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Generalizations of Cauchy-Schwarz inequality in unitary spaces

Journal of Inequalities and Applications20152015:201

  • Received: 10 December 2014
  • Accepted: 28 May 2015
  • Published:


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.


  • Cauchy-Schwarz inequality
  • unitary space
  • positive measure


  • 26D15
  • 62J10

1 Introduction

Let u and v be two vectors in a unitary space \(\mathbb{H}\). The Cauchy-Schwarz inequality is well known,
$$ \bigl\vert \langle u,v\rangle\bigr\vert \leq\|u\|\cdot\|v\|, $$
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
$$ \biggl(\int_{a}^{b}f\cdot g \, d\mu \biggr)^{2}\leq \biggl(\int_{a}^{b}f^{2} \, d\mu \biggr) \biggl(\int_{a}^{b}g^{2} \, d\mu \biggr). $$

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 [29]).

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
$$ \sum_{i=1}^{n}| x_{i} \otimes_{A}y_{i}|\leq \Biggl(\sum ^{n}_{i=1}\|x_{i}\|^{2} \Biggr)^{1/2} \Biggl(\sum^{n}_{i=1} \|y_{i}\|^{2} \Biggr)^{1/2}. $$
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
$$a=\frac{x_{1}+\cdots+x_{n}}{n}, \qquad b=\frac{y_{1}+\cdots+y_{n}}{n}, $$
$$\Biggl(\sum_{i=1}^{n}|x_{i}y_{i}-a y_{i} -b x_{i} | \Biggr)^{2}\leq \Biggl(\sum _{i=1}^{n}|x_{i}|^{2} \Biggr) \Biggl(\sum_{i=}^{n}|y_{i}|^{2} \Biggr) . $$
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
$$f\otimes_{A}g(x)=f(x)\cdot g(x)-f(x)\cdot\int_{\Omega}g\, d\mu-g(x)\cdot\int_{\Omega}f\, d\mu, $$
$$ \biggl(\int_{\Omega}|f\otimes_{A}g|\, d\mu \biggr)^{2}\leq \biggl(\int_{\Omega}f^{2}\, d \mu \biggr) \biggl(\int_{\Omega}g^{2}\, d\mu \biggr). $$
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
$$\begin{aligned} \sum_{i=1}^{n}\|y_{i}-2b \|^{2}&=\sum_{i=1}^{n} \langle y_{i}-2b,{y_{i}}-2{b}\rangle \\ &=\sum_{i=1}^{n} \bigl(4\|b \|^{2}-2\langle y_{i},{b}\rangle-2\langle b,{y_{i}}\rangle+\|y_{i}\|^{2} \bigr) \\ &=4n\|b\|^{2}-2\Biggl\langle \sum_{i=1}^{n}y_{i},{b} \Biggr\rangle -2\Biggl\langle b,{\sum_{i=1}^{n}y_{i}} \Biggr\rangle +\sum_{i=1}^{n} \|y_{i}\|^{2} \\ &=4n\|b\|^{2}-2n\|b\|^{2}-2n\|b\|^{2}+\sum _{i=1}^{n}\|y_{i}\|^{2} \\ &=\sum_{i=1}^{n}\|y_{i} \|^{2}. \end{aligned}$$
By (2.1), using the Cauchy-Schwarz inequality (1.1) and the discrete form of the Cauchy-Schwarz inequality, it follows that
$$\begin{aligned} \sum_{i=1}^{n}\bigl\vert \langle x_{i},y_{i}-2b\rangle\bigr\vert &\leq\sum _{i=1}^{n}\|x_{i}\|\cdot\| y_{i}-2b\| \\ &\leq \Biggl(\sum_{i=1}^{n} \|x_{i}\|^{2} \Biggr)^{\frac{1}{2}} \Biggl(\sum _{i=1}^{n}\| y_{i}-2b\|^{2} \Biggr)^{\frac{1}{2}} \\ &= \Biggl(\sum_{i=1}^{n}\|x_{i} \|^{2} \Biggr)^{\frac{1}{2}} \Biggl(\sum_{i=1}^{n} \|y_{i}\| ^{2} \Biggr)^{\frac{1}{2}}. \end{aligned}$$
Similarly to (2.2), we have
$$ \sum_{i=1}^{n}\bigl\vert \langle x_{i}-2a,y_{i}\rangle\bigr\vert \leq \Biggl(\sum ^{n}_{i=1}\|x_{i}\| ^{2} \Biggr)^{\frac{1}{2}} \Biggl(\sum^{n}_{i=1} \|y_{i}\|^{2} \Biggr)^{\frac{1}{2}}. $$
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:
$$ \int_{\Omega}\biggl\vert f \biggl(g-2\int_{\Omega}g\, d\mu \biggr)\biggr\vert \, d\mu\leq \biggl(\int_{\Omega}f^{2} \, d\mu \biggr)^{\frac{1}{2}} \biggl(\int_{\Omega}g^{2}\, d\mu \biggr)^{\frac{1}{2}}. $$
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
$$ \int_{\Omega}\biggl\vert g \biggl(f-2\int_{\Omega}f\, d\mu \biggr)\biggr\vert \, d\mu\leq \biggl(\int_{\Omega}f^{2}\, d\mu \biggr)^{\frac{1}{2}} \biggl(\int_{\Omega}g^{2}\, d\mu \biggr)^{\frac{1}{2}}. $$
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:
$$ \sum_{i=1}^{n}\bigl\vert \langle x_{i}-a,y_{i}-b\rangle-\langle a,b\rangle\bigr\vert \leq \|X\| \cdot\|Y\|, $$
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
$$ \bigl\vert \operatorname{Cov}(X,Y)-\langle a,b\rangle\bigr\vert \leq \frac{1}{n} \|X\|\cdot\|Y\|. $$



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

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (, which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

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


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