Generalizations of Cauchy-Schwarz inequality in unitary spaces

In this paper, we give a generalization of Cauchy-Schwarz inequality in unitary spaces and obtain its integral analogs. As an application, we establish an inequality for covariances.

Let u and v be two vectors in a unitary space $\mathbb{H}$. The Cauchy-Schwarz inequality is well known,

where $\langle\cdot,\cdot\rangle$ and $\|\cdot\|$ denote the inner product and norm in $\mathbb{H}$, respectively. Its integral form in the space of real-valued functions $L^{2}[a,b]$ is

The Cauchy-Schwarz inequality is one of the most important inequalities in mathematics. To date, a large number of generalizations and refinements of the inequalities (1.1) and (1.2) have been investigated in the literature (see [1] and references therein, also see [2–9]).

In this note, we will present some new generalizations of the Cauchy-Schwarz inequality (1.1).

Suppose that $\mathbb{H}$ is a unitary space (complex inner product space) with standard inner product $\langle\cdot,\cdot\rangle$ and norm $\|\cdot\|$, namely $\langle x,y\rangle=x^{T}\overline{y}$ and $\|x\| =\sqrt{\langle x,x\rangle}$ (see [10]). Let $X=( x_{1},x_{2},\ldots,x_{n})$ denote the n-tuple of vectors $x_{i} \in \mathbb{H}$, $i=1,\ldots,n$. For two n-tuples $X=(x_{1},\ldots,x_{n})$ and $Y=(y_{1},\ldots,y_{n})$ of $\mathbb{H}$, we define this A-product of vector $x_{i}$ and $y_{i}$ for X and Y by

where $a=\frac{x_{1}+\cdots+x_{n}}{n}$ and $b=\frac{y_{1}+\cdots+y_{n}}{n}$.

Our main results are the following theorems.

Theorem 1

Let $X=( x_{1},\ldots,x_{n})$ and $Y=(y_{1},\ldots ,y_{n})$ be two n-tuples of the unitary space $\mathbb{H}$, then

Equality holds if $y_{i}=(x_{i}-2a)\lambda$ ($i=1,\ldots,n$) for any $\{ x_{1},\ldots, x_{n}\}$, where λ is a non-negative constant.

In particular, if $n=1$, then (1.3) is the Cauchy-Schwarz inequality (1.1).

For complex numbers $\mathbb{C}$, by Theorem 1, we have the following.

Corollary 1

Suppose that $x_{1},\ldots,x_{n}$ and $y_{1},\ldots ,y_{n}$ are complex numbers. Set

then

Equality holds if $y_{i}=(x_{i}-2a)\lambda$ ($i=1,\ldots,n$) for any $\{x_{1},\ldots, x_{n}\}$, where λ is a non-negative constant.

Let $H\oplus H\oplus\cdots\oplus H $ denote the direct sum of n unitary space $\mathbb{H}$ with norm $\|X\|= (\sum_{i=1}^{n}\|x_{i}\|^{2} )^{\frac{1}{2}}$. Set $f(X,Y)=\sum_{i=1}^{n} x_{i}\otimes_{A}y_{i}$. Since $f(X,X)$ is not always non-negative, $f(X,Y)$ is not an inner product in the above direct sum. Hence, (1.3) is different from the Cauchy-Schwarz inequality in the above direct sum.

If we set $|X\otimes_{A} Y|=\sum_{i=1}^{n}| x_{i}\otimes_{A}y_{i}|$, then (1.3) can be restated as

Furthermore, we obtain the following integral form of (1.3) (only consider real-valued functions).

Theorem 2

Let μ be a positive measure such that $\mu(\Omega)=1$, f and g be real-valued functions in $L^{2}(\mu)$, and let

then

Equality holds if $g(x)=(f(x)-2 \int_{\Omega}f \, d\mu)\lambda $, where λ is a non-negative constant.

Proof of Theorem 1

Using the basic properties of the norm of a unitary space, we get

By (2.1), using the Cauchy-Schwarz inequality (1.1) and the discrete form of the Cauchy-Schwarz inequality, it follows that

Similarly to (2.2), we have

Combining (2.2) and (2.3), we infer that

This is the inequality (1.3), as desired. □

Proof of Theorem 2

We first prove the following inequality:

In fact, by the Cauchy-Schwarz inequality (1.2), we obtain

This is the inequality (2.4).

Similarly, we have

From (2.4) and (2.5), we find that

The inequality (1.4) follows. □

Let $(a_{1},b_{1}),\ldots,(a_{n},b_{n}) $ be n items of bivariate real data, $x=\{a_{1},\ldots,a_{n}\}$ and $y=\{b_{1},\ldots,b_{n}\}$, then their covariance $\operatorname{Cov}(x,y)$ is defined as [11]

where $a=\frac{a_{1}+\cdots+a_{n}}{n}$ and $b=\frac{b_{1}+\cdots+b_{n}}{n}$.

For the covariance $\operatorname{Cov}(x,y)$, it is well known that Pearson’s product moment inequality is

where $\operatorname{SD}(x)=\sqrt{\frac{1}{n}\sum_{i=1}^{n}(a_{i}-a)^{2}}$ and $\operatorname{SD}(y)=\sqrt{\frac{1}{n}\sum_{i=1}^{n}(b_{i}-b)^{2}}$.

Similarly, now we define this covariance of two n-tuples $X=( x_{1},x_{2},\ldots ,x_{n})$ and $Y=(y_{1},y_{2},\ldots,y_{n})$ of the unitary space $\mathbb{H}$ as

where $a=\frac{x_{1}+\cdots+x_{n}}{n}$ and $b=\frac{y_{1}+\cdots+y_{n}}{n}$.

Set $\alpha_{i}=x_{i}-a$ and $\beta_{i}=y_{i}-b$, $i=1,\ldots,n$. Note that

Hence, (1.3) can be written in the following form:

where $\|X\|= (\sum^{n}_{i=1}\|x_{i}\|^{2} )^{1/2}$ and $\|Y\|= (\sum^{n}_{i=1}\|y_{i}\|^{2} )^{1/2}$.

Using the triangle inequality on the left side of (3.1), we obtain

Finally, we can simply state the above result, as follows.

Theorem 3

Let $X=( x_{1},x_{2},\ldots,x_{n})$ and $Y=(y_{1},y_{2},\ldots,y_{n})$ be two n-tuples of the unitary space $\mathbb {H}$, then

This work was partially supported by the National Natural Science Foundation of China (Grant No. 11171053).

Affiliations

1. School of Computer Engineering and Science, Shanghai University, Shanghai, China

   • Leng Tuo

Corresponding author

Correspondence to Leng Tuo.

Competing interests

The author declares to have no competing interests.

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