text:
\begin{document}

\bibliographystyle{amsalpha}

\title{\bf Summing inclusion maps between symmetric sequence spaces}
\author{by \\[20pt] Andreas Defant, Mieczys{\l}aw Masty{\l}o\thanks{Research supported by KBN Grant 2 P03A
042 18}\,\, and Carsten Michels \\[10pt]  }
\date{}
  \maketitle
\begin{abstract}
In 1973/74 Bennett and (independently) Carl proved that for $1 \le u \le 2$ the identity map
id: $\ell_u \Id
\ell_2$ is absolutely $(u,1)$-summing, i.\,e. for every unconditionally summable sequence $(x_n)$
in $\ell_u$ the
scalar sequence $(\|x_n \|_{\ell_2})$ is contained in $\ell_u$,
 which improved upon well-known results of Littlewood and Orlicz. The following substantial
 extension is our main result:
 For a $2$-concave
symmetric Banach sequence space $E$ the identity map $\text{id}: E \Id \ell_2$ is absolutely
$(E,1)$-summing, i.\,e. for every unconditionally summable sequence $(x_n)$ in $E$ the scalar sequence
$(\|x_n \|_{\ell_2})$ is contained in
$E$. Various applications are given, e.\,g. to the theory of eigenvalue distribution of compact
operators where we show
that the sequence of
eigenvalues of an operator $T$ on $\ell_2$ with values in a $2$-concave symmetric Banach
sequence space $E$ is a
multiplier from $\ell_2$ into $E$. Furthermore,
 we prove an asymptotic formula for the $k$-th approximation number
 of the identity map $\text{id}: \ell_2^n \Id E_n$, where $E_n$ denotes the linear span of the
 first $n$ standard unit vectors in $E$, and apply it to
 Lorentz  and Orlicz sequence spaces.
\end{abstract}
\section{Introduction}
In 1930 Littlewood \cite{little} proved that for every bilinear and continuous operator
\mbox{$\varphi: c_0 \times c_0 \rightarrow \R$} the quantity
$\sum_{k,\ell=1}^\infty |\varphi(e_k,e_\ell)|^{4/3}$ is finite; this is
equivalent to the statement that for every unconditionally summable sequence $(x_n)$ in $\ell_1$
the scalar sequence $(\norm{x_n}_{\ell_{4/3}})$ is contained in $\ell_{4/3}$. Bennett \cite{bennett}
and (independently) Carl \cite{carl} extended Littlewood's result in the following way: For
$1 \le u \le v \le 2$ and every unconditionally summable sequence $(x_n)$ in $\ell_u$ the
sequence $(\norm{x_n}_{\ell_v})$ is contained in $\ell_r$, where $1/r =1/u-1/v+1/2$. Their result
has useful applications in various parts of analysis---in particular, in approximation theory
as well as for the theory of eigenvalue distribution of
compact operators, e.\,g. that for $1\le u <2$ every operator on $\ell_2$ with values in $\ell_u$
has absolutely $r$-summable eigenvalues, where $1/r=1/u-1/2$.
\par
The case $v=2$ in the Bennett--Carl result is crucial (for the proof as well as for applications).
 Motivated by applications
to interpolation theory (see e.\,g. \cite{ovchinni} and \cite{milman}) Maligranda and the
second named author in \cite{mastylo} proved that for an Orlicz function $\varphi$ for which the
map $t \mapsto \varphi(\sqrt{t})$ is equivalent to a concave function and for every unconditionally
summable sequence $(x_n)$ in the Orlicz sequence space $\ell_\varphi$ the sequence
$(\norm{x_n}_{\ell_2})$ is contained in $\ell_\varphi$. Moreover, based on complex interpolation, in
\cite{dm98} various commutative and non commutative variants were given.
\par
These results were the starting point for
the research on which this article is based upon. Developing and using
complex interpolation formulas for spaces of operators related to those of Kouba \cite{kouba},
our main result is a far reaching extension of the
 above results: For a $2$-concave
 symmetric Banach sequence space $E$  and every unconditionally summable sequence $(x_n)$ in
 $E$ the sequence $(\norm{x_n}_{\ell_2})$ is contained in $E$. In the language of
 $(E,1)$-summing operators (which we will recall later on) this means that the identity map
 \mbox{$\id : E \Id \ell_2$} is  $(E,1)$-summing. An example shows that the $2$-concavity of
 $E$ is not superfluous.  As in the classical case our result
  has some useful applications.  We show that the sequence of eigenvalues
   of an operator $T$ on $\ell_2$ with values in a $2$-concave symmetric Banach sequence space
   $E$ is a multiplier from $\ell_2$ into $E$, a result which for $E=\ell_u$, $1 \le u \le 2$, is
   well-known (note that the space of multipliers from $\ell_2$ into $\ell_u$ coincides with
 $\ell_r$, $1/r =1/u-1/2$). Furthermore, we prove for a $2$-concave
   symmetric Banach sequence space $E$ and
   $1 \le k \le n$ the asymptotic formula
$$
a_k(\id: \ell_2^n \Id E_n) \asymp \frac{\lambda_E(n-k+1)}{(n-k+1)^{1/2}},
$$
where $a_k(T)$ denotes the $k$-th approximation number of an operator $T$, $E_n$ stands for
the linear span of the first $n$ standard unit vectors in $E$ and $\lambda_E: \N \rightarrow
\R_+$ is the fundamental function of the sequence space $E$, and apply it to Lorentz and
Orlicz sequence spaces.
\section{Preliminaries}
For a positive number $a$ we denote by $\lfloor a \rfloor$ the largest integer less or equal than $a$.
If $(a_n)$ and $(b_n)$ are scalar sequences we write \mbox{$a_n \prec b_n$}
 whenever there is some $c \ge 0$ such that $a_n \le c \cdot b_n$ for all $n$,
  and
\mbox{$a_n \asymp b_n$} whenever \mbox{$a_n \prec b_n$} and
\mbox{$b_n \prec a_n$}. \par
 We  use standard notation and
 notions from Banach space theory, as presented e.\,g. in
 \cite{lt77}, \cite{lt} and \cite{tj}.
If $E$ is a Banach space, then
 $B_E$ is its (closed) unit ball and $E'$ its dual space.
\par
 Throughout the paper by a Banach sequence
space we mean a real Banach lattice $E$ modelled on the set of positive
integers $\N$ which contains an element $x$ with $\mbox{supp} \,x= \N$.
A Banach sequence space $E$ is said to be symmetric provided that
$\|(x_n)\|_{E} = \|(x_n^*)\|_{E}$, where $(x_n^*)$ denotes the decreasing
rearrangement of the sequence $(x_n)$, i.e.
$$
x_{n}^{*}:= \inf \{\sup_{i\in \N \setminus J} |x_i| \, | \,
J\subset \N, \, \mbox{card} (J) < n \}.
$$
 It is maximal if the unit ball $B_E$ is closed in the
pointwise convergence topology induced by the space $\omega$ of all real sequences. Note that this condition is
equivalent to $E^\times = E'$, where as usual $$ E^{\times}:= \{x=(x_n) \in \omega \, |\,  \Sigma_{n=1}^{\infty} |x_n y_n| <
\infty \, \mbox{for all} \, y=(y_n) \in E\} $$ is the K\"othe dual of $E$. Note that $E^{\times}$ is a maximal (symmetric,
provided that $E$ is) Banach sequence ´space under the norm $$ \|x\|:= \sup\{\Sigma_{n=1}^{\infty}|x_n y_n| \, | \,
\|y\|_{E} \leq 1\}. $$ The fundamental function of a symmetric Banach sequence space $E$ is defined by $$
\lambda_{E}(n):=\norm{\textstyle{\sum_{i=1}^{n}} e_i}_E, \quad n \in \N; $$ throughout the paper $(e_n)$ will denote
the standard unit vector basis in $c_0$ and $E_n$ the linear span of the first $n$ unit vectors.
 It is well-known that any symmetric Banach
sequence space $E$ is continuously embedded in the symmetric Marcinkiewicz sequence
space $m_{\lambda_E}$ of all sequences $x=(x_n)$ such that
$$
\|x\|_{\lambda_{E}}:=\sup_{n\geq 1} x_{n}^{**} \lambda_E(n)  < \infty,
$$
where $x_{n}^{**}:=\frac{1}{n} \sum_{k=1}^{n}\, x_{k}^{*}$.
 For the notions of $p$-convexity and $q$-concavity ($1 \le p,q \le \infty$) of a Banach lattice
 $X$ (the associated constants are denoted by $\convex{p}(X)$ and $\concave{q}(X)$, respectively) we refer
 to \cite[1.d.3]{lt}---but since the notion of $2$-concavity is crucial for our purposes recall
 that a Banach sequence space $E$ is called $2$-concave if there exists a constant
 $C>0$ such that for all
 $x_1, \ldots ,x_n \in E$
$$
 \left ( \sum_{i=1}^n \norm{x_i}_E^2 \right )^{1/2}
  \le C \cdot \bignorm{\left(\sum_{i=1}^n |x_i|^2 \right )^{1/2}}_E.
$$
It is well-known that this is equivalent to the notion of cotype~$2$ (see \cite[1.f.16]{lt});
recall that a Banach space $X$ has cotype $q$
($2\leq q <\infty $) if there is a constant $C>0$ such that for
finitely many $x_1, \ldots, x_n \in X$
\begin{eqnarray*}
\left ( \sum_{i=1}^n \norm{x_i}_X^q \right )^{1/q}
\le C \cdot \Big (\int_{0}^{1} \big \|\sum_{i=1}^n r_{i}(t) \cdot x_i
\big \|_{X}^{2}\, dt \Big )^{1/2}.
\end{eqnarray*}
Note that  $2$-concave symmetric Banach sequence spaces are separable and
maximal.
  An important tool for our purposes are powers of sequence spaces: Let $E$ be a (maximal) symmetric Banach sequence space
  and $0<r<\infty$ such that $\convex{\max(1,r)}(E)=1$. Then
$$
E^r := \{ x \in \ell_\infty \, | \, |x|^{1/r} \in E \}
$$
endowed with the norm
$$
\norm{x}_{E^r}:= \norm{|x|^{1/r}}_E^r, \quad x \in E^r
$$
is again a (maximal) symmetric Banach sequence space which is $1/\min(1,r)$-convex.
For two Banach sequence spaces $E$ and $F$ the space of multipliers $M(E,F)$
from $E$ into $F$ consists of all scalar sequences $x=(x_n) $ such that
the associated multiplication operator $(y_n) \mapsto (x_n\,y_n)$ is
defined and bounded from $E$ into $F$. $M(E,F)$ is a (maximal symmetric
provided that $E$ and $F$ are) Banach sequence space equipped with the norm
$$
\|x\|_{M(E,F)} :=\sup\{\|xy\|_F \, | \, y \in B_{E}\}.
$$
Note that if $E$ is a Banach sequence space then $M(E,\ell_{1})=E^\times$.
 In the case where $E=\ell_2$ and $F$ is
 $2$-concave with $\concave{2}(F)=1$ it can be
 easily seen that
\begin{equation}
\label{diagonalpower}
M(\ell_2,F) = (((F^\times)^2)^\times)^{1/2}
\end{equation}
holds isometrically. We will need that for any symmetric Banach sequence space $E \Id \ell_2$ not equivalent to
$\ell_2$
\begin{equation}
\label{czero}
M(\ell_2,E) \hookrightarrow c_0.
\end{equation}
In fact, for $F:=M(\ell_2,E)$, by the assumption we have
$$
 \lim_{n \rightarrow \infty}
 \lambda_F(n)=\sup_n \norm{\textstyle{\sum_1^n e_i}}_F =\sup_n \norm{\id : \ell_2^n \Id E_n}=
  \infty.
$$
Since for any $x=(x_n) \in F$ the estimate
$
x_n^*\cdot \lambda_F(n) \le \norm{x}_F
$
holds, the claim follows.
\par For all information on Banach operator ideals and $s$-numbers see \cite{djt}, \cite{koenig},
\cite{pietsch} and \cite{eigenvalues}.
As usual $\LL(E,F)$
 denotes the Banach space of all (bounded and linear) operators from
 $E$ into $F$ endowed with the operator norm $\norm{\cdot}$.
 For an operator $T: X \rightarrow Y$ between Banach spaces recall the definition of
the $k$-th approximation number
$$ a_k(T):= \inf \{ \norm{T-T_k} \, | \, T_k \in \LL(X,Y) \text{ has rank} \,<k\},$$
the $k$-th Weyl number
$$
x_k(T):= \sup \{a_k(TS) \, | \, S \in \LL(\ell_2,X) \text{ with } \norm{S}\le 1\}
$$
and the $k$-th Gelfand number
$$
c_k(T) := \inf \{\norm{T_{|G}} \, | \, G \subset X, \, \codim\, G <k\}.
$$
\par
Moreover, for an $s$-number function
 $s$ and a maximal symmetric Banach sequence space $E$ we denote by $\ui_E^s$ the Banach operator ideal
  of all operators $T$ with $(s_n(T)) \in E$,
 endowed with the norm $\norm{T}_{\ui_E^s}:= \norm{(s_n(T))}_E$; on $\ell_2$ and for fixed $E$ all
 these ideals coincide (isometrically)---for simplicity we then
  denote this space by $\ui_E$.
\par For basic results and notation from interpolation theory we refer to
\cite{bk} and \cite{BL}. We recall that a mapping $\cal F$ from (a subclass $\mathcal{C}$ of) the category of all
couples of Banach spaces into the category
 of all Banach spaces is said to be a method of interpolation (on $\mathcal{C}$)
if for any couple $(X_0,X_1)$ ($\in \mathcal{C}$), the Banach space
${\cal F}(X_0, X_1)$ is intermediate
with respect to $(X_0, X_1)$
(i.\,e. \mbox{$X_0 \cap X_1 \Id {\cal F}(X_0, X_1) \Id
X_0 +X_1$),} and  $T: {\cal F}(X_0, X_1) \to {\cal F}(Y_0, Y_1)$ for all
Banach couples $(X_0, X_1)$, $(Y_0, Y_1)$ ($\in \mathcal{C}$) and every $T: (X_0, X_1) \to
(Y_0, Y_1)$. Here as usual the notation $T: (X_0, X_1) \to (Y_0, Y_1)$
means that $T: X_0 +X_1 \to Y_0 + Y_1 $ is a linear operator such
that  for $j=0,1$ the restriction of $T$ to the space $X_{j}$ is a bounded operator from
$X_j$ into $Y_{j}$. If additionally
$$
\|T: {\cal F}(X_0, X_1) \to {\cal F}(Y_0, Y_1)\| \leq
\max \{ \|T :X_0 \to Y_0 \|, \, \|T: X_1 \to Y_1 \| \}
$$
holds, then ${\cal F}$ is called an exact method of interpolation (on $\mathcal{C}$).
Concrete examples of exact
interpolation methods are the real method of interpolation $(\cdot,\cdot)_{\theta,p}$,
$0<\theta<1$, $1 \le p \le \infty$ (see
e.\,g. \cite[Chapter~3]{BL}) defined on the class of all Banach couples and
the complex method of interpolation $[\cdot,\cdot]_\theta$, $0<\theta<1$
(see e.\,g. \cite[Chapter~4]{BL}) defined on the class of couples of complex Banach spaces.
Both methods are of power type $\theta$, i.\,e. if $\mathcal{F}=(\cdot,\cdot)_{\theta,p}$ or
$\mathcal{F}=[\cdot,\cdot]_\theta$ then for all
$T:(X_0,X_1) \rightarrow (Y_0,Y_1)$ it holds
\begin{equation}
\label{powerreal}
\norm{T:\mathcal{F}(X_0,X_1) \rightarrow \mathcal{F}(Y_0,Y_1)} \le
\norm{T:X_0 \to Y_0}^{1-\theta}
 \cdot \norm{T:X_1 \to Y_1}^\theta.
\end{equation}
In order to avoid misunderstandings,  if we
interpolate between real Banach spaces using the complex method of
interpolation we mean that we use any interpolation functor which
is an extension of the complex method. For such a functor we use the
original notation $[\cdot, \cdot]_{\theta}$.
\par In what follows we will often use the following well-known fact
(see e.\,g. \cite[2.5.1]{BL}) that for any interpolation space
$X$ with respect to $(X_0, X_1)$ there exists an exact interpolation
functor ${\cal F}$ such that ${\cal F}(X_0, X_1) =X$ up to equivalent
norms. An important class of interpolation spaces are $K$-spaces. Recall
that an intermediate Banach space $X$ with respect to a couple $(X_0, X_1)$
is called a relative $K$-space if, whenever $x\in X$ and $y\in X_0 + X_1$
satisfy
$$
K(t, y; X_0, X_1) \leq K(t, x; X_0, X_1) \, \, \, \mbox{for all } t>0,
$$
then it follows that $y \in X$, where
$$
K(t, x; X_0,X_1):= \inf \, \{ \|x_0 \|_{X_0} + t \|x_1 \|_{X_1} \,|
\, x = x_0 + x_1 \}, \quad  t >0
$$
is the Peetre $K$-functional.
\par A Banach couple $(X_0, X_1)$ is said to be a relative Calder\'on
couple if all interpolation spaces with respect to $(X_0, X_1)$ are
also relative $K$-spaces. This is equivalent to: For each pair of elements
$x \in X_0 + X_1$ and $y\in X_0 + X_1$ satisfying
$K(t,y; X_0, X_1) \leq K(t, x; X_0, X_1)$ for all $t>0$, there exists an
operator $T: (X_0, X_1) \to (X_0, X_1)$ such that $Tx=y$.
\section{$\pmb{(E,p)}$-summing operators}

The following definition is a natural extension of the notion of absolutely $(r,p)$-summing
 operators. For two Banach spaces $E$ and
 $F$ we mean by
 $E \Id F$ that $E$ is contained in $F$, and the natural identity map is continuous; in this case
 we put \mbox{$c_E^F:= \norm{\id: E \Id F}$}
 and $c_p^F := c_{\ell_p}^F$ whenever $\ell_p \Id F$. If $E$ and $F$ are Banach sequence spaces
 with  $\norm{e_n}_E=1$ for all $n$, then obviously
 $\ell_1 \Id E$ and $c_1^E=1$. Note also that $E^\times \Id M(E,F)$ with $c_{E^\times}^{M(E,F)}=1$.
\begin{defin}
\label{Epsumming} For $1 \le p < \infty$ let $E$ be a Banach sequence space such that
$\ell_p \Id E$ and $\norm{e_n}_E=1$ for all $n$.
Then an operator $T:X \rightarrow Y$ between
 Banach spaces $X$ and $Y$ is called $(E,p)$-summing (shortly: $T \in
\Pi_{E,p}$) if there exists a constant $C>0$ such that for all
$x_1, \ldots, x_n \in X$
$$
\norm{(\norm{Tx_i}_Y)_{i=1}^n}_E \le C \cdot c_p^E \cdot \sup_{x' \in B_{X'}} \left
( \sum_{i=1}^n |\langle x',x_i \rangle|^p \right )^{1/p},
$$
where in the sequel $(\xi_i)_{i=1}^n$ denotes the sequence $\sum_{i=1}^n \xi_i \cdot e_i$.
We write $\pi_{E,p}(T)$ for the smallest constant $C$ with the above property; in this way we
 obtain the
 Banach operator ideal $(\Pi_{E,p}, \pi_{E,p})$
(see also \cite{milman}), and for $E=\ell_r$ ($r\ge p$)
the well-known Banach operator ideal $(\Pi_{r,p},\pi_{r,p})$ of all absolutely $(r,p)$-summing
 operators.
\end{defin}
\par Let us collect some later needed observations which are all modelled along classical results
on $(r,p)$-summing operators.
We start with the following simple fact that for each maximal Banach
sequence space $E$ an operator  $T: X \to Y $ is $(E,p)$-summing if and only if the
induced linear operator
$$
\widehat{T}: \ell_p^w(X) \rightarrow E(Y), \quad \widehat{T}(x_n):=(Tx_n)
$$
is defined (and hence bounded). In this case,
$\|\widehat{T}: \ell_{p}^{w}(X) \to E(Y)\|=\pi_{E,p}(T)$ provided that
$c_{p}^{E}=1$. Here and in what follows for a given Banach space $X$,
$\ell_{p}^{w}(X)$ and $E(X)$ denotes the Banach space of all weakly
$p$-summable and absolutely $E$-summable sequences $x=(x_n)$ in $X$
equipped with the norms
$$
\|x\|_{\ell_{p}^{w}(X)}:=
\sup_{x'\in B_{X'}} \Big ( \sum_{n=1}^{\infty}
|\langle x', x_n \rangle|^{p}\Big )^{1/p}
$$
and
$$
\|x\|_{E(X)}:= \|(\|x_n\|_{X})\|_{E},
$$
respectively. It is well-known that the Pietsch Domination Theorem implies that any
\mbox{$p$-summing} operator $T: X \to Y$, $1 \le p <\infty$ is a Dunford-Pettis operator, i.\,e.
$T$ transforms weakly convergent sequences into norm convergent sequences,
and thus by Rosenthal's $\ell_1$-Theorem it is compact whenever $X$ does not contain
a copy of $\ell_{1}$. In general this is not true for $(r,p)$-summing
operators as has been noted by Bennett \cite{bennett}, namely the inclusion
map $\ell_{r} \Id \ell_{\infty}$ is $(r,1)$-summing for any $1<r<\infty$,
however not compact. But even in our  more general case the situation becomes
more favorable for operators acting between special Banach spaces (see also Corollary~\ref{calkin}).
\begin{samepage}
\begin{lemma}
\label{compactm} Let $Y$ be a Banach space and $E$  a Banach
sequence space with $\|e_n\|_E=1$ for all $n$. Then the following holds true:
\begin{enumerate}
\item [(a)] If $T \in \Pi_{E,p }(\ell_{p'}, Y)$ with $1<p<\infty$
and $\ell_p \Id E \Id c_{0}$, then
$T$ is a compact operator.
\item [(b)] If $T \in \Pi_{E, 1}(c_0, Y)$ with $\ell_1 \Id
E \Id c_0$, then $T$ is a compact operator.
\end{enumerate}
\end{lemma}
\end{samepage}
\proof
(a) Suppose $T$ is not compact. Then $T$ is no Dunford-Pettis operator
by the reflexivity of $\ell_{p'}$. Thus there exists a sequence
$(x_n)$ in $ \ell_{p'}$ such that $x_n \to 0$ weakly
and $\|Tx_n\|_{Y} \geq C$ for all $n$ with some constant $C>0$.
In consequence $Tx_n \to 0$ weakly in $Y$ and $\|x_n\|_{\ell_{p'}} \geq C/\|T\|$.
Passing to a subsequence, we may assume by the  Bessaga-Pe{\l}czy\'nski Selection
Theorem that $(x_{n})$ is equivalent to a block basis of the unit vector
basis in $\ell_{p'}$ and thus to the unit vector basis in $\ell_{p'}$.
Then $(x_n)$ is weakly $p$-summable in $\ell_{p'}$ since clearly $(e_n)$ is.
But $T: \ell_{p'} \to Y$ is $(E,p)$-summing, hence $(\|Tx_n\|_Y) \in E$, and in particular
$(\|Tx_n\|_Y) \in c_0$, a contradiction.
For (b) we similarly show that $T: c_0 \to Y$ is a Dunford-Pettis operator,
and thus compact since $c_0$ does not contain a copy of $\ell_1$. \qed
\par
 In the following three lemmas we fix $1 \le p < \infty$, and $E$
will always be a Banach sequence space such that $\ell_p \Id E$ and $\norm{e_n}_E=1$ for all $n$.
\begin{lemma}
\label{composition} For an operator $T:X \rightarrow Y$
 between Banach spaces the following are equivalent:
\begin{enumerate}[(a)]
\item $T \in \Pi_{E,p}$, and $\pi_{E,p}(T) \le C$.
\item For all $m$ the map $\Phi^m(T): \LL(\ell_2^m,X) \rightarrow E_m(Y), \quad S \mapsto (TSe_i)$ has norm
$\le C$.
\item $\pi_{E,p}(TS) \le C$ for all $m$ and $S \in \LL(\ell_{p'}^m,X) $ with $\norm{S} \le 1$.
\end{enumerate}
In particular, in this case,
\begin{equation}
\label{discrete}
\pi_{E,p}(T) = \sup_m \norm{\Phi^m(T)} = \sup_m \{ \pi_{E,p}(TS) \, | \, \norm{S: \ell_2^m
\rightarrow X} \le 1 \}.
\end{equation}
\em
The proof follows immediately from the definition and the standard observation
that for each \mbox{$S=\sum_{j=1}^m e_j \otimes x_j \in \LL(\ell_{p'}^m, X)$}
$$
\norm{S} = \sup_{x' \in B_{X'}} \left ( \sum_{j=1}^m |\langle x',x_j \rangle|^p \right
)^{1/p}.
$$
\end{lemma}
\par The following is an analogue of the well-known inclusion formulas (in the classical case due
to Kwapie\'{n} \cite{kwapien} and Tomczak-Jaegermann \cite{tj70}).
\begin{lemma}
\label{inclusion} For $1 \le p <q<\infty$ let $1 < r<\infty$ such that $1/r=1/p-1/q$. Then
 $$ \Pi_{E,p} \subset \Pi_{M(\ell_r,E),q},$$ and
for all $T \in \Pi_{E,p}$
$$\pi_{M(\ell_r,E),q}(T) \le c_p^E \cdot {c_q^{M(\ell_r,E)}}^{-1} \cdot \pi_{E,p}(T).$$
Moreover, if $X$ is a cotype~$2$ space, then for all Banach spaces $Y$
\begin{equation}
\label{inverse}
\Pi_{E,1}(X,Y)=\Pi_{M(\ell_2,E),2}(X,Y).
\end{equation}
\end{lemma}
\proof
 The first inclusion is easy: Let $T:X \rightarrow Y$ be $(E,p)$-summing. Then for
 \mbox{$x_1, \ldots, x_n \in X$} by the
H\"older inequality
\begin{align*}
\norm{(\norm{Tx_k})_1^n}_{M(\ell_r,E)}
&= \sup_{(\lambda_k)_1^n \in B_{\ell_r^n}}\norm{(\lambda_k \cdot\norm{Tx_k})_1^n}_E \\
& \le \pi_{E,p}(T) \cdot c_p^E \cdot \sup_{(\lambda_k)_1^n \in B_{\ell_r^n}}
\sup_{x' \in B_{X'}} \left ( \sum_1^n |\langle x',\lambda_k x_k \rangle|^p \right )^{1/p} \\
&= \pi_{E,p}(T) \cdot c_p^E \cdot   \sup_{x' \in B_{X'}}
\left ( \sum_1^n |\langle x',x_k \rangle|^q \right )^{1/q},
\end{align*}
which gives the claim. The reverse inclusion in the second part follows from the upcoming
Lemma~\ref{multi}:
By a well-known result of Maurey there exists a
constant $C>0$ such that for all $S \in \LL(\ell_\infty^n,X)$:
$\pi_2(S) \le C \cdot \norm{S}$ (see e.\,g. \cite[31.7]{df}). Then
for $T \in \Pi_{M(\ell_2,E),2}(X,Y)$ and $S \in \LL(\ell_\infty^n,X)$ with
$\norm{S}\le 1$ we obtain, together with Lemma~\ref{multi},
$$
\pi_{E,1}(TS) \le \pi_{M(\ell_2,E),2}(T) \cdot \pi_2(S)  \le
C \cdot \pi_{M(\ell_2,E),2}(T),
$$
which by Lemma~\ref{composition} implies $T \in \Pi_{E,1}$.
\qed
\\[15pt]
As announced it remains to prove the following:
\begin{lemma}
\label{multi}
For $1 \le p < q<\infty$ let $1<r<\infty$ such that $1/r=1/p-1/q$. Then
$$
\Pi_{M(\ell_r,E),q} \circ \Pi_r \subset \Pi_{E,p},
$$
and
$$
\pi_{E,p}(TS) \le \pi_{M(\ell_r,E),q}(T) \cdot \pi_r(S)
$$
for $S \in \Pi_r(X,Y)$ and $T \in \Pi_{M(\ell_r,E),q}(Y,Z)$.
\end{lemma}
\proof Let $S $ and $T$ be as in the proposition. By the Pietsch Domination
Theorem there exists a regular Borel probability measure $\mu$ on $B_{X'}$ such that for all $x \in X$
$$
\norm{Sx} \le \pi_r(S) \cdot \left ( \int_{B_{X'}} |\langle x',x \rangle|^r d\mu(x') \right )^{1/r}.
$$
Now take $0 \neq x_1, \ldots,x_n \in X$ and put for $k=1, \ldots,n$
$$
x_k^0 := \left ( \int_{B_{X'}} |\langle x',x_k \rangle|^p d\mu(x') \right )^{-1/r} \cdot x_k.
$$
Then by the H\"older Inequality (and $c_q^{M(\ell_r,E)} \le c_p^E$)
\begin{align*}
\norm{(\norm{TSx_k})_1^n}_E &\le \norm{(\norm{TSx_k^0})_1^n}_{M(\ell_r,E)} \cdot
\left ( \sum_1^n \int_{B_{X'}} |\langle x',x_k \rangle|^p d\mu(x') \right )^{1/r} \\
& \le \pi_{M(\ell_r,E),q}(T) \cdot c_p^E \cdot \sup_{y' \in B_{Y'}}
\left ( \sum_1^n |\langle y',Sx_k^0 \rangle|^q \right )^{1/q} \\
& \qquad \cdot \left ( \sum_1^n \int_{B_{X'}}
 |\langle x',x_k \rangle|^p d\mu(x') \right )^{1/r}.
\end{align*}
Now complete the proof exactly as in \cite{tj70}.
\qed
\par
As in the classical case of $(r,2)$-summing operators, the theory of $(F,2)$-summing operators is
deeply connected to the theory of $s$-numbers. In our case a crucial tool is an extension of an inequality
due to K\"onig, which can be proved exactly as in \cite[2.a.3]{koenig}.
\begin{prop}
\label{weyl}
Let $F$ be a maximal symmetric Banach sequence space such that $\ell_2 \Id F$.
Then $\Pi_{F,2}\subset \ui_F^x$. In particular, for all $T \in \Pi_{F,2}$ and $k$
\begin{equation}
\label{weyleq}
x_k(T) \le \lambda_F(k)^{-1} \cdot c_2^F \cdot \pi_{F,2}(T).
\end{equation}
\end{prop}
The above result allows to give a different proof of the Lemma~\ref{compactm} in the case $p=2$.
\begin{cor}
\label{calkin}
For any Banach space $Y$ any $(F,2)$-summing operator $T: \ell_2 \rightarrow Y$ is compact
whenever $ \ell_2 \Id F \Id c_0$.
\end{cor}
\proof By Proposition~\ref{weyl} we have $\Pi_{F,2}(\ell_2,Y) \subset \ui_F^x(\ell_2,Y) =
\ui_F^a(\ell_2,Y)$ which clearly gives the claim. \qed
\\[15pt]
See Section~6 for the fact that for $2$-convex $F$ the ideals $\Pi_{F,2}$ and
the unitary ideal $\ui_F$ coincide on Hilbert spaces.

 \section{$\pmb{(E,1)}$-summing identity maps}
The well-known results of Bennett \cite{bennett} and Carl \cite{carl}
(proved independently) assure that for $1 \le u \le 2$ the identity map
$\id: \ell_u \Id \ell_2$ is absolutely $(u,1)$-summing. In
 \cite{mastylo} an extension within the setting of Orlicz sequence spaces is presented.
\par
Using interpolation theory we prove as our main result the following proper extension:
\begin{theo}
\label{main}
Let $E$ be a $2$-concave symmetric Banach sequence space. Then the identity map
$\id: E \Id \ell_2$ is $(E,1)$-summing. In other words, for every unconditionally summable
sequence $(x_n)$ in $E$ the scalar sequence $(\norm{x_n}_{\ell_2})$ is contained in $E$.
\end{theo}
The following lemmas are essential:
\begin{lemma}
\label{power}
Let $(E_0,E_1)$ be a relative Calder\'{o}n couple of maximal symmetric Banach sequence spaces
and $E$ an interpolation
space with respect to $(E_0,E_1)$. Then $E^p$  for all $0<p<1$ is an interpolation space with
respect to $(E_0^p,E_1^p)$.
\end{lemma}
\proof It is enough to show that $E^{p}$ is a relative $K$-space with respect
to $(E_{0}^{p}, E_{1}^{p})$, i.e. if whenever $x\in E^{p}$ and
$y\in E_{0}^{p} + E_{1}^{p}$ satisfy
$$
K(t, y; E_{0}^p, E_{1}^p) \leq K(t, x; E_{0}^p, E_{1}^p) \, \, \,
\mbox{for all } t>0,
$$
then it follows that $y\in E^p$.
\par The claim follows from the well-known and easily verified equivalence
for the $K$-functionals, namely
$$
 K(t,x; E_0^{p},E_1^{p}) \asymp
K(t^{1/p},|x|^{1/p};E_0,E_1)^p
$$
for any $x \in E_0^p+E_1^p$ and $t>0$, and the fact that $E$ is a relative $K$-space with
respect to $(E_0, E_1)$.
\qed
\\[15pt]
As an immediate consequence we obtain
\begin{lemma}
\label{interpoldiagonal}
Let $E$ be a maximal symmetric Banach sequence space.
\begin{enumerate}[(a)]
\item If $E$ is $2$-convex, then it is an interpolation space with respect to the couple
$(\ell_2,\ell_\infty)$, i.\,e. there exists an exact interpolation functor $\cal{F}$
such that $E= \mathcal{F}(\ell_2,\ell_\infty)$.
\item
If $E$ is $2$-concave, then $M(\ell_2,E)$ is $2$-convex. In particular, $M(\ell_2,E)$ is an
interpolation space with respect to $(\ell_2,\ell_\infty)$.
\end{enumerate}
\end{lemma}
\proof (a) Without loss of generality we may assume that $\convex{2}(E)=1$. Then $E^2$ is a
 maximal symmetric Banach sequence space, and by Mitiagin~\cite{mitiagin}
(see also \cite[1.b.10]{koenig}) this implies that $E^2$ is an
interpolation space with respect to $(\ell_1,\ell_\infty)$. The claim now follows by the preceding
 lemma and the fact that $(\ell_1,\ell_\infty)$ is a relative Calder\'{o}n couple
 (see e.\,g. \cite[2.6.9]{bk}). \\
 (b) Without loss of generality we may assume that $\concave{2}(E)=1$. Then $E^\times$ is
 $2$-convex with $\convex{2}(E)=1$, hence $(E^\times)^2$ and therefore also
 $((E^\times)^2)^\times$ are normed.
Consequently, $M(\ell_2,E)=(((E^\times)^2)^\times)^{1/2}$ is $2$-convex.
\qed
\\[15pt]
For the sake of
 completeness  we give a proof of the following easy and well-known result:
\begin{lemma}
\label{vector}
Let $E$ and $F$ be Banach sequence spaces, $X$ a Banach space and
${\cal{F}}$ an exact interpolation functor. Then
$$
\norm{\id : {\cal{F}}(E_n(X),F_n(X)) \Id {\cal{F}}(E_n,F_n)(X)} \le 1.
$$
\end{lemma}
\proof For any given $x_1, \ldots,x_n \in X$ let $x_1', \ldots, x_n' \in X'$ be such that
$\norm{x_i'}=1$ and \mbox{$\langle x_i',x_i \rangle=\norm{x_i}$.} Then for
$T: \R^n(X) \rightarrow \R^n$
defined by $T((y_i)_1^n):= (\langle x_i',y_i \rangle)_1^n$ we obviously have $\norm{T: E_n(X) \rightarrow
E_n} \le 1$ and $\norm{T: F_n(X) \rightarrow
F_n} \le 1$, hence
$$
\norm{T: {\cal{F}}(E_n(X),F_n(X)) \rightarrow {\cal{F}}(E_n,F_n)} \le 1.
$$
\begin{samepage}
Thus
\begin{align*}
\norm{(x_i)_1^n}_{{\cal{F}}(E_n,F_n)(X)} &=
\norm{(\norm{x_i})_1^n}_{{\cal{F}}(E_n,F_n)} \\
& = \norm{(\langle x_i',x_i \rangle)_1^n}_{{\cal{F}}(E_n,F_n)} \\
& \le \norm{(x_i)_1^n}_{{\cal{F}}(E_n(X),F_n(X))}.
\end{align*}
\qed
\end{samepage}
\par The following lemma partially extends (in the lattice case) results of Pisier and Kouba on
the complex interpolation
 of spaces of operators (see \cite{kouba}, \cite{pisier90} and also \cite{dm99}).
 Recall that $\ell_2 = M(\ell_2,\ell_1)$ and $\ell_\infty=M(\ell_2,\ell_2)$;
 then the statement below says that under the given assumption the interpolation property of the
 spaces of multipliers (diagonal operators) can be transferred into the corresponding interpolation property
 of the associated spaces of bounded operators (at least in the finite-dimensional case). Note
 that a formula for the reverse inclusion holds whenever ${\cal{F}}(\ell_1,\ell_2) \Id E$.
\begin{lemma}
\label{kouba}
For a $2$-concave symmetric Banach sequence space $E$ let ${\cal{F}}$ be an exact interpolation
functor such that $M(\ell_2,E) \Id {\cal{F}}(\ell_2,\ell_\infty)$. Then
\begin{equation}
\label{koubaformula}
\sup_{m,n} \norm{\id: \LL(\ell_2^m,E_n) \Id {\cal{F}}(\LL(\ell_2^m,\ell_1^n),
\LL(\ell_2^m,\ell_2^n))}
\le \sqrt{2} \cdot c_{M(\ell_2,E)}^{\mathcal{F}(\ell_2,\ell_\infty)} \cdot \concave{2}(E).
\end{equation}
\end{lemma}
\proof Let $T \in \LL(\ell_2^m,E_n)$. By a variant of the Maurey--Rosenthal Factorization Theorem
(see
\cite[4.2]{defant} and also \cite{lust}) there exist an operator $R \in \LL(\ell_2^m,\ell_2^n)$ and $\lambda
\in \R^n$ such that
$$
\norm{R} \cdot \norm{\lambda}_{M(\ell_2^n,E_n)} \le \sqrt{2} \cdot \concave{2}(E) \cdot
\norm{T}
$$
and $T$ factorizes as follows:
\begin{center}
\resetparms
\Vtriangle[\ell_2^m`E_n`\ell_2^n;
T`R`M_\lambda].
\end{center}
Obviously the map $\Phi$ defined by
$$
\Phi (\mu) := M_{\mu} \circ R, \quad \mu \in \R^n
$$
 maps the couple
$(\ell_{2}^{n}, \ell_{\infty}^{n})$ into the couple
$
({\cal L}(\ell_{2}^{m}, \ell_{1}^{n}),
{\cal L}(\ell_{2}^{m}, \ell_{2}^{n}))
$
such that both restrictions have norm less or equal $\|R\|$. Hence by the interpolation property
and
the assumption
 the restriction map
$$
\Phi: M(\ell_2^n,E_n) \rightarrow {\cal{F}}(\LL(\ell_2^m,\ell_1^n),\LL(\ell_2^m,\ell_2^n))
$$
has norm $\le c_{M(\ell_2,E)}^{\mathcal{F}(\ell_2,\ell_\infty)} \cdot\norm{R}$. Thus we obtain
\begin{samepage}
\begin{align*}
\norm{T}_{{\cal{F}}(\LL(\ell_2^m,\ell_1^n),\LL(\ell_2^m,\ell_2^n))} & =
\norm{M_\lambda \circ R}_{{\cal{F}}(\LL(\ell_2^m,\ell_1^n),\LL(\ell_2^m,\ell_2^n))} \\
& \le c_{M(\ell_2,E)}^{\mathcal{F}(\ell_2,\ell_\infty)} \cdot \norm{R} \cdot \norm{\lambda}_{M(\ell_2^n,E_n)} \\
& \le \sqrt{2} \cdot c_{M(\ell_2,E)}^{\mathcal{F}(\ell_2,\ell_\infty)} \cdot \concave{2}(E) \cdot \norm{T}_{\LL(\ell_2^m,E_n)}.
\end{align*}
\qed
\end{samepage}
\\[20pt]
Now we are ready to give a proof of Theorem~\ref{main}: According to Lemma~\ref{interpoldiagonal}
let ${\cal{F}}$ be an interpolation
functor with $M(\ell_2,E)={\cal{F}}(\ell_2,\ell_\infty)$.
We consider the mapping
$$
\Phi^{m,n}: ({\cal L}(\ell_{2}^{m},\ell_{1}^{n}),
{\cal L}(\ell_{2}^{m}, \ell_{2}^{n})) \to
(\ell_{2}^{m}(\ell_{2}^{n}), \ell_{\infty}^{m}(\ell_{2}^{n}))
$$
defined by $\Phi^{m,n}(S):= (Se_i)_{1}^{m}$. By \eqref{discrete} we have
$$
\sup_{m} \|\Phi^{m,n}: {\cal L}(\ell_{2}^{m}, \ell_{1}^{n})
\to \ell_{2}^{m}(\ell_{2}^{n})\| = \pi_{2}(\mbox{id}: \ell_{1}^{n}
\hookrightarrow \ell_{2}^{n}\| =1
$$
and
$$
\sup_{m} \|\Phi^{m,n}: {\cal L}(\ell_{2}^{m}, \ell_{2}^{n})
\to \ell_{\infty}^{m}(\ell_{2}^{n})\| = \|\mbox{id}: \ell_{2}^{n}
\hookrightarrow \ell_{2}^{n}\| = 1.
$$
Then by the interpolation property we obtain that
$$
\Phi^{m,n}: {\cal{F}}(\LL(\ell_2^m,\ell_1^n),\LL(\ell_2^m,\ell_2^n)) \rightarrow
{\cal{F}}(\ell_2^m(\ell_2^n),\ell_\infty^m(\ell_2^n))
$$
also has norm $\le 1$. Now by the preceding lemma
$$
\norm{\Phi^{m,n}: \LL(\ell_2^m,E_n) \rightarrow M(\ell_2^m, E_m)(\ell_2^n)} \le \sqrt{2} \cdot
c_{M(\ell_2,E)}^{\mathcal{F}(\ell_2,\ell_\infty)} \cdot  \concave{2}(E).
$$
Hence, since $\sup_m \norm{\Phi^{m,n}} = \pi_{M(\ell_2,E),2}(\id: E_n \Id \ell_2^n)$,
$$
\pi_{M(\ell_2,E),2}(\id : E_n \Id \ell_2^n) \le \sqrt{2} \cdot
c_{M(\ell_2,E)}^{\mathcal{F}(\ell_2,\ell_\infty)}
\cdot \concave{2}(E),
$$
and since $\bigcup_n E_n$ is dense in $E$, this implies $(\id : E \Id \ell_2) \in
\Pi_{M(\ell_2,E),2}$. The final statement then follows from \eqref{inverse}. \qed
\par
Theorem~\ref{main} is best possible in the following sense:
\begin{cor}
\label{equivalent}
Let $E$ and $F$ be $2$-concave  symmetric Banach sequence spaces. Then
\begin{equation}
\pi_{F,1}(\id: E_n \Id \ell_2^n) \asymp \norm{\id: E_n \Id F_n}.
\end{equation}
In particular, $\id: E \Id \ell_2$ is $(F,1)$-summing if and only if $E \Id F$.
\end{cor}
\proof The upper estimate follows from Theorem~\ref{main} by factorization; for the lower estimate
 we may assume without loss of generality that $\concave{2}(E)=\concave{2}(F)=1$. Observe that
 for $\lambda \in \R^n$ one has $\norm{\lambda}_{M(\ell_2^n,F_n)} \le
 \pi_{M(\ell_2,F),2}(M_\lambda: \ell_2^n \rightarrow \ell_2^n)$ (simply take in the definition
 of $(M(\ell_2,F),2)$-summing $x_i=e_i$), hence, by Lemma~\ref{composition} and
 Lemma~\ref{inclusion} as well as
 \eqref{diagonalpower},
 \begin{samepage}
\begin{align*}
\pi_{F,1}(\id:E_n \Id \ell_2^n) & \ge \pi_{M(\ell_2,F),2}(\id:E_n \Id \ell_2^n) \\
&= \sup_m \sup_{\norm{S: \ell_2^m \rightarrow E_n} \le 1}
\pi_{M(\ell_2,F),2}(\ell_2^m \xrightarrow{S} E_n \xrightarrow{\id} \ell_2^n) \\
& \ge \sup_{\norm{\lambda}_{M(\ell_2^n,E_n)} \le 1} \pi_{M(\ell_2,F),2}(M_\lambda: \ell_2^n
\rightarrow \ell_2^n) \\
& \ge \sup_{\norm{\lambda}_{M(\ell_2^n,E_n)} \le 1} \norm{\lambda}_{M(\ell_2^n,F_n)} \\
& = \norm{\id : M(\ell_2^n, E_n) \Id M(\ell_2^n,F_n)} \\
& = \norm{\id : (((E_n^\times)^2)^\times)^{1/2} \Id (((F_n^\times)^2)^\times)^{1/2}} \\
& = \norm{\id : E_n \Id F_n}.
\end{align*}
\qed
\end{samepage}
\par
As a counterpart to Corollary~\ref{equivalent} we show that in Theorem~\ref{main} the Hilbert
space $\ell_2$ is minimal in the following sense:
\begin{cor}
\label{bestpossible}
Let $E$ and $F$ be maximal symmetric Banach sequence where E is $2$-concave. Then
\begin{equation}
\pi_{E,1}(\id: E_n \Id F_n) \asymp \norm{\ell_2^n \Id F_n}.
\end{equation}
In particular, $\id: E \Id F$ is $(E,1)$-summing if and only if $\ell_2 \Id F$.
\end{cor}
\proof Again the upper estimate obviously follows by factorization from Theorem~\ref{main}. For the
lower estimate note that by \cite[p.~237 (3)]{CD97} (which is also valid for
$\lfloor n/2 \rfloor +1$ instead of $[n/2]$)
$$
x_{\lfloor n/2 \rfloor +1}(\id : E_n \Id F_n) \ge \frac{1}{\sqrt{2}} \cdot
\frac{\norm{\id :\ell_2^n \Id F_n}}{\norm{\id: \ell_2^n \Id E_n}},
$$
hence, by Lemma~\ref{inclusion}, \eqref{weyleq} and \cite[p.~237 (1)]{CD97},
\begin{samepage}
\begin{align*}
\pi_{E,1}(\id : E_n \Id F_n) & \ge \pi_{M(\ell_2,E),2}(\id : E_n \Id F_n) \\
& \ge \frac{ \norm{\id: \ell_2^n \Id F_n}}{\sqrt{2}} \cdot \frac{\norm{\id : \ell_2^{\lfloor n/2 \rfloor+1} \Id
E_{\lfloor n/2 \rfloor+1}}}{\norm{\id: \ell_2^n \Id E_n}}  \\
&\ge \frac{\norm{\id: \ell_2^n \Id F_n}}{\sqrt{2}} \cdot \frac{\norm{\id : \ell_2^{n-\lfloor n/2 \rfloor} \Id
E_{n-\lfloor n/2 \rfloor}}}{\norm{\id: \ell_2^n \Id E_n}}   \\
& \ge \frac{\norm{\id: \ell_2^n \Id F_n}}{\sqrt{2}} \cdot
\frac{a_{\lfloor n/2 \rfloor +1}(\id: \ell_2^n \Id E_n)}{\norm{\id: \ell_2^n \Id E_n}}
 \\
&\ge \frac{\norm{\id: \ell_2^n \Id F_n}}{2}.
\end{align*}
\qed
\end{samepage}
\par We note that in general the assumption that a symmetric sequence
space $E$ is 2-concave is essential in Theorem~\ref{main}, even in the class
of Orlicz sequence spaces. This follows from the following proposition
and the fact that there is an example, constructed by Kalton \cite{kalton}
(see also \cite[4.c.3]{lt77}), of an Orlicz sequence space $\ell_{\varphi}$
such that the identity map $\mbox{id}: \ell_{\varphi} \hookrightarrow \ell_{2}$
is not a strictly singular operator, i.e. $\mbox{id}$ is an isomorphism
on some infinite dimensional closed subspace of $\ell_{\varphi}$.
\begin{prop}
\label{ss1}
Let $E \Id \ell_2$ be a Banach sequence
space not equivalent to $\ell_2$. Then  the identity map $\id: E \Id \ell_2$ is strictly singular
 whenever it is $(E,1)$-summing.
\end{prop}
\proof
Suppose that $\mbox{id}: E \hookrightarrow \ell_2$ is not strictly singular.
Thus there exists an infinite dimensional closed subspace
$X$ of $E$ such that the restriction of $\mbox{id}$ to $X$
is an isomorphism from $X$ into $\ell_{2}$. Let $P: \ell_{2} \to X$ be
a continuous linear projection. By assumption
$\mbox{id}: E \hookrightarrow \ell_{2}$ is $(E, 1)$-summing,
and thus by Lemma~\ref{inclusion}
$T=\mbox{id}\circ P: \ell_2 \to \ell_2$ is $(M(\ell_2, E), 2)$-summing.
Since $E \neq \ell_{2}$, we get $M(\ell_{2}, E) \hookrightarrow c_0$
(see \eqref{czero}). An application of Lemma~\ref{compact} yields that $T$ is compact which
  contradicts the fact that $T$ on $X$ is the identity.
\qed
\\[15pt] In view of Theorem~\ref{main}
 the following trivial consequence  seems to be of independent
interest.
\begin{cor}
\label{ss2}
If $E$ is a symmetric Banach sequence space not
equivalent to $\ell_2$ and such that the inclusion map
$\mbox{id}: E \hookrightarrow \ell_2$  is not strictly singular,
then $E$ does not have cotype $2$.
\\[15pt]
\em Combining Proposition~\ref{ss1} and Corollary~\ref{ss2} we see that Kalton's example $\ell_\varphi$
is not $2$-concave and $\id: \ell_\varphi \Id \ell_2$ is not $(\ell_\varphi,1)$-summing.
\end{cor}

\section{Applications to approximation numbers of identity operators}
Of
special interest for applications (e.\,g. in approximation theory)
are formulas for the asymptotic behavior of approximation numbers of
finite-dimensional identity
operators. One of the first well-known results in this direction
is due to Pietsch \cite{pietsch74}: For $1 \le k \le n$ and $1 \le p<q \le \infty$
\begin{equation}
\label{lpformula} a_k(\id: \ell_q^n \Id \ell_p^n) = (n-k+1)^{1/p-1/q}.
\end{equation}
For the special case $1 \le p < q=2$ let us rewrite this as follows:
\begin{equation}
\label{l2formula} a_k(\id: \ell_2^n \Id \ell_p^n) = \frac{\lambda_{\ell_p}(n-k+1)}{(n-k+1)^{1/2}}.
\end{equation}
 Using Theorem~\ref{main}
we show this formula---at
least asymptotically---for all $2$-concave symmetric Banach sequence spaces $E$
instead of $\ell_p$:
\begin{theo}
\label{formula1}
Let $E$ be a $2$-concave symmetric Banach sequence space. Then for all \mbox{$1 \le k \le n$}
\begin{equation}
\label{formula2}
a_k(\id: \ell_2^n \Id E_n) \asymp \frac{\lambda_E(n-k+1)}{(n-k+1)^{1/2}}.
\end{equation}
\end{theo}
\noindent
The proof needs the special case $k=1$, a result due to Szarek and
Tomczak-Jaegermann~\cite[Proposition~2.2]{stj}:
Under the assumption of the theorem
\begin{equation}
\label{stj}
a_1(\id : \ell_2^n \Id E_n) = \norm{\id: \ell_2^n \Id E_n} \asymp
\frac{\lambda_E(n)}{n^{1/2}}.
\end{equation}
Proof of Theorem~\ref{formula1}:
First we claim that it is enough to show
\begin{equation}
\label{formula3}
a_k(\id: \ell_2^n \Id E_n) \asymp \norm{\textstyle{\sum_1^{n-k+1} e_i}}_{M(\ell_2,E)}.
\end{equation}
Indeed, the right hand side in \eqref{formula3} is obviously equal to $\norm{\id: \ell_2^{n-k+1}
 \Id E_{n-k+1}}$, and by \eqref{stj} this is
 asymptotically equivalent to the right hand side in \eqref{formula2}.
\\[10pt]
The upper estimate in \eqref{formula3} is straightforward:
Put $\lambda:= \sum_1^{n-k+1} e_i \in \K^n$ and \mbox{$\mu:= \sum_{n-k+2}^n e_i \in \K^n$.}
Since the diagonal operator $M_\mu: \ell_2^n \rightarrow E_n$ has rank $k-1$, we obtain
$$
a_k(\id: \ell_2^n \Id E_n) \le \norm{\id-M_\mu}=\norm{M_\lambda}=
\norm{\textstyle{\sum_1^{n-k+1} e_i}}_{M(\ell_2,E)}.
$$
On the other hand, by a result of \cite{CD92}
$$
a_k(\id:\ell_2^n \Id E_n)=x_{n-k+1}(\id:E_n \Id \ell_2^n)^{-1},
$$
so that the lower estimate in \eqref{formula3} follows from
\begin{equation}
\label{formula4}
x_k(\id:E_n \Id \ell_2^n) \prec \norm{\textstyle{\sum_1^k e_i}}_{M(\ell_2,E)}^{-1}.
\end{equation}
In order to check \eqref{formula4} note that by Theorem~\ref{main} the identity map
$\id: E \Id \ell_2$ is $(E,1)$-summing. Hence by
Lemma~\ref{inclusion} it is also $(M(\ell_2,E),2)$-summing, and by the generalized
K\"onig inequality~\eqref{weyleq} we obtain
\begin{align*}
x_k(\id : E_n \Id \ell_2^n) &\le \norm{\textstyle{\sum_1^k e_i}}_{M(\ell_2,E)}^{-1} \cdot
\pi_{M(\ell_2,E),2}(\id: E_n \Id \ell_2^n) \\
&\le  \norm{\textstyle{\sum_1^k e_i}}_{M(\ell_2,E)}^{-1} \cdot
\pi_{M(\ell_2,E),2}(\id: E \Id \ell_2),
\end{align*}
which completes the proof. \qed \\[10pt]
To illustrate formula~\eqref{formula2} we consider Lorentz and Orlicz sequence spaces.
\begin{samepage}
\begin{cor}
\begin{enumerate}[(a)]
\item
Let $1 <p<2$ and $1 \le q \le 2$. Then for all $1 \le k \le n$
\begin{equation}
\label{lorentz}
a_k( \id : \ell_2^n \Id \ell_{p,q}^n) \asymp (n-k+1)^{1/p-1/2}.
\end{equation}
\item
\label{lorentz2}
Let $1<p<2$ and $w$ be a Lorentz sequence such that $n \cdot w_n^{2/(2-p)} \asymp \sum_1^n
w_i^{2/(2-p)}$. Then for all $1 \le k \le n$
\begin{equation}
a_k(\id: \ell_2^n \Id d_n(w,p)) \asymp (n-k+1)^{1/p-1/2} \cdot w_{n-k+1}^{1/p}.
\end{equation}
\item
\label{orlicz}
Let $\varphi$ be an Orlicz function such that the function $t \mapsto \varphi(\sqrt{t})$ is
equivalent to a concave function. Then
\begin{equation}
a_k(\id : \ell_2^n \Id \ell_\varphi^n) \asymp \frac{(\varphi^{-1}(1/(n-k+1)))^{-1}}{(n-k+1)^{1/2}}.
\end{equation}
 \end{enumerate}
\em Note that (a) is---asymptotical---the same result as for $\ell_p$ (see \eqref{lpformula});
although $\ell_{p,q}$ is
 ''very close`` to $\ell_p$, one may have expected an additional logarithmic term.
\end{cor}
\end{samepage}
\proof By Theorem~\ref{formula1} it is enough to ensure that all spaces considered in the corollary are
$2$-concave. For the Lorentz sequence spaces $\ell_{p,q}$ this is due to Creekmore~\cite{creekmore}
(see e.\,g. also \cite{defant}), for
the Lorentz sequence spaces $d(w,p)$ see Reisner~\cite{reisner}, and for Orlicz sequence spaces this
is contained in \cite{komarchev}.\qed
\section{Applications to eigenvalues of compact operators and unitary ideals}
By Pitt's Theorem every operator $T$ on $\ell_2$ with values in $\ell_u$, $1 \le u <2$, is
compact. The original Bennett--Carl result implies (see e.\,g. \cite[2.b.11]{koenig}) that its
sequence of singular numbers is contained in $\ell_r$, $1/r=1/u-1/2$, and by Weyl's Inequality
(see e.\,g. \cite[1.b.9]{koenig}) even its sequence of eigenvalues is contained in $\ell_r$.
Weyl's inequality also holds for arbitrary maximal symmetric Banach sequence spaces: If the
singular numbers of a compact operator on a Hilbert space is contained in a certain maximal symmetric
sequence space $F$, then the same is true for its sequence of eigenvalues (see \cite[1.b.10]{koenig}).
 Together with Theorem~\ref{main} this implies the following extension of the result mentioned
 above:
\begin{theo}
\label{compact}
Let $E \hookrightarrow \ell_2$ be a $2$-concave symmetric Banach sequence space not equivalent
 to $\ell_2$,
and $T \in \LL(\ell_2,\ell_2)$
be an operator with values in $E$. Then $T \in \ui_{M(\ell_2,E)}$. In particular, its
sequence of eigenvalues   $(\lambda_n(T))$ is contained in $M(\ell_2,E)$.
\end{theo}
\proof The assumption $E \not= \ell_2$ together with Corollary~\ref{equivalent} assures that
 the identity operator on $\ell_2$ is not contained in $\Pi_{M(\ell_2,E),2}$, and by a result
 of Calkin (see \cite[2.11.11]{eigenvalues}) it follows that every operator in
 $\Pi_{M(\ell_2,E),2}(\ell_2,\ell_2)$ is compact; alternatively one may directly use
 Lemma~\ref{compactm} together with \eqref{czero}. Now by Theorem~\ref{main} and the ideal property
 the operator $T : \ell_2 \xrightarrow{T} E \xrightarrow{\id} \ell_2$ is contained in
 $\Pi_{M(\ell_2,E),2}(\ell_2,\ell_2)$ and therefore compact, and by Proposition~\ref{weyl}
 the sequence of Weyl (=singular) numbers $(x_n(T))$ is contained in $M(\ell_2,E)$. The second
 claim now follows by Weyl's inequality mentioned above. \qed

\par
Next we discuss an alternative approach to Theorem~\ref{main} using interpolation
of unitary ideals; we first
illustrate our idea by considering the original result of Bennett and Carl:
\\[10pt]
Let $1\le u <2$. By Lemma~\ref{composition} and \eqref{inverse} the identity map
$I_u: \ell_u \Id \ell_2$ is absolutely $(u,1)$-summing whenever the composition $I_uS$
for any operator
$S: \ell_2 \rightarrow \ell_u$  is absolutely $(r,2)$-summing $(1/r=1/u-1/2)$.
By the (classical) Maurey--Rosenthal
Factorization Theorem there exist an operator $R \in \LL(\ell_2,\ell_2)$ and
 $\lambda \in \ell_r$ such that
 $S$ factorizes as follows:
\begin{center}
\resetparms
\Vtriangle[\ell_2`\ell_u`\ell_2;
S`R`M_\lambda].
\end{center}
Then obviously the operator $I_u M_\lambda: \ell_2 \rightarrow \ell_2$ is contained in the
Schatten-$r$-class $\ui_r$. By a
result of Mitiagin (see e.\,g. \cite[10.3]{djt}) $\ui_r = \Pi_{r,2}(\ell_2,\ell_2)$,
hence $I_u S = I_u M_\lambda R$ is absolutely $(r,2)$-summing which gives the claim. \qed
\\[10pt]
Mitiagin's result and its proof are of interpolative nature.
Alternatively, the inclusion \mbox{$\ui_r \subset \Pi_{r,2}(\ell_2,\ell_2)$} can be proved by complex interpolation of the
border cases $\Pi_{2,2}(\ell_2,\ell_2)=\ui_2$ and
$\Pi_{\infty,2}(\ell_2,\ell_2)=\LL(\ell_2,\ell_2)=\ui_\infty$: For $\theta:=2/r$
$$
\ui_r =[\ui_2,\ui_\infty]_\theta =[\Pi_{2,2}(\ell_2,\ell_2),\Pi_{\infty,2}(\ell_2,\ell_2)]_\theta
 \subset \Pi_{r,2}(\ell_2,\ell_2).
$$
The starting point for our alternative approach to Theorem~\ref{main} now is an
extension of Mitiagin's result for which we  need
 the following generalization of a result due to K\"onig (cf. \cite[2.c.10]{koenig}).
 \begin{samepage}
\begin{lemma}
\label{lemma35}
Let ${\cal F}$ be an interpolation functor
and $(E_0, E_1)$  a couple of Banach sequence spaces with
$\ell_{p} \hookrightarrow E_j$ and $\norm{e_n}_{E_j}=1$ for all $n$, $j=0,1$. Then for arbitrary
Banach spaces $X$ and $Y$, we have
$$
{\cal F} \big (\Pi_{E_0, p}(X, Y), \Pi_{E_1, p}(X,Y)\big)
\hookrightarrow \Pi_{{\cal F}(E_0, E_1), p}(X, Y).
$$
\end{lemma}
\end{samepage}
\proof For fixed vectors $x_1, \ldots, x_n \in X$ with $\sup_{x' \in B_{X'}}\sum_{j=1}^n
|\langle x',x_j \rangle|^p  \le 1$, we
define \mbox{$\Phi (T):= (Tx_j)_{j=1}^{n}$} for $T\in {\cal L}(X, Y)$.
Clearly
$$
\Phi : \big (\Pi_{E_0,p}(X, Y), \Pi_{E_1,p}(X,Y)\big)
\to (E_{0n}(Y), E_{1n}(Y))
$$
with norm $\le 1$.
By interpolation and Lemma~\ref{vector} we obtain that
$$
\Phi: {\cal F} \big (\Pi_{E_0, p}(X, Y), \Pi_{E_1, p}(X,Y)\big)
\to {\cal F}(E_{0n}, E_{1n})(Y)
$$
with norm $\le 1$. This yields that
$$
\pi_{{\cal F}(E_0, E_1),p}(T) \leq \norm{T}_{{\cal F}(\Pi_{E_0,p}(X,Y), \Pi_{E_1,p}(X,Y))}
$$
for any $T\in {\cal F}(\Pi_{E_0,p}(X,Y), \Pi_{E_1,p}(X,Y))$.
\qed
\\[10pt] The following theorem now admits the announced alternative proof of Theorem~\ref{main}
 exactly
 as it was done above for the original Bennett--Carl result---but it also seems to be of
 independent interest.
\begin{theo}
\label{interpolunitary}
Let $F$ be a $2$-convex maximal symmetric Banach sequence space. Then $\Pi_{F,2}(\ell_2,\ell_2)=
\ui_F$.
\end{theo}
\proof The inclusion $\Pi_{F,2}(\ell_{2}, \ell_2) \hookrightarrow
{\cal S}_{F}$ is contained in Proposition \ref{weyl}. For the reverse inclusion,
we note that if $E \hookrightarrow c_{0}$ is a maximal symmetric space, then
$E$ is an interpolation space with respect to $(\ell_{1}, c_{0})$.
Assume without loss of generality that $F \neq \ell_{\infty}$ and
\mbox{${\bf M_{(2)}}(F)=1$.} By the symmetry of $F$, it follows that
$F\hookrightarrow c_{0}$. In consequence $F^{2}$ is an interpolation
space with respect to $(\ell_{1}, c_{0})$, and thus by Lemma \ref{power}, $F$ is
an interpolation space with respect to $(\ell_{2}, c_0)$ (note that $(\ell_1,c_0)$ is a
relative Calder\'{o}n couple since $(\ell_1,\ell_\infty)$ is). Hence
there exists an exact interpolation functor ${\cal F}$ such that
$F={\cal F}(\ell_{2}, c_0)$. By applying Lemma \ref{lemma35}, we obtain
$$
{\cal F}(\Pi_{\ell_2, 2}(\ell_2, \ell_2),
\Pi_{c_0, 2}(\ell_{2}, \ell_{2})) \hookrightarrow \Pi_{F,2}(\ell_2, \ell_2)
$$
The claim now follows by the fact that ${\cal K}(\ell_{2}) \hookrightarrow
\Pi_{c_0, 2}(\ell_2, \ell_2)$ (${\cal K}(\ell_{2})$ denotes the space of compact
operators on $\ell_{2}$) and by a result on interpolation of unitary
ideals due to Arazy [Ara78]:
${\cal S}_{F} = {\cal S}_{{\cal F}(\ell_2, c_0)} =
{\cal F}({\cal S}_2, {\cal K}(\ell_2))$.
 \qed
\par Another nice application of Theorem~\ref{interpolunitary} is the following:
 \begin{cor}
Let $F$ be a maximal symmetric Banach sequence space such that
\mbox{$\ell_2 \Id F$.} Then
for every Banach space $X$ with $\dim X=n$
\begin{equation}
\label{id}
 \pi_{F,2}(\id_X) \ge C^{-1} \cdot \lambda_F(n) ,
\end{equation}
where $C>0$ is a constant depending on $F$ only. Moreover, if $F$ is $2$-convex, then even
\begin{equation}
\label{id2}
C^{-1} \cdot \lambda_F(n) \le \pi_{F,2}(\id_X) \le C \cdot \lambda_F(n).
\end{equation}
\end{cor}
\proof Let $G:=m_{\lambda_F}$ be the Marcinkiewicz sequence space associated to $F$.
By Proposition~\ref{weyl} and the fact that the continuous
inclusion $F \Id G$ is of norm one we get that
$$
\norm{(x_k(\id_X))_1^n}_G \le c_2^F \cdot \pi_{F,2}(\id_X).
$$
Since $G$ is a maximal symmetric Banach sequence space, it follows by
 the generalized Weyl Inequality \cite[2.a.8]{koenig} that
\begin{align*}
\norm{\textstyle{\sum_1^n e_i}}_F &= \sup_{1 \le k\le n} \norm{\textstyle{\sum_1^k e_i}}_F
\cdot 1 =
\norm{(\lambda_k(\id_X))_1^n}_G \\
&\le 2\sqrt{2e} \cdot \norm{(x_k(\id_X))_1^n}_G \le 2 \sqrt{2e} \cdot c_2^F \cdot \pi_{F,2}(\id_X),
\end{align*}
where $\lambda_k(\id_X)$ is the $k$-th eigenvalue of $\id_X$.
For the reverse estimate note that for an operator $T: Y \rightarrow Z$ of rank $n$ one has
$$
\pi_{F,2}(T)=\sup \{\pi_{F,2}(TS) \, | \,S \in \LL(\ell_2^n,Y), \quad \norm{S} \le 1 \}
$$
(check the proof of \cite[11.3 and 9.7]{tj}). Now let $S \in \LL(\ell_2^n,X)$. Then by
Theorem~\ref{interpolunitary}
$$
\pi_{F,2}(\id_X \circ S) \le \pi_{F,2}(\id_{\ell_2^n}) \cdot \norm{S} \le \tilde{C} \cdot
\norm{\textstyle{\sum_1^n e_i}}_F
\cdot \norm{S},
$$
where $\tilde{C}>0$ is a constant only depending on $F$.
\qed

\section{Complex interpolation in the range}

Based on the case $v=2$, Bennett and Carl also proved that for $1 \le u \le v \le 2$ the identity operator $\id: \ell_u
\Id \ell_v$ is absolutely $(r,2)$-summing whenever $1/r=1/u-1/v$. By using Theorem~\ref{main} and
complex interpolation in the range we obtain the following formal extension of our main result:
\begin{prop}
\label{complex}
Let $E$ be a $2$-concave symmetric Banach sequence space. Then for \mbox{$0 \le \theta<1$} the
identity operator $\id : E \hookrightarrow [\ell_2,E]_\theta$ is absolutely
$(M(\ell_2,E)^{1-\theta},2)$-summing.
\end{prop}
This now enables us to give an extension of the original Bennett--Carl result within the
framework of Lorentz sequence spaces:
\begin{cor}
\label{lorentzbc}
Let $1 <u_1 <v_1<2$ and $1\le u_2 \le v_2 \le 2$ be such that either $u_2=v_2=2$ or
$\frac{1/{v_1}-1/2}{1/{u_1}-1/2} = \frac{1/{v_2}-1/2}{1/{u_2}-1/2}$. Then
$$( \id: \ell_{u_1,u_2} \Id \ell_{v_1,v_2}) \in \Pi_{\ell_{r_1,r_2},2},$$
where $1/{r_1}=1/{u_1}-1/{v_1}$ and $1/{r_2}=1/{u_2}-1/{v_2}$.
\end{cor}
\proof This directly follows from the preceding proposition and the fact that for
$\theta:=\frac{1/{v_1}-1/2}{1/{u_1}-1/2}$ by the reiteration
theorem \cite[4.7.2]{BL} one has
$
[\ell_2, \ell_{u_1,u_2}]_\theta = \ell_{v_1,v_2};
$
finally,
$$M(\ell_2,\ell_{u_1,u_2})^{1-\theta}=\ell_{\tilde{u}_1,\tilde{u}_2}^{1-\theta}=\ell_{r_1,r_2},$$
where $1/{\tilde{u}_1}=1/{u_1}-1/2$ and $1/{\tilde{u}_2}=1/{u_2}-1/2$. \qed
\par
For our applications of this result we need the following two statements:
\begin{lemma}
Let $F$ be a maximal symmetric sequence space such that $\ell_2 \Id F$. Then for every invertible
operator $T: X \rightarrow Y$ between two $n$-dimensional Banach spaces and all $1 \le k \le n$
\begin{equation}
\label{gelfand}
c_k(T) \ge C^{-1} \cdot \frac{\lambda_F(n-k+1)}{\pi_{F,2}(T^{-1})},
\end{equation}
where $C:= 2 \sqrt{2e} \cdot c_2^F$.
\end{lemma}
\proof We copy the proof of \cite[p.~231]{CD97} for the $2$-summing norm. Take a subspace
$M \subset X$ with $\codim \, M <k$. Then
$$
n-k+1 \le \dim M,
$$
hence by \eqref{id}
$$
\norm{\textstyle{\sum_1^{n-k+1} e_i}}_F \le \norm{\textstyle{\sum_1^{\dim M} e_i}}_F \le
C \cdot \pi_{F,2}(\id_M).
$$
Clearly (by the injectivity of $\Pi_{F,2}$)
$$
\pi_{F,2}(\id_M) = \pi_{F,2}(\id: M \hookrightarrow X),
$$
therefore the commutative diagram
\begin{center}
\resetparms
\Vtriangle[M`X`Y;\id`T_{|M}`T^{-1}]
\end{center}
gives, as desired,
$
\norm{\textstyle{\sum_1^{n-k+1} e_i}}_F \le \norm{T_{|M}} \cdot C \cdot \pi_{F,2}(T^{-1}).
$
\qed
\\[15pt]
The following two results extend \eqref{formula2} and \eqref{lorentz}:
\begin{prop}
Let $E$ be a $2$-concave symmetric Banach sequence space. Then for \mbox{$0 \le \theta<1$} and all
$1 \le k \le n$
\begin{equation}
\label{ckinterpol}
a_k(\id: [\ell_2^n,E_n]_\theta \Id E_n) \asymp c_k(\id: [\ell_2^n,E_n]_\theta \Id E_n) \asymp
\left (\frac{\lambda_E(n-k+1)}{(n-k+1)^{1/2}}\right )^{1-\theta}.
\end{equation}
\end{prop}
\proof The estimate
$$
c_k(\id: [\ell_2^n,E_n]_\theta \Id E_n) \succ \left (
\frac{\lambda_E(n-k+1)}{(n-k+1)^{1/2}} \right )^{1-\theta}
$$
follows from Proposition~\ref{complex} and \eqref{gelfand} together with \eqref{stj}. Obviously
$$
c_k(\id: [\ell_2^n,E_n]_\theta \Id E_n) \le a_k(\id: [\ell_2^n,E_n]_\theta \Id E_n) \le
\norm{\id : [\ell_2^{n-k+1},E_{n-k+1}]_\theta \Id E_{n-k+1}},
$$
and by \eqref{powerreal} together with \eqref{stj}
$$
\norm{\id : [\ell_2^{n-k+1},E_{n-k+1}]_\theta \Id E_{n-k+1}}  \le \norm{\id: \ell_2^{n-k+1} \Id
E_{n-k+1}}^{1-\theta} \prec \left ( \frac{\lambda_E(n-k+1)}{(n-k+1)^{1/2}}
\right )^{1-\theta},
$$
which gives the claim. \qed
\begin{cor}
Let $1 <u_1 <v_1<2$ and $1\le u_2 \le v_2 \le 2$ be such that either $u_2=v_2=2$ or
$\frac{1/{v_1}-1/2}{1/{u_1}-1/2} = \frac{1/{v_2}-1/2}{1/{u_2}-1/2}$. Then for $1 \le k \le n$
\begin{equation}
\label{lorentzck}
c_k(\id: \ell_{v_1,v_2}^n \Id
\ell_{u_1,u_2}^n) \asymp a_k(\id: \ell_{v_1,v_2}^n \Id \ell_{u_1,u_2}^n)
 \asymp (n-k+1)^{1/{u_1}-1/{v_1}}.
\end{equation}
Moreover, formula \eqref{lorentzck} also holds in the case $1 <u_1<v_1<2$ and
$1 \le u_2 \le 2 \le v_2 \le \infty$.
\end{cor}
\proof The first part is clear (use the preceding proposition together with what was mentioned
in the proof of Corollary~\ref{lorentzbc}).
The lower estimates for the second part now follow by factorization:
$$
c_k(\id: \ell_{v_1,v_2}^n \Id \ell_{u_1,u_2}^n) \ge c_k(\id : \ell_{v_1,2}^n \Id \ell_{u_1,2}^n).
$$
The upper estimates are
again straightforward by real interpolation: Choose $0<\theta<1$ such that $1/{v_1}=(1-\theta)/2
+\theta/{u_1}$, then (with the help of \eqref{powerreal})
\begin{samepage}
\begin{align*}
c_k(\id: \ell_{v_1,v_2}^n \Id
\ell_{u_1,u_2}^n) &\le a_k(\id: \ell_{v_1,v_2}^n \Id \ell_{u_1,u_2}^n) \\
&\le \norm{\id : \ell_{v_1,v_2}^{n-k+1} \Id \ell_{u_1,u_2}^{n-k+1}} \\
& \asymp \norm{\id: (\ell_2^{n-k+1},\ell_{u_1,u_2}^{n-k+1})_{\theta,v_2} \Id \ell_{u_1,u_2}^{n-k+1}}
\\ & \le \norm{\id:\ell_2^{n-k+1} \Id \ell_{u_1,u_2}^{n-k+1}}^{1-\theta} \\
& \prec (n-k+1)^{1/{u_1}-1/{v_1}}.
\end{align*}
\qed
\end{samepage} \\[10pt]
We conjecture that formula \eqref{lorentzck} is true for all $1 <u_1 <v_1<2$ and
 $1 \le u_2 ,v_2 \le \infty$.

\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}

config: arxiv


TITLE: If $A$ is hermitian matrix s.t. $A^{99}=I_n$, show that $A=I_n$.
QUESTION [2 upvotes]: Let $A \in M_n(\mathbb{C})$ be a hermitian matrix that satisfies the property $A^{99}=I_n$. I want to show that $A=I_n$.
By definition, $A$ is a hermitian matrix if $a_{ij}=\overline{a_{ji}}$.
From the fact that $A^{99}=I_n$, we deduce that $A$ is invertible and its inverse is $A^{98}$. Does that help to get that $A=I_n$ ? Or do we use somehow the definition of a hermitian matrix?

REPLY [3 votes]: $A$ is hermitian so $\sigma(A) \subseteq \mathbb{R}$. On the other hand, the polynomial $z^{99} - 1$ annihilates $A$ so
$$\sigma(A) \subseteq \{\text{zeroes of }z^{99}-1\} \cap \mathbb{R} = \left\{e^{i\frac{2k\pi}{99}} : k \in \{0, \ldots, 98\}\right\} \cap \mathbb{R} = \{1\}$$
so $\sigma(A) = \{1\}$.
Since $A$ is diagonalizable, it diagonalizes to the indentity matrix so $A = I$.


TITLE: How to see this inequality
QUESTION [1 upvotes]: Consider
$$
\frac{dx}{dt}=\sqrt{x^2+1}+t^2,\qquad x\in\mathbb{R}.
$$
Then I do not see why it is
$$
\lvert\frac{dx}{dt}\rvert\leq 2\lvert x\rvert+t^2.
$$

REPLY [0 votes]: The steps that come to mind are
$$\begin{array}{}
\frac{dx}{dt}=\sqrt{x^2+1}+t^2 \\
|\frac{dx}{dt}|=|\sqrt{x^2+1}+t^2| \\
|\frac{dx}{dt}|\le|\sqrt{x^2+1}|+|t^2| & \qquad \text{Triangle Inequality} \\
|\frac{dx}{dt}|\le \sqrt{x^2+1} + t^2  \\
\end{array}$$
If we had $\sqrt{x^2+1} \le 2|x|$, then we could get your final result but
$$\begin{array}{}
\sqrt{x^2+1} \le 2|x| \\
x^2+1 \le 4x^2 \\
3x^2 \ge 1 \\
x^2 \ge \frac{1}{3} \\
|x| \ge \frac{1}{\sqrt3}
\end{array}$$
So we can say that
$$|\frac{dx}{dt}|\le \ 2|x| + t^2, \qquad \quad |x| \ge \frac{1}{\sqrt3}$$


text:
\begin{document}

\title{Empirical MSE Minimization to Estimate a Scalar Parameter}

\author{Cl\'{e}ment de Chaisemartin\thanks{University of California at Santa Barbara, clementdechaisemartin@ucsb.edu
} \and Xavier D'Haultf\oe{}uille
\thanks{de Chaisemartin: University of California at Santa Barbara (email: clementdechaisemartin@ucsb.edu); D'Haultf\oe uille: CREST-ENSAE (email: xavier.dhaultfoeuille@ensae.fr)}}

\maketitle ~\vspace{-1cm}

\begin{abstract}

We consider the estimation of a scalar parameter, when two estimators are available.  The first is always consistent. The second is inconsistent in general, but has  a smaller asymptotic variance than the first, and may be consistent if an assumption is satisfied. We propose to use the weighted sum of the two estimators with the lowest estimated mean-squared error (MSE). We show that this third estimator dominates the other two from a minimax-regret perspective: the maximum asymptotic-MSE-gain one may incur by using this estimator rather than one of the other estimators is larger than the maximum asymptotic-MSE-loss.
\end{abstract}
 \textbf{Keywords:} bias-variance trade-off, mean-squared error, consistent estimator, efficient estimator, statistical decision theory, minimax regret, local asymptotics.

\medskip
\textbf{JEL Codes:} C21, C23

\section{Introduction}

We consider the estimation of a scalar parameter $\beta_0$ when two estimators are available. The first, is $\sqrt{n}-$consistent. The second is inconsistent in general, but it has a smaller asymptotic variance than the first, and it may be $\sqrt{n}-$consistent if the data generating process satisfies an assumption $H_0$. Hereafter, those two estimators are respectively referred to as the consistent and efficient estimators.

\medskip
To fix ideas, we consider two of the many examples where this set-up we is applicable. In stratified randomized experiments, the parameter of interest is the average treatment effect (ATE). To estimate it, one may use the propensity score matching estimator \citep[see][]{hirano2003efficient}, which is $\sqrt{n}$-consistent and asymptotically normal under some assumptions. Alternatively, one may regress the outcome of interest on strata fixed effects and units' treatment status, and use the coefficient of the treatment in that regression. It follows from, e.g., Equation (3.3.7) in \cite{angrist2008mostly} that this estimator is $\sqrt{n}$-consistent and asymptotically normal for a weighted average of the effect of the treatment in each strata. One can also show that under some assumptions, the asymptotic variance of the strata fixed effects estimator is smaller than that of the propensity score matching estimator. So if the treatment effect is constant across strata, the strata fixed effects estimator is $\sqrt{n}$-consistent for the ATE, and it is more efficient than the propensity score matching estimator. But if the treatment effect is heterogeneous, the strata fixed effects estimator is inconsistent.

\medskip
Another example where our set-up is applicable is a linear and constant treatment effect model, where the treatment is potentially endogenous, but one has an instrumental variable at hand. Then, the 2SLS estimator is $\sqrt{n}-$consistent for the treatment effect. On the other hand, the OLS estimator is only $\sqrt{n}-$consistent if the treatment is actually exogenous, but its asymptotic variance is smaller than that of the 2SLS estimator.

\medskip
To estimate $\beta_0$, we propose to use $\widehat{\beta}_{MSE}$, the weighted sum of the consistent and efficient estimators with the lowest estimated mean-squared error (MSE). We show that this third estimator dominates the other two from a minimax-regret perspective: the maximum asymptotic-MSE-gain one may incur by using $\widehat{\beta}_{MSE}$ rather than one of the other estimators is larger than the maximum asymptotic-MSE-loss that one may incur by doing so.

\medskip
We also consider a family of pre-test estimators $(\widehat{\beta}_{MSE,\lambda})_{\lambda\geq 0}$, where $\lambda$ indexes the critical value used in the pre-test. First, we test whether the consistent and efficient estimators are equal. If the test is accepted, $\widehat{\beta}_{MSE,\lambda}$ is equal to the efficient estimator. If the test is rejected, $\widehat{\beta}_{MSE,\lambda}$ is equal to a convex combination of the consistent and efficient estimators. We show that such estimators have similar properties as $\widehat{\beta}_{MSE}$. However, $\widehat{\beta}_{MSE}$ dominates all of them from a minimax-regret perspective.

\medskip
We then extend the initial result by considering situations where one has two estimators at hand: one is $r_n-$consistent, where $r_n/\sqrt{n}\rightarrow 0$, and the other is inconsistent in general, but may be $\sqrt{n}-$consistent if the data generating process satisfies an assumption $H_0$. Such situations may for instance arise in regression discontinuity (RD) designs. Then, non-parametric estimators such as the one proposed by \cite{hahn2001identification} are $n^{2/5}-$consistent for the average treatment effect at the cut-off under weak conditions. On the other hand, the estimator using, say, linear regressions to the left and to the right of the cut-off without restricting the sample to observations in a narrow bandwidth around the cut-off is $\sqrt{n}-$consistent if the potential outcomes' CEFs are indeed linear in the running variable, but inconsistent otherwise. Again, we show that $\widehat{\beta}_{MSE}$ dominates the $r_n-$consistent estimator from a minimax-regret perspective, under mild assumptions.

\medskip
The idea of combining consistent and inconsistent, or unbiased and biased  estimators of a parameter has a long tradition in statistics and econometrics. \cite{green1991james} have proposed an estimator related to $\widehat{\beta}_{MSE}$, in the context of a normal model with known variances. Their estimator is related to the shrinkage estimator in \cite{charles1956inadmissibility} and \cite{james1961estimation}, and when the parameter of interest is of dimension greater than three, its MSE is lower than that of the unbiased estimator of the parameter of interest. \cite{judge2004semiparametric} and \cite{mittelhammer2005combining} have proposed an estimator similar to that in \cite{green1991james}, without assuming normality or that the variances are known, and study its asymptotic MSE, again when the parameter of interest is of dimension greater than three. \cite{cheng2019uniform} have considered an estimator related to $\widehat{\beta}_{MSE}$, in a GMM context with some valid and some potentially misspecified moment conditions. They show that asymptotically, the MSE of their estimator is uniformly smaller than that of the GMM estimator using only the valid moment conditions. Again, they focus on a multivariate parameter with a dimension greater than four.
In the context of a linear model with at least three endogenous variables and instruments, \cite{hansen2017stein} proposes to use a weighted average of the OLS and 2SLS estimators, with weights that depend on the Hausman-Wu statistic in a test of equality between the OLS and 2SLS estimators. He shows that this estimator has a lower asymptotic MSE than that of the 2SLS estimator. Finally, \cite{breusch2011fixed} have considered $\widehat{\beta}_{MSE}$ in a panel data context, in a case where the parameter of interest is univariate. However, they do not study the theoretical properties of that estimator.

\medskip
To the best of our knowledge, our paper is the first to study the theoretical properties of $\widehat{\beta}_{MSE}$ when the parameter of interest is univariate. This case is particularly relevant for policy evaluation and treatment choice. Even when one measures the effect of the policy on several outcomes, one is ultimately interested in summarizing those effects into a monetary assessment of the benefits of the policy, to be compared to its cost. Contrary to the previous literature, we do not find that the MSE of $\widehat{\beta}_{MSE}$ dominates uniformly that of the consistent estimator. However, we show that $\widehat{\beta}_{MSE}$ dominates the other two estimators from a minimax-regret perspective, thus giving a theoretical justification to its use to estimate a univariate parameter. Our paper also seems to be the first to consider the combination of estimators with different rates of convergence, which may be relevant in a number of contexts, such as RD designs.

\medskip
The remainder of the paper is organized as follows. Section 2 presents the set up and main results. Section 3 presents some extensions. Section 4 presents the proofs of the results.

\section{Set up and main results}

We are interested in a parameter $\beta_0\in \R$. To estimate $\beta_0$, we use a sample of size $n$. $\widehat{\beta}_C$ is $\sqrt{n}-$consistent and asymptotically normal for $\beta_0$. $\widehat{\beta}_E$ is $\sqrt{n}-$consistent and asymptotically normal for $\beta_E$. In general, $\beta_E\ne \beta_0$, but under an assumption on the data generating process $H_0$, $\beta_E=\beta_0$. The asymptotic variance of $\widehat{\beta}_E$ is smaller than that of $\widehat{\beta}_C$, so under $H_0$, $\widehat{\beta}_E$ is a more efficient estimator of $\beta_0$ than $\widehat{\beta}_C$. Moreover, the asymptotic variance of $\widehat{\beta}_E$ is smaller than that of any weighted sum of $\widehat{\beta}_E$ and $\widehat{\beta}_C$, which implies that the asymptotic covariance of $\widehat{\beta}_E$ and $\widehat{\beta}_C$ is equal to the asymptotic variance of $\widehat{\beta}_E$. Finally, we have estimators $\widehat{V}(\widehat{\beta}_C)$, $\widehat{V}(\widehat{\beta}_E)$, and $\widehat{\cov}(\widehat{\beta}_C,\widehat{\beta}_E)$ of the variances of $\widehat{\beta}_C$ and $\widehat{\beta}_E$ and of their covariance, that are such that $n\widehat{V}(\widehat{\beta}_C)$, $n\widehat{V}(\widehat{\beta}_E)$, and $n\widehat{\cov}(\widehat{\beta}_C,\widehat{\beta}_E)$ are consistent for their asymptotic variances and covariances. We summarize these conditions in Assumption \ref{hyp:setup} below:
\begin{hyp}\label{hyp:setup} (Set-up)
\begin{enumerate}
\item \label{hyp:setup_p1_asnorm} We have $$\sqrt{n}\left( \begin{array}{c}
\widehat{\beta}_C-\beta_0 \\
\widehat{\beta}_E-\beta_E
\end{array}
\right)\convNor{0,\left( \begin{array}{cc}
\sigma^2_C & \sigma^2_E  \\
\sigma^2_E & \sigma^2_E
\end{array}\right)},$$
with $\sigma^2_C<\sigma^2_E$.
\item \label{hyp:setup_p2_consistent_var_estimation} $n\widehat{V}(\widehat{\beta}_C)\convP\sigma^2_C$, $n\widehat{V}(\widehat{\beta}_E)\convP\sigma^2_E$, and $n\widehat{\cov}(\widehat{\beta}_C,\widehat{\beta}_E)\convP \sigma^2_E$.
\end{enumerate}
\end{hyp}
We consider the following estimator of $\beta_0$:
\begin{definition}\label{definition_betaMSE} (The empirical-MSE-minimizing estimator of $\beta_0$)

\medskip
Let
\begin{equation}\label{eq:betaMSE}
\widehat{\beta}_{MSE}=\widehat{p}\widehat{\beta}_E+\left(1-\widehat{p}\right)\widehat{\beta}_C,
\end{equation}
where
\begin{equation}\label{eq:popt}
\widehat{p}=\arg\min_{p\in\R} ~~p^2\left(\widehat{\beta}_E-\widehat{\beta}_C\right)^2+p^2\widehat{V}(\widehat{\beta}_E)+(1-p)^2\widehat{V}(\widehat{\beta}_C)+2p(1-p)\widehat{\cov}(\widehat{\beta}_C,\widehat{\beta}_E).
\end{equation}
\end{definition}
$\widehat{\beta}_{MSE}$ is the weighted sum of $\widehat{\beta}_E$ and $\widehat{\beta}_C$ with the lowest estimated mean-squared error (MSE).

Solving the problem in Equation \eqref{eq:popt} yields
\begin{equation}\label{eq:popt_closedform}
\widehat{p}=\frac{\widehat{V}(\widehat{\beta}_C)-\widehat{\cov}(\widehat{\beta}_C,\widehat{\beta}_E)}{\left(\widehat{\beta}_E-\widehat{\beta}_C\right)^2+\widehat{V}(\widehat{\beta}_E-\widehat{\beta}_C)}.
\end{equation}

Theorem \ref{thm_main} below gives the asymptotic distribution of $\widehat{\beta}_{MSE}$ and compares it with that of $\widehat{\beta}_C$. In particular, we compare the MSE of the asymptotic distribution of the two estimators.\footnote{If the second moments of the normalized estimators converge, our results provide a comparison of the asymptotic MSE of the estimators.  To avoid the issue that convergence in distribution does not imply convergence in $L^2$, one could consider instead, as \cite{cheng2019uniform}, a winsorized version of the square loss.}

\begin{thm}\label{thm_main} (Asymptotic distribution of $\widehat{\beta}_{MSE}$)

Suppose Assumption \ref{hyp:setup} holds.
\begin{enumerate}
	\item If $\beta_0\ne\beta_E$ and $\beta_0$ and $\beta_E$ do not depend on $n$, 	$\sqrt{n}\left(\widehat{\beta}_{MSE}-\beta_0\right)\convNor{0,\sigma^2_C}$.
	\item If $\beta_0=\beta_E$, $\sqrt{n}\left(\widehat{\beta}_{MSE}-\beta_0\right)\convD U_0$, where $U_0$ is such that $E(U_0)=0$ and $V(U_0)\in (\sigma^2_E, \sigma^2_C)$.
	\item If $\beta_E=\beta_0+h/\sqrt{n}$ for some $h\in \R$, 	$\sqrt{n}\left(\widehat{\beta}_{MSE}-\beta_0\right)\convD U_h$, where $(U_h)_{h\in \R}$ is such that
	$$\max_{h\in \R}\left[E(U_h^2)-\sigma^2_C\right]<\max_{h\in \R}\left[\sigma^2_C-E(U_h^2)\right].$$
\end{enumerate}
\end{thm}

\medskip
Point 1 of Theorem \ref{thm_main} shows that if $\beta_0\ne\beta_E$, $\widehat{\beta}_{MSE}$ and $\widehat{\beta}_{C}$ have the same asymptotic distribution. On the other hand, Point 2 shows that if $\beta_0=\beta_E$, their asymptotic distributions differ, and the MSE of the asymptotic distribution of $\widehat{\beta}_{MSE}$ is larger than that of $\widehat{\beta}_E$ but smaller than that of $\widehat{\beta}_{C}$. This comes from the fact that under $H_0$, $\widehat{p}$ converges in distribution to a nongenerate distribution. In other words, $\widehat{p}$ does not perform consistent ``model selection'': it does not converge to 1 if $H_0$ holds and to 0 otherwise.

\medskip
The asymptotic approximations under fixed values of $\beta_0$ and $\beta_E$ may not give good approximations of the finite sample behavior of the estimators. Instead, in Point 3 of the theorem, we compare the MSE of their asymptotic distributions under alternatives local to $\beta_0=\beta_E$, namely $\beta_E=\beta_0+h/\sqrt{n}$. We find that under this type of asymptotics, the MSE of the asymptotic distribution of $\widehat{\beta}_{MSE}$ is not always smaller than that of $\widehat{\beta}_{C}$. This phenomenon is reminiscent of Hodges' estimator, whose asymptotic distribution has a smaller MSE than that of the standard estimator if the true parameter is 0,  but whose maximal risk increases without bound as $n\rightarrow\infty$ \citep[see, e.g.][pp. 440-443]{lehmann1998}. An important difference with Hodges' estimator is that here, the maximum asymptotic-MSE-gain one may incur by using $\widehat{\beta}_{MSE}$ rather than $\widehat{\beta}_{C}$ is larger than the maximum asymptotic-MSE-loss. Thus, $\widehat{\beta}_{MSE}$ dominates $\widehat{\beta}_{C}$ from a minimax regret perspective \citep[see][]{savage1951theory}. It is straightforward to show that $\widehat{\beta}_{MSE}$ also dominates $\widehat{\beta}_{E}$ from a minimax regret perspective.

\section{Extensions}

\subsection{A family of ``pre-test'' estimators}
\label{sub:modified_versions_of_the_estimator}

When $\beta_0=\beta_E$, $\widehat{\beta}_{MSE}$ is not equivalent to $\widehat{\beta}_E$, and it has a higher asymptotic variance. This is because the estimated squared bias $(\widehat{\beta}_E-\widehat{\beta}_C)^2$ in \eqref{eq:popt} includes some noise and is not negligible even if $\beta_0=\beta_E$. We now consider a modified version of $\widehat{\beta}_{MSE}$ that uses a smaller estimator of the squared bias. Specifically, we replace \eqref{eq:popt} by
\begin{align*}
\widehat{p}_{\lambda}=\arg\min_{p\in\R} \quad & p^2\max\left[0,\left(\widehat{\beta}_E-\widehat{\beta}_C\right)^2 - \lambda \Vh(\widehat{\beta}_E-\widehat{\beta}_C) \right] \\
& + p^2\widehat{V}(\widehat{\beta}_E)+(1-p)^2\widehat{V}(\widehat{\beta}_C)+2p(1-p)\widehat{\cov}(\widehat{\beta}_C,\widehat{\beta}_E)
\end{align*}
for some $\lambda\geq 0$. We then let $\widehat{\beta}_{MSE,\lambda}=\widehat{p}_{\lambda} \widehat{\beta}_E + (1-\widehat{p}_{\lambda}) \widehat{\beta}_C$.

$\widehat{\beta}_{MSE,\lambda}$ can be viewed as a pre-test estimator. Assume that one uses $\widehat{V}(\widehat{\beta}_E)$ to estimate $\cov(\widehat{\beta}_C,\widehat{\beta}_E)$. Then, let $F_1$ denote the cdf of a $\chi^2_1$ distribution and let $\alpha=F_1(\lambda)$. To compute, $\widehat{\beta}_{MSE,\lambda}$, one first runs a level-$\alpha$ test of $H_0$, using the fact that $(\widehat{\beta}_E-\widehat{\beta}_C)^2/\Vh(\widehat{\beta}_E-\widehat{\beta}_C)\convD \chi^2_1$. If $H_0$ is accepted, then $\widehat{\beta}_{MSE,\lambda}=\widehat{\beta}_E$. If $H_0$ is rejected, $\widehat{\beta}_{MSE,\lambda}$ is equal to a convex combination between $\widehat{\beta}_E$ and $\widehat{\beta}_C$, where the weight assigned to $\widehat{\beta}_C$ depends on how far we are from accepting $H_0$.

\begin{prop}\label{prop:reduced_bias}
Suppose Assumption \ref{hyp:setup} holds and let $\lambda$ be a positive real number.
\begin{enumerate}
	\item If $\beta_0\ne\beta_E$ and $\beta_0$ and $\beta_E$ do not depend on $n$, $\sqrt{n}\left(\widehat{\beta}_{MSE,\lambda}-\beta_0\right)\convNor{0,\sigma^2_C}$.
	\item If $\beta_0=\beta_E$, $\sqrt{n}\left(\widehat{\beta}_{MSE,\lambda}-\beta_0\right)\convD U_{0,\lambda}$, where $U_{0,\lambda}$ is such that $E(U_{0,\lambda})=0$, $\lambda \mapsto V(U_{0,\lambda})$ is strictly decreasing and $\lim_{\lambda\to+\infty}  V(U_{0,\lambda})=\sigma^2_E$.
	\item If $\beta_E=\beta_0+h/\sqrt{n}$ for some $h\in \R$, $\sqrt{n}\left(\widehat{\beta}_{MSE,\lambda}-\beta_0\right)\convD U_{h,\lambda}$, where $(U_{h,\lambda})_{h\in \R}$ is such that for all $\lambda\neq 0$
\begin{equation}
\max_{h\in \R}\left[E(U_{h,\lambda}^2) - E(U_{h,0}^2) \right]>\max_{h\in \R}\left[E(U_{h,0}^2) - E(U_{h,\lambda}^2)\right],
		\label{eq:ineq_minimax}
\end{equation}
\end{enumerate}
\iffalse
, if $\lambda \leq 1.76$,
\begin{equation}
\max_{h\in \R}\left[E(U_{h,\lambda}^2) - \sigma^2_C\right]<\max_{h\in \R}\left[\sigma^2_C-E(U_{h,\lambda}^2)\right],
		\label{eq:ineq_minimax}
	\end{equation}
	but the reverse holds for $\lambda\geq 1.77$. Moreover,
$\lim_{\lambda\to\infty} \max_{h\in \R}E(U_{h,\lambda}^2)=+\infty$. Finally,
\begin{equation}
	0=\arg\max_{\lambda\geq 0} \left\{ \max_{h\in \R}\left[\sigma^2_C-E(U_{h,\lambda} ^2)\right] - \max_{h\in \R}\left[E(U_{h,\lambda} ^2)-\sigma^2_C\right]\right\}.
	\label{eq:opt_minimax}
\end{equation}\fi
\end{prop}

Points 1 and 2 in Proposition \ref{prop:reduced_bias} are similar to those in Theorem \ref{thm_main}, with the additional point that under $H_0$, the asymptotic, quadratic risk of $\widehat{\beta}_{MSE,\lambda}$ decreases and gets closer to that of $\widehat{\beta}_E$ as $\lambda$ increases. However, the third point shows that from a minimax regret perspective, such estimators are dominated by our intial estimator $\widehat{\beta}_{MSE}$. The reason is that the decrease of $\lambda\mapsto E(U_{0,\lambda}^2)$ does not compensate for the quick increase of  $\lambda \mapsto \max_{h\in \R}E(U_{h,\lambda}^2) - E(U_{h,0}^2)$.

This echoes the discussion in \cite{leeb2005}: as we move closer to an estimator based on a consistent model selection, the maximal asymptotic risk increases without bound.

\subsection{Averaging estimators with different rates of convergence}
\label{sub:averaging_estimators_with_different_rates_of_convergence}

In this subsection, we assume that $\widehat{\beta}_C$ is $r_n-$consistent for $\beta_0\in \R$, for some sequence $(r_n)_{n\in\N}$ such that $r_n\to\infty$, $r_n/n^{1/2}\to 0$. The estimator $\widehat{\beta}_E$ is still $\sqrt{n}-$consistent and asymptotically normal for $\beta_E$, which may be equal to $\beta_0$ under an assumption on the data generating process $H_0$. We also assume we have estimators $\widehat{V}(\widehat{\beta}_C)$, $\widehat{V}(\widehat{\beta}_E)$, and $\widehat{\cov}(\widehat{\beta}_C,\widehat{\beta}_E)$ of the variances of $\widehat{\beta}_C$ and $\widehat{\beta}_E$ and of their covariance, that are such that $r_n^2\widehat{V}(\widehat{\beta}_C)$, $n\widehat{V}(\widehat{\beta}_E)$, and $n^{1/2} r_n\widehat{\cov}(\widehat{\beta}_C,\widehat{\beta}_E)$ are consistent for their asymptotic variances and covariances. We summarize these conditions in Assumption \ref{hyp:setup_nonparam} below:
\begin{hyp}\label{hyp:setup_nonparam} (Set-up)
\begin{enumerate}
\item \label{hyp:setup_p1_asnorm_nonparam} There exists a sequence $(r_n)_{n\in\N}$, $r_n\to\infty$ and $r_n/n^{1/2}\to 0$, such that $$\left( \begin{array}{c}
r_n\left(\widehat{\beta}_C-\beta_0\right) \\
\sqrt{n}\left(\widehat{\beta}_E-\beta_E\right)
\end{array}
\right)\convNor{\left( \begin{array}{c}
\mu  \\
0
\end{array}
\right),\left( \begin{array}{cc}
\sigma^2_C & \rho  \\
\rho & \sigma^2_E
\end{array}
\right)}.$$
\item \label{hyp:setup_p2_consistent_var_estimation_nonparam} $r_n^2\widehat{V}(\widehat{\beta}_C)\convP\sigma^2_C$, $n\widehat{V}(\widehat{\beta}_E)\convP\sigma^2_E$, and $n^{1/2} r_n\widehat{\cov}(\widehat{\beta}_C,\widehat{\beta}_E)\convP \rho$.
\end{enumerate}
\end{hyp}

\medskip
Theorem \ref{thm_main_nonparam} below gives the asymptotic distribution of $\widehat{\beta}_{MSE}$ defined in Equation \ref{eq:betaMSE} above, under Assumption \ref{hyp:setup_nonparam} rather than Assumption \ref{hyp:setup}.

\begin{thm}\label{thm_main_nonparam} (Asymptotic distribution of $\widehat{\beta}_{MSE}$ under Assumption \ref{hyp:setup_nonparam})
Suppose Assumption \ref{hyp:setup_nonparam} holds.
\begin{enumerate}
\item If $\beta_0\ne\beta_E$ and $\beta_0$ and $\beta_E$ do not depend on $n$, $r_n\left(\widehat{\beta}_{MSE}-\beta_0\right)\convNor{0,\sigma^2_C}$.
\item If $\beta_0=\beta_E$, $r_n\left(\widehat{\beta}_{MSE}-\beta_0\right)\convD U_0$, where $U_0$ is such that $E(U_0^2)<E(V^2)$.

\item If $\beta_E=\beta_0+h/r_n$ for some $h\in \R$ and $|\mu/\sigma_C|\leq 0.4$, then $r_n\left(\widehat{\beta}_{MSE}-\beta_0\right)\convD U_h$, where $(U_h)_{h\in \R}$ is such that
$$\max_{h\in \R}\left[E(U_h^2)-(\mu^2+\sigma^2_C)\right]<\max_{h\in \R}\left[\mu^2+\sigma^2_C- E(U_h^2)\right].$$
    \end{enumerate}
\end{thm}

\medskip
Point 1 of Theorem \ref{thm_main_nonparam} shows that if $\beta_0\ne\beta_E$, $\widehat{\beta}_{MSE}$ and $\widehat{\beta}_{C}$ have the same asymptotic distribution. On the other hand, Point 2 shows that if $\beta_0=\beta_E$, their asymptotic distributions differ, and the MSE of the asymptotic distribution of $\widehat{\beta}_{MSE}$ is smaller than that of $\widehat{\beta}_{C}$. In Point 3 of the theorem, we compare the MSE of their asymptotic distributions under alternatives local to $\beta_0=\beta_E$, namely $\beta_E=\beta_0+h/r_n$. Under this type of asymptotics, the maximum asymptotic-MSE-gain one may incur by using $\widehat{\beta}_{MSE}$ rather than $\widehat{\beta}_{C}$ is larger than the maximum asymptotic-MSE-loss, provided the first-order bias is no greater in absolute value than 0.4$\sigma_C$. Again, $\widehat{\beta}_{MSE}$ dominates $\widehat{\beta}_{C}$ from a minimax regret perspective, provided the first-order bias of $\widehat{\beta}_{C}$ is not too large.

\section{Proofs}
\label{sec:proofs}

We use the folowing lemma below.

\begin{lemma}\label{lem:expect_symm}
	Suppose that $f$ is an odd function such that $f(x)>0$ for all $x>0$ and $Z$ has an even density $g$ that is strictly decreasing on $\R^+$. Then, $\sgn(E[f(x+Z)])=\sgn(x)$ for all $x\in\R$.
\end{lemma}

\textbf{Proof:} the result holds if $x=0$, because $E[f(Z)]=E[f(-Z)]=-E[f(Z)]$. For any $x<0$,
$E[f(Z+x)]=E[f(-Z+x)]=E[-f(Z-x)]$, so it suffices to show that $E[f(Z+x)]>0$ for $x>0$. We have
\begin{align*}
	E[f(Z+x)] & =\int_{\R} f(z+x)g(z)dz\\
	& = \int_{\R^+} f(z)g(z-x)dz + \int_{\R^-} f(z)g(z-x)dz \\
	& =  \int_{\R^+} f(z)g(z-x)dz - \int_{\R^-} f(-z)g(x-z)dz \\
	& =  \int_{\R^+} f(z)[g(z-x) - g(z+x)]dz.
\end{align*}
Now, for all $z\in (0,x]$, $|z-x|=x-z < x+z$ so $g(z-x) > g(z+x)$. If $z>x$, $|z-x|=z-x < z+x$ so again, $g(z-x) > g(z+x)$. The result follows since $f(z)>0$ on $(0,\infty)$.

\subsection{Theorem \ref{thm_main}}
\label{sub:theorem}

\subsubsection*{Proof of Point 1}

If $\beta_0\ne\beta_E$,  it follows from \eqref{eq:popt_closedform}, Assumption \ref{hyp:setup} and the continuous mapping theorem that
$$n\widehat{p}\convP \frac{\sigma^2_C- \sigma^2_E}{(\beta_E-\beta_0)^2}.$$
Moreover,
\begin{equation}\label{eq:diff_MSE_C}
\widehat{\beta}_{MSE}-\widehat{\beta}_C=\widehat{p}\left(\widehat{\beta}_E-\widehat{\beta}_C\right).
\end{equation}
Hence, by the continuous mapping theorem again, $$n\left(\widehat{\beta}_{MSE}-\widehat{\beta}_C\right)\convP\frac{\sigma^2_C- \sigma^2_E}{\beta_E-\beta_0}.$$
Therefore,
$$\sqrt{n}\left(\widehat{\beta}_{MSE}-\widehat{\beta}_C\right)=o_P(1).$$

\subsubsection*{Proof of Point 2}

Let $(V,W)$ be a normal vector with mean $(0,0)$, variances $(\sigma^2_C,\sigma^2_C-\sigma^2_E)$, and covariance $-\left(\sigma^2_C-\sigma^2_E\right)$. Let
\begin{equation}\label{eq:defU}
U_0=V+W\frac{\sigma^2_C-\sigma^2_E}{W^2+\sigma^2_C-\sigma^2_E}.
\end{equation}
\begin{align*}
&\sqrt{n}\left(\widehat{\beta}_{MSE}-\beta_0\right)\\
=&\sqrt{n}\left(\widehat{\beta}_{C}-\beta_0\right)\\
+&\frac{n\widehat{V}(\widehat{\beta}_C)-n\widehat{\cov}(\widehat{\beta}_C,\widehat{ \beta}_E)}{\left(\sqrt{n}\left(\widehat{\beta}_E-\beta_E-\left(\widehat{\beta}_C-\beta_0\right)\right)\right)^2+n\widehat{V}(\widehat{\beta}_E-\widehat{\beta}_C)}\sqrt{n}\left(\widehat{\beta}_E-\beta_E-\left(\widehat{\beta}_C-\beta_0\right)\right)\\
\convD&~U_0.
\end{align*}
The first equality follows from Equations \eqref{eq:diff_MSE_C} and  \eqref{eq:popt_closedform} and from $\beta_E=\beta_0$. The convergence in distribution arrow follows from Assumption \ref{hyp:setup}, the Slutsky lemma, and the continuous mapping theorem.

\medskip
$E(U_0)=0$ as $\phi: w\mapsto w\frac{\sigma^2_C-\sigma^2_E}{w^2+\sigma^2_C-\sigma^2_E}$ is such that $\phi(-w)=-\phi(w)$ and the pdf of $W$ is symmetric around 0.

\medskip
Let $\Psi=V+W$. The vector $(\Psi,W)$ is normally distributed with $E(\Psi)=0$ and $\cov(\Psi,W)=0$. Hence, $\Psi\indep W$ and we have
$$U_0 = \Psi + W\left(\frac{\sigma^2_C - \sigma^2_E}{W^2+\sigma^2_C-\sigma^2_E} -1 \right),$$
Hence, $V(U_0)>V(\Psi)=\sigma^2_E$. Moreover,
\begin{align*}
V(U_0)-\sigma^2_C=&E\left(U_0^2\right)-E\left(V^2\right)\\
=&E((U_0-V)(U_0+V))\\
=&E\left(W\frac{\sigma^2_C-\sigma^2_E}{W^2+\sigma^2_C-\sigma^2_E}\left(2V+W\frac{\sigma^2_C-\sigma^2_E}{W^2+\sigma^2_C-\sigma^2_E}\right)\right)\\
=&E\left(W\frac{\sigma^2_C-\sigma^2_E}{W^2+\sigma^2_C-\sigma^2_E}\left(-2W+2\Psi+W\frac{\sigma^2_C-\sigma^2_E}{W^2+\sigma^2_C-\sigma^2_E}\right)\right)\\
=&E\left(W^2\frac{\sigma^2_C-\sigma^2_E}{W^2+\sigma^2_C-\sigma^2_E}\left(-2+\frac{\sigma^2_C-\sigma^2_E}{W^2+\sigma^2_C-\sigma^2_E}\right)\right)\\
<&0.
\end{align*}
The first equality follows from $E(U_0)=E(V)=0$, the third from Equation \eqref{eq:defU}, the fourth  from $V=-W+\Psi$ and the fifth from $\Psi\indep W$ and $E(\Psi)=0$. The inequality holds since $(\sigma^2_C-\sigma^2_E)/(W^2+\sigma^2_C-\sigma^2_E) < 1$ with probability 1, as $\sigma^2_C>\sigma^2_E$.

\subsubsection*{Proof of Point 3}

Let $(V,W_h)$ be a normal vector with means $(0,h)$, variances $(\sigma^2_C,\sigma^2_C-\sigma^2_E)$, and covariance $-\left(\sigma^2_C-\sigma^2_E\right)$. Let $U_h=V+W_h(\sigma^2_C-\sigma^2_E)/(W_h^2+\sigma^2_C-\sigma^2_E)$. We have
\begin{align*}
&\sqrt{n}\left(\widehat{\beta}_{MSE}-\beta_0\right)\\
=&\sqrt{n}\left(\widehat{\beta}_{C}-\beta_0\right)\\
+&\frac{n\widehat{V}(\widehat{\beta}_C)-n\widehat{\cov}(\widehat{\beta}_C,\widehat{ \beta}_E)}{\left(\sqrt{n}\left(\widehat{\beta}_E-\beta_E-\left(\widehat{\beta}_C-\beta_0\right)\right)+h\right)^2+n\widehat{V}(\widehat{\beta}_E-\widehat{\beta}_C)}\left(\sqrt{n}\left(\widehat{\beta}_E-\beta_E-\left(\widehat{\beta}_C-\beta_0\right)\right)+h\right)\\
\convD&~U_h.
\end{align*}
The first equality follows from Equations \eqref{eq:diff_MSE_C} and  \eqref{eq:popt_closedform} and from $\beta_E=\beta_0+h/\sqrt{n}$. The convergence in distribution arrow follows from Assumption \ref{hyp:setup}, the Slutsky lemma, and the continuous mapping theorem.

\medskip
Let $\Psi_h=V+W_h$. The vector $(\Psi_h,W_h)$ is normally distributed, $\cov(\Psi_h,W_h)=0$ so $\Psi_h\indep W_h$. \iffalse One has
\begin{equation}\label{eq:decompV_local}
V=-W_h+\Psi_h.
\end{equation} \fi
Let $g=h/\sqrt{\sigma^2_C-\sigma^2_E}$, and let $N_{g}=\frac{W_h}{\sqrt{\sigma^2_C-\sigma^2_E}}$, so that $N_{g}\sim\mathcal{N}(g,1)$. We have
\begin{align}\label{eq:comparison_MSE_local_alt}
&E(U_h^2)-\sigma^2_C \nonumber\\
=&E((U_h-V)(U_h+V))\nonumber\\
=&E\left(W_h\frac{\sigma^2_C-\sigma^2_E}{W_h^2+\sigma^2_C-\sigma^2_E}\left(2V+W_h\frac{\sigma^2_C-\sigma^2_E}{W_h^2+\sigma^2_C-\sigma^2_E}\right)\right)\nonumber\\
=&E\left(W_h\frac{\sigma^2_C-\sigma^2_E}{W_h^2+\sigma^2_C-\sigma^2_E}\left(-2W_h+2\Psi_h+W_h\frac{\sigma^2_C-\sigma^2_E}{W_h^2+\sigma^2_C-\sigma^2_E}\right)\right)\nonumber\\
=&-2\left(E\left(W_h\frac{\sigma^2_C-\sigma^2_E}{W_h^2+\sigma^2_C-\sigma^2_E}W_h\right)-E\left(W_h\frac{\sigma^2_C-\sigma^2_E}{W_h^2+\sigma^2_C-\sigma^2_E}\right)E\left(\Psi_h\right)\right)\nonumber\\
+&E\left(\left(W_h\frac{\sigma^2_C-\sigma^2_E}{W_h^2+\sigma^2_C-\sigma^2_E}\right)^2\right)\nonumber\\

=&-2\left(E\left(W_h\frac{\sigma^2_C-\sigma^2_E}{W_h^2+\sigma^2_C-\sigma^2_E}W_h\right)-E\left(W_h\frac{\sigma^2_C-\sigma^2_E}{W_h^2+\sigma^2_C-\sigma^2_E}\right)E\left(W_h\right)\right)\nonumber\\
+&E\left(\left(W_h\frac{\sigma^2_C-\sigma^2_E}{W_h^2+\sigma^2_C-\sigma^2_E}\right)^2\right)\nonumber\\
=&-2\cov\left(W_h\frac{\sigma^2_C-\sigma^2_E}{W_h^2+\sigma^2_C-\sigma^2_E},W_h\right)+E\left(\left(W_h\frac{\sigma^2_C-\sigma^2_E}{W_h^2+\sigma^2_C-\sigma^2_E}\right)^2\right)\nonumber\\
=&\left(\sigma^2_C-\sigma^2_E\right)\left\{E\left[\left(\frac{N_{g}}{N_{g}^2+1}\right)^2\right]-2\cov\left(\frac{N_{g}}{N_{g}^2+1},N_{g}\right)\right\}.

\end{align}
Let
\begin{equation}\label{eq:def_Delta(g)}
\Delta(g)=E\left[\left(\frac{N_{g}}{N_{g}^2+1}\right)^2\right]-2\cov\left(\frac{N_{g}}{N_{g}^2+1},N_{g}\right).
\end{equation}
Since $N_{-g}\sim -N_{g}$, we have
$$\Delta(-g)=E\left[\left(\frac{-N_{g}}{(-N_{g})^2+1}\right)^2\right]-2\cov\left(\frac{-N_{g}}{(-N_{g})^2+1},-N_{g}\right)=\Delta(g).$$
Moreover,

we obtain through numerical simulations that
\iffalse
 use Monte-Carlo simulations with 1,000,000 observations to evaluate $\Delta(g)$ for every $g\in\{0,0.01,0.02,...,10\}$. The result is shown in Figure \ref{fig_diff_mse} below.
\begin{figure}[H]
\begin{center}
\includegraphics[scale = 0.8]{diff_mse}
\caption{$g\mapsto \Delta(g)$}
\label{fig_diff_mse}
\end{center}
\end{figure}
Those simulations show that
\begin{align}\label{eq:minmax_Delta(g)}
&\underset{g\in\R}{\min}~\Delta(g)\simeq -0.53\nonumber \\
&\underset{g\in\R}{\max}~\Delta(g)\simeq 0.25
\end{align}\fi
$\min_{g\in\R^+}\;\Delta(g)\simeq -0.53$ and $\max_{g\in\R^+}\;\Delta(g)\simeq 0.25$. The inequality
$$\max_{h\in \R}\left[(E(U_h))^2+V(U_h)-\sigma^2_C\right]<\max_{h\in \R}\left[\sigma^2_C-\left((E(U_h))^2+V(U_h)\right)\right]$$
follows from these last results, Equations \eqref{eq:comparison_MSE_local_alt} and $\sigma^2_C>\sigma^2_E$.

\subsection{Proof of Proposition \ref{prop:reduced_bias}}
\label{sub:proof_of_proposition_ref_prop_reduced_bias}

\subsubsection*{Proof of Point 1}

First, we have
$$\widehat{p}_{\lambda}=\frac{\widehat{V}(\widehat{\beta}_C)-\widehat{\cov}(\widehat{\beta}_C,\widehat{\beta}_E)}{\max\left[0,\left(\widehat{\beta}_E-\widehat{\beta}_C\right)^2-\lambda\Vh (\widehat{\beta}_E-\widehat{\beta}_C) \right]+\widehat{V}(\widehat{\beta}_E-\widehat{\beta}_C)}.$$
Then, if $\beta_0\neq\beta_E$, Assumption \ref{hyp:setup} and the continuous mapping theorem yields
$$n\widehat{p}_{\lambda}\convP \frac{\sigma^2_C - \sigma^2_E}{(\beta_E-\beta_C)^2}.$$
The result follows as in the previous proof.

\subsubsection*{Proof of Point 2}

Instead of considering $U_0$ defined by \eqref{eq:defU}, we now consider $U_{0,\lambda}$ defined by
$$U_{0,\lambda} = V + W\frac{\sigma^2_C- \sigma^2_E}{\max\left[0,W^2- \lambda(\sigma^2_C- \sigma^2_E)\right] + \sigma^2_C- \sigma^2_E}.$$
Then, the proofs of the convergence in distribution, of $E(U_{0,\lambda})=0$ and $E\left(U_{0,\lambda}^2\right)<\sigma^2_C$ are identical to those above. We  now show that $\lambda\mapsto V(U_{0,\lambda})$ is decreasing. Let  $Z=W/(\sigma^2_C-\sigma^2_E)^{1/2}$ Using the same definition of $\Psi$ as above, we have
$$U_{0,\lambda} =\Psi +  (\sigma^2_C-\sigma^2_E)^{1/2} Z\left(\frac{1}{\max(0,Z^2-\lambda)+1} - 1\right),$$
with $\Psi\indep Z$. Then, for any two $\lambda>\lambda'$,
\begin{align*}
	 V(U_{0,\lambda})-V(U_{0,\lambda'}) = & E\left[(U_{0,\lambda} +U_{0,\lambda'})(U_{0,\lambda}- U_{0,\lambda'})\right] \\
	 = & (\sigma^2_C-\sigma^2_E) E\left[Z^2\left(\frac{1}{\max(0,Z^2-\lambda)+1} + \frac{1}{\max(0,Z^2-\lambda')+1} - 2 \right) \right.\\
	 & \left. \hspace{2.2cm}\times \left(\frac{1}{\max(0,Z^2-\lambda)+1} - \frac{1}{\max(0,Z^2-\lambda')+1} \right)\right].
\end{align*}
Moreover,
$$\frac{1}{\max(0,Z^2-\lambda)+1} + \frac{1}{\max(0,Z^2-\lambda')+1} - 2 \leq 0,$$
and the inequality is strict for $|Z|>\sqrt{\lambda'}$. Also,
$$\frac{1}{\max(0,Z^2-\lambda)+1} - \frac{1}{\max(0,Z^2-\lambda')+1} \geq 0,$$
and again, the inequality is strict for $|Z|>\sqrt{\lambda'}$. Hence, $V(U_{0,\lambda})<V(U_{0,\lambda'})$. Finally, as $\lambda\to+\infty$, $U_{0,\lambda}$ converges almost surely to $\Psi$. Moreover, $|U_{0,\lambda}|\leq |\Psi| +  (\sigma^2_C-\sigma^2_E)^{1/2} |Z|$, so by the dominated convergence theorem, $V(U_{0,\lambda})\to V(\Psi)=\sigma^2_E$.

\subsubsection*{Proof of Point 3}

With the same notation as above, let
$$U_{h,\lambda} = \Psi_h + W_h \left(\frac{\sigma^2_C - \sigma^2_E}{\max(0,W_h^2 - \lambda(\sigma^2_C - \sigma^2_E)) + \sigma^2_C - \sigma^2_E} - 1\right).$$
Using a reasoning similar to that in the proof of Point 3 of Theorem \ref{thm_main},
\begin{align*}
&\sqrt{n}\left(\widehat{\beta}_{MSE,\lambda}-\beta_0\right)\convD~U_{h,\lambda}.
\end{align*}
By the same reasoning as in the proof of Point 3 of Theorem \ref{thm_main},
\begin{align*}
E(U_{h,\lambda}^2)- E(U_{h,0}^2) = (\sigma^2_C-\sigma^2_E)
 & \left\{ E\left[\left(\frac{N_g}{\max(0,N_{g}^2-\lambda)+1}\right)^2  - \left(\frac{N_g}{N_{g}^2+1}\right)^2\right]\right.  \\
 & \; \left. -2\cov\left[N_{g}\left(\frac{1}{\max(0,N_{g}^2-\lambda)+1} - \frac{1}{N_g^2+1}\right),N_{g}\right]\right\}.
\end{align*}
Then, simulations show that \eqref{eq:ineq_minimax} holds.

\subsection{Proof of Theorem \ref{thm_main_nonparam}}
\label{sub:proof_of_theorem_ref}

\subsubsection*{Proof of Point 1}

We have
\begin{equation}\label{eq:diff_MSE_C2}
\widehat{\beta}_{MSE}-\widehat{\beta}_C=\widehat{p}\left(\widehat{\beta}_E-\widehat{\beta}_C\right).
\end{equation}
If $\beta_0\ne\beta_E$, it follows from Assumption \ref{hyp:setup_nonparam} and the continuous mapping theorem that
$$r_n^2\left(\widehat{\beta}_{MSE}-\widehat{\beta}_C\right)\convP\frac{\sigma^2_C}{\beta_E-\beta_0}.$$
Therefore,
$$r_n\left(\widehat{\beta}_{MSE}-\widehat{\beta}_C\right)=o_P(1).$$

\subsubsection*{Proof of Point 2}

Let $V$ be a normal variable with mean $\mu$ and variance $\sigma^2_C$.
Let
\begin{equation}\label{eq:defU_nonparam}
U_0=V\left(1-\frac{\sigma^2_C}{V^2+\sigma^2_C}\right).
\end{equation}
If $\beta_0=\beta_E$,
\begin{align*}
&r_n\left(\widehat{\beta}_{MSE}-\beta_0\right)\\
=&r_n\left(\widehat{\beta}_{C}-\beta_0\right)\\
+&\frac{r_n^2\widehat{V}(\widehat{\beta}_C)-r_n^2\widehat{\cov}(\widehat{\beta}_C,\widehat{ \beta}_E)}{\left(r_n\left(\widehat{\beta}_E-\beta_E-\left(\widehat{\beta}_C-\beta_0\right)\right)\right)^2+r_n^2\widehat{V}(\widehat{\beta}_E-\widehat{\beta}_C)}r_n\left(\widehat{\beta}_E-\beta_E-\left(\widehat{\beta}_C-\beta_0\right)\right)\\
\convD&~U_0.
\end{align*}
The first equality follows from Equations \eqref{eq:diff_MSE_C2} and  \eqref{eq:popt_closedform} and from $\beta_E=\beta_0$. The convergence in distribution arrow follows from Assumption \ref{hyp:setup_nonparam}, the Slutsky lemma, and the continuous mapping theorem. Further,
\begin{align*}
E\left(U_0^2\right)-E\left(V^2\right)=&E((U_0-V)(U_0+V))\\
=&E\left(-\frac{\sigma^2_C}{V^2+\sigma^2_C}V\left(2V-\frac{\sigma^2_C}{V^2+\sigma^2_C}V\right)\right)\\
=&E\left(-\frac{\sigma^2_C}{V^2+\sigma^2_C}V^2\left(2-\frac{\sigma^2_C}{V^2+\sigma^2_C}\right)\right)\\
<&0.
\end{align*}
The third equality follows from Equation \eqref{eq:defU_nonparam}. The inequality follows from the fact
$1>\frac{\sigma^2_C}{V^2+\sigma^2_C}$
with probability 1.

\subsubsection*{Proof of Point 3}

Let
\begin{equation}\label{eq:defUh_nonparam}
U_h=V + (h-V)\frac{\sigma^2_C}{(h-V)^2+\sigma^2_C}.
\end{equation}
\begin{align*}
&r_n\left(\widehat{\beta}_{MSE}-\beta_0\right)\\
=&r_n\left(\widehat{\beta}_{C}-\beta_0\right)\\
+&\frac{r_n^2\widehat{V}(\widehat{\beta}_C)-r_n^2\widehat{\cov}(\widehat{\beta}_C,\widehat{ \beta}_E)}{\left(r_n\left(\widehat{\beta}_E-\beta_E-\left(\widehat{\beta}_C-\beta_0\right)\right)+h\right)^2+r_n^2\widehat{V}(\widehat{\beta}_E-\widehat{\beta}_C)}\left(r_n\left(\widehat{\beta}_E-\beta_E-\left(\widehat{\beta}_C-\beta_0\right)\right)+h\right)\\
\convD&~U_h.
\end{align*}
The first equality follows from Equations \eqref{eq:diff_MSE_C2} and  \eqref{eq:popt_closedform} and from $\beta_E=\beta_0+h/r_n$. The convergence in distribution arrow follows from Assumption \ref{hyp:setup_nonparam}, the Slutsky lemma, and the continuous mapping theorem.

\medskip
Let $g=\frac{h-\mu}{\sigma_C}$, $N_g=\frac{h-V}{\sigma_C}$, and $\mu_{sd}=\frac{\mu}{\sigma_C}$. Then $N_g\sim \mathcal{N}(g,1)$ and we have:
\begin{align}\label{eq:comparison_MSE_local_alt_non_param}
E\left(U_h^2\right)-E\left(V^2\right)=&E((U_h-V)(U_h+V))\nonumber\\
=&E\left((h-V)\frac{\sigma^2_C}{(h-V)^2+\sigma^2_C}\left(2V+(h-V)\frac{\sigma^2_C}{(h-V)^2+\sigma^2_C}\right)\right)\nonumber\\
=&\sigma^2_CE\left(\frac{N_g}{N_g^2+1}\left(2\left(g+\mu_{sd}-N_g\right)+\frac{N_g}{N_g^2+1}\right)\right)\nonumber\\
=&\sigma^2_C\left(\Delta(g)+2\mu_{sd}E\left(\frac{N_g}{N_g^2+1}\right)\right),
\end{align}
with $\Delta(g)$ defined as in Equation \eqref{eq:def_Delta(g)}.
Let $\Lambda(g,\mu_{sd})=\Delta(g)+2\mu_{sd}E\left(\frac{N_g}{N_g^2+1}\right)$. The function $x\mapsto x/(x^2+1)$ and the density of a $\mathcal{N}(0,1)$ satisfy the conditions of Lemma \ref{lem:expect_symm}. Thus, $E\left(\frac{N_g}{N_g^2+1}\right)\geq 0$ if $g\geq 0$, and $E\left(\frac{N_g}{N_g^2+1}\right)<0$ otherwise. Then, if $\mu$ and $g$ are not of the same sign, $\Lambda(g,\mu_{sd})\leq \Delta(g)$, so it follows from Point 3 of Theorem \ref{thm_main} that for every $\mu_{sd}$, the minimum of $\Lambda(g,\mu_{sd})$ with respect to $g$ is greater in absolute value than its maximum. We can then restrict attention to cases where $\mu$ and $g$ are of the same sign. As $\Lambda(-g,-\mu_{sd})=\Lambda(g,\mu_{sd})$, we can further restrict attention to cases where $g\geq 0$ and $\mu_{sd}\geq 0$. We use Monte-Carlo simulations with $10^7$ Halton draws to approximate $\Lambda(g,\mu_{sd})$ for every $g\in\{0,0.1,0.2,...,10\}$ and  $\mu_{sd}\in[0,0.41]$,
the absolute value of the minimum of $\Lambda(g,\mu_{sd})$ with respect to $g$ is larger than the absolute value of its maximum. This proves the result, together with Equation \eqref{eq:comparison_MSE_local_alt_non_param}.

\newpage
\bibliography{biblio}

\end{document}


TITLE: How to find the form of coordinates on coincident planes
QUESTION [1 upvotes]: I have three simultaneous equations:

$5x-2y+z=28$
$3x-4y-5z=0$
$-2x+3y+4z=2$
When I represent these equations as a 3x3 matrix and try to solve them, I find that $[x,y,z]=(1/0)*[0,0,0]$. I think this means the planes are coincident i.e. the same plane, therefore there are infinite solutions to the equations. My textbook gives the answer as $x=λ, y=2λ, z=8-λ$. I think this means that coordinates on the plane(s) will be of the form $(λ, 2λ, 8-λ)$, but I'm unsure how to reach this conclusion.
Could someone please explain how to find the form of the coordinates from the equations in the context of matrices?
Here is a picture of the textbook question:
Question
And here is a picture of my working:
Working

REPLY [1 votes]: No, when the determinant is zero, as is the case here, there need not be three coincident planes. Any two coincident planes, or one plane expressible as a linear combination of the other two (as we happen to have here),  will lead to a 1-parameter family of solutions.
Your textbook's answer is wrong, or perhaps you looked up the wrong answer; the right answer starting from $x=\lambda$ is $y=-10-2\lambda, z = 8-\lambda$.
Let's proceed as in Gauss-Jordan elimination via elementary operations, and see where we get in trying to solve these equations:
$$
\pmatrix{5&-2&1\\3&-4&-5\\-2&3&4} \pmatrix{x\\y\\z} = \pmatrix{28\\0\\2}
$$
$$
\pmatrix{1&-2/5&1/5\\3&-4&-5\\-2&3&4} \pmatrix{x\\y\\z} = \pmatrix{28/5\\0\\2}
$$
$$
\pmatrix{1&-2/5&1/5\\0&-14/5&-28/5\\0&11/5&22/5} \pmatrix{x\\y\\z} = \pmatrix{28/5\\-84/5\\66/5
}
$$
$$
\pmatrix{1&-2/5&1/5\\0&1&2\\0&11/5&22/5} \pmatrix{x\\y\\z}= \pmatrix{28/5\\6\\0
}
$$
$$
\pmatrix{1&0&1\\0&1&2\\0&0&0} \pmatrix{x\\y\\z}= \pmatrix{8\\6\\0
}
$$
and now we are "stuck" because there is no way to eliminate the $1$ and $2$ in the top two rows of the third column so there is no way to determine $z$.  So let's make $z$  some arbitrary number, say $\alpha$. Then that last equation says:
$$
x+\alpha = 8\\y +2\alpha = 6\\z=\alpha
$$
Or we could have insisted that $x=\lambda$ in which case $\alpha = 8-\lambda$ and
$$
x = \lambda \\y =6-2\alpha = -10 - 2\lambda\\z=8-\lambda
$$