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\fancyhead[LO]{\textbf{Improved Cauchy-Schwarz-Kaplansky Inequality}}
\fancyhead[RE]{\textbf{K. Mahesh Krishna}}


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{\bf\large{{Improved Cauchy-Schwarz-Kaplansky Inequality}}}\\
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\textbf{K. Mahesh Krishna}\\
School of Mathematics and Natural Sciences\\
Chanakya University Global Campus\\
NH-648, Haraluru Village\\
Devanahalli Taluk, 	Bengaluru  North District\\
Karnataka State, 562 110, India\\
Email: kmaheshak@gmail.com\\

Date: \today
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\textbf{Abstract}: 
We make an improvement and give a new proof for  Cauchy-Schwarz-Kaplansky inequality derived by Kaplansky [\textit{Am. J. Math., 1953}].


\textbf{Keywords}: Cauchy-Schwarz-Kaplansky inequality, C*-algebra, Lagrange identity.


\textbf{Mathematics Subject Classification (2020)}:  26D15, 46L05.

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\section{Introduction}
Cauchy-Schwarz angle-length inequality, derived in 1821,  is the following. 
\begin{theorem}\cite{STEELE, MITRINOVIC, MITRINOVICPECARICFINK, HARDYLITTLEWOODPOLYA, VENKATACHALA, ENGEL, BECKENBACHBELLMAN}
	(\textbf{Cauchy-Schwarz Angle-Length Inequality}) \label{CSI}
Let $n \in \mathbb{N}$. Then 
\begin{align*}
	\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.
\end{align*}
\end{theorem}
We clearly have the following sequential version of Theorem \ref{CSI}.
\begin{theorem}\cite{STEELE, MITRINOVIC, MITRINOVICPECARICFINK, HARDYLITTLEWOODPOLYA, VENKATACHALA, ENGEL, BECKENBACHBELLMAN}
	(\textbf{Cauchy-Schwarz Angle-Length Inequality}) 
	Let $\{a_j\}_{j=1}^\infty, \{b_j\}_{j=1}^\infty \subseteq \mathbb{C}$ be such that 
\begin{align*}
	\sum_{j=1}^{\infty}|a_j|^2 \in \mathbb{R}, \quad 	\sum_{j=1}^{\infty}|b_j|^2 \in \mathbb{R}.
\end{align*}
	 Then 
	\begin{align*}
		\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}.
	\end{align*}
\end{theorem}
In 1953, Kaplansky extended Cauchy-Schwarz inequality to commutative C*-algebras \cite{KAPLANSKY}.
\begin{theorem}\cite{KAPLANSKY}	(\textbf{Cauchy-Schwarz-Kaplansky Inequality}) 
Let $\mathcal{A}$ be a commutative C*-algebra. Let $n \in \mathbb{N}$. Then 
\begin{align*}
	\left[\left(\sum_{j=1}^n a^*_jb_j\right)\left(\sum_{k=1}^n a^*_kb_k\right)^*\right]^\frac{1}{2}
	\leq \left(\sum_{j=1}^n a^*_ja_j\right)^\frac{1}{2}\left(\sum_{k=1}^n b^*_kb_k\right)^\frac{1}{2}, \quad \forall (a_j)_{j=1}^n, (b_j)_{j=1}^n \in \mathcal{A}^n.
\end{align*}
In particular, 
\begin{align*}
	\left\|\sum_{j=1}^n a^*_jb_j\right\|\leq  \left\|\sum_{j=1}^n a^*_ja_j\right\|^\frac{1}{2}\left\|\sum_{k=1}^n b^*_kb_k\right\|^\frac{1}{2}.
\end{align*}
\end{theorem}
\begin{theorem}\cite{KAPLANSKY}	(\textbf{Cauchy-Schwarz-Kaplansky Inequality}) 
Let $\mathcal{A}$ be a commutative C*-algebra. 	Let $\{a_j\}_{j=1}^\infty, \{b_j\}_{j=1}^\infty \subseteq \mathcal{A}$ be such that 
\begin{align*}
	\sum_{j=1}^{\infty}a_j^*a_j \in \mathcal{A}, \quad 	\sum_{j=1}^{\infty}b_j^*b_j \in \mathcal{A}.
\end{align*}
Then 
\begin{align*}
	\left[\left(\sum_{j=1}^\infty a^*_jb_j\right)\left(\sum_{k=1}^\infty a^*_kb_k\right)^*\right]^\frac{1}{2}
	\leq \left(\sum_{j=1}^\infty a^*_ja_j\right)^\frac{1}{2}\left(\sum_{k=1}^\infty b^*_kb_k\right)^\frac{1}{2}.
\end{align*}
In particular, 
\begin{align*}
	\left\|\sum_{j=1}^\infty a^*_jb_j\right\|\leq  \left\|\sum_{j=1}^\infty a^*_ja_j\right\|^\frac{1}{2}\left\|\sum_{k=1}^\infty b^*_kb_k\right\|^\frac{1}{2}.
\end{align*}	
\end{theorem}
In this note, we improve Cauchy-Schwarz-Kaplansky inequality.



\section{Improved Cauchy-Schwarz-Kaplansky  Inequality}
Following result is improvement of Theorem \ref{CSI}.
\begin{theorem}(\textbf{Improved Cauchy-Schwarz-Kaplansky Inequality}) \label{K}
	Let $\mathcal{A}$ be a commutative C*-algebra. Let $n \in \mathbb{N}$. Then 
\begin{align*}
	\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)\\
	&\leq \left(\sum_{j=1}^n a_j^*a_j\right)\left(\sum_{k=1}^n b^*_kb_k\right), \quad \forall (a_j)_{j=1}^n, (b_j)_{j=1}^n \in \mathcal{A}^n.
\end{align*}
In particular, 
\begin{align*}
	\left[\left(\sum_{j=1}^n a^*_jb_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}\\
	&\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}\\
	&=\left(\sum_{j=1}^n a^*_ja_j\right)^\frac{1}{2}\left(\sum_{k=1}^n b^*_kb_k\right)^\frac{1}{2}, \quad \forall (a_j)_{j=1}^n, (b_j)_{j=1}^n \in \mathcal{A}^n.
\end{align*}
\end{theorem}
\begin{proof}
Our proof is motivated from the arguments for the scalar case by Cauchy using Lagrange identity  \cite{STEELE, MALIGRANDA, MITRINOVIC, MITRINOVICPECARICFINK, GIDEANICULESCU, GUASTI, WRIGHT}. From Lagrange identity, we have,
\begin{align*}
	\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)\\
	&=\left(\sum_{j=1}^n a^*_ja_j\right)\left(\sum_{k=1}^n b^*_kb_k\right)-\left(\sum_{j=1}^n a^*_jb_j\right)\left(\sum_{k=1}^n a^*_kb_k\right)^*.
\end{align*}
Rearranging, we get 
\begin{align*}
	\left(\sum_{j=1}^n a^*_jb_j\right)\left(\sum_{k=1}^n a^*_kb_k\right)^*&=\left(\sum_{j=1}^n a^*_ja_j\right)\left(\sum_{k=1}^n b^*_kb_k\right)-\sum_{1\leq j<k \leq n}(a_jb_k-a_kb_j)^*(a_jb_k-a_kb_j)\\
	&\leq \left(\sum_{j=1}^n a_j^*a_j\right)\left(\sum_{k=1}^n b^*_kb_k\right).
\end{align*}
Since square root respects the order of positive elements in a C*-algebra, we have the result. 
\end{proof}
Following extension of  Theorem \ref{K} is clear. 
\begin{theorem}
(\textbf{Improved Cauchy-Schwarz-Kaplansky Inequality}) 
Let $\mathcal{A}$ be a commutative C*-algebra. Let $\{a_j\}_{j=1}^\infty, \{b_j\}_{j=1}^\infty \subseteq \mathcal{A}$ be such that 
\begin{align*}
	\sum_{j=1}^{\infty}a_j^*a_j \in \mathcal{A}, \quad 	\sum_{j=1}^{\infty}b_j^*b_j \in \mathcal{A}.
\end{align*}
 Then 
\begin{align*}
	\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)\\
	&\leq \left(\sum_{j=1}^\infty a_j^*a_j\right)\left(\sum_{k=1}^\infty b^*_kb_k\right).
\end{align*}
In particular, 
\begin{align*}
	\left[\left(\sum_{j=1}^\infty a^*_jb_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}\\
	&\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}\\
	&=\left(\sum_{j=1}^\infty a^*_ja_j\right)^\frac{1}{2}\left(\sum_{k=1}^\infty b^*_kb_k\right)^\frac{1}{2}.
\end{align*}	
\end{theorem}




\section{An Open Problem}
For non-commutative C*-algebras, the following result has been obtained independently by Paschke \cite{PASCHKE} and Rieffel \cite{RIEFFEL}.
\begin{theorem}\cite{PASCHKE, RIEFFEL} (\textbf{Cauchy-Schwarz-Paschke-Rieffel Inequality})\label{PR}
Let $\mathcal{A}$ be a C*-algebra. Let $n \in \mathbb{N}$. Then 
\begin{align*}
	\left[\left(\sum_{j=1}^n a^*_jb_j\right)\left(\sum_{k=1}^n a^*_kb_k\right)^*\right]^\frac{1}{2}
	\leq \left(\sum_{j=1}^n a^*_ja_j\right)^\frac{1}{2}\left\|\sum_{k=1}^n b^*_kb_k\right\|^\frac{1}{2}, \quad \forall (a_j)_{j=1}^n, (b_j)_{j=1}^n \in \mathcal{A}^n.
\end{align*}
In particular, 
\begin{align*}
	\left\|\sum_{j=1}^n a^*_jb_j\right\|\leq  \left\|\sum_{j=1}^n a^*_ja_j\right\|^\frac{1}{2}\left\|\sum_{k=1}^n b^*_kb_k\right\|^\frac{1}{2}, \quad \forall (a_j)_{j=1}^n, (b_j)_{j=1}^n \in \mathcal{A}^n.
\end{align*}	
\end{theorem}
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 \ref{PR} to noncommutative C*-algebras. 
Note that the Cauchy-Binet identity
\begin{align*}
\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\leq j<k \leq n}(a_jb_k-b_ja_k)(c_jd_k-d_jc_k),\\
\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
\end{align*}
 seems not to improve Cauchy-Schwarz-Paschke-Rieffel inequality (Cauchy-Binet identity holds only for commutative rings).

\section{Conclusions}
\begin{enumerate}
	\item In 1821, Cauchy extended Lagrange identity and proved Cauchy-Schwarz inequality \cite{MALIGRANDA}.
	\item In 1953, Kaplansky extended Cauchy-Schwarz inequality to commutative C*-algebras \cite{KAPLANSKY}.
	\item In this note, we improved Cauchy-Schwarz-Kaplansky inequality.
\end{enumerate}

	

 \section{Acknowledgments}
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