Generalizations of Cauchy-Schwarz inequality in unitary spaces

Abstract

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.

1 Introduction

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

$$ \bigl\vert \langle u,v\rangle\bigr\vert \leq\|u\|\cdot\|v\|, $$

(1.1)

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

$$ \biggl(\int_{a}^{b}f\cdot g \, d\mu \biggr)^{2}\leq \biggl(\int_{a}^{b}f^{2} \, d\mu \biggr) \biggl(\int_{a}^{b}g^{2} \, d\mu \biggr). $$

(1.2)

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

$$x_{i}\otimes_{A} y_{i}=\langle x_{i},y_{i}\rangle-\langle x_{i},b\rangle- \langle a,y_{i}\rangle, $$

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

$$ \sum_{i=1}^{n}| x_{i} \otimes_{A}y_{i}|\leq \Biggl(\sum ^{n}_{i=1}\|x_{i}\|^{2} \Biggr)^{1/2} \Biggl(\sum^{n}_{i=1} \|y_{i}\|^{2} \Biggr)^{1/2}. $$

(1.3)

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

$$a=\frac{x_{1}+\cdots+x_{n}}{n}, \qquad b=\frac{y_{1}+\cdots+y_{n}}{n}, $$

then

$$\Biggl(\sum_{i=1}^{n}|x_{i}y_{i}-a y_{i} -b x_{i} | \Biggr)^{2}\leq \Biggl(\sum _{i=1}^{n}|x_{i}|^{2} \Biggr) \Biggl(\sum_{i=}^{n}|y_{i}|^{2} \Biggr) . $$

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

$$|X\otimes_{A} Y|\leq\|X\|\cdot\|Y\|. $$

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

$$f\otimes_{A}g(x)=f(x)\cdot g(x)-f(x)\cdot\int_{\Omega}g\, d\mu-g(x)\cdot\int_{\Omega}f\, d\mu, $$

then

$$ \biggl(\int_{\Omega}|f\otimes_{A}g|\, d\mu \biggr)^{2}\leq \biggl(\int_{\Omega}f^{2}\, d \mu \biggr) \biggl(\int_{\Omega}g^{2}\, d\mu \biggr). $$

(1.4)

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

2 The proofs of the theorems

Proof of Theorem 1

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

$$\begin{aligned} \sum_{i=1}^{n}\|y_{i}-2b \|^{2}&=\sum_{i=1}^{n} \langle y_{i}-2b,{y_{i}}-2{b}\rangle \\ &=\sum_{i=1}^{n} \bigl(4\|b \|^{2}-2\langle y_{i},{b}\rangle-2\langle b,{y_{i}}\rangle+\|y_{i}\|^{2} \bigr) \\ &=4n\|b\|^{2}-2\Biggl\langle \sum_{i=1}^{n}y_{i},{b} \Biggr\rangle -2\Biggl\langle b,{\sum_{i=1}^{n}y_{i}} \Biggr\rangle +\sum_{i=1}^{n} \|y_{i}\|^{2} \\ &=4n\|b\|^{2}-2n\|b\|^{2}-2n\|b\|^{2}+\sum _{i=1}^{n}\|y_{i}\|^{2} \\ &=\sum_{i=1}^{n}\|y_{i} \|^{2}. \end{aligned}$$

(2.1)

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

$$\begin{aligned} \sum_{i=1}^{n}\bigl\vert \langle x_{i},y_{i}-2b\rangle\bigr\vert &\leq\sum _{i=1}^{n}\|x_{i}\|\cdot\| y_{i}-2b\| \\ &\leq \Biggl(\sum_{i=1}^{n} \|x_{i}\|^{2} \Biggr)^{\frac{1}{2}} \Biggl(\sum _{i=1}^{n}\| y_{i}-2b\|^{2} \Biggr)^{\frac{1}{2}} \\ &= \Biggl(\sum_{i=1}^{n}\|x_{i} \|^{2} \Biggr)^{\frac{1}{2}} \Biggl(\sum_{i=1}^{n} \|y_{i}\| ^{2} \Biggr)^{\frac{1}{2}}. \end{aligned}$$

(2.2)

Similarly to (2.2), we have

$$ \sum_{i=1}^{n}\bigl\vert \langle x_{i}-2a,y_{i}\rangle\bigr\vert \leq \Biggl(\sum ^{n}_{i=1}\|x_{i}\| ^{2} \Biggr)^{\frac{1}{2}} \Biggl(\sum^{n}_{i=1} \|y_{i}\|^{2} \Biggr)^{\frac{1}{2}}. $$

(2.3)

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

$$\begin{aligned} \sum_{i=1}^{n} \vert x_{i} \otimes_{A}y_{i}\vert &=\frac{1}{2}\sum _{i=1}^{n}\bigl\vert 2\langle x_{i},y_{i} \rangle-\langle x_{i},2b\rangle-\langle2a,y_{i}\rangle\bigr\vert \\ &=\frac{1}{2}\sum_{i=1}^{n}\bigl\vert \langle x_{i},y_{i}-2b\rangle+\langle x_{i}-2a,y_{i}\rangle\bigr\vert \\ &\leq\frac{1}{2}\sum_{i=1}^{n}\bigl\vert \langle x_{i},y_{i}-2b\rangle\bigr\vert + \frac{1}{2}\sum_{i=1}^{n}\bigl\vert \langle x_{i}-2a,y_{i}\rangle\bigr\vert \\ &\leq \Biggl(\sum_{i=1}^{n} \|x_{i}\|^{2} \Biggr)^{\frac{1}{2}} \Biggl(\sum _{i=1}^{n}\| y_{i}\|^{2} \Biggr)^{\frac{1}{2}}. \end{aligned}$$

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

Proof of Theorem 2

We first prove the following inequality:

$$ \int_{\Omega}\biggl\vert f \biggl(g-2\int_{\Omega}g\, d\mu \biggr)\biggr\vert \, d\mu\leq \biggl(\int_{\Omega}f^{2} \, d\mu \biggr)^{\frac{1}{2}} \biggl(\int_{\Omega}g^{2}\, d\mu \biggr)^{\frac{1}{2}}. $$

(2.4)

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

$$\begin{aligned}& \int_{\Omega}\biggl\vert f \biggl(g-2\int_{\Omega}g\, d\mu \biggr)\biggr\vert \,d\mu \\& \quad \leq \biggl(\int_{\Omega}f^{2} \,d\mu \biggr)^{\frac{1}{2}} \biggl(\int_{\Omega}\biggl(g-2\int_{\Omega}g\, d\mu \biggr)^{2}\,d\mu \biggr)^{\frac{1}{2}} \\& \quad = \biggl(\int_{\Omega}f^{2} \,d\mu \biggr)^{\frac{1}{2}}\cdot \biggl(\int_{\Omega}\biggl(4 \biggl( \int_{\Omega}g \,d\mu \biggr)^{2}-4 \biggl(\int _{\Omega}g \,d\mu \biggr)\cdot g+g^{2} \biggr)\,d\mu \biggr)^{\frac{1}{2}} \\& \quad = \biggl(\int_{\Omega}f^{2} \,d\mu \biggr)^{\frac{1}{2}}\cdot \biggl(4 \biggl(\int_{\Omega}g \,d\mu \biggr)^{2}\mu(\Omega)-4 \biggl(\int_{\Omega}g \,d\mu \biggr)^{2}+\int_{\Omega}g^{2} \,d\mu \biggr)^{\frac{1}{2}} \\& \quad = \biggl(\int_{\Omega}f^{2} \,d\mu \biggr)^{\frac{1}{2}} \biggl(\int_{\Omega}g^{2} \,d\mu \biggr)^{\frac{1}{2}}. \end{aligned}$$

This is the inequality (2.4).

Similarly, we have

$$ \int_{\Omega}\biggl\vert g \biggl(f-2\int_{\Omega}f\, d\mu \biggr)\biggr\vert \, d\mu\leq \biggl(\int_{\Omega}f^{2}\, d\mu \biggr)^{\frac{1}{2}} \biggl(\int_{\Omega}g^{2}\, d\mu \biggr)^{\frac{1}{2}}. $$

(2.5)

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

$$\begin{aligned}& \int_{\Omega}|f\otimes_{A}g| \, d\mu \\& \quad = \frac{1}{2}\int_{\Omega}\biggl\vert \biggl(g-2\int _{\Omega}g\, d\mu \biggr)f+ \biggl(f-2\int_{\Omega}f\, d\mu \biggr)g\biggr\vert \, d\mu \\& \quad \leq\frac{1}{2} \biggl(\int_{\Omega}\biggl\vert f \biggl(g-2\int_{\Omega}g\, d\mu \biggr)\biggr\vert \, d\mu + \int_{\Omega}\biggl\vert g \biggl(f-2\int_{\Omega}f\, d\mu \biggr)\biggr\vert \, d\mu \biggr) \\& \quad \leq \biggl(\int_{\Omega}f^{2} \, d\mu \biggr)^{\frac{1}{2}} \biggl(\int_{\Omega}g^{2} \, d\mu \biggr)^{\frac{1}{2}}. \end{aligned}$$

The inequality (1.4) follows. □

3 An application

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]

$$\operatorname{Cov}(x,y)=\frac{1}{n}\sum_{i=1}^{n}(a_{i}-a) (b_{i}-b), $$

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

$$\bigl\vert \operatorname{Cov}(x,y)\bigr\vert \leq \operatorname{SD}(x)\cdot \operatorname{SD}(y), $$

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

$$\operatorname{Cov}(X,Y)=\frac{1}{n} \sum_{i=1}^{n} \langle x_{i}-a,y_{i}-b\rangle, $$

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

$$\begin{aligned} x_{i}\otimes_{A}y_{i}&=\langle \alpha_{i}+a,\beta_{i}+b\rangle-\langle\alpha _{i}+a,b\rangle-\langle a,\beta_{i}+b\rangle \\ &=\langle\alpha_{i},\beta_{i}\rangle-\langle a,b\rangle \\ &=\langle x_{i}-a,y_{i}-b\rangle-\langle a,b\rangle. \end{aligned}$$

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

$$ \sum_{i=1}^{n}\bigl\vert \langle x_{i}-a,y_{i}-b\rangle-\langle a,b\rangle\bigr\vert \leq \|X\| \cdot\|Y\|, $$

(3.1)

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

$$\bigl\vert n \operatorname{Cov}(X,Y)-n\langle a,b\rangle\bigr\vert \leq\|X\| \cdot\|Y\|. $$

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

$$ \bigl\vert \operatorname{Cov}(X,Y)-\langle a,b\rangle\bigr\vert \leq \frac{1}{n} \|X\|\cdot\|Y\|. $$

(3.2)