About the Cauchy–Bunyakovsky–Schwarz Inequality for Hilbert Space Operators

: The symmetric shape of some inequalities between two sequences of real numbers generates inequalities of the same shape in operator theory. In this paper, we study a new reﬁnement of the Cauchy–Bunyakovsky–Schwarz inequality for Euclidean spaces and several inequalities for two bounded linear operators on a Hilbert space, where we mention Bohr’s inequality and Bergström’s inequality for operators. We present an inequality of the Cauchy–Bunyakovsky–Schwarz type for bounded linear operators, by the technique of the monotony of a sequence. We also prove a reﬁnement of the Aczél inequality for bounded linear operators on a Hilbert space. Finally, we present several applications of some identities for Hermitian operators.


Introduction
Let (H, ·, · ) be a complex Tv, v ≥ 0 for every v ∈ H is equivalent to the following condition: T is self-adjoint, and σ(T) ⊂ [0, ∞), where σ(T) = {λ : T − λI is not invertible}; and I is the identity operator [1]. If T ≥ 0, then there exists a unique T 0 ≥ 0 such that T = T 2 0 . The absolute value or modulus of the operator T ∈ B(H) is given by |T| = (T * T) 1/2 , so |T| 2 = T * T. It is easy to see that |T| is always positive and |T| = 0 if only if T = 0. We write T 1 ≥ T 2 if T 1 and T 2 are self-adjoint operators and if T 1 − T 2 ≥ 0. In [2], we found several inequalities for absolute value operators.
Many results in the theory of inequalities, probability and statistics, Hilbert spaces theory, and numerical and complex analysis are given by using the Cauchy-Bunyakovsky-Schwarz inequality (the C-B-S inequality).
The C-B-S inequality is defined as follows: let (a 1 , a 2 , ..., a n ) and (b 1 , b 2 , ..., b n ) be two sequences of real numbers, then: with equality if and only if sequences (a 1 , a 2 , ..., a n ) and (b 1 , b 2 , ..., b n ) are proportional. In [7], for arbitrary complex sequences a = (a 1 , a 2 , ..., a n ) and b = (b 1 , b 2 , ..., b n ), we have: with equality if and only if sequences a and b are proportional. We remark that the symmetric shape of some inequalities for real numbers indicates ideas for extending these inequalities in operator theory.
In Inequality (2), for n = 2, p, q > 1, 1/p + 1/q = 1, a 1 = √ pa, a 2 = √ pb, b 1 = , we obtain the classical Bohr inequality [9], given by the following: where a, b ∈ C, with equality if and only if (p − 1)a = b. In [10], Hirzallah established an extention of Bohr's inequality to B(H); thus: with T 1 , T 2 ∈ B(H) and q ≥ p > 1 with 1/p + 1/q = 1. Zhang, in [11], studied the operator inequalities of the Bohr type. An important consequence of the C-B-S inequality is Aczél's inequality. Several methods in the theory of functional equations in one variable were studied in [12] by Aczél and showed the following inequality: let A and B be two positive real numbers, and let a = (a 1 , a 2 , ..., a n ) and b = (b 1 , b 2 , ..., b n ) be two sequences of positive real numbers such that: Then: Equality holds if and only if the sequences a and b are proportional. This inequality has many applications in non-Euclidean geometry, in the theory of functional equations, and in operators theory (see [13][14][15][16]).
Popoviciu [17] presented a generalized form of the inequality of Aczél, as follows: let A and B be two positive real numbers, and let p, q > 1 be such that 1 p + 1 q = 1 and a = (a 1 , a 2 , ..., a n ), b = (b 1 , b 2 , ..., b n ) be two sequences of positive real numbers such that: Then: Equality holds if and only if sequences a and b are proportional.
In the special case p = q = 2, we deduce the classical Aczél inequality. In [18], we found an approach of some bounds for several statistical indicators with the Aczél inequality, and in [19], we found a proof of the Aczél inequality given with tools of the Lorentz-Finsler geometry.
Motivated by the above results, in Section 2, we study a new refinement of the C-B-S inequality for the Euclidean space and several inequalities for two bounded linear operators on the Hilbert space H, where we mention Bohr's inequality and Bergström's inequality for operators. We also show an inequality of the Cauchy-Bunyakovsky-Schwarz type for bounded linear operators, by the technique of the monotony of a sequence. Finally, we prove a refinement of the Aczél inequality for bounded linear operators on the Hilbert space H. In Section 3, we present some identities for real numbers obtained from some identities for Hermitian operators.
This work is important because it extends a series of inequalities for real numbers to inequalities that are true for different classes of operators. This development is not easy in most cases. We also obtain new inequalities between operators, which by choosing a particular case, can generate new inequalities for real numbers and for matrices.

Results on the Cauchy-Bunyakovsky-Schwarz inequality and on the Aczél Inequality for Operators
The symmetric shape of some inequalities between two sequences of real numbers suggests inequalities of the same shape in operator theory.
Proof. We use the technique of the monotony of a sequence given in [20]. This technique is given for real numbers, but we study its application in broader contexts. Therefore, we consider the sequence: To study the monotony of the sequence S k , k ≤ n, we evaluate the difference of two consecutive terms of the sequence. Therefore, we have: For two vectors x 1 , x 2 ∈ H and for real numbers a 1 , a 2 = 0, with a 1 + a 2 = 0, the following equality holds: If a 1 a 2 (a 1 + a 2 ) > 0 in the above equality, then we have: Using the inequality from Relation (8), we have: This means that S k+1 − S k ≥ 0, so sequence S k is increasing. Therefore, we obtain S n ≥ S n−1 ≥ ... ≥ S 2 ≥ S 1 = 0. However, and taking into account that we can rearrange the terms of the sequence S n , we have the inequality: Multiplying the above inequality by we deduce the inequality of the statement.
for any vectors x 1 , x 2 , ..., x n in a Euclidean space H and for every n-tuple of real numbers λ 1 , λ 2 , ..., λ n . This inequality is an inequality of the Cauchy-Bunyakovsky-Schwarz type for Euclidean spaces.
Next, we study the problem of the existence of such relations, as above, for the bounded linear operators on a Hilbert space. Next, we present several results related to the bounded linear operators on a Hilbert space. Lemma 1. Let T 1 , T 2 ∈ B(H). Then, for real numbers a, b, the following identity holds: Proof. In the left part of the identity, we have: Therefore, the statement is true. (12), if we replace b by −b, we deduce the equality:

Remark 3. In Relation
for all T 1 , T 2 ∈ B(H) and for every a, b ∈ R. Replacing T 2 by −T 2 in Relation (13), we obtain: for all T 1 , T 2 ∈ B(H) and for every a, b ∈ R. If in Equality (14), we choose a = 1 + t and b = 1 + 1 t , where t ∈ R * = R − {0}, we deduce an identity given by Fuji and Zuo [21]: for all T 1 , T 2 ∈ B(H) and for every t ∈ R * . Proposition 1. Let T 1 , T 2 ∈ B(H). We assume that p, q > 1, with 1 p + 1 q = 1, then the following identity holds: Proof. For a = 1 q and b = 1 p in Relation (13), we obtain the statement.
Using Equality (16), we obtain a reverse and an improvement of Bohr's inequality for operators given in [23,24]; thus, for p ≥ 2, we have: and for 1 < p ≤ 2, we deduce: Replacing in Relation (14) b by 1 − a, we find an identity from [23], given by: for all T 1 , T 2 ∈ B(H) and for every a ∈ R.
Then, the following inequality holds: Equality holds when a = 1 2 Proof. Since we have the inequality: for any a ∈ [0, 1], using Identity (20), the inequality of the statement follows. Because, if we obtain the equality of the statement. For T 1 = λT 2 , we deduce: so we need to have λ = 1.

Corollary 2.
Let T 1 , T 2 ∈ B(H) and a, b > 0. Then, we have: Proof. Because a, b > 0 implies 1 > a a + b > 0, therefore, replacing a by a a + b in Inequality (22), we obtain the inequality of the statement.
Next, we obtain another improvement of Bohr's inequality.
Proof. We replace in Relation (22) T 1 by pT 1 , T 2 by pT 2 , and a by 1 p . Thus, by simple calculations, we obtain the inequality of the statement.
Let T 1 , T 2 ∈ B ++ (H); we have the following operators: called the geometric mean of operators T 1 and T 2 (see [27]).
Proof. Since we replace in Relation (22) T 1 by T 1 ∇ λ T 2 and T 2 by T 1 λ T 2 , the inequality of the statement follows.
Proof. Since the term |p 2 T 1 − p 1 T 2 | 2 p 1 p 2 (p 1 + p 2 ) is positive and uses Inequality (28), we obtain the inequality of the statement.
This inequality can be viewed as Bergström's inequality for operators, the classical Bergström inequality can be found in [8].
Proof. We consider sequence F n , n ≥ 1, given by: We study the monotony of sequence F k , k ≤ n. Therefore, we have the difference between two consecutive terms of the sequence: Using Bergström's inequality for operators, for two terms, we have: This means that F k+1 − F k ≥ 0, so sequence F k is increasing. Therefore, we obtain F n ≥ F n−1 ≥ ... ≥ F 2 ≥ F 1 = 0. However, and taking into account that we can rearrange the terms of the sequences, we have the inequality: for all i, j ∈ {1, ..., n}, n ≥ 2. Multiplying the above inequality by n ∑ i=1 λ 2 i > 0, we deduce the inequality of the statement.

Remark 5. This inequality represents an improvement of the C-B-S inequality for operators:
for n ≥ 1 and for any operators T 1 , T 2 , ..., T n in B(H) and for arbitrary real numbers λ 1 , λ 2 , ..., λ n .

Theorem 3.
For any operators T 1 , T 2 , ..., T n in B(H) and for complex numbers b 1 , b 2 , ..., b n ∈ C, such that there is at least one b i = 0, we have: Proof. For ∑ n i=1 |b i | 2 = 0, we have the following: (33), then we deduce the relation: which proves the equality of the statement.

Lemma 2.
Let T 1 , T 2 , T 3 in B(H). Then, the equality holds: Proof. Using the properties of the modulus operator, we have the following calculations: Consequently, the proof is complete. Proposition 6. Let T 1 , T 2 , T 3 ∈ B(H). Then, for real numbers b 1 , b 2 , b 3 = 0, the following equality holds: Proof. From Equality (28), we obtain the following three equalities: Adding the above relations, we deduce: Therefore, using Lemma 17, we obtain the equality of the statement.
The identity from Proposition 18 suggests a general result for n operators, namely: Theorem 4. For any operators T 1 , T 2 , ..., T n in B(H) and for real numbers b 1 , b 2 , ..., b n ∈ R * , n ≥ 2, we have: Proof. We use the mathematical induction to prove the relation of the statement. We consider the following proposition: For n = 2, the proposition is true, taking into account the relation from Proposition 12. Assume that P(n) is true; we will prove that P(n + 1) is true.
Therefore, from the principle of mathematical induction, we deduce the statement.

Remark 7.
If T i ∈ B(H) and r i > 1, i ∈ {1, ..., n}, with ∑ n i=1 1 r i = 1, then for b i = 1 r i in the equality from Theorem 19, we obtain: the identity given by Fuji and Zuo in [21].
Now, we will focus on a general result related to Aczél's inequality for operators, namely: for all j ∈ {1, ..., n}.

Proof.
We use the technique of monotony to sequence A n , which is defined as follows: For k ≤ n, we have: Using Bergström's inequality for operators, for two terms, we have: This means that A k+1 − A k ≥ 0, so sequence A k is increasing. Therefore, we obtain A n ≥ A n−1 ≥ ... ≥ A 2 ≥ A 1 . However, and taking into account that we can rearrange the terms of the two sequences, we have the inequality: we deduce the inequality of the statement.
Remark 8. Inequality (42) gives an inequality of the Aczél type for operators; thus:

Applications of some Identities of Hermitian Operators
If we choose various particular cases for different classes of operators, then we deduce a series of known identities. Therefore, we have the following: (1) If we take T i = a i I, where I is the identity operator and a i ∈ R, then using Relation (40), we find: which means that: However, when qI = 0, q ∈ R implies q = 0, because 0 = 0, x = qIx, x = q x, x = q x 2 , for every x ∈ H, so q = 0. Therefore, we obtain the Lagrange identity: (2) If we take T i = T − a i I in Relation (40), where T is the Hermitian operator, T = 0, and a i ∈ R, then we deduce: For all i ∈ {1, 2, ..., n}, we have: From the above relations, we deduce that: and it follows that: because from qT = 0, for a nonzero Hermitian operator T and q ∈ R, we deduce q = 0. In the same way, if in Relation (40), we take T i = c i T − a i I, we obtain the following identity: In Relation (47) for b i = 1, for all i ∈ {1, 2, ..., n}, we have: This proves Chebyshev's inequality [7]; thus: (a) If a 1 ≥ a 2 ≥ ... ≥ a n and c 1 ≥ c 2 ≥ ... ≥ c n , then we deduce: (b) If a 1 ≥ a 2 ≥ ... ≥ a n and c 1 ≤ c 2 ≤ ... ≤ c n , then we have: In Equality (47), if b i > 0, for all i ∈ {1, 2, ..., n}, then we replace b i by √ b i , c i by √ b i c i , and a i by √ b i a i , and then, we have: In Equality (47), if α i , γ i > 0, for all i ∈ {1, 2, ..., n}, then we take and then, we have: which is in fact the Cauchy-Binet formula [7]. In this paper, some inequalities that characterize the bounds of the variance of a random variable in the discrete case can be identified, where the variance is an important statistical indicator that measures the degree of data dispersion. In this sense, Inequalities (10) and (11) can be seen.
Funding: This research received no external funding.
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