slug: hayajneh-2018-unitarily-invariant-norm
bibkey: hayajneh2018uinorm
doi: 10.7153/mia-2018-21-81
url: https://doi.org/10.7153/mia-2018-21-81
triage: theorem
motivation_gids:
  - D5/S3/Quantum/Matrix/HayajnehConjectureRefutation.result

Hayajneh–Hayajneh–Kittaneh: mixed-power monotonicity for unitarily invariant norms

Problem

M. Hayajneh, S. Hayajneh and F. Kittaneh, On some classical trace inequalities and a new Hilbert–Schmidt norm inequality, Mathematical Inequalities & Applications 21(4) (2018), 1175–1183, DOI 10.7153/mia-2018-21-81, §1, Conjecture 1, printed p. 1177. See hayajneh2018uinorm.

CONJECTURE 1. Let A be a positive semidefinite matrix and B be a Hermitian matrix. Then for 1/2 ⩽ s ⩽ t ⩽ 1 and for every unitarily invariant norm, we have |||A^s B + BA^{1−s}||| ⩽ |||A^t B + BA^{1−t}|||.

The formal statement retains every quantifier: every natural dimension $n$, all $A,B\in M_n(\mathbb{C})$ with $A$ positive semidefinite and $B$ Hermitian, all real $1/2\le s\le t\le 1$, and every unitarily invariant norm on $M_n(\mathbb{C})$. Real powers are interpreted by continuous functional calculus. No lower spectral bound $A\ge I$ is part of the conjecture.

Motivation

The source proves this monotonicity for the Hilbert–Schmidt norm and proves the endpoint $s=1/2,t=1$ for every unitarily invariant norm. Conjecture 1 asks whether these special cases extend to every allowed parameter pair and every such norm.

Gap

Hilbert–Schmidt control does not force the sum of singular values to increase. The trace norm detects a strict decrease even for positive definite $A$ and $B$ in dimension two.

Route

Set

$$A=\left(\begin{array}{cc}1/81& 0\\ 0& 1\end{array}\right),B=\left(\begin{array}{cc}1& 3/4\\ 3/4& 1\end{array}\right),s=\frac{1}{2},t=\frac{3}{4}.$$

Their eigenvalues are $1/81,1$ and $1/4,7/4$, so the required positivity and Hermitian conditions hold. Diagonal functional calculus yields powers $\mathrm{diag}(1/9,1)$, $\mathrm{diag}(1/3,1)$ and $\mathrm{diag}(1/27,1)$ at exponents $1/2,1/4,3/4$. Therefore

$$L=A^{1/2}B+B{A}^{1/2}=\left(\begin{array}{cc}2/9& 5/6\\ 5/6& 2\end{array}\right),R=A^{3/4}B+B{A}^{1/4}=\left(\begin{array}{cc}10/27& 7/9\\ 1& 2\end{array}\right).$$

For every complex two by two matrix $T$,

$$\Vert T{\Vert }_1^2=\mathrm{tr}(T^{\ast }T)+2|\det T|.$$

Using $\det L=-1/4$ and $\det R=-1/27$ gives

$$\Vert L{\Vert }_1^2=\frac{481}{81},\Vert R{\Vert }_1^2=\frac{4240}{729},\Vert L{\Vert }_1^2-\Vert R{\Vert }_1^2=\frac{89}{729}>0.$$

Falsifier

The trace norm is a unitarily invariant norm and both norms are nonnegative. Hence $\Vert L{\Vert }_1>\Vert R{\Vert }_1$, contradicting the asserted inequality at $n=2$, $s=1/2$, $t=3/4$. No approximation is required for this strict comparison.

Evidence

D5/S3/Quantum/Matrix/HayajnehConjectureRefutation.result : ¬ conjecture targets the complete universally quantified conjecture. D5/S3/Quantum/Schatten/TraceNorm supplies the trace-norm instance of the unitarily invariant norm interface and the two-dimensional identity used in the comparison. The singular-value sum is the full prefix in dimension two, so the same counterexample refutes the corresponding weak-majorization claim.

Triage

PROVED — Refutation of Conjecture 1. The displayed positive definite pair reverses the trace-norm inequality at $s=1/2,t=3/4$ by an exact difference of squares $89/729$.

PROVED — The Hilbert–Schmidt case remains valid. This case is proved in the source, §2; the trace-norm counterexample does not contradict it.

PROVED — The endpoint $s=1/2,t=1$ remains valid for every unitarily invariant norm. This case is proved in the source, p. 1179; the counterexample has $t=3/4$.

COMPUTED — Schatten $p\in \{1.5,2,3,\infty \}$ do not fail for this pair. Numerical values $(\Vert L{\Vert }_p,\Vert R{\Vert }_p)$ are approximately $(2.34487039,2.39704819)$, $(2.33201021,2.39627054)$, $(2.32961548,2.39622090)$ and $(2.32953957,2.39622069)$, respectively. These are finite numerical checks, not universal proofs.

OPEN — Schatten $p\ge 2$ on arbitrary admissible inputs. The displayed finite computation does not decide this restricted universal inequality.

OPEN — The operator norm on arbitrary admissible inputs. The operator-norm comparison for this pair does not establish the universal statement.

PROVED — Weak majorization fails at $k=2$. The sum of both singular values is the trace norm, and the exact reverse inequality violates the full-prefix comparison.

ASSUMED-UNVERIFIED

The literature check covered the DOI and title with conjecture/counterexample/proof; “Hayajneh” “Conjecture 1”; and A^s B + BA^{1−s} with unitarily invariant norm. Related work checked: S. Hayajneh, Filomat 38(19) (2024), 6891–6896, on a related scalar trace inequality; and Bottazzi–Elencwajg–Larotonda–Varela, arXiv:1403.7472, on A^tB^{1−t} + B^tA^{1−t}. No settlement of Conjecture 1 was found within that scope. Citation indexes were not exhaustively checked. Absence of subsequent settlements and worldwide priority are not established. The universal Schatten $p\ge 2$ and operator-norm restrictions are not settled by this result.