Isometries of perfect norm ideals of compact operators

TL;DR

This paper characterizes all surjective linear isometries and 2-local isometries on perfect Banach symmetric ideals of compact operators, showing they are essentially implemented by unitary conjugations or transpositions, with implications for operator symmetry.

Contribution

It provides a complete description of surjective isometries and 2-local isometries on perfect Banach symmetric ideals, extending understanding of their structure beyond previous partial results.

Findings

Surjective linear isometries are implemented by unitary operators or transpositions.

Any surjective 2-local isometry is a linear isometry.

Results apply to all perfect Banach symmetric ideals except $\\mathcal{C}_2$.

Abstract

It is proved that for every surjective linear isometry $V$ on a perfect Banach symmetric ideal $\mathcal C_E\neq \mathcal C_2$ of compact operators, acting in a complex separable infnite-dimensional Hilbert space $\mathcal H$ there exist unitary operators $u$ and $v$ on $\mathcal H$ such that $V(x)=uxv$ or $V(x) = ux^tv$ for all $x\in \mathcal C_E$, where $x^t$ is the transpose of an operator $x$ with respect to a fixed orthonormal basis in $\mathcal H$. In addition, it is shown that any surjective 2-local isometry on a perfect Banach symmetric ideal $\mathcal C_E \neq \mathcal C_2$ is a linear isometry on $\mathcal C_E$.

Equations

$$x\in {\mathcal{C}}_E⟺\{s_n(x){\}}_{n=1}^{\infty }\in E$$

$$\Vert x{\Vert }_{{\mathcal{C}}_E}=\Vert \{s_n(x){\}}_{n=1}^{\infty }{\Vert }_E,$$

$$V(x)=uxv(\text{or}V(x)=u{x}^{t}v)$$

$$x^{\ast }=\{{\xi }_n^{\ast }\},\text{where}{\xi }_n^{\ast }:=\underset{|F|<n}{\inf }\underset{n\notin F}{\sup }|{\xi }_n|,F\text{is a finite subset of}\mathbb{N}.$$

$$\mathcal{F}(\mathcal{H})\subset \mathcal{I}\subset \mathcal{K}(\mathcal{H})$$

$${\mathcal{C}}_E:=\{x\in \mathcal{K}(\mathcal{H}):\{s_n(x){\}}_{n=1}^{\infty }\in E\}$$

$${\mathcal{C}}_1\subset {\mathcal{C}}_E\subset \mathcal{K}(\mathcal{H})\text{and}\Vert x{\Vert }_{{\mathcal{C}}_E}\le \Vert x{\Vert }_{{\mathcal{C}}_1},\Vert y{\Vert }_{\infty }\le \Vert y{\Vert }_{{\mathcal{C}}_E}$$

$$E^{\times }=\{\xi =\{{\xi }_n{\}}_{n=1}^{\infty }\in l_{\infty }:\xi \eta \in l_1\text{for all}\eta \in E\}$$

$$(\text{respectively},{\mathcal{C}}_E^{\times }=\{x\in B(H):xy\in {\mathcal{C}}_1\text{for all}y\in {\mathcal{C}}_E\})$$

$$\Vert \xi {\Vert }_{E^{\times }}=\sup \{\sum _{n=1}^{\infty }|{\xi }_n{\eta }_n|:\eta =\{{\eta }_n{\}}_{n=1}^{\infty }\in E,\Vert \eta {\Vert }_E\le 1\}\text{if}\xi \in E^{\times }$$

$$(\text{respectively},\Vert x{\Vert }_{{\mathcal{C}}_E^{\times }}=\sup \{tr(|xy|):y\in {\mathcal{C}}_E,\Vert y{\Vert }_{{\mathcal{C}}_E}\le 1\}\text{if}x\in {\mathcal{C}}_E^{\times }),$$

$$({\mathcal{C}}_E^{\times },\Vert \cdot {\Vert }_{{\mathcal{C}}_E^{\times }})=({\mathcal{C}}_{E^{\times }},\Vert \cdot {\Vert }_{{\mathcal{C}}_{E^{\times }}}).$$

$$0\le a_k\le a_{k+1},a_k\in E\text{(respectively},a_k\in {\mathcal{C}}_E)\text{for all}k\in \mathbb{N}$$

$$\text{ and }\underset{k\ge 1}{\sup }\Vert a_k{\Vert }_E<\infty \text{(respectively},\underset{k\ge 1}{\sup }\Vert a_k{\Vert }_{{\mathcal{C}}_E}<\infty )$$

$${\mathcal{C}}_E\text{is perfect}\iff {\mathcal{C}}_E={\mathcal{C}}_E^{\times \times }\iff E=E^{\times \times }\iff$$

$$E\text{has the Fatou property}\iff {\mathcal{C}}_E\text{has Fatou property}.$$

$$0\le x=\sum _{j=1}^{\infty }{\lambda }_j{p}_j\in {\mathcal{C}}_E\setminus \mathcal{F}(\mathcal{H}),$$

$$\mathcal{F}(\mathcal{H})\ni x-y_n\stackrel{\sigma ({\mathcal{C}}_E,{\mathcal{C}}_E^{\times })}{⟶}x.$$

$$f(T(x_{\alpha }))=T^{\ast }(f)(x_{\alpha })\to T^{\ast }(f)(x)=f(T(x))$$

$$\Vert T{\Vert }_{\mathcal{B}({\mathcal{C}}_E)}=\underset{\Vert x{\Vert }_{{\mathcal{C}}_E}\le 1}{\sup }\Vert T(x){\Vert }_{{\mathcal{C}}_E}.$$

$$\Vert T_n^{\ast }(f)-T^{\ast }(f){\Vert }_{{\mathcal{C}}_E^{\ast }}\le \Vert T_n^{\ast }-T^{\ast }{\Vert }_{\mathcal{B}({\mathcal{C}}_E^{\ast })}\Vert f{\Vert }_{{\mathcal{C}}_E^{\ast }}=\Vert T_n-T{\Vert }_{\mathcal{B}({\mathcal{C}}_E)}\Vert f{\Vert }_{{\mathcal{C}}_E^{\ast }}\to 0,$$

$$B(x,\epsilon )=\{y\in X:\Vert y-x{\Vert }_X\le \epsilon \},\epsilon >0,$$

$$X\setminus \bigcup _{i=1}^{n}B(x_i,{\epsilon }_i),x_i\in X,\Vert x_0-x{\Vert }_X>{\epsilon }_i,i=1,...,n,n\in \mathbb{N}.$$

$$s_n=p_{n1}{r}_1+p_{n2}{r}_2+\dots +p_{nn}{r}_n$$

$$y_n=p_{n1}{x}_1+p_{n2}{x}_2+\dots +p_{nk(n)}{x}_{k(n)},\text{where}\sum _{i=1}^{k(n)}|p_{ni}|=1;$$

$$q_n=(p_{n1},p_{n2}\dots ,p_{nk(n)},0,0\dots )\in \prod _{i=1}^{\infty }[-1,1].$$

$$\underset{N}{\sup }\Vert \sum _{n=1}^N{t}_n{x}_n{\Vert }_X\le C\Vert \{t_n{\}}_{n=1}^{\infty }{\Vert }_{\infty }$$

$$\sum _{n=1}^m|f(x_n-x_{n+1})|=f(\sum _{n=1}^m(x_n-x_{n+1}))=f(x_1-x_{m+1})\le f(x_1)$$

$$u(x)=Ref(x)=\frac{f(x)+\overline{f(x)}}{2},v(x)=Imf(x)=\frac{f(x)-\overline{f(x)}}{2i},x\in {\mathcal{C}}_E.$$

$$\sum _{n=1}^{\infty }(V{x}_n-V{x}_{n+1})$$

Full text

1. Introduction

Let $\mathcal{H}$ be a complex separable infinite-dimensional Hilbert space. Let $(E,\|\cdot\|_{E})\subset c_{0}$ be a real Banach symmetric sequence space. Consider an ideal $\mathcal{C}_{E}$ of compact linear operators in $\mathcal{H}$, which is defined by the relations

[TABLE]

and

[TABLE]

where $\{s_{n}(x)\}_{n=1}^{\infty}$ are the singular values of $x$ (i.e. the eigenvalues of $(x^{*}x)^{1/2}$ in decreasing order). In the paper [10] it is shown that $\|\cdot\|_{\mathcal{C}_{E}}$ is a Banach norm on $\mathcal{C}_{E}$. In addition, $\mathcal{C}_{1}:=\mathcal{C}_{l_{1}}\subset\mathcal{C}_{E}.$

In the case when $(E,\|\cdot\|_{E})$ is a separable Banach space, the Banach ideal $(\mathcal{C}_{E},\|\cdot\|_{\mathcal{C}_{E}})$ is a minimal Banach ideal in the terminology of Schatten [16], i.e. the set of finite rank operators is dense in $\mathcal{C}_{E}$.

It is known [18] that for every surjective linear isometry $V$ on a minimal Banach ideal $\mathcal{C}_{E}\neq\mathcal{C}_{2}$, where $\mathcal{C}_{2}:=\mathcal{C}_{l_{2}}$, there exist unitary operators $u$ and $v$ on $\mathcal{H}$ such that

[TABLE]

for all $x\in\mathcal{C}_{E}$, where $x^{t}$ is the transpose of the operator $x$ with respect to a fixed orthonormal basis in $\mathcal{H}$. In the case of a Banach ideal $\mathcal{C}_{1}$, a description of surjective linear isometries of the form of (1) was obtained in [15], and for the ideals $\mathcal{C}_{p}:=\mathcal{C}_{l_{p}},\ 1\leq p\leq\infty,\ p\neq 2$, in [2].

In this paper we show that, in the case a Banach symmetric sequence space $E\neq l_{2}$ with Fatou property, every surjective linear isometry $V$ on $\mathcal{C}_{E}$ has the form (1). In addition, it is proved that in this case any 2-local surjective isometry on $\mathcal{C}_{E}$ also is of the form (1).

2. Preliminaries

Let $l_{\infty}$ (respectively, $c_{0}$) be a Banach lattice of all bounded (respectively, converging to zero) sequences $\{\xi_{n}\}_{n=1}^{\infty}$ of real numbers with respect to the norm $\|\{\xi_{n}\}\|_{\infty}=\sup\limits_{n\in\mathbb{N}}|\xi_{n}|$, where $\mathbb{N}$ is the set of natural numbers. If $x=\{\xi_{n}\}_{n=1}^{\infty}\in l_{\infty}$, then a non-increasing rearrangement of $x$ is defined by

[TABLE]

A non-zero linear subspace $E\subset l_{\infty}$ with a Banach norm $\|\cdot\|_{E}$ is called Banach symmetric sequences space if the conditions $y\in E$, $x\in l_{\infty},\ x^{*}\leq y^{*}$, imply that $x\in E$ and $\|x\|_{E}\leq\|y\|_{E}$. In this case, the inequality $\|x\|_{\infty}\leq\|x\|_{E}$ follows for each $x\in E$.

Let $\mathcal{H}$ be a complex separable infinite-dimensional Hilbert space and let $\mathcal{B}(\mathcal{H})$ (respectively, $\mathcal{K}(\mathcal{H}),\mathcal{F}(\mathcal{H})$) be the $\ast$-algebra of all bounded (respectively, compact, finite rank) linear operators in $\mathcal{H}$. It is well known that

[TABLE]

for any proper two-sided ideal $\mathcal{I}$ in $\mathcal{B}(\mathcal{H})$ (see for example, [17, Proposition 2.1]).

If $(E,\|\cdot\|_{E})\subset c_{0}$ is a Banach symmetric sequence space, then the set

[TABLE]

is a proper two-sided ideal in $\mathcal{B}(\mathcal{H})$ ([3]; see also [17, Theorem 2.5]). In addition, $(\mathcal{C}_{E},\|\cdot\|_{\mathcal{C}_{E}})$) is a Banach space with respect to the norm $\|x\|_{\mathcal{C}_{E}}=\|\{s_{n}(x)\}_{n=1}^{\infty}\|_{E}$ [10], and the norm $\|\cdot\|_{\mathcal{C}_{E}}$ has the following properties

1. $\|xzy\|_{\mathcal{C}_{E}}\leq\|x\|_{\infty}\|y\|_{\infty}\|z\|_{\mathcal{C}_{E}}$ for all $x,y\in\mathcal{B}(\mathcal{H})$ and $z\in\mathcal{C}_{E}$, where $\|\cdot\|_{\infty}$ is the usual operator norm in $\mathcal{B}(\mathcal{H})$;

2. $\|x\|_{\mathcal{C}_{E}}=\|x\|_{\infty}$ if $x\in\mathcal{C}_{E}$ is of rank 1.

In this case we say that $(\mathcal{C}_{E},\|\cdot\|_{\mathcal{C}_{E}})$) is a Banach symmetric ideal (cf. [9, Ch. III], [17, Ch. 1, §1.7]).  It is clear that $\|uxv\|_{\mathcal{C}_{E}}=\|x\|_{\mathcal{C}_{E}}$ for all unitary operators $u,v\in\mathcal{B}(\mathcal{H})$ and $x\in{\mathcal{C}_{E}}$. Besides,

[TABLE]

for all $x\in\mathcal{C}_{1},\ y\in\mathcal{C}_{E}$.

A Banach symmetric sequence space $(E,\|\cdot\|_{E})$ (respectively, a Banach symmetric ideal $(\mathcal{C}_{E},\|\cdot\|_{\mathcal{C}_{E}})$) is said to have order continuous norm if $\|a_{k}\|_{E}\downarrow 0$ (respectively, $\|a_{k}\|_{\mathcal{C}_{E}}\downarrow 0$) whenever $a_{k}\in E$ (respectively, $a_{k}\in\mathcal{C}_{E})$ and $a_{k}\downarrow 0$. It is known that $(\mathcal{C}_{E},\|\cdot\|_{\mathcal{C}_{E}}))$ has order continuous norm if and only if $(E,\|\cdot\|_{E})$ has order continuous norm (see, for example, [5, Proposition 3.6]). Moreover, every Banach symmetric ideal with order continuous norm is a minimal norm ideal, i.e. the subspace $\mathcal{F}(\mathcal{H})$ is dense in $\mathcal{C}_{E}$.

If $(E,\|\cdot\|_{E})$ is a Banach symmetric sequence space (respectively, $(\mathcal{C}_{E},\|\cdot\|_{\mathcal{C}_{E}})$ is a Banach symmetric ideal), then the Köthe dual $E^{\times}$ (respectively, $\mathcal{C}_{E}^{\times}$) is defined as

[TABLE]

[TABLE]

and

[TABLE]

[TABLE]

where $tr(\cdot)$ is the standard trace on $\mathcal{B}(\mathcal{H})$.

It is known that $(E^{\times},\|\cdot\|_{E^{\times}})$ (respectively, $(\mathcal{C}_{E}^{\times},\|\cdot\|_{\mathcal{C}_{E}^{\times}})$ is a Banach symmetric sequence space (respectively, a Banach symmetric ideal). In addition, $\mathcal{C}_{1}^{\times}=\mathcal{B}(\mathcal{H})$ and if $\mathcal{C}_{E}\neq\mathcal{C}_{1}$, then $\mathcal{C}_{E}^{\times}\subset\mathcal{K}(\mathcal{H})$ [7, Proposition 7]. We also note the following useful property [5, Theorem 5.6]:

[TABLE]

A Banach symmetric ideal $\mathcal{C}_{E}$ is said to be perfect if $\mathcal{C}_{E}=\mathcal{C}_{E}^{\times\times}$. It is clear that $\mathcal{C}_{E}$ is perfect if and only if $E=E^{\times\times}$.

A Banach symmetric sequence space $(E,\|\cdot\|_{E})$ (respectively, a Banach symmetric ideal $(\mathcal{C}_{E},\|\cdot\|_{\mathcal{C}_{E}})$) is said to possess Fatou property if the conditions

[TABLE]

[TABLE]

imply that there exists $a\in E\ (a\in\mathcal{C}_{E})$ such that $a_{k}\uparrow a$ and $\|a\|_{E}=\sup\limits_{k\geq 1}\|a_{k}\|_{E}$ (respectively, $\|a\|_{\mathcal{C}_{E}}=\sup\limits_{k\geq 1}\|a_{k}\|_{\mathcal{C}_{E}}$).

It is known that $(E,\|\cdot\|_{E})$ (respectively, $(\mathcal{C}_{E},\|\cdot\|_{\mathcal{C}_{E}})$) has the Fatou property if and only if $E=E^{\times\times}$ [12, Vol. II, Ch. 1, Section a] (respectively, $\mathcal{C}_{E}=\mathcal{C}_{E}^{\times\times}$ [5, Theorem 5.14]). Therefore

[TABLE]

[TABLE]

If $y\in\mathcal{C}_{E}^{\times}$, then a linear functional $f_{y}(x)=tr(xy)=tr(yx),\ x\in\mathcal{C}_{E}$, is continuous on $\mathcal{C}_{E}$. In addition, $\|f_{y}\|_{\mathcal{C}_{E}^{\ast}}=\|y\|_{\mathcal{C}_{E}^{\times}}$, where $\mathcal{C}_{E}^{*}$ is the dual of the Banach space $(\mathcal{C}_{E},\|\cdot\|_{\mathcal{C}_{E}})$ (see, for example, [7]). Identifying an element $y\in\mathcal{C}_{E}^{\times}$ and the linear functional $f_{y}$, we may assume that $\mathcal{C}_{E}^{\times}$ is a closed linear subspace in $\mathcal{C}_{E}^{*}$. Since $\mathcal{F}(\mathcal{H})\subset\mathcal{C}_{E}^{\times}$, it follows that $\mathcal{C}_{E}^{\times}$ is a total subspace in $\mathcal{C}_{E}^{*}$. Thus, the weak topology $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$ is a Hausdorff topology.

Proposition 1**.**

[4, Proposition 2.8]** A linear functional $f\in\mathcal{C}_{E}^{*}$ is continuous with respect to weak topology $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$ (i.e. $f\in\mathcal{C}_{E}^{\times}$) if and only if $f(x_{n})\to 0$ for every sequence $x_{n}\downarrow 0,\ x_{n}\in\mathcal{C}_{E},\ n\in\mathbb{N}$.

It is clear that $\mathcal{C}_{E}^{h}=\{x\in\mathcal{C}_{E}:x=x^{\ast}\}$ is a real Banach space with respect to the norm $\|\cdot\|_{\mathcal{C}_{E}}$; besides, the cone $\mathcal{C}_{E}^{+}=\{x\in\mathcal{C}_{E}^{h}:x\geq 0\}$ is closed in $\mathcal{C}_{E}^{h}$ and $\mathcal{C}_{E}^{h}=\mathcal{C}_{E}^{+}-\mathcal{C}_{E}^{+}$. If $0\leq x_{n}\leq\ x_{n+1}\leq x,\ x,x_{n}\in\mathcal{C}_{E}^{h},\ n\in\mathbb{N}$, then there exists $\sup_{n\geq 1}x_{n}\in\mathcal{C}_{E}^{h}$, i.e. $\mathcal{C}_{E}^{h}$ is a quasi-$(o)$-complete space [1]. Hence, by [1, Theorem 1], we have the following proposition.

Proposition 2**.**

Every linear functional $f\in(\mathcal{C}_{E}^{h})^{*}$ is a difference of two positive linear functionals from $(\mathcal{C}_{E}^{h})^{*}$.

We need the following property of the weak topology $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$.

Proposition 3**.**

If $x\in\mathcal{C}_{E}$, then there exists a sequence $\{x_{n}\}\subset\mathcal{F}(\mathcal{H})$ such that $x_{n}\stackrel{{\scriptstyle\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})}}{{\longrightarrow}}x$.

Proof.

It suffices to establish the validity of Proposition 3 for

[TABLE]

where $\lambda_{j}\geq 0$ are the eigenvalues of the compact operator $x$ and $p_{j}\in\mathcal{F}(\mathcal{H})$ are finite-dimensional projectors for all $j\in\mathbb{N}$ (the series converges with respect to the norm $\|\cdot\|_{\infty}$). If $y_{n}=\sum\limits_{j=n}^{\infty}\lambda_{j}p_{j}$, then $y_{n}\downarrow 0$ and, by Proposition, 1 we have $f(y_{n})\to 0$ for all $f\in\mathcal{C}_{E}^{\times}$. Therefore

[TABLE]

∎

Let $T$ be a bounded linear operator acting in a Banach symmetric ideal $(\mathcal{C}_{E},\|\cdot\|_{\mathcal{C}_{E}})$, and let $T^{*}$ be its adjoint operator. If $T^{*}(\mathcal{C}_{E}^{\times})\subset\mathcal{C}_{E}^{\times}$, then

[TABLE]

for every net $x_{\alpha}\stackrel{{\scriptstyle\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})}}{{\longrightarrow}}x,\ x_{\alpha},x\in\mathcal{C}_{E}$, and for all $f\in\mathcal{C}_{E}^{\times}$. Thus $T$ is a $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-continuous operator.

Inversely, if the operator $T$ is a $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-continuous and $f\in\mathcal{C}_{E}^{\times}$, then the linear functional $(T^{*}f)(x)=f(T(x))$ is also $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-continuous, i.e. $T^{*}(f)\in\mathcal{C}_{E}^{\times}$.

Therefore, a linear bounded operator $T\colon\mathcal{C}_{E}\to\mathcal{C}_{E}$ is $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-continuous if and only if $T^{*}(f)\in\mathcal{C}_{E}^{\times}$ for every $f\in\mathcal{C}_{E}^{\times}$. Thus, for any $\lambda\in\mathbb{C}$ and linear bounded operators $T,S:\mathcal{C}_{E}\to\mathcal{C}_{E}$, continuous with respect to a $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-topology, the operators $T+S$, $TS$ and $\lambda T$ are $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-continuous.

Let $\mathcal{B}(\mathcal{C}_{E})$ be the Banach space of bounded linear operators $T$ acting in $(\mathcal{C}_{E},\|\cdot\|_{\mathcal{C}_{E}})$ with the norm

[TABLE]

Proposition 4**.**

If $T_{n}\in\mathcal{B}(\mathcal{C}_{E})$ are $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-continuous operators, $T\in\mathcal{B}(\mathcal{C}_{E})$, and $\|T_{n}-T\|_{\mathcal{B}(\mathcal{C}_{E})}\to 0$ as $n\to\infty$, then $T$ is also $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-continuous.

Proof.

It suffices to show that $T^{*}(f)\in\mathcal{C}_{E}^{\times}$ for any functional $f\in\mathcal{C}_{E}^{\times}$. Since $T_{n}\in\mathcal{B}(\mathcal{C}_{E})$ are $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-continuous operators, it follows that  $T_{n}^{*}(f)\in\mathcal{C}_{E}^{\times}$  for all  $f\in\mathcal{C}_{E}^{\times},\ n\in\mathbb{N}$. Considering the closed subspace $\mathcal{C}_{E}^{\times}$ in $(\mathcal{C}_{E}^{*},\|\cdot\|_{\mathcal{C}_{E}^{*}})$ and noting that

[TABLE]

we conclude that $T^{*}(f)\in\mathcal{C}_{E}^{\times}$. ∎

We also need the following well-known properties of perfect symmetrically normed ideals.

Theorem 1**.**

Let $(\mathcal{C}_{E},\|\cdot\|_{\mathcal{C}_{E}})\subset c_{0}$ be a Banach symmetric sequence space with Fatou property. Then

$(i)$. [5, Theorem 5.11]. $\|x\|_{\mathcal{C}_{E}}=\sup\limits_{y\in\mathcal{C}_{E}^{\times},\|y\|_{\mathcal{C}_{E}^{\times}}\leq 1}|tr(xy)|$  for every  $x\in\mathcal{C}_{E}$;

$(ii)$. [4, Theorem 3.5]. A Banach space $\mathcal{C}_{E}$ is $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-sequentially complete, i.e. if for $x_{n}\in\mathcal{C}_{E},\ n\in\mathbb{N}$, and for every $f\in\mathcal{C}_{E}^{\times}$ there is $\lim\limits_{n\to\infty}f(x_{n})$, then there exists $x\in\mathcal{C}_{E}$ such that $x_{n}\stackrel{{\scriptstyle\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})}}{{\longrightarrow}}x$.

3. The ball topology in norm ideals of compact operators

Let $(X,\|\cdot\|_{X})$ be a real Banach space and let $b_{X}$ be the ball topology in $X$, i.e. $b_{X}$ is the coarsest topology such that every closed ball

[TABLE]

is closed in $b_{X}$ [8]. The family

[TABLE]

is a base of neighborhoods of the point $x_{0}\in X$ in $b_{X}$. Therefore $x_{\alpha}\stackrel{{\scriptstyle b_{X}}}{{\longrightarrow}}x,\ x_{\alpha},x\in X$, if and only if $\lim\inf\|x_{\alpha}-x\|_{X}\geq\|x-y\|_{X}$ for all $y\in X$ [8]. In particular, every surjective isometry $V$ in $(X,\|\cdot\|_{X})$ is continuous with respect to the ball topology $b_{X}$.

Let us note that $b_{X}$ is not a Hausdorff topology. The following theorem provides a sufficient condition for $T_{2}$-axiom of $b_{X}$ on subsets of $X$. Recall that $A$ is a Rosenthal subset of $(X,\|\cdot\|_{X})$ if every sequence in $A$ has a weakly Cauchy subsequence.

Theorem 2**.**

[8, Theorem 3.3]**. Let $(X,\|\cdot\|_{X})$ be a real Banach space, and let $A$ be a bounded absolutely convex Rosenthal subset of $X$. Then $(A,b_{X})$ is a Hausdorff space.

In the proof of required properties of the ball topology (see Theorem 3 below), we utilize the following well-known proposition.

Proposition 5**.**

[14, section 3, Ch. 1, §5]**. Let $p_{nk}$ be real numbers, $n,k\in\mathbb{N}$, such that $\sum\limits_{k=1}^{n}|p_{nk}|=1$ for all $n\in\mathbb{N}$, and let the limit $\lim\limits_{n\to\infty}p_{nk}=p_{k}$ exist for every fixed $k\in\mathbb{N}$. Then the sequence

[TABLE]

converges for every convergent sequence $\{r_{n}\}$.

Theorem 3**.**

Let $(\mathcal{C}_{E},\|\cdot\|_{\mathcal{C}_{E}})$ be a Banach symmetric ideal, $x_{n}\in\mathcal{C}_{E}^{+}$, and let $x_{n}\downarrow 0$. Then the sequence $\{x_{n}\}$ can converge with respect to the topology $b_{\mathcal{C}_{E}}$ to no more than one element.

Proof.

Consider $\mathcal{C}_{E}$ as a real Banach space. Let $A$ be the absolutely convex hull of the sequence $\{0,x_{1},...,x_{n},...\}$. Since $x_{n}\downarrow 0$ and the norm $\|\cdot\|_{\mathcal{C}_{E}}$ is monotone, it follows that $A$ is a bounded subset of $\mathcal{C}_{E}$. Let us show that $A$ is a Rosenthal subset.

Using the inequality $0\leq x_{n+1}\leq x_{n}$ and a decomposition of linear functional $f\in(\mathcal{C}_{E}^{h})^{*}$ as the difference of two positive functionals from $(\mathcal{C}_{E}^{h})^{*}$ (see Proposition 2), we conclude that there exists $\lim\limits_{n\to\infty}f(x_{n})=r_{n}$ for every functional $f\in(\mathcal{C}_{E}^{h})^{*}$. For any sequence $\{y_{n}\}_{n=1}^{\infty}\subset A$, we have

[TABLE]

in particular, $p_{ni}\in[-1,1]$ for all $i\in 1,\ldots,k(n)$. Consider a sequence

[TABLE]

By Tychonoff Theorem, a set $\prod\limits_{i=1}^{\infty}[-1,1]$ is compact with respect to the product topology. Moreover, by [11, ch. 4, theorem 17], the set $\prod\limits_{i=1}^{\infty}[-1,1]$ is a metrizable compact. Hence, the sequence $\{q_{n}\}$ has a convergent subsequence $\{q_{n_{i}}\}_{i=1}^{\infty}$. In particular, there are the limits $\lim\limits_{n_{i}\to\infty}p_{n_{i}k}$ for all $k\in\mathbb{N}$. Besides, the sequence $f(y_{n_{i}})=\sum\limits_{j=1}^{k}p_{{n_{i}}j}f(x_{j}),\ f\in(\mathcal{C}_{E}^{h})^{*}$, satisfies all conditions of Proposition 5, which implies its convergence. Therefore, the sequence $\{y_{n}\}_{n=1}^{\infty}\subset A$ has a weakly Cauchy subsequence $\{y_{n_{i}}\}$. This means that $A$ is a Rosenthal subset of $(\mathcal{C}_{E}^{h},\|\cdot\|_{\mathcal{C}_{E}})$.

By Theorem 2, the topological space $(A,b_{\mathcal{C}_{E}^{h}})$ is Hausdorff, hence the sequence $\{x_{n}\}_{n=1}^{\infty}\subset A$ could not have more than one limit with respect to the topology $b_{\mathcal{C}_{E}^{h}}$. Since $\mathcal{C}_{E}^{h}$ is a closed subspace in $(\mathcal{C}_{E},\|\cdot\|_{\mathcal{C}_{E}})$ (we assume that $\mathcal{C}_{E}$ is a real space), it follows that the restriction $(b_{\mathcal{C}_{E}})|_{\mathcal{C}_{E}^{h}}$ is finer than $b_{\mathcal{C}_{E}^{h}}$. Therefore, the sequence $\{x_{n}\}_{n=1}^{\infty}$ can have no more than one limit with respect to the topology $b_{\mathcal{C}_{E}}$. ∎

Proposition 6**.**

If $E\subset c_{0}$ is a Banach symmetric sequence space with Fatou property, then $b_{\mathcal{C}_{E}}\leq\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$.

Proof.

It suffices to show that $B(0,1)=\{x\in\mathcal{C}_{E}:\|x\|_{\mathcal{C}_{E}}\leq 1\}$ is closed with respect to the weak topology $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$. Let $x_{\alpha}\in B(0,1)$  and  $x_{\alpha}\stackrel{{\scriptstyle\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})}}{{\longrightarrow}}x\in\mathcal{C}_{E}$. Assume that $x\notin B(0,1)$, i.e. $\|x\|_{\mathcal{C}_{E}}=q>1$, and let $\epsilon>0$ be such that $q-\epsilon>1$. By Theorem 1 $(i)$, there exists $y\in\mathcal{C}_{E}^{\times}$ such that $\|y\|_{\mathcal{C}_{E}^{\times}}\leq 1$ and $q\geq|tr(xy)|>q-\epsilon$.

On the other hand, $x_{n}\stackrel{{\scriptstyle\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})}}{{\longrightarrow}}x$ implies that $|tr(x_{n}y)|\to|tr(xy)|$. Since $x_{n}\in B(0,1)$, it follows that $|tr(x_{n}y)|\leq 1$ and therefore $|tr(xy)|\leq 1$, which is impossible. Thus $B(0,1)$ is a closed set in $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$. ∎

4. Weak continuity of Hermitian operators in the perfect ideals of compact operators

In this section, we establish $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-continuity of the Hermitian operator acting in a perfect Banach symmetric ideal $\mathcal{C}_{E}$ of compact operators. For that we need $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-continuity of a surjective isometry on $\mathcal{C}_{E}$.

Recall that the series $\sum\limits_{n=1}^{\infty}x_{n}$ converges weakly unconditionally in a Banach space $X$ if a numerical series $\sum\limits_{n=1}^{\infty}f(x_{n})$ converges absolutely for every $f\in X^{*}$ [19, Ch. 2, §3].

Proposition 7**.**

[19*, Ch. 2, §3]**. Let $(X,\|\cdot\|_{X})$ be a Banach space, $x_{n}\in X,\ n\in\mathbb{N}$. Then the following conditions are equivalent:

$(i)$ A series $\sum\limits_{n=1}^{\infty}x_{n}$ converges weakly unconditionally;

$(ii)$ There exists a constant $C>0$, such that*

[TABLE]

for all $\{t_{n}\}_{n=1}^{\infty}\in l_{\infty}.$

Proposition 7 implies the following.

Corollary 4**.**

If $V$ is a surjective linear isometry on a Banach space $X$ and a series $\sum\limits_{n=1}^{\infty}x_{n}$ converges weakly unconditionally in $X$, then the series $\sum\limits_{n=1}^{\infty}V(x_{n})$ also converges weakly unconditionally in $X$.

Now we can show that every surjective linear isometry of $\mathcal{C}_{E}$ is $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-continuous.

Proposition 8**.**

Let $\mathcal{C}_{E}$ be a perfect Banach symmetric ideal of compact operators and $V$ a surjective linear isometry on $\mathcal{C}_{E}$. Then $V$ is $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-continuous.

Proof.

It suffices to show that $V^{*}(f)\in\mathcal{C}_{E}^{\times}$ for each functional $f\in\mathcal{C}_{E}^{\times}$. According to Proposition 1, it should be established that $x_{n}\downarrow 0,\ \{x_{n}\}_{n=1}^{\infty}\subset\mathcal{C}_{E}^{+}$, implies $f(V(x_{n}))=V^{*}(f)(x_{n})\to 0$.

Let $\{x_{n}\}_{n=1}^{\infty}\subset\mathcal{C}_{E}$  and  $x_{n}\downarrow 0$. We will show that the series $\sum\limits_{n=1}^{\infty}(x_{n}-x_{n+1})$   converges weakly unconditionally in $\mathcal{C}_{E}$. If $f\in(\mathcal{C}_{E}^{h})^{*}$ is a positive linear functional, then

[TABLE]

for all $m\in\mathbb{N}$. Hence the numerical series $\sum\limits_{n=1}^{\infty}f(x_{n}-x_{n+1})$ converges absolutely. Since every functional $f\in(\mathcal{C}_{E}^{h})^{*}$ is the difference of two positive functionals from $(\mathcal{C}_{E}^{h})^{*}$ (see Proposition 2), the series $\sum\limits_{n=1}^{\infty}(x_{n}-x_{n+1})$ converges weakly unconditionally in $\mathcal{C}_{E}^{h}$.

Let $f\in\mathcal{C}_{E}^{*}$. Denote

[TABLE]

It is clear that $u,v\in(\mathcal{C}_{E}^{h})^{*}$. Therefore, the series $\sum\limits_{n=1}^{\infty}u(x_{n}-x_{n+1})$ and $\sum\limits_{n=1}^{\infty}v(x_{n}-x_{n+1})$ converge absolutely. Thus, the series $\sum\limits_{n=1}^{\infty}f(x_{n}-x_{n+1})$ also converges absolutely. This means that a series $\sum\limits_{n=1}^{\infty}(x_{n}-x_{n+1})$ converges weakly unconditionally in $\mathcal{C}_{E}$.

By Corollary 4, the series

[TABLE]

converges weakly unconditionally in $\mathcal{C}_{E}$. Hence the numerical series

[TABLE]

converges for every $f\in\mathcal{C}_{E}^{*}$. In particular, the limit $\lim\limits_{n\to\infty}f(Vx_{n})$ exists. Since $\mathcal{C}_{E}$ is a $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-sequentially complete set (see Theorem 1$(ii)$) and $V$ is a bijection, it follows that there exists $x_{0}\in\mathcal{C}_{E}$ such that $V(x_{n})\stackrel{{\scriptstyle\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})}}{{\longrightarrow}}V(x_{0})$ as $n\to\infty$.

It remains to show that $V(x_{0})=0$. By Proposition 6,

[TABLE]

Since the isometry $V^{-1}$ is continuous with respect to the topology $b_{\mathcal{C}_{E}}$, it follows that $x_{n}\stackrel{{\scriptstyle b_{\mathcal{C}_{E}}}}{{\longrightarrow}}x_{0}$. Now, taking into account that $x_{n}\downarrow 0$, we obtain $x_{n}\stackrel{{\scriptstyle\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})}}{{\longrightarrow}}0$ (see Proposition 1). Then, by Proposition 6, $x_{n}\stackrel{{\scriptstyle b_{\mathcal{C}_{E}}}}{{\longrightarrow}}0$, and Theorem 3 implies that $x_{0}=0$. ∎

Let $(X,\|\cdot\|_{X})$ be a complex Banach space. A bounded linear operator $T\colon X\to X$ is called Hermitian, if the operator  $e^{itT}=\sum\limits_{n=0}^{\infty}\frac{(itT)^{n}}{n!}$  is an isometry of the space $X$ for all $t\in\mathbb{R}$ (see, for example, [6, Ch. 5, §2]).

We show that a Hermitian operator acting in a perfect Banach symmetric ideal $\mathcal{C}_{E}$ is $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-continuous.

Theorem 5**.**

Let $E\subset c_{0}$ be a Banach symmetric sequence space with Fatou property, and let $T$ be a Hermitian operator acting in $\mathcal{C}_{E}$ . Then $T$ is $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-continuous.

Proof.

Consider the non-negative continuous function $\alpha(t)=\|e^{itT}-I\|_{\mathcal{B}(\mathcal{C}_{E})}$, where $t\in\mathbb{R}$. As $\alpha(0)=0$, it follows that there exists $t_{0}\in\mathbb{R}$ such that $\alpha(t_{0})<1$. Since the operator $T$ is Hermitian, it follows that the operator $V=e^{it_{0}T}$ (and hence $V^{-1}=e^{-it_{0}T}$) is an isometry on $\mathcal{C}_{E}$. By Proposition 8, $S=V-I$ is a $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-continuous operator. In addition, $\|S\|=\alpha(t_{0})<1$. Now, since

[TABLE]

Proposition 4 implies that $it_{0}T$  is a $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-continuous operator. Therefore, $T$ is a $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-continuous operator. ∎

If $a,b$ are self-adjoint operators in $\mathcal{B}(\mathcal{H})$, $x\in\mathcal{C}_{E}$, and

[TABLE]

for all $x\in\mathcal{C}_{E}$, then $T$ is a Hermitian operator acting in $\mathcal{C}_{E}$ [18]. In the following theorem, utilizing the method of proof of Theorem 1 in [18], we show that every Hermitian operators acting in a perfect norm ideal $(\mathcal{C}_{E},\|\cdot\|_{\mathcal{C}_{E}})\neq C_{2}$ has the form (2).

Theorem 6**.**

Let $E\subset c_{0}$ be a Banach symmetric sequence space with Fatou property,  $E\neq l_{2}$, and let $T$ be a Hermitian operator acting in a Banach symmetric ideal $\mathcal{C}_{E}$. Then there are self-adjoint operators $a,b\in\mathcal{B}(\mathcal{H})$ such that $T(x)=ax+xb$ for all $x\in\mathcal{C}_{E}$.

Proof.

As in the proof of Theorem 1 from [18], we have that there are self-adjoint operators $a,b\in\mathcal{B}(\mathcal{H})$ such that $T(x)=ax+xb$ for all $x\in\mathcal{F}(\mathcal{H})$. Fix $x\in\mathcal{C}_{E}$. By Proposition 3, there is a sequence $\{x_{n}\}\subset\mathcal{F}(\mathcal{H})$ such that $x_{n}\stackrel{{\scriptstyle\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})}}{{\longrightarrow}}x$. If $y\in\mathcal{C}_{E}^{\times}$, then

[TABLE]

[TABLE]

Therefore

[TABLE]

Since $T$ is a Hermitian operator, it follows that $T$ is $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-continuous (see Theorem 5). Therefore

[TABLE]

∎

5. Isometries of a Banach symmetric ideal

In this section, we prove our main result, Theorem 7. The proof of Theorem 7 is similar to the proof of Theorem 2 in [18]. We use a version of Theorem 1 in [18] for a perfect Banach symmetric ideal $\mathcal{C}_{E}$ (Theorem 6) as well as $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-continuity of every isometry on $\mathcal{C}_{E}$ (Proposition 8) and $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-density of the space $\mathcal{F}(\mathcal{H})$ in $\mathcal{C}_{E}$ (Proposition 3).

Recall that $x^{t}$ stands for the transpose of an operator $x\in\mathcal{K}(\mathcal{H})$ with respect to a fixed orthonormal basis in $\mathcal{H}$.

Theorem 7**.**

Let $E\subset c_{0}$ be a Banach symmetric sequence space with Fatou property,  $E\neq l_{2}$, and let $V$ be a surjective linear isometry on the Banach symmetric ideal $\mathcal{C}_{E}$. Then there are unitary operators $u$ and $v$ on $\mathcal{H}$ such that

[TABLE]

for all $x\in\mathcal{C}_{E}$

Note that each linear operator of the form (3) is an isometry on every Banach symmetric ideal $\mathcal{C}_{E}$.

Proof.

Let $y\in\mathcal{B}(\mathcal{H})$, and let $l_{y}(x)=yx$ (respectively, $r_{y}(x)=xy$) for all $x\in\mathcal{C}_{E}$. It is clear that  $l_{y}$  and  $r_{y}$  are bounded linear operators acting in $\mathcal{C}_{E}$. Using Theorem 6 and repeating the proof of the Theorem 2 [18], we conclude that there are unitary operators $u,v\in\mathcal{B}(\mathcal{H})$ such that $Vl_{y}V^{-1}=l_{uyu^{*}}$ and $Vr_{y}V^{-1}=r_{v^{*}yv}$  for any $y\in\ \mathcal{B}(\mathcal{H})$.

In the case $Vl_{y}V^{-1}=l_{uyu^{*}}$, we define an isometry $V_{0}$ on $\mathcal{C}_{E}$ by the equation $V_{0}(x)=u^{*}V(x)v^{*}$. As in the proof of Theorem 2 [18], we get that $V_{0}(x)=\lambda x$ for every $x\in\mathcal{F}(\mathcal{H})$ and some $\lambda\in\mathbb{C}$. Since the isometry $V_{0}$ is $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$-continuous (Proposition 8) and the space $\mathcal{F}(\mathcal{H})$ is $\sigma(\mathcal{C}_{E},\mathcal{C}_{E}^{\times})$- dense in $\mathcal{C}_{E}$ (Proposition 3), it follows that $V_{0}(x)=\lambda x$ for every $x\in\mathcal{C}_{E}$. Now, since $V_{0}$ is an isometry on $\mathcal{C}_{E}$, we have $\lambda=1$, i.e. $V(x)=uxv$ for all $x\in\mathcal{C}_{E}$.

In the case $Vl_{y}V^{-1}=r_{v^{*}yv}$, as in the proof of Theorem 2 [18], we use the above to derive that there are unitary operators $u,v\in\mathcal{B}(\mathcal{H})$ such that $V(x)=ux^{t}v$ for all $x\in\mathcal{C}_{E}$. ∎

Corollary 8**.**

Let $E\subset c_{0}$ be a Banach symmetric sequence space with Fatou property,  $E\neq l_{2}$, and let $V$ be a surjective linear isometry on $\mathcal{C}_{E}$. Then $V(yx^{*}y)=V(y)(V(x))^{*}V(y)$ for all $x,y\in\mathcal{C}_{E}$.

Proof.

By Theorem 7, there are unitary operators $u$ and $v$ on $\mathcal{H}$ such that $V(x)=uxv\ \ \text{or}\ \ V(x)=ux^{t}v$ for all $x\in\mathcal{C}_{E}$. If $V(x)=uxv$, then

[TABLE]

Since $(x^{t})^{*}=(x^{*})^{t},\ x\in\mathcal{B}(\mathcal{H})$, in the case $V(x)=ux^{t}v$, we have

[TABLE]

[TABLE]

∎

Let $(X,\|\cdot\|_{X})$ be an arbitrary complex Banach space. A surjective (not necessarily linear) mapping $T\colon X\to X$ is called a surjective 2-local isometry [13], if for any $x,y\in X$ there exists a surjective linear isometry $V_{x,y}$ on $X$ such that $T(x)=V_{x,y}(x)$ and $T(y)=V_{x,y}(y)$. It is clear that every surjective linear isometry on $X$ is automatically a surjective 2-local isometry on $X$. In addition,

[TABLE]

for any $x\in X$ and $\lambda\in\mathbb{C}$. In particular,

[TABLE]

Thus, in order to establish linearity of a 2-local isometry $T$, it is sufficient to show that $T(x+y)=T(x)+T(y)$ for all $x,y\in X$.

Note also that

[TABLE]

for any $x,y\in X$. Therefore, in the case a real Banach space $X$,  from (4), (5) and Mazur-Ulam Theorem (see, for example, [6, Chapter I, §1.3, Theorem 1.3.5.]) it follows that every surjective 2-local isometry on $X$ is a linear. For complex Banach spaces, this fact is not valid.

Using the description of isometries on a minimal Banach symmetric ideal $\mathcal{C}_{E}$ from [18], L. Molnar proved that every surjective 2-local isometry on a minimal Banach symmetric ideal $\mathcal{C}_{E}$ is necessarily linear [13, Corallary 5].

The following Theorem is a version of Molnar’s result for a perfect Banach symmetric ideal.

Theorem 9**.**

Let $E\subset c_{0}$ be a Banach symmetric sequence space with Fatou property,  $E\neq l_{2}$, and let $V$ be a surjective 2-local isometry on a Banach symmetric ideal $\mathcal{C}_{E}$. Then $V$ is a linear isometry on $\mathcal{C}_{E}$.

Proof.

Fix $x,y\in\mathcal{F}(\mathcal{H})$ and let $V_{x,y}\colon\mathcal{C}_{E}\to\mathcal{C}_{E}$ be a surjective isometry such that $V(x)=V_{x,y}(x)$  and  $V(y)=V_{x,y}(y)$. By Theorem 7,  there are unitary operators $u$ and $v$ on $\mathcal{H}$ such that $V(x)=uxv$  or  $V(x)=ux^{t}v$ (respectively, $V(y)=uyv$  or  $V(y)=uy^{t}v$). Then we have

[TABLE]

for every $x,y\in\mathcal{F}(\mathcal{H})$. In addition, $V(x)\in\mathcal{F}(\mathcal{H})$ and $V$ is a bijective mapping on $\mathcal{F}(\mathcal{H})$.

If $x,y,z\in\mathcal{F}(\mathcal{H})$, then

[TABLE]

[TABLE]

Therefore

[TABLE]

for all $z\in\mathcal{F}(\mathcal{H})$. Taking $z=x+y$,  $z=x$ and $z=y$, we obtain

[TABLE]

that is, $V(x+y)=V(x)+V(y)$ for all $x,y\in\mathcal{F}(\mathcal{H})$. Hence $V$ is a bijective linear isometry on a normed space $(\mathcal{F}(\mathcal{H}),\|\cdot\|_{\mathcal{C}_{E}})$.

Let $\mathcal{I}$ be the closure of the subspace $\mathcal{F}(\mathcal{H})$ in the Banach space $(\mathcal{C}_{E},\|\cdot\|_{\mathcal{C}_{E}})$. It is clear that $(\mathcal{I},\|\cdot\|_{\mathcal{C}_{E}})$ is a minimal Banach symmetric ideal. Since $V$ is a surjective linear isometry on $(\mathcal{F}(\mathcal{H}),\|\cdot\|_{\mathcal{C}_{E}})$, it follows that $V$ is a surjective linear isometry on $(\mathcal{I},\|\cdot\|_{\mathcal{C}_{E}})$.

By Theorem 2 [18], there are unitary operators $u_{1}$ and $v_{1}$ on $\mathcal{H}$ such that $V(x)=u_{1}xv_{1}$  or  $V(x)=u_{1}x^{t}v_{1}$ for all $x\in\mathcal{I}$. Repeating the ending of the proof of Theorem 1 in [13], we conclude that $V$ is a linear isometry on $\mathcal{C}_{E}$. ∎