An Operator Inequality for Bounded Linear Maps Between $C^*$-Algebras

Abstract

In this paper, we prove that if $\Phi : \mathscr {A} \rightarrow \mathscr {B}$ is a bounded linear map between $C^*$-algebras, then

$$\begin{aligned} \left\| \sum _{k=1}^n \left\{ |\Phi (A_k)|^2+|\Phi (A_k^*)|^2\right\} \right\| \le 2 \Vert \Phi \Vert ^2 \left\| \sum _{k=1}^n \left( |A_k|^2+|A_k^*|^2\right) \right\| \end{aligned}$$

for all $n\in \mathbb {N}$ and all $A_1,\ldots , A_n\in \mathscr {A}$. This improves the Haagerup–Pisier–Ringrose (H–P–R) inequality for Hermitian bounded maps $\Phi$. In addition, we present a refinement of the H–P–R inequality and give an example to support it. Moreover, we establish several versions of the H–P–R inequality involving unitarily invariant norms. Further, we present some useful inequalities in the setting of Hilbert $C^*$-modules and then apply it to get several H–P–R-type inequalities.

1 Introduction

Ringrose [11] proved the following theorem when $\mathscr {A}$ or $\mathscr {B}$ is commutative and conjectured that it is valid in the general case.

Theorem

(Haagerup–Pisier–Ringrose Inequality) If $\Phi$ is a bounded linear map from a $C^*$-algebra $\mathscr {A}$ into a $C^*$-algebra $\mathscr {B}$, then

$$\begin{aligned} \left\| \sum _{j=1}^n\left\{ \Phi (A_j)^* \Phi (A_j)+\Phi (A_j) \Phi (A_j)^*\right\} \right\| \le K\Vert \Phi \Vert ^2 \left\| \sum _{j=1}^n\left( A_j^*A_j+A_jA_j^*\right) \right\| \end{aligned}$$

(1.1)

holds for some K and for any finite family $\{A_1, \ldots , A_n\}$ of elements of $\mathscr {A}$.

Haagerup [5] showed that if $\Phi$ is a bounded linear map from a $C^*$-algebra $\mathscr {A}$ into a Hilbert space $\mathscr {H}$, then there exist states $\rho$ and $\rho '$ of $\mathscr {A}$ such that

$$\begin{aligned} \Vert \Phi (A)\Vert ^2\le \Vert \Phi \Vert ^2(\rho (A^*A)+\rho '(AA^*))\qquad (A \in \mathscr {A}). \end{aligned}$$

(1.2)

In fact, Pisier [10] proved (1.1) with $K=6$, and then Haagerup [5] proved it with $K=4$. We call inequality (1.1) the Haagerup–Pisier–Ringrose (H–P–R) inequality.

In his proof of inequality (1.1), Kadison [6] assumed $\Phi$ as a nonzero bounded linear map of a unital $C^*$-algebra $\mathscr {A}$ into a Hilbert space $(\mathscr {H}, \langle \cdot ,\cdot \rangle )$ and proved that there is a sequence of unitary elements $\{U_n\}$ in $\mathscr {A}$ such that $\Vert \Phi (U_n)\Vert \rightarrow \Vert \Phi \Vert$ and there weak*-limit points (states) of $\rho$ and $\rho '$ of $\{\Vert \Phi \Vert ^{-2}\rho _n\}$ and $\{\Vert \Phi \Vert ^{-2}\rho '_n\}$, respectively, such that

$$\begin{aligned} \rho _n(A)=\langle \Phi (U_nA),\Phi (U_n)\rangle \qquad \rho '_n(A)=\langle \Phi (AU_n),\Phi (U_n)\rangle \qquad (A \in \mathscr {A}) \end{aligned}$$

and (1.2) holds. He then used this inequality and constructed the state $\rho _0(A)=(\rho (A)+\rho '(A))/2$ satisfying $\rho (A)\le 2\rho _0(A)$ and $\rho '(A)\le 2\rho _0(A)$ to prove the H–P–R inequality by using the same reasoning as that of Haagerup.

One may notice that if $\Phi (UV)=\Phi (VU)$ for all unitaries $U, V \in \mathscr {A}$, then $\rho =\rho '$ and we no longer need to consider $\rho _0$ and sharpen the constant 4 in the H–P–R inequality. Clearly validity of $\Phi (UV)=\Phi (VU)$ for all unitaries $U, V\in \mathscr {A}$ is equivalent to that of $\Phi (AB)=\Phi (BA)$ for all $A, B \in \mathscr {A}$ (or equivalently $\Phi (A^*A)=\Phi (AA^*)$ for all $A\in \mathscr {A}$). Such maps are called tracial; see [4] and references therein. Clearly if $\mathscr {A}$ is commutative, then one obtains the result of Ringrose [11] with a refined constant $K=2$ in (1.1). It is remarkable to notice that Choi and Tsui [3] showed that a tracial positive linear map on a $C^*$-algebra is completely positive. If $\Phi$ is a tracial positive linear map, then $-\Phi$ is a bounded and tracial map but is not positive.

An interesting problem is to characterize tracial bounded linear maps on some classes of $C^*$-algebras. For example, any tracial linear map $\Phi :\mathbb {M}_n\rightarrow \mathscr {B}$ has the form $\Phi (A)={{\,\mathrm{tr}\,}}(A)\Phi (I)$. This follows from the observation that for the canonical matrix unit system $\{E_{ij}\}\subseteq \mathbb {M}_n$, we have $\Phi (E_{ij})=\Phi (E_{ij}E_{jj})=\Phi (E_{jj}E_{ij})=\Phi (0)=0$ if $i\ne j$, and $\Phi (E_{ii})=\Phi (E_{ij}E_{jj}E_{ji})=\Phi (E_{ji}E_{ij}E_{jj})=\Phi (E_{jj})$ for any i, j. Similar reasoning shows that if $\mathscr {A}=\mathbb {K}({\mathscr {H}})$ or $\mathscr {A}=\mathbb {B}({\mathscr {H}})$, then the only tracial linear map $\Phi : \mathscr {A}\rightarrow \mathscr {B}$ is the zero map.

In this paper, we improve the H–P–R inequality for Hermitian bounded maps $\Phi$ by showing that K in (1.1) can be chosen to be 2. In addition, we present a refinement of the H–P–R inequality and present an example to support it. Moreover, we establish some versions of the H–P–R inequality involving unitarily invariant norms on matrix algebras and discuss the H–P–R inequality in several various directions as well. In the last section, we present some useful inequalities in the setting of Hilbert $C^*$-modules and then apply it to get several H–P–R-type inequalities.

2 Preliminaries

Let ${{\mathbb {B}}}({{\mathscr {H}}})$ denote the $C^*$-algebra of all bounded linear operators on a complex Hilbert space $({\mathscr {H}}, \langle \cdot , \cdot \rangle )$ equipped with the operator norm $\Vert \cdot \Vert$, and I be its identity. If $\dim \mathscr {H}=n$, we can identify $\mathbb {B}(\mathscr {H})$ with the matrix algebra $\mathbb {M}_n$ of all $n \times n$ matrices with entries in the complex field $\mathbb {C}$ and denote its identity by $I_n$.

For self-adjoint operators (Hermitian matrices) A, B, we write $A\ge B$ if $\langle A x,x \rangle \ge \langle Bx,x\rangle$ for all $x\in \mathscr {H}$. In particular, if $A \ge 0$, then A is called positive. For a linear map $\Phi : {{\mathscr {A}}} \rightarrow {\mathscr {B}}$ between $*$-algebras one can define the map $\Phi ^*: {\mathscr {A}} \rightarrow {{\mathscr {B}}}$ by $\Phi ^*(A)=\Phi (A^*)^*$, and it is called a Hermitian (self-adjoint) map if $\Phi ^*=\Phi$. A map $\Phi$ between $C^*$-algebras is said to be positive if $\Phi (A) \ge 0$ whenever $A \ge 0$. We say that $\Phi$ is unital if ${{\mathscr {A}}}, {\mathscr {B}}$ are unital $C^*$-algebras and $\Phi (I_{\mathscr {A}})=I_{\mathscr {B}}$, where $I_{\mathscr {A}}$ stands for the unit of $\mathscr {A}$. Any linear map $\Phi : \mathscr {A} \rightarrow \mathscr {B}$ induces a linear map $\Phi _n$ from $\mathbb {M}_n(\mathscr {A})$ to $\mathbb {M}_n(\mathscr {B})$ defined by $\Phi _n([A_{ij}]_{n\times n})=[\Phi (A_{ij})]_{n \times n}$, where $\mathbb {M}_n (\mathscr {A})$ stands for the $n \times n$ block matrix with entries in $\mathscr {A}$. We say that $\Phi$ is n-positive if the map $\Phi _{n}$ is positive, and $\Phi$ is completely positive if the maps $\Phi _{n}$ are positive for all $n=1,2,\ldots$.

For any operator A in the algebra $\mathbb {K}(\mathscr {H})$ of all compact operators, we denote by $\{s_j(A)\}$ the sequence of singular values of A, i.e., the eigenvalues $\lambda _j(|A|)$, where $|A|=(A^*A)^{1\over {2}}$, in decreasing order and repeated according to multiplicity. If $A \in \mathbb {M}_n$, we take $s_k(A)=0$ for any $k>n$. A norm $\left| \left| \left| \cdot \right| \right| \right|$ defined on a certain ideal $\mathcal {J}$ of $\mathbb {K}(\mathscr {H})$ is called unitarily invariant if $\left| \left| \left| UAV\right| \right| \right| =\left| \left| \left| A\right| \right| \right|$ for all unitary matrices $U,V\in \mathbb {B}(\mathscr {H})$ and all $A\in \mathcal {J}$. The inequalities involving unitarily invariant norms have been of special interest; see, e.g., [7]. The Ky Fan norms, defined by $\Vert A\Vert _{(k)}=\sum _{j=1}^{k}s_{j}(A)$ for $k=1, 2, \ldots , n$, give rise to a family of unitarily invariant norms. The Ky Fan dominance theorem [1, Theorem IV.2.2] states that $\Vert A\Vert _{(k)}\le \Vert B\Vert _{(k)}\,\,(k=1,2,\ldots )$ if and only if $|||A||| \le |||B|||$ for all unitarily invariant norms $|||\cdot |||$. Weyl’s monotonicity theorem says that if $A, B\in \mathbb {K}(\mathscr {H})$ are self-adjoint operators and $A\le B$, then $\lambda _j(A) \le \lambda _j(B), j=1, 2, \ldots$. It is known that the Schatten p-norms $\Vert A\Vert _p=\left( \sum _{j=1}^\infty s_j^p(A)\right) ^{1/p}$ are also unitarily invariant for $1 \le p$; cf. [1, Section IV.2]. We use the notation $A\oplus B$ for the diagonal block matrix $\mathrm{diag} (A,B)$. It is clear that

$$\begin{aligned} \Vert A\oplus B\Vert =\max \{\Vert A\Vert ,\Vert B\Vert \} \quad \mathrm{and}\quad \Vert A\oplus B\Vert _p=(\Vert A\Vert _p^p+\Vert B\Vert _p^p)^{1/p}\,. \end{aligned}$$

(2.1)

In finite-dimensional cases, we assume that a unitarily invariant norm $|||\cdot |||$ is given on a two-sided ideal of bounded linear operators acting on a separable Hilbert space and then the norms $|||\cdot |||$ on matrix algebras $\mathbb {M}_n$ for all finite values of n are induced by it via $|||A|||=|||A\oplus 0|||$. Thus, we indeed deal with a system of unitarily invariant norms $\{|||\cdot |||_s\}$ on algebras $\mathbb {M}_s,\,\, s \le n$ or on all algebras $\mathbb {M}_s,\,\, s \ge 1$ satisfying the relation

$$\begin{aligned} |||A|||_s=|||A\oplus 0_{(t-s)(t-s)}|||_t \qquad (A \in \mathbb {M}_s, t>s) \end{aligned}$$

(2.2)

between norms of matrices of different sizes. In particular, the normalized condition $|||\mathrm{diag}(1, 0, \ldots , 0)|||=1$ is assumed to be satisfied. For a linear map $\Phi :{\mathbb {M}}_n\rightarrow {\mathbb {M}}_k$, we write $|||\Phi |||$ for the operator norm of $\Phi$ with respect to the norms $|||\cdot |||$ on ${\mathbb {M}}_n$ and ${\mathbb {M}}_k$.

3 Generalizations of H–P–R Inequality

We start our work with the following lemma in whose proof we adopt the proof of [5, Lemma 3.1].

Lemma 3.1

Let $\mathscr {A}$ be a unital $C^*$-algebra and $\mathscr {H}$ be a Hilbert space. Let $T: \mathscr {A} \rightarrow \mathscr {H}$ be a bounded linear map such that $\Vert T\Vert =\Vert T(U)\Vert$ for some unitary operator $U\in \mathscr {A}$. Then, there exists a state $\varphi$ on $\mathscr {A}$ such that

$$\begin{aligned} \Vert T(X)\Vert ^2+\Vert T(X^*)\Vert ^2\le 2\Vert T\Vert ^2 \varphi (X^*X+XX^*) \end{aligned}$$

for each $X\in \mathscr {A}$.

Proof

If $U=I$, the proof of [5, Lemma 3.1] shows that there exists a state $\varphi _1$ on $\mathscr {A}$ such that $\varphi _1(X)=\langle T(X),T(I)\rangle$ for each $X\in \mathscr {A}$ and

$$\begin{aligned} \Vert T(X)\Vert ^2+\Vert T(X^*)\Vert ^2\le 2\Vert T\Vert ^2\varphi _1(X^*X+XX^*). \end{aligned}$$

If $\Vert T\Vert = \Vert T(U)\Vert = 1$ for some unitary operator $U\in \mathscr {A}$, then put $S(X)= T(UX)$. As $\Vert S\Vert = \Vert S(I)\Vert = 1$, by the first case,

$$\begin{aligned} \Vert S(X)\Vert ^2+\Vert S(X^*)\Vert ^2\le \varphi _1(X^*X+XX^*) \end{aligned}$$

for some state $\varphi _1$ and for each $X\in \mathscr {A}$. Consider $\varphi _2(X)= \varphi _1(U^*XU)$ for each $X\in \mathscr {A}$ and set $\varphi =1/2(\varphi _1+\varphi _2)$. Then

$$\begin{aligned} \Vert T(X)\Vert ^2 + \Vert T(X^*)\Vert ^2= & {} \Vert S(U^*X)\Vert ^2 + \Vert S(U^*X^*)\Vert ^2\\\le & {} \varphi _1(X^*X+U^*XX^*U) + \varphi _1(XX^*+U^*X^*XU)\\= & {} \varphi _1(X^*X+XX^*) + \varphi _1(U^*(X^*X+X^*X)U)\\= & {} \varphi _1(X^*X+XX^*) + \varphi _2(X^*X+XX^*)\\= & {} 2\varphi (X^*X+XX^*). \end{aligned}$$

$\square$

By a similar argument as in [5, Lemma 3.2], we have the following lemma.

Lemma 3.2

Let $\mathscr {A}$ be a $C^*$-algebra and $\mathscr {H}$ be a Hilbert space. If $T: \mathscr {A} \rightarrow \mathscr {H}$ is a bounded linear map, then there exists an state $\varphi$ on $\mathscr {A}$ such that

$$\begin{aligned} \Vert T(X)\Vert ^2+\Vert T(X^*)\Vert ^2\le 2\Vert T\Vert ^2 \varphi (X^*X+XX^*). \end{aligned}$$

for each $X\in \mathscr {A}$.

Theorem 3.3

Let $\mathscr {A}$ and $\mathscr {B}$ be $C^*$-algebras, and let $\Phi : \mathscr {A} \rightarrow \mathscr {B}$ be a bounded linear map. Then

$$\begin{aligned} \left\| \sum _{k=1}^n\left\{ |\Phi (A_k)|^2+|\Phi (A_k^*)|^2\right\} \right\| \le 2 \Vert \Phi \Vert ^2 \left\| \sum _{k=1}^n \left( |A_k|^2+|A_k^*|^2\right) \right\| \end{aligned}$$

for all $n\in \mathbb {N}$ and for all $A_1,\ldots , A_n\in \mathscr {A}$.

Proof

Assume that $\mathscr {B}\subset \mathbb {B}(\mathscr {H})$, for some Hilbert space $\mathscr {H}$. Let $x\in \mathscr {H}$ be a unit vector. Assume that $T(X)=\Phi (X)x$. Then, $T:\mathscr {A}\rightarrow \mathscr {H}$ is a bounded linear map and $\Vert T\Vert \le \Vert \Phi \Vert$. So by Lemma 3.2, we have

$$\begin{aligned} \Vert T(A_k)\Vert ^2+\Vert T(A_k^*)\Vert ^2\le 2\Vert T\Vert ^2 \ \varphi (A_k^*A_k+A_kA_k^*), \end{aligned}$$

for each $1\le k\le n$. Hence

$$\begin{aligned} \sum _{k=1}^n \left\{ \Vert \Phi (A_k)x\Vert ^2+\Vert \Phi (A_k^*)x\Vert ^2\right\}\le & {} 2 \Vert \Phi \Vert ^2\ \varphi \left( \sum _{k=1}^n \left( A_k^*A_k+A_kA_k^*\right) \right) \\\le & {} 2 \Vert \Phi \Vert ^2 \left\| \sum _{k=1}^n \left( A_k^*A_k+A_kA_k^*\right) \right\| . \end{aligned}$$

Thus

$$\begin{aligned}&\left\| \sum _{k=1}^n \left\{ \Phi (A_k)^*\Phi (A_k)+\Phi (A_k^*)^*\Phi (A_k^*)\right\} \right\| \\&\quad =\sup _{\Vert x\Vert =1}\left( \sum _{k=1}^n \left\{ \langle \Phi (A_k)^*\Phi (A_k)x,x \rangle +\Phi (A_k^*)^*\Phi (A_k^*)x,x \rangle \right\} \right) \\&\quad =\sup _{\Vert x\Vert =1}\left( \sum _{k=1}^n\left\{ \Vert \Phi (A_k)x\Vert ^2+\Vert \Phi (A_k^*)x\Vert ^2 \right\} \right) \\&\quad \le 2 \Vert \Phi \Vert ^2 \left\| \sum _{k=1}^n \left( A_k^*A_k+A_kA_k^*\right) \right\| . \end{aligned}$$

$\square$

It follows from Theorem 3.3 that if $\Phi : \mathscr {A} \rightarrow \mathscr {B}$ is a Hermitian bounded linear map, then we can prove the H–P–R inequality (1.1) with constant $K = 2$.

Corollary 3.4

Let $\mathscr {A}$ and $\mathscr {B}$ be $C^*$-algebras, and let $\Phi : \mathscr {A} \rightarrow \mathscr {B}$ be a Hermitian bounded linear map. Then

$$\begin{aligned} \left\| \sum _{k=1}^n \left\{ \Phi (A_k)^*\Phi (A_k)+\Phi (A_k)\Phi (A_k)^*\right\} \right\| \le 2 \Vert \Phi \Vert ^2 \left\| \sum _{k=1}^n\left( A_k^*A_k+A_kA_k^*\right) \right\| \end{aligned}$$

for all $n\in \mathbb {N}$ and for all $A_1,\ldots ,A_n\in \mathscr {A}$.

Moreover, Theorem 3.3 provides a refinement of the H–P–R inequality (1.1) as follows:

Corollary 3.5

Let $\mathscr {A}, \mathscr {B}$ be $C^*$-algebras, and let $\Phi : \mathscr {A} \rightarrow \mathscr {B}$ be a bounded linear map. Then

$$\begin{aligned}&\left\| \sum _{k=1}^n \left\{ \Phi (A_k)^*\Phi (A_k)+\Phi (A_k)\Phi (A_k)^*\right\} \right\| \\&\quad \le \left\| \sum _{k=1}^n \left\{ \Phi (A_k)^*\Phi (A_k)+\Phi (A_k)\Phi (A_k)^*+\Phi (A_k^*)^*\Phi (A_k^*)+\Phi (A_k^*)\Phi (A_k^*)^* \right\} \right\| \\&\quad \le 4 \Vert \Phi \Vert ^2 \left\| \sum _{k=1}^n\left( A_k^*A_k+ A_kA_k^*\right) \right\| \end{aligned}$$

for all $n\in \mathbb {N}$ and for all $A_1,\ldots , A_n\in \mathscr {A}$.

To see that Corollary 3.5 actually refines the H–P–R inequality (1.1), assume that the bounded linear mapping $\Phi :\mathbb {M}_2\rightarrow \mathbb {M}_2$ is defined by $\Phi (A)=AX$, where $X=\left( \begin{array}{cc} 2 &{} 0 \\ 0 &{} 0 \end{array}\right)$. Then, $\Vert \Phi \Vert =2$. With $n=1$ and $A_1 =\left( \begin{array}{cc} a &{} 0 \\ 0 &{} 0 \end{array}\right)$, it is easy to see that

$$\begin{aligned} \left\| \Phi (A )^*\Phi (A )+\Phi (A )\Phi (A )^* \right\| =8a^2 \end{aligned}$$

and

$$\begin{aligned} \left\| \Phi (A)^*\Phi (A )+\Phi (A)\Phi (A )^*+\Phi (A^*)^*\Phi (A^*)+\Phi (A^*)\Phi (A^*)^* \right\| =16a^2 \end{aligned}$$

and

$$\begin{aligned} 4 \Vert \Phi \Vert ^2 \left\| A^*A+ AA^*\right\| =32a^2. \end{aligned}$$

Recall that the Kadison inequality says that if $\Phi$ is a unital positive linear map, then $\Phi (A)^2 \le \Phi (A^2)$ for any Hermitian A. Now, let $\Phi : \mathscr {A} \rightarrow \mathbb {B}$ be a unital positive linear map. It is stated in [9, Exercise 4.1] that

$$\begin{aligned} \Phi (A^*A+AA^*)\ge \Phi (A)^* \Phi (A)+\Phi (A) \Phi (A)^*. \end{aligned}$$

(3.1)

Indeed, if $B=A+A^*$ and $C=\mathrm{i}(A-A^*)$, then by employing the Kadison inequality we get

$$\begin{aligned} \Phi (A^2+(A^*)^2+AA^*+A^*A)= & {} \Phi (B^*B)\\\ge & {} \Phi (B)^*\Phi (B)\\= & {} \Phi (A)^2+\Phi (A^*)^2+\Phi (A)\Phi (A)^*+\Phi (A)^*\Phi (A). \end{aligned}$$

Similarly

$$\begin{aligned} \Phi (-A^2-(A^*)^2+AA^*+A^*A)= & {} \Phi (C^*C)\\\ge & {} \Phi (C)^*\Phi (C)\\= & {} -\Phi (A)^2-\Phi (A^*)^2+\Phi (A)\Phi (A)^*+\Phi (A)^*\Phi (A). \end{aligned}$$

Hence

$$\begin{aligned} \Phi (A^*A+AA^*)\ge \Phi (A)^* \Phi (A)+\Phi (A) \Phi (A)^*. \end{aligned}$$

We can state the following corollary.

Corollary 3.6

If $\Phi : \mathbb {M}_n \rightarrow \mathbb {M}_k$ is a unital positive linear map and $|||\cdot |||$ is a system of unitarily invariant norms, then

$$\begin{aligned} \left| \left| \left| \sum _{j=1}^m\left\{ \Phi (A_j)^*\Phi (A_j)+\Phi (A_j)\Phi (A_j)^*\right\} \right| \right| \right|&\le |||\Phi |||\,\left| \left| \left| \sum _{j=1}^m(A_j^*A_j+A_jA_j^*)\right| \right| \right| \nonumber \\&\le k\,\left| \left| \left| \sum _{j=1}^m(A_j^*A_j+A_jA_j^*)\right| \right| \right| . \end{aligned}$$

(3.2)

for all $A_1, \ldots , A_m\in \mathbb {M}_n$ .

Proof

The first inequality follows from 3.1. Indeed,

$$\begin{aligned} \sum _{j=1}^m\left\{ \Phi (A_j)^*\Phi (A_j)+\Phi (A_j)\Phi (A_j)^*\right\}&\le \Phi \left( \sum _{j=1}^m\left( A_j^*A_j+A_jA_j^*\right) \right) . \end{aligned}$$

Now, By [1, p. 93],

$$\begin{aligned} |||\Phi |||&:=\sup _{X\ne 0}\frac{|||\Phi (X)|||}{|||X|||}\\&\le \sup _{X\ne 0}\frac{\Vert \Phi (X)\Vert _1}{\Vert X\Vert }\quad (\mathrm{since~} |||\Phi (X)|||\le \Vert \Phi (X)\Vert _1 \mathrm{~and~} |||X|||\ge \Vert X\Vert )\\&\le k \quad (\mathrm{since~} \Vert \Phi (X)\Vert _1\le k\Vert \Phi (X)\Vert \mathrm{~and~} \Vert \Phi (X)\Vert \le \Vert X\Vert ). \end{aligned}$$

$\square$

Remark 3.7

If $\Phi$ is a unital positive linear map, then $\Vert \Phi \Vert =1$, by the Russo–Dye theorem.

In the infinite-dimensional case, the problem of finding a unitarily invariant norm version of H–P–R inequality may be complicated since there are some $C^*$-algebras of Hilbert space operators having no compact operators and so there are no unitarily invariant norm except the operator norm. An example is $\mathbb {C}p+\mathbb {C}q$ where p, q are two orthogonal infinite rank projections on a Hilbert space.

Utilizing a standard technique with the polar decomposition (see [2]), we have the following observation.

Proposition 3.8

Let $\mathcal {J}_i$ be ideals of $\mathbb {B}(\mathscr {H}_i)$ associated with the unitarily invariant norms $|||\cdot |||_i$ for $i=1,2$, let $\mathscr {A}\subseteq \mathbb {B}(\mathscr {H}_1)$, and let $\pi : \mathscr {A} \rightarrow \mathbb {B}(\mathscr {H}_2)$ be a bounded representation (not necessarily a Hermitian map) of a von Neumann algebra $\mathscr {A}$ such that $\pi (\mathcal {J}_1\cap \mathscr {A}) \subseteq \mathcal {J}_2$. Then

$$\begin{aligned} \left| \left| \left| \sum _{j=1}^n\left\{ \pi (A_j)^* \pi (A_j)+\pi (A_j) \pi (A_j)^* \right\} \right| \right| \right| _2\le K \Vert \pi \Vert ^2 \,\left| \left| \left| \sum _{j=1}^n \left( A_j^*A_j+A_jA_j^*\right) \right| \right| \right| _1 \end{aligned}$$

(3.3)

for some K and all $A_1, \ldots , A_n\in \mathcal {J}_1\cap \mathscr {A}$ if and only if

$$\begin{aligned} \left| \left| \left| \sum _{j=1}^n \pi (A_j)^* \pi (A_j) \right| \right| \right| _2\le K' \Vert \pi \Vert ^2 \,\left| \left| \left| \sum _{j=1}^n A_j^*A_j\right| \right| \right| _1 \end{aligned}$$

(3.4)

and

$$\begin{aligned} \left| \left| \left| \sum _{j=1}^n \pi (A_j) \pi (A_j)^* \right| \right| \right| _2\le K' \Vert \pi \Vert ^2 \,\left| \left| \left| \sum _{j=1}^n A_jA_j^*\right| \right| \right| _1\,. \end{aligned}$$

(3.5)

for some $K'$ and all $A_1, \ldots , A_n\in \mathcal {J}_1\cap \mathscr {A}$.

Proof

Let (3.3) holds and $A_j=U_j|A_j|$ be the polar decomposition of $A_j$, where $U_j$ is a partial isometry (and therefore of norm 1). It follows from the continuous functional calculus that if $A\ge 0$, then $A\le MI$ if and only if $\Vert A\Vert \le M$. We therefore have

$$\begin{aligned} \left| \left| \left| \sum _{j=1}^n \pi (A_j)^* \pi (A_j) \right| \right| \right| _2&=\left| \left| \left| \sum _{j=1}^n \pi (|A_j|)^*\pi (U_j)^*\pi (U_j)\pi (A_j) \right| \right| \right| _2\\&\le \Vert \pi \Vert ^2\left| \left| \left| \sum _{j=1}^n \pi (|A_j|)^*\pi (|A_j|) \right| \right| \right| _2\\&\quad (\mathrm{by~} \pi (|A_j|)^*\pi (U_j)^*\pi (U_j)\pi (|A_j|) \le \Vert \pi \Vert ^2\Vert U_j\Vert ^2\pi (|A_j|)^*\pi (|A_j|))\\&=\Vert \pi \Vert ^2\left| \left| \left| \sum _{j=1}^n \left\{ \pi (|A_j|)^*\pi (|A_j|)+\pi (|A_j|)\pi (|A_j|)^* \right\} \right| \right| \right| _2\\&\quad (\mathrm{by~Weyl's~monotone~theorem})\\&\le K \Vert \pi \Vert ^2 \,\left| \left| \left| \sum _{j=1}^n \left( |A_j|^*\,|A_j|+|A_j|\,|A_j|^*\right) \right| \right| \right| _1\\&\quad (\mathrm{by~the~assumption})\\&\le 2K \Vert \pi \Vert ^2 \,\left| \left| \left| \sum _{j=1}^n A_j^*A_j\right| \right| \right| _1\,. \end{aligned}$$

Furthermore, we have the polar decomposition of $A_j^*=U_j^*|A_j^*|$. Hence, $A_j=|A_j^*|U_j$ and using the same reasoning as above, we can prove (3.5).

Conversely, assume that (3.4) and (3.5) hold. Then

$$\begin{aligned}&\left| \left| \left| \sum _{j=1}^n \left\{ \pi (A_j)^* \pi (A_j)+\pi (A_j) \pi (A_j)^* \right\} \right| \right| \right| _2\\&\quad \le \left| \left| \left| \sum _{j=1}^n \pi (A_j)^* \pi (A_j) \right| \right| \right| _2+\left| \left| \left| \sum _{j=1}^n \pi (A_j) \pi (A_j)^* \right| \right| \right| _2\\&\quad \le K \Vert \pi \Vert ^2 \left( \left| \left| \left| \sum _{j=1}^n A_j^*A_j\right| \right| \right| _1+\left| \left| \left| \sum _{j=1}^n A_jA_j^*\right| \right| \right| _1\right) \\&\quad \le 2K \Vert \pi \Vert ^2 \left| \left| \left| \sum _{j=1}^n \left( A_j^*A_j + A_jA_j^*\right) \right| \right| \right| _1 \\&\qquad (\mathrm{by~Weyl's~monotone~theorem}). \end{aligned}$$

$\square$

Remark 3.9

It is remarkable that if $\Phi : \mathscr {A} \rightarrow \mathscr {B}$ is 3-positive linear and unitary preserving map, then it should be a homomorphism. This can be easily followed from the fact [9, p. 41] that if $U, V, X \in \mathscr {A}$ with U, V unitary, then

$$\begin{aligned} \left( \begin{array}{ccc}I_\mathscr {A}&{}U&{}X\\ U^*&{}I_\mathscr {A}&{}V\\ X^*&{}V^*&{}I_\mathscr {A}\end{array}\right) \ge 0 \mathrm{~if~and~only~if~} X=UV\,. \end{aligned}$$

An example of a map that is not 3-positive but still preserves unitaries is the transpose map $A \mapsto A^{tr}$. It is not a homomorphism, but is an antihomomorphism.

The following lemma can be proved by employing the Weyl’s monotonicity theorem.

Lemma 3.10

[1, p. 94] Let $A, X, B\in \mathbb {M}_n$. Then

$$\begin{aligned} s_{j}(AXB)\le \Vert A\Vert \,s_{j}(X)\,\Vert B\Vert \quad (j=1,2,\ldots , n). \end{aligned}$$

Theorem 3.11

Let $\mathscr {A}$ be a $C^*$-algebra of Hilbert space operators, $\Phi : \mathscr {A} \rightarrow \mathbb {B}(\mathscr {H})$ be a completely bounded linear map with $\Vert \Phi \Vert _{cb}=\sup _n\Vert \Phi _n\Vert =1$, let $X_1, \ldots , X_n$, $Y_1, \ldots , Y_n$ be compact operators, which are mapped by $\Phi$ into the ideal of compact operators, and let $A_1, \ldots , A_n\in \mathscr {A}$. Then

$$\begin{aligned}&s_j\left( \sum _{j=1}^n\left\{ \Phi (A_j)^*\Phi (X_j)\Phi (A_j)+\Phi (A_j)\Phi (Y_j)\Phi (A_j)^*\right\} \right) \\&\quad \le \left\| \sum _{j=1}^n(A_j^*A_j+A_jA_j^*)\right\| s_j\Big (\displaystyle {\oplus _{k=1}^n} \left( \Phi (X_k)\oplus \Phi (Y_k)\right) \Big ) \end{aligned}$$

for all $j=1, 2, \ldots$.

Proof

It is easy to see that $\Vert \Phi _{2n}\Vert _{cb}\le 1$. Let

$$\begin{aligned} {\widetilde{X}}=\mathrm{diag}(X_1, Y_1, \ldots , X_n, Y_n) \mathrm{~and~}{\widetilde{A}}=\left( \begin{array}{cccc}A_1 &{} 0 &{}\cdots &{}0\\ A_1^*&{} &{} &{}\\ \vdots &{} \vdots &{} &{}\vdots \\ A_n &{} &{}&{}\\ A_n^* &{} 0 &{}\cdots &{}0\end{array}\right) . \end{aligned}$$

Then

$$\begin{aligned}&s_j\left( \sum _{j=1}^n\left\{ \Phi (A_j)^*\Phi (X_j)\Phi (A_j)+\Phi (A_j)\Phi (Y_j)\Phi (A_j)^*\right\} \right) \\&\quad =s_j\left( \Phi _{2n}(\widetilde{A})^*\Phi _{2n}(\widetilde{X})\Phi _{2n}(\widetilde{A})\right) \\&\quad \le \left\| \Phi _{2n}(\widetilde{A})^*\right\| s_j\left( \Phi _{2n}(\widetilde{X})\right) \left\| \Phi _{2n}(\widetilde{A})\right\| \qquad \qquad (\mathrm{by~Lemma~}~3.10)\\&\quad =\left\| \Phi _{2n}(\widetilde{A})^* \Phi _{2n}(\widetilde{A})\right\| \, s_j\left( \Phi _{2n}(\widetilde{X})\right) \\&\quad = \left\| \Phi _{2n}(\widetilde{A})\right\| ^2\ s_j\left( \Phi _{2n}(\widetilde{X})\right) \\&\quad \le \left\| \widetilde{A} \right\| ^2\, s_j\left( \Phi _{2n}(\widetilde{X})\right) \\&\quad =\left\| \widetilde{A}^*\widetilde{A}\right\| \, s_j\left( \Phi _{2n}(\widetilde{X})\right) \\&\quad = \left\| \sum _{j=1}^n(A_j^*A_j+A_jA_j^*)\right\| \,s_j\Big (\displaystyle {\oplus _{k=1}^n} \left( \Phi (X_k)\oplus \Phi (Y_k)\right) \Big ) \end{aligned}$$

for $j=1, 2, \ldots$. $\square$

Passing to the operator norm and matrix and using the Stinespring theorem, we reach the following result.

Corollary 3.12

Let $\Phi : \mathbb {M}_m \rightarrow \mathbb {M}_k$ be a unital completely positive linear map and let $X_1, \ldots , X_n$, $Y_1, \ldots , Y_n, A_1, \ldots , A_n\in \mathbb {M}_m$. Then

$$\begin{aligned}&\left\| \sum _{j=1}^n \left\{ \Phi (A_j)^*\Phi (X_j)\Phi (A_j)+\Phi (A_j)\Phi (Y_j)\Phi (A_j)^*\right\} \right\| \\&\quad \le \left\| \sum _{j=1}^n \left( A_j^*A_j+A_jA_j^*\right) \right\| \max \{\Vert X_k\Vert : 1\le k\le n\}\max \{\Vert Y_k\Vert : 1\le k\le n\}. \end{aligned}$$

4 Generalized H–P–R Inequality in Hilbert $C^*$-Modules

Hilbert $C^*$-modules are generalizations of Hilbert spaces by allowing inner products to take values in some $C^*$-algebra instead of the complex field $\mathbb {C}$. Although Hilbert $C^*$-modules behave like Hilbert spaces in some ways, some fundamental Hilbert space properties do not hold in general. Let $\mathscr {A}$ be a unital $C^*$-algebra, and let $(\mathcal {H}, \langle \cdot ,\cdot \rangle )$ be a Hilbert $C^*$-module over $\mathscr {A}$ with the norm $\Vert X\Vert =\Vert \langle X,X\rangle ^{1/2}\Vert$. For each $X\in \mathcal {H}$, we set $|X|=\langle X,X\rangle ^{1/2}$. Evidently, $|X_1+X_2|^2+|X_1-X_2|^2=2|X_1|^2+2|X_2|^2$ for all $X_1,X_2\in \mathcal {H}$.

The main result of this section reads as follows by adapting some arguments of [8].

Theorem 4.1

Let $\mathscr {A}$ and $\mathscr {B}$ be unital $C^*$-algebras, and let $\mathcal {H}$ be a Hilbert $\mathscr {B}$- module. Let $\Phi : \mathscr {A} \rightarrow \mathcal {H}$ be a bounded linear operator such that $|\Phi (A)|=\Vert \Phi \Vert I_{\mathscr {B}}$ for some unit vector $A\in \mathscr {A}$. Then, there exist unital positive linear maps $\Psi _1,\Psi _2,\Psi :\mathscr {A}\rightarrow \mathscr {B}$ such that

$$\begin{aligned} |\Phi (AX)|^2 +|\Phi (AX^*)|^2\le & {} \Vert \Phi \Vert ^2 \Psi _1(X^*X+XX^*), \end{aligned}$$

(4.1)

$$\begin{aligned} |\Phi (XA)|^2 +|\Phi (X^*A)|^2\le & {} \Vert \Phi \Vert ^2 \Psi _2(X^*X+XX^*) \end{aligned}$$

(4.2)

$$\begin{aligned} |\Phi (AX+XA)|^2+|\Phi (AX^*+X^*A)|^2\le & {} 4 \Vert \Phi \Vert ^2 \Psi (X^*X+XX^*), \end{aligned}$$

(4.3)

for each $X\in \mathscr {A}$.

Proof

Without loss of generality, assume that $\Vert \Phi \Vert =1$. Let $B\in \mathscr {A}$ be a self-adjoint operator and $t\in (0,\infty )$. As $e^{\sqrt{2}\mathrm{i}tB}$ is unitary, we have $\Vert e^{\sqrt{2}\mathrm{i}tB}A\Vert \le \Vert A\Vert$. Moreover

$$\begin{aligned} e^{\sqrt{2}\mathrm{i}tB}A=A+\sqrt{2}\mathrm{i}tBA-t^2B^2A+o(t^3)X_1, \end{aligned}$$

where $X_1\in \mathscr {A}$. Hence

$$\begin{aligned} A-t^2B^2A+\sqrt{2}\mathrm{i}tBA=e^{\sqrt{2}\mathrm{i}tB}A-o(t^3)X_1. \end{aligned}$$

Similarly,

$$\begin{aligned} A-t^2B^2A-\sqrt{2}\mathrm{i}tBA=e^{-\sqrt{2}\mathrm{i}tB}A-o(t^3)X_2 \end{aligned}$$

for some $X_2\in \mathscr {A}$. We have

$$\begin{aligned}&|\Phi (A-t^2B^2A)|^2+2t^2|\Phi (BA)|^2\\&\quad =1/2\left( |\Phi (A-t^2B^2A+\sqrt{2}\mathrm{i}tBA)|^2+|\Phi (A-t^2B^2A-\sqrt{2}\mathrm{i}tBA)|^2\right) \\&\quad \le 1/2\left( \Vert \ |\Phi (A-t^2B^2A+\sqrt{2}\mathrm{i}tBA)|^2\ \Vert +\Vert \ |\Phi (A-t^2B^2A-\sqrt{2}\mathrm{i}tBA)|^2 \ \Vert \right) I_{\mathscr {B}} \\&\quad \le 1/2\left( \Vert A-t^2B^2A+\sqrt{2}\mathrm{i}tBA\Vert ^2+\Vert A-t^2B^2A-\sqrt{2}\mathrm{i}tBA\Vert ^2\right) I_{\mathscr {B}} \\&\quad \le 1/2\left( \Vert e^{\sqrt{2}\mathrm{i}tB}A- o(t^3)X_1\Vert ^2+\Vert e^{-\sqrt{2}\mathrm{i}tB}A- o(t^3)X_2\Vert ^2\right) I_{\mathscr {B}}\\&\quad \le (1+o(t^3))I_{\mathscr {B}}. \end{aligned}$$

Therefore,

$$\begin{aligned} 2t^2|\Phi (BA)|^2\le & {} I_{\mathscr {B}}-|\Phi (A-t^2B^2A)|^2+o(t^3)I_{\mathscr {B}}\\= & {} I_{\mathscr {B}}-\langle \Phi (A-t^2B^2A),\Phi (A-t^2B^2A)\rangle +o(t^3)I_{\mathscr {B}}\\= & {} I_{\mathscr {B}}-\langle \Phi (A),\Phi (A)\rangle +2t^2{\mathfrak {R}}(\langle \Phi (A),\Phi (B^2A)\rangle )+o(t^3)X_3\\= & {} 2t^2{\mathfrak {R}}(\langle \Phi (A),\Phi (B^2A)\rangle )+o(t^3)X_3. \end{aligned}$$

for some $X_3\in \mathscr {B}$, where ${\mathfrak {R}(Y)}$ denotes the real part of $Y\in \mathscr {B}$. So

$$\begin{aligned} 2|\Phi (BA)|^2\le & {} 2{\mathfrak {R}}(\langle \Phi (A),\Phi (B^2A)\rangle )+o(t)X_3. \end{aligned}$$

Letting $t\rightarrow 0$, we get

$$\begin{aligned} |\Phi (BA)|^2 \le {\mathfrak {R}}(\langle \Phi (A),\Phi (B^2A)\rangle ) \end{aligned}$$

Put $\Psi _1(X)=\langle \Phi (A),\Phi (XA)\rangle$ for each $X\in \mathscr {A}$. Then $\Psi _1(I_{\mathscr {A}})=I_{\mathscr {B}}$ and by the Cauchy–Schwarz inequality, $\Vert \Psi _1\Vert = \Vert \Psi _1(I_{\mathscr {A}})\Vert =1.$ Hence, $\Psi _1$ is a unital positive linear map and

$$\begin{aligned} {\mathfrak {R}}(\langle \Phi (A),\Phi (B^2A)\rangle )={\mathfrak {R}}(\Psi _1(B^2))=\Psi _1(B^2). \end{aligned}$$

If X is arbitrary, then $\mathrm{i}(X-X^*)$ and $X+X^*$ are self-adjoint operators and we have

$$\begin{aligned} |\Phi (XA)|^2 +|\Phi (X^*A)|^2= & {} 1/2\left( |\Phi ((X+X^*)A)|^2+|\Phi ((X-X^*)A)|^2\right) \\= & {} 1/2\left( |\Phi ((X+X^*)A)|^2+|\Phi (\mathrm{i}(X-X^*)A)|^2\right) \\\le & {} 1/2\left( \Psi _1((X+X^*)^2)+\Psi _1(-(X-X^*)^2))\right) \\= & {} \Psi _1(X^*X+XX^*). \end{aligned}$$

By a similar argument, there exists a positive linear map $\Psi _2:\mathscr {A}\rightarrow \mathscr {B}$ such that

$$\begin{aligned} |\Phi (AX)|^2 +|\Phi (AX^*)|^2\le \Vert \Phi \Vert ^2 \Psi _2(X^*X+XX^*). \end{aligned}$$

for all $X\in \mathscr {A}$. Consider $\Psi =1/2(\Psi _1+\Psi _2)$. Then

$$\begin{aligned}&|\Phi (AX+XA)|^2+|\Phi (AX^*+X^*A)|^2\\&\quad = |\Phi (AX)+\Phi (XA)|^2+ |\Phi (AX^*)+\Phi (X^*A)|^2\\&\quad \le 2|\Phi (AX)|^2+2|\Phi (XA)|^2+2|\Phi (AX^*)|^2+2|\Phi (X^*A)|^2\\&\quad \le 2\Psi _1(X^*X+XX^*)+2\Psi _2(X^*X+XX^*)\\&\quad = 4\Psi (X^*X+XX^*). \end{aligned}$$

$\square$

If $A=I_\mathscr {A}$ in the previous theorem, then $\Psi$ is a unital positive linear map and

$$\begin{aligned} |\Phi (X)|^2 +|\Phi (X^*)|^2\le \Psi (X^*X+XX^*), \end{aligned}$$

(4.4)

for each $X\in \mathscr {A}$. Moreover, if A is unitary, then we get the following corollary whose proof is similar to that of Lemma 3.1.

Corollary 4.2

Let $\mathscr {A}$ and $\mathscr {B}$ be unital $C^*$-algebras, and let $\mathcal {H}$ be a Hilbert $\mathscr {B}$-module. Let $\Phi : \mathscr {A} \rightarrow \mathcal {H}$ be a norm one linear operator $\langle \Phi (U),\Phi (U)\rangle =I_{\mathscr {B}}$ for some unitary map $U\in \mathscr {A}$. Then, there exists a unital positive linear map $\Psi :\mathscr {A}\rightarrow \mathscr {B}$ such that

$$\begin{aligned} |\Phi (X)|^2 +|\Phi (X^*)|^2\le 2\Psi (X^*X+XX^*). \end{aligned}$$

for each $X\in \mathscr {A}$.

Let $\mathscr {B}$ be a unital $C^*$-algebra. Then, $\mathscr {B}$ is a $C^*$-module over itself via $\langle B_1,B_2\rangle =B_1^*B_2$. The next result follows from Theorem 4.1.

Corollary 4.3

Let $\mathscr {A}$ and $\mathscr {B}$ be unital $C^*$-algebras and $\Phi : \mathscr {A} \rightarrow \mathscr {B}$ be a bounded linear map such that $|\Phi (A)|=\Vert A\Vert \ \Vert \Phi \Vert$ for some $A\in \mathscr {A}$. Then, there exist unital positive linear maps $\Psi _1,\Psi _2,\Psi :\mathscr {A}\rightarrow \mathscr {B}$ such that

$$\begin{aligned} |\Phi (AX)|^2 +|\Phi (AX^*)|^2\le & {} \Vert A\Vert ^2\Vert \Phi \Vert ^2\Psi _1(X^*X+XX^*),\\ |\Phi (XA)|^2 +|\Phi (X^*A)|^2\le & {} \Vert A\Vert ^2\Vert \Phi \Vert ^2\Psi _2(X^*X+XX^*)\\ |\Phi (AX+XA)|^2+|\Phi (AX^*+X^*A)|^2\le & {} 4\Vert A\Vert ^2\Vert \Phi \Vert ^2\Psi (X^*X+XX^*), \end{aligned}$$

for each $X\in \mathscr {A}$.

Corollary 4.4

Let $\mathscr {A}$ and $\mathscr {B}$ be unital $C^*$-algebras and $\Phi : \mathscr {A} \rightarrow \mathscr {B}$ be a norm one linear map $\Phi (U)=V$ for some unitary operators $U\in \mathscr {A}$ and $V\in \mathscr {B}$. Then

$$\begin{aligned} \left| \left| \left| \sum _{i=1}^n \ |\Phi (X_i)|^2 +|\Phi (X_i^*)|^2 \right| \right| \right| \le 2\left| \left| \left| \Phi \right| \right| \right| \ \sum _{i=1}^n |||X_i^*X_i+X_iX_i^*||| \end{aligned}$$

for any $X_1,\ldots ,X_n \in \mathscr {A}$ and any unitarily invariant norm $|||\cdot |||$.

Proof

By Corollary 4.2, there exists a unital positive linear map $\Psi :\mathscr {A}\rightarrow \mathscr {B}$ such that

$$\begin{aligned} |\Phi (X_i)|^2 +|\Phi (X_i^*)|^2\le 2\Psi (X_i^*X_i+X_iX_i^*). \end{aligned}$$

for each $1\le i \le n$. Hence

$$\begin{aligned} \left| \left| \left| \sum _{i=1}^n \ |\Phi (X_i)|^2 +|\Phi (X_i^*)|^2 \right| \right| \right|\le & {} 2\left| \left| \left| \Psi \left( \sum _{i=1}^n X_i^*X_i+X_iX_i^*\right) \right| \right| \right| \\\le & {} 2 \left| \left| \left| \Psi \right| \right| \right| \ \sum _{i=1}^n |||X_i^*X_i+X_iX_i^*|||. \end{aligned}$$

Proof of Theorem 4.1 implies that $\Psi (X)=\langle \Phi (U),\Phi (XU)\rangle =V^*\Phi (XU)$ for each $X\in \mathscr {A}$. Therefore, $\left| \left| \left| \Psi \right| \right| \right| \le \left| \left| \left| \Phi \right| \right| \right|$. $\square$

Corollary 4.5

Let $\Phi :\mathbb {M}_m\rightarrow \mathbb {M}_k$ be a norm one linear map such that $|\Phi (A)|=I_k$ for some contraction $A\in \mathbb {M}_m$. Then

$$\begin{aligned} \left| \left| \left| \sum _{i=1}^n \ |\Phi (X_i)|^2 +|\Phi (X_i^*)|^2\right| \right| \right| \ \le 2|||\Phi ||| \sum _{i=1}^n |||X_i^*X_i+X_iX_i^*|||, \end{aligned}$$

for any $X_1,\ldots ,X_n \in \mathbb {M}_m$ and any unitarily invariant norm $|||\cdot |||$.

Proof

First note that $\Phi (A)$ is a unitary matrix. So, $\Phi (A)$ is an extreme point of the unit ball of $\mathbb {M}_k$. There exist unitary matrices $U,V\in \mathbb {M}_n$ such that $A=\frac{1}{2}(U+V)$. Since $\Vert \Phi \Vert = 1$, we have $\Vert \Phi (U)\Vert ,\Vert \Phi (V)\Vert \le 1$. As $\Phi (A)=\frac{1}{2}(\Phi (U)+\Phi (V))$, we infer that $\Phi (U)$ is an extreme point of the unit ball of $\mathbb {M}_k$ and so it is a unitary matrix. By Corollary 4.4, the proof is complete. $\square$