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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.

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

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


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

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Correspondence to Leng Tuo.

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Tuo, L. Generalizations of Cauchy-Schwarz inequality in unitary spaces. J Inequal Appl 2015, 201 (2015).

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