Selberg’s Inequality and Selberg Operator Bounds in Hilbert Spaces with Applications

Abstract

In the present work, we give a new proof of the well-known Selberg’s inequality in complex Hilbert spaces from an operator-theoretic perspective, establishing its fundamental equivalence with the Cauchy–Bunyakovsky–Schwarz inequality. We also derive several lower and upper bounds for the Selberg operator, including its norm estimates, refining classical results such as de Bruijn’s and Bohr’s inequalities. Additionally, we revisit a recent claim in the literature, providing a clarification of the conditions under which Selberg’s inequality extends to abstract bilinear forms.

1. Introduction

Mathematical inequalities are vital tools in mathematics, supporting results in analysis, geometry, and operator theory [1,2,3,4,5]. They offer bounds and relationships that drive theoretical and applied advances, as seen in studies of Schwarz, Bessel, and Selberg inequalities [6,7]. This study, motivated by the role of inequalities in linking mathematical concepts, presents a new proof of Selberg’s inequality and explores extensions to enrich inequality theory [8,9].

In 1821, Augustin-Louis Cauchy introduced an inequality for real numbers [10]:

$$a_1{b}_1+a_2{b}_2+\cdots +a_n{b}_n\le \sqrt{a_1^2+a_2^2+\cdots +a_n^2}\sqrt{b_1^2+b_2^2+\cdots +b_n^2},$$

with equality if and only if there exists $r\in \mathbb{R}$ such that $a_k=r{b}_k$ for each $k=1,\dots ,n$. In 1859, V. Y. Bunyakovsky gave its integral form [11]. Later, H. A. Schwarz extended it to inner-product spaces, known as Schwarz’s inequality.

Let $\mathcal{H}$ be a complex Hilbert space with inner product $\langle ·,·\rangle$ and norm $\parallel ·\parallel$. The Cauchy–Bunyakovsky–Schwarz inequality (CBSI) states

$$|\langle u,v\rangle |\le \parallel u\parallel \parallel v\parallel ,$$

for all $u,v\in \mathcal{H}$, with equality if and only if there exists $\mu \in \mathbb{C}$ such that $u=\mu v$. This inequality is widely used in mathematics [7].

Many extensions of the CBSI exist. In [12], M. L. Buzano proved an extension called Buzano’s inequality (BuI):

Lemma 1

([12,13,14]). For any $x,y,z\in \mathcal{H}$, the following holds:

$$\left|\langle x,z\rangle \langle z,y\rangle \right|\le {\displaystyle \frac{1}{2}}{\parallel z\parallel }^2\left(\right|\langle x,y\rangle |+\parallel x\parallel \parallel y\parallel ).$$

(1)

If $\{x,y\}$ is linearly independent, equality in (1) holds for $z\in \mathcal{H}$ if and only if $z=\alpha (\parallel y\parallel x+e^{i\theta }\parallel x\parallel y)$ for some scalar α, where $\theta =arg\langle x,y\rangle$. If $\{x,y\}$ is linearly dependent, equality holds for $z=\alpha x$ for some scalar α.

For more on the CBSI and its extensions, see [7,15,16,17,18,19,20,21,22,23,24].

Another inequality related to the CBSI is the well-known Bessel’s inequality (BeI). For any orthonormal vectors $\{e_1,e_2,\dots ,e_n\}$ in $\mathcal{H}$ and any $x\in \mathcal{H}$ [4]

$${\displaystyle \sum _{j=1}^n}|\langle x,e_j\rangle {|}^2\le {\parallel x\parallel }^2.$$

Selberg’s inequality (SI), a generalization of BeI, was introduced by A. Selberg [4]. For vectors $x,z_1,\dots ,z_n\in \mathcal{H}$ with $z_i\ne 0$ for all $i=1,\dots ,n$, it states

$${\displaystyle \sum _{i=1}^n}{\displaystyle \frac{|\langle x,z_i\rangle {|}^2}{{\sum }_{j=1}^n\left|\langle z_i,z_j\rangle \right|}}\le {\parallel x\parallel }^2.$$

(2)

Furuta described the equality case in [25]: equality holds if and only if $x={\sum }_{i=1}^n{a}_i{z}_i$ for complex scalars $a_1,\dots ,a_n$, with $\langle z_i,z_j\rangle =0$ or $|a_i|=|a_j|$ and $\langle a_i{z}_i,a_j{z}_j\rangle \ge 0$ for $i\ne j$.

The paper is organized as follows. In Section 2, we recall the necessary background and present a new proof of Selberg’s inequality (see (2)), by deriving an operator-norm estimate for the associated Selberg operator acting on a Hilbert space. In Section 3, we develop several alternative lower and upper bounds for the left-hand side of (2). Section 4 is devoted to applications of our main estimates to two classical results: de Bruijn’s inequality and Bohr’s inequality. Finally, in Section 5, we offer a brief digression toward a vector-space generalization, extending Selberg’s framework to bilinear functionals on abstract vector spaces.

2. A New Proof of Selberg’s Inequality

In this section, we assume that $\mathcal{H}$ is a complex, infinite-dimensional Hilbert space. The $C^{*}$-algebra of all bounded linear operators on $\mathcal{H}$ is denoted by $\mathcal{B}\left(\mathcal{H}\right)$, equipped with the usual operator norm, defined for each $T\in \mathcal{B}\left(\mathcal{H}\right)$ by

$$\parallel T\parallel =\underset{\begin{array}{c}x\in \mathcal{H}\\ \parallel x\parallel =1\end{array}}{sup}\parallel Tx\parallel .$$

Given an operator $T\in \mathcal{B}\left(\mathcal{H}\right)$, its adjoint is denoted by $T^{*}$, and the identity operator on $\mathcal{H}$ is denoted by I. An operator T is said to be self-adjoint if $T=T^{*}$. In particular, a positive operator is a self-adjoint operator that satisfies $\langle Tx,x\rangle \ge 0$ for all $x\in \mathcal{H}$; we write $T\ge 0$ in this case. Moreover, for self-adjoint operators T and S, we write $T\ge S$ if $T-S\ge 0$, that is, if $T-S$ is a positive operator. It is well known that if T is a self-adjoint operator, then

$$\parallel T\parallel =\underset{\begin{array}{c}x\in \mathcal{H}\\ \parallel x\parallel =1\end{array}}{sup}\left|\langle Tx,x\rangle \right|.$$

Lemma 2.

Let $T\in \mathcal{B}\left(\mathcal{H}\right)$ such that $T\ge 0$, then the operator Cauchy–Bunyakovsky–Schwarz inequality holds:

$$\left|\langle Tx,y\rangle \right|\le {\langle Tx,x\rangle }^{\frac{1}{2}}{\langle Ty,y\rangle }^{\frac{1}{2}},$$

for any $x,y\in \mathcal{H}$. Additionally, we can derive the following consequence:

$${\parallel Tx\parallel }^2\le \parallel T\parallel \langle Tx,x\rangle ,$$

which holds for any positive operator T and $x\in \mathcal{H}$.

Proposition 1.

Let $T\in \mathcal{B}\left(\mathcal{H}\right)$ be a self-adjoint operator such that there exists a positive number C

$${\parallel Tx\parallel }^2\le C\left|\langle Tx,x\rangle \right|,$$

(3)

for any $x\in \mathcal{H}$, then $\parallel T\parallel \le C.$

Proof. 

From the inequality (3), we have that

$$\langle Tx,Tx\rangle ={\parallel Tx\parallel }^2\le C\left|\langle Tx,x\rangle \right|,$$

or equivalent as T is a self-adjoint operator, such that

$$\langle T^{2}x,x\rangle \le C\left|\langle Tx,x\rangle \right|.$$

Taking the supremun over all $x\in \mathcal{H}$ with $\parallel x\parallel =1$, we obtain

$${\parallel T\parallel }^2=\parallel TT\parallel =\underset{x\in \mathcal{H},\parallel x\parallel =1}{sup}\langle T^{2}x,x\rangle \le \underset{x\in \mathcal{H},\parallel x\parallel =1}{sup}C|\langle Tx,x\rangle |=C\parallel T\parallel .$$

Then, we conclude that $\parallel T\parallel \le C.$ □

For the subsequent discussion, it is important to recall that the expression $x\otimes y$ represents a rank one operator defined by $(x\otimes y)\left(z\right)=\langle z,y\rangle x$, where x, y, and z are vectors in the space $\mathcal{H}$. Now, we introduce the Selberg operator defined as follows:

Definition 1.

Given a subset $\mathcal{Z}=\{z_i:i=1,\dots ,n\}$ of nonzero vectors in the space $\mathcal{H}$, the Selberg operator $S_{\mathcal{Z}}$ is defined by

$$S_{\mathcal{Z}}={\displaystyle \sum _{i=1}^n}{\displaystyle \frac{z_i\otimes z_i}{{\sum }_{j=1}^n\left|\langle z_i,z_j\rangle \right|}}\in \mathcal{B}\left(\mathcal{H}\right).$$

We now offer a novel derivation of (SI), which also illustrates the optimality of certain norm estimates for positive operators. The approach is based on the properties of the operator $S_{\mathcal{Z}}$ together with Proposition 1, yielding a concise and elegant argument.

Theorem 1.

Let $\mathcal{Z}=\{z_i:i=1,\dots ,n\}$ be a subset of nonzero vectors in the Hilbert space $\mathcal{H}$. Then, the following inequality holds:

$${\displaystyle \sum _{i=1}^n}{\displaystyle \frac{|\langle x,z_i\rangle {|}^2}{{\sum }_{j=1}^n\left|\langle z_i,z_j\rangle \right|}}\le {\parallel x\parallel }^2,$$

for every $x\in \mathcal{H}$.

Proof. 

Using the Selberg operator, we can rewrite (SI) in the form

$$0\le \langle S_{\mathcal{Z}}x,x\rangle ={\displaystyle \sum _{i=1}^n}{\displaystyle \frac{|\langle x,z_i\rangle {|}^2}{{\sum }_{j=1}^n\left|\langle z_i,z_j\rangle \right|}}\le \langle x,x\rangle ,$$

for any $x\in \mathcal{H}$.

To establish (SI), it is, therefore, sufficient to show that the Selberg operator associated with $\mathcal{Z}$ is a positive contraction; that is,

$$0\le S_{\mathcal{Z}}\le I$$

for any subset $\mathcal{Z}\subseteq \mathcal{H}$ consisting of nonzero vectors.

Observe that the positivity of the Selberg operator $S_{\mathcal{Z}}$ is immediate since it is defined as a sum of positive operators.

On the other hand, for any $x\in \mathcal{H}$, we have

$$\begin{array}{ccc}\parallel S_{\mathcal{Z}}{x\parallel }^2\hfill & =\hfill & {∥{\displaystyle {\displaystyle \sum _{i=1}^n}}{\displaystyle \frac{\langle x,z_i\rangle z_i}{{\sum }_{j=1}^n\left|\langle z_i,z_j\rangle \right|}}∥}^2\hfill \\ \hfill & \le \hfill & {\displaystyle {\displaystyle \sum _{i,k=1}^n}}{\displaystyle \frac{|\langle x,z_i\rangle |}{{\sum }_{j=1}^n\left|\langle z_i,z_j\rangle \right|}}{\displaystyle \frac{|\langle x,z_k\rangle |}{{\sum }_{j=1}^n\left|\langle z_k,z_j\rangle \right|}}\left|\langle z_i,z_k\rangle \right|.\hfill \end{array}$$

(4)

Note that for non-negative real numbers $a_1,\dots ,a_n$ and $b_{ik}\ge 0$ for all $1\le i,k\le n$, the following identity holds:

$${\displaystyle \sum _{i=1}^n}{a}_i^2{\displaystyle \sum _{k=1}^n}{b}_{ik}-{\displaystyle \sum _{i,k=1}^n}{a}_i{a}_k{b}_{ik}={\displaystyle \frac{1}{2}}{\displaystyle \sum _{i,k=1}^n}{(a_i-a_k)}^2{b}_{ik}\ge 0.$$

In particular, if we consider the quantities $a_i$ and $b_{ik}$ defined by

$$a_i={\displaystyle \frac{|\langle x,z_i\rangle |}{{\sum }_{j=1}^n\left|\langle z_i,z_j\rangle \right|}},b_{ik}=\left|\langle z_i,z_k\rangle \right|,$$

then we obtain

$${\displaystyle \sum _{i,k=1}^n}{\displaystyle \frac{|\langle x,z_i\rangle |}{{\sum }_{j=1}^n\left|\langle z_i,z_j\rangle \right|}}·{\displaystyle \frac{|\langle x,z_k\rangle |}{{\sum }_{j=1}^n\left|\langle z_k,z_j\rangle \right|}}·|\langle z_i,z_k\rangle |\le {\displaystyle \sum _{i=1}^n}{\left({\displaystyle \frac{|\langle x,z_i\rangle |}{{\sum }_{j=1}^n\left|\langle z_i,z_j\rangle \right|}}\right)}^2{\displaystyle \sum _{k=1}^n}\left|\langle z_i,z_k\rangle \right|$$

(5)

By applying (5) to inequality (4), we obtain

$$\begin{array}{ccc}\parallel S_{\mathcal{Z}}{x\parallel }^2\hfill & \le \hfill & {\displaystyle {\displaystyle \sum _{i,k=1}^n}}{\displaystyle \frac{|\langle x,z_i\rangle |}{{\sum }_{j=1}^n\left|\langle z_i,z_j\rangle \right|}}·{\displaystyle \frac{|\langle x,z_k\rangle |}{{\sum }_{j=1}^n\left|\langle z_k,z_j\rangle \right|}}·\left|\langle z_i,z_k\rangle \right|\hfill \\ \hfill & \le \hfill & {\displaystyle {\displaystyle \sum _{i=1}^n}}{\left({\displaystyle \frac{|\langle x,z_i\rangle |}{{\sum }_{j=1}^n\left|\langle z_i,z_j\rangle \right|}}\right)}^2{\displaystyle \sum _{k=1}^n}\left|\langle z_i,z_k\rangle \right|\hfill \\ \hfill & =\hfill & {\displaystyle {\displaystyle \sum _{i=1}^n}}{\displaystyle \frac{|\langle x,z_i\rangle {|}^2}{{\sum }_{j=1}^n\left|\langle z_i,z_j\rangle \right|}}=\langle S_{\mathcal{Z}}x,x\rangle .\hfill \end{array}$$

(6)

Using Proposition 1, we conclude that $\parallel S_{\mathcal{Z}}\parallel \le 1$. Since $S_{\mathcal{Z}}$ is a positive operator, we have

$$\parallel S_{\mathcal{Z}}\parallel =\underset{\begin{array}{c}x\in \mathcal{H}\\ \parallel x\parallel =1\end{array}}{sup}\langle S_{\mathcal{Z}}x,x\rangle ,$$

and, therefore, $\langle S_{\mathcal{Z}}x,x\rangle \le {\parallel x\parallel }^2$ for all $x\in \mathcal{H}$, which completes the proof. □

Remark 1.

It is immediately evident that the classical Cauchy–Bunyakovsky–Schwarz inequality (CBSI) in an inner product space can be derived from the Selberg inequality (SI). However, from the previous proof, we observe that the converse also holds: the CBSI implies the SI. In conclusion, both inequalities are equivalent and can be derived from one another. This equivalence was previously observed in [26] [Theorem 23], although established there using a completely different argument.

3. More Bounds for Selberg Operator

In this section, we use the operator-theoretic approach to study the Selberg inequality (SI). From this viewpoint, we find several upper and lower bounds for the norm of the Selberg operator, showing the usefulness and depth of this method. This analysis helps us better understand the inequality and opens the door to more applications in the study of positive operators and their norms.

Theorem 2.

Let $\mathcal{Z}=\{z_i:i=1,\dots ,n\}$ be a set of nonzero vectors in a Hilbert space $\mathcal{H}$, and let $x\in \mathcal{H}$. Then, the following upper bounds hold:

$$\langle S_{\mathcal{Z}}x,x\rangle \le {\displaystyle \frac{{\sum }_{i=1}^n{\left|\langle x,z_i\rangle \right|}^2}{{\displaystyle \underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|\langle z_i,z_j\rangle \right|}}}\le {\displaystyle \frac{{\sum }_{i=1}^n{\left|\langle x,z_i\rangle \right|}^2}{{\displaystyle \underset{1\le i\le n}{min}{\parallel z_i\parallel }^2}}}.$$

Moreover, if $\langle z_i,Z_n\rangle \ne 0$ for all $i=1,\dots ,n$, where $Z_n:={\sum }_{j=1}^n{z}_j$, then the following bounds hold:

$$\begin{array}{cc}\hfill {\displaystyle \frac{{\sum }_{i=1}^n{\left|\langle x,z_i\rangle \right|}^2}{{\displaystyle \underset{1\le i\le n}{min}\left|\langle z_i,Z_n\rangle \right|}}}\ge {\displaystyle \frac{{\sum }_{i=1}^n{\left|\langle x,z_i\rangle \right|}^2}{{\displaystyle \underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|\langle z_i,z_j\rangle \right|}}}& \ge \langle S_{\mathcal{Z}}x,x\rangle \hfill \\ & \ge {\displaystyle \frac{1}{{\displaystyle \underset{1\le i\le n}{max}{\displaystyle \sum _{j=1}^n}\left|\langle z_i,z_j\rangle \right|}}}{\displaystyle \sum _{i=1}^n}{\left|\langle x,z_i\rangle \right|}^2\hfill \\ \hfill & \ge {\displaystyle \frac{1}{{\displaystyle \left({\displaystyle \sum _{j=1}^n}\parallel z_j\parallel \right)\underset{1\le i\le n}{max}\parallel z_i\parallel }}}{\displaystyle \sum _{i=1}^n}{\left|\langle x,z_i\rangle \right|}^2\hfill \\ \hfill & \ge {\displaystyle \frac{1}{{\displaystyle n\underset{1\le i\le n}{max}{\parallel z_i\parallel }^2}}}{\displaystyle \sum _{i=1}^n}{\left|\langle x,z_i\rangle \right|}^2.\hfill \end{array}$$

Proof. 

In the proof of Theorem 1 (see inequality (6)), we showed, among other results, that

$${\displaystyle {∥S_{\mathcal{Z}}x∥}^2\le {\displaystyle \sum _{i=1}^n}\frac{{\left|\langle x,z_i\rangle \right|}^2}{{\sum }_{j=1}^n\left|\langle z_i,z_j\rangle \right|}=\langle S_{\mathcal{Z}}x,x\rangle \le {\parallel x\parallel }^2,}$$

for all $x\in \mathcal{H}$.

Since

$$\langle S_{\mathcal{Z}}x,x\rangle ={\displaystyle \sum _{i=1}^n}{\displaystyle \frac{{\left|\langle x,z_i\rangle \right|}^2}{{\sum }_{j=1}^n\left|\langle z_i,z_j\rangle \right|}}\le {\displaystyle \frac{{\sum }_{i=1}^n{\left|\langle x,z_i\rangle \right|}^2}{{\displaystyle \underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|\langle z_i,z_j\rangle \right|}}},$$

and noting that

$${\displaystyle \sum _{j=1}^n}\left|\langle z_i,z_j\rangle \right|\ge \left|\langle z_i,z_i\rangle \right|={\parallel z_i\parallel }^2,i=1,\dots ,n,$$

it follows that

$$\underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|\langle z_i,z_j\rangle \right|\ge \underset{1\le i\le n}{min}{\parallel z_i\parallel }^2.$$

Hence, we obtain the simpler upper bound

$$\langle S_{\mathcal{Z}}x,x\rangle \le {\displaystyle \frac{{\displaystyle {\displaystyle \sum _{i=1}^n}{\left|\langle x,z_i\rangle \right|}^2}}{{\displaystyle \underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|\langle z_i,z_j\rangle \right|}}}\le {\displaystyle \frac{{\displaystyle {\displaystyle \sum _{i=1}^n}{\left|\langle x,z_i\rangle \right|}^2}}{{\displaystyle \underset{1\le i\le n}{min}{\parallel z_i\parallel }^2}}},x\in \mathcal{H},$$

(7)

where $z_i\ne 0$ for all $i=1,\dots ,n$.

Next, by the triangle inequality for the modulus, we have

$${\displaystyle \sum _{j=1}^n}\left|\langle z_i,z_j\rangle \right|\ge \left|{\displaystyle \sum _{j=1}^n}\langle z_i,z_j\rangle \right|=\left|\langle z_i,Z_n\rangle \right|,$$

where $Z_n={\sum }_{j=1}^n{z}_j$.

This implies

$$\underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|\langle z_i,z_j\rangle \right|\ge \underset{1\le i\le n}{min}\left|\langle z_i,Z_n\rangle \right|.$$

If $\langle z_i,Z_n\rangle \ne 0$ for all $i=1,\dots ,n$, then we also have the upper bound

$$\langle S_{\mathcal{Z}}x,x\rangle \le {\displaystyle \frac{{\displaystyle {\displaystyle \sum _{i=1}^n}{\left|\langle x,z_i\rangle \right|}^2}}{{\displaystyle \underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|\langle z_i,z_j\rangle \right|}}}\le {\displaystyle \frac{{\sum }_{i=1}^n{\left|\langle x,z_i\rangle \right|}^2}{{\displaystyle \underset{1\le i\le n}{min}\left|\langle z_i,Z_n\rangle \right|}}},x\in \mathcal{H}.$$

Furthermore, by the (CBSI) in $\mathcal{H}$, we have

$${\displaystyle \sum _{j=1}^n}\left|\langle z_i,z_j\rangle \right| \le \parallel z_i\parallel {\displaystyle \sum _{j=1}^n}\parallel z_j\parallel ,$$

which yields

$$\underset{1\le i\le n}{max}{\displaystyle \sum _{j=1}^n}\left|\langle z_i,z_j\rangle \right|\le \left(\underset{1\le i\le n}{max}\parallel z_i\parallel \right){\displaystyle \sum _{j=1}^n}\parallel z_j\parallel \le n\underset{1\le i\le n}{max}{\parallel z_i\parallel }^2.$$

Hence, the following lower bound holds:

$$\begin{array}{cc}\hfill \langle S_{\mathcal{Z}}x,x\rangle & \ge {\displaystyle \frac{1}{{\displaystyle \underset{1\le i\le n}{max}{\displaystyle \sum _{j=1}^n}\left|\langle z_i,z_j\rangle \right|}}}{\displaystyle \sum _{i=1}^n}{\left|\langle x,z_i\rangle \right|}^2\hfill \\ \hfill & \ge {\displaystyle \frac{1}{{\displaystyle \left({\displaystyle \underset{1\le i\le n}{max}\parallel z_i\parallel }\right){\displaystyle \sum _{j=1}^n}\parallel z_j\parallel }}}{\displaystyle \sum _{i=1}^n}{\left|\langle x,z_i\rangle \right|}^2\hfill \\ \hfill & \ge {\displaystyle \frac{1}{{\displaystyle n\underset{1\le i\le n}{max}{\parallel z_i\parallel }^2}}}{\displaystyle \sum _{i=1}^n}{\left|\langle x,z_i\rangle \right|}^2,\hfill \end{array}$$

(8)

for any $x\in \mathcal{H}$.

□

We can provide now other upper bounds as follows:

Theorem 3.

Let $\mathcal{Z}=\{z_i:i=1,\dots ,n\}$ be a set of nonzero vectors in a Hilbert space $\mathcal{H}$, and let $x\in \mathcal{H}$. Then, the following upper bounds hold:

$${∥S_{\mathcal{Z}}x∥}^2\le {\displaystyle \frac{\underset{1\le i\le n}{max}{\left|〈x,z_i〉\right|}^2}{{\left({\displaystyle \underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|〈z_i,z_j〉\right|}\right)}^2}}\left[{\displaystyle \sum _{i=1}^n}{∥z_i∥}^2+\sum _{1\le i\ne j\le n}\left|〈z_i,z_j〉\right|\right]$$

and

$$\begin{array}{cc}\hfill {∥S_{\mathcal{Z}}x∥}^2& \le {\displaystyle \frac{{\left({\displaystyle {\displaystyle \sum _{i=1}^n}}{\left|〈x,z_i〉\right|}^{2p}\right)}^{\frac{1}{p}}}{{\left({\displaystyle \underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|〈z_i,z_j〉\right|}\right)}^2}}{\beta }_{n,q}\hfill \end{array}$$

(9)

where

$${\beta }_{n,q}=\left[{\left({\displaystyle \sum _{i=1}^n}{\parallel z_i\parallel }^{2q}\right)}^{\frac{1}{q}}+{(n-1)}^{\frac{1}{p}}{\left({\displaystyle \sum _{\begin{array}{c}i,j=1\\ i\ne j\end{array}}^n}{\left|\langle z_i,z_j\rangle \right|}^q\right)}^{\frac{1}{q}}\right],$$

with $p>1$ and q such that $\frac{1}{p}+\frac{1}{q}=1$.

Also, we have that

$$\begin{array}{cc}\hfill {∥S_{\mathcal{Z}}x∥}^2& \le {\displaystyle \frac{1}{{\left({\displaystyle \underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|〈z_i,z_j〉\right|}\right)}^2}}{\displaystyle \sum _{i=1}^n}{\left|〈x,z_i〉\right|}^2\hfill \\ \hfill & \times \left\{\underset{1\le i\le n}{max}{∥z_i∥}^2+{\left(\sum _{1\le i\ne j\le n}{\left|\left(z_i,z_j\right)\right|}^2\right)}^{\frac{1}{2}}\right\}.\hfill \end{array}$$

(10)

Proof. 

We start by using the following inequality, which states that for any scalars ${\alpha }_i\in \mathbb{C}$ and vectors $z_i\in \mathcal{H}$, $i=1,\dots ,n$, the following holds:

$$\begin{array}{c}\hfill {∥{\displaystyle \sum _{i=1}^n}{\alpha }_i{z}_i∥}^2\le \underset{1\le i\le n}{max}{\left|{\alpha }_i\right|}^2\left[{\displaystyle \sum _{i=1}^n}\parallel z_i{\parallel }^2+{\displaystyle \sum _{\begin{array}{c}i,j=1\\ i\ne j\end{array}}^n}\left|\langle z_i,z_j\rangle \right|\right].\end{array}$$

Choosing ${\alpha }_i={\displaystyle \frac{\langle x,z_i\rangle }{{\sum }_{j=1}^n\left|\langle z_i,z_j\rangle \right|}},i=1,\dots ,n,$ and applying the inequality above, we obtain

$$\begin{array}{cc}\hfill {∥{\displaystyle \sum _{i=1}^n}{\displaystyle \frac{\langle x,z_i\rangle }{{\sum }_{j=1}^n\left|\langle z_i,z_j\rangle \right|}}{z}_i∥}^2& \le \underset{1\le i\le n}{max}{\left|{\displaystyle \frac{\langle x,z_i\rangle }{{\sum }_{j=1}^n\left|\langle z_i,z_j\rangle \right|}}\right|}^2\left[{\displaystyle \sum _{i=1}^n}\parallel z_i{\parallel }^2+{\displaystyle \sum _{\begin{array}{c}i,j=1\\ i\ne j\end{array}}^n}\left|\langle z_i,z_j\rangle \right|\right]\hfill \\ \hfill & \le {\displaystyle \frac{{\displaystyle \underset{1\le i\le n}{max}{\left|\langle x,z_i\rangle \right|}^2}}{{\left({\displaystyle \underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|\langle z_i,z_j\rangle \right|}\right)}^2}}\left[{\displaystyle \sum _{i=1}^n}\parallel z_i{\parallel }^2+{\displaystyle \sum _{\begin{array}{c}i,j=1\\ i\ne j\end{array}}^n}\left|\langle z_i,z_j\rangle \right|\right].\hfill \end{array}$$

In the same reference [27], the author also established a related inequality valid for any $p>1$ with $\frac{1}{p}+\frac{1}{q}=1$:

$$\begin{array}{c}\hfill {∥{\displaystyle \sum _{i=1}^n}{\alpha }_i{z}_i∥}^2\le {\left({\displaystyle \sum _{i=1}^n}{\left|{\alpha }_i\right|}^{2p}\right)}^{\frac{1}{p}}\left[{\left({\displaystyle \sum _{i=1}^n}{\parallel z_i\parallel }^{2q}\right)}^{\frac{1}{q}}+{(n-1)}^{\frac{1}{p}}{\left({\displaystyle \sum _{\begin{array}{c}i,j=1\\ i\ne j\end{array}}^n}{\left|\langle z_i,z_j\rangle \right|}^q\right)}^{\frac{1}{q}}\right].\end{array}$$

Again, inserting ${\alpha }_i={\displaystyle \frac{\langle x,z_i\rangle }{{\sum }_{j=1}^n\left|\langle z_i,z_j\rangle \right|}},i=1,\dots ,n,$ into this inequality yields

$$\begin{array}{cc}\hfill \parallel S_{\mathcal{Z}}{x\parallel }^2& \le {\left({\displaystyle \sum _{i=1}^n}{\displaystyle \frac{|\langle x,z_i\rangle {|}^{2p}}{{\left({\sum }_{j=1}^n\left|\langle z_i,z_j\rangle \right|\right)}^{2p}}}\right)}^{\frac{1}{p}}{\beta }_{n,q}\hfill \\ \hfill & \le {\displaystyle \frac{{\left({\displaystyle {\displaystyle \sum _{i=1}^n}{\left|\langle x,z_i\rangle \right|}^{2p}}\right)}^{\frac{1}{p}}}{{\left({\displaystyle \underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|\langle z_i,z_j\rangle \right|}\right)}^2}}{\beta }_{n,q},\hfill \end{array}$$

which establishes inequality (9).

Moreover, from another inequality found in [27]

$$\begin{array}{c}\hfill {∥{\displaystyle \sum _{i=1}^n}{\alpha }_i{z}_i∥}^2\le {\displaystyle \sum _{i=1}^n}{\left|{\alpha }_i\right|}^2\left\{\underset{1\le i\le n}{max}{\parallel z_i\parallel }^2+{\left({\displaystyle \sum _{\begin{array}{c}i,j=1\\ i\ne j\end{array}}^n}{\left|\langle z_i,z_j\rangle \right|}^2\right)}^{\frac{1}{2}}\right\},\end{array}$$

we obtain

$$\begin{array}{cc}\hfill \parallel S_{\mathcal{Z}}{x\parallel }^2& \le {\displaystyle \sum _{i=1}^n}{\displaystyle \frac{|\langle x,z_i\rangle {|}^2}{{\left({\sum }_{j=1}^n\left|\langle z_i,z_j\rangle \right|\right)}^2}}\left\{\underset{1\le i\le n}{max}{\parallel z_i\parallel }^2+{\left({\displaystyle \sum _{\begin{array}{c}i,j=1\\ i\ne j\end{array}}^n}{\left|\langle z_i,z_j\rangle \right|}^2\right)}^{\frac{1}{2}}\right\}\hfill \\ \hfill & \le {\displaystyle \frac{1}{{\left({\displaystyle \underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|\langle z_i,z_j\rangle \right|}\right)}^2}}{\displaystyle \sum _{i=1}^n}{\left|\langle x,z_i\rangle \right|}^2\hfill \\ \hfill & \times \left\{\underset{1\le i\le n}{max}{\parallel z_i\parallel }^2+{\left({\displaystyle \sum _{\begin{array}{c}i,j=1\\ i\ne j\end{array}}^n}{\left|\langle z_i,z_j\rangle \right|}^2\right)}^{\frac{1}{2}}\right\}.\hfill \end{array}$$

This finishes the proof. □

Remark 2.

We observe that for $p=q=2$, we obtain

$$\begin{array}{cc}\hfill {∥S_{\mathcal{Z}}x∥}^2& \le {\displaystyle \frac{{\left({\displaystyle {\displaystyle \sum _{i=1}^n}}{\left|〈x,z_i〉\right|}^4\right)}^{\frac{1}{4}}}{{\left({\displaystyle \underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|〈z_i,z_j〉\right|}\right)}^2}}\hfill \\ \hfill & \times \left[{\left({\displaystyle \sum _{i=1}^n}{∥z_i∥}^4\right)}^{\frac{1}{4}}+{\left(n-1\right)}^{\frac{1}{2}}{\left(\sum _{1\le i\ne j\le n}{\left|\left(z_i,z_j\right)\right|}^2\right)}^{\frac{1}{2}}\right].\hfill \end{array}$$

Following [28], we introduce the hypo-q-norms on the Cartesian product of inner product spaces. For any $\mathbf{x}=(x_1,\dots ,x_n)\in {\mathcal{H}}^n$, these norms are defined by

$${\parallel \mathbf{x}\parallel }_{h,n,q}:=\underset{\parallel y\parallel =1}{sup}{\left({\displaystyle \sum _{j=1}^n}{\left|\langle x_j,y\rangle \right|}^q\right)}^{\frac{1}{q}},$$

where $q\ge 1$. Additionally, we define the norm

$${\parallel \mathbf{x}\parallel }_{h,n,\infty }:=\underset{1\le j\le n}{max}\parallel x_j\parallel .$$

In particular, the hypo-Euclidean norm corresponds to the case $q=2$, given by

$${\parallel \mathbf{x}\parallel }_{h,e}:=\underset{\parallel y\parallel =1}{sup}{\left({\displaystyle \sum _{j=1}^n}{\left|\langle x_j,y\rangle \right|}^2\right)}^{\frac{1}{2}}.$$

Corollary 1.

Let $\mathcal{Z}=\{z_i:i=1,\dots ,n\}$ be a set of nonzero vectors in a Hilbert space $\mathcal{H}$. Then, the following inequalities hold:

$${\displaystyle \frac{{\parallel \mathbf{z}\parallel }_{h,e}^2}{{\displaystyle \underset{1\le i\le n}{max}{\displaystyle \sum _{j=1}^n}\left|\langle z_i,z_j\rangle \right|}}}\le \parallel S_{\mathcal{Z}}\parallel \le {\displaystyle \frac{{\parallel \mathbf{z}\parallel }_{h,e}^2}{{\displaystyle \underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|\langle z_i,z_j\rangle \right|}}},$$

(11)

and

$$\begin{array}{cc}\hfill \parallel S_{\mathcal{Z}}\parallel & \le {\displaystyle \frac{{\parallel \mathbf{z}\parallel }_{h,2p}}{{\displaystyle \underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|\langle z_i,z_j\rangle \right|}}}{\gamma }_{n,p},\hfill \end{array}$$

(12)

where

$${\gamma }_{n,p}={\left[{\left({\displaystyle \sum _{i=1}^n}{\parallel z_i\parallel }^{2q}\right)}^{\frac{1}{q}}+{(n-1)}^{\frac{1}{p}}{\left(\sum _{\begin{array}{c}1\le i,j\le n\\ i\ne j\end{array}}{\left|\langle z_i,z_j\rangle \right|}^q\right)}^{\frac{1}{q}}\right]}^{\frac{1}{2}},$$

$\mathbf{z}=(z_1,\dots ,z_n)$, with $p>1$ and q such that $\frac{1}{p}+\frac{1}{q}=1$.

In particular,

$$\begin{array}{cc}\hfill \parallel S_{\mathcal{Z}}\parallel & \le {\displaystyle \frac{{\parallel \mathbf{z}\parallel }_{h,4}}{{\displaystyle \underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|\langle z_i,z_j\rangle \right|}}}{\gamma }_{n,2}.\hfill \end{array}$$

Furthermore,

$$\begin{array}{cc}\hfill {∥S_{\mathcal{Z}}∥}^2& \le {\displaystyle \frac{{∥\mathbf{z}∥}_{h,e}^2}{{\left({\displaystyle \underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|〈z_i,z_j〉\right|}\right)}^2}}\left\{\underset{1\le i\le n}{max}{∥z_i∥}^2+{\left(\sum _{1\le i\ne j\le n}{\left|\left(z_i,z_j\right)\right|}^2\right)}^{\frac{1}{2}}\right\},\hfill \end{array}$$

Proof. 

If we take the supremum over all $x\in \mathcal{H}$ with $\parallel x\parallel =1$ in (7), then we obtain

$$∥S_{\mathcal{Z}}∥=\underset{∥x∥=1}{sup}〈S_{\mathcal{Z}}x,x〉\le {\displaystyle \frac{{\displaystyle \underset{∥x∥=1}{sup}{\displaystyle \sum _{i=1}^n}{\left|〈x,z_i〉\right|}^2}}{\underset{1\le i\le n}{min}{\sum }_{j=1}^n\left|〈z_i,z_j〉\right|}}={\displaystyle \frac{{∥\mathbf{z}∥}_{h,e}^2}{{\displaystyle \underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|〈z_i,z_j〉\right|}}},$$

which gives the second inequality in (11). The first inequality follows in a similar way from (8).

Taking the supremum over all $x\in \mathcal{H}$ with $\parallel x\parallel =1$ in (9), we obtain

$$\begin{array}{cc}\hfill {∥S_{\mathcal{Z}}∥}^2& \le {\displaystyle \frac{{\displaystyle \underset{∥x∥=1}{sup}{\left({\displaystyle \sum _{i=1}^n}{\left|〈x,z_i〉\right|}^{2p}\right)}^{\frac{1}{p}}}}{{\left({\displaystyle \underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|〈z_i,z_j〉\right|}\right)}^2}}{\gamma }_{n,p}={\displaystyle \frac{{∥\mathbf{z}∥}_{h,2p}^2}{{\left({\displaystyle \underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|〈z_i,z_j〉\right|}\right)}^2}}{\gamma }_{n,p},\hfill \end{array}$$

which gives (12).

Finally, from (10), we obtain

$$\begin{array}{cc}\hfill {∥S_{\mathcal{Z}}∥}^2& \le {\displaystyle \frac{{∥\mathbf{z}∥}_{h,e}^2}{{\left({\displaystyle \underset{1\le i\le n}{min}{\displaystyle \sum _{j=1}^n}\left|〈z_i,z_j〉\right|}\right)}^2}}\left\{\underset{1\le i\le n}{max}{∥z_i∥}^2+{\left(\sum _{1\le i\ne j\le n}{\left|\left(z_i,z_j\right)\right|}^2\right)}^{\frac{1}{2}}\right\},\hfill \end{array}$$

which finishes the proof. □

The following result also holds.

Theorem 4.

Let $\mathcal{Z}=\{z_i:i=1,\dots ,n\}$ be a set of nonzero vectors in a Hilbert space $\mathcal{H}$ and $x\in \mathcal{H}$, we have

$${∥S_{\mathcal{Z}}x∥}^2\le \left\{\begin{array}{c}\underset{1\le i\le n}{max}{\left|〈x,z_i〉\right|}^2{\displaystyle \frac{{\displaystyle {\displaystyle \sum _{i,j=1}^n}\left|\left(z_i,z_j\right)\right|}}{{\left({\displaystyle \underset{1\le i,j\le n}{min}{\displaystyle \sum _{j=1}^n}\left|〈z_i,z_j〉\right|}\right)}^2}};\hfill \\ \\ {\left({\displaystyle {\displaystyle \sum _{i=1}^n}{\left|〈x,z_i〉\right|}^p}\right)}^{\frac{2}{p}}{\displaystyle \frac{{\left({\displaystyle {\displaystyle \sum _{i,j=1}^n}{\left|\left(z_i,z_j\right)\right|}^q}\right)}^{\frac{1}{q}}}{{\left({\displaystyle \underset{1\le i,j\le n}{min}{\displaystyle \sum _{j=1}^n}\left|〈z_i,z_j〉\right|}\right)}^2}},\hfill \\ \\ {\left({\displaystyle {\displaystyle \sum _{i=1}^n}\left|〈x,z_i〉\right|}\right)}^2{\displaystyle \frac{\underset{1\le i,j\le n}{max}\left|\left(z_i,z_j\right)\right|}{{\left({\displaystyle \underset{1\le i,j\le n}{min}{\displaystyle \sum _{j=1}^n}\left|〈z_i,z_j〉\right|}\right)}^2}},\hfill \end{array}\right.$$

(13)

with $p>1$ and q such that $\frac{1}{p}+\frac{1}{q}=1$. In particular,

$${∥S_{\mathcal{Z}}x∥}^2\le {\displaystyle \sum _{i=1}^n}{\left|〈x,z_i〉\right|}^2{\displaystyle \frac{{\left({\displaystyle {\displaystyle \sum _{i,j=1}^n}{\left|\left(z_i,z_j\right)\right|}^2}\right)}^{\frac{1}{2}}}{{\left({\displaystyle \underset{1\le i,j\le n}{min}{\displaystyle \sum _{j=1}^n}\left|〈z_i,z_j〉\right|}\right)}^2}}.$$

Proof. 

Let $z_1,\dots ,z_n\in \mathcal{H}$ and ${\alpha }_1,\dots ,{\alpha }_n\in \mathbb{C}$. In [29], the author obtained the following inequalities for the norm of a linear combination of vectors:

$${∥{\displaystyle \sum _{i=1}^n}{\alpha }_i{z}_i∥}^2\le \left\{\begin{array}{c}\underset{1\le i\le n}{max}{\left|{\alpha }_i\right|}^2{\displaystyle {\sum }_{i,j=1}^n}\left|\left(z_i,z_j\right)\right|;\hfill \\ \\ {\left({\displaystyle {\sum }_{i=1}^n}{\left|{\alpha }_i\right|}^p\right)}^{\frac{2}{p}}{\left({\displaystyle {\sum }_{i,j=1}^n}{\left|\left(z_i,z_j\right)\right|}^q\right)}^{\frac{1}{q}},\hfill \\ \\ {\left({\displaystyle {\sum }_{i=1}^n}\left|{\alpha }_i\right|\right)}^2\underset{1\le i,j\le n}{max}\left|\left(z_i,z_j\right)\right|.\hfill \end{array}\right.$$

(14)

where $p>1,\frac{1}{p}+\frac{1}{q}=1$.

Now, if we take in (14) ${\alpha }_i={\displaystyle \frac{〈x,z_i〉}{{\sum }_{j=1}^n\left|〈z_i,z_j〉\right|}},$ for any $i=1,\cdots ,n$ then we obtain

$$\begin{array}{ccc}\hfill {∥{\displaystyle \sum _{i=1}^n}{\displaystyle \frac{〈x,z_i〉}{{\sum }_{j=1}^n\left|〈z_i,z_j〉\right|}}{z}_i∥}^2& \le & \left\{\begin{array}{c}\underset{1\le i\le n}{max}{\left|{\displaystyle \frac{〈x,z_i〉}{{\sum }_{j=1}^n\left|〈z_i,z_j〉\right|}}\right|}^2{\displaystyle {\sum }_{i,j=1}^n}\left|\left(z_i,z_j\right)\right|;\hfill \\ \\ {\left({\displaystyle {\sum }_{i=1}^n}{\left|{\displaystyle \frac{〈x,z_i〉}{{\sum }_{j=1}^n\left|〈z_i,z_j〉\right|}}\right|}^p\right)}^{\frac{2}{p}}{\left({\displaystyle {\sum }_{i,j=1}^n}{\left|\left(z_i,z_j\right)\right|}^q\right)}^{\frac{1}{q}},\hfill \\ \\ {\left({\displaystyle {\sum }_{i=1}^n}\left|{\displaystyle \frac{〈x,z_i〉}{{\sum }_{j=1}^n\left|〈z_i,z_j〉\right|}}\right|\right)}^2\underset{1\le i,j\le n}{max}\left|\left(z_i,z_j\right)\right|;\hfill \end{array}\right.\hfill \\ & \le & \left\{\begin{array}{c}\underset{1\le i\le n}{max}{\left|〈x,z_i〉\right|}^2{\displaystyle \frac{{\sum }_{i,j=1}^n\left|\left(z_i,z_j\right)\right|}{{\left(\underset{1\le i,j\le n}{min}{\sum }_{j=1}^n\left|〈z_i,z_j〉\right|\right)}^2}};\hfill \\ \\ {\left({\displaystyle {\sum }_{i=1}^n}{\left|〈x,z_i〉\right|}^p\right)}^{\frac{2}{p}}{\displaystyle \frac{{\left({\sum }_{i,j=1}^n{\left|\left(z_i,z_j\right)\right|}^q\right)}^{\frac{1}{q}}}{{\left(\underset{1\le i,j\le n}{min}{\sum }_{j=1}^n\left|〈z_i,z_j〉\right|\right)}^2}},\hfill \\ \\ {\left({\displaystyle {\sum }_{i=1}^n}\left|〈x,z_i〉\right|\right)}^2{\displaystyle \frac{\underset{1\le i,j\le n}{max}\left|\left(z_i,z_j\right)\right|}{{\left(\underset{1\le i,j\le n}{min}{\sum }_{j=1}^n\left|〈z_i,z_j〉\right|\right)}^2}};\hfill \end{array}\right.\hfill \end{array}$$

with $p>1$ and q such that $\frac{1}{p}+\frac{1}{q}=1$.

Finally, since ${\displaystyle S_{\mathcal{Z}}x={\displaystyle \sum _{i=1}^n}\frac{〈x,z_i〉}{{\sum }_{j=1}^n\left|〈z_i,z_j〉\right|}{z}_i,}$ inequality (13) is then established.

□

Corollary 2.

Let $\mathcal{Z}=\{z_i:i=1,\dots ,n\}$ be a set of nonzero vectors in a Hilbert space $\mathcal{H}$. Then, the following inequality holds:

$$∥S_{\mathcal{Z}}∥\le \left\{\begin{array}{c}\underset{1\le i\le n}{max}∥z_i∥{\displaystyle \frac{{\left({\sum }_{i,j=1}^n\left|\left(z_i,z_j\right)\right|\right)}^{1/2}}{\underset{1\le i,j\le n}{min}{\sum }_{j=1}^n\left|〈z_i,z_j〉\right|}},\hfill \\ \\ {∥\mathbf{z}∥}_{h,p}{\displaystyle \frac{{\left({\sum }_{i,j=1}^n{\left|\left(z_i,z_j\right)\right|}^q\right)}^{\frac{1}{2q}}}{\underset{1\le i,j\le n}{min}{\sum }_{j=1}^n\left|〈z_i,z_j〉\right|}},\hfill \\ \\ {∥\mathbf{z}∥}_{h,1}{\displaystyle \frac{\underset{1\le i,j\le n}{max}{\left|\left(z_i,z_j\right)\right|}^{1/2}}{\underset{1\le i,j\le n}{min}{\sum }_{j=1}^n\left|〈z_i,z_j〉\right|}},\hfill \end{array}\right.$$

where $\mathbf{z}=\left(z_1,\dots ,z_n\right)$, $p>1$ and q such that $\frac{1}{p}+\frac{1}{q}=1$.

In particular,

$$∥S_{\mathcal{Z}}∥\le {∥\mathbf{z}∥}_{h,2}{\displaystyle \frac{{\left({\sum }_{i,j=1}^n{\left|\left(z_i,z_j\right)\right|}^2\right)}^{\frac{1}{4}}}{\underset{1\le i,j\le n}{min}{\sum }_{j=1}^n\left|〈z_i,z_j〉\right|}}.$$

Proof. 

The result follows immediately from Theorem 4 by taking the supremum over all $x\in \mathcal{H}$ with $\parallel x\parallel =1$. □

4. Applications: Refinements of de Bruijn’s and Bohr’s Inequalities

In this section, we highlight the strength and significance of the inequality—and its generalization—recently established by some of the authors of this work. In particular, we show that this result enables the recovery of a refined version of the Cauchy–Schwarz inequality, originally introduced by de Bruijn in 1960 [30].

We begin by recalling a result that serves as a simultaneous extension of both the Selberg and Buzano inequalities, obtained by Fujii et al. [14] [Theorem 2.3]. Let $\mathcal{Z}=\{z_i:i=1,\dots ,n\}$ be a collection of nonzero vectors in a Hilbert space $\mathcal{H}$, and let $u,v\in \mathcal{H}$ be vectors orthogonal to each element of $\mathcal{Z}$; that is, $\langle u,z_i\rangle =\langle v,z_i\rangle =0$ for all $i=1,\dots ,n$. Under these assumptions, the following inequality holds for all $w\in \mathcal{H}$:

$$\left|\langle u,w\rangle \langle w,v\rangle \right|+B(u,v)\langle S_{\mathcal{Z}}w,w\rangle \le B(u,v){\parallel w\parallel }^2,$$

(15)

where the constant $B(u,v)$ is defined by

$$B(u,v)={\displaystyle \frac{1}{2}}\left(\parallel u\parallel \parallel v\parallel +|\langle u,v\rangle |\right).$$

Now, we derive a refinement of de Bruijn’s inequality. To this end, we begin by presenting a preparatory result that will play a key role in establishing a refined version of de Bruijn’s inequality. This auxiliary statement, which builds on additional orthogonality assumptions, provides a sharper variant of (CBSI).

Lemma 3.

Let $u,v,w\in \mathcal{H}$ be elements of a Hilbert space such that $\parallel u\parallel =\parallel v\parallel$ and $|\langle w,v\rangle |=|\langle u,w\rangle |$. Let $\mathcal{Z}=\{z_i:i=1,\dots ,n\}$ be a collection of nonzero vectors in $\mathcal{H}$ satisfying $\langle u,z_i\rangle =\langle v,z_i\rangle =0$ for all $i=1,\dots ,n$. Then, the following inequality holds:

$${\left|\langle u,w\rangle \right|}^2\le {\left|\langle u,w\rangle \right|}^2+B(u,v)\langle S_{\mathcal{Z}}w,w\rangle \le {\displaystyle \frac{1}{2}}{\parallel w\parallel }^2\left[{\parallel u\parallel }^2+\left|\langle u,v\rangle \right|\right].$$

(16)

Furthermore, if the set $\{u,v\}$ is linearly independent, the equality

$${\left|\langle u,w\rangle \right|}^2={\displaystyle \frac{1}{2}}{\parallel w\parallel }^2{[\parallel u\parallel }^2+\left|\langle u,v\rangle \right|]$$

(17)

is satisfied for $w\in \mathcal{H}$ if and only if $w=\alpha \parallel u\parallel (u+e^{i\theta }v)$ for some scalar α, where $\theta =arg〈u,v〉$. Conversely, if $\{u,v\}$ is linearly dependent, the equality in (17) holds for $w\in \mathcal{H}$ if and only if $w=\alpha u$ for some scalar α.

Proof. 

By (15), we have that

$$\left|\langle u,w\rangle \langle w,v\rangle \right|+B(u,v)\langle S_{\mathcal{Z}}w,w\rangle \le B(u,v){\parallel w\parallel }^2.$$

Using the hypothesis, we conclude that

$$\begin{array}{c}\hfill {\left|\langle u,w\rangle \right|}^2\le {\left|\langle u,w\rangle \right|}^2+B(u,v)\langle S_{\mathcal{Z}}w,w\rangle \le {\displaystyle \frac{1}{2}}{\parallel w\parallel }^2{[\parallel u\parallel }^2+|\langle u,v\rangle \left|\right].\end{array}$$

Finally, we obtain the equality in the previous inequality if and only if we have the equality in Buzano’s inequality. Then, we complete the proof from Lemma 1. □

We present a new proof of de Bruijn’s inequality, together with a refinement based on the preceding result.

Corollary 3.

Let $w_1,\dots ,w_n$ be a sequence of real numbers and $u_1,\dots ,u_n$ a sequence of complex numbers. Suppose that $\mathcal{Z}=\{{\mathbf{z}}_i:i=1,\dots ,n\}$ is a collection of nonzero vectors in ${\mathbb{C}}^n$ such that

$${\displaystyle \sum _{k=1}^n}{u}_k\overline{{\left(z_i\right)}_k}=0={\displaystyle \sum _{k=1}^n}\overline{u_k}\overline{{\left(z_i\right)}_k}\mathit{for}\mathit{all}i=1,\dots ,n.$$

Then, the following inequality holds:

$$\begin{array}{cc}\hfill {\left|{\displaystyle \sum _{k=1}^n}{w}_k{u}_k\right|}^2& \le {\left|{\displaystyle \sum _{k=1}^n}{w}_k{u}_k\right|}^2+{\displaystyle \frac{1}{2}}\left[{\displaystyle \sum _{k=1}^n}{\left|u_k\right|}^2+\left|{\displaystyle \sum _{k=1}^n}{u}_k^2\right|\right]{\displaystyle \sum _{i=1}^n}{\displaystyle \frac{{\left|{\sum }_{k=1}^n{w}_k\overline{{\left(z_i\right)}_k}\right|}^2}{{\sum }_{j=1}^n\left|{\sum }_{k=1}^n{\left(z_i\right)}_k\overline{{\left(z_j\right)}_k}\right|}}\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}{\displaystyle \sum _{k=1}^n}{w}_k^2\left[{\displaystyle \sum _{k=1}^n}{\left|u_k\right|}^2+\left|{\displaystyle \sum _{k=1}^n}{u}_k^2\right|\right].\hfill \end{array}$$

Moreover, equality

$${\left|{\displaystyle \sum _{k=1}^n}{w}_k{u}_k\right|}^2={\displaystyle \frac{1}{2}}{\displaystyle \sum _{k=1}^n}{w}_k^2\left[{\displaystyle \sum _{k=1}^n}{\left|u_k\right|}^2+\left|{\displaystyle \sum _{k=1}^n}{u}_k^2\right|\right],$$

(18)

holds if and only if there exists $\lambda \in \mathbb{C}$ such that $w_k=Re\left(\lambda u_k\right)$ for all $k=1,\dots ,n$, and ${\sum }_{k=1}^n{\lambda }^2{u}_k^2\ge 0$.

Proof. 

We consider ${\mathbb{C}}^n$ as a Hilbert space equipped with the standard inner product

$$\langle \mathbf{x},\mathbf{y}\rangle ={\displaystyle \sum _{k=1}^n}{x}_k\overline{y_k},$$

where $\mathbf{x}=(x_1,\dots ,x_n)$ and $\mathbf{y}=(y_1,\dots ,y_n)$ are elements of ${\mathbb{C}}^n$.

Let $\mathbf{w}=(w_1,\dots ,w_n)$, $\mathbf{u}=(u_1,\dots ,u_n)$, and $\mathbf{v}=(\overline{u_1},\dots ,\overline{u_n})$ be vectors in ${\mathbb{C}}^n$. Then, it holds that

$$\parallel \mathbf{u}\parallel =\parallel \mathbf{v}\parallel ={\left({\displaystyle \sum _{k=1}^n}{\left|u_k\right|}^2\right)}^{1/2},$$

and the inner products satisfy

$$\left|\langle \mathbf{u},\mathbf{w}\rangle \right|=\left|{\displaystyle \sum _{k=1}^n}{w}_k{u}_k\right|=\left|{\displaystyle \sum _{k=1}^n}{w}_k\overline{u_k}\right|=\left|\langle \mathbf{v},\mathbf{w}\rangle \right|=\left|\langle \mathbf{w},\mathbf{v}\rangle \right|.$$

Now, we obtain

$$\left|\langle \mathbf{u},\mathbf{w}\rangle \right|=\left|{\displaystyle \sum _{k=1}^n}{w}_k{u}_k\right|,B(\mathbf{u},\mathbf{v})={\displaystyle \frac{1}{2}}\left[{\displaystyle \sum _{k=1}^n}{\left|u_k\right|}^2+\left|{\displaystyle \sum _{k=1}^n}{u}_k^2\right|\right],$$

and for each $i=1,\dots ,n$,

$$\left|\langle \mathbf{u},{\mathbf{z}}_i\rangle \right|=\left|\langle \mathbf{v},{\mathbf{z}}_i\rangle \right|=0,$$

as well as

$$\langle S_{\mathcal{Z}}\mathbf{w},\mathbf{w}\rangle ={\displaystyle \sum _{i=1}^n}{\displaystyle \frac{{\left|{\sum }_{k=1}^n{w}_k\overline{{\left(z_i\right)}_k}\right|}^2}{{\sum }_{j=1}^n\left|{\sum }_{k=1}^n{\left(z_i\right)}_k\overline{{\left(z_j\right)}_k}\right|}}.$$

Substituting all of this into inequality (16), we obtain

$$\begin{array}{cc}\hfill {\left|{\displaystyle \sum _{k=1}^n}{w}_k{u}_k\right|}^2& \le {\left|{\displaystyle \sum _{k=1}^n}{w}_k{u}_k\right|}^2+{\displaystyle \frac{1}{2}}\left[{\displaystyle \sum _{k=1}^n}{\left|u_k\right|}^2+\left|{\displaystyle \sum _{k=1}^n}{u}_k^2\right|\right]{\displaystyle \sum _{i=1}^n}{\displaystyle \frac{{\left|{\sum }_{k=1}^n{w}_k\overline{{\left(z_i\right)}_k}\right|}^2}{{\sum }_{j=1}^n\left|{\sum }_{k=1}^n{\left(z_i\right)}_k\overline{{\left(z_j\right)}_k}\right|}}\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}{\displaystyle \sum _{k=1}^n}{w}_k^2\left[{\displaystyle \sum _{k=1}^n}{\left|u_k\right|}^2+\left|{\displaystyle \sum _{k=1}^n}{u}_k^2\right|\right].\hfill \end{array}$$

On the other hand, by Lemma 3, we conclude that the equality (18) holds if and only if $\mathbf{w}=\alpha \parallel \mathbf{u}\parallel (\mathbf{u}+e^{i\theta }\mathbf{v})$ for some scalar $\alpha$, where $\theta =arg〈\mathbf{u},\mathbf{v}〉$, if the set $\{\mathbf{v},\mathbf{u}\}$ is linearly independent. Conversely, if $\{\mathbf{u},\mathbf{v}\}$ is linearly dependent, the equality in (18) holds if and only if $\mathbf{w}=\alpha \mathbf{u}$ for some scalar $\alpha$.

We will consider the following cases to obtain a clearer characterization of the de Bruijn equality.

• Suppose that $\{\mathbf{u},\mathbf{v}\}$ is linearly dependent. Then, the equality in (18) holds if and only if $\mathbf{w}=\alpha \mathbf{u}$ for some scalar $\alpha \in \mathbb{C}$.
  Since $\mathbf{w}\in {\mathbb{R}}^n$, for each $k\in \{1,\dots ,n\}$, we obtain

  $$w_k={\displaystyle \frac{w_k+\overline{w_k}}{2}}={\displaystyle \frac{\alpha u_k+\overline{\alpha u_k}}{2}}=\mathrm{Re}\left(\alpha u_k\right),$$

  which implies that $w_k=\mathrm{Re}\left(\alpha u_k\right)$.
  Moreover, the inner product of $\mathbf{w}$ with itself yields

  $$0\le \langle \mathbf{w},\mathbf{w}\rangle ={\displaystyle \sum _{k=1}^n}{w}_k^2={\displaystyle \sum _{k=1}^n}{\alpha }^2{u}_k^2.$$
  In conclusion, taking $\lambda =\alpha$ satisfies the desired condition for the equality in (18) to hold in the case where $\{\mathbf{u},\mathbf{v}\}$ is linearly dependent.
• We assume that $\{\mathbf{v},\mathbf{u}\}$ is linearly independent. Thus, for any $k=1,\dots ,n$, we have

  $$\begin{array}{c}\hfill w_k=\parallel \mathbf{u}\parallel (\alpha u_k+\overline{\overline{\alpha }{e}^{-i\theta }{u}_k})\in \mathbb{R},\end{array}$$

  then

  $$\alpha u_k+\overline{\overline{\alpha }{e}^{-i\theta }{u}_k}=\overline{(\alpha u_k+\overline{\overline{\alpha }{e}^{-i\theta }{u}_k})}=\overline{\alpha u_k}+\overline{\alpha }{e}^{-i\theta }{u}_k,$$

  and this implies,

  $$(\alpha -\overline{\alpha }{e}^{-i\theta })u_k+(\overline{\alpha }{e}^{i\theta }-\alpha )\overline{u_k}=0.$$
  From the linear indepedence of the set $\{\mathbf{v},\mathbf{u}\}$, we obtain that

  $$\alpha =\overline{\alpha }{e}^{-i\theta }=\overline{\alpha }{e}^{i\theta },$$

  (19)

  and

  $$w_k=\parallel \mathbf{u}\parallel (\alpha u_k+\overline{\alpha u_k})=2\parallel \mathbf{u}\parallel \mathrm{Re}\left(\alpha u_k\right)=\mathrm{Re}\left(\lambda z_k\right),$$

  where $\lambda =2\parallel \mathbf{u}\parallel \alpha .$
  Let $\alpha =\left|\alpha \right|e^{i\gamma }$ with $\gamma \in [0,2\pi )$. From the equality (19), we obtain

  $$e^{i\theta }={\displaystyle \frac{e^{-i\gamma }\left|\alpha \right|}{e^{i\gamma }\left|\alpha \right|}}=e^{-2i\gamma },$$

  this means that $\gamma =-{\displaystyle \frac{\theta }{2}}=-{\displaystyle \frac{arg〈\mathbf{u},\mathbf{v}〉}{2}}$.
  Finally, we conclude that

  $$\begin{array}{ccc}\hfill {\lambda }^2{\displaystyle \sum _{k=1}^n}{u}_k& =& {4\parallel \mathbf{u}\parallel }^2{\alpha }^2{\displaystyle \sum _{k=1}^n}{u}_k={4\parallel \mathbf{u}\parallel }^2{e}^{-i arg〈\mathbf{u},\mathbf{v}〉}{\left|\alpha \right|}^2〈\mathbf{u},\mathbf{v}〉\hfill \\ & =& {4\parallel \mathbf{u}\parallel }^2{\left|\alpha \right|}^2\left|〈\mathbf{u},\mathbf{v}〉\right|\ge 0,\hfill \end{array}$$

  and this completes the proof.

□

One of the earliest and most celebrated estimates in complex analysis is Bohr’s inequality [31]. It asserts that whenever $u_1,u_2\in \mathbb{C}$ and $w_1,w_2>1$ satisfy ${\displaystyle \frac{1}{w_1}}+{\displaystyle \frac{1}{w_2}}=1,$ then

$$|u_1+u_2{|}^2\le w_1|u_1{|}^2+w_2{\left|u_2\right|}^2.$$

(20)

Moreover, equality in (20) occurs precisely when $u_2=(w_1-1)u_1$.

In a natural extension to n terms, Archbold [32] showed that if $u_1,\dots ,u_n\in \mathbb{C}$ and positive weights $w_1,\dots ,w_n$ satisfy ${\sum }_{k=1}^n{\displaystyle \frac{1}{w_k}}=1$, then

$${\left|{\displaystyle \sum _{k=1}^n}{u}_k\right|}^2\le {\displaystyle \sum _{k=1}^n}{w}_k{\left|u_k\right|}^2.$$

(21)

Building on these classical results, we have recently obtained (via de Bruijn’s inequality) a sharper bound that refines (21) (see [33] [Theorem 1]). As a direct consequence of our new Corollary, no further proof is required to establish the following statement:

Corollary 4.

Let $u_1,\dots ,u_n$ be a sequence of complex numbers and $w_1,\dots ,w_n$ be a sequence of positive numbers such that ${\sum }_{k=1}^n{\displaystyle \frac{1}{w_k}}=1$. Suppose that $\mathcal{Z}=\{{\mathbf{z}}_i:i=1,\dots ,n\}$ is a collection of nonzero vectors in ${\mathbb{C}}^n$ such that

$${\displaystyle \sum _{k=1}^n}{u}_k\overline{{\left(z_i\right)}_k}=0={\displaystyle \sum _{k=1}^n}\overline{u_k}\overline{{\left(z_i\right)}_k}foralli=1,\dots ,n.$$

Then,

$$\begin{array}{cc}\hfill {\left|{\displaystyle \sum _{k=1}^n}{u}_k\right|}^2& \le {\left|{\displaystyle \sum _{k=1}^n}{u}_k\right|}^2+{\displaystyle \frac{1}{2}}\left[{\displaystyle \sum _{k=1}^n}{w}_k{\left|u_k\right|}^2+\left|{\displaystyle \sum _{k=1}^n}{w}_k{u}_k^2\right|\right]{\displaystyle \sum _{i=1}^n}{\displaystyle \frac{{\left|{\sum }_{k=1}^n\frac{1}{\sqrt{w_k}}\overline{{\left(z_i\right)}_k}\right|}^2}{{\sum }_{j=1}^n\left|{\sum }_{k=1}^n{\left(z_i\right)}_k\overline{{\left(z_j\right)}_k}\right|}}\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left[{\displaystyle \sum _{k=1}^n}{w}_k{\left|u_k\right|}^2+\left|{\displaystyle \sum _{k=1}^n}{w}_k{u}_k^2\right|\right]\le {\displaystyle \sum _{k=1}^n}{w}_k{\left|u_k\right|}^2.\hfill \end{array}$$

Proof. 

Let us begin by noting that

$${\left|{\displaystyle \sum _{k=1}^n}{u}_k\right|}^2={\left|{\displaystyle \sum _{k=1}^n}{\displaystyle \frac{1}{\sqrt{w_k}}}\sqrt{w_k}{u}_k\right|}^2.$$

(22)

Now, by replacing each ${\displaystyle \frac{1}{\sqrt{w_k}}}$ with $w_k$ and each $\sqrt{w_k}{u}_k$ with $u_k$, we are effectively under the assumptions of Corollary 3. Furthermore, since ${\sum }_{k=1}^n{\displaystyle \frac{1}{w_k}}=1$, it follows that for any $z\in \mathcal{Z}$,

$$\begin{array}{cc}\hfill {\left|{\displaystyle \sum _{k=1}^n}{\displaystyle \frac{1}{\sqrt{w_k}}}\sqrt{w_k}{u}_k\right|}^2& \le {\left|{\displaystyle \sum _{k=1}^n}{u}_k\right|}^2+{\displaystyle \frac{1}{2}}\left[{\displaystyle \sum _{k=1}^n}{w}_k{\left|u_k\right|}^2+\left|{\displaystyle \sum _{k=1}^n}{w}_k{u}_k^2\right|\right]{\displaystyle \sum _{i=1}^n}{\displaystyle \frac{{\left|{\sum }_{k=1}^n\frac{1}{\sqrt{w_k}}\overline{{\left(z_i\right)}_k}\right|}^2}{{\sum }_{j=1}^n\left|{\sum }_{k=1}^n{\left(z_i\right)}_k\overline{{\left(z_j\right)}_k}\right|}}\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left[{\displaystyle \sum _{k=1}^n}{w}_k{\left|u_k\right|}^2+\left|{\displaystyle \sum _{k=1}^n}{w}_k{u}_k^2\right|\right]\le {\displaystyle \sum _{k=1}^n}{w}_k{\left|u_k\right|}^2.\hfill \end{array}$$

□

5. A Critical Note on the Selberg Inequality Involving Bilinear Functionals

A natural idea in extending classical inequalities from inner product spaces to more general contexts is to consider replacing the inner product with a bilinear functional. In particular, one might attempt to generalize the Selberg inequality by formulating it in terms of a bilinear functional, possibly subject to additional assumptions such as boundedness, symmetry, or positivity. However, since many of the key inequalities in Hilbert spaces are deeply rooted in the specific properties of the inner product—such as conjugate symmetry and positive definiteness—this type of generalization is not straightforward and warrants a critical examination.

Recently, some authors have proposed a generalization of the Selberg inequality within this framework by extending its formulation to the setting of bilinear functionals. While the motivation behind such an extension is understandable, we shall show that the proposed inequality does not hold in full generality, and, in fact, its validity can be critically questioned under the assumptions given.

To that end, we begin by recalling the basic notions involved in such formulations (see [2] [Section 4.3]).

Definition 2.

A bilinear functional φ on a complex vector space E is a function $\phi :E\times E\to \mathbb{C}$ that satisfies linearity in the first argument and conjugate linearity in the second argument. More precisely, for any scalars $\alpha ,\beta$ and vectors $x,x_1,x_2,y,y_1,y_2\in E$, the following holds:

1.

$\phi (\alpha x_1+\beta x_2,y)=\alpha \phi (x_1,y)+\beta \phi (x_2,y)$.

2.

$\phi (x,\alpha y_1+\beta y_2)=\overline{\alpha }\phi (x,y_1)+\overline{\beta }\phi (x,y_2)$.

Furthermore,

3.

φ is called symmetric if $\phi (x,y)=\overline{\phi (y,x)}$ for all $x,y\in E$.

4.

φ is called positive if $\phi (x,x)\ge 0$ for any $x\in E.$

Let us recall the result recently obtained by Izadi et al. regarding the Selberg-type inequality in the context of a vector space endowed with a bilinear functional; see [34] [Theorem 3.1].

Statement. 

Let E be a vector space. Then, for all $x\in E$ and nonzero vectors $y_j\in E$, the inequality

$${\displaystyle \sum _{j=1}^n}{\displaystyle \frac{\phi (x,y_j)}{{\sum }_{k=1}^n\phi (y_j,y_k)}}\le \phi (x,x)$$

(23)

holds. Equality occurs if and only if

$$x={\displaystyle \sum _{j=1}^n}{\alpha }_j{y}_j,{\alpha }_j\in \mathbb{C},$$

and for each pair $(j,k)$, $j\ne k$,

$$\phi (y_j,y_k)=0\mathit{or}|{\alpha }_j|=|{\alpha }_k|\mathit{and}\phi ({\alpha }_j{y}_j,{\alpha }_k{y}_k)\ge 0.$$

However, it is important to point out that both the statement and the proof presented by the authors in [34] closely follow an argument originally due to Furuta [25], without due acknowledgement. More significantly, there are several issues in the formulation and justification of the inequality as stated.

To begin with, if one assumes that E is an inner product space and that $\langle ·,·\rangle =\phi (·,·)$, then the inequality stated in (23) does not align with the classical formulation of Selberg’s inequality. While part of the discrepancy can be attributed to typographical errors—such as the omission of modulus signs or squared terms in the expression—there are also more fundamental issues arising from the structural properties of the bilinear functional $\phi$, which we now examine.

In addition, for inequality (23) to be meaningful and for both sides to be directly comparable, all involved terms must be real numbers. This, in particular, implies that the bilinear functional $\phi$ must be symmetric; see [2] [Theorem 4.3.9].

Moreover, upon examining the proof provided in [34], one notices that the authors begin by assuming that $\Phi \left(x\right)=\phi (x,x)\ge 0$ for all $x\in E$; that is, they implicitly assume that $\phi$ is a positive (or positive semi-definite) bilinear form.

In light of these observations, the correct version of the result—both in mathematical substance and logical accuracy—requires that $\phi$ be a symmetric and positive bilinear form. The properly stated inequality is presented below. Its proof follows from the original argument given by Furuta or from the approach proposed by Izadi et al., with the necessary corrections and clarifications incorporated.

Theorem 5.

Let E be a vector space, and let φ be a symmetric, positive bilinear functional on E. Suppose $\mathcal{Z}=\{z_1,\dots ,z_n\}\subseteq E$ is a finite collection of vectors such that $\phi (z_j,z_j)\ne 0$ for any $j\in \{1,\dots ,n\}$. Then, for all $x\in E$,

$${\displaystyle \sum _{j=1}^n}{\displaystyle \frac{{\left|\phi (x,z_j)\right|}^2}{{\sum }_{k=1}^n\left|\phi (z_j,z_k)\right|}}\le \phi (x,x).$$

Proof. 

Let ${\alpha }_1,\dots ,{\alpha }_n\in \mathbb{C}$ be arbitrary complex coefficients. Then, we consider the following quadratic form:

$$\begin{array}{cc}\hfill 0& \le \Phi \left(x-{\displaystyle \sum _{j=1}^n}{\alpha }_j{z}_j\right)=\phi \left(x-{\displaystyle \sum _{j=1}^n}{\alpha }_j{z}_j,x-{\displaystyle \sum _{k=1}^n}{\alpha }_k{z}_k\right)\hfill \\ \hfill & =\phi (x,x)-{\displaystyle \sum _{k=1}^n}\overline{{\alpha }_k}\phi (x,z_k)-{\displaystyle \sum _{j=1}^n}{\alpha }_j\phi (z_j,x)+{\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{\alpha }_j\overline{{\alpha }_k}\phi (z_j,z_k)\hfill \\ \hfill & =\phi (x,x)-2\Re \left({\displaystyle \sum _{j=1}^n}{\alpha }_j\phi (z_j,x)\right)+{\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{\alpha }_j\overline{{\alpha }_k}\phi (z_j,z_k).\hfill \end{array}$$

By applying the triangle inequality and estimating the last double sum in absolute value, we obtain

$$\begin{array}{cc}\hfill \Phi \left(x-{\displaystyle \sum _{j=1}^n}{\alpha }_j{z}_j\right)& \le \phi (x,x)-2\Re \left({\displaystyle \sum _{j=1}^n}{\alpha }_j\phi (z_j,x)\right)+{\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}|{\alpha }_j||{\alpha }_k||\phi (z_j,z_k)|\hfill \\ \hfill & \le \phi (x,x)-2\Re \left({\displaystyle \sum _{j=1}^n}{\alpha }_j\phi (z_j,x)\right)+{\displaystyle \sum _{j=1}^n}|{\alpha }_j{|}^2{\displaystyle \sum _{k=1}^n}\left|\phi (z_j,z_k)\right|.\hfill \end{array}$$

In the second inequality, we used the following well-known identity: for any finite sequence of complex (or real) numbers ${\left\{a_i\right\}}_{i=1}^n$ and any symmetric matrix $W=\left[w_{ij}\right]$ with non-negative real entries (i.e., $w_{ij}=w_{ji}\ge 0$ for all $i,j$), it holds that

$${\displaystyle \sum _{i=1}^n}|a_i{|}^2{\displaystyle \sum _{j=1}^n}{w}_{ij}-{\displaystyle \sum _{i,j=1}^n}|a_i||a_j|w_{ij}={\displaystyle \frac{1}{2}}{\displaystyle \sum _{i,j=1}^n}\left(\right|a_i|-|a_j{\left|\right)}^2{w}_{ij}\ge 0.$$

Now, we consider

$${\alpha }_j:={\displaystyle \frac{\phi (x,z_j)}{{\sum }_{k=1}^n\left|\phi (z_j,z_k)\right|}}.$$

This is a valid definition since, by assumption, there exists at least one pair $(j,k)$ with $\phi (z_j,z_k)\ne 0$, so at least one denominator is positive.

Substituting this choice into the inequality, we find

$$\begin{array}{cc}\hfill 0& \le \phi (x,x)-2\Re \left({\displaystyle \sum _{j=1}^n}{\displaystyle \frac{\phi (x,z_j)\phi (z_j,x)}{{\sum }_{k=1}^n\left|\phi (z_j,z_k)\right|}}\right)+{\displaystyle \sum _{j=1}^n}{\displaystyle \frac{|\phi (x,z_j){|}^2}{{\left({\sum }_{k=1}^n\left|\phi (z_j,z_k)\right|\right)}^2}}{\displaystyle \sum _{k=1}^n}\left|\phi (z_j,z_k)\right|\hfill \\ \hfill & =\phi (x,x)-{\displaystyle \sum _{j=1}^n}{\displaystyle \frac{|\phi (x,z_j){|}^2}{{\sum }_{k=1}^n\left|\phi (z_j,z_k)\right|}}.\hfill \end{array}$$

Consequently, we conclude the desired inequality:

$${\displaystyle \sum _{j=1}^n}{\displaystyle \frac{|\phi (x,z_j){|}^2}{{\sum }_{k=1}^n\left|\phi (z_j,z_k)\right|}}\le \phi (x,x).$$

This finishes the proof. □

We now emphasize that the equality condition stated in [34] [Theorem 3.1] does not necessarily hold.

Example 1.

Consider the sesquilinear form φ, defined by

$$\phi (\left\{a_n\right\},\left\{b_n\right\})={\displaystyle \sum _{n=1}^{\infty }}{a}_{2n}\overline{b_{2n}},$$

and let $\mathcal{Z}=\{z_1,z_2\}$, where

$$z_1=(1,1,0,0,\dots ),z_2=(0,1,0,0,\dots ),x=z_1-z_2=(1,0,0,0,\dots ).$$

We compute

$$\phi (z_1,z_2)={\left(z_1\right)}_2\overline{{\left(z_2\right)}_2}=1\ne 0,$$

and

$$\phi (x,z_1)=x_2\overline{{\left(z_1\right)}_2}=0,\phi (x,z_2)=x_2\overline{{\left(z_2\right)}_2}=0,$$

so that

$${\displaystyle \sum _{j=1}^2}{\displaystyle \frac{|\phi (x,z_j){|}^2}{{\sum }_{k=1}^2\left|\phi (z_j,z_k)\right|}}=0=\phi (x,x).$$

However, this example does not satisfy the characterization of equality claimed in (23), which states that the equality

$$\phi (x,x)={\displaystyle \sum _{j=1}^n}{\displaystyle \frac{|\phi (x,z_j){|}^2}{{\sum }_{k=1}^n\left|\phi (z_j,z_k)\right|}}$$

holds if and only if x can be written as a linear combination

$$x={\displaystyle \sum _{j=1}^n}{\alpha }_j{z}_j,{\alpha }_j\in \mathbb{C},$$

and for each pair $(j,k)$, with $j\ne k$, either

$$\phi (z_j,z_k)=0,\mathit{or}|{\alpha }_j|=|{\alpha }_k|\mathit{and}\phi ({\alpha }_j{z}_j,{\alpha }_k{z}_k)\ge 0.$$

In this case, although $x=z_1-z_2$ is a linear combination of $z_1$ and $z_2$ with coefficients ${\alpha }_1=1$, ${\alpha }_2=-1$, we find

$$\phi (z_1,z_2)=1\ne 0,\mathit{and}\phi ({\alpha }_1{z}_1,{\alpha }_2{z}_2)=\phi (z_1,-z_2)=-1<0,$$

which violates the non-negativity condition. Therefore, this example shows that the equality in (23) may hold even when the stated necessary and sufficient conditions fail. This illustrates that the characterization claimed in [34] [Theorem 3.1] does not hold in general.

Moreover, since the form φ is positive but not positive definite, one can find a nonzero vector x outside of the linear span of $\{z_1,z_2\}$ such that the equality still holds. For instance, consider

$$x=(1,0,1,0,1,0,\dots ),$$

i.e., the vector having 1 in every odd-indexed position and 0 elsewhere. Clearly, $x\notin span\{z_1,z_2\}$ since it has nonzero entries beyond positions 1 and 2. Moreover, since all the even-indexed entries of x vanish, we have

$$\phi (x,x)={\displaystyle \sum _{n=1}^{\infty }}{x}_{2n}\overline{x_{2n}}=0,\mathit{and}\phi (x,z_j)=0\mathit{for}j=1,2.$$

Hence,

$${\displaystyle \sum _{j=1}^2}{\displaystyle \frac{|\phi (x,z_j){|}^2}{{\sum }_{k=1}^2\left|\phi (z_j,z_k)\right|}}=0=\phi (x,x),$$

but $x\notin span\{z_1,z_2\}$. This confirms that in the presence of a nontrivial kernel of φ, the equality can hold even when x is not in the span of the given family.

Before proceeding with an refinement of the Selberg inequality (SI) to the setting of bilinear functionals, let us first recall some classical results that hold in this context. For the reader’s convenience, we include the proof of one such result.

Lemma 4.

Let E be a vector space, and let φ be a symmetric, positive, bilinear functional on E. Then, for any $x,y,z\in E$, the following holds:

1.

$$\left|\phi (x,y)\right|\le \phi {(x,x)}^{\frac{1}{2}}\phi {(y,y)}^{\frac{1}{2}}.$$

2.

$$\left|\phi (x,z)\phi (z,y)\right|\le {\displaystyle \frac{1}{2}}{\phi (z,z)[\phi (x,x)}^{\frac{1}{2}}\phi {(y,y)}^{\frac{1}{2}}+\left|\phi (x,y)\right|].$$

Proof. 

For inequality (1), we refer the reader to [2] [4.12, Exercise 5].

We prove only part (2). First, observe that if $\phi (z,z)=0$, then the inequality holds trivially. Indeed, by (1), this implies $\phi (x,z)=\phi (z,y)=0$, and hence both sides of the inequality vanish.

Thus, we may assume $\phi (z,z)\ne 0$ and define $z_0={\displaystyle \frac{z}{\phi {(z,z)}^{1/2}}}$. Then,

$$\begin{array}{cc}\hfill |\phi (x,z_0)\phi (z_0,y)|& =\left|\phi \left(\phi (x,z_0)z_0-{\displaystyle \frac{1}{2}}x,y\right)+{\displaystyle \frac{1}{2}}\phi (x,y)\right|\hfill \\ \hfill & \le \left|\phi \left(\phi (x,z_0)z_0-{\displaystyle \frac{1}{2}}x,y\right)\right|+\left|{\displaystyle \frac{1}{2}}\phi (x,y)\right|\hfill \\ \hfill & \le {\left[\phi \left(\phi (x,z_0)z_0-{\displaystyle \frac{1}{2}}x,\phi (x,z_0)z_0-{\displaystyle \frac{1}{2}}x\right)\right]}^{1/2}\phi {(y,y)}^{1/2}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}\left|\phi (x,y)\right|(\mathrm{by}(1))\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\phi {(x,x)}^{1/2}\phi {(y,y)}^{1/2}+{\displaystyle \frac{1}{2}}\left|\phi (x,y)\right|.\hfill \end{array}$$

□

It is worth noting that the inequalities in Lemma 4 extend both the classical (CBSI) and (BuI) to the framework of a vector space endowed with a bilinear functional, subject to the conditions stated therein.

We conclude this article by presenting a simultaneous extension of the classical Buzano and Selberg inequalities in the context of bilinear functionals. This result is obtained by adapting the general strategy developed in [14]. For conciseness, and since the proof follows closely the ideas therein, we omit the details.

Proposition 2.

Let E be a vector space and φ a semi-inner product defined by E, $\mathcal{Y}=\{y_1,\cdots ,y_n\}\subseteq E$ such that $\phi (y_j,y_j)\ne 0$ for any $j\in \{1,\dots ,n\}$ and $z_1,z_2\in E$ with $\phi (z_k,y_j)=0$ for any $k=1,2$ and $j=1,\cdots ,n$. Then, for all $x\in E$, we have

$$\begin{array}{c}\hfill |\phi (z_1,x)\phi (x,z_2)|+\mathcal{B}\left(z_1,z_2\right){\displaystyle \sum _{j=1}^n}{\displaystyle \frac{{\left|\phi \left(x,y_j\right)\right|}^2}{{\sum }_{k=1}^n\left|\phi \left(y_j,y_k\right)\right|}}\le \mathcal{B}\left(z_1,z_2\right)\phi (x,x),\end{array}$$

where $\mathcal{B}\left(z_1,z_2\right)={\displaystyle \frac{1}{2}}[\phi {(z_1,z_1)}^{\frac{1}{2}}\phi {(z_2,z_2)}^{\frac{1}{2}}+|\phi (z_1,z_2)\left|\right].$

Remark 3.

Finally, we would like to emphasize that our previous results for symmetric, positive bilinear functional are not fruitless generalizations of the inner product case but generating some interesting results for operators in Hilbert space.

Consider the self-adjoint operator P such that $P\ge kI,$$k>0$ in the operator order of $\mathcal{B}\left(\mathcal{H}\right)$. Define $\phi :\mathcal{H}\times \mathcal{H}\to \mathbb{C}$ by

$$\phi \left(x,y\right):=〈Px,y〉,x,y\in \mathcal{H}.$$

We observe that φ is a symmetric, positive bilinear functional on $H.$

From the second inequality in Lemma 4, we obtain the following generalized version of Buzano’s inequality:

$$\left|〈Px,z〉〈Pz,y〉\right|\le {\displaystyle \frac{1}{2}}〈Pz,z〉\left[{〈Px,x〉}^{1/2}{〈Py,y〉}^{1/2}+\left|〈Px,y〉\right|\right]$$

for all $x,y,z\in \mathcal{H}.$ This is a sharp inequality since for $P=I$, we recapture the classical Buzano result.

Now, if we take instead of y, $B^{*}y$, we obtain

$$\left|〈Px,z〉〈Pz,B^{*}y〉\right|\le {\displaystyle \frac{1}{2}}〈Pz,z〉\left[{〈Px,x〉}^{1/2}{〈P{B}^{*}y,B^{*}y〉}^{1/2}+\left|〈Px,B^{*}y〉\right|\right]$$

for all $x,y,z\in \mathcal{H},$ or, equivalently,

$$\left|〈Pz,x〉〈BPz,y〉\right|\le {\displaystyle \frac{1}{2}}〈Pz,z〉\left[{〈Px,x〉}^{1/2}{〈BP{B}^{*}y,y〉}^{1/2}+\left|〈BPx,y〉\right|\right]$$

for all $x,y,z\in \mathcal{H}.$

Further, if we take the supremum over $∥x∥=∥y∥=1$ and observe that

$$\underset{∥x∥=∥y∥=1}{sup}\left|〈Pz,x〉〈BPz,y〉\right|=\underset{∥x∥=1}{sup}\left|〈Pz,x〉\right|\underset{∥x∥=1}{sup}\left|〈BPz,y〉\right|=∥Pz∥∥BPz∥\ge k∥BPz∥$$

and

$$\underset{∥x∥=∥y∥=1}{sup}\left[{〈Px,x〉}^{1/2}{〈BP{B}^{*}y,y〉}^{1/2}+\left|〈BPx,y〉\right|\right]\le {∥P∥}^{1/2}{∥BP{B}^{*}∥}^{1/2}+∥BP∥,$$

then we obtain the following vector norm inequality

$$∥BPz∥\le {\displaystyle \frac{1}{2k}}〈Pz,z〉\left[{∥P∥}^{1/2}{∥BP{B}^{*}∥}^{1/2}+∥BP∥\right]$$

for all $z\in \mathcal{H}.$ This is a kind of vector norm operator version of Buzano’s inequality.

Moreover, if we take the supremum over $∥z∥=1,$ then we obtain the norm operator inequality

$${\displaystyle \frac{∥BP∥}{∥P∥}}\le {\displaystyle \frac{1}{2k}}\left[{∥P∥}^{1/2}{∥BP{B}^{*}∥}^{1/2}+∥BP∥\right],$$

which, in the case that $∥P∥<2k,$ gives

$$\left({\displaystyle \frac{2k-∥P∥}{2k∥P∥}}\right)∥BP∥\le {\displaystyle \frac{1}{2k}}\left[{∥P∥}^{1/2}{∥BP{B}^{*}∥}^{1/2}\right],$$

namely,

$$∥BP∥\le {\displaystyle \frac{{∥P∥}^{3/2}}{2k-∥P∥}}{∥BP{B}^{*}∥}^{1/2}.$$