New proofs on two recent inequalities for unitarily invariant norms

In this short note, we provide alternative proofs for several recent results due to Audenaert (Oper. Matrices 9:475–479, 2015) and Zou (J. Math. Inequal. 10:1119–1122, 2016; Linear Algebra Appl. 552:154–162, 2019).


Introduction
Let M n be the set of n × n complex matrices. For A ∈ M n , the singular values and eigenvalues of A are denoted by σ i (A) and λ i (A), respectively, i = 1, . . . , n. The singular values σ 1 (A), σ 2 (A), . . . , σ n (A) of a matrix A are the eigenvalues of |A| = (A * A) 1 2 arranged in decreasing order and repeated according to multiplicity. The Ky Fan k-norm, a particular unitarily invariant norm, is defined as · (k) = k j=1 σ j (A), 1 ≤ k ≤ n. If A is Hermitian, then all eigenvalues of A are real and ordered as λ 1 (A) ≥ · · · ≥ λ n (A).
Let A, B ∈ M n . Bhatia and Kittaneh [8] proved an arithmetic-geometric mean inequality for unitarily invariant norms As a generalization of (1), Bhatia and Davis [7] proved that for A, X, B ∈ M n . Albadawi [3] obtained a stronger version of the Hölder inequality for unitarily invariant norms. Let A, X, B ∈ M n and 1 p + 1 q = 1, p, q > 1, r ≥ 0. Then which is a generalization of Horn and Zhan's result [10] (also called the Hölder inequality) Recently, Audenaert [5] proved that if A, B ∈ M n and 1 p + 1 q = 1, p, q > 1, r ≥ 0, α ∈ [0, 1], then which is a unification of inequalities (1) and (4). By setting r = 1 and p = p = 2 in (5) we have Lin [12] gave a new proof of inequality (6). Zou and Jiang [16] generalized it to the following inequality: Let A, B, X ∈ M n and q ∈ [0, 1]. Then Al-khlyleh and Kittaneh [2, Theorem 2.5] presented an inequality that refines inequality (7) for the particular unitarily invariant norm, Hilbert-Schmidt norm. For more results on interpolation between the arithmetic-geometric mean inequality and the Cauchy-Schwarz inequality for matrices, see [1].
In this paper, we provide alternative proofs of inequalities (5) and (7), which provide new perspectives to the elegant results.

Main results
For presenting the new proofs, we need the following several lemmas.

Lemma 2.2 (see [6, p. 41]) Let A, B ∈ M n and suppose that f is convex and increasing on
Lemma 2.5 (see [4] and [13, p. 228]) Let A, B ∈ M n be positive semidefinite and 0 ≤ q ≤ 1. Then Audenaert [5] proved the following theorem. We give a different proof of the result.

Theorem 2.6 Let A, B ∈ M n and
Proof By Fan's dominance theorem (see [11,Theorem 1.4
First, let us show this inequality for the Ky Fan 1-norm, that is, the spectral norm: which means that Second, using a standard argument via the antisymmetric product (see [5, p. 18]), (13) yields . . , n. Since weak log-majorization implies weak majorization (see, [9, p. 174]), by (10) we have The left-hand side is |A * B| r (k) . By the Hölder inequality the right-hand side is bounded from above by Thus (12) holds, and so does the conclusion. This completes the proof.
In fact, by a similar technique used in the theorem, we may present a new proof of the following result due to Zou [10], which is a unified version of inequalities (2) and (3).
Proof There is a subtle difference between the proof of (14) and that of the previous theorem although most techniques are similar. For the readers' convenience, we present the proof simply. By Fan's dominance theorem (14) is equivalent to for all k = 1, . . . , n. If X is a positive semidefinite matrix, then for Ky Fan 1-norm, we have which means that Using a standard argument via the antisymmetric product (see [5, p. 18 for k = 1, . . . , n. The left-hand side is |A * XB| 2r (k) . By the Hölder inequality the right-hand side is bounded from above by by (8) and Lemma 2.2 by Lemmas 2.2 and 2.3 (16) Next, we consider the case where X is any matrix. By the singular value decomposition we know that there exist unitary matrices U and V such that X = UDV * , and then by (16) we have where the last equality is due to the fact that |U * 1 PU 2 | r = |P| r for any P ∈ M n and unitary matrices U 1 , U 2 . This completes the proof.
Finally, we give an alterative proof of (7) due to Zou Proof First, consider the special case where A, B, X are Hermitian and A = B. Then Similarly, Thus by (17) and (18) which is just the desired inequality in this particular case. Next, consider the more general situation where A and B are Hermitian and X is any matrix. Let Then by the particular case considered before Multiplying out the block-matrices, we have Hence we obtain the following inequality from (19): which means that So by (17) we have The following inequality can be proved in exactly the same way: In this case, from (20) and (21) we have AXB * 2 ≤ qA * AX + (1q)XB * B × (1q)A * AX + qXB * B .
Finally, Let A = UA 1 and B = VB 1 be polar decompositions of A and B. Then So the theorem follows from inequality (22). This completes the proof.