On the Symmetry of Heinz Means and Equality Cases in Symmetrically Normed Ideals

Abstract

The Heinz mean is well known for its symmetry. This property naturally extends to operators through the function $f\left(\kappa \right)=\left|\right||A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\left|\right||,0\le \kappa \le 1.$ Since this function is convex and symmetric, it reaches its minimum at $\kappa =\frac{1}{2}$ and its maximum at the endpoints $\kappa =0$ and $\kappa =1$. In this paper, we exploit this symmetry together with the strict convexity of symmetrically normed ideals to characterize the equality cases in the Heinz inequality and related Young-type inequalities. Most of our results do not require the operators A and B to be invertible. For the cases that do require invertibility, such as those involving the logarithmic mean, we state this assumption explicitly. We also show new connections between equality in operator means, the relation $AX=XB$, and the Sylvester equation. Finally, for the trace-class ideal, we fully describe when the maximum and minimum values of $f\left(\kappa \right)$ are equal, using trace properties and Birkhoff–James orthogonality.

1. Introduction

Mathematical inequalities are fundamental tools in functional analysis and operator theory. They help elucidate the deep connections between various mathematical objects. A prominent example is the inequality between the arithmetic and geometric means. For a self-improvement of this scalar inequality, we refer the reader to Aldaz [1], and for related operator inequalities, see Kosaki [2].

Another well-known example is Young’s inequality. Ando [3] studied its matrix version. Since then, numerous authors have explored this topic. For instance, Furuichi [4] presented further improvements of Young’s inequality, and Al-Manasrah and Kittaneh [5] generalized two refined Young inequalities.

Among the classical means for two nonnegative real numbers $a,b\ge 0$, the Heinz mean is particularly significant. It is symmetric and provides a continuous path between the arithmetic and geometric means. For $0\le \kappa \le 1$, the Heinz mean is defined by

$$H_{\kappa }(a,b)={\displaystyle \frac{a^{\kappa }{b}^{1-\kappa }+a^{1-\kappa }{b}^{\kappa }}{2}}.$$

It satisfies the symmetry relations

$$H_{\kappa }(a,b)=H_{\kappa }(b,a)=H_{1-\kappa }(a,b),$$

implying that the function $\kappa \mapsto H_{\kappa }(a,b)$ is symmetric around $\kappa =\frac{1}{2}$. Consequently, $H_{\kappa }(a,b)$ attains its minimum at $\kappa =\frac{1}{2}$ (yielding the geometric mean $\sqrt{ab}$) and its maximum at the endpoints $\kappa =0,1$ (yielding the arithmetic mean $\frac{a+b}{2}$). Thus, we have

$$\sqrt{ab}\le H_{\kappa }(a,b)\le {\displaystyle \frac{a+b}{2}},0\le \kappa \le 1,$$

with equality in both bounds if and only if $a=b$. The symmetry and convexity of the Heinz mean make it a natural candidate for extension to the operator setting.

Given their importance, researchers have extensively studied the Heinz and Young inequalities for matrices. For instance, Kittaneh and Manasrah [6] proved improved Young and Heinz inequalities for matrices, and they later [7] explored reverse Young and Heinz inequalities for matrices. Building on this, Zhao and Wu [8] studied operator inequalities involving the improved Young inequality and its reverses, while Al-Manasrah and Kittaneh [9] provided further generalizations, refinements, and reverses of both the Young and Heinz inequalities.

Let $\mathcal{H}$ be a complex separable Hilbert space, and let $\mathcal{L}\left(\mathcal{H}\right)$ be the algebra of all bounded linear operators on $\mathcal{H}$. For $T\in \mathcal{L}\left(\mathcal{H}\right)$, its adjoint is denoted by $T^{\ast }$. We say that T is positive, and write $T\ge 0$, if $\langle Tx,x\rangle \ge 0$ for all vectors $x\in \mathcal{H}$. Every positive operator T has a unique positive square root, denoted by $T^{\frac{1}{2}}$. Using continuous functional calculus, we can also define $T^p$ for any real number $p>0$, which yields another positive operator. The modulus (or absolute value) of T is defined as $\left|T\right|={\left(T^{\ast }T\right)}^{\frac{1}{2}}$, which is always positive.

We now recall the concept of symmetrically normed ideals. Let $\mathcal{J}$ be a two-sided ideal in $\mathcal{L}\left(\mathcal{H}\right)$ equipped with a norm $\left|\right||·|\left|\right|$. It is called a symmetrically normed ideal (or unitarily invariant norm ideal) if it satisfies the following conditions:

(i)

$\mathcal{F}\left(\mathcal{H}\right)\subset \mathcal{J}\subset \mathcal{K}\left(\mathcal{H}\right)$, where $\mathcal{F}\left(\mathcal{H}\right)$ is the ideal of finite-rank operators and $\mathcal{K}\left(\mathcal{H}\right)$ is the ideal of compact operators on $\mathcal{H}$. Additionally, $\mathcal{F}\left(\mathcal{H}\right)$ is dense in $\mathcal{J}$ with respect to the norm $\left|\right||·|\left|\right|$.

(ii)

$\left|\right|\left|AXB\right|\left|\right|\le \parallel A\parallel \left|\right|\left|X\right|\left|\right|\parallel B\parallel$ for all operators $A,B\in \mathcal{L}\left(\mathcal{H}\right)$ and $X\in \mathcal{J}$.

(iii)

There exists a symmetric norming function $\Phi$ on the cone of finite nonnegative sequences $c_{00}^{+}$, such that

$$\left|\right|\left|X\right|\left|\right|=\Phi \left(s\left(X\right)\right),X\in \mathcal{F}\left(\mathcal{H}\right),$$

where $s\left(X\right)={\left\{s_n\left(X\right)\right\}}_{n\ge 1}$ is the sequence of singular values of X. These singular values are the eigenvalues of $\left|X\right|$, arranged in descending order and counted with their multiplicities.

We write ${\mathcal{J}}_{\left|\right||·|\left|\right|}$ to denote the ideal $\mathcal{J}$ equipped with this norm. Typical examples of symmetrically normed ideals include the Schatten p-classes ${\mathfrak{S}}_p\left(\mathcal{H}\right)$, as well as the Lorentz, Orlicz, and Marcinkiewicz ideals. For more details on this topic, see [10].

The Schatten p-class is defined as

$${\mathfrak{S}}_p\left(\mathcal{H}\right)=\left\{X\in \mathcal{L}\left(\mathcal{H}\right):s\left(X\right)\in {\ell }^p\right\},1\le p<\infty ,$$

with the norm

$${\parallel X\parallel }_p={\left(\sum _{n=1}^{\infty }{s}_n{\left(X\right)}^p\right)}^{\frac{1}{p}}={\left({\mathrm{tr}\left|X\right|}^p\right)}^{\frac{1}{p}}.$$

When $p=1$, this represents the trace class. When $p=2$, it corresponds to the Hilbert–Schmidt class, which forms a Hilbert space with the inner product ${\langle A,B\rangle }_2=\mathrm{tr}\left(A{B}^{\ast }\right)$. It is well established that the norm ${\parallel ·\parallel }_p$ is strictly convex for $1<p<\infty$, whereas the classes ${\mathfrak{S}}_1\left(\mathcal{H}\right)$ and ${\mathfrak{S}}_{\infty }\left(\mathcal{H}\right)$ lack this property.

For positive operators $A,B\in \mathcal{L}\left(\mathcal{H}\right)$, any operator X, and any unitarily invariant norm $\left|\right||·|\left|\right|$, we define the function

$$f\left(\kappa \right)=\left|\right||A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\left|\right||,0\le \kappa \le 1.$$

The function $\kappa \mapsto f\left(\kappa \right)$ is convex on $[0,1]$ and symmetric around $\kappa =\frac{1}{2}$, meaning $f\left(\kappa \right)=f(1-\kappa )$ for all $\kappa \in [0,1]$. Consequently, f attains its minimum at $\kappa =\frac{1}{2}$ and its maximum at the endpoints $\kappa =0$ and $\kappa =1$. This extends the symmetry and extremal properties of the scalar Heinz mean to the operator setting. For a detailed proof of the aforementioned properties of the function $f\left(\kappa \right)$, we refer the reader to [11]. This foundation leads to the famous Heinz inequalities, which provide sharp connections between the arithmetic and geometric means of positive operators using unitarily invariant norms. Specifically, for every $\kappa \in [0,1]$, we have

$$2\left|\right||A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\left|\right||\le \left|\right||A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\left|\right||\le \left|\right||AX+XB\left|\right||.$$

(1)

In this paper, we establish new results for the Schatten p-ideals. We present refinements utilizing spectral separation and Rosenblum’s theorem. These results provide a unified framework linking symmetry, convexity, and equality cases in operator inequalities.

Recent developments in applied analysis and systems theory have also emphasized the role of operator-theoretic techniques and homogeneity-based structures in the study of nonlinear models and dynamical systems. In particular, homogeneity principles and operator formulations appear naturally in the analysis of PDE–ODE coupled systems and in modern statistical learning frameworks. For instance, Xu and Li [12] investigated stabilization mechanisms for parabolic PDE–ODE systems with input saturation, while Li et al. [13] studied homogeneity pursuit methods in functional–coefficient quantile regression models for panel data. Although arising in different contexts, these works share structural features related to symmetry, convexity, and homogeneity, which are also fundamental in the operator framework underlying the Heinz mean considered in the present paper. This perspective highlights the broader relevance of operator inequalities and interpolation structures beyond the classical setting of functional analysis.

The paper is organized as follows. Section 1 serves as an introduction, explaining the symmetry of the scalar Heinz mean and its extension to operators with unitarily invariant norms. In Section 2, we review preliminary concepts, including strict convexity, equality conditions for the triangle inequality, and foundational results by Kittaneh. Section 3 investigates symmetrically normed ideals with strictly convex norms, where we characterize the equality cases in Heinz-type inequalities and explore the consequences of intertwining relations. Finally, Section 4 applies these results to the Schatten p-classes. This section features refinements based on Rosenblum’s theorem for the Sylvester equation and provides further equivalent conditions for the equality cases.

2. Preliminaries

In this section, we present key definitions and results that will be utilized throughout the paper. We begin with the concept of a strictly convex normed space.

Definition 1.

A normed space $(\mathcal{X},\parallel ·\parallel )$ is said to be strictly convex if for any two distinct elements $x,y\in \mathcal{X}$ with $\parallel x\parallel =\parallel y\parallel =1$, the following inequality holds:

$$∥{\displaystyle \frac{x+y}{2}}∥<1.$$

Equivalently, the unit sphere in $\mathcal{X}$ contains no nontrivial line segments.

A fundamental geometric property of strictly convex normed spaces is that equality in the triangle inequality holds only when the involved vectors are positively linearly dependent. This property plays a central role in our study of operator inequalities, as it allows us to precisely characterize the cases of equality in norms defined on symmetrically normed ideals.

Lemma 1

([14], Theorem 1.6). Let $(\mathcal{X},\parallel ·\parallel )$ be a strictly convex normed space. If $x,y\in \mathcal{X}$ satisfy

$$\parallel x+y\parallel =\parallel x\parallel +\parallel y\parallel ,$$

then x and y are linearly dependent and point in the same direction; that is, there exists $\lambda \ge 0$ such that

$$x=\lambda y.$$

Lemma 2

([15], Theorem 2.7). For $1<p<\infty$, the Schatten class $({\mathfrak{S}}_p\left(\mathcal{H}\right),\parallel ·{\parallel }_p)$ is strictly convex.

Next, we recall several results by Kittaneh that characterize the equality cases for the various terms of the Heinz inequality for the Schatten p-norm for $1<p<\infty$.

Lemma 3

([16]). Let $A,B,X\in \mathcal{L}\left(\mathcal{H}\right)$ such that A and B are positive, and let $1<p<\infty$. Then

$$2{\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\parallel }_p={\parallel AX+XB\parallel }_p$$

if and only if $AX=XB$.

Lemma 4.

Let $A,B,X\in \mathcal{L}\left(\mathcal{H}\right)$ such that A and B are positive, and let $1<p<\infty$. Then

$${\parallel A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\parallel }_p={\parallel AX+XB\parallel }_p$$

for some $0<\kappa <1$ if and only if $AX=XB$.

Lemma 5

([17], Theorem 5). Let $A,B,X\in \mathcal{L}\left(\mathcal{H}\right)$ such that A and B are positive and invertible, and let $1<p<\infty$. Then

$$2{\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\parallel }_p={\parallel A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\parallel }_p$$

for some $0<\kappa <1$, $\kappa \ne \frac{1}{2}$, if and only if $AX=XB$.

In connection with the preceding results, we present the following characterization of the kernel of an elementary operator $E\left(X\right)=AXB-CXD$, intersected with the Schatten p-class:

Lemma 6

([18], Theorem 17). Let $A,B,C,D\in \mathcal{L}\left(\mathcal{H}\right)$ be commuting normal operators such that

$$kerA\cap kerC=\left\{0\right\}\mathrm{and}kerB\cap kerD=\left\{0\right\}.$$

Define the elementary operator

$$E:{\mathfrak{S}}_p\left(\mathcal{H}\right)\to {\mathfrak{S}}_p\left(\mathcal{H}\right),E\left(X\right)=AXB-CXD,$$

where $1<p<\infty$. For a fixed $S\in {\mathfrak{S}}_p\left(\mathcal{H}\right)$, consider the functional

$$F_E\left(X\right)={\parallel E\left(X\right)+S\parallel }_p,X\in {\mathfrak{S}}_p\left(\mathcal{H}\right).$$

Then, the following statements are equivalent:

(i) 

$F_E$ attains a global minimum at S;

(ii) 

$S\perp ran\left(E\right)$;

(iii) 

$S\in ker\left(E\right)$.

We conclude this section by presenting a result showing that if the Sylvester equation

$$AX-XB=0$$

admits a nontrivial solution, then the subalgebra generated by A, B, and this solution consists entirely of solutions of the same equation.

Assume that $A,B,X\in \mathcal{L}\left(\mathcal{H}\right)$ with $X\ne 0$. For these operators, we define the iterated product sequence acting on $\mathcal{H}:$

$${\left(AXB\right)}^n:=A^{n}X{B}^n,n\in {\mathbb{N}}_0,$$

where ${\mathbb{N}}_0$ denotes the set of non-negative integers.

We consider the operator family

$${\mathcal{A}}_{A,X,B}:=\overline{\left\{p\right(AXB):p\mathrm{is}\mathrm{a}\mathrm{complex}\mathrm{polynomial}\}}.$$

This set is the smallest unital subalgebra of $\mathcal{L}\left(\mathcal{H}\right)$ that contains $AXB$ and is closed under polynomial combinations. Equipped with the inherited operator norm $\parallel ·\parallel$ and standard operations, the set ${\mathcal{A}}_{A,X,B}$ constitutes a separable Banach subspace of $\mathcal{L}\left(\mathcal{H}\right)$. Furthermore, it forms a commutative Banach algebra with the identity element X.

The following lemma highlights a fundamental invariance property of the elements in ${\mathcal{A}}_{A,X,B}$ that will be useful throughout our analysis.

Lemma 7

([19], Theorem 3.4). Let $A,B,X\in \mathcal{L}\left(\mathcal{H}\right)$ satisfying $AX=XB$. Then, for every $Y\in {\mathcal{A}}_{A,X,B}$, it holds that

$$AY=YB.$$

In other words, every element of ${\mathcal{A}}_{A,X,B}$ is a solution to the homogeneous Sylvester equation

$$AY-YB=0.$$

3. Symmetrically Normed Ideals

In this section, we focus on symmetrically normed ideals endowed with strictly convex norms. Strict convexity plays a central role in the study of operator inequalities, since it yields uniqueness in equality cases of convexity and triangle-type inequalities and strengthens equality conditions in convexity and interpolation formulas. We present our main results, which characterize the equality cases in the Heinz inequality for strictly convex symmetrically normed ideals.

Theorem 1.

Let $A,B\in \mathcal{L}\left(\mathcal{H}\right)$ be positive and invertible operators, and let $X\in {\mathcal{J}}_{\left|\right||·|\left|\right|}$, where $\left|\right||·|\left|\right|$ is a strictly convex norm. Then

$$2\left|\right||A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\left|\right||=\left|\right||AX+XB\left|\right||$$

if and only if $2A^{{\displaystyle \frac{1}{2}}}X{B}^{{\displaystyle \frac{1}{2}}}=AX+XB.$

Proof. 

Suppose that

$$2\left|\right||A^{{\displaystyle \frac{1}{2}}}X{B}^{{\displaystyle \frac{1}{2}}}\left|\right||=\left|\right||AX+XB\left|\right||.$$

Then, by applying the Heinz inequality, an inequality from [20] (Theorem 3.9), and the triangle inequality, we obtain for every $\frac{1}{2}\le \alpha <1$ that

$$\begin{array}{ccc}\hfill \left|\left|\left|{\displaystyle \frac{AX+XB}{2}}\right|\right|\right|& =& \left|\left|\left|A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\right|\right|\right|\hfill \\ & \le & \left|\left|\left|(1-\alpha )A^{\frac{1}{2}}X{B}^{\frac{1}{2}}+\alpha {\displaystyle \frac{AX+XB}{2}}\right|\right|\right|\hfill \\ & \le & (1-\alpha )\left|\left|\left|A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\right|\right|\right|+\alpha \left|\left|\left|{\displaystyle \frac{AX+XB}{2}}\right|\right|\right|\hfill \\ & \le & \left|\left|\left|{\displaystyle \frac{AX+XB}{2}}\right|\right|\right|.\hfill \end{array}$$

Consequently, all the above inequalities must hold as equalities, yielding

$$\left|\left|\left|(1-\alpha )A^{\frac{1}{2}}X{B}^{\frac{1}{2}}+\alpha {\displaystyle \frac{AX+XB}{2}}\right|\right|\right|=(1-\alpha )\left|\left|\left|A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\right|\right|\right|+\alpha \left|\left|\left|{\displaystyle \frac{AX+XB}{2}}\right|\right|\right|.$$

Setting $\alpha =\frac{1}{2}$ gives

$$\left|\left|\left|A^{\frac{1}{2}}X{B}^{\frac{1}{2}}+{\displaystyle \frac{AX+XB}{2}}\right|\right|\right|=\left|\left|\left|A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\right|\right|\right|+\left|\left|\left|{\displaystyle \frac{AX+XB}{2}}\right|\right|\right|.$$

(2)

This demonstrates that equality holds in the triangle inequality. Because the norm is strictly convex, there exists $\lambda \ge 0$ such that

$$A^{\frac{1}{2}}X{B}^{\frac{1}{2}}=\lambda {\displaystyle \frac{AX+XB}{2}}.$$

Using the norm equality again, we conclude that $\lambda =1$, and therefore

$$2A^{\frac{1}{2}}X{B}^{\frac{1}{2}}=AX+XB.$$

□

Remark 1.

In Theorem 1, the invertibility of A and B is required in order to apply [20] (Theorem 3.9).

It is currently unknown whether this assumption can be dropped in the general framework of symmetrically normed ideals. However, when restricting to the Schatten ideals ${\mathfrak{S}}_p\left(\mathcal{H}\right)$ with $1<p<\infty$, the invertibility hypothesis is not needed, and the same conclusion remains valid (see Lemma 3).

For symmetric normed ideals beyond the Schatten classes, the invertibility condition appears to be essential, at least with the techniques currently available.

The following example demonstrates that the strict convexity assumption in Theorem 1 is essential.

Example 1.

Consider the operators

$$A=B=\left[\begin{array}{cc}1& 0\\ 0& 4\end{array}\right],X=\left[\begin{array}{cc}1& 1\\ 1& 2\end{array}\right],$$

acting on ${\mathbb{C}}^2$ endowed with the standard inner product. Let us define

$$M_1=A^{\frac{1}{2}}X{B}^{\frac{1}{2}},M_2=AX+XB.$$

Since $A^{\frac{1}{2}}=diag(1,2)$, we obtain

$$M_1=\left[\begin{array}{cc}1& 2\\ 2& 8\end{array}\right],M_2=\left[\begin{array}{cc}2& 5\\ 5& 16\end{array}\right].$$

Both matrices are symmetric with positive traces and positive determinants:

$$tr\left(M_1\right)=9,det\left(M_1\right)=4>0,tr\left(M_2\right)=18,det\left(M_2\right)=7>0,$$

and therefore, both matrices are positive definite. For positive operators, the trace norm equals the trace; consequently,

$${\parallel M_1\parallel }_1=9,{\parallel M_2\parallel }_1=18.$$

Hence,

$$2{\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\parallel }_1=2{\parallel M_1\parallel }_1=18={\parallel M_2\parallel }_1={\parallel AX+XB\parallel }_1,$$

so the equality condition in the theorem’s premise is satisfied.

However, the identity

$$2A^{\frac{1}{2}}X{B}^{\frac{1}{2}}=AX+XB$$

does not hold. Indeed, a direct computation yields

$$2A^{\frac{1}{2}}X{B}^{\frac{1}{2}}=\left[\begin{array}{cc}2& 4\\ 4& 16\end{array}\right],AX+XB=\left[\begin{array}{cc}2& 5\\ 5& 16\end{array}\right].$$

Based on the inequality derived for the function $f\left(\kappa \right)$—a consequence of its convexity on $[0,1]$ and its symmetry around $\kappa =\frac{1}{2}$—along with Theorem 1, we establish the following result.

Theorem 2.

Let $A,B\in \mathcal{L}\left(\mathcal{H}\right)$ be positive and invertible operators, and let $X\in {\mathcal{J}}_{\left|\right||·|\left|\right|}$, where $\left|\right||·|\left|\right|$ is a strictly convex norm. Then, for some $0<\kappa <1$, it holds that

$$\left|\right||A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\left|\right||=\left|\right||AX+XB\left|\right||,$$

(3)

if and only if

$$2A^{\frac{1}{2}}X{B}^{\frac{1}{2}}=AX+XB.$$

Proof. 

Recall that by [17] (Inequality (2.13)), for every $0<\mu <1$ we have

$$\begin{array}{cc}\hfill 2r_0\left(\left|\right||AX+XB\left|\right||-2\left|\right||A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\left|\right||\right)\le & \\ \hfill \left|\right||AX+XB\left|\right||-\left|\right||A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }\left|\right||,\end{array}$$

where $r_0=min\{\mu ,1-\mu \}$.

If (3) holds for some $0<\kappa <1$, the right-hand side of the above inequality vanishes, yielding

$$2\left|\right||A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\left|\right||=\left|\right||AX+XB\left|\right||.$$

Applying Theorem 1 gives

$$2A^{\frac{1}{2}}X{B}^{\frac{1}{2}}=AX+XB.$$

The converse implication follows directly by substituting this identity into (1), completing the proof. □

The preceding results immediately yield the following corollary.

Corollary 1.

Let $A,B\in \mathcal{L}\left(\mathcal{H}\right)$ be positive and invertible operators, and let $X\in {\mathcal{J}}_{\left|\right||·|\left|\right|}$, where $\left|\right||·|\left|\right|$ is a strictly convex norm. If, for some $0<\kappa <1$, it holds that

$$\left|\right||A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\left|\right||=\left|\right||AX+XB\left|\right||,$$

then

$$2\left|\right||A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\left|\right||=\left|\right||AX+XB\left|\right||,$$

and consequently,

$$\left|\right||A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }\left|\right||=\left|\right||AX+XB\left|\right||\mathrm{for}\mathrm{all}\mu \in [0,1].$$

Proof. 

Assume there exists $\kappa \in (0,1)$ such that

$$\left|\right||A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\left|\right||=\left|\right||AX+XB\left|\right||.$$

Then, by Theorem 2, we have

$$2A^{\frac{1}{2}}X{B}^{\frac{1}{2}}=AX+XB.$$

Applying Theorem 1, it follows that

$$2\left|\right||A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\left|\right||=\left|\right||AX+XB\left|\right||.$$

This demonstrates that the function

$$f\left(\mu \right)=\left|\right||A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }\left|\right||$$

is constant on $[0,1]$, as its maximum and minimum values coincide. Therefore,

$$\left|\right||A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }\left|\right||=\left|\right||AX+XB\left|\right||\mathrm{for}\mathrm{all}\mu \in [0,1].$$

□

We begin by recalling a result established by Kittaneh, which serves as a key tool for deriving sufficient conditions for equality in various forms of the Heinz inequality.

Lemma 8

([21], Corollary 2). Let $A,B\in \mathcal{L}\left(\mathcal{H}\right)$ be positive operators. Then, for every positive real number r,

$$A^{r}X=X{B}^r\mathit{if}\mathit{and}\mathit{only}\mathit{if}AX=XB.$$

Proposition 1.

Let $A,B,X\in \mathcal{L}\left(\mathcal{H}\right)$ such that A and B are positive, and

$$AX=XB.$$

Then, for every positive real number r, we have

$$A^{r}X{B}^r=A^{2r}X=X{B}^{2r},$$

and, in addition,

$$A^{s}X{B}^{1-s}=AX=XB,\mathit{for}\mathit{all}s\in (0,1).$$

Proof. 

Because A and B are positive, continuous functional calculus ensures that the powers $A^r$ and $B^r$ are well defined for any real $r>0$. By Lemma 8, the equality $AX=XB$ implies

$$A^{r}X=X{B}^r,\mathrm{for}\mathrm{all}r>0.$$

Right-multiplying both sides of this identity by $B^r$ yields

$$A^{r}X{B}^r=X{B}^{2r}.$$

Conversely, applying Lemma 8 again with the same hypothesis but replacing r with $2r$ gives

$$A^{2r}X=X{B}^{2r}.$$

Hence,

$$A^{r}X{B}^r=A^{2r}X=X{B}^{2r},$$

which proves the first assertion.

For the second identity, let $s\in (0,1)$. By the assumption $AX=XB$ and Lemma 8, we have

$$A^{s}X=X{B}^s.$$

Right-multiplying this equation by $B^{1-s}$ yields

$$A^{s}X{B}^{1-s}=XB.$$

Applying the hypothesis $AX=XB$ again, we conclude

$$A^{s}X{B}^{1-s}=AX=XB,\mathrm{for}\mathrm{all}s\in (0,1).$$

This completes the proof.  □

Proposition 1 shows that the relation $AX=XB$ is sufficient for equality in all terms of the Heinz inequality.

Proposition 2.

Let $A,B\in \mathcal{L}\left(\mathcal{H}\right)$ be positive operators, and let $X\in {\mathcal{J}}_{\left|\right||·|\left|\right|}$ satisfy $AX=XB$. Then, for every $\kappa \in (0,1)$,

$$2\left|\right||A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\left|\right||=\left|\right||A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\left|\right||=\left|\right||AX+XB\left|\right||.$$

Next, we turn our attention to the Young-type inequality for operators. T. Ando [3] observed that while Young’s inequality holds for scalars, it generally fails for the operator norm. Nevertheless, he established a weaker but closely related estimate: for all positive operators $A,B\in \mathcal{L}\left(\mathcal{H}\right)$, any $X\in {\mathcal{J}}_{\left|\right||·|\left|\right|}$, and every $0<\kappa <1$, we have

$$\left|\right||A^{\kappa }X{B}^{1-\kappa }\left|\right||\le \kappa \left|\right|\left|AX\right|\left|\right|+(1-\kappa )\left|\right|\left|XB\right|\left|\right|.$$

(4)

Note that if the intertwining relation $AX=XB$ holds, Proposition 1 immediately yields equality in (4).

Proposition 3.

Let $A,B\in \mathcal{L}\left(\mathcal{H}\right)$ be positive operators, and let $X\in {\mathcal{J}}_{\left|\right||·|\left|\right|}$ satisfy $AX=XB$. Then, for every $\kappa \in (0,1)$, the following equality holds:

$$\left|\right||A^{\kappa }X{B}^{1-\kappa }\left|\right||=\kappa \left|\right|\left|AX\right|\left|\right|+(1-\kappa )\left|\right|\left|XB\right|\left|\right|=\left|\right|\left|AX\right|\left|\right|.$$

(5)

Proof. 

Because $AX=XB$, we have $A^{\kappa }X{B}^{1-\kappa }=AX=XB$ for every $\kappa \in (0,1)$. Consequently,

$$\left|\right||A^{\kappa }X{B}^{1-\kappa }\left|\right||=\left|\right|\left|AX\right|\left|\right|=\left|\right|\left|XB\right|\left|\right|,$$

and thus the asserted identity

$$\left|\right||A^{\kappa }X{B}^{1-\kappa }\left|\right||=\kappa \left|\right|\left|AX\right|\left|\right|+(1-\kappa )\left|\right|\left|XB\right|\left|\right|=\left|\right|\left|AX\right|\left|\right|$$

follows immediately. □

Before stating the general result, we first explain the argument in the Schatten p-norm setting, where the mechanism can be seen explicitly. Let $1<p<\infty$ and let $A,B\in \mathcal{L}\left(\mathcal{H}\right)$ be positive. Assume that

$${\parallel AX\parallel }_p+{\parallel XB\parallel }_p=2{\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\parallel }_p.$$

Then, by the Heinz and triangle inequalities, we obtain

$${\parallel AX\parallel }_p+{\parallel XB\parallel }_p=2{\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\parallel }_p\le {\parallel AX+XB\parallel }_p\le {\parallel AX\parallel }_p+{\parallel XB\parallel }_p.$$

Consequently, all the above inequalities must hold as equalities, leading to

$$2{\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\parallel }_p={\parallel AX+XB\parallel }_p.$$

By Lemma 4, which characterizes the equality cases in the Heinz inequality, we conclude that $AX=XB$.

The previous argument relies only on strict convexity. We now formulate the general result.

Theorem 3.

Let $A,B\in \mathcal{L}\left(\mathcal{H}\right)$ be positive operators, and let $X\in {\mathcal{J}}_{\left|\right||·|\left|\right|}$, where $\left|\right||·|\left|\right|$ is a strictly convex norm. Then

$$\left|\right||A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\left|\right||=\frac{1}{2}\left|\right|\left|AX\right|\left|\right|+\frac{1}{2}\left|\right|\left|XB\right|\left|\right|$$

(6)

if and only if $AX=XB$.

Proof. 

Suppose equality holds in (6). Then,

$$\left|\right|\left|AX\right|\left|\right|+\left|\right|\left|XB\right|\left|\right|=2\left|\right||A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\left|\right||\le \left|\right||AX+XB\left|\right||\le \left|\right|\left|AX\right|\left|\right|+\left|\right|\left|XB\right|\left|\right|.$$

Hence,

$$\left|\right||AX+XB|\left|\right|=\left|\right|\left|AX\right|\left|\right|+\left|\right|\left|XB\right|\left|\right|,$$

indicating that equality is achieved in the triangle inequality. Because the norm $\left|\right||·|\left|\right|$ is strictly convex, there exists a scalar $\lambda \ge 0$ such that

$$AX=\lambda XB.$$

Moreover, by [22] (Theorem 2.3), we have $\left|\right|\left|AX\right|\left|\right|=\left|\right|\left|XB\right|\left|\right|$, which forces $\lambda =1$. Therefore,

$$AX=XB.$$

Conversely, if $AX=XB$, equality in (6) follows directly from Proposition 1. □

Finally, the triangle inequality gives

$$\left|\right||A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\left|\right||\le \left|\right||A^{\kappa }X{B}^{1-\kappa }\left|\right||+\left|\right||A^{1-\kappa }X{B}^{\kappa }\left|\right||,$$

we establish the following characterization.

Theorem 4.

Let $A,B\in \mathcal{L}\left(\mathcal{H}\right)$ be positive operators, and let $X\in {\mathcal{J}}_{\left|\right||·|\left|\right|}$, where $\left|\right||·|\left|\right|$ is a strictly convex norm. Then, for some $\kappa \in (0,1)$, the equality

$$\left|\right||A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\left|\right||=\left|\right|\left|AX\right|\left|\right|+\left|\right|\left|XB\right|\left|\right|$$

(7)

holds if and only if $AX=XB$.

Proof. 

From the hypothesis, we obtain

$$\left|\right|\left|AX\right|\left|\right|+\left|\right|\left|XB\right|\left|\right|=\left|\right||A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\left|\right||\le \left|\right||AX+XB\left|\right||.$$

Hence, equality holds in the triangle inequality. Because the norm $\left|\right||·|\left|\right|$ is strictly convex, there exists $\lambda \ge 0$ such that

$$AX=\lambda XB.$$

Additionally, by [22] (Corollary 3.4), we have $\left|\right|\left|AX\right|\left|\right|=\left|\right|\left|XB\right|\left|\right|$, implying $\lambda =1$. Therefore,

$$AX=XB.$$

The reverse implication is straightforward. □

To complete this section, we recall a well-known chain of inequalities relating the Heinz mean to the logarithmic mean for positive operators and unitarily invariant norms. These classical estimates, obtained through a sequence of contributions by Bhatia, Davis, Hiai, and Kosaki, provide a natural context for several of our earlier results. We summarize the optimal form of these inequalities below.

For $\frac{1}{4}\le \kappa \le \frac{3}{4}$, it holds that

$$\begin{array}{cc}\hfill \left|\right||A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\left|\right||& \le \frac{1}{2}\left|\right||A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\left|\right||\hfill \\ \hfill & \le \left|\left|\left|{\int }_0^1{A}^{1-s}X{B}^{s}ds\right|\right|\right|\le {\displaystyle \frac{1}{2}}\left|\right||AX+XB\left|\right||.\hfill \end{array}$$

(8)

The sequence of inequalities above was established progressively in [11,23,24], with each contribution relying on different methods. Subsequently, Drissi [25] proved that the range $\left[\frac{1}{4},\frac{3}{4}\right]$ is indeed the sharp interval for the intermediate estimate.

We include (8) to highlight its connection to the symmetry properties of the Heinz mean and the equality conditions investigated in the present work.

Theorem 5.

Let $A,B\in \mathcal{L}\left(\mathcal{H}\right)$ be positive and invertible operators, and let $X\in {\mathcal{J}}_{\left|\right||·|\left|\right|}$, where $\left|\right||·|\left|\right|$ is a strictly convex norm. Then

$$2\left|\left|\left|{\int }_0^1{A}^{1-s}X{B}^{s}ds\right|\right|\right|=\left|\right||AX+XB\left|\right||$$

if and only if $2A^{\frac{1}{2}}X{B}^{\frac{1}{2}}=AX+XB.$

Proof. 

Assume that equality holds among the norms. By [26] (Corollary 3.10), we have

$$\begin{array}{cc}\hfill \pi \left|\left|\left|{\displaystyle \frac{AX+XB}{2}}\right|\right|\right|& =\pi \left|\left|\left|{\int }_0^1{A}^{1-s}X{B}^{s}ds\right|\right|\right|\hfill \\ \hfill & \le \pi \left|\left|\left|(1-\alpha )A^{\frac{1}{2}}X{B}^{\frac{1}{2}}+\alpha {\displaystyle \frac{AX+XB}{2}}\right|\right|\right|\hfill \\ \hfill & \le \pi (1-\alpha )\left|\left|\left|A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\right|\right|\right|+\pi \alpha \left|\left|\left|{\displaystyle \frac{AX+XB}{2}}\right|\right|\right|\hfill \\ \hfill & \le \pi \left|\left|\left|{\displaystyle \frac{AX+XB}{2}}\right|\right|\right|,\hfill \end{array}$$

(9)

for every $\frac{1}{2}\le \alpha <1$.

Thus, equality must occur in the triangle inequality. In particular, setting $\alpha =\frac{1}{2}$ yields

$$\left|\left|\left|A^{\frac{1}{2}}X{B}^{\frac{1}{2}}+{\displaystyle \frac{AX+XB}{2}}\right|\right|\right|=\left|\left|\left|A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\right|\right|\right|+\left|\left|\left|{\displaystyle \frac{AX+XB}{2}}\right|\right|\right|.$$

(10)

By Lemma 1, there exists $\lambda \ge 0$, such that

$$A^{\frac{1}{2}}X{B}^{\frac{1}{2}}=\lambda {\displaystyle \frac{AX+XB}{2}}.$$

Substituting this identity into (9) with $\alpha =\frac{1}{2}$, we obtain

$$\begin{array}{cc}\hfill \pi \left|\left|\left|{\displaystyle \frac{AX+XB}{2}}\right|\right|\right|& =\pi \left|\left|\left|{\int }_0^1{A}^{1-s}X{B}^{s}ds\right|\right|\right|\hfill \\ \hfill & ={\displaystyle \frac{\pi }{2}}\left|\left|\left|A^{\frac{1}{2}}X{B}^{\frac{1}{2}}+{\displaystyle \frac{AX+XB}{2}}\right|\right|\right|\hfill \\ \hfill & ={\displaystyle \frac{\pi }{2}}\left|\left|\left|(1+\lambda ){\displaystyle \frac{AX+XB}{2}}\right|\right|\right|,\hfill \end{array}$$

(11)

which forces $1+\lambda =2$, or equivalently,

$$2A^{\frac{1}{2}}X{B}^{\frac{1}{2}}=AX+XB.$$

Conversely, if

$$2A^{\frac{1}{2}}X{B}^{\frac{1}{2}}=AX+XB,$$

then Theorem 1 dictates that

$$2\left|\right||A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\left|\right||=\left|\right||AX+XB\left|\right||.$$

Using (8), it follows that

$$2\left|\left|\left|{\int }_0^1{A}^{1-s}X{B}^{s}ds\right|\right|\right|=\left|\right||AX+XB\left|\right||.$$

□

4. $\mathit{p}$-Schatten Ideals

Among the best-known examples of symmetric ideals endowed with strictly convex norms are the Schatten ideals ${\mathfrak{S}}_p\left(\mathcal{H}\right)$ for $1<p<\infty$. Recall that if $A,B\in \mathcal{L}\left(\mathcal{H}\right)$ are positive operators and $X\in {\mathfrak{S}}_p\left(\mathcal{H}\right)$ with $1\le p\le \infty$, then for all $0\le \kappa \le 1$,

$$2\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}{\parallel }_p\le \parallel A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }{\parallel }_p\le {\parallel AX+XB\parallel }_p.$$

(12)

Combining Lemmas 3–5, we deduce the equivalence of the following statements:

(i)

$2\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}{\parallel }_p={\parallel AX+XB\parallel }_p$;

(ii)

$AX=XB$;

(iii)

${\parallel A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\parallel }_p={\parallel AX+XB\parallel }_p$, for any $\kappa \in (0,1)$;

(iv)

${\parallel A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\parallel }_p={\parallel AX+XB\parallel }_p$, for some $\kappa \in (0,1)$.

If A and B are additionally invertible, the preceding conditions are also equivalent to:

(v)

$2\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}{\parallel }_p={\parallel A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\parallel }_p$ for every $\kappa \in (0,1)$;

(vi)

$2\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}{\parallel }_p={\parallel A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\parallel }_p$ for some $\kappa \in (0,1)$, $\kappa \ne \frac{1}{2}$.

We note that any of the above equalities hold if and only if $AX=XB$. This condition characterizes the equality case in the arithmetic–geometric mean inequality for the p-norm with $1<p<\infty$, is closely related to the homogeneous Sylvester equation. Recall that the Sylvester equation plays a central role in linear algebra and operator theory, finding numerous applications in control theory, signal processing, and numerical linear algebra.

In its classical form, the Sylvester equation is given by

$$AX-XB=C,$$

(13)

where $A,B,C\in \mathcal{L}\left(\mathcal{H}\right)$ are given operators and $X\in \mathcal{L}\left(\mathcal{H}\right)$ is the unknown. Equation (13) is solvable if there exists $\tilde{X}\in \mathcal{L}\left(\mathcal{H}\right)$ such that $A\tilde{X}-\tilde{X}B=C$.

In [27], Rosenblum proved that if $\sigma \left(A\right)\cap \sigma \left(B\right)=\varnothing$, where $\sigma \left(A\right)$ denotes the spectrum of A, then (13) admits a unique solution $\tilde{X}\in \mathcal{L}\left(\mathcal{H}\right)$ for every C. An elegant proof of Rosenblum’s theorem can be found in [28], and for a comprehensive overview, we refer the reader to the excellent survey on the Sylvester equation in [29].

Combining Kittaneh’s characterizations with those established in this work, we obtain

$$AX=XB\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}2A^{\frac{1}{2}}X{B}^{\frac{1}{2}}=AX+XB,$$

whenever $X\in {\mathfrak{S}}_p\left(\mathcal{H}\right)$ with $1<p<\infty$, and both A and B are positive, invertible operators.

We formalize this new characterization in the following theorem.

Theorem 6.

Let $A,B\in \mathcal{L}\left(\mathcal{H}\right)$ be positive and invertible operators, and let $X\in {\mathfrak{S}}_p\left(\mathcal{H}\right)$ for $1<p<\infty$. Then the following conditions are equivalent:

(a) 

$2\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}{\parallel }_p={\parallel AX+XB\parallel }_p;$

(b) 

$2A^{\frac{1}{2}}X{B}^{\frac{1}{2}}=AX+XB;$

(c) 

$AX=XB.$

The preceding theorem, together with Lemma 6, yields the following corollary.

Corollary 2.

Let $A,B\in \mathcal{L}\left(\mathcal{H}\right)$ be positive and invertible operators, and let $X\in {\mathfrak{S}}_p\left(\mathcal{H}\right)$ for $1<p<\infty$. Then the following statements are equivalent:

(a) 

$2\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}{\parallel }_p={\parallel AX+XB\parallel }_p;$

(b) 

$2A^{\frac{1}{2}}X{B}^{\frac{1}{2}}=AX+XB;$

(c) 

$AX=XB;$

(d) 

$F_E$ attains a global minimum at X, where $F_E\left(Z\right)={\parallel AZ-ZB+X\parallel }_p$ for $Z\in {\mathfrak{S}}_p\left(\mathcal{H}\right)$;

(e) 

$X\perp ran\left(E\right)$, where $E\left(Z\right)=AZ-ZB$ for $Z\in {\mathfrak{S}}_p\left(\mathcal{H}\right)$.

Corollary 3.

Let $A,B\in \mathcal{L}\left(\mathcal{H}\right)$ be positive operators, and let $X\in {\mathfrak{S}}_p\left(\mathcal{H}\right)$ with $1<p<\infty$ be such that

$$2\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}{\parallel }_p={\parallel AX+XB\parallel }_p.$$

Then, for every $Y\in {\mathcal{A}}_{A,X,B}$ and any $\kappa \in (0,1)$, we have

$$2\parallel A^{\frac{1}{2}}Y{B}^{\frac{1}{2}}{\parallel }_p={\parallel AY+YB\parallel }_p={\parallel A^{\kappa }Y{B}^{1-\kappa }+A^{1-\kappa }Y{B}^{\kappa }\parallel }_p.$$

Kapil and Singh [20] (Theorems 3.7 and 3.8) established a refinement of the Heinz inequality. Building upon Lemma 3 and this refinement, we provide an alternative proof of Lemma 4 for cases where the parameter $\kappa$ is restricted to the interval $\frac{1}{4}\le \kappa \le \frac{3}{4}$.

Lemma 9.

Let $A,B$, and X be operators such that A and B are positive, and let $1<p<\infty$. Then

$${\parallel A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\parallel }_p={\parallel AX+XB\parallel }_p$$

for some κ with $\frac{1}{4}\le \kappa \le \frac{3}{4}$ if and only if $AX=XB$.

Proof. 

Assume that for some $\frac{1}{4}\le \kappa \le \frac{3}{4}$ we have

$${\parallel A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\parallel }_p={\parallel AX+XB\parallel }_p.$$

Observe that the claim trivially holds when $\kappa =\frac{1}{2}$ by Lemma 3; thus, we may assume $\kappa \ne \frac{1}{2}$.

Then, by the inequality established in [20] (Theorem 3.8) and the triangle inequality, we obtain for every $\frac{1}{2}\le \alpha <1$ that

$$\begin{array}{ccc}\hfill {\displaystyle \frac{1}{2}}{\parallel AX+XB\parallel }_p& =& {\displaystyle \frac{1}{2}}{\parallel A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\parallel }_p\hfill \\ & \le & {∥(1-\alpha )A^{\frac{1}{2}}X{B}^{\frac{1}{2}}+\alpha {\displaystyle \frac{AX+XB}{2}}∥}_p\hfill \\ & \le & (1-\alpha )\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}{\parallel }_p+\alpha {∥{\displaystyle \frac{AX+XB}{2}}∥}_p\hfill \\ & \le & \frac{1}{2}{\parallel AX+XB\parallel }_p.\hfill \end{array}$$

Therefore, all the above inequalities must hold as equalities, which means

$$(1-\alpha )\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}{\parallel }_p=(1-\alpha ){∥{\displaystyle \frac{AX+XB}{2}}∥}_p.$$

Because $(1-\alpha )\ne 0$, it follows that

$$2\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}{\parallel }_p={\parallel AX+XB\parallel }_p.$$

By Lemma 3, this equality implies $AX=XB$. The converse implication is immediate and is thus omitted. □

Combining the Heinz inequality (12) with Rosenblum’s uniqueness theorem for the Sylvester equation yields refined versions of the Heinz inequality within the Schatten p-ideals, provided the spectra of the positive operators A and B are disjoint.

Corollary 4.

Let $A,B\in \mathcal{L}\left(\mathcal{H}\right)$ be positive operators such that $\sigma \left(A\right)\cap \sigma \left(B\right)=\varnothing$, and let $X\in {\mathfrak{S}}_p\left(\mathcal{H}\right)\setminus \left\{0\right\}$ for $1<p<\infty$. Then, for every $0<\kappa <1$, we have

$${\parallel A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\parallel }_p<{\parallel AX+XB\parallel }_p.$$

(14)

Proof. 

The proof is immediate. By Rosenblum’s theorem, the homogeneous Sylvester equation admits a unique solution, which must be the zero operator. Since $X\ne 0$, $AX\ne XB$. Therefore, Lemma 4 dictates that

$${\parallel A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\parallel }_p<{\parallel AX+XB\parallel }_p.$$

□

Under the additional assumption that both A and B are invertible, the following refinement of the previous corollary holds (the proof is omitted).

Corollary 5.

Let $A,B\in \mathcal{L}\left(\mathcal{H}\right)$ be positive and invertible operators such that $\sigma \left(A\right)\cap \sigma \left(B\right)=\varnothing$, and let $X\in {\mathfrak{S}}_p\left(\mathcal{H}\right)\setminus \left\{0\right\}$ for $1<p<\infty$. Then, for every $0<\kappa <1$, with $\kappa \ne {\displaystyle \frac{1}{2}}$ we have

$$2\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}{\parallel }_p<\parallel A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }{\parallel }_p<{\parallel AX+XB\parallel }_p.$$

(15)

Remark 2.

The strict inequalities established in Corollaries 4 and 5 can alternatively be derived from (12) combined with [30] (Theorem 1). Indeed, suppose that for some $p\in (1,\infty )$ and $\kappa \in (0,1)$ we have

$${\parallel A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\parallel }_p={\parallel AX+XB\parallel }_p.$$

Then, Lemma 4 implies

$$AX-XB=0\in {\mathfrak{S}}_p\left(\mathcal{H}\right).$$

Furthermore, because $dist\left(\sigma \right(A),\sigma (B\left)\right)=\delta >0$ due to the disjointness and compactness of the spectra of A and B in $\mathbb{C}$, we obtain

$${\parallel X\parallel }_p\le \alpha {\parallel AX-XB\parallel }_p=0,$$

which contradicts the assumption that $X\ne 0$. A similar argument establishes the strictness of the remaining inequality in (15).

We conclude by characterizing the case in which the maximum and minimum values of the function $f\left(\kappa \right)$ coincide within the trace class ideal.

Theorem 7.

Let $A,B\in \mathcal{L}\left(\mathcal{H}\right)$ be positive and invertible operators, and let $X\in {\mathfrak{S}}_1\left(\mathcal{H}\right)$. If

$$2\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}{\parallel }_1={\parallel AX+XB\parallel }_1$$

then there exists a partial isometry $U\in \mathcal{L}\left(\mathcal{H}\right)$ such that

$$\begin{array}{cc}& U\left(A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\right)=|A^{\frac{1}{2}}X{B}^{\frac{1}{2}}|,U(AX+XB)=|AX+XB|,\hfill \\ & \mathit{and}\mathrm{tr}\left(U\left(2A^{\frac{1}{2}}X{B}^{\frac{1}{2}}-(AX+XB)\right)\right)=0.\hfill \end{array}$$

Proof. 

Mirroring the argument from Theorem 1 and recalling (2), we obtain

$${∥A^{\frac{1}{2}}X{B}^{\frac{1}{2}}+{\displaystyle \frac{AX+XB}{2}}∥}_1={\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\parallel }_1+{∥{\displaystyle \frac{AX+XB}{2}}∥}_1.$$

Applying [31] (Theorem 2.3), we deduce the existence of a partial isometry $U\in \mathcal{L}\left(\mathcal{H}\right)$ such that

$$U\left(A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\right)=\left|A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\right|\mathrm{and}U\left({\displaystyle \frac{AX+XB}{2}}\right)=\left|{\displaystyle \frac{AX+XB}{2}}\right|.$$

Hence,

$$\begin{array}{ccc}\hfill \mathrm{tr}\left(U2A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\right)& =& \mathrm{tr}\left(2\left|A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\right|\right)=2{\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\parallel }_1\hfill \\ & =& {\parallel AX+XB\parallel }_1=\mathrm{tr}\left(\right|AX+XB\left|\right)=tr\left(U(AX+XB)\right).\hfill \end{array}$$

This demonstrates that

$$\mathrm{tr}\left(U2A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\right)=\mathrm{tr}\left(U(AX+XB)\right)<\infty ,$$

and therefore,

$$\mathrm{tr}\left(U\left(2A^{\frac{1}{2}}X{B}^{\frac{1}{2}}-(AX+XB)\right)\right)=0.$$

The converse follows immediately.  □

We verify that Example 1, introduced above, is fully consistent with the characterization established in Theorem 7. Moreover, we emphasize that this result does not contradict the equivalence obtained for the Schatten p-ideals with $1<p<\infty$, where the strict convexity of the norm plays a crucial role. As is well known, the Schatten 1-norm is not strictly convex, which explains the different behavior observed in this case.

Example 2.

As demonstrated in Example 1, we have

$$2\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}{\parallel }_1={\parallel AX+XB\parallel }_1,$$

satisfying the equality in the theorem’s premise. Because both $M_1$ and $M_2$ are positive operators, their polar decompositions reduce to

$$M_1=I|M_1|,M_2=I\left|M_2\right|,$$

Thus, the common partial isometry in both decompositions is $U=I$. In particular,

$$tr\left(U(2M_1-M_2)\right)=tr(2M_1-M_2)=0,$$

showing that all conditions of the theorem are fulfilled. However,

$$2A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\ne AX+XB.$$

This does not contradict Theorem 1 (nor Lemma 3), as the Schatten 1-norm is not strictly convex.

Restricting our attention to the matrix setting—particularly the case where the identity operator belongs to the trace class—Theorem 7 admits a new characterization involving the notion of orthogonality. Recall that for two $n\times n$ matrices A and B, A is said to be orthogonal to B with respect to the trace norm (denoted by $A{\perp }_{BJ}^{1}B$) if and only if

$${\parallel A+\lambda B\parallel }_1\ge {\parallel A\parallel }_1,\mathrm{for}\mathrm{all}\lambda \in \mathbb{C}.$$

For a comprehensive discussion of Birkhoff–James orthogonality, we refer the reader to [32,33].

Theorem 8.

Let $A,B$, and X be $n\times n$ matrices such that A and B are positive and invertible. Then, the following statements are equivalent:

1. 

$2\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}{\parallel }_1={\parallel AX+XB\parallel }_1$.

2. 

There exists a partial isometry $U\in \mathcal{L}\left(\mathcal{H}\right)$ such that

$$U\left(A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\right)=|A^{\frac{1}{2}}X{B}^{\frac{1}{2}}|,U(AX+XB)=|AX+XB|,$$

and

$$\mathrm{tr}\left(U\left(2A^{\frac{1}{2}}X{B}^{\frac{1}{2}}-(AX+XB)\right)\right)=0.$$

3. 

$I{\perp }_{BJ}^1\left(2|A^{\frac{1}{2}}X{B}^{\frac{1}{2}}|-|AX+XB|\right)$.

Proof. 

The equivalence between items (1) and (2) was established in Theorem 7. Suppose condition (2) holds. Then,

$$\mathrm{tr}\left(2\left|A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\right|-|AX+XB|\right)=0.$$

Hence, by [34] (Corollary 3.10), we conclude that

$$I{\perp }_{BJ}^1\left(2\left|A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\right|-|AX+XB|\right).$$

Conversely, if condition (3) holds, reapplying [34] (Corollary 3.10) yields

$$\mathrm{tr}\left(2\left|A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\right|-|AX+XB|\right)=0,$$

or equivalently,

$$2\parallel A^{\frac{1}{2}}X{B}^{\frac{1}{2}}{\parallel }_1=\mathrm{tr}\left(2\left|A^{\frac{1}{2}}X{B}^{\frac{1}{2}}\right|\right)=\mathrm{tr}\left(|AX+XB|\right)={\parallel AX+XB\parallel }_1.$$

□

5. Conclusions

The symmetry of the function

$$f\left(\kappa \right)=\left|\right||A^{\kappa }X{B}^{1-\kappa }+A^{1-\kappa }X{B}^{\kappa }\left|\right||$$

around $\kappa =\frac{1}{2}$ plays a fundamental structural role in the analysis of Heinz-type operator inequalities. Combined with the strict convexity of the underlying symmetrically normed ideal, this symmetry enables a precise description of the equality cases in both the Heinz inequality and several Young-type refinements.

Throughout the paper, we observed that some equality statements hold for general positive operators, whereas others require the additional assumption that A and B are invertible. This distinction is crucial when invoking identities involving the logarithmic mean or expressions of the form $A^{\kappa }X{B}^{1-\kappa }$, as the functional calculus for noninvertible positive operators may fail to preserve the relations necessary to propagate equality from a single parameter value to the entire interval $[0,1]$.

A central conclusion of our analysis is that, provided the assumptions guarantee the validity of the associated functional calculus, equality in any intermediate form of the Heinz inequality necessitates the intertwining relation $AX=XB$. Once established, this relation dictates the behavior of all operator means of the form $A^{r}X{B}^r$ and $A^{s}X{B}^{1-s}$, rendering the function $f\left(\kappa \right)$ constant on $[0,1]$ if equality is attained at even a single point.

For the Schatten p-ideals ($1<p<\infty$), these conclusions integrate naturally with Kittaneh’s characterizations, offering alternative formulations for the equality case in the arithmetic–geometric mean inequality. Furthermore, when the spectra of A and B are disjoint, Rosenblum’s theorem guarantees strict inequalities, highlighting the rigidity of the equality scenario.

Finally, in the trace-class setting, the trace functional provides additional geometric insight. We identified the structural conditions under which the extremal values of $f\left(\kappa \right)$ coincide within this ideal, thereby completing the description of the equality cases in the trace-class framework.

In summary, this work reveals a unified picture in which symmetry, strict convexity, and the Sylvester equation govern equality phenomena across a broad class of operator means. Possible directions for future research include extending these ideas to other symmetric means, exploring settings where invertibility cannot be assumed, and investigating nonlinear variants of the Sylvester equation.