2}).

\bigskip
{\sc Acknowledgement.}  The authors are grateful to the anonymous referee for making valuable
comments which improved the exposition of the paper.


TITLE: Question about simplifying absolute value/exponents
QUESTION [1 upvotes]: Does $|x^n|^{\frac{1}{n}} = |x|$? Also n is a natural number. Sorry that this is such a stupid question, I'm just simplifying something and trying to make sure I'm doing it right.

REPLY [3 votes]: $$\quad \left|x^n\right|=\left|x\right|^n\quad \mathrm{so} \quad\sqrt[n]{\left|x^n\right|}=\sqrt[n]{\left|x\right|^n}=|x|$$

REPLY [2 votes]: Yes (provided $n\neq0$), since $|x^n|=|x|^n$, and $(r^a)^b=\exp(ab\ln r)=r^{ab}$ holds for all real $r>0$ and all $a,b$, while $(0^a)^b=0^{ab}$ also holds in the here relevant case that $a=n$ and $b=\frac1n$.
(Indeed $(0^a)^b=0^{ab}$ holds in all cases where both sides are defined, i.e., for real $a,b\geq0$: both sides are $0$ unless $ab=0$ in which case both sides are $1$. But I'm sure someone is going to attack me for that last case.)

REPLY [0 votes]: You are correct! you can also show the result by using the definition of $|x|$ and splitting the cases.


text:
\section{Proofs of Section~\ref{sec:basic-properties}}
\label{sec:proof-basic-properties}

\begin{proof}[Proof of Theorem~\ref{thm:existence-of-PR}]
  We first prove the former claim.

 In \cite{ikeda2018finding},
 it was proven that $\ca{L}_H$ is a maximal monotone operator. (We omit the definition here because we do not need it).
  Then, by the general theory of maximal monotone operators~\cite{miyadera1992nonlinear,komura1967nonlinear}, the operator $I + \lambda\ca{L}_H$ is injective for any $\lambda>0$.
  Hence, its inverse $J_\lambda := {(I+\lambda \ca{L}_H)}^{-1}$ is single-valued, and is called a \emph{resolvent} of $\ca{L}_H$.
  Because the image of $\ca{L}_H$ is $\RR^V$, for any $\bm{s}\in \RR^V$, there exists a unique vector $J_\lambda(\bm{s})$, which is a solution to~\eqref{eq:hypergraph-pagerank-via-laplacian} when
  $\lambda = (1-\alpha)/2\alpha > 0$.

  We now prove the latter claim.
  For $\bm{p} = \pr_\alpha(\bm{s})$, we have
  \[
    \bm{p} + \frac{1-\alpha}{2\alpha}\ca{L}_{H_{\bm{p}}}\bm{p} = \bm{s},
  \]
  where $H_{\bm{p}}$ is the graph obtained from $H$ and $\bm{p}$, as defined in Section~\ref{subsec:pagerank-hypergraph}.
  This implies that $\bm{p}$ is a classical PPR on $H_{\bm{p}}$, and hence the claim holds by the Perron--Frobenius theorem for
  the matrix $\alpha\bm{s}\1^\top + (1-\alpha)W_{H_{\bm{p}}}$.
\end{proof}

\begin{proof}[Proof of Propostion~\ref{prop:continuity}]
As mentioned in the proof of Theorem~\ref{thm:existence-of-PR}, the PPR $\bm{\pr}_\alpha(\bm{s})$ is equal to a resolvent $J_\lambda(\bm{s})$ for $\lambda = (1-\alpha)/2\alpha$.
Thus, it suffices to prove the continuity of the resolvent $J_\lambda(\bm{s})$ with respect to $\lambda>0$.
By~\cite{miyadera1992nonlinear},

for $\lambda \geq \mu>0$,
the resolvents satisfy the following inequality:
\[
 J_\lambda(\bm{s}) = J_{\mu}\left( \frac{\mu}{\lambda}\bm{s}
 + \frac{\lambda-\mu}{\lambda} J_{\lambda}(\bm{s})  \right).
\]
On the other hand, it is well known that $J_\lambda$ is a non-expansive map, i.e., for any $\bm{s}, \bm{t} \in \RR^V$ and any
$\lambda > 0$,
$ \|J_\lambda(\bm{s}) - J_\lambda(\bm{t}) \|_{D_H^{-1}} \leq
 \|\bm{s} - \bm{t}\|_{D_H^{-1}}$,
where $\| \bm{x} \|_{D_H^{-1}} =
\sqrt{\bm{x}^\top D_H^{-1} \bm{x}}$.

By these relations, we can show the following:
\begin{align*}
 & \|J_\lambda(\bm{s}) - J_\mu(\bm{s}) \|_{D_H^{-1}}
 \leq \left\|J_\mu\left(\frac{\mu}{\lambda}\bm{s} + \frac{\lambda-\mu}{\lambda}J_\lambda(\bm{s})\right) - J_\mu(\bm{s})  \right\|_{D_H^{-1}}
 \\
 &\leq \left\|\frac{\mu}{\lambda}\bm{s} + \frac{\lambda-\mu}{\lambda}J_\lambda(\bm{s}) - \bm{s}  \right\|_{D_H^{-1}}
 \leq \frac{\lambda-\mu}{\lambda} \|J_\lambda(\bm{s}) - \bm{s}
 \|_{D_H^{-1}}.
\end{align*}
Hence, $\lim_{\lambda \to \mu} J_\lambda(\bm{s}) = J_\mu(\bm{s})$.
\qedhere
\end{proof}

To prove Proposition~\ref{prop:bounds-of-values-of-PPR},
we prepare the following lemma:
\begin{lem}\label{lem:pagerank-in-terms-of-beta}
  For $\beta:=2\alpha/(1+\alpha)$, we have
  \[
    \beta (\bm{s} - \bm{\pr}_\alpha(\bm{s})) - (1-\beta) \ca{L}_H (\bm{\pr}_\alpha(\bm{s})) = \bm{0}.
  \]
\end{lem}

\begin{proof}[Proof of Lemma~\ref{lem:pagerank-in-terms-of-beta} ]
  Let $\bm{p} = \pr_\alpha(\bm{s})$.
  Note that $(1-\alpha)/2\alpha = 1/\beta-1$.
  Then by~\eqref{eq:hypergraph-pagerank-via-laplacian} and easy calculation, we have

  \begin{align*}
    \left(I + \frac{1-\alpha}{2\alpha}\ca{L}_H\right)(\bm{p}) \ni \bm{s}
    \Leftrightarrow\; \beta (\bm{s} - \bm{p}) - (1-\beta) \ca{L}_H (\bm{p}) \ni \bm{0}.
  \end{align*}
  As $\bm{p}$ is unique by Theorem~\ref{thm:existence-of-PR}, we can replace the inclusion with equality.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{prop:bounds-of-values-of-PPR}]
  Let $\bm{p} = \pr_\alpha(\bm{\chi}_u)$.
  Because $\bm{p}$ is a distribution by the latter claim of Theorem~\ref{thm:existence-of-PR}, we have
  $\bm{p}(v) \geq 0$ for any $v\in V$.
  Then by Lemma~\ref{lem:pagerank-in-terms-of-beta}, we have
  \[
   \bm{p} = \bm{\chi}_u - \frac{1-\beta}{\beta}
   \ca{L}_{H_{\bm{p}}} \bm{p} \geq \bm{0}.
  \]

  Hence for $v \neq u$, we have $-\ca{L}_{H_{\bm{p}}}\bm{p}(v) \geq 0$.
  Note that $-\ca{L}_{H_{\bm{p}}}\bm{p}(v)$ is equal to the difference between the heat diffusing into $v$ and that diffusing out from $v$ in the heat equation:
  \[
   \frac{d\bm{\rho}_t}{dt} = -\ca{L}_{H_{\bm{p}}}\bm{\rho}_t.
  \]

  Because the heat diffuses to the stationary distribution $\bm{\pi}_V$, the non-negativity of $-\ca{L}_{H_{\bm{p}}}\bm{p}(v)$ implies $\bm{p}(v) \leq \bm{\pi}_V(v)$.

  When $v = u$, we have $\bm{p}(u)-\bm{\chi}_u(u) \leq 0$.
  By a similar argument, we have the desired inequality.
\end{proof}

config: arxiv


TITLE: What is $\sum_{k, l}$?
QUESTION [1 upvotes]: For example, when you have an expression like this,
$\sum_{k, l} g^{kl}$,
what is this sum if one wrote it out?

REPLY [4 votes]: It's usually shorthand notation for
$$\sum\limits_{l}\sum\limits_k g^{kl} = \sum\limits_{k}\sum\limits_l g^{kl}$$
These sums are equal as long as there are finite indices.


TITLE: Is the restriction of a graded automorphism linearizable in characteristic zero?
QUESTION [8 upvotes]: This question follows up a previous one which was answered by Todd Leason. I want to impose two new requirements on the setup.
Let $k$ be a characteristic zero field. Let $A=k[x_1,\dots,x_n]$ be the polynomial algebra with the usual grading. Let $g$ be a graded automorphism of $A$ and let $B$ be a graded subring of $A$ such that:

$A$ is integral over $B$.
$B$ is fixed setwise by $g$.
$B$ is itself a polynomial algebra.

Is $g$'s restriction to $B$ linearizable?

By linearizable I mean that there exists a set of algebra generators $f_1,\dots,f_n$ of $B$ such that the $k$-vector space $V = \langle f_1,\dots,f_n\rangle_k$ is invariant under $g$. (So that $B$ can be seen as the symmetric algebra over $V$ and $g|_B$ the automorphism induced on the symmetric algebra by $g|_V$.)
The two new requirements are integrality and characteristic zero.
Todd Leason's response to the previous question shows that without the characteristic zero assumption the answer is no. I think it's probably no in general, since if $B$ is generated in distinct degrees then $g$ must act on its generators diagonally in order to be linearizable, and that seems a lot to ask. But Todd's example used the characteristic $p$-ness in an essential way, so I remain curious.

REPLY [2 votes]: Gregor Kemper answered a related question with a technique that can be used to answer this one affirmatively in the case that $g$ has finite order. If $g$ does not have finite order and we drop the integrality assumption, the answer is negative.
If $g$'s restriction to $B$ has finite order, then its action on $B$ is linearizable.
Proof: Note that $B$ is connected (i.e., its degree 0 component is just $k$) since $A$ is. Let $I$ be $B$'s positively-graded ideal. By the graded Nakayama lemma, any set of homogeneous elements of $I$ that generate the $k$-vector space $I/I^2$ will also generate $I$ as an ideal in $B$. By a standard induction argument, any homogeneous ideal generators for $I\triangleleft B$ are actually algebra generators for $B$. (To express an arbitrary homogeneous element $f$ in $B$, which without loss of generality can be taken to have positive degree since $B$ is connected, as a polynomial in these generators, express it first as a linear combination of them with coefficients in $B$. This is possible since $f$ has positive degree, and is therefore in $I$, and $I$ is an ideal. Since $B$ is graded, this linear relation can be taken to be homogeneous just by discarding all terms of degree different than $f$. But then the coefficients are all homogeneous of lower degree, so apply the induction hypothesis.) Since $B$ is presumed to be a polynomial algebra, $\dim_k I/I^2 = \dim_{\text{Krull}}B$. Thus any homogeneous lifts to $I$ of a $k$-basis for $I/I^2$ will automatically be polynomial-algebra generators for $B$.
Since $g$ acts as a graded automorphism on $B$, it preserves $I$. Therefore, it also preserves $I^2$. Since $k$ is of characteristic zero and $g$ has finite order, we have access to Maschke's theorem, which asserts that $I^2$ has a $g$-invariant complement $V$ in $I$. Because the action of $g$ on $B$ respects the grading, and $I$ and therefore $I^2$ are graded ideals, $V$ is even a graded vector space, i.e., it is the direct sum of its intersections with the graded components of $B$. (Indeed, for any $d\in\mathbb{N}$, we can apply Maschke's theorem in $I\cap B_d$ to find a $g$-invariant complement $V_d$ for $I^2\cap B_d$, and then take $V=\bigoplus V_d$.) Therefore, it has a homogeneous basis $\mathcal{B}$. Because $I=I^2\oplus V$, this basis descends to a basis of $I/I^2$. By the work in the previous paragraph, $\mathcal{B}$ is a set of polynomial algebra generators for $B$, so $V$ is the desired subspace.
Remark: This argument actually also works if $k$ has positive characteristic, as long as the order of $g$'s action on $B$ is not divisible by the characteristic. The argument also makes no use of the assumption that $A$ is integral over $B$, although if we know integrality, then we know that $\dim_{\text{Krull}}B = n$, so we know what to expect for the dimension of $I/I^2$.
If $g$'s restriction to $B$ has infinite order, and we drop the hypothesis that $A$ is integral over $B$, then $g$'s action on $B$ may not be linearizable.
Let $A=\mathbb{C}[x,y]$ and let $B=\mathbb{C}[x,xy]$. Let $g$ act on $A$ by $x\mapsto x$, $y\mapsto x+y$. Then $gB\subset B$ since $xy\mapsto x^2+xy\in B$, and $B\subset gB$ since $xy\in \mathbb{C}[x,x^2+xy]=gB$. Thus $B$ is fixed setwise by $G$. But $B$ is algebra-generated in distinct degrees and $g|_B$ does not act diagonally on the generators, so it is not linearizable.
The previous argument fails in this situation because the conclusion of Maschke's theorem fails: while $I^2$ is still a $g$-invariant subspace of $I$, it does not have a $g$-invariant complement.
But note that in this situation, $A$ is not integral over $B$, as $y$ is not integral over $B$.
I do not know if there is an example of infinite order $g$ and $A/B$ integral in which $g|_B$ is not linearizable.


TITLE: If $\ $ $ X\ =\ \left\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5}........\right\}$ will $ P(X)$ be countable?
QUESTION [1 upvotes]: Question:
If $\ $ $ X\ =\ \left\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5}........\right\}$ will $ P(X)$ be countable?
My attempt:
I think $X$ is countable since you can list all the elements. But I am not sure if that implies $P(X)$ is countable.

REPLY [5 votes]: $X$ is clearly countably infinite, so $P(X)$ has the same cardinality as $P(\mathbb N)$, which is uncountable by Cantor's theorem https://en.wikipedia.org/wiki/Cantor%27s_theorem.


TITLE: Snails and Sums
QUESTION [0 upvotes]: At the beginning of a $10\,\mathrm m$ long rubber band sits a snail.
Every day it crawls one meter ahead. Every night, when the snail is resting, an evil man stretches the tape evenly by $10\,\mathrm m$ out. (In the morning of the second
Day the band is $20\,\mathrm m$ long and the snail has made $2\,\mathrm m$.) Will the snail ever reach the end of the tape?
My attempt:
If this continues forever, we have $d_{\text{Snail}}=\sum_{n=0}^{\infty}{1}$ and $d_{\text{Tape}}=\sum_{n=0}^{\infty}{10}$. They  are both equal to $\infty$ since $a\cdot \infty = \infty$. But how does that prove it?

REPLY [1 votes]: Instead of measuring the snail's (or snake's?) progress in meters, measure it in fractions of the entire rubber band, since that is a measure that stays the same during the nightly stretching.
Then on day $n$ it makes a progress of $1/10n$ rubber bands. So you want find out whether there's a $k$ such that
$$ \sum_{n=1}^{k} \frac1{10n} \ge 1 $$
and we know this to be the case since the harmonic series diverges.

A more vivid argument: Start by painting 20 dots on the rubber band at equal intervals. The rubber between two dots stretches by 50 cm every night, but since the snail crawls at 100 cm a day, each morning it will be at least 50 cm closer to the next dot than it was the previous morning (unless it passed a dot yesterday). So it will inevitably reach the next dot. And all it needs to do to get all the way is to "reach the next dot" 20 times in a row.
(Any fencepost errors in this argument are left for the reader to think around.)


text:
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\title{Lecture Notes in Computer Science:\\Authors' Instructions
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TITLE: Multiplicatively closed set of matrices containing $A$ such that $X\ \mapsto\ AXA^3$ is surjective contains the identity?
QUESTION [3 upvotes]: Let $n\geq 2$ and let $U$ be a set of $n\times n$-matrices, such that
1) $\forall X,Y \in U:\ XY \in U$,
2) $\exists A \in U$ such that $f: U \longrightarrow U:\ X\ \longmapsto AXA^3 $ is surjective.
Show that $I_n \in U$, where $I_n$ is the $n\times n$ identity matrix.
I believe this to be false, but I was not able to find a counterexample.I tried setting, for example $AXA^3 = A^3$ and thus $A^3 = AXAXA^3 =\ldots= (AX)^n A^3$ and other tricks like that, but to no avail.
Edit: I think I have a counterexample: If $U$ consists only of $\operatorname{diag}(1,0)$, this satisfies the rules. Also $U = \left\{\operatorname{diag}(1,0,0),\operatorname{diag}(1,1,0)\right\}$ with $A = \operatorname{diag}(1,1,0)$ satisfies the rules. Are these counterexamples right?

REPLY [3 votes]: You have indeed found a whole collection of counterexamples. It is easy to verify that for $n=2$, setting
$$X:=\begin{pmatrix}1&0\\0&0\end{pmatrix},$$
the singleton set $U:=\{X\}$ is a counterexample. Indeed $X^2=X\in U$ and the map
$$f:\ U\ \longrightarrow\ U:\ X\ \longmapsto\ X\cdot X\cdot X^3=X^5=X,$$
is surjective. But clearly $I_2\notin U$.
However, if $A$ is required to be invertible then the statement is true: Let $X\in U$ be such that $f(X)=A^4$, i.e. such that $AXA^3=A^4$. Because $A$ is invertible it follows that $X=I$ and hence that  $I\in U$.


TITLE: Which gets you first when you are falling into a black hole, the black hole singularity or the cosmic background radiation?
QUESTION [12 upvotes]: If you look up while you are falling into a black hole you see the universe blue shifted, that is, you see the universe moving quickly forward in time compared to your local time. Since this effect increases as you get closer to the singularity, a star aimed at you could blast you with a billion years of radiation in maybe a second. Of course, no star will be perfectly aligned for billions of years. On the other hand, there is lots of cosmic background radiation coming at you from everywhere in the sky and it will be blue shifted too. So, how bad is it? Would this radiation cook you before you hit the singularity?

REPLY [0 votes]: I remember calculating the amount of time one would take to fall radially inward with zero energy (so as to maximize the time taken) starting at the schwarzschild radius of a black hole of n solar masses to be
$$ n(1.55 \times 10^{-5} ) \text{ seconds}$$
And you would never be able to tell that you had crossed it in the first place, the event horizon is simply a coordinate singularity (the Ricci scalar is very well behaved at the event horizon indicating large but finite curvature).
But since the curvature isn't infinite it doesn't make sense to say things are blue shifted to infinity for an inward falling observer, the only place where the curvature diverges is at the singularity and there isn't any understanding at that point.
Now you would have a problem if you attempted to remain stationary just outside the event horizon.


TITLE: Let $(0,3)\rightarrow S^1$ be a map such that $x\in (0,3)\rightarrow e^{2\pi i x}$.
QUESTION [0 upvotes]: Let $(0,3)\rightarrow S^1$ be a map such that $x\in (0,3)\rightarrow e^{2\pi i x}$.
Is this a covering?

REPLY [3 votes]: Let $f$ denote your map. If it were a covering, then all fibers $f^{-1}(z)$ would have the same cardinality because $S^1$ is connected. But $f^{-1}(1) = \mathbb Z \cap (0,3) = \{1,2\}$ and $f^{-1}(-1) = \{n + 1/2 \mid n \in \mathbb Z \} \cap (0,3) = \{1/2, 3/2, 5/2 \}$.
Note that $f$ is a surjective local homeomorphism.


TITLE: Polar body of a convex body that avoids a lattice
QUESTION [32 upvotes]: Let $K \subset {\bf R}^d$ be a symmetric convex body (an open bounded convex neighbourhood of the origin with $K = -K$) with the property that $K + {\bf Z}^d \neq {\bf R}^d$, i.e. the projection of $K$ to the standard torus ${\bf R}^d/{\bf Z}^d$ is not surjective, or equivalently $K$ is disjoint from some coset $x + {\bf Z}^d$ of the standard lattice.  My question is: what does this say about the polar body
$$K^* := \{ \xi \in {\bf R}^d: \xi \cdot x < 1 \hbox{ for all } x \in K \}?$$
Intuitively, the property $K + {\bf Z}^d \neq {\bf R}^d$ is a "smallness" condition on K, and is thus a "largeness" condition on $K^*$.
If $K^*$ contains a non-trivial element $n$ of $2 {\bf Z}^d$, then $K$ is contained in the strip $\{ x: |n \cdot x| < 1/2 \}$, and will thus avoid the coset $x+{\bf Z}^d$ whenever $x \cdot n = 1/2$.  So this is a sufficient condition for $K + {\bf Z}^d \neq {\bf R}^d$, but it is not necessary.  Indeed, if one takes $K$ to be the octahedron
$$K := \{ (x_1,\ldots,x_d) \in {\bf R}^d: |x_1|+\ldots+|x_d| < d/2 \}$$
then $K$ avoids $(1/2,\ldots,1/2)+{\bf Z}^d$, but the dual body
$$ K^* = \{ (\xi_1,\ldots,\xi_d) \in {\bf R}^d: |\xi_1|,\ldots,|\xi_d| < 2/d \}$$
is quite far from reaching a non-trivial element of $2 {\bf Z}^d$.
On the other hand, by using the theory of Mahler bases or Fourier analysis one can show that if $K + {\bf Z}^d \neq {\bf R}^d$, then $K^*$ must contain a non-trivial element of $\varepsilon_d {\bf Z}^d$ for some $\varepsilon_d > 0$ depending only on $d$.  However the bounds I can get here are exponentially poor in $d$.
Based on the octahedron example (which intuitively seems to be the "biggest" convex set that still avoids a coset of ${\bf Z}^d$), one might tentatively conjecture that if $K + {\bf Z}^d \neq {\bf R}^d$, then the closure of $K^*$ contains a non-trivial element of $\frac{2}{d} {\bf Z}^d$.  I do not know how to prove or disprove this conjecture (though I think the $d=2$ case might be worked out by ad hoc methods, and the $d=1$ case is trivial), so I am posing it here as a question.

REPLY [3 votes]: This is not really an answer, but a comment on a natural variation of the original problem: assume a convex body in $\mathbb{R}^n$ that is not necessarily centrally symmetric contains only one lattice point as an interior point. What can be said about the dual of $K$ defined with respect to the unique lattice point in its interior?
In general, you can say that the volume of the dual body cannot be too small. In this paper Balacheff, Tzanev, and I conjecture that the volume of the dual cannot be less than $(n+1)/n!$. We can prove this in dimension two under the following equivalent guise:
Theorem. The area of a convex body in the plane that intersects every
integer line $mx + ny = 1$ is at least $3/2$. Moreover, equality holds only
for the triangle with vertices $(1, 0)$, $(0, 1)$, $(−1,−1)$ and its images under
(integer) unimodular transformations.
In higher dimensions, we rely (heavily!) on Greg Kuperberg's lower bound for the volume product to prove.
Theorem. The volume of a convex body in $\mathbb{R}^n$ that intersects every
integer hyperplane $m_1x_1 + · · · + m_nx_n = 1$ is at least $(\pi/8)^n(n + 1)/n!$.
But the fun part of the paper is that is also shows that these inequalities are (in the case of the first theorem) or should be extended to inequalities about Finsler metrics or even contact manifolds.


text:
\begin{document}

\title{Truncated theta series and \\partitions into distinct parts}
\author{Mircea Merca\\

	\footnotesize Department of Mathematics, University of Craiova, 200585 Craiova, Romania\\
	\footnotesize Academy of Romanian Scientists, Ilfov 3, Sector5, Bucharest, Romania\\
	\footnotesize mircea.merca@profinfo.edu.ro

}
\date{}
\maketitle

\begin{abstract}
	Linear inequalities involving Euler’s partition function $p(n)$ have been the subject of recent studies.
	In this article, we consider the partition function $Q(n)$ counting the partitions of $n$ into distinct parts.
	Using truncated theta series, we provide four infinite families of linear inequalities for $Q(n)$ and partition theoretic interpretations for these results.
\\
\\
{\bf Keywords:} inequalities, partitions, recurrences, theta series
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{\bf MSC 2010:}  05A17, 11P81, 11P83
\end{abstract}

\section{Introduction}

In this paper, in order to simplify the expressions, we denote  the $n$th triangular number by
$$T_n=n(n+1)/2$$
and the $n$th generalized pentagonal number by
$$G_n=T_n-T_{\lfloor n/2 \rfloor},$$
for any nonnegative integer $n$.

In \cite{Andrews12}, G.E. Andrews and M. Merca considered Euler's pentagonal number theorem
\begin{equation}\label{eq:1}
\sum_{n=0}^{\infty} (-1)^{T_n} q^{G_n} = (q;q)_\infty
\end{equation}
and obtained the following truncated version:
\begin{equation}\label{eq:2}
\frac{1}{(q;q)_\infty} \sum_{n=0}^{2k-1} (-1)^{T_n} q^{G_n}
= 1+(-1)^{k-1} \sum_{n=1}^{\infty} \frac{q^{\binom{k}{2}+(k+1)n}}{(q;q)_n}
\begin{bmatrix}
n-1\\k-1
\end{bmatrix},
\end{equation}
where
\begin{align*}
& (a;q)_n = \begin{cases}
1, & \text{for $n=0$,}\\
(1-a)(1-aq)\cdots(1-aq^{n-1}), &\text{for $n>0$,}
\end{cases}\\
& (a;q)_\infty = \lim_{n\to\infty} (a;q)_n,\\
&
\begin{bmatrix}
n\\k
\end{bmatrix}
=
\begin{bmatrix}
n\\k
\end{bmatrix}_{q}
=
\begin{cases}
0, & \text{ if $k<0$ or $k>n$}\\
\dfrac{(q;q)_n}{(q;q)_k(q;q)_{n-k}}, &\text{otherwise.}
\end{cases}
\end{align*}
From \eqref{eq:2}, they derived an infinite family of inequalities for Euler's partition function $p(n)$:
\begin{equation}\label{eq:3}
(-1)^{k-1}\sum_{j=0}^{2k-1} (-1)^{T_j} p(n-G_j) \geqslant 0,
\end{equation}
with strict inequality if $n>G_{2k}$.

Inspired by these results, V.J.W. Guo and J. Zeng \cite{Guo} considered two other classical theta identities, usually attributed to
Gauss \cite[p.23, eqs. (2.2.12) and (2.2.13)]{Andrews76}, i.e.,
\begin{equation}\label{eq:4}
1+2\sum_{n=1}^{\infty} (-q)^{n^2} = \frac {(q;q)_\infty} {(-q;q)_\infty},
\end{equation}
\begin{equation}\label{eq:5}
\sum_{n=0}^{\infty} (-q)^{T_j} = \frac {(q^2;q^2)_\infty} {(-q;q^2)_\infty},
\end{equation}
and proved new truncated forms of these:
\allowdisplaybreaks{
	\begin{align}
	& \frac{(-q;q)_{\infty}}{(q;q)_{\infty}} \left(1+2\sum_{j=1}^{k} (-q)^{j^2} \right) \label{eq:6}\\
	& \qquad\qquad = 1+(-1)^k \sum_{n=k+1}^{\infty} \frac{(-q;q)_k (-1;q)_{n-k} q^{(k+1)n}}{(q;q)_n}
	\begin{bmatrix}
	n-1\\k-1
	\end{bmatrix}, \notag \\
	& \frac{(-q;q^2)_{\infty}}{(q^2;q^2)_{\infty}} \sum_{j=0}^{2k-1} (-q)^{T_j} \label{eq:7}\\
	& \qquad\qquad = 1+(-1)^k \sum_{n=k}^{\infty} \frac{(-q;q^2)_k (-q;q^2)_{n-k} q^{2(k+1)n-k}}{(q^2;q^2)_n}
	\begin{bmatrix}
	n-1\\k-1
	\end{bmatrix}_{q^2}, \notag
	\end{align}
}respectively.
As an immediate consequence of \eqref{eq:6}, they deduced an infinite family of inequalities for the number of overpartitions of $n$, $\overline{p}(n)$:
\begin{equation}\label{eq:8}
(-1)^k \left(\overline{p}(n) +2 \sum_{j=1}^{k} (-1)^j \overline{p}(n-j^2)\right) \geqslant 0,
\end{equation}
with strict inequality if $n\geqslant (k+1)^2$. Similarly, from \eqref{eq:7} Guo and Zeng obtained an infinite family of inequalities for $pod(n)$, the  number of partitions of $n$ in which odd parts are not repeated:
\begin{equation}\label{eq:9}
(-1)^{k-1} \sum_{j=0}^{2k-1} (-1)^{T_j} pod \left( n-T_j \right)  \geqslant 0,
\end{equation}
with strict inequality if $n\geqslant (2k+1)k$.

Very recently, Andrews and Merca \cite{Andrews17} have revealed that \eqref{eq:2}, \eqref{eq:6} and \eqref{eq:7}
are essentially corollaries of the Rogers-Fine identity  \cite[p.15]{Rogers}:
\begin{equation*}
\sum_{n=0}^{\infty} \frac{(\alpha;q)_n \tau^n }{(\beta;q)_n}
= \sum_{n=0}^{\infty} \frac{(\alpha;q)_n (\alpha\tau q/\beta;q)_n \beta^n \tau^n q^{n^2-n}(1-\alpha\tau q^{2n})}{(\beta;q)_n (\tau;q)_{n+1}}.
\end{equation*}
A new infinite family of inequalities for Euler's partition function $p(n)$ is proved in this context:
if at least one of $n$ and $k$ is odd,
\begin{equation}\label{eq:1.10}
(-1)^{k-1} \sum_{j=0}^{2k-1} (-1)^{T_j}  p(n-T_j)  \geqslant 0.
\end{equation}

Other recent investigations on the truncated theta series  can be found in several papers by Chan, Ho and Mao \cite{Chan},
Chern \cite{Chern}, He, Ji and Zang \cite{He},   Kolitsch \cite{Kolitsch},  Kolitsch and Burnette \cite{KolitschB}, Mao \cite{Mao,Mao17}, Merca \cite{Merca16},
Merca, Wang and Yee \cite{MWY},
Wang and Yee \cite{WY1,WY2}, and Yee \cite{Yee}.

In this paper, motivated by these results, we shall provide similar families of inequalities for the partition function $Q(n)$ which counts the number of partition of $n$ into distinct parts. We consider $Q(0)=1$ and $Q(n)=0$ for all negative $n$.

By the truncated pentagonal number theorem \eqref{eq:2}, with $q$ replaced by $q^2$, we easily deduce the following result.

\begin{theorem}\label{Th:1.1}
	For a positive integer $k$, we have
	\begin{align*}
	& (-q;q)_\infty \sum_{n=0}^{2k-1} (-1)^{T_n} q^{2G_n}\\
	& \qquad 	= \frac{(q^2;q^2)_\infty}{(q;q^2)_\infty}+(-1)^{k-1} \frac{(q^2;q^2)_\infty}{(q;q^2)_\infty} \sum_{n=1}^{\infty} \frac{q^{k(k-1)+2(k+1)n}}{(q^2;q^2)_n}
	\begin{bmatrix}
	n-1\\k-1
	\end{bmatrix}_{q^2}.
	\end{align*}
\end{theorem}

An immediate consequence owing to the positivity of the sum on the right is given by the following inequality  where $\delta_{i,j}$ is the Kronecker delta function.

\begin{corollary}\label{Cor:1.2}
	For $m,n\geqslant 0$, $k\geqslant 1$,
	$$(-1)^{k-1} \left( \sum_{j=0}^{2k-1} (-1)^{T_j} Q\left(n-2G_j\right) - \delta_{n,T_m} \right)  \geqslant 0,$$
	with strict inequality if and only if  $n\geqslant 2G_{2k}$.
\end{corollary}

As we can see in \cite[Corollary 4.5]{Merca}, the recurrence relation given by the limiting case $k\to\infty$ of this corollary is known:
\begin{equation}\label{rec:1}
\sum_{j=0}^\infty (-1)^{T_j} Q(n-2G_j)
=\begin{cases}
1, & \text{if $n=T_m$, $m\in\mathbb{N}_0$,}\\
0, & \text{otherwise}.
\end{cases}
\end{equation}
The number of terms in this recurrence relation is about $\sqrt{4n/3}$.
By this recurrence relation, we can deduce the following parity result.

\begin{corollary}\label{Cor:1.3}
	Let $n$ be a positive integer.
	The number of representations of $n$ as the sum of a generalized pentagonal number and a twice generalized pentagonal number is odd if and only if $n$ is a triangular number.
\end{corollary}

In analogy with \eqref{eq:6}, we have a stronger result than Theorem \ref{Th:1.1}.

\begin{theorem}\label{Th:1.4}
	For the positive integers $k$ and $r$, we have:
	\begin{align*}
	& (-q;q)_\infty \left( 1+2\sum_{j=1}^k (-1)^j q^{r\cdot j^2} \right)\\
	& \quad = \frac{(-q;q)_\infty (q^r;q^r)_\infty}{(-q^r;q^r)_\infty} + 2(-1)^k q^{r(k+1)^2} \frac{(q^r;q^{2r})_\infty}{(q;q^2)_\infty}
	\sum_{j=0}^{\infty}\frac{q^{(2k+2j+3)rj}}{(q^{2r};q^{2r})_j(q^r;q^{2r})_{k+j+1}}.
	\end{align*}
\end{theorem}

Considering Euler's pentagonal number theorem \eqref{eq:1} and the following two identities
\begin{align*}
& \frac{(-q;q)_\infty (q^2;q^2)_\infty}{(-q^2;q^2)_\infty} =  \sum\limits_{j=0}^\infty (-1)^{T_{\lfloor j/2 \rfloor}} q^{G_j},\\
& \frac{(-q;q)_\infty (q^3;q^3)_\infty}{(-q^3;q^3)_\infty} = \sum\limits_{j=0}^\infty q^{G_j},
\end{align*}
the positivity of
$$
\sum_{j=0}^{\infty}\frac{q^{(2k+2j+3)rj}}{(q^{2r};q^{2r})_j(q^r;q^{2r})_{k+j+1}}
$$
allows us to deduce the following families of inequalities for the partition function $Q(n)$.

\begin{corollary}\label{cor:1.5}
	For $m,n\geqslant 0$, $k\geqslant 1$,
	\begin{enumerate}
		\item[a)] $\displaystyle{ (-1)^{k}\left( Q(n)+2 \sum_{j=1}^{k} (-1)^j Q(n-j^2)-(-1)^{T_m} \delta_{n,G_m}\right) \geqslant 0, }$
		\item[] with strict inequality if and only if  $n\geqslant (k+1)^2$.
		\item[b)] $\displaystyle{(-1)^{k}\left( Q(n)+2 \sum_{j=1}^{k} (-1)^j Q(n-2j^2)-(-1)^{T_{\lfloor m/2 \rfloor}}\delta_{n,G_m}\right) \geqslant 0},$
		\item[] with strict inequality if and only if  $n\geqslant 2(k+1)^2$.
		\item[c)] $\displaystyle{(-1)^{k}\left( Q(n)+2 \sum_{j=1}^{k} (-1)^j Q(n-3j^2)-\delta_{n,G_m}\right) \geqslant 0},$
		\item[] with strict inequality if and only if  $n\geqslant 3(k+1)^2$.
	\end{enumerate}
\end{corollary}

There is a substantial amount of numerical evidence to conjecture that the sums in this corollary satisfy the following inequalities.

\begin{conjecture}\label{C:1}
	For $m,n\geqslant 0$, $k\geqslant 1$,
	\begin{align*}
	& (-1)^{k}\left( Q(n)+2 \sum_{j=1}^{k} (-1)^j Q(n-j^2)-(-1)^{T_m} \delta_{n,G_m}\right) \\
	& \qquad \geqslant (-1)^{k}\left( Q(n)+2 \sum_{j=1}^{k} (-1)^j Q(n-2j^2)-(-1)^{T_{\lfloor m/2 \rfloor}}\delta_{n,G_m}\right) \\
	& \qquad \geqslant (-1)^{k}\left( Q(n)+2 \sum_{j=1}^{k} (-1)^j Q(n-3j^2)-\delta_{n,G_m}\right).
	\end{align*}
\end{conjecture}

As we can see in \cite[Corollary 4.6]{Merca}, the recurrence relation given by the limiting case $k\to\infty$ of Corollary \ref{cor:1.5}.a) is known:
\begin{equation}\label{rec:2}
Q(n)+2 \sum_{j=1}^{\infty} (-1)^j Q(n-j^2)
=\begin{cases}
(-1)^{T_m}, & \text{if $n=G_m$, $m\in\mathbb{N}_0$,}\\
0, & \text{otherwise}.
\end{cases}
\end{equation}
The number of terms in this recurrence relation is about $\sqrt{n}$.

The limiting case $k\to\infty$ of Corollary \ref{cor:1.5}.b) gives a new linear recurrence relation more efficient than \eqref{rec:2}:
\begin{equation}\label{rec:3}
Q(n)+2 \sum_{j=1}^{\infty} (-1)^j Q(n-2j^2)
=\begin{cases}
(-1)^{T_{\lfloor m/2 \rfloor}}, & \text{if $n=G_m$, $m\in\mathbb{N}_0$,}\\
0, & \text{otherwise}.
\end{cases}
\end{equation}
We see that the number of terms in this recurrence relation is about $\sqrt{n/2}$.

Corollary \ref{cor:1.5}.c) provides a new and extremely efficient algorithm for computing the partition function $Q(n)$:
\begin{equation}\label{rec:4}
Q(n)+2 \sum_{j=1}^{\infty} (-1)^j Q(n-3j^2)
=\begin{cases}
1, & \text{if $n=G_m$, $m\in\mathbb{N}_0$,}\\
0, & \text{otherwise}.
\end{cases}
\end{equation}
The number of terms in this recurrence relation is about $\sqrt{n/3}$.

As consequence of the recurrence relations \eqref{rec:2}-\eqref{rec:4}, we remark the following parity result.

\begin{corollary}\label{cor:1.6}
	Let $n$ be a positive integer.
	\begin{enumerate}
		\item[a)] 	The number of representations of $n$ as the sum of a generalized pentagonal number and a square number is odd if and only if $n$ is a generalized pentagonal number.
		\item[b)] 	The number of representations of $n$ as the sum of a generalized pentagonal number and a twice square number is odd if and only if $n$ is a generalized pentagonal number.
		\item[c)] 	The number of representations of $n$ as the sum of a generalized pentagonal number and a thrice square number is odd if and only if $n$ is a generalized pentagonal number.
	\end{enumerate}
\end{corollary}

Theorems \ref{Th:1.1} and \ref{Th:1.4} are good reasons to look for new infinite families of linear inequalities for the partition function $Q(n)$.
The rest of this paper is organized as follows. We will first prove Theorem \ref{Th:1.4} in Section \ref{S2}. In Section \ref{S3}, we provide combinatorial interpretations for Corollaries \ref{Cor:1.2} and \ref{cor:1.5}. Two infinite families of linear inequalities for $Q(n)$ which we have experimentally discovered are introduced in Section \ref{S4}.

\section{Proof of Theorems \ref{Th:1.4}}\label{S2}

The result follows directly from

\begin{lemma}\label{L:2.1}
	For a positive integer $k$,
	$$\sum_{j=0}^\infty (-1)^j q^{j^2+2j(k+1)} = (q^{2k+3};q^2)_\infty \sum_{j=0}^\infty \frac{q^{j(2j+2k+3)}}{(q^2;q^2)_j(q^{2k+3};q^2)_j}.$$
\end{lemma}

\begin{proof}
	To prove the lemma, we consider the second identity by Heine's transformation  of ${_{2}}\phi_{1}$ series \cite[(III.2)]{gasper}, namely
	$$
	{_{2}}\phi_{1}\bigg(\begin{matrix}a, b\\ c\end{matrix}\,;q,z\bigg)
	= \frac{(c/b;q)_\infty(bz;q)_\infty}{(c;q)_\infty(z;q)_\infty}
	{_{2}}\phi_{1}\bigg(\begin{matrix}abz/c, b\\ bz\end{matrix}\,;q,c/b\bigg).
	$$
	We have
	\begin{align*}
	&\sum_{j=0}^\infty (-1)^j q^{j^2+2j(k+1)}\\
	&\qquad = \lim_{\tau\to 0} \sum_{j=0}^\infty \frac{(q^{2k+3}/\tau;q^2)_j}{(\tau;q^2)_j}\tau^j\\
	&\qquad = \lim_{\tau\to 0} {_{2}}\phi_{1}\bigg(\begin{matrix}q^{2},  q^{2k+3}/\tau\\ \tau\end{matrix}\,;q^{2},\tau\bigg)\\
	&\qquad = \lim_{\tau\to 0} \frac{(\tau^2/q^{2k+3};q^2)_\infty(q^{2k+3};q^2)_\infty}{(\tau;q^2)^2_\infty} {_{2}}\phi_{1}\bigg(\begin{matrix}q^{2k+5}/\tau,  q^{2k+3}/\tau\\ q^{2k+3}\end{matrix}\,;q^{2},\tau^2/q^{2k+3}\bigg)\\
	&\qquad = (q^{2k+3};q^2)_\infty \lim_{\tau\to 0}  \sum_{j=0}^\infty \frac{(q^{2k+5}/\tau;q^2)_j(q^{2k+3}/\tau;q^2)_j}{(q^2;q^2)_j(q^{2k+3};q^2)_j}\left( \frac{\tau^2}{q^{2k+3}}\right)^j\\
	&\qquad = (q^{2k+3};q^2)_\infty \lim_{\tau\to 0} \sum_{j=0}^\infty \frac{(-1)^j q^{j(j+1)}(q^{2k+3}/\tau;q^2)_j}{(q^2;q^2)_j(q^{2k+3};q^2)_j}\tau^j\\
	&\qquad = (q^{2k+3};q^2)_\infty \sum_{j=0}^\infty \frac{q^{j(2j+2k+3)}}{(q^2;q^2)_j(q^{2k+3};q^2)_j}
	\end{align*}
	and the proof of this lemma is finished.
\end{proof}

With $q$ replaced by $q^r$, the Gauss identity \eqref{eq:4} becomes
$$1+2\sum_{n=1}^k (-1)^n q^{rn^2} = \frac{(q^r;q^r)_\infty}{(-q^r;q^r)_\infty}-2\sum_{n=k+1}^\infty (-1)^n q^{rn^2}.$$
Multiplying both sides of this identity by $(-q;q)_\infty$, we get
\allowdisplaybreaks{
	\begin{align*}
	&(-q;q)_\infty \left( 1+2\sum_{j=1}^k (-1)^j q^{rj^2} \right) \\
	&\quad = \frac{(-q;q)_\infty(q^r;q^r)_\infty}{(-q^r;q^r)_\infty} - 2(-q;q)_\infty \sum_{j=k+1}^{\infty} (-1)^j q^{rj^2}\\
	&\quad = \frac{(-q;q)_\infty(q^r;q^r)_\infty}{(-q^r;q^r)_\infty} + 2(-1)^k  \frac{q^{r(k+1)^2}}{(q;q^2)_\infty} \sum_{j=0}^{\infty} (-1)^j q^{rj^2+2rj(k+1)}\\

	&\quad =\frac{(-q;q)_\infty(q^r;q^r)_\infty}{(-q^r;q^r)_\infty} \\
	&\qquad + 2(-1)^k q^{r(k+1)^2} \frac{(q^{r(2k+3)};q^{2r})_\infty}{(q;q^2)_{\infty}}\sum_{j=0}^{\infty}\frac{q^{2rj^2+(2k+3)rj}}{(q^{2r};q^{2r})_j(q^{r(2k+3)};q^{2r})_j}
	\tag{By Lemma \ref{L:2.1}, with $q$ replaced by $q^2$.}\\
	&\quad =\frac{(-q;q)_\infty(q^r;q^r)_\infty}{(-q^r;q^r)_\infty}\\
	&\qquad + 2(-1)^k q^{r(k+1)^2} \frac{(q^{r};q^{2r})_\infty}{(q;q^2)_{\infty}(q^r;q^{2r})_{k+1}}\sum_{j=0}^{\infty}\frac{q^{(2k+2j+3)rj}}{(q^{2r};q^{2r})_j(q^{r(2k+3)};q^{2r})_j}\\
	&\quad =\frac{(-q;q)_\infty(q^r;q^r)_\infty}{(-q^r;q^r)_\infty} + 2(-1)^k q^{(k+1)^2}
	\frac{(q^{r};q^{2r})_\infty}{(q;q^2)_{\infty}} \sum_{j=0}^{\infty}\frac{q^{(2k+2j+3)rj}}{(q^{2r};q^{2r})_j(q^r;q^{2r})_{k+j+1}}
	\end{align*}
}
and the proof is finished.

\section{Combinatorial interpretations}
\label{S3}

Regarding the inequality \eqref{eq:3}, we recall the following
partition theoretic interpretation given by Andrews and Merca \cite[Theorem 1]{Andrews12}:
\begin{equation*}
(-1)^{k-1}\sum_{j=0}^{2k-1} (-1)^{T_j} p(n-G_j) = M_k(n),
\end{equation*}
where $M_k(n)$ is the number of partitions of $n$ in which
$k$ is the least integer that is not a part and there are
more parts $>k$ than there are $<k$. In \cite{Yee} has
given a combinatorial proof of this result. Considering the generating function for $M_k(n)$, i.e.,
$$\sum_{n=0}^{\infty} M_k(n) q^n = \sum_{n=1}^{\infty} \frac{q^{\binom{k}{2}+(k+1)n}}{(q;q)_n}
\begin{bmatrix}
n-1\\k-1
\end{bmatrix},
$$
Theorem \ref{Th:1.1} can be written as:
\begin{align*}
(-q;q)_\infty \sum_{n=0}^{2k-1} (-1)^{T_n} q^{2G_n}- \frac{(q^2;q^2)_\infty}{(q;q^2)_\infty}
= (-1)^{k-1} \frac{(q^2;q^2)_\infty}{(q;q^2)_\infty} \sum_{n=0}^{\infty} M_k(n) q^{2n}
\end{align*}
or
\begin{align*}
& \left( \sum_{n=0}^\infty Q(n) q^n\right)  \left( \sum_{n=0}^{2k-1} (-1)^{T_n} q^{2G_n}\right) - \sum_{n=0}^\infty q^{T_n} \\
& \qquad\qquad\qquad\qquad= (-1)^{k-1} \left( \sum_{n=0}^\infty q^{T_n}\right) \left(  \sum_{n=0}^{\infty} M_k(n) q^{2n}\right),
\end{align*}
where we have invoked the theta identity \eqref{eq:5}.  In this way, we derive the following identity.

\begin{corollary}\label{cor:3.1}
	For $m,n\geqslant 0$, $k\geqslant 1$,
	$$(-1)^{k-1} \left( \sum_{j=0}^{2k-1} (-1)^{T_j} Q\left(n-2G_j\right) - \delta_{n,T_m} \right)
	=\sum_{j=0}^\infty M_k\left(n/2-T_j/2 \right),
	$$
	where $M_k(x)=0$ if $x$ is not an integer.
\end{corollary}

More explicitly, the right hand side of this identity can be rewritten as:
$$\sum_{j=0}^\infty M_k\left(n-T_j/2 \right)
=\sum_{j=-\infty}^\infty M_k\big(n-j(4j-1) \big),
$$
and
$$\sum_{j=0}^\infty M_k\big((2n+1)/2-T_j/2 \big)
=\sum_{j=-\infty}^\infty M_k\big(n-j(4j-3) \big).
$$

Very recently, Merca, Wang and Yee \cite{MWY} provided the following partition-theoretic interpretation of the sum in Theorem \ref{Th:1.4}.

\begin{theorem} \label{Th:3.2}
	For a fixed $k\geqslant 1$,
	\begin{equation*}
	\sum_{n=0}^{\infty} {M_{o,k}}(n) q^n =q^{(k+1)^2} \sum_{j=0}^{\infty}\frac{q^{(2k+2j+3)j}}{(q^2;q^2)_j(q;q^2)_{k+j+1}},
	\end{equation*}
	where ${M_{o,k}}(n)$ counts partitions of $n$ into odd parts such that all odd numbers less than or equal to $2k+1$ occur as parts at least once and the parts below the $(k+2)$-Durfee rectangle in the $2$-modular graph are strictly less than the width of the rectangle.
\end{theorem}

For $r>1$, we denote by  $Q_{r}(n)$ the number of partitions of $n$ into distinct parts not congruent to $0$ modulo $r$. The generating function of $Q_{r}(n)$ is given by
$$\sum_{n=0}^\infty Q_r(n) q^n = \frac{(q^r;q^{2r})_\infty}{(q;q^2)_\infty}.$$
Considering Theorems \ref{Th:1.4} and \ref{Th:3.2}, we immediately deduced the following partition-theoretic interpretation of Corollary \ref{cor:1.5}.

\begin{corollary}\label{cor:3.2}
	For $m,n\geqslant 0$, $k\geqslant 1$,
	\begin{enumerate}
		\item[a)] $\displaystyle{ (-1)^{k}\left( Q(n)+2 \sum_{j=1}^{k} (-1)^j Q(n-j^2)-(-1)^{T_m} \delta_{n,G_m}\right) = 2M_{o,k}(n) },$
		\item[b)] $\displaystyle{ (-1)^{k}\left( Q(n)+2 \sum_{j=1}^{k} (-1)^j Q(n-2j^2)-(-1)^{T_{\lfloor m/2 \rfloor}}\delta_{n,G_m}\right)  }$
		\item[] $\displaystyle{\qquad = 2\sum_{j=0}^{\lfloor n/2 \rfloor} M_{o,k}(j) Q_{2}(n-2j)},$

		\item[c)] $\displaystyle{ (-1)^{k}\left( Q(n)+2 \sum_{j=1}^{k} (-1)^j Q(n-3j^2)-\delta_{n,G_m}\right)  }$
		\item[] $\displaystyle{\qquad = 2\sum_{j=0}^{\lfloor n/3 \rfloor} M_{o,k}(j) Q_{3}(n-3j)}.$

	\end{enumerate}

\end{corollary}

We can rewrite Conjecture \ref{C:1} in the following equivalent form.

\begin{conjecture}\label{Cj:2}
	For $n\geqslant 0$, $k\geqslant 1$,
	$$\sum_{j=0}^{\lfloor n/3 \rfloor} M_{o,k}(j) Q_{3}(n-3j) \leqslant
	\sum_{j=0}^{\lfloor n/2 \rfloor} M_{o,k}(j) Q_{2}(n-2j) \leqslant
	M_{o,k}(n).$$
\end{conjecture}

Related to this conjecture, we remark that $Q_3(n)$ counts the partitions of $n$ into odd parts in which no part appears more than twice. More details about this combinatorial interpretation can be found in \cite{Alladi,Andrews00}. On the other hand, $Q_2(n)$ counts the partitions of $n$ into distinct odd parts. For $n\geqslant 0$, it is clear that
$$Q_3(n)\geqslant Q_2(n).$$
This inequality makes Conjecture \ref{Cj:2} more interesting.

Combinatorial proofs of Corollaries \ref{cor:3.1} and \ref{cor:3.2} would be very interesting.

\section{Two open problems}
\label{S4}

Related to the sum in the right hand side  of the truncated pentagonal number theorem \eqref{eq:2}, we remark
that there is a substantial amount of numerical evidence to state
the following conjecture.

\begin{conjecture}
	For $k\geqslant 1$, the theta series
	$$(-q;q)_\infty \sum_{n=0}^\infty (-1)^{T_n} q^{G_{n+2k}} = (q^2;q^2)_\infty \sum_{n=1}^{\infty} \frac{q^{\binom{k}{2}+(k+1)n}}{(q;q)_n}
	\begin{bmatrix}
	n-1\\k-1
	\end{bmatrix}$$
	has non-negative coefficients.
\end{conjecture}

Assuming this conjecture, we derive the following family of linear inequalities.

\begin{inequality}\label{I:4.1}
	For $m,n\geqslant 0$, $k\geqslant 1$,
	$$(-1)^{k-1} \left( \sum_{j=0}^{2k-1} (-1)^{T_j} Q\left(n-G_j\right) - (-1)^{T_m}\delta_{n,2G_m} \right) \geqslant 0$$
	with strict inequality if and only  if $n\geqslant G_{2k}$.
\end{inequality}

We notice that this inequality is equivalent with: for $n\geqslant 0$, $k\geqslant 1$,
$$\sum_{j=0}^\infty (-1)^{T_j} M_k(n-2G_j) \geqslant 0$$
with strict inequality if and only  if $n\geqslant G_{2k}$.

In \cite{Andrews17}, Andrews and Merca provide the following revision of \eqref{eq:7}:
\begin{align*}
& \frac{(-q;q^2)_\infty}{(q^2;q^2)_\infty} \sum_{j=0}^{2k-1}(-q)^{T_j}  \\
& \qquad = 1 - (-1)^k \frac{(-q;q^2)_k}{(q^2;q^2)_{k-1}}
\sum_{j=0}^{\infty} \frac{q^{k(2j+2k+1)}(-q^{2j+2k+3};q^2)_{\infty}}{(q^{2k+2j+2};q^2)_{\infty}},
\end{align*}
with
$$\sum_{n=0}^\infty MP_k(n) q^n = \frac{(-q;q^2)_k}{(q^2;q^2)_{k-1}}
\sum_{j=0}^{\infty} \frac{q^{k(2j+2k+1)}(-q^{2j+2k+3};q^2)_{\infty}}{(q^{2k+2j+2};q^2)_{\infty}},$$
where $MPk(n)$ is the number of partitions of $n$ in which the first part larger than $2k-1$ is odd and appears exactly $k$ times; all other odd parts appear at most once. A purely combinatorial proof of this result can be found in \cite{Ballantine}.
Related to the sum in the right hand side  of these identities, we remark

the following conjecture.

\begin{conjecture}
	For $k\geqslant 1$, the theta series
	$$(-q;q)_\infty \sum_{n=0}^\infty (-1)^{T_n} q^{T_{n+2k}} = (q^4;q^4)_\infty \frac{(-q;q^2)_k}{(q^2;q^2)_{k-1}}
	\sum_{j=0}^{\infty} \frac{q^{k(2j+2k+1)}(-q^{2j+2k+3};q^2)_{\infty}}{(q^{2k+2j+2};q^2)_{\infty}}$$
	has non-negative coefficients.
\end{conjecture}

Assuming this conjecture, we derive a new infinity family of linear inequalities.

\begin{inequality}\label{I:4.2}
	For $m,n\geqslant 0$, $k\geqslant 1$,
	$$(-1)^{k-1} \left( \sum_{j=0}^{2k-1} (-1)^{T_j} Q\left(n-T_j\right) - (-1)^{T_m}\cdot \delta_{n,4G_m} \right)  \geqslant 0,$$
	with strict inequality if and only if  $n\geqslant T_{2k}$.
\end{inequality}

We remark that this inequality can be rewritten in terms of $MP_k(n)$ as follows: for $n\geqslant 0$, $k\geqslant 1$,
$$\sum_{j=0}^\infty (-1)^{T_j} MP_{k} (n-4G_j) \geqslant 0,$$
with strict inequality if and only if  $n\geqslant T_{2k}$.

The limiting case $k\to\infty$ of Identities \ref{I:4.1} and \ref{I:4.2} give the following linear recurrence relations:
\begin{align*}
&\sum_{j=0}^\infty (-1)^{T_j} Q(n-G_j)
=\begin{cases}
(-1)^{T_{m}}, & \text{if $n=2G_m$, $m\in\mathbb{N}_0$,}\\
0, & \text{otherwise,}
\end{cases}\\
&\sum_{j=0}^\infty (-1)^{T_j} Q(n-T_j)
=\begin{cases}
(-1)^{T_{m}}, & \text{if $n=4G_m$, $m\in\mathbb{N}_0$,}\\
0, & \text{otherwise.}
\end{cases}
\end{align*}

The effectiveness of these recurrence relations can not be called into  question because the number of terms in each relations is greater than $\sqrt{n/3}$.
As consequences of these recurrence relations, we can derive the following parity results.

\begin{corollary}
	Let $n$ be a positive integer.
	\begin{enumerate}
		\item[a)] The number of representations of $n$ as the sum of two generalized pentagonal numbers is odd if and only if $n$ is a twice generalized pentagonal number.
		\item[b)] The number of representations of $n$ as the sum of a generalized pentagonal number and a triangular number is odd if and only if $n$ is a four times generalized pentagonal number.
	\end{enumerate}
\end{corollary}

Finally, we remark that the linear recurrence relations presented in this paper can be easily derived considering the Jacobi triple product identity.
It is still an open problem to give partition interpretations for our truncated sums in Inequalities \ref{I:4.1} and \ref{I:4.2}.