Introduction
Cauchy-Schwarz angle-length inequality, derived in 1821, is the following.
Theorem 1.1. [1 , 2 , 5 , 8 , 9 , 12 , 13 ] (Cauchy-Schwarz Angle-Length Inequality ) Let n ∈ N n \in\mathbb{N} n ∈ N
∣ ∑ j = 1 n a j b j ∣ ≤ ( ∑ j = 1 n ∣ a j ∣ 2 ) 1 2 ( ∑ k = 1 n ∣ b k ∣ 2 ) 1 2 , ∀ ( a j ) j = 1 n , ( b j ) j = 1 n ∈ C n . \left|\sum_{j=1}^{n}a_jb_j\right|\leq\left(\sum_{j=1}^{n}|a_j|^2\right)^{\frac{1}{2}}\left(\sum_{k=1}^{n}|b_k|^2\right)^{\frac{1}{2}},\quad\forall(a_j)_{j=1}^{n},(b_j)_{j=1}^{n}\in\mathbb{C}^{n}. j = 1 ∑ n a j b j ≤ ( j = 1 ∑ n ∣ a j ∣ 2 ) 2 1 ( k = 1 ∑ n ∣ b k ∣ 2 ) 2 1 , ∀ ( a j ) j = 1 n , ( b j ) j = 1 n ∈ C n . We clearly have the following sequential version of Theorem 1.1.
Theorem 1.2. [1 , 2 , 5 , 8 , 9 , 12 , 13 ] (Cauchy-Schwarz Angle-Length Inequality ) Let { a j } j = 1 ∞ , { b j } j = 1 ∞ ⊆ C \{a_j\}_{j=1}^{\infty},\{b_j\}_{j=1}^{\infty}\subseteq\mathbb{C} { a j } j = 1 ∞ , { b j } j = 1 ∞ ⊆ C
∑ j = 1 ∞ ∣ a j ∣ 2 ∈ R , ∑ j = 1 ∞ ∣ b j ∣ 2 ∈ R . \sum_{j=1}^{\infty}|a_j|^2\in\mathbb{R},\qquad\sum_{j=1}^{\infty}|b_j|^2\in\mathbb{R}. j = 1 ∑ ∞ ∣ a j ∣ 2 ∈ R , j = 1 ∑ ∞ ∣ b j ∣ 2 ∈ R . Then
∣ ∑ j = 1 ∞ a j b j ∣ ≤ ( ∑ j = 1 ∞ ∣ a j ∣ 2 ) 1 2 ( ∑ k = 1 ∞ ∣ b k ∣ 2 ) 1 2 . \left|\sum_{j=1}^{\infty}a_jb_j\right|\leq\left(\sum_{j=1}^{\infty}|a_j|^2\right)^{\frac{1}{2}}\left(\sum_{k=1}^{\infty}|b_k|^2\right)^{\frac{1}{2}}. j = 1 ∑ ∞ a j b j ≤ ( j = 1 ∑ ∞ ∣ a j ∣ 2 ) 2 1 ( k = 1 ∑ ∞ ∣ b k ∣ 2 ) 2 1 . In 1953, Kaplansky extended Cauchy-Schwarz inequality to commutative C*-algebras [6 ] .
Theorem 1.3. [6 ] (Cauchy-Schwarz-Kaplansky Inequality ) Let A \mathcal{A} A n ∈ N n\in\mathbb{N} n ∈ N
[ ( ∑ j = 1 n a j ∗ b j ) ( ∑ k = 1 n a k ∗ b k ) ∗ ] 1 2 ≤ ( ∑ j = 1 n a j ∗ a j ) 1 2 ( ∑ k = 1 n b k ∗ b k ) 1 2 , ∀ ( a j ) j = 1 n , ( b j ) j = 1 n ∈ A n . \left[\left(\sum_{j=1}^{n}a_j^{*}b_j\right)\left(\sum_{k=1}^{n}a_k^{*}b_k\right)^{*}\right]^{\frac{1}{2}}\leq\left(\sum_{j=1}^{n}a_j^{*}a_j\right)^{\frac{1}{2}}\left(\sum_{k=1}^{n}b_k^{*}b_k\right)^{\frac{1}{2}},\quad\forall(a_j)_{j=1}^{n},(b_j)_{j=1}^{n}\in\mathcal{A}^{n}. [ ( j = 1 ∑ n a j ∗ b j ) ( k = 1 ∑ n a k ∗ b k ) ∗ ] 2 1 ≤ ( j = 1 ∑ n a j ∗ a j ) 2 1 ( k = 1 ∑ n b k ∗ b k ) 2 1 , ∀ ( a j ) j = 1 n , ( b j ) j = 1 n ∈ A n . In particular,
∥ ∑ j = 1 n a j ∗ b j ∥ ≤ ∥ ∑ j = 1 n a j ∗ a j ∥ 1 2 ∥ ∑ k = 1 n b k ∗ b k ∥ 1 2 . \left\|\sum_{j=1}^{n}a_j^{*}b_j\right\|\leq\left\|\sum_{j=1}^{n}a_j^{*}a_j\right\|^{\frac{1}{2}}\left\|\sum_{k=1}^{n}b_k^{*}b_k\right\|^{\frac{1}{2}}. j = 1 ∑ n a j ∗ b j ≤ j = 1 ∑ n a j ∗ a j 2 1 k = 1 ∑ n b k ∗ b k 2 1 . Theorem 1.4. [6 ] (Cauchy-Schwarz-Kaplansky Inequality ) Let A \mathcal{A} A C ∗ C^* C ∗ { a j } j = 1 ∞ , { b j } j = 1 ∞ ⊆ A \{a_j\}_{j=1}^{\infty},\{b_j\}_{j=1}^{\infty}\subseteq\mathcal{A} { a j } j = 1 ∞ , { b j } j = 1 ∞ ⊆ A
∑ j = 1 ∞ a j ∗ a j ∈ A , ∑ j = 1 ∞ b j ∗ b j ∈ A . \sum_{j=1}^{\infty}a_j^*a_j\in\mathcal{A},\qquad\sum_{j=1}^{\infty}b_j^*b_j\in\mathcal{A}. j = 1 ∑ ∞ a j ∗ a j ∈ A , j = 1 ∑ ∞ b j ∗ b j ∈ A . Then
[ ( ∑ j = 1 ∞ a j ∗ b j ) ( ∑ k = 1 ∞ a k ∗ b k ) ∗ ] 1 2 ≤ ( ∑ j = 1 ∞ a j ∗ a j ) 1 2 ( ∑ k = 1 ∞ b k ∗ b k ) 1 2 . \left[\left(\sum_{j=1}^{\infty}a_j^*b_j\right)\left(\sum_{k=1}^{\infty}a_k^*b_k\right)^*\right]^{\frac{1}{2}}\leq\left(\sum_{j=1}^{\infty}a_j^*a_j\right)^{\frac{1}{2}}\left(\sum_{k=1}^{\infty}b_k^*b_k\right)^{\frac{1}{2}}. [ ( j = 1 ∑ ∞ a j ∗ b j ) ( k = 1 ∑ ∞ a k ∗ b k ) ∗ ] 2 1 ≤ ( j = 1 ∑ ∞ a j ∗ a j ) 2 1 ( k = 1 ∑ ∞ b k ∗ b k ) 2 1 . In particular,
∥ ∑ j = 1 ∞ a j ∗ b j ∥ ≤ ∥ ∑ j = 1 ∞ a j ∗ a j ∥ 1 2 ∥ ∑ k = 1 ∞ b k ∗ b k ∥ 1 2 . \left\|\sum_{j=1}^{\infty}a_j^*b_j\right\|\leq\left\|\sum_{j=1}^{\infty}a_j^*a_j\right\|^{\frac{1}{2}}\left\|\sum_{k=1}^{\infty}b_k^*b_k\right\|^{\frac{1}{2}}. j = 1 ∑ ∞ a j ∗ b j ≤ j = 1 ∑ ∞ a j ∗ a j 2 1 k = 1 ∑ ∞ b k ∗ b k 2 1 . In this note, we improve Cauchy-Schwarz-Kaplansky inequality.
Improved Cauchy-Schwarz-Kaplansky Inequality
Following result is improvement of Theorem 1.1.
Theorem 2.1. (Improved Cauchy-Schwarz-Kaplansky Inequality ) Let A \mathcal{A} A C ∗ C^* C ∗ n ∈ N n\in\mathbb{N} n ∈ N
( ∑ j = 1 n a j ∗ b j ) ( ∑ k = 1 n a k ∗ b k ) ∗ = ( ∑ j = 1 n a j ∗ a j ) ( ∑ k = 1 n b k ∗ b k ) − ∑ 1 ≤ j < k ≤ n ( a j b k − a k b j ) ∗ ( a j b k − a k b j ) \left(\sum_{j=1}^{n}a_j^*b_j\right)\left(\sum_{k=1}^{n}a_k^*b_k\right)^*=\left(\sum_{j=1}^{n}a_j^*a_j\right)\left(\sum_{k=1}^{n}b_k^*b_k\right)-\sum_{1\leq j<k\leq n}(a_jb_k-a_kb_j)^*(a_jb_k-a_kb_j) ( j = 1 ∑ n a j ∗ b j ) ( k = 1 ∑ n a k ∗ b k ) ∗ = ( j = 1 ∑ n a j ∗ a j ) ( k = 1 ∑ n b k ∗ b k ) − 1 ≤ j < k ≤ n ∑ ( a j b k − a k b j ) ∗ ( a j b k − a k b j ) ≤ ( ∑ j = 1 n a j ∗ a j ) ( ∑ k = 1 n b k ∗ b k ) , ∀ ( a j ) j = 1 n , ( b j ) j = 1 n ∈ A n . \leq\left(\sum_{j=1}^{n}a_j^*a_j\right)\left(\sum_{k=1}^{n}b_k^*b_k\right),\qquad\forall(a_j)_{j=1}^{n},(b_j)_{j=1}^{n}\in\mathcal{A}^{n}. ≤ ( j = 1 ∑ n a j ∗ a j ) ( k = 1 ∑ n b k ∗ b k ) , ∀ ( a j ) j = 1 n , ( b j ) j = 1 n ∈ A n . In particular,
[ ( ∑ j = 1 n a j ∗ b j ) ( ∑ k = 1 n a k ∗ b k ) ∗ ] 1 2 = [ ( ∑ j = 1 n a j ∗ a j ) ( ∑ k = 1 n b k ∗ b k ) − ∑ 1 ≤ j < k ≤ n ( a j b k − a k b j ) ∗ ( a j b k − a k b j ) ] 1 2 \left[\left(\sum_{j=1}^{n}a_j^*b_j\right)\left(\sum_{k=1}^{n}a_k^*b_k\right)^*\right]^{\frac{1}{2}}=\left[\left(\sum_{j=1}^{n}a_j^*a_j\right)\left(\sum_{k=1}^{n}b_k^*b_k\right)-\sum_{1\leq j<k\leq n}(a_jb_k-a_kb_j)^*(a_jb_k-a_kb_j)\right]^{\frac{1}{2}} [ ( j = 1 ∑ n a j ∗ b j ) ( k = 1 ∑ n a k ∗ b k ) ∗ ] 2 1 = ( j = 1 ∑ n a j ∗ a j ) ( k = 1 ∑ n b k ∗ b k ) − 1 ≤ j < k ≤ n ∑ ( a j b k − a k b j ) ∗ ( a j b k − a k b j ) 2 1 ≤ [ ( ∑ j = 1 n a j ∗ a j ) ( ∑ k = 1 n b k ∗ b k ) ] 1 2 \leq\left[\left(\sum_{j=1}^{n}a_j^*a_j\right)\left(\sum_{k=1}^{n}b_k^*b_k\right)\right]^{\frac{1}{2}} ≤ [ ( j = 1 ∑ n a j ∗ a j ) ( k = 1 ∑ n b k ∗ b k ) ] 2 1 = ( ∑ j = 1 n a j ∗ a j ) 1 2 ( ∑ k = 1 n b k ∗ b k ) 1 2 , ∀ ( a j ) j = 1 n , ( b j ) j = 1 n ∈ A n . =\left(\sum_{j=1}^{n}a_j^*a_j\right)^{\frac{1}{2}}\left(\sum_{k=1}^{n}b_k^*b_k\right)^{\frac{1}{2}},\qquad\forall(a_j)_{j=1}^{n},(b_j)_{j=1}^{n}\in\mathcal{A}^{n}. = ( j = 1 ∑ n a j ∗ a j ) 2 1 ( k = 1 ∑ n b k ∗ b k ) 2 1 , ∀ ( a j ) j = 1 n , ( b j ) j = 1 n ∈ A n . Proof. Our proof is motivated from the arguments for the scalar case by Cauchy using Lagrange identity [3 , 4 , 7 , 8 , 9 , 12 , 14 ] . From Lagrange identity, we have,
∑ 1 ≤ j < k ≤ n ( a j b k − a k b j ) ∗ ( a j b k − a k b j ) = 1 2 ∑ j = 1 n ∑ k = 1 n ( a j b k − a k b j ) ∗ ( a j b k − a k b j ) \sum_{1\leq j<k\leq n}(a_jb_k-a_kb_j)^*(a_jb_k-a_kb_j)=\frac{1}{2}\sum_{j=1}^{n}\sum_{k=1}^{n}(a_jb_k-a_kb_j)^*(a_jb_k-a_kb_j) 1 ≤ j < k ≤ n ∑ ( a j b k − a k b j ) ∗ ( a j b k − a k b j ) = 2 1 j = 1 ∑ n k = 1 ∑ n ( a j b k − a k b j ) ∗ ( a j b k − a k b j ) = ( ∑ j = 1 n a j ∗ a j ) ( ∑ k = 1 n b k ∗ b k ) − ( ∑ j = 1 n a j ∗ b j ) ( ∑ k = 1 n a k ∗ b k ) ∗ . =\left(\sum_{j=1}^{n}a_j^*a_j\right)\left(\sum_{k=1}^{n}b_k^*b_k\right)-\left(\sum_{j=1}^{n}a_j^*b_j\right)\left(\sum_{k=1}^{n}a_k^*b_k\right)^*. = ( j = 1 ∑ n a j ∗ a j ) ( k = 1 ∑ n b k ∗ b k ) − ( j = 1 ∑ n a j ∗ b j ) ( k = 1 ∑ n a k ∗ b k ) ∗ . Improved Cauchy-Schwarz-Kaplansky Inequality
Rearranging, we get
( ∑ j = 1 n a j ∗ b j ) ( ∑ k = 1 n a k ∗ b k ) ∗ = ( ∑ j = 1 n a j ∗ a j ) ( ∑ k = 1 n b k ∗ b k ) − ∑ 1 ≤ j < k ≤ n ( a j b k − a k b j ) ∗ ( a j b k − a k b j ) ≤ ( ∑ j = 1 n a j ∗ a j ) ( ∑ k = 1 n b k ∗ b k ) . \left(\sum_{j=1}^{n}a_j^*b_j\right)\left(\sum_{k=1}^{n}a_k^*b_k\right)^*=\left(\sum_{j=1}^{n}a_j^*a_j\right)\left(\sum_{k=1}^{n}b_k^*b_k\right)-\sum_{1\leq j<k\leq n}(a_jb_k-a_kb_j)^*(a_jb_k-a_kb_j)
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\leq\left(\sum_{j=1}^{n}a_j^*a_j\right)\left(\sum_{k=1}^{n}b_k^*b_k\right). ( j = 1 ∑ n a j ∗ b j ) ( k = 1 ∑ n a k ∗ b k ) ∗ = ( j = 1 ∑ n a j ∗ a j ) ( k = 1 ∑ n b k ∗ b k ) − 1 ≤ j < k ≤ n ∑ ( a j b k − a k b j ) ∗ ( a j b k − a k b j ) ≤ ( j = 1 ∑ n a j ∗ a j ) ( k = 1 ∑ n b k ∗ b k ) . Since square root respects the order of positive elements in a C*-algebra, we have the result.
Following extension of Theorem 2.1 is clear.
Theorem 2.2. (Improved Cauchy-Schwarz-Kaplansky Inequality) Let A \mathcal{A} A { a j } j = 1 ∞ , { b j } j = 1 ∞ ⊆ A \{a_j\}_{j=1}^{\infty}, \{b_j\}_{j=1}^{\infty}\subseteq\mathcal{A} { a j } j = 1 ∞ , { b j } j = 1 ∞ ⊆ A
∑ j = 1 ∞ a j ∗ a j ∈ A , ∑ j = 1 ∞ b j ∗ b j ∈ A . \sum_{j=1}^{\infty}a_j^*a_j\in\mathcal{A},\qquad\sum_{j=1}^{\infty}b_j^*b_j\in\mathcal{A}. j = 1 ∑ ∞ a j ∗ a j ∈ A , j = 1 ∑ ∞ b j ∗ b j ∈ A . Then
( ∑ j = 1 ∞ a j ∗ b j ) ( ∑ k = 1 ∞ a k ∗ b k ) ∗ = ( ∑ j = 1 ∞ a j ∗ a j ) ( ∑ k = 1 ∞ b k ∗ b k ) − ∑ 1 ≤ j < k < ∞ ( a j b k − a k b j ) ∗ ( a j b k − a k b j ) ≤ ( ∑ j = 1 ∞ a j ∗ a j ) ( ∑ k = 1 ∞ b k ∗ b k ) . \left(\sum_{j=1}^{\infty}a_j^*b_j\right)\left(\sum_{k=1}^{\infty}a_k^*b_k\right)^*=\left(\sum_{j=1}^{\infty}a_j^*a_j\right)\left(\sum_{k=1}^{\infty}b_k^*b_k\right)-\sum_{1\leq j<k<\infty}(a_jb_k-a_kb_j)^*(a_jb_k-a_kb_j)
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\leq\left(\sum_{j=1}^{\infty}a_j^*a_j\right)\left(\sum_{k=1}^{\infty}b_k^*b_k\right). ( j = 1 ∑ ∞ a j ∗ b j ) ( k = 1 ∑ ∞ a k ∗ b k ) ∗ = ( j = 1 ∑ ∞ a j ∗ a j ) ( k = 1 ∑ ∞ b k ∗ b k ) − 1 ≤ j < k < ∞ ∑ ( a j b k − a k b j ) ∗ ( a j b k − a k b j ) ≤ ( j = 1 ∑ ∞ a j ∗ a j ) ( k = 1 ∑ ∞ b k ∗ b k ) . In particular,
[ ( ∑ j = 1 ∞ a j ∗ b j ) ( ∑ k = 1 ∞ a k ∗ b k ) ∗ ] 1 2 = [ ( ∑ j = 1 ∞ a j ∗ a j ) ( ∑ k = 1 ∞ b k ∗ b k ) − ∑ 1 ≤ j < k < ∞ ( a j b k − a k b j ) ∗ ( a j b k − a k b j ) ] 1 2 ≤ [ ( ∑ j = 1 ∞ a j ∗ a j ) ( ∑ k = 1 ∞ b k ∗ b k ) ] 1 2 = ( ∑ j = 1 ∞ a j ∗ a j ) 1 2 ( ∑ k = 1 ∞ b k ∗ b k ) 1 2 . \left[\left(\sum_{j=1}^{\infty}a_j^*b_j\right)\left(\sum_{k=1}^{\infty}a_k^*b_k\right)^*\right]^{\frac{1}{2}}=\left[\left(\sum_{j=1}^{\infty}a_j^*a_j\right)\left(\sum_{k=1}^{\infty}b_k^*b_k\right)-\sum_{1\leq j<k<\infty}(a_jb_k-a_kb_j)^*(a_jb_k-a_kb_j)\right]^{\frac{1}{2}}
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\leq\left[\left(\sum_{j=1}^{\infty}a_j^*a_j\right)\left(\sum_{k=1}^{\infty}b_k^*b_k\right)\right]^{\frac{1}{2}}
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=\left(\sum_{j=1}^{\infty}a_j^*a_j\right)^{\frac{1}{2}}\left(\sum_{k=1}^{\infty}b_k^*b_k\right)^{\frac{1}{2}}. [ ( j = 1 ∑ ∞ a j ∗ b j ) ( k = 1 ∑ ∞ a k ∗ b k ) ∗ ] 2 1 = ( j = 1 ∑ ∞ a j ∗ a j ) ( k = 1 ∑ ∞ b k ∗ b k ) − 1 ≤ j < k < ∞ ∑ ( a j b k − a k b j ) ∗ ( a j b k − a k b j ) 2 1 ≤ [ ( j = 1 ∑ ∞ a j ∗ a j ) ( k = 1 ∑ ∞ b k ∗ b k ) ] 2 1 = ( j = 1 ∑ ∞ a j ∗ a j ) 2 1 ( k = 1 ∑ ∞ b k ∗ b k ) 2 1 . An Open Problem
For non-commutative C*-algebras, the following result has been obtained independently by Paschke [10 ] and Rieffel [11 ] .
Theorem 3.1. [10 , 11 ] (Cauchy-Schwarz-Paschke-Rieffel Inequality) Let A \mathcal{A} A n ∈ N n\in\mathbb{N} n ∈ N
[ ( ∑ j = 1 n a j ∗ b j ) ( ∑ k = 1 n a k ∗ b k ) ∗ ] 1 2 ≤ ( ∑ j = 1 n a j ∗ a j ) 1 2 ∥ ∑ k = 1 n b k ∗ b k ∥ 1 2 , ∀ ( a j ) j = 1 n , ( b j ) j = 1 n ∈ A n . \left[\left(\sum_{j=1}^{n}a_j^*b_j\right)\left(\sum_{k=1}^{n}a_k^*b_k\right)^*\right]^{\frac{1}{2}}\leq\left(\sum_{j=1}^{n}a_j^*a_j\right)^{\frac{1}{2}}\left\lVert\sum_{k=1}^{n}b_k^*b_k\right\rVert^{\frac{1}{2}},\qquad\forall(a_j)_{j=1}^{n},(b_j)_{j=1}^{n}\in\mathcal{A}^{n}. [ ( j = 1 ∑ n a j ∗ b j ) ( k = 1 ∑ n a k ∗ b k ) ∗ ] 2 1 ≤ ( j = 1 ∑ n a j ∗ a j ) 2 1 k = 1 ∑ n b k ∗ b k 2 1 , ∀ ( a j ) j = 1 n , ( b j ) j = 1 n ∈ A n . In particular,
∥ ∑ j = 1 n a j ∗ b j ∥ ≤ ∥ ∑ j = 1 n a j ∗ a j ∥ 1 2 ∥ ∑ k = 1 n b k ∗ b k ∥ 1 2 , ∀ ( a j ) j = 1 n , ( b j ) j = 1 n ∈ A n . \left\lVert\sum_{j=1}^{n} a_j^*b_j \right\rVert\le\left\lVert\sum_{j=1}^{n} a_j^*a_j \right\rVert^{\frac{1}{2}} \left\lVert\sum_{k=1}^{n} b_k^*b_k \right\rVert^{\frac{1}{2}}, \qquad\forall(a_j)_{j=1}^{n},(b_j)_{j=1}^{n} \in A^n. j = 1 ∑ n a j ∗ b j ≤ j = 1 ∑ n a j ∗ a j 2 1 k = 1 ∑ n b k ∗ b k 2 1 , ∀ ( a j ) j = 1 n , ( b j ) j = 1 n ∈ A n . Note that Cauchy-Schwarz-Paschke-Rieffel inequality is not a direct generalization of Cauchy-Schwarz-Kaplansky inequality. We don’t know how to generalize Theorem 3.1 to noncommutative C ∗ C^* C ∗
Note that the Cauchy-Binet identity
( ∑ j = 1 n a j c j ) ( ∑ k = 1 n b k d k ) − ( ∑ j = 1 n a j d j ) ( ∑ k = 1 n b k c k ) = ∑ 1 ≤ j < k ≤ n ( a j b k − b j a k ) ( c j d k − d j c k ) , \left(\sum_{j=1}^{n}a_jc_j\right)\left(\sum_{k=1}^{n}b_kd_k\right)-\left(\sum_{j=1}^{n}a_jd_j\right)\left(\sum_{k=1}^{n}b_kc_k\right)=\sum_{1\le j<k\le n}(a_jb_k-b_ja_k)(c_jd_k-d_jc_k), ( j = 1 ∑ n a j c j ) ( k = 1 ∑ n b k d k ) − ( j = 1 ∑ n a j d j ) ( k = 1 ∑ n b k c k ) = 1 ≤ j < k ≤ n ∑ ( a j b k − b j a k ) ( c j d k − d j c k ) , ∀ ( a j ) j = 1 n , ( b j ) j = 1 n , ( c j ) j = 1 n , ( d j ) j = 1 n ∈ C n \forall(a_j)_{j=1}^{n},(b_j)_{j=1}^{n},(c_j)_{j=1}^{n},(d_j)_{j=1}^{n}\in\mathbb{C}^n ∀ ( a j ) j = 1 n , ( b j ) j = 1 n , ( c j ) j = 1 n , ( d j ) j = 1 n ∈ C n seems not to improve Cauchy-Schwarz-Paschke-Rieffel inequality (Cauchy-Binet identity holds only for commutative rings).
Conclusions
In 1821, Cauchy extended Lagrange identity and proved Cauchy-Schwarz inequality [7 ] .
In 1953, Kaplansky extended Cauchy-Schwarz inequality to commutative C ∗ C^* C ∗ [6 ] .
In this note, we improved Cauchy-Schwarz-Kaplansky inequality.
Acknowledgments
AI is used for the literature survey.
