TITLE: Finding a linear map $T:\mathbb R^3 \to \mathbb R^4$ that satisfy conditions about the kernel and image
QUESTION [0 upvotes]: Find a linear map $T:\mathbb R^3 \to \mathbb R^4$ that satisfy the following conditions (or explain why there can't be such linear map):

$\ker(T)=sp\{(1,-1,0)^T,(3,2,-5)^T  \} $
$\ker(T)=sp\{(1,-1,0)^T,(3,2,-5)^T  \},\operatorname{im}(T)=sp\{(2,1,1,-3)^T,(4,3,0,1)^T  \}$
$\operatorname{im}(T)=sp\{(2,1,1,-3)^T,(4,3,0,1)^T$

I don't know how to use the given information about the kernel and image to make a linear map so any help would be appreciated.

REPLY [1 votes]: Some Background
Definitely knowing only the kernel of the map $T$ is incomplete information to find a map, unless the whole domain is the kernel i.e. $Ker \; T = \mathbb{R}^3$ in which case the map is the zero map. ($T \equiv 0$).
So in your questions there is no unique map but many if that exists. Because, if $Ker \; T$ is a proper subspace of the domain, then you can extend the basis of $Ker \; T$ to the basis of domain $\mathbb{R}^3$. (In all the above cases the $\mbox{ dim of } Ker \; T = 2$ which makes it a proper subspace of $\mathbb{R}^3$ whose dim is 3, so we will get just one and only more vector while extending.)
After that take the vectors outside $Ker \; T$ to arbitrary vectors if $Im \; T$ is not specified.(In this case I can take the third basis vector outside $Ker \; T$ to arbitrary image in $R^4$.)
The image of $T$ if specified brings in a little more constraint, but then we also have the freedom to take the third basis vector to arbitrary vector in the given image space of $T$.
But then as you pointed out : Rank - Nullity theorem guarantees that we must have $ dim \; Ker \; T +  dim \; Im \; T = 3 $ so ANSWER TO Question 2 is impossible to find. (Another way of looking at it is because the new basis has only one vector outside $Ker \; T$, the image of $T$ is only span of this third basis vector's image as all the vectors are linear combination of basis vectors.)
Extension of vectors : Well you might be thinking how will I guess the third vector, as there are a bunch of vectors which can complete the basis by being the third basis vector. To overcome this problem note we have an inner product structure on $\mathbb{R}^n$ here we have $\mathbb{R}^3$.
So if you apply Gram-Schmidt orthonormalization on the basis of $Ker \; T$ you will get two orthogonal vectors. Now in $R^3$ there is only one unique orthonormal vector to two given orthonormal vectors. (Prove it!)
Note that the only thing that connects the $Ker \; T$ and $Im \; T$ in different vector spaces, is the Rank - Nullity theorem. So the only thing that we can do best is check dimension compatibility.
Also a linear transformation is completely determined by its action on basis. So defining where $T$ takes its basis we actually spell out the map $T$. That's what we are going to do next.

ANSWERS
If you have understood the background finding the answers is routine now. By inspection $(1,0,0)$ is linearly independent to Kernel Bases given in 1,2 questions.(I don't use Gram-Schmidt Orthonormalization as I readily obtain the the third vector from inspection.)
Because we are asked to find a map we will only find a map :
1) $T(1,0,0) = (1,0,0,0), \; T(v) = (0,0,0,0)$ if $v \not\in <(1,0,0)>$.
2) Is not possible as Rank Nullity Theorem is violated.
3) Choose the simplest basis of $\mathbb{R}^3$ i.e. $(1,0,0), (0,1,0), (0,0,1)$, take define $T$ to be :
$T(1,0,0) = (0,0,0,0), T(0,1,0) = (2,1,1,-3), T(0,0,1) = (4,3,0,1)$.
@ GinKin : So I hope this answers all your questions completely.
@ Prafulla Koushik : I really liked Linear Algebra as a an undergraduate, and my favorite book is "Linear Algebra : A geometric viewpoint" whose author teaches you. So you must love the subject more ! :)


text:
\begin{document}

\allowdisplaybreaks

\begin{abstract}
By considering the $M_2$-rank of an overpartition as well as a residual crank,
we give another combinatorial refinement of the congruences
$\sptBar{2}{3n}\equiv \sptBar{2}{3n+1}\equiv 0\pmod{3}$. Here $\sptBar{2}{n}$ is the total number
of occurrences of the smallest parts among the overpartitions of $n$ where the
smallest part is even and not overlined.
Our proof depends on Bailey's Lemma
and the rank difference formulas of Lovejoy and Osburn for the
$M_2$-rank of an overpartition. This congruence, along with a modulo $5$
congruence, has previously been refined using the rank of an overpartition.
\end{abstract}

\maketitle

\section{Introduction and statement of Results}

\allowdisplaybreaks

We recall a partition of a positive integer $n$ is a non-increasing sequence
of positive integers that sum to $n$. As an example, the partitions of $4$
are $4$, $3+1$, $2+2$, $2+1+1$, and $1+1+1+1$. Similar to this is the idea
of an overpartition. An overpartition of $n$ is a partition of $n$ where the
first occurrence of a part may (or may not) be overlined. For example, the
overpartitions of $4$ are $4$, $\overline{4}$, $3+1$, $3+\overline{1}$,
$\overline{3}+1$, $\overline{3}+\overline{1}$, $2+2$,
$\overline{2}+2$, $2+1+1$, $2+\overline{1}+1$, $\overline{2}+1+1$,
$\overline{2}+\overline{1}+1$, $1+1+1+1$, and $\overline{1}+1+1+1$.

We have a weighted count on partitions, and overpartitions, given by counting a
partition by the number of times the smallest occurs. We use the convention  of
not counting overpartitions where the smallest part is overlined. We let
$\spt{n}$ denote the total number of occurrences of the smallest parts among
the partitions of $n$. We let $\sptBar{}{n}$ denote the total number of
occurrences of the smallest parts among the overpartitions of $n$ without smallest
part overlined. The function $\spt{n}$ was introduced by Andrews in
\cite{Andrews} and the function $\sptBar{}{n}$ was introduced by Bringmann,
Lovejoy, and Osburn in \cite{BLO2}. Two restrictions of $\sptBar{}{n}$
are $\sptBar{1}{n}$ and $\sptBar{2}{n}$, where restrict to overpartitions
where the smallest part is odd and even respectively. We see
$\spt{4}=10$, $\sptBar{}{4}=13$, $\sptBar{1}{4}=10$, and $\sptBar{2}{4}=3$.

Similar to the work of Andrews, Garvan, and Liang in \cite{AGL} for
$\spt{n}$, in \cite{GarvanJennings} Garvan and the author gave combinatorial
refinements of congruences satisfied by $\sptBar{}{n}$, $\sptBar{1}{n}$, and
$\sptBar{2}{n}$. The idea is to introduce an extra variable into the generating
function of each spt-function to get a crank type statistic. This statistic can
then be shown in certain cases to equally split up the numbers $\sptBar{}{n}$,
$\sptBar{1}{n}$, and $\sptBar{2}{n}$ based on the residue class of the
statistic. We explain this in more detail shortly.

For $\sptBar{2}{n}$ we have the congruences
\begin{align}\label{EqTheMod3Congruences1}
	\sptBar{2}{3n} \equiv 0 \pmod{3},
	\\\label{EqTheMod3Congruences2}
	\sptBar{2}{3n+1} \equiv 0 \pmod{3},
	\\\label{EqTheMod5Congruences}
	\sptBar{2}{5n+3} \equiv 0 \pmod{5}
.
\end{align}
In this paper we give another proof of the modulo $3$ congruences.

To start, by summing according to the smallest part, we find a generating
function for $\sptBar{2}{n}$ to be given by
\begin{align}
	\sum_{n=1}^\infty \sptBar{2}{n}q^n
	&=
	\sum_{n=1}^\infty \frac{q^{2n} \aqprod{-q^{2n+1}}{q}{\infty}}
	{(1-q^{2n})^2\aqprod{q^{2n+1}}{q}{\infty}}.
\end{align}
Here we use the standard product notation,
\begin{align}
	\aqprod{a}{q}{n} &= \prod_{j=0}^{n-1}(1-aq^j)
	,\\
	\aqprod{a_1,a_2,\dots,a_k}{q}{n}
	&=
	\aqprod{a_1}{q}{n}\aqprod{a_2}{q}{n}\dots\aqprod{a_k}{q}{n}
	,\\
	\aqprod{a}{q}{\infty} &= \prod_{j=0}^{\infty}(1-aq^j)
	,\\
	\aqprod{a_1,a_2,\dots,a_k}{q}{\infty}
	&=
	\aqprod{a_1}{q}{\infty}\aqprod{a_2}{q}{\infty}\dots\aqprod{a_k}{q}{\infty}
.
\end{align}

In \cite{GarvanJennings} we considered the two variable generalization given
by
\begin{align}
	\overline{\mbox{S}}_2(z,q)
	&=
	\sum_{n=1}^\infty
		\frac{q^{2n}\aqprod{-q^{2n+1}}{q}{\infty}\aqprod{q^{2n+1}}{q}{\infty}}
			{\aqprod{zq^{2n}}{q}{\infty}\aqprod{z^{-1}q^{2n}}{q}{\infty}}
.
\end{align}
We note setting $z=1$ gives the generating function for $\sptBar{2}{n}$.
It turns out $\overline{\mbox{S}}_2(z,q)$ can be expressed in terms of the Dyson rank of an
overpartition and a residual crank from \cite{BLO2}. In \cite{LO1}
Lovejoy and Osburn determined difference formulas for the Dyson rank
of an overpartition, these formulas are essential to the proofs in
\cite{GarvanJennings}. In the same paper, we also used the difference formulas
for the $M_2$-rank of a partition without repeated odd parts, these formulas
were also determined by Lovejoy and Osburn in \cite{LO2}. Lovejoy
and Osburn also found difference formulas for the $M_2$-rank
of an overpartition \cite{LO3}, but we did not use these formulas in
\cite{GarvanJennings}.

In \cite{CJS1} the
author gave higher order generalizations of $\sptBar{}{n}$ and
$\sptBar{2}{n}$ and noted that one could use the $M_2$-rank and another
residual crank from \cite{BLO2}, as used in that paper in working with $\sptBar{2}{n}$,
to explain the modulo $3$ congruences for $\sptBar{2}{n}$. Here we give the proof.
In this paper we instead use
\begin{align}
	\SB(z,q) &= \sum_{n=1}^\infty
	\frac{q^{2n}\aqprod{-q^{2n+1}}{q}{\infty}\aqprod{q^{2n+1}}{q}{\infty}}
		{\aqprod{zq^{2n},z^{-1}q^{2n}}{q^2}{\infty}\aqprod{q^{2n+1}}{q^2}{\infty}^2}
	\\
	&=
	\sum_{n=1}^\infty \sum_{m=-\infty}^\infty N_{\SB}(m,n)z^mq^n
.
\end{align}
Again setting $z=1$ gives the generating function for $\sptBar{2}{n}$. This is
not the same series $S(z,q)$ used in \cite{GarvanJennings}, however we do not
want to overly complicate matters with additional notation.

For a positive integer $t$ we let
\begin{align}
	N_{\SB}(k,t,n) = \sum_{m\equiv k \pmod{t}}N_{\SB}(m,n).
\end{align}
We then have
\begin{align}
	\sptBar{2}{n} &= \sum_{m=-\infty}^\infty N_{\SB}(m,n)
	= \sum_{k=0}^{t-1} N_{\SB}(k,t,n)
.
\end{align}
Additionally we see if $\zeta$ is a $t^{th}$ root of unity, then
\begin{align}
	\SB(\zeta,q)
	&=
	\sum_{n=1}^\infty \left(\sum_{k=0}^{t-1} N_{\SB}(k,t,n)\zeta^k\right)q^n
.
\end{align}
We consider when $\zeta=\zeta_3$ is a primitive third root of unity. Here the minimal
polynomial for $\zeta_3$ is $x^2+x+1$. If
$N_{\SB}(0,t,N)+N_{\SB}(1,t,N)\zeta_3+N_{\SB}(2,t,N)\zeta_3^2=0$ then we must in fact have
$N_{\SB}(0,t,N)=N_{\SB}(1,t,N)=N_{\SB}(2,t,N)$. That is to say, if
the coefficient of $q^N$ in $\SB(\zeta_3,q)$ is zero, then
$\sptBar{2}{N} = 3\cdot N_{\SB}(0,t,N)$ and so
$\sptBar{2}{N} \equiv 0 \pmod{3}$.

Our proof of $\sptBar{2}{3n}\equiv\sptBar{2}{3n+1}\equiv 0$ is to find the $3$-dissection of
$\SB(\zeta_3,q)$ with the $q^{3n}$ and $q^{3n+1}$ terms being all zero.
This is the same idea that was used in \cite{GarvanJennings}, we are just using
$\SB(z,q)$ rather than $\overline{\mbox{S}}_2(z,q)$.

\begin{theorem}\label{SptBarCrankDissection}
\begin{align}
	\SB(\zeta_3,q)
	&=  A_0(q^3) + qA_1(q^3) + q^2A_2(q^3)
,
\end{align}
where
\begin{align}
	A_0(q) &= 0,
	\\
	A_1(q) &= 0,
	\\
	A_2(q)
	&=
	\frac{\aqprod{q^6}{q^6}{\infty}^4}
		{\aqprod{q^2}{q^2}{\infty}\aqprod{q^3}{q^3}{\infty}^2}
	+\frac{2q\aqprod{-q^3}{q^3}{\infty}}{\aqprod{q^3}{q^3}{\infty}}
	\sum_{n=-\infty}^\infty \frac{(-1)^n q^{3n^2+6n}}{1-q^{6n+2}}
.
\end{align}
\end{theorem}

We prove these formulas by relating $\SB(z,q)$ to a certain rank and crank
and using dissections of these related functions.
We recall the $M_2$-rank of an overpartition $\pi$ is given by
\begin{align}
	M_2\mbox{-rank} = \left\lceil\frac{l(\pi)}{2} \right\rceil - \#(\pi)
		+\#(\pi_o) - \chi(\pi)
,
\end{align}
where $l(\pi)$ is the largest part of $\pi$, $\#(\pi)$ is the number of
parts of $\pi$, $\#(\pi_o)$ is the number of non-overlined odd parts,
and $\chi(\pi)=1$ if the largest part of $\pi$ is odd and non-overlined
and otherwise  $\chi(\pi)=0$. The $M_2$-rank was introduced by
Lovejoy in \cite{LO3}. We let $\overline{N2}(m,n)$ denote the number of
overpartitions of $n$ with $M_2$-rank $m$. Lovejoy found
the generating function for $\overline{N2}$ is given by
\begin{align}\label{RankDef}
	\sum_{n=0}^\infty\sum_{m=-\infty}^\infty \overline{N2}(m,n)z^mq^n
	&=
	\sum_{n=0}^\infty \frac{\aqprod{-1}{q}{2n}q^n}
			{\aqprod{zq^2,z^{-1}q^2}{q^2}{n}}
	=
	\frac{\aqprod{-q}{q}{\infty}}{\aqprod{q}{q}{\infty}}
		\Parans{1+
			2\sum_{n=1}^\infty\frac{(1-z)(1-z^{-1})(-1)^nq^{n^2+2n}}
				{(1-zq^{2n})(1-z^{-1}q^{2n})}
		}
.
\end{align}

We also use a residual crank first defined \cite{BLO2}. We define
\begin{align}\label{CrankDef}
	\sum_{n=0}^\infty\sum_{m=-\infty}^\infty \overline{M2}(m,n)z^mq^n
	&= \frac{\aqprod{-q}{q}{\infty}\aqprod{q^2}{q^2}{\infty}}
		{\aqprod{q}{q^2}{\infty}\aqprod{zq^2}{q^2}{\infty}\aqprod{z^{-1}q^2}{q^2}{\infty}}.
\end{align}
The interpretation of this is as follows.
For an overpartition $\pi$ of $n$ we take
the crank of the partition $\frac{\pi_e}{2}$ obtained by taking the
subpartition $\pi_e$, of the even non-overlined parts of
$\pi$, and halving each part of $\pi_e$.
Then $\overline{M2}(m,n)$ is the number of overpartitions $\pi$ of $n$
and such that the partition $\frac{\pi_e}{2}$ has crank $m$.
However this interpretation fails when considering overpartitions
whose only even non-overlined parts are a single two, as the
corresponding interpretation of
$\frac{\aqprod{q}{q}{\infty}}{\aqprod{zq,q/z}{q}{\infty}}$
as the generating function of the crank for ordinary partitions fails for
the partition of $1$.

\begin{theorem}\label{SptBar2RankCrank}
\begin{align}
	\SB(z,q)
	&=
	\frac{1}{(1-z)(1-z^{-1})}
	\sum_{n=0}^\infty\sum_{m=-\infty}^\infty
		\Parans{\overline{N2}(m,n)-\overline{M2}(m,n)}z^mq^n
.
\end{align}
\end{theorem}

Using the rank difference formulas derived by Lovejoy and Osburn in
\cite{LO3}, we have
\begin{theorem}\label{DissectionForRank}
\begin{align}
	\sum_{n=0}^\infty\sum_{m=-\infty}^\infty
	\overline{N2}(m,n)\zeta_3^mq^n
	&= \overline{N2}_{0}(q^3)
		+ q\overline{N2}_{1}(q^3)
		+ q^2\overline{N2}_{2}(q^3)
,
\end{align}
where
\begin{align}
	\overline{N2}_{0}(q)
	&=
	\frac{\aqprod{-q}{q}{\infty}\aqprod{q^3}{q^3}{\infty}^2}
		{\aqprod{q}{q}{\infty}\aqprod{-q^3}{q^3}{\infty}^2}
	,
	\\
	\overline{N2}_{1}(q)
	&=
	\frac{2\aqprod{q^3}{q^3}{\infty}\aqprod{q^6}{q^6}{\infty}}
		{\aqprod{q}{q}{\infty}}
	,\\
	\overline{N2}_{2}(q)
	&=
	\frac{4\aqprod{q^6}{q^6}{\infty}^4}
		{\aqprod{q^2}{q^2}{\infty}\aqprod{q^3}{q^3}{\infty}^2}
	+
	\frac{6q\aqprod{-q^3}{q^3}{\infty}}{\aqprod{q^3}{q^3}{\infty}}
	\sum_{n=-\infty}^\infty \frac{(-1)^n q^{3n^2+6n}}{1-q^{6n+2}}
.
\end{align}
\end{theorem}

\begin{theorem}\label{M2CrankDissection}
\begin{align}
	\frac{\aqprod{-q}{q}{\infty}\aqprod{q^2}{q^2}{\infty}}
		{\aqprod{q}{q^2}{\infty}\aqprod{\zeta_3q^2,\zeta_3^{-1}q^2}{q^2}{\infty}}
	&=
	\sum_{n=0}^\infty\sum_{m=-\infty}^\infty
		\overline{M2}(m,n)\zeta_3^mq^n
	= \overline{M2}_{0}(q^3)
		+ q\overline{M2}_{1}(q^3)
		+ q^2\overline{M2}_{2}(q^3)
,
\end{align}
where
\begin{align}
	\overline{M2}_{0}(q)
	&=
	\frac{\aqprod{-q}{q}{\infty}\aqprod{q^3}{q^3}{\infty}^2}
		{\aqprod{q}{q}{\infty}\aqprod{-q^3}{q^3}{\infty}^2}
	,
	\\
	\overline{M2}_{1}(q)
	&=
	2\frac{\aqprod{q^3}{q^3}{\infty}\aqprod{q^6}{q^6}{\infty}}
		{\aqprod{q}{q}{\infty}}
	,\\
	\overline{M2}_{2}(q)
	&=
	\frac{\aqprod{q^6}{q^6}{\infty}^4}
		{\aqprod{q^2}{q^2}{\infty}\aqprod{q^3}{q^3}{\infty}^2}
.
\end{align}
\end{theorem}

We see Theorem \ref{SptBarCrankDissection} follows from Theorems
\ref{SptBar2RankCrank}, \ref{DissectionForRank}, and \ref{M2CrankDissection}, noting
$\frac{1}{(1-\zeta_3)(1-\zeta_3^{-1})}=\frac{1}{3}$.
We give the proofs of Theorems \ref{SptBar2RankCrank} and
\ref{M2CrankDissection} in the next section. In Section 3 we give brief
combinatorial interpretations of the coefficients $N_{\SB}(m,n)$, in
particular they are non-negative, and in Section 4 we end with a few remarks.

\section{The Proofs}

\begin{proof}[Proof of Theorem \ref{SptBar2RankCrank}]

We recall a pair of sequences  $(\alpha_n,\beta_n)$,
is a Bailey pair relative to $(a,q)$ if
\begin{align}
	\beta_n &= \sum_{r=0}^n \frac{\alpha_r}
		{\aqprod{q}{q}{n-r}\aqprod{aq}{q}{n+r}}
.
\end{align}
A limiting case of Bailey's Lemma gives for a Bailey pair
$(\alpha_n,\beta_n)$ that
\begin{align}
	\sum_{n=0}^\infty
	\aqprod{\rho_1,\rho_2}{q}{n} \Parans{\frac{aq}{\rho_1\rho_2}}^n  \beta_n
	&=
	\frac{\aqprod{aq/\rho_1,aq/\rho_2}{q}{\infty}}
		{\aqprod{aq,aq/\rho_1\rho_2}{q}{\infty}}
	\sum_{n=0}^\infty
	\frac{\aqprod{\rho_1,\rho_2}{q}{n} (\frac{aq}{\rho_1\rho_2})^n \alpha_n}
		{\aqprod{aq/\rho_1,aq/\rho_2}{q}{n}}
.
\end{align}

As in \cite{CJS1} the Bailey pair connecting the $M_2$-rank of
an overpartition and the residual crank is
\begin{align}
	\alpha_n &= \PieceTwo{1}{(-1)^n2q^{n^2}}{n=0}{n\ge 1}
	,\\
	\beta_n &= \frac{\aqprod{q}{q^2}{n}^2}{\aqprod{q^2}{q^2}{2n}}
.
\end{align}
This is a Bailey pair with respect to $(1,q^2)$.

We note that
\begin{align}
	\frac{\aqprod{q^2}{q^2}{\infty}}{\aqprod{z,z^{-1}}{q^2}{\infty}\aqprod{q}{q^2}{\infty}^2}
	\cdot
	\frac{\aqprod{zq^2,z^{-1}q^2}{q^2}{\infty}}{\aqprod{q^2}{q^2}{\infty}^2}
	&=
	\frac{\aqprod{q^2}{q^2}{\infty}}{(1-z)(1-z^{-1})\aqprod{q}{q}{\infty}^2}
	=
	\frac{\aqprod{-q}{q}{\infty}}{(1-z)(1-z^{-1})\aqprod{q}{q}{\infty}}
	.
\end{align}
With this Bailey pair we have
\begin{align}
	\SB(z,q)
	&=
	\sum_{n=1}^\infty
	\frac{q^{2n}\aqprod{-q^{2n+1}}{q}{\infty}\aqprod{q^{2n+1}}{q}{\infty}}
		{\aqprod{zq^{2n},z^{-1}q^{2n}}{q^2}{\infty}\aqprod{q^{2n+1}}{q^2}{\infty}^2}
	\nonumber\\
	&=
	\sum_{n=1}^\infty
	\frac{q^{2n}\aqprod{q^{4n+2}}{q^2}{\infty}}
		{\aqprod{zq^{2n},z^{-1}q^{2n}}{q^2}{\infty}\aqprod{q^{2n+1}}{q^2}{\infty}^2}
	\nonumber\\
	&=
	\frac{\aqprod{q^2}{q^2}{\infty}}{\aqprod{z,z^{-1}}{q^2}{\infty}\aqprod{q}{q^2}{\infty}^2}
	\sum_{n=1}^\infty
		q^{2n}\aqprod{z,z^{-1}}{q^2}{n}\beta_n
	\nonumber\\
	&=
	\frac{\aqprod{q^2}{q^2}{\infty}}{\aqprod{z,z^{-1}}{q^2}{\infty}\aqprod{q}{q^2}{\infty}^2}
	\sum_{n=0}^\infty
		q^{2n}\aqprod{z,z^{-1}}{q^2}{n}\beta_n
	-
	\frac{\aqprod{q^2}{q^2}{\infty}}{\aqprod{z,z^{-1}}{q^2}{\infty}\aqprod{q}{q^2}{\infty}^2}
	\nonumber\\
	&=
	\frac{\aqprod{-q}{q}{\infty}}{(1-z)(1-z^{-1})\aqprod{q}{q}{\infty}}
	\sum_{n=0}^\infty
	\frac{q^{2n}\aqprod{z,z^{-1}}{q^2}{n}  \alpha_n}
		{\aqprod{zq^2,z^{-1}q^2}{q^2}{n}}
	-
	\frac{\aqprod{q^2}{q^2}{\infty}}{\aqprod{z,z^{-1}}{q^2}{\infty}\aqprod{q}{q^2}{\infty}^2}
	\nonumber\\
	&=
	\frac{\aqprod{-q}{q}{\infty}}{(1-z)(1-z^{-1})\aqprod{q}{q}{\infty}}
	\sum_{n=0}^\infty
	\frac{q^{2n}(1-z)(1-z^{-1})\alpha_n}
		{(1-zq^{2n})(1-z^{-1}q^{2n})}
	-
	\frac{\aqprod{-q}{q^2}{\infty}\aqprod{q^2}{q^2}{\infty}}
		{\aqprod{q}{q^2}{\infty}\aqprod{z,z^{-1}}{q^2}{\infty}}
	\nonumber\\
	&=
	\frac{\aqprod{-q}{q}{\infty}}{(1-z)(1-z^{-1})\aqprod{q}{q}{\infty}}
	\left(
		1+
		2\sum_{n=1}^\infty
		\frac{(1-z)(1-z^{-1})(-1)^nq^{n^2+2n}}
			{(1-zq^{2n})(1-z^{-1}q^{2n})}
	\right)
	-
	\frac{\aqprod{-q}{q^2}{\infty}\aqprod{q^2}{q^2}{\infty}}
		{\aqprod{q}{q^2}{\infty}\aqprod{z,z^{-1}}{q^2}{\infty}}
.
\end{align}
By equations (\ref{RankDef}) and (\ref{CrankDef}) we then have
\begin{align}
	\SB(z,q)
	&=
	\frac{1}{(1-z)(1-z^{-1})}
	\sum_{n=0}^\infty\sum_{m=-\infty}^\infty
		\Parans{\overline{N2}(m,n)-\overline{M2}(m,n)}z^mq^n
.
\end{align}
\end{proof}

\begin{proof}[Proof of Theorem \ref{M2CrankDissection}]
We begin by noting
\begin{align}
	\frac{\aqprod{-q}{q}{\infty}\aqprod{q^2}{q^2}{\infty}}
		{\aqprod{q}{q^2}{\infty}\aqprod{\zeta_3q^2,\zeta_3^{-1}q^2}{q^2}{\infty}}
	&=
	\frac{\aqprod{q^2}{q^2}{\infty}^2}{\aqprod{q}{q^2}{\infty}^2\aqprod{q^6}{q^6}{\infty}}
.
\end{align}
By Gauss we have
\begin{align}
	\frac{\aqprod{q^2}{q^2}{\infty}}{\aqprod{q}{q^2}{\infty}}
	&=
	\sum_{n=0}^\infty q^{n(n+1)/2}
	=
	\frac{1}{2}	\sum_{n=-\infty}^\infty q^{n(n+1)/2}
.
\end{align}

By the Jacobi Triple Product Identity we then have
\begin{align}
	\frac{\aqprod{q^2}{q^2}{\infty}}{\aqprod{q}{q^2}{\infty}}
	&=
	\frac{1}{2}\sum_{k=0}^2\sum_{n=-\infty}^\infty q^{(3n+k)(3n+k+1)/2}
	\nonumber\\
	&=
	\frac{1}{2}\sum_{n=-\infty}^\infty q^{(9n^2+3n)/2}
	+\frac{1}{2}\sum_{n=-\infty}^\infty q^{(9n^2+9n)/2}
	+\frac{1}{2}\sum_{n=-\infty}^\infty q^{(9n^2+15n)/2}
	\nonumber\\
	&=
	\frac{1}{2}\aqprod{-q^6,-q^3,q^9}{q^9}{\infty}
	+\frac{1}{2}q\aqprod{-1,-q^9,q^9}{q^9}{\infty}
	+\frac{1}{2}\aqprod{-q^{-3},-q^{12},q^9}{q^9}{\infty}
	\nonumber\\
	&=
	\aqprod{-q^6,-q^3,q^9}{q^9}{\infty}
	+q\aqprod{-q^9,-q^9,q^9}{q^9}{\infty}
.
\end{align}

Using the above to expand $\frac{\aqprod{q^2}{q^2}{\infty}^2}{\aqprod{q}{q^2}{\infty}^2}$
and dividing by $\aqprod{q^6}{q^6}{\infty}$ then gives
\begin{align}
	\overline{M2}_{0}(q)
	&=
	\frac{\aqprod{-q,-q^2,q^3}{q^3}{\infty}^2}{\aqprod{q^2}{q^2}{\infty}}
	,
	\\
	\overline{M2}_{1}(q)
	&=
	2\frac{\aqprod{-q,-q^2,-q^3,-q^3,q^3,q^3}{q^3}{\infty}}{\aqprod{q^2}{q^2}{\infty}}
	\\
	\overline{M2}_{2}(q)
	&=
	\frac{\aqprod{-q^3,-q^3,q^3}{q^3}{\infty}^2}{\aqprod{q^2}{q^2}{\infty}}
.
\end{align}
However these products easily reduce to those in the statement of the Theorem
\ref{M2CrankDissection}.
\end{proof}

\section{Combinatorial Interpretations}

We see $N_{\SB}(m,n)$ can be interpreted in terms of vector partitions.
We let $\overline{V} = \mathcal{D}\times\mathcal{P}\times\mathcal{P}\times\mathcal{D}$,
where $\mathcal{P}$ denotes the set of all partitions and $\mathcal{D}$
denotes the set of all partitions into distinct parts. For a partition
$\pi$ we let $s(\pi)$ denote the smallest part of $\pi$
(with the convention that the empty partition has smallest part $\infty$),
$\#(\pi)$ the number of parts in $\pi$,
$\#(\pi_e)$ the number of even parts in $\pi$,
and $|\pi|$ the sum of the parts of $\pi$.
For $\vec{\pi} = (\pi^1,\pi^2,\pi^3,\pi^4) \in\overline{V}$, we define the weight
$\omega(\vec{\pi}) = (-1)^{\#(\pi^1)-1 }$, the
$\mbox{crank}(\vec{\pi}) = \#(\pi^2_e) - \#(\pi^3_e)$,
and the norm $|\vec{\pi}|=|\pi^1|+|\pi^2|+|\pi^3|+|\pi^4|$. We say $\vec{\pi}$ is
a vector partition of $n$ if $|\vec{\pi}|=n$.

By writing the summands of $\SB(z,q)$ as
\begin{align}
	&q^{2n}\aqprod{q^{2n+1}}{q}{\infty}
	\cdot
	\frac{1}{\aqprod{zq^{2n}}{q^2}{\infty}\aqprod{q^{2n+1}}{q^2}{\infty}}
	\cdot
	\frac{1}{\aqprod{z^{-1}q^{2n}}{q^2}{\infty}\aqprod{q^{2n+1}}{q^2}{\infty}}
	\cdot
	\aqprod{q^{2n+1}}{q}{\infty}
,
\end{align}
we see $N_{\SB}(m,n)$ is the number of vector partitions
$(\pi^1,\pi^2,\pi^3,\pi^4)$ from $\overline{V}$ of $n$ with the additional constraints
that $\pi^1$ is non-empty,
$s(\pi^1)\le s(\pi^2)$, $s(\pi^1)\le s(\pi^3)$, $s(\pi^1)< s(\pi^4)$,
and $s(\pi^1)$ is even, but counted with the weight $\omega$.

However, this interpretation hides the fact that each $N_{\SB}(m,n)$ is
non-negative, as we are counting with a weight that may be negative.
We can also interpret $N_{\SB}(m,n)$ in terms of partition pairs.
This interpretation makes the non-negativity clear. Using the $q$-binomial theorem
we have
\begin{align}
	&\sum_{n=1}^\infty
	\frac{q^{2n}\aqprod{q^{4n+2}}{q^2}{\infty}}
		{\aqprod{zq^{2n},z^{-1}q^{2n}}{q^2}{\infty}\aqprod{q^{2n+1}}{q^2}{\infty}^2}
	\nonumber\\
	&=
	\sum_{n=1}^\infty
	\frac{q^{2n}}{\aqprod{zq^{2n}}{q^2}{\infty}\aqprod{q^{2n+1}}{q^2}{\infty}^2}
	\sum_{k=0}^\infty
	\frac{\aqprod{zq^{2n+2}}{q^2}{k} z^{-k}q^{2nk}}
		{\aqprod{q^2}{q^2}{k}}
	\nonumber\\
	&=
	\sum_{n=1}^\infty
	\frac{q^{2n}}{\aqprod{zq^{2n}}{q^2}{\infty}\aqprod{q^{2n+1}}{q^2}{\infty}^2}
	+
	\sum_{n=1}^\infty\sum_{k=1}^\infty
	\frac{z^{-k}q^{2n+2nk}}
		{(1-zq^{2n})\aqprod{zq^{2n+2k+2}}{q^2}{\infty}\aqprod{q^2}{q^2}{k}\aqprod{q^{2n+1}}{q^2}{\infty}^2}
	\nonumber\\
	&=
	\sum_{n=1}^\infty
	\frac{q^{2n}}{\aqprod{zq^{2n}}{q^2}{\infty}\aqprod{q^{2n+1}}{q^2}{\infty}^2}
	\nonumber\\&\quad
		+
		\sum_{n=1}^\infty\sum_{k=1}^\infty
		\frac{q^{2n}\aqprod{q^2}{q^2}{n}}
			{(1-zq^{2n})\aqprod{q^2}{q^2}{n+k}\aqprod{zq^{2n+2k+2}}{q^2}{\infty}\aqprod{q^{2n+1}}{q^2}{\infty}}
		\cdot
		\frac{z^{-k}q^{2nk}\aqprod{q^2}{q^2}{n+k}}
			{\aqprod{q^2}{q^2}{k}\aqprod{q^2}{q^2}{n}\aqprod{q^{2n+1}}{q^2}{\infty}}
	\nonumber\\
	&=
	\sum_{n=1}^\infty
	\frac{q^{2n}}{\aqprod{zq^{2n}}{q^2}{\infty}\aqprod{q^{2n+1}}{q^2}{\infty}^2}
	\nonumber\\\label{EqPairCrankSeries}&\quad
		+
		\sum_{n=1}^\infty\sum_{k=1}^\infty
		\frac{q^{2n}}
			{(1-zq^{2n})\aqprod{q^{2n+2}}{q^2}{k}\aqprod{zq^{2n+2k+2}}{q^2}{\infty}
			\aqprod{q^{2n+1}}{q^2}{\infty}}
		\cdot
		\frac{z^{-k}q^{2nk}\aqprod{q^2}{q^2}{n+k}}
			{\aqprod{q^2}{q^2}{k}\aqprod{q^2}{q^2}{n}\aqprod{q^{2n+1}}{q^2}{\infty}}
.
\end{align}

We can now give the partition pair interpretation. We let $PP_2$ denote the set of
partition pairs $(\pi^1,\pi^2)$ such that $\pi^1$ is non-empty, $s(\pi^1)$ is
even, $s(\pi^1)\le s(\pi^2)$, and the even parts of $\pi^2$
are at most $2s(\pi^1)$. For such a partition pair we
let $k(\pi^1,\pi^2)$ denote the number of even parts of $\pi^1$ that are either
the smallest part or are larger than $s(\pi_1)+2\#(\pi^2_e)$. We note when
$\pi^2$ contains no even parts that $k(\pi_1,\pi_2)$ reduces to $\#(\pi^1_e)$.
We define a crank on the elements of $PP_2$ by
\begin{align}
	c(\pi_1,\pi_2)
	&=
		k(\pi^1,\pi^2) - \#(\pi^2_e) 	 - 1
.
\end{align}
We claim $N_{\SB}(m,n)$ is also the number of partitions pairs of $n$ from
$PP_2$ with $c(\pi^1,\pi^2)=m$.

For this we note the first series in (\ref{EqPairCrankSeries}) gives the cases
when $\pi^2$ has no
even parts. The second series in (\ref{EqPairCrankSeries}) gives the cases when $\pi^2$ has even
parts, since
 $\frac{q^{2nk}\aqprod{q^2}{q^2}{n+k}}{\aqprod{q^2}{q^2}{k}\aqprod{q^2}{q^2}{n}}$
is the generating function for partitions into even parts with exactly $k$ parts
and each part between $2n$ and $4n$ (inclusive).

It may be possible to define a bijection from these partition pairs to
marked overpartitions with smallest part even, and through that determine
a crank defined on marked overpartitions. However, we do not pursue that here.

\section{Remarks}
While $\SB(\zeta_3,q)$ can be used to prove
$\sptBar{2}{3n}\equiv 0 \pmod{3}$
and $\sptBar{2}{3n+1}\equiv 0 \pmod{3}$, we cannot use $\SB(\zeta_5,q)$ to prove
$\sptBar{2}{5n+3}\equiv 0 \pmod{5}$. In particular we find the coefficient
of $q^8$ in $\SB(\zeta_5,q)$ to be
$z^3+z^2+3z+5+3z^{-1}+z^{-2}+z^{-3}$. That is to say,
$N_{\SB}(0,5,8) = 5$,
$N_{\SB}(1,5,8) = 3$,
$N_{\SB}(2,5,8) = 2$,
$N_{\SB}(3,5,8) = 2$, and
$N_{\SB}(4,5,8) = 3$.

However $\sptBar{2}{5n+3}\equiv 0 \pmod{5}$ does follow by considering
$\overline{\mbox{S}}_2(\zeta_5,q)$.
This can be compared with the rank of a partition explaining the congruences for
$p(5n+4)$ and $p(7n+5)$ but not $p(11n+6)$, whereas the crank of an partition
does explain all three.

\bibliographystyle{abbrv}
\bibliography{alternateSptBar2Mod3Ref}

\end{document}

config: arxiv


text:
\begin{document}

\title{Quantum information aspects of approximate position measurement}
\author{A. S. Holevo, V. I. Yashin \\

Steklov Mathematical Institute, RAS, Moscow, Russia}
\date{}
\maketitle

\begin{abstract}
We perform a quantum information analysis for multi-mode Gaussian
 approximate position measurements, underlying
 noisy homodyning in quantum optics. The
\textquotedblleft Gaussian maximizer\textquotedblright\ property is
established for the entropy reduction of these measurements which provides
explicit formulas for computations including their entanglement-assisted capacity.
The case of one mode is discussed in detail.

Keywords: approximate position measurement, entropy reduction, Gaussian maximizers, energy constraint, entanglement-assisted capacity,  continuous variable system
\end{abstract}

\section{\protect\bigskip Introduction}\label{s0}

Among quantitative characteristics of quantum measurement, the entropy
reduction and the entanglement-assisted classical information capacity
(called also measurement strength or information gain, depending on different operational
interpretations) play a significant role. We
refer a reader to \cite{winmas}, \cite{Sh-ERQM}, \cite{h5}, \cite{H1}, \cite
{H2} and to \cite{BRW} where one can find also a detailed survey and further
references. In \cite{H3} we presented a study of these quantities for a
class of multi-mode Gaussian approximate joint
position-momentum measurements (or, in the context of quantum optics, \textquotedblleft
noisy heterodyning\textquotedblright ). In particular, for this class of
measurements we established the \textquotedblleft Gaussian
maximizer\textquotedblright\ property of the entropy reduction which allowed
for an explicit computation of this quantity and of the  related
entanglement-assisted capacity. In the present paper we perform a similar
analysis for another important class of multi-mode Gaussian measurements,
namely, of approximate position measurements underlying the
\textquotedblleft noisy homodyning\textquotedblright\ which is in a sense
opposite to the \textquotedblleft heterodyning\textquotedblright\  (see \cite
{caves}, \cite{hall}, \cite{sera} for a detailed physical description of the relevant
measurement processes). The \textquotedblleft Gaussian
maximizer\textquotedblright\ property is established in theorem \ref{t3}
which also gives the explicit formula for computations. The
entanglement-assisted capacity is considered in sec. \ref{s3} and the case
of one mode is discussed in detail in sec. \ref{s4}.

\section{\protect\bigskip Entropy reduction of quantum observable}\label{s1}

Let $\mathcal{H}$ be the separable Hilbert space of a quantum system, and
let $\mathfrak{S}(\mathcal{H})$ be the convex set of quantum states (density
operators on $\mathcal{H}$). Let $(\mathcal{X},\mathcal{F})$ be a standard
measurable space of the outcomes of a measurement described by an \emph{
completely positive (c.p.) instrument} $\mathcal{M=}\left\{ \mathcal{M}
(A)[\cdot ],A\in \mathcal{F}\right\} ,$ i.e. operation-valued measure where
for each $A$ the map $\rho \rightarrow \mathcal{M}(A)[\rho ],\,\rho \in
\mathfrak{S}(\mathcal{H}),$ is c.p. trace-nonincreasing, and $\mathcal{M}(
\mathcal{X})$ is trace-preserving (t.p.) (see, e.g. \cite{BL-2}). Let $
M=\{M(A),A\in \mathcal{F}\}$ be the associated \emph{\ observable} i.e. a
probability operator-valued measure (POVM) on $(\mathcal{X},\mathcal{F})$
such that $M(A)=\mathcal{M}^{\ast }(A)[I],$ where $I$ \ is the identity
operator on $\mathcal{H}$. \emph{The probability distribution} of the
observable $M$ in the state $\rho \in \mathfrak{S}(\mathcal{H})$ is given by
the formula
\begin{equation*}
P_{\rho }(A)=\Tr \rho M(A),\quad A\in \mathcal{F}.
\end{equation*}
As shown in \cite{Oz} for each $\rho \in \mathfrak{S}(\mathcal{H})$ there is
a family of \emph{posterior states} $\left\{ \hat{\rho}(x)\right\} $ such
that
\begin{equation*}
\mathcal{M}(A)\left[ \rho \right] =\int_{A}\hat{\rho}(x)P_{\rho }(dx).
\end{equation*}
for any state $\rho \in \mathfrak{S}(\mathcal{H})$. The \emph{entropy
reduction} of the instrument is defined as
\begin{equation*}
ER(\rho ,\mathcal{M})=\mathrm{H}(\rho )-\int_{\mathcal{X}}\,\mathrm{H}(\hat{
\rho}(x))P_{\rho }(dx),
\end{equation*}
where $\mathrm{H}(\rho )=-\Tr \rho \log \rho ,$ provided $\mathrm{H}
(\rho )<\infty $ \cite{Oz}, \cite{Sh-ERQM}.

We will deal with the special class of instrument, the observables of which
have bounded \emph{operator-valued density}, more precisely,
\begin{equation}
M(A)=\int_{A}m(x)\,\mu (dx),\quad A\in \mathcal{F},  \label{den}
\end{equation}
where $\mu $ is a $\sigma $-finite measure on the $\sigma $-algebra $
\mathcal{F}$ such that $m(x)=V(x)^{\ast }V(x)$, where $V(x)$ is a weakly
measurable function with values in the algebra of bounded operators on $
\mathcal{H}$ satisfying
\begin{equation*}
\int_{\mathcal{X} }V({x})^{\ast }V({x})\mu (dx)=I,
\end{equation*}
and the integral weakly converges \cite{H1}. With any measurable
factorization of $m(x)$ one can associate an \emph{efficient instrument}
(see \cite{Oz}, \cite{Sh-ERQM})
\begin{equation*}
\mathcal{M}(A)\left[ \rho \right] =\int_{A}V({x}) \rho V({x})^{\ast }\mu
(dx).
\end{equation*}
Then the probability distribution $P_{\rho }(dx)$ has the density
\begin{equation}
p_{\rho }(x)=\Tr \rho V({x})^{\ast }V({x})  \label{opd}
\end{equation}
with respect to the measure $\mu ,$ while the family of posterior states is
\begin{equation}
\hat{\rho}(x)=p_{\rho }(x)^{-1}V(x)\rho V(x)^{\ast }  \label{post2}
\end{equation}

The entropy reduction of the efficient instrument is then
\begin{equation}
ER(\rho ,M)=\mathrm{H}(\rho )-\int_{\Omega }p_{\rho }(x)\,\mathrm{H}(\hat{
\rho}(x))\mu (dx),  \label{er-1}
\end{equation}
where the posterior states are given by (\ref{post2}).
In \cite{Oz} it was shown that the entropy reduction of an efficient
instrument is nonnegative. In \cite{Sh-ERQM} the entropy reduction was
related to the quantum mutual information of the instrument and hence it is a
concave, subadditive, lower semicontinuous function of $\rho .$

An essential observation made in \cite{h5}, \cite{H3} is that $\mathrm{H}(
\hat{\rho}(x))$ in the entropy reduction (\ref{er-1}) depend only on $m(${$x$
}$)$ (i.e. on the observable $M$) and not on the way of its measurable
factorization (i.e. the choice of an efficient instrument), which justifies
the notation $ER(\rho ,M).$

\section{\protect\bigskip Gaussian maximizers for the entropy reduction}\label{s2}

Our framework will be a bosonic system with $s$ modes described by the
canonical position-momentum operators $q=[q_{1},\dots
,q_{s}]^{t},\,p=[p_{1},\dots ,p_{s}]^{t}$ (see e.g. \cite{QSCI}, \cite{sera}
). It will be convenient to take the Schr\"{o}dinger (position)
representation space $\mathcal{H}=L^{2}(\mathbb{R}^{s})$ with the operators $
q_{j}=\xi _{j},\,p_{j}=i^{-1}\frac{\partial }{\partial \xi _{j}}.$ We denote
by
\begin{equation*}
D(x,y)=\mathrm{e}^{-ix^{t}y/2}\mathrm{e}^{iy^{t}q}\mathrm{e}
^{-ix^{t}p},\quad x,y\in \mathbb{R}^{s}
\end{equation*}
the unitary displacement operators.

We will be interested in approximate position measurement in $s$ modes. In
quantum optics this underlies the multi-mode noisy homodyning, as measurement of any quadrature
of the radiation field can be reduced to the measurement of position by a Bogoljubov transformation.
In this case $\mathcal{X}=\mathbb{R}^{s}$ and the POVM is
\begin{eqnarray}
M(d^{s}x) &=&\exp \left[ -\frac{1}{2}\left( q-x\right) ^{t}\beta ^{-1}\left(
q-x\right) \right] \frac{d^{s}x}{\sqrt{\left( 2\pi \right) ^{s}\det \beta }}
\notag \\
&=&D(x)\exp \left[ -\frac{1}{2}q^{t}\beta ^{-1}q\right] D(x)^{\ast }\frac{
d^{s}x}{\sqrt{\left( 2\pi \right) ^{s}\det \beta }} ,  \label{apprqs1}
\end{eqnarray}
where $\beta $ is a real positive definite covariance matrix of
the measurement noise, and $D(x)=D(x,0)=\exp \left( -ix^{t}p\right) $ is the
position displacement operator. Notice that POVM $M$ has the form (\ref{den}
) with the bounded operator-valued density $m(x)$, hence we are in the
situation where the formulas (\ref{post2}) and (\ref{er-1}) are applicable.

Denote
\begin{equation*}
V(x)=V^{\ast }(x)=\sqrt{m(x)}=\left[ \left( 2\pi \right) ^{s}\det \beta
\right] ^{-1/4}\exp \left[ -\frac{1}{4}\left( q-x\right) ^{t}\beta
^{-1}\left( q-x\right) \right] .
\end{equation*}
Let $\rho $ be a state and $\rho _{u,v}=D(u,v)\rho D(u,v)^{\ast }$ a
displaced state. Then
\begin{eqnarray*}
&& V(x)\rho _{u,v}V(x)=\\ &=&D(u,v)\left( D(u,v)^{\ast }V(x)D(u,v)\right) \rho
\left( D(u,v)^{\ast }V(x)\rho D(u,v)\right) D(u,v)^{\ast } \\
&=&D(u,v)V(x+u)\rho V(x+u)D(u,v)^{\ast }.
\end{eqnarray*}
It follows that
\begin{equation*}
p_{\rho _{u,v}}(x)=p_{\rho }(x+u),\quad \hat{\rho}_{u,v}(x)=D(u,v)\hat{\rho}
(x+u)D(u,v)^{\ast },
\end{equation*}
and
\begin{equation}
ER(\rho _{u,v},M)=ER(\rho ,M)  \label{zerom}
\end{equation}
by the unitary invariance of the entropy. Therefore in what follows we can
restrict to centered states $\rho$ ($\Tr \,\rho \,q_{j}=\mathrm{
\Tr }\,\rho \,p_{j}=0$).

Consider the quantum state $\rho _{\alpha }$ $\in \mathfrak{S}(\mathcal{H})$
which is centered Gaussian density operator in $\mathcal{H}$ with the
covariance $2s\times 2s-$matrix
\begin{equation}
\alpha =\left[
\begin{array}{cc}
\alpha _{qq} & \alpha _{qp} \\
\alpha _{pq} & \alpha _{pp}
\end{array}
\right] ,  \label{cm}
\end{equation}
where
\begin{equation*}
\alpha _{qq}=\alpha _{qq}^{t}=\left[ \Tr \,\rho _{\alpha
}q_{j}q_{k}\right] _{j,k=1,\dots s},\quad \quad \alpha _{pp}=\alpha
_{pp}^{t}=\left[ \Tr \,\rho _{\alpha }p_{j}p_{k}\right]
_{j,k=1,\dots s},
\end{equation*}
\begin{equation*}
\alpha _{qp}=\alpha _{pq}^{t}=\left[ \frac{1}{2}\Tr \,\rho
_{\alpha }\left( q_{j}p_{k}+p_{k}q_{j}\right) \right] _{j,k=1,\dots s}.
\end{equation*}
It is explicitly given by the kernel in the Schr\"{o}dinger representation
\begin{equation}
\langle \xi |\rho _{\alpha }|\xi ^{\prime }\rangle =\frac{1}{\sqrt{\left(
2\pi \right) ^{s}\det \alpha _{qq}}}\exp \left[ -\frac{1}{2}\theta \left(
\xi ,\xi ^{\prime }\right) \right] ,  \label{mm0}
\end{equation}
where
\begin{eqnarray}
\theta \left( \xi ,\xi ^{\prime }\right) &=&\frac{1}{4}\left( \xi ^{\prime
}+\xi \right) ^{t}\alpha _{qq}^{-1}\left( \xi ^{\prime }+\xi \right) +\left(
\xi ^{\prime }-\xi \right) ^{t}\left( \alpha _{pp}-\alpha _{pq}\alpha
_{qq}^{-1}\alpha _{qp}\right) \left( \xi ^{\prime }-\xi \right)  \notag \\
&+&i\left( \xi ^{\prime }+\xi \right) ^{t}\alpha _{qq}^{-1}\alpha
_{qp}\left( \xi ^{\prime }-\xi \right) ,  \label{theta}
\end{eqnarray}
see Appendix.

Let
\begin{equation*}
\rho _{\alpha }(x)=V(x)\rho _{\alpha }V(x),
\end{equation*}
then the probability distribution of the observable (\ref{apprqs1})\ has the
density
\begin{equation}
p_{\alpha }(x)=\Tr \,\rho _{\alpha }(x)  \label{pd1}
\end{equation}
and the posterior states are
\begin{equation}
\hat{\rho}_{\alpha }(x)=\rho _{\alpha }(x)/p_{\alpha }(x).  \label{post1}
\end{equation}

\begin{lemma}
\label{l1} \emph{The probability density (\ref{pd1}) is Gaussian:
\begin{equation}
p_{\alpha }(x)=\frac{1}{\sqrt{\left( 2\pi \right) ^{s}\det \left( \alpha
_{qq}+\beta \right) }}\exp \left[ -\frac{1}{2}x^{t}\left( \alpha _{qq}+\beta
\right) ^{-1}x\right] ,  \label{pd2}
\end{equation}
and the posterior states (\ref{post1}) are Gaussian
\begin{equation}
\hat{\rho}_{\alpha }(x)=D(K_{q}x,K_{p}x)\rho _{\tilde{\alpha}
}D(K_{q}x,K_{p}x)^{\ast },  \label{mean}
\end{equation}
where
\begin{equation*}
K_{q}=\alpha _{qq}\left( \alpha _{qq}+\beta \right)
^{-1},\quad K_{p}=\alpha _{pq}\left( \alpha _{qq}+\beta \right) ^{-1} ,
\end{equation*}
 and $\rho_{\hat{\alpha}}$ is centered Gaussian state with the covariance matrix
\begin{equation*}
\hat{\alpha}=\left[
\begin{array}{cc}
\hat{\alpha}_{qq} & \hat{\alpha}_{qp} \\
\hat{\alpha}_{pq} & \hat{\alpha}_{pp}
\end{array}
\right]
\end{equation*}
with}
\begin{eqnarray}
\hat{\alpha}_{qq} &=&\left( \alpha _{qq}^{-1}+\beta ^{-1}\right) ^{-1},\quad
\notag \\
\hat{\alpha}_{pp} &=&\left( \alpha _{pp}-\alpha _{pq}(\alpha _{qq}+\beta
)^{-1}\alpha _{qp}\right) +\frac{1}{4}\beta ^{-1},  \label{corr} \\
\hat{\alpha}_{qp} &=&\left( \alpha _{qq}^{-1}+\beta ^{-1}\right) ^{-1}\alpha
_{qq}^{-1}\alpha _{qp}.  \notag
\end{eqnarray}
\end{lemma}

\textit{Proof}. From the quantum characteristic function of $\rho _{\alpha }$
we have
\begin{equation*}
\Tr \,\rho _{\alpha }\exp \left( iu^{t}q\right) =\exp \left(
-\frac{1}{2}u^{t}\alpha _{qq} u\right) ,
\end{equation*}
hence the diagonal value of the kernel of $\rho _{\alpha }$ in the position
representation is the inverse Fourier transform
\begin{equation*}
\langle \xi |\rho _{\alpha }|\xi \rangle =\frac{1}{\sqrt{\left( 2\pi \right)
^{s}\det \alpha _{qq}}}\exp \left[ -\frac{1}{2}\xi ^{t}\alpha _{qq}^{-1}\xi
\right] ,
\end{equation*}
and
\begin{equation*}
p_{\alpha }(x)=\Tr \,\rho _{\alpha }(x)=\int \langle \xi
|\rho _{\alpha }|\xi \rangle \exp \left[ -\frac{1}{2}\left( \xi -x\right)
^{t}\beta ^{-1}\left( \xi -x\right) \right] \frac{d^{s}\xi }{\sqrt{\left(
2\pi \right) ^{s}\det \beta }}
\end{equation*}
is the convolution of the two Gaussian probability densities giving the
right-hand side of (\ref{pd2}).

The posterior state (\ref{post1}) is Gaussian since it has the Gaussian
kernel in the Schr\"{o}dinger representation
\begin{eqnarray*}
\langle \xi |\hat{\rho}_{\alpha }(x)|\xi ^{\prime }\rangle &=&\frac{1}{\sqrt{
\left( 2\pi \right) ^{s}\det \beta }} \\
&\times &\exp \left[ -\frac{1}{4}\left( \xi -x\right) ^{t}\beta ^{-1}\left(
\xi -x\right) -\frac{1}{4}\left( \xi ^{\prime }-x\right) ^{t}\beta
^{-1}\left( \xi ^{\prime }-x\right) \right] \\
&\times &\langle \xi |\rho _{\alpha }|\xi ^{\prime }\rangle /p_{\alpha }(x),
\end{eqnarray*}
where $\langle \xi |\rho _{\alpha }|\xi ^{\prime }\rangle $ is given by (\ref
{mm0}) and $p_{\alpha }(x)$ -- by (\ref{pd2}). Substituting (\ref{mm0}
) and (\ref{pd2}), we obtain
\begin{equation*}
\langle \xi |\hat{\rho}_{\alpha }(x)|\xi ^{\prime }\rangle =\sqrt{\frac{\det
\left( \alpha _{qq}+\beta \right) }{\left( 2\pi \right) ^{s}\det \alpha
_{qq}\det \beta }}\exp \left[ \hat{\mu}\left( x;\xi ,\xi ^{\prime }\right) -
\frac{1}{2}\hat{\theta}\left( \xi ,\xi ^{\prime }\right) \right] ,
\end{equation*}
where
\begin{equation*}
\hat{\mu}\left( x;\xi ,\xi ^{\prime }\right) =\frac{1}{2}x^{t}\beta ^{-1}\xi
+\frac{1}{2}x^{t}\beta ^{-1}\xi ^{\prime }-\frac{1}{2}x^{t}\beta ^{-1}x+
\frac{1}{2}x^{t}\left( \alpha _{qq}+\beta \right) ^{-1}x,
\end{equation*}
\begin{equation*}
\hat{\theta}\left( \xi ,\xi ^{\prime }\right) =\frac{1}{2}\xi ^{t}\beta
^{-1}\xi +\frac{1}{2}\left( \xi ^{\prime }\right) ^{t}\beta ^{-1}\xi
^{\prime }+\theta \left( \xi ,\xi ^{\prime }\right)
\end{equation*}
and $\theta \left( \xi ,\xi ^{\prime }\right) $ is given by (\ref{theta}).
Let us first consider the terms independent of $\ x.$ We have
\begin{eqnarray*}
\hat{\theta}\left( \xi ,\xi ^{\prime }\right) &=&\frac{1}{4}\left( \xi
^{\prime }+\xi \right) ^{t}\beta ^{-1}\left( \xi ^{\prime }+\xi \right) +
\frac{1}{4}\left( \xi ^{\prime }-\xi \right) ^{t}\beta ^{-1}\left( \xi
^{\prime }-\xi \right) +\theta \left( \xi ,\xi ^{\prime }\right) \\
&=&\frac{1}{4}\left( \xi ^{\prime }+\xi \right) ^{t}\hat{\alpha}
_{qq}^{-1}\left( \xi ^{\prime }+\xi \right) +\left( \xi ^{\prime }-\xi
\right) ^{t}(\hat{\alpha}_{pp}-\hat{\alpha}_{pq}\hat{\alpha}_{qq}^{-1}\hat{
\alpha}_{qp})\left( \xi ^{\prime }-\xi \right) \\
&+&i\left( \xi ^{\prime }+\xi \right) ^{t}\hat{\alpha}_{qq}^{-1}\hat{\alpha}
_{qp}\left( \xi ^{\prime }-\xi \right) ,
\end{eqnarray*}
where
\begin{eqnarray*}
\hat{\alpha}_{qq}^{-1} &=&\alpha _{qq}^{-1}+\beta ^{-1}, \\
\hat{\alpha}_{qq}^{-1}\hat{\alpha}_{qp} &=&\alpha _{qq}^{-1}\alpha _{qp}, \\
\hat{\alpha}_{pp}-\hat{\alpha}_{pq}\hat{\alpha}_{qq}^{-1}\hat{\alpha}_{qp}
&=&\left( \alpha _{pp}-\alpha _{pq}\alpha _{qq}^{-1}\alpha _{qp}\right) +
\frac{1}{4}\beta ^{-1},
\end{eqnarray*}
resulting in the elements of the posterior covariance matrix (\ref{corr}).

To find the posterior mean values $\hat{m}_{q},\hat{m}_{p}$, we have
\begin{eqnarray*}
\hat{\mu}\left( x;\xi ,\xi ^{\prime }\right) &=&\frac{1}{2}x^{t}\beta
^{-1}\left( \xi +\xi ^{\prime }\right) -\frac{1}{2}x^{t}\beta ^{-1}\left(
\alpha _{qq}^{-1}+\beta ^{-1}\right) ^{-1}\beta ^{-1}x \\
&=&\frac{1}{2}\left( K_{q}x\right) ^{t}\hat{\alpha}_{qq}^{-1}\left( \xi +\xi
^{\prime }\right) -\frac{1}{2}\left( K_{q}x\right) ^{t}\hat{\alpha}
_{qq}^{-1}\left( K_{q}x\right) ,
\end{eqnarray*}
where $K_{q}=\alpha _{qq}\left( \alpha _{qq}+\beta \right) ^{-1}.$ Comparing
with the corresponding terms under the exponent in (\ref{mm}) (see Appendix), we should
have for the posterior mean values$:$
\begin{equation*}
\hat{m}_{q}=K_{q}x,\quad \hat{m}_{p}=\hat{\alpha}_{pq}\hat{\alpha}_{qq}^{-1}
\hat{m}_{q},
\end{equation*}
where the second relation follows from the fact that the second term in (\ref
{mm}) is zero in our case. Thus $\hat{m}_{p}=K_{p}x,$ where $K_{p}=\alpha
_{pq}\left( \alpha _{qq}+\beta \right) ^{-1}.$ $\square $

Let $\mathfrak{S}(\alpha )$ be the set of all (not necessarily centered)
states $\rho $ with the covariance matrix $\alpha .$ We will study the
following entropic characteristic of the Gaussian measurement $M$
\begin{equation}
ER(M;\alpha )=\sup_{\rho \in \mathfrak{S}(\alpha )}ER(\rho ,M),
\label{cchi1}
\end{equation}
which is strictly related to the entanglement-assisted capacity of $M$ (see
sec. \ref{s3}). Due to (\ref{zerom}), in (\ref{cchi1}) we can restrict to
centered states $\rho $ for which $\alpha $ coincides with the matrix of
second moments. We denote $$g(x)=(x+1)\log (x+1)-x\log x .$$
\begin{theorem}
\label{t3} \emph{The supremum in (\ref{cchi1}) is attained on the Gaussian
state $\rho _{\alpha }$ and it is equal to
\begin{equation}
ER(M;\alpha )=\frac{1}{2}\left[ \mathrm{Sp}\,g\left( \mathrm{abs}(\Delta
^{-1}\alpha )-\frac{I_{2s}}{2}\right) -\mathrm{Sp}\,g\left( \mathrm{abs}
(\Delta ^{-1}\hat{\alpha})-\frac{I_{2s}}{2}\right) \right] ,  \label{cchi2}
\end{equation}
where $\hat{\alpha}$ is given by (\ref{corr}), and }$\mathrm{abs}(\Delta
^{-1}\alpha )$ \emph{denotes the matrix with eigenvalues equal to modulus of
eigenvalues of } $\Delta ^{-1}\alpha $ \ \emph{and with the same eigenvectors
$.$ }
\end{theorem}

\emph{Proof.} Let $\rho $ be a centered density operator from $\mathfrak{S}
(\alpha )$, then denote $\rho (x)=V(x)\,\rho \,V(x)$ and $p(x)=\Tr
\rho (x).$ Also introduce the channel $\mathcal{E}\,[\rho ]=\,\left\{
V(x)\,\rho \,V(x)\right\} $ with quantum input and hybrid \textit{
classical-quantum} (cq)-output. For the Gaussian state  $\rho _{\alpha }$ we
have
\begin{eqnarray}
&&ER(\rho _{\alpha },M)-ER(\rho ,M)  \label{difer} \\
&=&\mathrm{H}(\rho \Vert \rho _{\alpha })+\Tr (\rho -\rho _{\alpha
})\log \rho_{\alpha }   \notag \\
&-&\mathrm{H}_{cq}(\mathcal{E}\,[\rho ]\Vert \mathcal{E}\,[\rho _{\alpha
}])\,\,+\mathrm{H}_{c}(p\Vert p_{\alpha })  \notag \\
&+&\int \Tr (\rho _{\alpha }(x)-\rho (x))\log \,\hat{\rho}_{\alpha
}(x)d^{s}x.  \notag
\end{eqnarray}
Here
\begin{equation*}
\mathrm{H}_{c}(p\Vert p_{\alpha })=\int p(x)\log \left( \frac{p(x)}{
p_{\alpha }(x)}\right) d^{s}x
\end{equation*}
is the classical relative entropy between $p$, $p_{\alpha }$ and
\begin{equation*}
\mathrm{H}_{cq}(\mathcal{E}\,[\rho ]\Vert \mathcal{E}\,[\rho _{\alpha
}])=\int \Tr \,\rho (x)\,\left( \log \rho (x)-\log \rho
_{\alpha }(x)\right) d^{s}x,
\end{equation*}
is the relative entropy of the cq-states (see Eq. (3) in \cite{BL-2}).

Monotonicity of the relative entropy for cq-states (\cite{BL-2}, theorem 1)
then implies
\begin{equation*}
\mathrm{H}_{cq}(\mathcal{E}\,[\rho ]\Vert \mathcal{E}\,[\rho _{\alpha
}])\leq \mathrm{H}(\rho \Vert \rho _{\alpha }),
\end{equation*}
hence we have for the first three terms in the right-hand side of (\ref
{difer})
\begin{equation}
\mathrm{H}(\rho \Vert \rho _{\alpha })-\mathrm{H}_{cq}(\mathcal{E}\,[\rho
]\Vert \mathcal{E}\,[\rho _{\alpha }])+\mathrm{H}_{c}(p\Vert p_{\alpha
})\geq 0.  \label{first}
\end{equation}

Without loss of generality we can assume that $\rho _{\alpha }$ is
non-degenerate so that $\log \rho _{\alpha }$ exists and it is a polynomial
in $q,p$ of the second order$.$ This follows from the exponential form of
the density operator (theorem 12.23 in \cite{QSCI}). Since the first and
second moments of the states $\rho $ and $\rho _{\alpha }$ coincide, we have
\begin{equation}
\Tr (\rho -\rho _{\alpha })\log \rho _{\alpha }=0.  \label{second}
\end{equation}
It remains to show that also
\begin{equation}
\int \Tr (\rho _{\alpha }(x)-\rho (x))\log \,\hat{\rho}_{\alpha
}(x)\,d^{s}x=0.  \label{res}
\end{equation}

Substituting the posterior state (\ref{mean}) into the right-hand side of (
\ref{res}), we obtain
\begin{eqnarray}
\int \Tr (\rho _{\alpha }(x)-\rho (x))\log \hat{\rho}_{\alpha
}(x)d^{s}x &=&\Tr \,\Phi _{M}\,[\rho _{\alpha }-\rho ]\,\log \rho _{
\hat{\alpha}}  \label{res3} \\
&=&\Tr \,\,(\rho _{\alpha }-\rho )\,\Phi _{M}^{\ast }[\log \rho _{
\hat{\alpha}}],  \label{res3a}
\end{eqnarray}
where we have introduced the channel
\begin{equation}
\Phi _{M}[\sigma ]=\int D(K_q x, K_p x)^{\ast }V(x)\,\sigma \,V(x)D(K_q x, K_p x)d^{s}x.
\label{M-add}
\end{equation}
The channel $\Phi _{M}$ is Gaussian. First, it is a c.p.t.p. map. Complete positivity is apparent from the structure of the
map (\ref{M-add}). Trace preservation follows from
\begin{equation*}
\Phi _{M}^{\ast }[I]=\int V(x)^{2}\,d^{s}x=\int M(\,d^{s}x)=I.
\end{equation*}
A routine calculation shows that if $\sigma $ is a Gaussian state then $\Phi
_{M}[\sigma ]$ is again Gaussian. Then by result of \cite{dpgh}, $\Phi _{M}$
is a Gaussian channel.

Since $\rho _{\hat{\alpha}}$ is Gaussian state, $\log \rho _{\hat{\alpha}}$
is a polynomial in $q,p$ of the second order. The dual Gaussian channel $
\Phi _{M}^{\ast }$ takes it into another polynomial in $q,p$ of the second
order (see sec. 12.4.2 of \cite{QSCI}). Since the first and second moments of the states $\rho $ and $\rho
_{\alpha }$ coincide, one obtains (\ref{res}). Then (\ref{difer}), (\ref
{first}), (\ref{second}) and (\ref{res}) imply
\begin{equation*}
ER(\rho _{\alpha },M)-ER(\rho ,M)\geq 0
\end{equation*}
for arbitrary $\rho \in \mathfrak{S}(\alpha )$ proving that the supremum (
\ref{cchi1}) is attained on the Gaussian state $\rho _{\alpha }$.

Taking into account the formula (\ref{mean}) for the posterior states, \ we
obtain
\begin{eqnarray*}
ER(M;\alpha ) &=&ER(\rho _{\alpha },M) \\
&=&\mathrm{H}(\rho _{\alpha })-\int_{\Omega }p_{\alpha }(x)\,\mathrm{H}(\hat{
\rho}_{\alpha }(x))d^{s}x \\
&=&\mathrm{H}(\rho _{\alpha })-\int p_{\alpha }(x)\mathrm{H}(D(K_q x, K_p x)^{\ast
}\rho _{\hat{\alpha}}D(K_q x, K_p x))d^{s}x \\
&=&\mathrm{H}(\rho _{\alpha })-\mathrm{H}(\rho _{\hat{\alpha}}).
\end{eqnarray*}
Then (\ref{cchi2}) follows by applying the formula for the entropy of a
Gaussian state
\begin{equation}
\mathrm{H}(\rho _{\alpha })=\frac{1}{2}\mathrm{Sp}\,g\left( \mathrm{abs}
(\Delta ^{-1}\alpha )-\frac{I_{2s}}{2}\right) ,  \label{entro}
\end{equation}
given in \cite{QSCI}, Eq. (12.110). $\square $

\section{The entanglement-assisted capacity}
\label{s3}

Consider a quadratic Hamiltonian of the form
\begin{equation*}
H=\sum_{j,k=1}^{s}\left( q^{t}\epsilon _{qq}q+q^{t}\epsilon
_{qp}p+p^{t}\epsilon _{pq}q+p^{t}\epsilon _{pp}p\right) ,
\end{equation*}
where
\begin{equation*}
\epsilon =\left[
\begin{array}{cc}
\epsilon _{qq} & \epsilon _{qp} \\
\epsilon _{pq} & \epsilon _{pp}
\end{array}
\right]
\end{equation*}
is a real symmetric positive definite energy matrix. The importance of the
quantity (\ref{cchi1}) is that it underlies the energy-constrained
entanglement-assisted capacity, given by the formula (see e.g. \cite{H3})
\begin{equation*}
C_{ea}(M,H,E)=\max_{\rho :~\Tr \rho H\leq E}ER(\rho ,M),
\end{equation*}
where $E$ is the energy level. For any state $\rho \in \mathfrak{S}(\alpha )$
the mean energy\footnote{
We denote $\mathrm{Sp}$ trace of matrices as distinct from the trace of
operators in $\mathcal{H}$.}
\begin{equation*}
\Tr \rho H=\mathrm{Sp}\,\epsilon \alpha _{2}\geq \mathrm{Sp}
\,\epsilon \alpha ,
\end{equation*}
where $\alpha _{2}$ is the matrix of second moments of $\rho ,$ with the
equality attained for centered states, therefore
\begin{equation}
C_{ea}(M;H,E)=\max_{\alpha :\mathrm{Sp}\,\epsilon \alpha \leq E}ER(M;\alpha
).  \label{ceag}
\end{equation}
It is easy to check that
\begin{equation*}
\sup_{\alpha }ER(M;\alpha )=+\infty ,
\end{equation*}
therefore theorem 1 of \cite{H3} applies showing that the maximum in (\ref{ceag}) is
attained on $\alpha $ satisfying $\mathrm{Sp}\,\epsilon \alpha =E.$

In the case of the oscillator-type Hamiltonian
\begin{equation*}
H=\sum_{j,k=1}^{s}\left( q^{t}\epsilon _{qq}q+p^{t}\epsilon _{pp}p\right) ,
\end{equation*}
the energy constraint is
\begin{equation}
\mathrm{Sp}\,\epsilon _{qq}\alpha _{qq}+\mathrm{Sp}\,\epsilon _{pp}\alpha
_{pp}\leq E.  \label{ed}
\end{equation}
Then the maximization in (\ref{ceag}) can be taken over only block-diagonal
covariance matrices $\alpha \ ($satisying (\ref{ed}) with equality). To see
this consider the transformation $T:q_j\rightarrow q_j,\,p_j\rightarrow -p_j$ which
changes the sign of the commutators between $q_j$ and $p_j$ and which is
implemented by anti-unitary operator of complex conjugation in the Schr\"odinger representation. If $\rho $
is a centered state with the covariance matrix \footnote{
We denote by $I_{s}$ the unit $s\times s-$matrix.}
\begin{equation*}
\alpha =\left[
\begin{array}{cc}
\alpha _{qq} & \alpha _{qp} \\
\alpha _{pq} & \alpha _{pp}
\end{array}
\right] \geq \pm \frac{i}{2}\left[
\begin{array}{cc}
0 & I_{s} \\
I_{s} & 0
\end{array}
\right] ,
\end{equation*}
then $T[\rho ]$ has the covariance matrix
\begin{equation*}
\tilde{\alpha} =\left[
\begin{array}{cc}
\alpha _{qq} & -\alpha _{qp} \\
-\alpha _{pq} & \alpha _{pp}
\end{array}
\right] \geq \mp \frac{i}{2}\left[
\begin{array}{cc}
0 & I_{s} \\
I_{s} & 0
\end{array}
\right] ,
\end{equation*}
because it is the covariance matrix for $q_j,\,-p_j.$ The mixture $\frac{1}{2}
\left( \rho +T[\rho ]\right) $ has the covariance matrix
\begin{equation*}
\left[
\begin{array}{cc}
\alpha _{qq} & 0 \\
0 & \alpha _{pp}
\end{array}
\right] =\frac{1}{2}\left( \alpha +\tilde{\alpha}\right) \geq \pm \frac{i}{2}
\left[
\begin{array}{cc}
0 & I_{s} \\
I_{s} & 0
\end{array}
\right] ,
\end{equation*}
while
\begin{equation*}
ER(\frac{1}{2}\left( \rho +T[\rho ]\right) ,M)\geq \frac{1}{2}\left[ ER(\rho
,M)+ER(T[\rho ],M)\right] =ER(\rho ,M)
\end{equation*}
by the concavity of the entropy reduction and its invariance under the
transformation $T$.

\section{The case of one mode}
\label{s4}

\begin{figure}[t]
\center{\includegraphics{C_ea}}
\caption{(color online) Plot of the assisted capacity $C_{ea}(E)$ for $\beta=0, 1, 5, 10$.}
\label{Fig1}
\end{figure}

Consider the approximate measurement of position for one bosonic mode $q,p$.
The corresponding POVM is
\begin{equation}
M(dx)=\exp \left[ -\frac{\left( q-x\right) ^{2}}{2\beta }\right] \frac{dx}{
\sqrt{2\pi \beta }}=D(x)\mathrm{e}^{-q^{2}/2\beta }D(x)^{\ast }\frac{dx}{
\sqrt{2\pi \beta }},  \label{apprq}
\end{equation}
where $\beta >0.$ Probability distribution for the Gaussian state $\rho
_{\alpha }$
\begin{equation}
\Tr \,\rho _{\alpha }M(dx)=\exp \left[ -\frac{x^{2}}{2\left(
\alpha _{qq}+\beta \right) }\right] \frac{dx}{\sqrt{2\pi \left( \alpha
_{qq}+\beta \right) }}=p_{\alpha }(x)\,dx,  \label{pd}
\end{equation}
with the covariance $2\times 2-$matrix (\ref{cm}), where $\alpha
_{qq},\alpha _{qp},\alpha _{pp}$ are real numbers satisfying
\begin{equation}
\alpha _{qq}\alpha _{pp}-\alpha _{qp}^{2}\geq 1/4.  \label{ur}
\end{equation}
The formula (\ref{entro}) for the entropy amounts to (see Example 12.25 in
\cite{QSCI})
\begin{equation*}
\mathrm{H}(\rho _{\alpha })=g\left( \sqrt{\alpha _{qq}\alpha _{pp}-\alpha
_{qp}^{2}}-\frac{1}{2}\right) .
\end{equation*}

According to eq. (\ref{corr}) posterior states have the covariance $2\times 2$-matrix $
\hat{\alpha}$ with the elements
\begin{equation*}
\hat{\alpha}_{qq}=\frac{\alpha _{qq}\beta }{\alpha _{qq}+\beta },\quad \hat{
\alpha}_{pp}=\left( \alpha _{pp}-\frac{\alpha _{qp}^{2}}{\alpha _{qq}+\beta }
\right) +1/\left( 4\beta \right) ,\quad \hat{\alpha}_{qp}=\frac{\alpha
_{qp}\beta }{\alpha _{qq}+\beta }.
\end{equation*}
Then theorem \ref{t3} gives
\begin{eqnarray}
ER(M;\alpha ) &=&g\left( \sqrt{\alpha _{qq}\alpha _{pp}-\alpha _{qp}^{2}}-
\frac{1}{2}\right) -g\left( \sqrt{\hat{\alpha}_{qq}\hat{\alpha}_{pp}-\hat{
\alpha}_{qp}^{2}}-\frac{1}{2}\right)  \notag \\
&=&g\left( \sqrt{\alpha _{qq}\alpha _{pp}-\alpha _{qp}^{2}}-\frac{1}{2}
\right)  \notag \\
&-&g\left( \sqrt{\frac{\alpha _{qq}(\beta \alpha _{pp}+1/4)-\beta \alpha
_{qp}^{2}}{\alpha _{qq}+\beta }}-\frac{1}{2}\right) .  \label{erm}
\end{eqnarray}

For the oscillator Hamiltonian $H=\frac{1}{2}\left( q^{2}+p^{2}\right) ,$
the energy constraint has the form $\alpha _{qq}+\alpha _{pp}\leq 2E$ and
the entanglement-assisted capacity (\ref{ceag}) is equal to
\begin{equation}
C_{ea}(M;H,E)=  \label{ceab}
\end{equation}
\begin{equation*}
=\max \left[ g\left( \sqrt{\alpha
_{qq}\,\alpha _{pp}}-\frac{1}{2}\right) -g\left( \sqrt{\frac{\alpha
_{qq}(\beta \alpha _{pp}+1/4)}{\alpha _{qq}+\beta }}-\frac{1}{2}\right)
\right] ,
\end{equation*}
where the maximum over positive $\alpha _{qq},\alpha _{pp}$ satisfying $
\alpha _{qq}+\alpha _{pp}= 2E,$ $\alpha _{qq}\alpha _{pp}\geq 1/4$, and $\alpha_{qp}$ is taken to be $0$ (see Fig. \ref{Fig1}).

Let us now consider the limit $\beta \rightarrow 0$ corresponding to the
exact measurement of position, where
\begin{equation}
M_{ex}(dx)=|x\rangle \langle x|dx  \label{exactq}
\end{equation}
is the spectral measure of the operator $q$. Notice that strictly speaking
it is not of the form (\ref{den}), i.e. it does not have a bounded
operator-valued density. In the limit $\beta \rightarrow 0$ we have
\begin{equation}
ER(M;\alpha )\rightarrow g\left( \sqrt{\alpha _{qq}\,\alpha _{pp}}-\frac{1}{2
}\right) .  \label{ER2}
\end{equation}
We obtain the entanglement-assisted capacity
\begin{equation}
C_{ea}\left( M_{ex},H,E\right) =\max_{}g\left( \sqrt{\alpha _{qq}\,\alpha
_{pp}}-\frac{1}{2}\right) ,  \label{cerm}
\end{equation}
where the maximum over positive $\alpha _{qq},\alpha _{pp}$ satisfying $
\alpha _{qq}+\alpha _{pp}= 2E,$ $\alpha _{qq}\alpha _{pp}\geq 1/4.$ By
the concavity of $g(x)$, this maximum is attained for $\alpha _{qq}=\alpha
_{pp}=E$ and is equal to
\begin{equation}
C_{ea}\left( M_{ex},H,E\right) =g(E-1/2).  \label{ceaex}
\end{equation}

In \cite{H2} the POVM (\ref{exactq}) was considered as associated with the
instrument
\begin{equation*}
\mathcal{M}_{ex}[\rho ](dx)=|e\rangle \langle x|\rho |x\rangle \langle
e|\,dx,
\end{equation*}
where $|e\rangle $ is an arbitrary unit vector. Notice that here $
V(x)=|e\rangle \langle x|$ is an unbounded and even non-closable operator.
Then the posterior state $|e\rangle \langle e|$ is pure and has zero
entropy, resulting in $ER(M_{ex};\rho )=\mathrm{H}(\rho ).$ Thus
\begin{equation}
ER(M_{ex};\alpha )=\mathrm{H}(\rho _{\alpha })=g\left( \sqrt{\alpha
_{qq}\,\alpha _{pp}}-\frac{1}{2}\right) ,  \label{ER1}
\end{equation}
which agrees with (\ref{ER2}) and leads again to the capacity (\ref{ceaex}).

Let us consider another realization of the exact measurement -- the squeezed
joint measurement of $q,p,$ described by POVM
\begin{equation*}
M_{sq}(dx\,dy)=|x,y\rangle _{\beta }\langle x,y|\frac{dx\,dy}{2\pi }
=D(x,y)\rho _{sq,\beta }\,D(x,y)^{\ast }\frac{dx\,dy}{2\pi },
\end{equation*}
where $\rho _{sq,\beta }$ is the squeezed vacuum with the covariance matrix $
\left[
\begin{array}{cc}
\beta  & 0 \\
0 & 1/\left( 4\beta \right)
\end{array}
\right] .$ Then
\begin{equation}
\int_{y}M_{sq}(dx\,dy)=M(dx).  \label{ave}
\end{equation}
To see this it is sufficient to show that
\begin{equation}
\int D(x,y)\rho _{sq,\beta }D(x,y)^{\ast }\frac{\,dy}{2\pi }=\frac{1}{\sqrt{
2\pi \beta }}\exp \left[ -\frac{\left( q-x\right) ^{2}}{2\beta }\right] .
\label{int}
\end{equation}
Indeed, in the position representation
\begin{eqnarray*}
&&\langle \xi |\int D(x,y)\rho _{sq,\beta }D(x,y)^{\ast }\frac{\,dy}{2\pi }
|\xi ^{\prime }\rangle  \\
&=&\delta (\xi -\xi ^{\prime })\langle \xi -x|\rho _{sq,\beta }|\xi
-x\rangle  \\
&=&\langle \xi |\xi ^{\prime }\rangle \langle \xi -x|\rho _{sq,\beta }|\xi
-x\rangle .
\end{eqnarray*}
The classical characteristic function is
\begin{equation*}
\int \langle \xi |\rho _{sq,\beta }|\xi \rangle \mathrm{e}^{i\xi \lambda
}d\xi =\Tr \,\rho _{sq,\beta }\mathrm{e}^{iq\lambda }=\exp \left( -
\frac{1}{2}\beta \lambda ^{2}\right),
\end{equation*}
whence
\begin{equation*}
\langle \xi -x|\rho _{sq,\beta }|\xi -x\rangle =\frac{1}{\sqrt{2\pi \beta }}
\exp \left[ -\frac{\left( \xi -x\right) ^{2}}{2\beta }\right] .
\end{equation*}
which proves (\ref{int}).

Similarly to (\ref{ER1}) we have
\begin{equation*}
ER(M_{sq};\alpha )=\mathrm{H}(\rho _{\alpha })=g\left( \sqrt{\alpha
_{qq}\,\alpha _{pp}}-\frac{1}{2}\right) .
\end{equation*}
The difference between this expression and (\ref{erm}) reflects the loss of
information due to the averaging (\ref{ave}) over the values of the momentum.

\textbf{Acknowledgment}. The work was supported by the grant of Russian
Science Foundation (project No 19-11-00086).

\section{Appendix}

\emph{The kernel of a Gaussian density operator $\rho _{m,\alpha }$ with the mean $
m_{q},m_{p}$ and the covariance matrix (\ref{cm}) in the Schr\"{o}dinger
representation has the form}
\begin{equation}
\langle \xi |\rho _{\alpha }|\xi ^{\prime }\rangle =\frac{1}{\sqrt{\left(
2\pi \right) ^{s}\det \alpha _{qq}}}\exp \left[ \mu \left( m;\xi ,\xi
^{\prime }\right) -\frac{1}{2}\theta \left( \xi ,\xi ^{\prime }\right)
\right] ,  \label{mm}
\end{equation}
\emph{where}
\begin{equation}
\mu \left( m;\xi ,\xi ^{\prime }\right) =\frac{1}{2}m_{q}^{t}\alpha
_{qq}^{-1}\left( \xi ^{\prime }+\xi \right) -i\left( m_{p}-\alpha
_{pq}\alpha _{qq}^{-1}m_{q}\right) ^{t}\left( \xi ^{\prime }-\xi \right) -
\frac{1}{2}m_{q}^{t}\alpha _{qq}^{-1}m_{q},  \label{mu}
\end{equation}
\begin{eqnarray}
\theta \left( \xi ,\xi ^{\prime }\right)  &=&\frac{1}{4}\left( \xi ^{\prime
}+\xi \right) ^{t}\alpha _{qq}^{-1}\left( \xi ^{\prime }+\xi \right) +\left(
\xi ^{\prime }-\xi \right) ^{t}\left( \alpha _{pp}-\alpha _{pq}\alpha
_{qq}^{-1}\alpha _{qp}\right) \left( \xi ^{\prime }-\xi \right)   \notag \\
&+&i\left( \xi ^{\prime }+\xi \right) ^{t}\alpha _{qq}^{-1}\alpha
_{qp}\left( \xi ^{\prime }-\xi \right) .  \label{teta}
\end{eqnarray}

\emph{Proof.} The quantum characteristic function of $\rho _{m,\alpha }$ is
\begin{equation*}
\varphi (x,y)=\Tr \rho _{m,\alpha }W(x,y)=\exp [im(x,y)-\frac{1}{2}
\alpha (x,y)],
\end{equation*}
where
\begin{equation*}
\begin{aligned} m(x,y) &= m_q x + m_p y ,\\ \alpha(x,y) &= x^t \alpha_{qq} x
+ x^t \alpha_{qp} y + y^t \alpha_{pq} x + y^t \alpha_{pp} y \end{aligned}
\end{equation*}
and $W(x,y)=e^{ix^{t}y/2}e^{ix^{t}q}e^{iy^{t}p}$ is the Weyl operator. The
kernel of the Weyl operator is
\begin{equation*}
\bra{\xi}W(x,y)\ket{\xi'}=\exp \left[ {i}\left( \frac{\xi +\xi ^{\prime }}{2}
\right) ^{t}x\right] \delta (y-(\xi ^{\prime }-\xi ))=\exp \left(
iu^{t}x\right) \delta (y-v).
\end{equation*}
Here we introduced the variables $u=\frac{\xi +\xi ^{\prime }}{2}$, $v=\xi ^{\prime }-\xi $.
Using the inversion formula quantum Fourier transform, we  readily compute
the kernel of the Gaussian state:
\begin{multline*}
\bra{\xi}\rho _{m,\alpha }\ket{\xi'}=\int \frac{d^{s}xd^{s}y}{(2\pi )^{s}}
\varphi (x,y)\bra{\xi}W(-x,-y)\ket{\xi'}= \\
=\int \frac{d^{s}x}{(2\pi )^{s}}\exp \left[ -iu^{t}x+im(x,-v)-\frac{1}{2}
\alpha (x,-v)\right] ,
\end{multline*}
the expression under the exponent expands as
\begin{equation*}
-im_{p}^{t}v-\frac{1}{2}v^{t}\alpha _{pp}v-i(u^{t}-m_{q}^{t}+iv^{t}\alpha
_{pq})x-\frac{1}{2}x^{t}\alpha _{qq}x.
\end{equation*}
Taking the $s$-dimensional Gaussian integral yields
\begin{multline*}
\bra{\xi}\rho _{\alpha }\ket{\xi'}=\frac{e^{-im_{p}^{t}v-\frac{1}{2}
v^{t}\alpha _{pp}v}}{\sqrt{(2\pi )^{s}\det {\alpha _{qq}}}}\exp \left[ -
\frac{1}{2}(u-m_{q}+i\alpha _{qp}v)^{t}\alpha _{qq}^{-1}(u-m_{q}+i\alpha
_{qp}v)\right] = \\
=\frac{1}{\sqrt{(2\pi )^{s}\det {\alpha _{qq}}}}\exp \left[ \mu (m;\xi ,\xi
^{\prime })-\frac{1}{2}\theta (\xi ,\xi ^{\prime })\right] ,
\end{multline*}
where $\mu (m;\xi ,\xi ^{\prime })$ and  $\theta (\xi ,\xi ^{\prime })$ are
given by (\ref{mu}), (\ref{teta}). $\square $


TITLE: Radiofrequency acceleration of ions in a collider
QUESTION [0 upvotes]: Reading about the radiofrequency acceleration in circular colliders (https://home.cern/about/engineering/radiofrequency-cavities) made me wonder, this is how it is performed for particles going in one direction, but at the same time the particles going on the other direction need to be accelerated. Are the beams synchronized in order for one of them to be inside the cylindrical electrodes while the one running in the opposite direction is crossing the cavity and being accelerated?
What about for linear colliders? Is the acceleration performed on the same way?

REPLY [1 votes]: The particles going in each direction are in separate beamlines. They both have their own set of RF cavities, as well as dipole and quadrupole magnets.
This needs to be the case, since both beams contain positively charged particles, so they need an opposite direction magnetic field to stay in the beam lines. Since this forces them to be in separate beamlines, they can't share RF cavities.
Linear colliders are the same way (usually, at least), since you need to have two separate beamlines if you want to collide them at the end.


TITLE: A trigonometric-integral inequality
QUESTION [5 upvotes]: This problem comes from a discussion with one of my friends:

Prove that: $$\displaystyle \lim_{n \to \infty}\int_{1}^{n}{\sin (x)\sin(x^2)}\,{\mathrm dx}< \lim_{n \to \infty}\int_{1}^{n}{\sin(x^2)}\,{\mathrm dx}$$

I wonder if there is a reasonable way to somehow solve such a problem. Thanks.

REPLY [1 votes]: The idea goes as follows. Consider your inequality
$$
\displaystyle \lim_{n \to \infty}\int_{1}^{n}{\sin (x)\sin(x^2)}\,{\mathrm dx}< \lim_{n \to \infty}\int_{1}^{n}{\sin(x^2)}\,{\mathrm dx}
$$
but both integrals are meaningful in the given limit being standard values of Fresnel functions. So, you have
$$
\int_{1}^{\infty}{\sin (x)\sin(x^2)}\,{\mathrm dx}< \int_{1}^{\infty}{\sin(x^2)}\,{\mathrm dx}.
$$
The rhs can be immediately evaluated to give
$$
  \int_{1}^{\infty}{\sin(x^2)}\,{\mathrm dx}=\frac{1}{2}\sqrt{\frac{\pi}{2}}\left[1-2S\left(\sqrt{2}{\pi}\right)\right].
$$
I have introduced here a Fresnel function. These are defined in the following way
$$
   S(x)=\int_0^x dt\sin\left(\frac{\pi}{2}t^2\right)\qquad C(x)=\int_0^x dt\cos\left(\frac{\pi}{2}t^2\right)
$$
and $\frac{\pi}{2}$ factor that you see in the integrals is the reason why your integral is evaluated at that value for $S$. For your case is $S\left(\sqrt{2}{\pi}\right)\sim 0.247558$ and so the rhs evaluates to 0.316389.
On the lhs you have to do some more work. You will have
$$
  \sin(x)\sin(x^2)=\frac{1}{2}\left[\cos(x-x^2)-\cos(x+x^2)\right]
$$
and so you have to evaluate
$$
  \int_{1}^{\infty}{\sin (x)\sin(x^2)}\,{\mathrm dx}=\frac{1}{2}\int_{1}^{\infty}\left[\cos(x-x^2)-\cos(x+x^2)\right]\,{\mathrm dx}.
$$
In order to evaluate these integrals we firstly note that $x^2+x=\left(x+\frac{1}{2}\right)^2-\frac{1}{4}$ and  $x^2-x=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}$ and so
$$
  \frac{1}{2}\int_{1}^{\infty}\left[\cos(x-x^2)-\cos(x+x^2)\right]\,{\mathrm dx}=
   \frac{1}{2}\int_{\frac{1}{2}}^{\infty}\cos\left(y^2-\frac{1}{4}\right)\,{\mathrm dy}-\frac{1}{2}\int_{\frac{3}{2}}^{\infty}\cos\left(y^2-\frac{1}{4}\right)\,{\mathrm dy}
$$
after an obvious change of variables in the integrals. But we recognize immediately that
$$
   \int_{1}^{\infty}{\sin (x)\sin(x^2)}\,{\mathrm dx}= \frac{1}{2}\int_{\frac{1}{2}}^{\frac{3}{2}}\cos\left(y^2-\frac{1}{4}\right)\,{\mathrm dy}
$$
and expanding the cosine we realize that
$$
   \int_{\frac{1}{2}}^{\frac{3}{2}}\cos\left(y^2-\frac{1}{4}\right)\,{\mathrm dy}=
\int_{\frac{1}{2}}^{\frac{3}{2}}\cos(y^2)\cos\left(\frac{1}{4}\right)\,{\mathrm dy}+\int_{\frac{1}{2}}^{\frac{3}{2}}\sin(y^2)\sin\left(\frac{1}{4}\right)\,{\mathrm dy}
$$
and we are back to Fresnel functions. Already at this stage, you should be able to realize that the inequality holds as you are evaluating Frensel functions on a smaller interval and these are also multiplied by factors that are lesser than unity. But we can go on with numerical evaluation obtaining 0.286035 that is lower than 0.316389 as it should.


TITLE: AUTO Software for ODE's: references or forums?
QUESTION [3 upvotes]: I'm learning the AUTO software that does numerical continuation of ODE's by following these two references:

The official manual found here www.dam.brown.edu/.../auto07p.pdf
Lecture notes found at indy.cs.concordia.ca/auto/notes.pdf

The examples are understandable, but going from there to defining your own problem presents a very steep learning curve.
So I'm looking for other good references, or perhaps even an online forum for people using this software. Any suggestions?
Cheers!

REPLY [0 votes]: I'm learning AUTO-07p too, in my experience the best way to define your own problems is editing some demo according to your problem. I actually don't know more references that the official manual, that's one big problem with AUTO, there is not so much documentation. The software XPPAUT http://www.math.pitt.edu/~bard/xpp/xpp.html is another way to run AUTO in a more simple way, maybe you should try it.


TITLE: How can a irreducible polynomial be used in order to extend a field?
QUESTION [1 upvotes]: I'm interested in getting to know a bit more about coding theory, so I've been studying a bit about finite fields. I haven't been able to understand how the following topics work:

If we have a field $\mathbb{F}_p$ ($p$ is prime), we can use an irreducible polynomial (irreducible in $\mathbb{F}_p[x]$) $f$ of positive degree $n$ to generate an extension field $\mathbb{F}_q$, where $q = p^n$.
We can also use that very polynomial to provide unique representations of the elements in $\mathbb{F}_q$.

I'm not really trained in maths, so simpler explanations are very welcome. Also, examples are very much appreciated.
Thanks in advance!

REPLY [3 votes]: The idea is that if you have a polynomial $q(x)$ which is irreducible you can take $y$ to be one of the roots of the polynomial so that $q(y)=0$
Let's take $q(x)=x^3+x+1$ and suppose that $p=5$ to make it concrete. Note that this polynomial has no roots modulo $5$ - so being a cubic polynomial without a root, it is irreducible. And simply assume that there is a $y$ with $q(y)=0$.
What happens when we look at expressions involving the elements of $\mathbb F_5$, which we can call $0,1,2,3,4$ and $y$ - well we get polynomials involving $y$ with coefficients chosen from the field. And then we can divide one polynomial by another.
In theory the degree of these polynomials can be as large as you like. But notice that because $y^3+y+1=0$ or $y^3=4y+4$ (modulo $5$ remember) whenever we have a term of degree greater than $2$ in $y$ we can fund an equivalent polynomial of degree at most $2$ (just substitute for $y^3$ until you are done - there are easier ways, but this is the easiest way to see it is true).
So the expressions in $y$ that we get are $ay^2+by+c$. And how many do we get, well there are five possibilities for $a$, five for $b$ and five for $c$ so that gives $125=5^3$ elements.
It is quite obvious that we can add and multiply these, but what happens when we divide one polynomial by another? Well we can't divide by a multiple of $q(y)$, because this is zero by assumption. But if $r(x)$ is not divisible by $q(x)$ it turns out that there are polynomials such that $$r(x)s(x)+q(x)t(x)=1$$ and substituting in $y$ gives is $s(y)=\frac 1{r(y)}$ so that dividing by $r(y)$ is the same as multiplying by $s(y)$.
This is proved using the Euclidean algorithm for polynomials - it works the same way as for coprime integers, if you know it.
So we have arithmetic with all the familiar properties.

Just to show one example, what is $\frac 1y$ in the example case?
Well $y^3+y=4$ so that $4y^3+4y=1$ and $\frac 1y=4y^2+4$
($4\times 4 = 16\equiv 1 \bmod 5$)

And that is probably what you need to know to get hold of the basics. You can see how the polynomial functions to enable us to write expressions in the larger context.
To be a little more explicit, if we take expressions $ay^2+by+c$ with $a=b=0$ we find we have five elements $0,1,2,3,4$ which behave just like the field $\mathbb F_5$ that we started with. Because this field is concealed within the larger one, we call the larger one an extension. Because every expression is made up of (at most) three terms, it is of degree $3$ - the elements $1, y, y^2$ can be regarded as a basis of the larger field with coefficients in the smaller one (technically we have a vector space of dimension $3$ - but because we can multiply and divide, there is more structure than that).
The proofs that all this works in the general case belong to field theory - the role of $p$ being a prime and the polynomial being irreducible then become clear (really this is to make sure that division always works when you need it). The existence of a suitable $y$, which we assumed, can also be shown to be justified.


text:
\begin{document}

\keywords{derived projective limit functor $\cdot$ slowly increasing functions $\cdot$ rapidly decreasing distributions $\cdot$ sequence space representations}
\subjclass[2000]{Primary 46M18 $\cdot$ 46F05 ; Secondary 46A08 $\cdot$ 46A45}
\title[Bornologicity of $\mathcal{O}_M$]{A new proof for the bornologicity of the space of slowly increasing functions}
\author{Julian Larcher}
\address{Institut für Mathematik, Universität Innsbruck, A-6020 Innsbruck, Austria}
\email{julian.larcher@uibk.ac.at}
\author{Jochen Wengenroth}
\address{Universität Trier, FB IV -- Mathematik, 54286 Trier, Germany
}
\email{wengenroth@uni-trier.de}

\begin{abstract}
A. Grothendieck proved at the end of his thesis that the space $\om$ of slowly increasing functions and the space $\ocs$ of rapidly decreasing distributions are bornological.
Grothendieck's proof relies on the isomorphy of these spaces to a sequence space and we present the first proof that does not utilize this fact by using homological methods
and, in particular, the derived projective limit functor.
\end{abstract}

\maketitle

\section{Introduction and notation}

In \cite[p.~243]{SCHW} L. Schwartz introduced the space of multipliers of temperate distributions, i.e.,
the space of slowly increasing functions
\[\om=\{f\in\mathcal{C^\infty}(\R^d)\,;\, \forall \alpha\in\N_0^d\,\exists N\in\N:\, \langle x\rangle^{-N}\partial^\alpha f\in L^\infty\},\]
where $\mathcal{C^\infty}(\R^d)$
is the space of complex valued, infinitely differentiable functions on $\R^d$, $\langle x \rangle=1+|x|^2$, $\partial^\alpha$ is the partial derivative, and

$L^\infty$ is the Lebesgue space of bounded functions.

The dual $\oms$ of $\om$ is the space of very rapidly decreasing distributions.
\par
Schwartz also introduced the space of convolutors of temperate distributions, i.e., the space $\ocs$ of rapidly decreasing distributions,
which is the dual of the space
\[\oc=\{f\in\mathcal{C}^\infty(\R^d)\,;\, \exists N \,\forall \alpha\in\N_0^d:\, \langle x\rangle^{-N}\partial^\alpha f\in L^\infty\}\]
of very slowly increasing functions. These spaces are related as in the diagram
\begin{large}
\begin{center}
\begin{tabular}{c c c c c c c c c c c c c}
$\oc$ & $\subseteq$ &$\om$  \\
    \rotatebox[origin=c]{90}{$\cong$}  & & \rotatebox[origin=c]{90}{$\cong$} \\
$\oms$& $\subseteq$&$\ocs$
\end{tabular}
\end{center}
\end{large}
\vspace{8pt}
where in both cases the Fourier transform can be taken as the isomorphism. \\\par
It is comparatively easy to see that the four spaces are nuclear and semi-reflexive, that $\om$ and $\ocs$ are complete and that $\oc$ and $\oms$ are
(LF)-spaces and hence bornological. But the completeness of $\oc$ and $\oms$ and the bornologicity of $\om$ and $\ocs$ are not trivial (which was even
asserted by Grothendieck, \cite[Chap.~II, p.~130]{GRO}). Since the dual of a bornological space is complete and the dual of a complete nuclear space is bornological,
these two problems are equivalent (for the definitions of these topological properties and relations between them see \cite[Section 424]{EDM}). \par
Grothendieck proved that $\om$ is bornological by showing that it is isomorphic to a complemented subspace of the sequence space
$s\ptp s'$ \cite[Chap.~II, Lemme 18, p.\,132]{GRO} and verified ``directly'' that the space $s\ptp s'$ is bornological \cite[Chap.II, Prop.~15, p.\,125, Cor.~2, p.~128]{GRO}.
We will find out more about this isomorphy in Section \ref{sec2} and also give a homological proof of the bornologicity of $s\ptp s'$. \\\par
In \cite{KUC85}, J. Ku\v{c}era claimed to have presented a new (and simple) proof for the main properties of the space $\om$.
That Ku\v{c}era's proof contains severe mistakes and that it is based on incorrect propositions is clarified in \cite{LAR12}, where also the lack of a proof of
the bornologicity of $\om$, that does not use the isomorphy $\om\cong s\ptp s'$, is pointed out. In Section \ref{sec3} we will give such a proof.

\section{Projective limits and the space $s\ptp s'$}\label{sec2}

Since quotients (and, in particular, complemented subspaces) of bornological spaces are bornological, it was sufficient for Grothendieck to prove that $\om$ is isomorphic to a
complemented subspace of $s\ptp s'$, where $s$ is the space of rapidly decreasing sequences
\[ s=\{(x_j)_{j\in\N}\in \C^{\N}\,;\, \forall k: \sup_{j\in\N} j^k |x_j|<\infty\} \]
and $s'$ is its dual, the space of slowly increasing sequences
\[s'=\{(x_j)_{j\in\N}\in \C^{\N}\,;\, \exists k: \sup_{j\in\N} j^{-k} |x_j|<\infty\}.\]
By $s{\otimes}_\pi s'$ we denote the completed projective tensor product of these spaces. E.g., by \cite[Remark~1, p.~321]{BAR12a}, this space $s\ptp s'$ is canonically isomorphic to
\[
s\ptp s'\cong \{x\in\C^{\N\times\N}\,;\, \forall n \,\exists N :\,  \sup_{i,j}i^n j^{-N} |x_{i,j}|<\infty\}.
\]
In \cite{VAL}, M.\ Valdivia proved that $\om$ is even isomorphic to $s\ptp s'$ itself which answered a question
posed in \cite[Chap.~II, p.~134]{GRO}. C.\ Bargetz used this fact, the bornologicity of $s\ptp s'$, and methods of the theory
of topological tensor products to obtain the isomorphy $\oc\cong s\hat{\otimes}_\iota s'$ \cite[Prop.~1, p.~318]{BAR12a}.
\\\par
The descriptions of the spaces $\om$ and $s\hat{\otimes}_\pi s'$ already indicate how they can be written as projective limits of
LB-spaces (countable inductive limits of Banach spaces)
\begin{gather}\label{om}
\om= \bigcap_{n\in\N} X_{n}=\bigcap_{n\in\N} \bigcup_{N\in\N} X_{n,N},\\\label{s}
s\hat{\otimes}_\pi s'=\bigcap_{n\in\N} Y_{n}=\bigcap_{n\in\N} \bigcup_{N\in\N} Y_{n,N},
\end{gather}
where $X_{n,N}$ and $Y_{n,N}$ are the Banach spaces
\begin{gather*}
X_{n,N}=\{f\in\mathcal{C}^n(\R^d)\,;\,\Vert f\Vert_{n,N}=\sup_{x\in\R^d,|\alpha|\leq n}\langle x \rangle^{-N}| \partial^\alpha f(x)|<\infty\},\\
Y_{n,N}=\{x\in\C^{\N\times\N}\,;\, \| x\|_{n,N}=\sup_{i,j}i^n j^{-N} |x_{i,j}|<\infty
\}.
\end{gather*}
These representations as projective limits of LB-spaces are not only natural but also extremely useful since there are very good criteria for checking bornologicity.
They are related to the derived projective limit functor $\text{Proj}^1 \XX$
(which can be defined as the cokernel of the map $\prod X_n \to \prod X_n$, $(x_n)_n\mapsto (x_n-\varrho_{n+1}^n(x_{n+1}))_n$
where $\varrho_m^n$ are the connecting maps of the projective spectrum $\XX$, in our cases, $\varrho_m^n$ are just inclusions).
Indeed, an unbublished theorem of D.\ Vogt (his proof reproduced in \cite[Th.\ 3.3.4]{WLN}) says that $\Proj \XX$ is bornological
whenever $\Proj^1 \XX=0$. Moreover, there is a variety of evaluable conditions ensuring $\Proj^1\XX=0$. We are going to apply the following results of
Palamodov-Retakh \cite{Pal} and the second named author, respectively:
\begin{quote}
  A specturm $\XX$ of LB-spaces satisfies $\Proj^1\XX=0$ if and only if there are Banach discs $D_n$ in $X_n$ with $\varrho_m^n(D_m)\subseteq D_n$ and
  \[ \forall\,  n\in\N \;\exists\, m\ge n\; \forall \, k\ge m:\; \varrho_m^n(X_m)\subseteq \varrho_k^n(X_k) + D_n.\]
\end{quote}
The requirement $\varrho_m^n(D_m)\subseteq D_n$ is sometimes very easy to fulfil but in many cases it is very inconvenient. It can be omitted if either all steps $X_n$
are LS-spaces (i.e., the inclusions $X_{n,N}\hookrightarrow X_{n,N+1}$ are compact) or if a slightly stronger condition of Palamodov-Retakh type is required. Denoting by
$\varrho_\infty^n: \Proj\XX \to X_n$ the obvious map we have:
\begin{quote}
  A specturm $\XX$ of LB-spaces satisfies $\Proj^1\XX=0$ if and only if, for every $n\in\N$, there are a Banach discs $D_n$ in $X_n$ and $m\ge n$ with
  \[ \varrho_m^n(X_m)\subseteq \varrho_\infty^n(\Proj \XX) + D_n.\]
\end{quote}
We refer to \cite{WLN} for the proofs of these characterization and much more information about derived functors.
Typically, the decompositions required in conditions of Retakh-Palamodov type are quite easy to produce in the case of spaces of sequences (or matrices) since one can write
$x= \chi\, x + (1-\chi)x$ where $\chi$ is the indicator function of a suitably chosen set. We want to exemplify this by giving a very short proof for the bornologicity of
$s\hat{\otimes}_\pi s'$
(which is similar to
Vogt's proof of $\text{Ext}^1(s,s)=0$ \cite[Lemma 2.1, p.~359]{VOG83}).

\begin{proposition}\label{prop:s}
The space $s\hat{\otimes}_\pi s'$ is bornological.
\end{proposition}
\begin{proof}
We keep the notation $s\hat{\otimes}_\pi s'\cong\bigcap_{n\in\N} Y_{n}=\bigcap_{n\in\N} \bigcup_{N\in\N} Y_{n,N}$ from above
and we will verify the Palamodov-Retakh condition for the unit balls $D_n$ of $Y_{n,0}$ which trivially satisfy $D_{n+1}\subseteq D_n$.
For $n\in\N$ we take $m=n+1$ and fix $x\in Y_n$ as well as $k\ge n+1$. Since $x\in Y_{m,M}$ for some $M\in \N$ we have
\[
  \|x\|_{m,M} =\sup_{i,j} i^m j^{-N} |x_{i,j}|=c <\infty.
\]
We set $y_{i,j}=x_{i,j}$ if $i < cj^M$ and $y_{i,j}=0$ else, as well as $z=x-y$.
For $i<cj^m$ we have $z_{i,j}=0$ and for $i\ge cj^M$ we estimate
\[
 i^nj^{-0} |z_{i,j}| = i^m j^{-M} |z_{i,j}|\, j^M/i \le \|x\|_{m,M}/c=1
\]
which proves $z\in D_n$. It remains to show $y\in Y_{k,K}$ for $K$ sufficienly large. Indeed, for $K=M(k-m+1)$ we have $y_{i,j}=0$ if $i\ge cj^M$ and if $i< cj^M$ we estimate
\[
  i^kj^{-K}|y_{i,j}| =i^m j^{-M} |y_{i,j}| i^{k-m}j^{M-K} \le \|x\|_{m,M} c^{k-m} j^{(k-m)M + M- K} = c^{k-m+1}.
\]
This proves $\|y\|_{k,K}<\infty$, as required.
\end{proof}

\section{The new proof}\label{sec3}

Now we want to prove $\text{Proj}^1\XX=0$ for the spectrum $\XX=(X_n)_{n\in\N}$ in (\ref{om}) in order
to obtain that $\om$ is bornological.
Splitting up a given function $f\in X_m$  as $f=\chi f+ (1-\chi) f$ with a cut-off function $\chi$ (as in the proof of Proposition \ref{prop:s})
does not work in this case. But we will see how $f$ can be ``split up'' in the following proof of Grothendieck's result.

\begin{proposition} The space $\om$ is bornological.
\end{proposition}
\begin{proof}
To obtain $\text{Proj}^1\XX=0$ we will show
\begin{gather}\label{hom}
\forall n\,\exists m,N:\,\,X_{m}\subseteq \om + B_{n,N}
\end{gather}  where $B_{n,N}$ is the unit ball of $X_{n,N}$.
This condition means that we have to approximate every $f\in X_m$ with respect to the norm $\|\cdot\|_{n,N}$ by elements of $\om$.
To achieve such an approximation we use a kernel $K\in\om(\R^d\times\R^d)$ satisfying

\begin{align*}
& K\ge 0,\; \int_{\R^d} K(t,x)\,dt=1\text{ for all }x\in\R^d,\, \text{ and} \\
&\supp K(\cdot,x)\subseteq \prod\limits_{j=1}^{d}[x_j,x_j+\varepsilon \langle x \rangle^{-\mu}]=:A_x \text{ for all }x\in\R^d
\end{align*}
where we will see later how $\varepsilon$ and $\mu$ have to be chosen in dependence on $f\in X_m$. We can obtain such a kernel by defining
\[
K(t,x)=\varepsilon^{-d}\langle x \rangle^{\mu d} \varphi(\varepsilon^{-1}\langle x \rangle^\mu(t-x))
\]
for a positive test function $\varphi\in\mathcal{C}^\infty(\R^d)$ with support in $[0,1]^d$ and $\int_{\R^d} \varphi(t) dt=1$
(the conditions above can be checked easily and $K\in\om$ since every derivative of $K$ can be estimated by a polynomial). \par
We start with the one-dimensional case $d=1$ where we can take $m=n+1$ and $N=0$. So let $f\in X_{n+1,M}$ for some $M\in \N$.
We want to find $g\in \om$ such that $f-g\in B_{n,0}$. At first we set
\[
g_n(x)=\int_{\R}f^{(n)}(t) K(t,x)\,dt
\]
and show that this is a good approximation to $f^{(n)}$. Since for $l\in\N_0$
\[
\left| g_n^{(l)}(x) \right|=\left|\int_{A_x}f^{(n)}(t) \partial_{x}^{l} K(t,x)\,dt\right|\leq\int_{A_x} |P(t)|\, |Q(t,x)| \,dt\leq |R(x)|
\] for some polynomials $P$, $Q$, $R$, the function $g_n$ is contained in $\om$. Furthermore we can estimate in virtue of Taylor's formula \begin{gather*}
\left| f^{(n)}(t)-f^{(n)}(x)\right|\leq |t-x| \langle \xi(t,x) \rangle^M \Vert f\Vert_{n+1,M}
\end{gather*} with a point $\xi(t,x)$ between $t$ and $x$. For $\varepsilon$ small enough the inequality $\langle \xi(t,x)\rangle\leq 2\langle x \rangle$ holds for every $x\in\R$ and $t\in A_x$. We obtain
\begin{equation}\label{est}
\begin{aligned}
|g_n(x)-f^{(n)}(x)|&=\left|\int_{\R}\left(f^{(n)}(t)-f^{(n)}(x)\right) K(t,x)\,
dt\right|\\
&\leq\int_{A_x}\left|f^{(n)}(t)-f^{(n)}(x)\right| K(t,x)\,dt\\
&\leq \int_{A_x}|t-x|\langle \xi(t,x) \rangle^M \Vert f\Vert_{n+1,M}K(t,x)\,dt\\
&\leq\varepsilon \,2^M\, \langle x\rangle^{M-\mu} \Vert f\Vert_{n+1,M}\int_{A_x} K(t,x)\,dt\\
&=\varepsilon\,2^M\,\langle x\rangle^{M-\mu}\Vert f\Vert_{n+1,M}.
\end{aligned}
\end{equation}
Now if

\begin{gather*}T:\om(\R)\rightarrow\om(\R), h\mapsto \left(x\mapsto \int_{0}^{x}h(t)\,dt\right),
\end{gather*}
we can set
\[g(x)=\sum_{j=0}^{n-1}\frac{f^{(j)}(0)}{j !}x^j +(T^n g_n)(x).\]
Then $g\in\om$ and since
\[(T^n f^{(n)})(x)=f(x)-\sum_{j=0}^{n-1}\frac{f^{(j)}(0)}{j !}x^j,\]
integrating (\ref{est}) (the integral starting at 0) yields
\[
| g^{(l)}(x)-f^{(l)}(x)|\leq 1,\,x\in\R^d,l\leq n
\]
for $\varepsilon$ small enough and $\mu$ large enough. Hence $g-f\in B_{n,0}$ and the proof is complete for the one-dimensional case.
\par
Now we will prove the two-dimensional case  $d=2$. We set $m=2n+1$ and $N=n-1$ in (\ref{hom}). So let $f\in X_{2n+1,M}$ for some $M$. With the help of a kernel
$K\in\om(\R^2\times \R^2)$ like above, we set \[g_n(x)=\int_{\R^2}\partial^{(n,n)}f(t)K(t,x)\,dt\] in order to approximate
$\partial^{(n,n)}f$ by $g_n$. Similar to the one-dimensional case we have
\begin{gather*}
\left| \partial^{(n,n)}f(t)-\partial^{(n,n)}f(x)\right|\leq c\cdot|t-x| \langle \xi(t,x) \rangle^M \Vert f\Vert_{2n+1,M}
\end{gather*} and $\langle \xi(t,x)\rangle\leq 2\langle x \rangle$ for $t\in A_x$ and $\varepsilon$ small enough and thus
\begin{equation}\label{est2}
\begin{aligned}
|g_n(x)-\partial^{(n,n)}f(x)|&\leq c \int_{A_x}|t-x|\langle \xi(t,x) \rangle^M \Vert f\Vert_{2n+1,M}K(t,x)\,dt\\
&\leq \tilde{c}\,\varepsilon\,\langle x\rangle^{M-\mu}\Vert f\Vert_{2n+1,M}.
\end{aligned}
\end{equation}
Let us denote $T_j$ the integral with respect to the $j$-th component (the integral starting at 0). Applying $T_{1}\circ T_{2}$ $n$-times to $\partial^{(n,n)}f(x)$ yields
\begin{multline*}
(T_{1}^{n}T_{2}^{n}f)(x)=\\f(x)+ \sum_{\alpha<(n,n)}\partial^\alpha f (0,0) \frac{x^\alpha}{\alpha !} -
\sum_{j=0}^{n-1}\partial^{(j,0)} f(0,x_2) \frac{x_1^j}{j !}- \sum_{j=0}^{n-1}\partial^{(0,j)} f(x_1,0) \frac{x_2^j}{j !}.
\end{multline*}
As in the one-dimensional case we can choose
$g^1_0,\ldots g^1_{n-1},g^2_0,\ldots,g^2_{n-1}\in\om(\R)$ such that $\Vert g^1_j-\partial^{(0,j)} f(\cdot,0)\Vert_{n,0}
\leq \varepsilon$ and $\Vert g^2_j-\partial^{(j,0)} f(\cdot,0)\Vert_{n,0}\leq \varepsilon$. Defining
\[
g(x)=(T_{1}^{n}T_{2}^{n})g_n(x)- \sum_{\alpha<(n,n)}\partial^\alpha f (0,0) \frac{x^\alpha}{\alpha !} +
\sum_{j=0}^{n-1}g^2_j(x_2) \frac{x_1^j}{j !}+ \sum_{j=0}^{n-1}g^1_j(x_1) \frac{x_2^j}{j !}
\]
and applying $T_{1}^{n}T_{2}^{n}$ to (\ref{est2}) yields
\begin{multline*}
|g(x)-f(x)|\leq \\\varepsilon +\sum_{j=0}^{n-1}\left( \left|g^1_j(x_1)-\partial^{(0,j)} f(x_1,0)\right| \frac{|x_2|^j}{j !}
+\left| g^2_j(x_2)-\partial^{(j,0)} f(0,x_2)\right|\frac{|x_1|^j}{j !}\right)
\end{multline*}
for $\mu$ large enough which implies
\begin{align*}
|g(x)-f(x)|\leq \varepsilon +\varepsilon \sum_{j=0}^{n-1}\frac{|x_2|^j}{j !} +\varepsilon \sum_{j=0}^{n-1}\frac{|x_1|^j}{j !}\leq \varepsilon \,c \,\langle x \rangle^{n-1}
\end{align*}
for some $c>1$. Since similar estimates also hold for $|\partial^\alpha g(x)-\partial^\alpha f(x)|$, $|\alpha|\leq n $, we obtain $g-f \in B_{n,n-1}$ and the proof is complete for $d=2$.
\par
The general case $d\in\N$ is very similar. Inductively we want to show $$X_{dn+1}\subseteq \om+B_{n,(d-1)(n-1)}$$ and start by
approximating $\partial^{(n,\ldots,n)}f$ by $g_n(x):=\int_{R^d}\partial^{(n,\ldots,n)}f(t)K(t,x)\,dt$.
Then we integrate the estimate of $g_n-\partial^{(n,\ldots,n)}f$ $n$-times with respect to each component.
The integral $T_1^n\cdots T_d^n \partial^{(n,\ldots,n)}f$ contains $f$ as a summand and terms that are the
product of a derivative of $f$ that only depends on less than $d$ components and a polynomial in less than $d$ components
with exponents less than $n$. But we can estimate the functions that only depend on less than $d$ variables by the induction
hypothesis and hence we can obtain $g\in\om$ with $g-f\in B_{n,(d-1)(n-1)}$.
\end{proof}


TITLE: Strong convergence to zero
QUESTION [1 upvotes]: Let $(a^m)_{m\ge 1}$ be a sequence in a Hilbert space and also $$a^m=\sum_{n=1}^\infty a^m_n e_n$$ with $a^m_n\xrightarrow{m\rightarrow\infty}0,$ for any $n\ge 1.$
Does $a^m$ converges strongly to $0$ ?

REPLY [1 votes]: I'm assuming that $(e_n)_n$ is an orthonormal set in $H$.
The answer is no. Consider $a^m = me_m$. We have $a^m_n = m\delta_{mn} \xrightarrow{m\to\infty} 0$ for all $n \in \mathbb{N}$. However, $(a^m)_m$ is not even a bounded sequence.

However, there is this result:

Let $(e_n)_n$ be an orthonormal basis for $H$. Let $(a_m)_m$ be a bounded sequence in $H$ such that there exists $a \in H$ with $\langle a_m, e_n\rangle \xrightarrow{m\to\infty} \langle a, e_n\rangle$ for all $n \in \mathbb{N}$. Then $(a_m)_m$ converges weakly to $a$.