---
abstract: 'Reverses of Schwarz, triangle and Bessel inequalities in inner product spaces that improve some earlier results are pointed out. They are applied to obtain new Grüss type inequalities in inner product spaces. Some natural applications for integral inequalities are also pointed out.'
address: |
    School of Computer Science and Mathematics\
    Victoria University of Technology\
    PO Box 14428, MCMC 8001\
    Victoria, Australia.
author:
- 'S.S. Dragomir'
date: 'August 04, 2003.'
title: 'Reverses of Schwarz, Triangle and Bessel Inequalities in Inner Product Spaces'
---

Introduction\[s1\]
==================

Let $\left( H;\left\langle \cdot ,\cdot \right\rangle \right) $ be an inner product over the real or complex number field $\mathbb{K}$. The following inequality is known in the literature as *Schwarz’s inequality*:$$\left\vert \left\langle x,y\right\rangle \right\vert ^{2}\leq \left\Vert
x\right\Vert ^{2}\left\Vert y\right\Vert ^{2},\ \ \ \ x,y\in H;  \label{1.1}$$where $\left\Vert z\right\Vert ^{2}=\left\langle z,z\right\rangle ,$ $z\in
H. $ The equality occurs in (\[1.1\]) if and only if $x$ and $y$ are linearly dependent.

In [@SSD1], the following *reverse* of Schwarz’s inequality has been obtained:$$0\leq \left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}-\left\vert
\left\langle x,y\right\rangle \right\vert ^{2}\leq \frac{1}{4}\left\vert
A-a\right\vert ^{2}\left\Vert y\right\Vert ^{4},  \label{1.2}$$provided $x,y\in H$ and $a,A\in \mathbb{K}$ are so that either$$\func{Re}\left\langle Ay-x,x-ay\right\rangle \geq 0,  \label{1.3}$$or, equivalently,$$\left\Vert x-\frac{a+A}{2}\cdot y\right\Vert \leq \frac{1}{2}\left\vert
A-a\right\vert \left\Vert y\right\Vert ,  \label{1.4}$$holds. The constant $\frac{1}{4}$ is best possible in (\[1.2\]) in the sense that it cannot be replaced by a smaller constant.

If $x,y,A,a$ satisfy either (\[1.3\]) or (\[1.4\]), then the following reverse of Schwarz’s inequality also holds [@SSD2]$$\begin{aligned}
\left\Vert x\right\Vert \left\Vert y\right\Vert & \leq \frac{1}{2}\cdot 
\frac{\func{Re}\left[ A\overline{\left\langle x,y\right\rangle }+\overline{a}%
\left\langle x,y\right\rangle \right] }{\left[ \func{Re}\left( \overline{a}%
A\right) \right] ^{\frac{1}{2}}}  \label{1.5} \\
& \leq \frac{1}{2}\cdot \frac{\left\vert A\right\vert +\left\vert
a\right\vert }{\left[ \func{Re}\left( \overline{a}A\right) \right] ^{\frac{1%
}{2}}}\left\vert \left\langle x,y\right\rangle \right\vert ,  \notag\end{aligned}$$provided that, the complex numbers $a$ and $A$ satisfy the condition $\func{%
Re}\left( \overline{a}A\right) >0.$ In both inequalities in (\[1.5\]), the constant $\frac{1}{2}$ is best possible.

An additive version of (\[1.5\]) may be stated as well (see also [SSD3]{})$$0\leq \left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}-\left\vert
\left\langle x,y\right\rangle \right\vert ^{2}\leq \frac{1}{4}\cdot \frac{%
\left( \left\vert A\right\vert -\left\vert a\right\vert \right) ^{2}+4\left[
\left\vert Aa\right\vert -\func{Re}\left( \overline{a}A\right) \right] }{%
\func{Re}\left( \overline{a}A\right) }\left\vert \left\langle
x,y\right\rangle \right\vert ^{2}.  \label{1.6}$$In this inequality, $\frac{1}{4}$ is the best possible constant.

It has been proven in [@SSD4], that$$0\leq \left\Vert x\right\Vert ^{2}-\left\vert \left\langle x,y\right\rangle
\right\vert ^{2}\leq \frac{1}{4}\left\vert \phi -\varphi \right\vert
^{2}-\left\vert \frac{\phi +\varphi }{2}-\left\langle x,e\right\rangle
\right\vert ^{2};  \label{1.7}$$provided, either $$\func{Re}\left\langle \phi e-x,x-\varphi e\right\rangle \geq 0,  \label{1.8}$$or, equivalently,$$\left\Vert x-\frac{\phi +\varphi }{2}e\right\Vert \leq \frac{1}{2}\left\vert
\phi -\varphi \right\vert ,  \label{1.9}$$where $e=H,$ $\left\Vert e\right\Vert =1.$ The constant $\frac{1}{4}$ in [1.7]{} is also best possible.

If we choose $e=\frac{y}{\left\Vert y\right\Vert },$ $\phi =\Gamma
\left\Vert y\right\Vert ,$ $\varphi =\gamma \left\Vert y\right\Vert $ $%
\left( y\neq 0\right) ,$ $\Gamma ,\gamma \in \mathbb{K}$, then by (\[1.8\]), (\[1.9\]) we have,$$\func{Re}\left\langle \Gamma y-x,x-\gamma y\right\rangle \geq 0,
\label{1.10}$$or, equivalently,$$\left\Vert x-\frac{\Gamma +\gamma }{2}y\right\Vert \leq \frac{1}{2}%
\left\vert \Gamma -\gamma \right\vert \left\Vert y\right\Vert ,  \label{1.11}$$imply the following reverse of Schwarz’s inequality:$$0\leq \left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}-\left\vert
\left\langle x,y\right\rangle \right\vert ^{2}\leq \frac{1}{4}\left\vert
\Gamma -\gamma \right\vert ^{2}\left\Vert y\right\Vert ^{4}-\left\vert \frac{%
\Gamma +\gamma }{2}\left\Vert y\right\Vert ^{2}-\left\langle
x,y\right\rangle \right\vert ^{2}.  \label{1.12}$$The constant $\frac{1}{4}$ in (\[1.12\]) is sharp.

Note that this inequality is an improvement of (\[1.2\]), but it might not be very convenient for applications.

Now, let $\left\{ e_{i}\right\} _{i\in I}$ be a finite or infinite family of orthornormal vectors in the inner product space $\left( H;\left\langle \cdot
,\cdot \right\rangle \right) ,$ i.e., we recall that $$\left\langle e_{i},e_{j}\right\rangle =\left\{ 
\begin{array}{ll}
0 & \text{if \ }i\neq j \\ 
&  \\ 
1 & \text{if \ }i=j%
\end{array}%
\right. ,\ \ \ i,j\in I.$$In [@SSD5], we proved that if $\left\{ e_{i}\right\} _{i\in I}$ is as above, $F\subset I$ is a finite part of $I$ such that either$$\func{Re}\left\langle \sum_{i\in F}\phi _{i}e_{i}-x,x-\sum_{i\in F}\varphi
_{i}e_{i}\right\rangle \geq 0,  \label{1.13}$$or, equivalently,$$\left\Vert x-\sum_{i\in F}\frac{\phi _{i}+\varphi _{i}}{2}e_{i}\right\Vert
\leq \frac{1}{2}\left( \sum_{i\in F}\left\vert \phi _{i}-\varphi
_{i}\right\vert ^{2}\right) ^{\frac{1}{2}},  \label{1.14}$$holds, where $\left( \phi _{i}\right) _{i\in I},$ $\left( \varphi
_{i}\right) _{i\in I}$ are real or complex numbers, then we have the following reverse of *Bessel’s inequality:*$$\begin{aligned}
0& \leq \left\Vert x\right\Vert ^{2}-\sum_{i\in F}\left\vert \left\langle
x,e_{i}\right\rangle \right\vert ^{2}  \label{1.15} \\
& \leq \frac{1}{4}\cdot \sum_{i\in F}\left\vert \phi _{i}-\varphi
_{i}\right\vert ^{2}-\func{Re}\left\langle \sum_{i\in F}\phi
_{i}e_{i}-x,x-\sum_{i\in F}\varphi _{i}e_{i}\right\rangle  \notag \\
& \leq \frac{1}{4}\cdot \sum_{i\in F}\left\vert \phi _{i}-\varphi
_{i}\right\vert ^{2}.  \notag\end{aligned}$$The constant $\frac{1}{4}$ in both inequalities is sharp. This result improves an earlier result by N. Ujević obtained only for real spaces [@NU].

In [@SSD4], by the use of a different technique, another reverse of Bessel’s inequality has been proven, namely:$$\begin{aligned}
0& \leq \left\Vert x\right\Vert ^{2}-\sum_{i\in F}\left\vert \left\langle
x,e_{i}\right\rangle \right\vert ^{2}  \label{1.16} \\
& \leq \frac{1}{4}\cdot \sum_{i\in F}\left\vert \phi _{i}-\varphi
_{i}\right\vert ^{2}-\sum_{i\in F}\left\vert \frac{\phi _{i}+\varphi _{i}}{2}%
-\left\langle x,e_{i}\right\rangle \right\vert ^{2}  \notag \\
& \leq \frac{1}{4}\cdot \sum_{i\in F}\left\vert \phi _{i}-\varphi
_{i}\right\vert ^{2},  \notag\end{aligned}$$provided that $\left( e_{i}\right) _{i\in I},$ $\left( \phi _{i}\right)
_{i\in I},$ $\left( \varphi _{i}\right) _{i\in I},$ $x$ and $F$ are as above.

Here the constant $\frac{1}{4}$ is sharp in both inequalities.

It has also been shown that the bounds provided by (\[1.15\]) and ([1.16]{}) for the Bessel’s difference $\left\Vert x\right\Vert ^{2}-\sum_{i\in
F}\left\vert \left\langle x,e_{i}\right\rangle \right\vert ^{2}$ cannot be compared in general, meaning that there are examples for which one is smaller than the other [@SSD4].

Finally, we recall another type of reverse for Bessel inequality that has been obtained in [@SSD6]:$$\left\Vert x\right\Vert ^{2}\leq \frac{1}{4}\cdot \frac{\sum_{i\in F}\left(
\left\vert \phi _{i}\right\vert +\left\vert \varphi _{i}\right\vert \right)
^{2}}{\sum_{i\in F}\func{Re}\left( \phi _{i}\overline{\varphi _{i}}\right) }%
\sum_{i\in F}\left\vert \left\langle x,e_{i}\right\rangle \right\vert ^{2};
\label{1.17}$$provided $\left( \phi _{i}\right) _{i\in I},$ $\left( \varphi _{i}\right)
_{i\in I}$ satisfy (\[1.13\]) (or, equivalently (\[1.14\])) and $%
\sum_{i\in F}\func{Re}\left( \phi _{i}\overline{\varphi _{i}}\right) >0.$ Here the constant $\frac{1}{4}$ is also best possible.

An additive version of (\[1.17\]) is $$\begin{aligned}
0& \leq \left\Vert x\right\Vert ^{2}-\sum_{i\in F}\left\vert \left\langle
x,e_{i}\right\rangle \right\vert ^{2}  \label{1.18} \\
& \leq \frac{1}{4}\cdot \frac{\sum_{i\in F}\left\{ \left( \left\vert \phi
_{i}\right\vert -\left\vert \varphi _{i}\right\vert \right) ^{2}+4\left[
\left\vert \phi _{i}\varphi _{i}\right\vert -\func{Re}\left( \phi _{i}%
\overline{\varphi _{i}}\right) \right] \right\} }{\sum_{i\in F}\func{Re}%
\left( \phi _{i}\overline{\varphi _{i}}\right) }.  \notag\end{aligned}$$The constant $\frac{1}{4}$ is best possible.

It is the main aim of the present paper to point out new reverse inequalities to Schwarz’s, triangle and Bessel’s inequalities.

Some results related to Grüss’ inequality in inner product spaces are also pointed out. Natural applications for integrals are also provided.

Some Reverses of Schwarz’s Inequality\[s2\]
===========================================

The following result holds.

\[t2.1\]Let $\left( H;\left\langle \cdot ,\cdot \right\rangle \right) $ be an inner product space over the real or complex number field $\mathbb{K}$ $\left( \mathbb{K}=\mathbb{R},\ \mathbb{K}=\mathbb{C}\right) $ and $x,a\in
H, $ $r>0$ are such that$$x\in \overline{B}\left( a,r\right) :=\left\{ z\in H|\left\Vert
z-a\right\Vert \leq r\right\} .  \label{2.1}$$

1.  If $\left\Vert a\right\Vert >r,$ then we have the inequality$$0\leq \left\Vert x\right\Vert ^{2}\left\Vert a\right\Vert ^{2}-\left\vert
    \left\langle x,a\right\rangle \right\vert ^{2}\leq \left\Vert x\right\Vert
    ^{2}\left\Vert a\right\Vert ^{2}-\left[ \func{Re}\left\langle
    x,a\right\rangle \right] ^{2}\leq r^{2}\left\Vert x\right\Vert ^{2}.
    \label{2.2}$$The constant $C=1$ in front of $r^{2}$ is best possible in the sense that it cannot be replaced by a smaller one.

2.  If $\left\Vert a\right\Vert =r,$ then$$\left\Vert x\right\Vert ^{2}\leq 2\func{Re}\left\langle x,a\right\rangle
    \leq 2\left\vert \left\langle x,a\right\rangle \right\vert .  \label{2.3}$$The constant $2$ is best possible in both inequalities.

3.  If $\left\Vert a\right\Vert <r,$ then$$\left\Vert x\right\Vert ^{2}\leq r^{2}-\left\Vert a\right\Vert ^{2}+2\func{Re%
    }\left\langle x,a\right\rangle \leq r^{2}-\left\Vert a\right\Vert
    ^{2}+2\left\vert \left\langle x,a\right\rangle \right\vert .  \label{2.4}$$Here the constant $2$ is also best possible.

Since $x\in \overline{B}\left( a,r\right) ,$ then obviously $\left\Vert
x-a\right\Vert ^{2}\leq r^{2},$ which is equivalent to $$\left\Vert x\right\Vert ^{2}+\left\Vert a\right\Vert ^{2}-r^{2}\leq 2\func{Re%
}\left\langle x,a\right\rangle .  \label{2.5}$$

1.  If $\left\Vert a\right\Vert >r,$ then we may divide (\[2.5\]) by $\sqrt{\left\Vert a\right\Vert ^{2}-r^{2}}>0$ getting $$\frac{\left\Vert x\right\Vert ^{2}}{\sqrt{\left\Vert a\right\Vert ^{2}-r^{2}}%
    }+\sqrt{\left\Vert a\right\Vert ^{2}-r^{2}}\leq \frac{2\func{Re}\left\langle
    x,a\right\rangle }{\sqrt{\left\Vert a\right\Vert ^{2}-r^{2}}}.  \label{2.6}$$Using the elementary inequality$$\alpha p+\frac{1}{\alpha }q\geq 2\sqrt{pq},\ \ \ \alpha >0,\ \ p,q\geq 0,$$we may state that$$2\left\Vert x\right\Vert \leq \frac{\left\Vert x\right\Vert ^{2}}{\sqrt{%
    \left\Vert a\right\Vert ^{2}-r^{2}}}+\sqrt{\left\Vert a\right\Vert ^{2}-r^{2}%
    }.  \label{2.7}$$Making use of (\[2.6\]) and (\[2.7\]), we deduce$$\left\Vert x\right\Vert \sqrt{\left\Vert a\right\Vert ^{2}-r^{2}}\leq \func{%
    Re}\left\langle x,a\right\rangle .  \label{2.8}$$Taking the square in (\[2.8\]) and re-arranging the terms, we deduce the third inequality in (\[2.2\]). The others are obvious.

    To prove the sharpness of the constant, assume, under the hypothesis of the theorem, that, there exists a constant $c>0$ such that$$\left\Vert x\right\Vert ^{2}\left\Vert a\right\Vert ^{2}-\left[ \func{Re}%
    \left\langle x,a\right\rangle \right] ^{2}\leq cr^{2}\left\Vert x\right\Vert
    ^{2},  \label{2.9}$$provided $x\in \overline{B}\left( a,r\right) $ and $\left\Vert a\right\Vert
    >r.$

    Let $r=\sqrt{\varepsilon }>0,$ $\varepsilon \in \left( 0,1\right) ,$ $a,e\in
    H$ with $\left\Vert a\right\Vert =\left\Vert e\right\Vert =1$ and $a\perp e.$ Put $x=a+\sqrt{\varepsilon }e.$ Then obviously $x\in \overline{B}\left(
    a,r\right) ,$ $\left\Vert a\right\Vert >r$ and $\left\Vert x\right\Vert
    ^{2}=\left\Vert a\right\Vert ^{2}+\varepsilon \left\Vert e\right\Vert
    ^{2}=1+\varepsilon $, $\func{Re}\left\langle x,a\right\rangle =\left\Vert
    a\right\Vert ^{2}=1,$ and thus $\left\Vert x\right\Vert ^{2}\left\Vert
    a\right\Vert ^{2}-\left[ \func{Re}\left\langle x,a\right\rangle \right]
    ^{2}=\varepsilon .$ Using (\[2.9\]), we may write that$$\varepsilon \leq c\varepsilon \left( 1+\varepsilon \right) ,\ \ \varepsilon
    >0$$giving $$c+c\varepsilon \geq 1\text{ \ for any }\varepsilon >0  \label{2.10}$$Letting $\varepsilon \rightarrow 0+,$ we get from (\[2.10\]) that $c\geq 1,
    $ and the sharpness of the constant is proved.

2.  The inequality (\[2.3\]) is obvious by (\[2.5\]) since $%
    \left\Vert a\right\Vert =r.$ The best constant follows in a similar way to the above.

3.  The inequality (\[2.3\]) is obvious. The best constant may be proved in a similar way to the above. We omit the details.

The following reverse of Schwarz’s inequality holds.

\[t2.2\]Let $\left( H;\left\langle \cdot ,\cdot \right\rangle \right) $ be an inner product space over $\mathbb{K}$ and $x,y\in H,$ $\gamma ,\Gamma
\in \mathbb{K}$ such that either$$\func{Re}\left\langle \Gamma y-x,x-\gamma y\right\rangle \geq 0,
\label{2.11}$$or, equivalently,$$\left\Vert x-\frac{\Gamma +\gamma }{2}y\right\Vert \leq \frac{1}{2}%
\left\vert \Gamma -\gamma \right\vert \left\Vert y\right\Vert ,  \label{2.12}$$holds.

1.  If $\func{Re}\left( \Gamma \overline{\gamma }\right) >0,$ then we have the inequalities$$\begin{aligned}
    \left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}& \leq \frac{1}{4}%
    \cdot \frac{\left\{ \func{Re}\left[ \left( \overline{\Gamma }+\overline{%
    \gamma }\right) \left\langle x,y\right\rangle \right] \right\} ^{2}}{\func{Re%
    }\left( \Gamma \overline{\gamma }\right) }  \label{2.13} \\
    & \leq \frac{1}{4}\cdot \frac{\left\vert \Gamma +\gamma \right\vert ^{2}}{%
    \func{Re}\left( \Gamma \overline{\gamma }\right) }\left\vert \left\langle
    x,y\right\rangle \right\vert ^{2}.  \notag\end{aligned}$$The constant $\frac{1}{4}$ is best possible in both inequalities.

2.  If $\func{Re}\left( \Gamma \overline{\gamma }\right) =0,$ then $$\left\Vert x\right\Vert ^{2}\leq \func{Re}\left[ \left( \overline{\Gamma }+%
    \overline{\gamma }\right) \left\langle x,y\right\rangle \right] \leq
    \left\vert \Gamma +\gamma \right\vert \left\vert \left\langle
    x,y\right\rangle \right\vert .  \label{2.14}$$

3.  If $\func{Re}\left( \Gamma \overline{\gamma }\right) <0,$ then $$\begin{aligned}
    \left\Vert x\right\Vert ^{2}& \leq -\func{Re}\left( \Gamma \overline{\gamma }%
    \right) \left\Vert y\right\Vert ^{2}+\func{Re}\left[ \left( \overline{\Gamma 
    }+\overline{\gamma }\right) \left\langle x,y\right\rangle \right]
    \label{2.15} \\
    & \leq -\func{Re}\left( \Gamma \overline{\gamma }\right) \left\Vert
    y\right\Vert ^{2}+\left\vert \Gamma +\gamma \right\vert \left\vert
    \left\langle x,y\right\rangle \right\vert .  \notag\end{aligned}$$

The proof of the equivalence between the inequalities (\[2.11\]) and ([2.12]{}) follows by the fact that in an inner product space $\func{Re}%
\left\langle Z-x,x-z\right\rangle \geq 0$ for $x,z,Z\in H$ is equivalent with $\left\Vert x-\frac{z+Z}{2}\right\Vert \leq \frac{1}{2}\left\Vert
Z-z\right\Vert $ (see for example [@SSD3]).

Consider, for $y\neq 0,$ $a=\frac{\gamma +\Gamma }{2}y$ and $r=\frac{1}{2}%
\left\vert \Gamma -\gamma \right\vert \left\Vert y\right\Vert ^{2}.$ Then$$\left\Vert a\right\Vert ^{2}-r^{2}=\frac{\left\vert \Gamma +\gamma
\right\vert ^{2}-\left\vert \Gamma -\gamma \right\vert ^{2}}{4}\left\Vert
y\right\Vert ^{2}=\func{Re}\left( \Gamma \overline{\gamma }\right)
\left\Vert y\right\Vert ^{2}.$$

1.  If $\func{Re}\left( \Gamma \overline{\gamma }\right) >0,$ then the hypothesis of (i) in Theorem \[t2.1\] is satisfied, and by the second inequality in (\[2.2\]) we have$$\left\Vert x\right\Vert ^{2}\frac{\left\vert \Gamma +\gamma \right\vert ^{2}%
    }{4}\left\Vert y\right\Vert ^{2}-\frac{1}{4}\left\{ \func{Re}\left[ \left( 
    \overline{\Gamma }+\overline{\gamma }\right) \left\langle x,y\right\rangle %
    \right] \right\} ^{2}\leq \frac{1}{4}\left\vert \Gamma -\gamma \right\vert
    ^{2}\left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}$$from where we derive$$\frac{\left\vert \Gamma +\gamma \right\vert ^{2}-\left\vert \Gamma -\gamma
    \right\vert ^{2}}{4}\left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert
    ^{2}\leq \frac{1}{4}\left\{ \func{Re}\left[ \left( \overline{\Gamma }+%
    \overline{\gamma }\right) \left\langle x,y\right\rangle \right] \right\}
    ^{2},$$giving the first inequality in (\[2.13\]).

    The second inequality is obvious.

    To prove the sharpness of the constant $\frac{1}{4},$ assume that the first inequality in (\[2.13\]) holds with a constant $c>0,$ i.e., $$\left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}\leq c\cdot \frac{%
    \left\{ \func{Re}\left[ \left( \overline{\Gamma }+\overline{\gamma }\right)
    \left\langle x,y\right\rangle \right] \right\} ^{2}}{\func{Re}\left( \Gamma 
    \overline{\gamma }\right) },  \label{2.16}$$provided $\func{Re}\left( \Gamma \overline{\gamma }\right) >0$ and either (\[2.11\]) or (\[2.12\]) holds.

    Assume that $\Gamma ,\gamma >0,$ and let $x=\gamma y.$ Then (\[2.11\]) holds and by (\[2.16\]) we deduce$$\gamma ^{2}\left\Vert y\right\Vert ^{4}\leq c\cdot \frac{\left( \Gamma
    +\gamma \right) ^{2}\gamma ^{2}\left\Vert y\right\Vert ^{4}}{\Gamma \gamma }$$giving$$\Gamma \gamma \leq c\left( \Gamma +\gamma \right) ^{2}\text{ \ for any \ }%
    \Gamma ,\gamma >0.  \label{2.17}$$Let $\varepsilon \in \left( 0,1\right) $ and choose in (\[2.17\]), $\Gamma
    =1+\varepsilon ,$ $\gamma =1-\varepsilon >0$ to get $1-\varepsilon ^{2}\leq
    4c$ for any $\varepsilon \in \left( 0,1\right) .$ Letting $\varepsilon
    \rightarrow 0+,$ we deduce $c\geq \frac{1}{4},$ and the sharpness of the constant is proved.

    \(ii) and (iii) are obvious and we omit the details.

We observe that the second bound in (\[2.13\]) for $\left\Vert
x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}$ is better than the second bound provided by (\[1.5\]).

The following corollary provides a reverse inequality for the additive version of Schwarz’s inequality.

\[c2.3\]With the assumptions of Theorem \[t2.2\] and if $\func{Re}%
\left( \Gamma \overline{\gamma }\right) >0,$ then we have the inequality:$$0\leq \left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}-\left\vert
\left\langle x,y\right\rangle \right\vert ^{2}\leq \frac{1}{4}\cdot \frac{%
\left\vert \Gamma -\gamma \right\vert ^{2}}{\func{Re}\left( \Gamma \overline{%
\gamma }\right) }\left\vert \left\langle x,y\right\rangle \right\vert ^{2}.
\label{2.18}$$The constant $\frac{1}{4}$ is best possible in (\[2.18\]).

The proof is obvious from (\[2.13\]) on subtracting in both sides the same quantity $\left\vert \left\langle x,y\right\rangle \right\vert ^{2}.$ The sharpness of the constant may be proven in a similar manner to the one incorporated in the proof of (i), Theorem \[t2.2\]. We omit the details.

It is obvious that the inequality (\[2.18\]) is better than (\[1.6\]) obtained in [@SSD3].

For some recent results in connection to Schwarz’s inequality, see [@ADR], [@DM] and [@GH].

Reverses of the Triangle Inequality\[s3\]
=========================================

The following reverse of the triangle inequality holds.

\[p2.4\]Let $\left( H;\left\langle \cdot ,\cdot \right\rangle \right) $ be an inner product space over the real or complex number field $\mathbb{K}$ $\left( \mathbb{K}=\mathbb{R},\mathbb{C}\right) $ and $x,a\in H,$ $r>0$ are such that $$\left\Vert x-a\right\Vert \leq r<\left\Vert a\right\Vert .  \label{2.19}$$Then we have the inequality$$0\leq \left\Vert x\right\Vert +\left\Vert a\right\Vert -\left\Vert
x+a\right\Vert \leq \sqrt{2}r\cdot \sqrt{\frac{\func{Re}\left\langle
x,a\right\rangle }{\sqrt{\left\Vert a\right\Vert ^{2}-r^{2}}\left( \sqrt{%
\left\Vert a\right\Vert ^{2}-r^{2}}+\left\Vert a\right\Vert \right) }}.
\label{2.20}$$

Using the inequality (\[2.8\]), we may write that$$\left\Vert x\right\Vert \left\Vert a\right\Vert \leq \frac{\left\Vert
a\right\Vert \func{Re}\left\langle x,a\right\rangle }{\sqrt{\left\Vert
a\right\Vert ^{2}-r^{2}}},$$giving$$\begin{aligned}
0& \leq \left\Vert x\right\Vert \left\Vert a\right\Vert -\func{Re}%
\left\langle x,a\right\rangle  \label{2.21} \\
& \leq \func{Re}\left\langle x,a\right\rangle \frac{\left\Vert a\right\Vert -%
\sqrt{\left\Vert a\right\Vert ^{2}-r^{2}}}{\sqrt{\left\Vert a\right\Vert
^{2}-r^{2}}}  \notag \\
& =\frac{r^{2}\func{Re}\left\langle x,a\right\rangle }{\sqrt{\left\Vert
a\right\Vert ^{2}-r^{2}}\left( \sqrt{\left\Vert a\right\Vert ^{2}-r^{2}}%
+\left\Vert a\right\Vert \right) }.  \notag\end{aligned}$$Since$$\left( \left\Vert x\right\Vert +\left\Vert a\right\Vert \right)
^{2}-\left\Vert x+a\right\Vert ^{2}=2\left( \left\Vert x\right\Vert
\left\Vert a\right\Vert -\func{Re}\left\langle x,a\right\rangle \right) ,$$then by (\[2.21\]), we have$$\begin{aligned}
\left\Vert x\right\Vert +\left\Vert a\right\Vert & \leq \sqrt{\left\Vert
x+a\right\Vert ^{2}+\frac{2r^{2}\func{Re}\left\langle x,a\right\rangle }{%
\sqrt{\left\Vert a\right\Vert ^{2}-r^{2}}\left( \sqrt{\left\Vert
a\right\Vert ^{2}-r^{2}}+\left\Vert a\right\Vert \right) }} \\
& \leq \left\Vert x+a\right\Vert +\sqrt{2}r\cdot \sqrt{\frac{\func{Re}%
\left\langle x,a\right\rangle }{\sqrt{\left\Vert a\right\Vert ^{2}-r^{2}}%
\left( \sqrt{\left\Vert a\right\Vert ^{2}-r^{2}}+\left\Vert a\right\Vert
\right) }},\end{aligned}$$giving the desired inequality (\[2.20\]).

The following proposition providing a simpler reverse for the triangle inequality also holds.

\[p2.5\]Let $\left( H;\left\langle \cdot ,\cdot \right\rangle \right) $ be an inner product space over $\mathbb{K}$ and $x,y\in H,$ $M>m>0$ such that either$$\func{Re}\left\langle My-x,x-my\right\rangle \geq 0,  \label{2.22}$$or, equivalently,$$\left\Vert x-\frac{M+m}{2}\cdot y\right\Vert \leq \frac{1}{2}\left(
M-m\right) \left\Vert y\right\Vert ,  \label{2.23}$$holds. Then we have the inequality$$0\leq \left\Vert x\right\Vert +\left\Vert y\right\Vert -\left\Vert
x+y\right\Vert \leq \frac{\sqrt{M}-\sqrt{m}}{\sqrt[4]{mM}}\sqrt{\func{Re}%
\left\langle x,y\right\rangle }.  \label{2.24}$$

Choosing in (\[2.8\]), $a=\frac{M+m}{2}y,$ $r=\frac{1}{2}\left( M-m\right)
\left\Vert y\right\Vert $ we get$$\left\Vert x\right\Vert \left\Vert y\right\Vert \sqrt{Mm}\leq \frac{M+m}{2}%
\func{Re}\left\langle x,y\right\rangle$$giving $$0\leq \left\Vert x\right\Vert \left\Vert y\right\Vert -\func{Re}\left\langle
x,y\right\rangle \leq \frac{\left( \sqrt{M}-\sqrt{m}\right) ^{2}}{2\sqrt{mM}}%
\func{Re}\left\langle x,y\right\rangle .$$Following the same arguments as in the proof of Proposition \[p2.4\], we deduce the desired inequality (\[2.24\]).

For some results related to triangle inequality in inner product spaces, see [@JBDFTM], [@SMK], [@PMM] and [@DKR].

Some Grüss Type Inequalities\[s4\]
==================================

We may state the following result.

\[t4.1\]Let $\left( H;\left\langle \cdot ,\cdot \right\rangle \right) $ be an inner product space over the real or complex number field $\mathbb{K}$ $\left( \mathbb{K}=\mathbb{R},\mathbb{K}=\mathbb{C}\right) $ and $x,y,e\in H$ with $\left\Vert e\right\Vert =1.$ If $r_{1},r_{2}\in \left( 0,1\right) $ and $$\left\Vert x-e\right\Vert \leq r_{1},\ \ \ \ \left\Vert y-e\right\Vert \leq
r_{2},  \label{4.1}$$then we have the inequality$$\left\vert \left\langle x,y\right\rangle -\left\langle x,e\right\rangle
\left\langle e,y\right\rangle \right\vert \leq r_{1}r_{2}\left\Vert
x\right\Vert \left\Vert y\right\Vert .  \label{4.2}$$The inequality (\[4.2\]) is sharp in the sense that the constant $c=1$ in front of $r_{1}r_{2}$ cannot be replaced by a smaller constant.

Apply Schwarz’s inequality in $\left( H;\left\langle \cdot ,\cdot
\right\rangle \right) $ for the vectors $x-\left\langle x,e\right\rangle e,$ $y-\left\langle y,e\right\rangle e,$ to get (see also [@SSD3])$$\left\vert \left\langle x,y\right\rangle -\left\langle x,e\right\rangle
\left\langle e,y\right\rangle \right\vert ^{2}\leq \left( \left\Vert
x\right\Vert ^{2}-\left\vert \left\langle x,e\right\rangle \right\vert
^{2}\right) \left( \left\Vert y\right\Vert ^{2}-\left\vert \left\langle
y,e\right\rangle \right\vert ^{2}\right) .  \label{4.3}$$Using Theorem \[t2.1\] for $a=e,$ we may state that$$\left\Vert x\right\Vert ^{2}-\left\vert \left\langle x,e\right\rangle
\right\vert ^{2}\leq r_{1}^{2}\left\Vert x\right\Vert ^{2},\ \ \ \ \ \
\left\Vert y\right\Vert ^{2}-\left\vert \left\langle y,e\right\rangle
\right\vert ^{2}\leq r_{2}^{2}\left\Vert y\right\Vert ^{2}.  \label{4.4}$$Utilizing (\[4.3\]) and (\[4.4\]), we deduce$$\left\vert \left\langle x,y\right\rangle -\left\langle x,e\right\rangle
\left\langle e,y\right\rangle \right\vert ^{2}\leq
r_{1}^{2}r_{2}^{2}\left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2},
\label{4.5}$$which is clearly equivalent to the desired inequality (\[4.2\]).

The sharpness of the constant follows by the fact that for $x=y,$ $%
r_{1}=r_{2}=r,$ we get from (\[4.2\])$$\left\Vert x\right\Vert ^{2}-\left\vert \left\langle x,e\right\rangle
\right\vert ^{2}\leq r^{2}\left\Vert x\right\Vert ^{2}  \label{4.6}$$provided $\left\Vert e\right\Vert =1$ and $\left\Vert x-e\right\Vert \leq
r<1.$ The inequality (\[4.6\]) is sharp, as shown in Theorem \[t2.1\], and the theorem is thus proved.

Another companion of the Grüss inequality may be stated as well.

\[t4.2\]Let $\left( H;\left\langle \cdot ,\cdot \right\rangle \right) $ be an inner product space over $\mathbb{K}$ and $x,y,e\in H$ with $%
\left\Vert e\right\Vert =1.$ Suppose also that $a,A,b,B\in \mathbb{K}$ $%
\left( \mathbb{K}=\mathbb{R},\mathbb{C}\right) $ such that $\func{Re}\left( A%
\overline{a}\right) ,$ $\func{Re}\left( B\overline{b}\right) >0.$ If either$$\func{Re}\left\langle Ae-x,x-ae\right\rangle \geq 0,\ \ \func{Re}%
\left\langle Be-y,y-be\right\rangle \geq 0,\   \label{4.7}$$or, equivalently,$$\left\Vert x-\frac{a+A}{2}e\right\Vert \leq \frac{1}{2}\left\vert
A-a\right\vert ,\ \ \left\Vert y-\frac{b+B}{2}e\right\Vert \leq \frac{1}{2}%
\left\vert B-b\right\vert ,  \label{4.8}$$holds, then we have the inequality$$\left\vert \left\langle x,y\right\rangle -\left\langle x,e\right\rangle
\left\langle e,y\right\rangle \right\vert \leq \frac{1}{4}\cdot \frac{%
\left\vert A-a\right\vert \left\vert B-b\right\vert }{\sqrt{\func{Re}\left( A%
\overline{a}\right) \func{Re}\left( B\overline{b}\right) }}\left\vert
\left\langle x,e\right\rangle \left\langle e,y\right\rangle \right\vert .
\label{4.9}$$The constant $\frac{1}{4}$ is best possible.

We know, by (\[4.3\]), that$$\left\vert \left\langle x,y\right\rangle -\left\langle x,e\right\rangle
\left\langle e,y\right\rangle \right\vert ^{2}\leq \left( \left\Vert
x\right\Vert ^{2}-\left\vert \left\langle x,e\right\rangle \right\vert
^{2}\right) \left( \left\Vert y\right\Vert ^{2}-\left\vert \left\langle
y,e\right\rangle \right\vert ^{2}\right) .  \label{4.10}$$If we use Corollary \[c2.3\], then we may state that$$\left\Vert x\right\Vert ^{2}-\left\vert \left\langle x,e\right\rangle
\right\vert ^{2}\leq \frac{1}{4}\cdot \frac{\left\vert A-a\right\vert ^{2}}{%
\func{Re}\left( A\overline{a}\right) }\left\vert \left\langle
x,e\right\rangle \right\vert ^{2}  \label{4.11}$$and$$\left\Vert y\right\Vert ^{2}-\left\vert \left\langle y,e\right\rangle
\right\vert ^{2}\leq \frac{1}{4}\cdot \frac{\left\vert B-b\right\vert ^{2}}{%
\func{Re}\left( B\overline{b}\right) }\left\vert \left\langle
y,e\right\rangle \right\vert ^{2}.  \label{4.12}$$Utilizing (\[4.10\]) – (\[4.12\]), we deduce$$\left\vert \left\langle x,y\right\rangle -\left\langle x,e\right\rangle
\left\langle e,y\right\rangle \right\vert ^{2}\leq \frac{1}{16}\cdot \frac{%
\left\vert A-a\right\vert ^{2}\left\vert B-b\right\vert ^{2}}{\func{Re}%
\left( A\overline{a}\right) \func{Re}\left( B\overline{b}\right) }\left\vert
\left\langle x,e\right\rangle \left\langle e,y\right\rangle \right\vert ^{2},$$which is clearly equivalent to the desired inequality (\[4.9\]).

The sharpness of the constant follows from Corollary \[c2.3\], and we omit the details.

With the assumptions of Theorem \[t4.2\] and if $\left\langle
x,e\right\rangle ,\left\langle y,e\right\rangle \neq 0$ (that is actually the interesting case), one has the inequality$$\left\vert \frac{\left\langle x,y\right\rangle }{\left\langle
x,e\right\rangle \left\langle e,y\right\rangle }-1\right\vert \leq \frac{1}{4%
}\cdot \frac{\left\vert A-a\right\vert \left\vert B-b\right\vert }{\sqrt{%
\func{Re}\left( A\overline{a}\right) \func{Re}\left( B\overline{b}\right) }}.
\label{4.13}$$The constant $\frac{1}{4}$ is best possible.

The inequality (\[4.9\]) provides a better bound for the quantity $$\left\vert \left\langle x,y\right\rangle -\left\langle x,e\right\rangle
\left\langle e,y\right\rangle \right\vert$$than (2.3) of [@SSD3].

For some recent results on Grüss type inequalities in inner product spaces, see [@SSD0], [@SSD00] and [@PFR].

Reverses of Bessel’s Inequality\[s5\]
=====================================

Let $\left( H;\left\langle \cdot ,\cdot \right\rangle \right) $ be a real or complex infinite dimensional Hilbert space and $\left( e_{i}\right) _{i\in 
\mathbb{N}}$ an orthornormal family in $H$, i.e., we recall that $%
\left\langle e_{i},e_{j}\right\rangle =0$ if $i,j\in \mathbb{N}$, $i\neq j$ and $\left\Vert e_{i}\right\Vert =1$ for $i\in \mathbb{N}$.

It is well known that, if $x\in H,$ then the sum $\sum_{i=1}^{\infty
}\left\vert \left\langle x,e_{i}\right\rangle \right\vert ^{2}$ is convergent and the following inequality, called *Bessel’s inequality*$$\sum_{i=1}^{\infty }\left\vert \left\langle x,e_{i}\right\rangle \right\vert
^{2}\leq \left\Vert x\right\Vert ^{2},  \label{5.1}$$holds.

If $\ell ^{2}\left( \mathbb{K}\right) :=\left\{ \mathbf{a}=\left(
a_{i}\right) _{i\in \mathbb{N}}\subset \mathbb{K}\left\vert
\sum_{i=1}^{\infty }\left\vert a_{i}\right\vert ^{2}\right. <\infty \right\}
,$ where $\mathbb{K}=\mathbb{C}$ or $\mathbb{K}=\mathbb{R}$, is the Hilbert space of all complex or real sequences that are $2$-summable and $\mathbf{%
\lambda }=\left( \lambda _{i}\right) _{i\in \mathbb{N}}\in \ell ^{2}\left( 
\mathbb{K}\right) ,$ then the sum $\sum_{i=1}^{\infty }\lambda _{i}e_{i}$ is convergent in $H$ and if $y:=\sum_{i=1}^{\infty }\lambda _{i}e_{i}\in H,$ then $\left\Vert y\right\Vert =\left( \sum_{i=1}^{\infty }\left\vert \lambda
_{i}\right\vert ^{2}\right) ^{\frac{1}{2}}.$

We may state the following result.

\[t5.1\]Let $\left( H;\left\langle \cdot ,\cdot \right\rangle \right) $ be an infinite dimensional Hilbert space over the real or complex number field $\mathbb{K}$, $\left( e_{i}\right) _{i\in \mathbb{N}}$ an orthornormal family in $H,$ $\mathbf{\lambda }=\left( \lambda _{i}\right) _{i\in \mathbb{N%
}}\in \ell ^{2}\left( \mathbb{K}\right) $ and $r>0$ with the property that$$\sum_{i=1}^{\infty }\left\vert \lambda _{i}\right\vert ^{2}>r^{2}.
\label{5.2}$$If $x\in H$ is such that$$\left\Vert x-\sum_{i=1}^{\infty }\lambda _{i}e_{i}\right\Vert \leq r,
\label{5.3}$$then we have the inequality$$\begin{aligned}
\left\Vert x\right\Vert ^{2}& \leq \frac{\left( \sum_{i=1}^{\infty }\func{Re}%
\left[ \overline{\lambda _{i}}\left\langle x,e_{i}\right\rangle \right]
\right) ^{2}}{\sum_{i=1}^{\infty }\left\vert \lambda _{i}\right\vert
^{2}-r^{2}}  \label{5.4} \\
& \leq \frac{\left\vert \sum_{i=1}^{\infty }\overline{\lambda _{i}}%
\left\langle x,e_{i}\right\rangle \right\vert ^{2}}{\sum_{i=1}^{\infty
}\left\vert \lambda _{i}\right\vert ^{2}-r^{2}}  \notag \\
& \leq \frac{\sum_{i=1}^{\infty }\left\vert \lambda _{i}\right\vert ^{2}}{%
\sum_{i=1}^{\infty }\left\vert \lambda _{i}\right\vert ^{2}-r^{2}}%
\sum_{i=1}^{\infty }\left\vert \left\langle x,e_{i}\right\rangle \right\vert
^{2};  \notag\end{aligned}$$and$$\begin{aligned}
0 &\leq &\left\Vert x\right\Vert ^{2}-\sum_{i=1}^{\infty }\left\vert
\left\langle x,e_{i}\right\rangle \right\vert ^{2}  \label{5.5} \\
&\leq &\frac{r^{2}}{\sum_{i=1}^{\infty }\left\vert \lambda _{i}\right\vert
^{2}-r^{2}}\sum_{i=1}^{\infty }\left\vert \left\langle x,e_{i}\right\rangle
\right\vert ^{2}.\end{aligned}$$

Applying the third inequality in (\[2.2\]) for $a=\sum_{i=1}^{\infty
}\lambda _{i}e_{i}\in H,$ we have$$\left\Vert x\right\Vert ^{2}\left\Vert \sum_{i=1}^{\infty }\lambda
_{i}e_{i}\right\Vert ^{2}-\left[ \func{Re}\left\langle x,\sum_{i=1}^{\infty
}\lambda _{i}e_{i}\right\rangle \right] ^{2}\leq r^{2}\left\Vert
x\right\Vert ^{2}  \label{5.6}$$and since$$\begin{aligned}
\left\Vert \sum_{i=1}^{\infty }\lambda _{i}e_{i}\right\Vert ^{2}&
=\sum_{i=1}^{\infty }\left\vert \lambda _{i}\right\vert ^{2}, \\
\func{Re}\left\langle x,\sum_{i=1}^{\infty }\lambda _{i}e_{i}\right\rangle &
=\sum_{i=1}^{\infty }\func{Re}\left[ \overline{\lambda _{i}}\left\langle
x,e_{i}\right\rangle \right] ,\end{aligned}$$then by (\[5.6\]) we deduce$$\left\Vert x\right\Vert ^{2}\sum_{i=1}^{\infty }\left\vert \lambda
_{i}\right\vert ^{2}-\left[ \func{Re}\left\langle x,\sum_{i=1}^{\infty
}\lambda _{i}e_{i}\right\rangle \right] ^{2}\leq r^{2}\left\Vert
x\right\Vert ^{2},$$giving the first inequality in (\[5.4\]).

The second inequality is obvious by the modulus property.

The last inequality follows by the Cauchy-Bunyakovsky-Schwarz inequality$$\left\vert \sum_{i=1}^{\infty }\overline{\lambda _{i}}\left\langle
x,e_{i}\right\rangle \right\vert ^{2}\leq \sum_{i=1}^{\infty }\left\vert
\lambda _{i}\right\vert ^{2}\sum_{i=1}^{\infty }\left\vert \left\langle
x,e_{i}\right\rangle \right\vert ^{2}.$$The inequality (\[5.5\]) follows by the last inequality in (\[5.4\]) on subtracting in both sides the quantity $\sum_{i=1}^{\infty }\left\vert
\left\langle x,e_{i}\right\rangle \right\vert ^{2}<\infty .$

The following result provides a generalization for the reverse of Bessel’s inequality obtained in [@SSD6].

\[t5.2\]Let $\left( H;\left\langle \cdot ,\cdot \right\rangle \right) $ and $\left( e_{i}\right) _{i\in \mathbb{N}}$ be as in Theorem \[t5.1\]. Suppose that $\mathbf{\Gamma }=\left( \Gamma _{i}\right) _{i\in \mathbb{N}%
}\in \ell ^{2}\left( \mathbb{K}\right) ,$ $\mathbf{\gamma }=\left( \gamma
_{i}\right) _{i\in \mathbb{N}}\in \ell ^{2}\left( \mathbb{K}\right) $ are sequences of real or complex numbers such that$$\sum_{i=1}^{\infty }\func{Re}\left( \Gamma _{i}\overline{\gamma _{i}}\right)
>0.  \label{5.7}$$If $x\in H$ is such that either$$\left\Vert x-\sum_{i=1}^{\infty }\frac{\Gamma _{i}+\gamma _{i}}{2}%
e_{i}\right\Vert \leq \frac{1}{2}\left( \sum_{i=1}^{\infty }\left\vert
\Gamma _{i}-\gamma _{i}\right\vert ^{2}\right) ^{\frac{1}{2}}  \label{5.8}$$or, equivalently,$$\func{Re}\left\langle \sum_{i=1}^{\infty }\Gamma
_{i}e_{i}-x,x-\sum_{i=1}^{\infty }\gamma _{i}e_{i}\right\rangle \geq 0
\label{5.9}$$holds, then we have the inequalities$$\begin{aligned}
\left\Vert x\right\Vert ^{2}& \leq \frac{1}{4}\cdot \frac{\left(
\sum_{i=1}^{\infty }\func{Re}\left[ \left( \overline{\Gamma _{i}}+\overline{%
\gamma _{i}}\right) \left\langle x,e_{i}\right\rangle \right] \right) ^{2}}{%
\sum_{i=1}^{\infty }\func{Re}\left( \Gamma _{i}\overline{\gamma _{i}}\right) 
}  \label{5.10} \\
& \leq \frac{1}{4}\cdot \frac{\left\vert \sum_{i=1}^{\infty }\left( 
\overline{\Gamma _{i}}+\overline{\gamma _{i}}\right) \left\langle
x,e_{i}\right\rangle \right\vert ^{2}}{\sum_{i=1}^{\infty }\func{Re}\left(
\Gamma _{i}\overline{\gamma _{i}}\right) }  \notag \\
& \leq \frac{1}{4}\cdot \frac{\sum_{i=1}^{\infty }\left\vert \Gamma
_{i}+\gamma _{i}\right\vert ^{2}}{\sum_{i=1}^{\infty }\func{Re}\left( \Gamma
_{i}\overline{\gamma _{i}}\right) }\sum_{i=1}^{\infty }\left\vert
\left\langle x,e_{i}\right\rangle \right\vert ^{2}.  \notag\end{aligned}$$The constant $\frac{1}{4}$ is best possible in all inequalities in ([5.10]{}).

We also have the inequalities:$$0\leq \left\Vert x\right\Vert ^{2}-\sum_{i=1}^{\infty }\left\vert
\left\langle x,e_{i}\right\rangle \right\vert ^{2}\leq \frac{1}{4}\cdot 
\frac{\sum_{i=1}^{\infty }\left\vert \Gamma _{i}-\gamma _{i}\right\vert ^{2}%
}{\sum_{i=1}^{\infty }\func{Re}\left( \Gamma _{i}\overline{\gamma _{i}}%
\right) }\sum_{i=1}^{\infty }\left\vert \left\langle x,e_{i}\right\rangle
\right\vert ^{2}.  \label{5.11}$$Here the constant $\frac{1}{4}$ is also best possible.

Since $\mathbf{\Gamma }$, $\mathbf{\gamma }\in \ell ^{2}\left( \mathbb{K}%
\right) ,$ then also $\frac{1}{2}\left( \mathbf{\Gamma }+\mathbf{\gamma }%
\right) \in \ell ^{2}\left( \mathbb{K}\right) ,$ showing that the series$$\sum_{i=1}^{\infty }\left\vert \frac{\Gamma _{i}+\gamma _{i}}{2}\right\vert
^{2},\ \sum_{i=1}^{\infty }\left\vert \frac{\Gamma _{i}-\gamma _{i}}{2}%
\right\vert ^{2}\text{ and}\ \sum_{i=1}^{\infty }\func{Re}\left( \Gamma _{i}%
\overline{\gamma _{i}}\right)$$are convergent. Also, the series $$\sum_{i=1}^{\infty }\Gamma _{i}e_{i},\text{ }\sum_{i=1}^{\infty }\gamma
_{i}e_{i}\text{ and }\sum_{i=1}^{\infty }\frac{\gamma _{i}+\Gamma _{i}}{2}%
e_{i}$$ are convergent in the Hilbert space $H.$

The equivalence of the conditions (\[5.8\]) and (\[5.9\]) follows by the fact that in an inner product space we have, for $x,z,Z\in H,$ $\func{Re}%
\left\langle Z-x,x-z\right\rangle \geq 0$ is equivalent to $\left\Vert x-%
\frac{z+Z}{2}\right\Vert \leq \frac{1}{2}\left\Vert Z-z\right\Vert ,$ and we omit the details.

Now, we observe that the inequalities (\[5.10\]) and (\[5.11\]) follow from Theorem \[t5.1\] on choosing $\lambda _{i}=\frac{\gamma _{i}+\Gamma
_{i}}{2},$ $i\in \mathbb{N}$ and $r=\frac{1}{2}\left( \sum_{i=1}^{\infty
}\left\vert \Gamma _{i}-\gamma _{i}\right\vert ^{2}\right) ^{\frac{1}{2}}.$

The fact that $\frac{1}{4}$ is the best constant in both (\[5.10\]) and (\[5.11\]) follows from Theorem \[t2.2\] and Corollary \[c2.3\], and we omit the details.

Note that (\[5.10\]) improves (\[1.17\]) and (\[5.11\]) improves ([1.18]{}), that have been obtained in [@SSD6].

For some recent results related to Bessel inequality, see [@SSD01], [SSDJS]{}, [@HXC] and [@GH1].

Some Grüss Type Inequalities for Orthonormal Families\[s6\]
===========================================================

The following result related to Grüss inequality in inner product spaces, holds.

\[t6.1\]Let $\left( H;\left\langle \cdot ,\cdot \right\rangle \right) $ be an infinite dimensional Hilbert space over the real or complex number field $\mathbb{K}$, and $\left( e_{i}\right) _{i\in \mathbb{N}}$ an orthornormal family in $H.$ Assume that $\mathbf{\lambda }=\left( \lambda
_{i}\right) _{i\in \mathbb{N}},\ \mathbf{\mu }=\left( \mu _{i}\right) _{i\in 
\mathbb{N}}\in \ell ^{2}\left( \mathbb{K}\right) $ and $r_{1},r_{2}>0$ with the properties that$$\sum_{i=1}^{\infty }\left\vert \lambda _{i}\right\vert ^{2}>r_{1}^{2},\ \ \
\sum_{i=1}^{\infty }\left\vert \mu _{i}\right\vert ^{2}>r_{2}^{2}.
\label{6.1}$$If $x,y\in H$ are such that$$\left\Vert x-\sum_{i=1}^{\infty }\lambda _{i}e_{i}\right\Vert \leq r_{1},\ \
\ \ \ \ \left\Vert y-\sum_{i=1}^{\infty }\mu _{i}e_{i}\right\Vert \leq r_{2},
\label{6.2}$$then we have the inequalities$$\begin{aligned}
&&\left\vert \left\langle x,y\right\rangle -\sum_{i=1}^{\infty }\left\langle
x,e_{i}\right\rangle \left\langle e_{i},y\right\rangle \right\vert 
\label{6.3} \\
&\leq &\frac{r_{1}r_{2}}{\sqrt{\sum_{i=1}^{\infty }\left\vert \lambda
_{i}\right\vert ^{2}-r_{1}^{2}}\sqrt{\sum_{i=1}^{\infty }\left\vert \mu
_{i}\right\vert ^{2}-r_{2}^{2}}}\cdot \sqrt{\sum_{i=1}^{\infty }\left\vert
\left\langle x,e_{i}\right\rangle \right\vert ^{2}\sum_{i=1}^{\infty
}\left\vert \left\langle y,e_{i}\right\rangle \right\vert ^{2}}  \notag \\
&\leq &\frac{r_{1}r_{2}\left\Vert x\right\Vert \left\Vert y\right\Vert }{%
\sqrt{\sum_{i=1}^{\infty }\left\vert \lambda _{i}\right\vert ^{2}-r_{1}^{2}}%
\sqrt{\sum_{i=1}^{\infty }\left\vert \mu _{i}\right\vert ^{2}-r_{2}^{2}}}. 
\notag\end{aligned}$$

Applying Schwarz’s inequality for the vectors $x-\sum_{i=1}^{\infty
}\left\langle x,e_{i}\right\rangle e_{i},$ $y-\sum_{i=1}^{\infty
}\left\langle y,e_{i}\right\rangle e_{i},$ we have$$\begin{gathered}
\left\vert \left\langle x-\sum_{i=1}^{\infty }\left\langle
x,e_{i}\right\rangle e_{i},y-\sum_{i=1}^{\infty }\left\langle
y,e_{i}\right\rangle e_{i}\right\rangle \right\vert ^{2}  \label{6.4} \\
\leq \left\Vert x-\sum_{i=1}^{\infty }\left\langle x,e_{i}\right\rangle
e_{i}\right\Vert ^{2}\left\Vert y-\sum_{i=1}^{\infty }\left\langle
y,e_{i}\right\rangle e_{i}\right\Vert ^{2}.\end{gathered}$$Since$$\left\langle x-\sum_{i=1}^{\infty }\left\langle x,e_{i}\right\rangle
e_{i},y-\sum_{i=1}^{\infty }\left\langle y,e_{i}\right\rangle
e_{i}\right\rangle =\left\langle x,y\right\rangle -\sum_{i=1}^{\infty
}\left\langle x,e_{i}\right\rangle \left\langle e_{i},y\right\rangle$$and$$\left\Vert x-\sum_{i=1}^{\infty }\left\langle x,e_{i}\right\rangle
e_{i}\right\Vert ^{2}=\left\Vert x\right\Vert ^{2}-\sum_{i=1}^{\infty
}\left\vert \left\langle x,e_{i}\right\rangle \right\vert ^{2},$$then by (\[5.5\]) applied for $x$ and $y,$ and from (\[6.4\]), we deduce the first part of (\[6.3\]).

The second part follows by Bessel’s inequality.

The following Grüss type inequality may be stated as well.

\[t6.2\]Let $\left( H;\left\langle \cdot ,\cdot \right\rangle \right) $ be an infinite dimensional Hilbert space and $\left( e_{i}\right) _{i\in 
\mathbb{N}}$ an orthornormal family in $H.$ Suppose that $\left( \Gamma
_{i}\right) _{i\in \mathbb{N}},$ $\left( \gamma _{i}\right) _{i\in \mathbb{N}%
},$ $\left( \phi _{i}\right) _{i\in \mathbb{N}},$ $\left( \Phi _{i}\right)
_{i\in \mathbb{N}}\in \ell ^{2}\left( \mathbb{K}\right) $ are sequences of real and complex numbers such that$$\sum_{i=1}^{\infty }\func{Re}\left( \Gamma _{i}\overline{\gamma _{i}}\right)
>0,\ \ \ \sum_{i=1}^{\infty }\func{Re}\left( \Phi _{i}\overline{\phi _{i}}%
\right) >0.  \label{6.5}$$If $x,y\in H$ are such that either$$\begin{aligned}
\left\Vert x-\sum_{i=1}^{\infty }\frac{\Gamma _{i}+\gamma _{i}}{2}\cdot
e_{i}\right\Vert & \leq \frac{1}{2}\left( \sum_{i=1}^{\infty }\left\vert
\Gamma _{i}-\gamma _{i}\right\vert ^{2}\right) ^{\frac{1}{2}}  \label{6.6} \\
\left\Vert y-\sum_{i=1}^{\infty }\frac{\Phi _{i}+\phi _{i}}{2}\cdot
e_{i}\right\Vert & \leq \frac{1}{2}\left( \sum_{i=1}^{\infty }\left\vert
\Phi _{i}-\phi _{i}\right\vert ^{2}\right) ^{\frac{1}{2}}  \notag\end{aligned}$$or, equivalently,$$\begin{aligned}
\func{Re}\left\langle \sum_{i=1}^{\infty }\Gamma
_{i}e_{i}-x,x-\sum_{i=1}^{\infty }\gamma _{i}e_{i}\right\rangle & \geq 0,
\label{6.7} \\
\func{Re}\left\langle \sum_{i=1}^{\infty }\Phi
_{i}e_{i}-y,y-\sum_{i=1}^{\infty }\phi _{i}e_{i}\right\rangle & \geq 0, 
\notag\end{aligned}$$holds, then we have the inequality$$\begin{aligned}
& \left\vert \left\langle x,y\right\rangle -\sum_{i=1}^{\infty }\left\langle
x,e_{i}\right\rangle \left\langle e_{i},y\right\rangle \right\vert 
\label{6.8} \\
& \leq \frac{1}{4}\cdot \frac{\left( \sum_{i=1}^{\infty }\left\vert \Gamma
_{i}-\gamma _{i}\right\vert ^{2}\right) ^{\frac{1}{2}}\left(
\sum_{i=1}^{\infty }\left\vert \Phi _{i}-\phi _{i}\right\vert ^{2}\right) ^{%
\frac{1}{2}}}{\left( \sum_{i=1}^{\infty }\func{Re}\left( \Gamma _{i}%
\overline{\gamma _{i}}\right) \right) ^{\frac{1}{2}}\left(
\sum_{i=1}^{\infty }\func{Re}\left( \Phi _{i}\overline{\phi _{i}}\right)
\right) ^{\frac{1}{2}}}  \notag \\
& \times \left( \sum_{i=1}^{\infty }\left\vert \left\langle
x,e_{i}\right\rangle \right\vert ^{2}\right) ^{\frac{1}{2}}\left(
\sum_{i=1}^{\infty }\left\vert \left\langle y,e_{i}\right\rangle \right\vert
^{2}\right) ^{\frac{1}{2}}  \notag \\
& \leq \frac{1}{4}\cdot \frac{\left( \sum_{i=1}^{\infty }\left\vert \Gamma
_{i}-\gamma _{i}\right\vert ^{2}\right) ^{\frac{1}{2}}\left(
\sum_{i=1}^{\infty }\left\vert \Phi _{i}-\phi _{i}\right\vert ^{2}\right) ^{%
\frac{1}{2}}}{\left[ \sum_{i=1}^{\infty }\func{Re}\left( \Gamma _{i}%
\overline{\gamma _{i}}\right) \right] ^{\frac{1}{2}}\left[
\sum_{i=1}^{\infty }\func{Re}\left( \Phi _{i}\overline{\phi _{i}}\right) %
\right] ^{\frac{1}{2}}}\left\Vert x\right\Vert \left\Vert y\right\Vert  
\notag\end{aligned}$$The constant $\frac{1}{4}$ is best possible in the first inequality.

Follows by (\[5.11\]) and (\[6.4\]).

The best constant follows from Theorem \[t4.2\], and we omit the details.

We note that the inequality (\[6.8\]) is better than the inequality (3.3) in [@SSD6]. We omit the details.

Integral Inequalities\[s7\]
===========================

Let $\left( \Omega ,\Sigma ,\mu \right) $ be a measurable space consisting of a set $\Omega ,$ a $\sigma -$algebra of parts $\Sigma $ and a countably additive and positive measure $\mu $ on $\Sigma $ with values $\mathbb{R\cup 
}\left\{ \infty \right\} .$ Let $\rho \geq 0$ be a $g-$measurable function on $\Omega $ with $\int_{\Omega }\rho \left( s\right) d\mu \left( s\right)
=1.$ Denote by $L_{\rho }^{2}\left( \Omega ,\mathbb{K}\right) $ the Hilbert space of all real or complex valued functions defined on $\Omega $ and $%
2-\rho -$integrable on $\Omega ,$ i.e.,$$\int_{\Omega }\rho \left( s\right) \left\vert f\left( s\right) \right\vert
^{2}d\mu \left( s\right) <\infty .  \label{7.1}$$It is obvious that the following inner product$$\left\langle f,g\right\rangle _{\rho }:=\int_{\Omega }\rho \left( s\right)
f\left( s\right) \overline{g\left( s\right) }d\mu \left( s\right) ,
\label{7.2}$$generates the norm $\left\Vert f\right\Vert _{\rho }:=\left( \int_{\Omega
}\rho \left( s\right) \left\vert f\left( s\right) \right\vert ^{2}d\mu
\left( s\right) \right) ^{\frac{1}{2}}$ of $L_{\rho }^{2}\left( \Omega ,%
\mathbb{K}\right) ,$ and all the above results may be stated for integrals.

It is important to observe that, if $$\func{Re}\left[ f\left( s\right) \overline{g\left( s\right) }\right] \geq 0%
\text{ \ for }\mu -\text{a.e. }s\in \Omega ,  \label{7.3}$$then, obviously,$$\begin{aligned}
\func{Re}\left\langle f,g\right\rangle _{\rho }& =\func{Re}\left[
\int_{\Omega }\rho \left( s\right) f\left( s\right) \overline{g\left(
s\right) }d\mu \left( s\right) \right]  \label{7.4} \\
& =\int_{\Omega }\rho \left( s\right) \func{Re}\left[ f\left( s\right) 
\overline{g\left( s\right) }\right] d\mu \left( s\right) \geq 0.  \notag\end{aligned}$$The reverse is evidently not true in general.

Moreover, if the space is real, i.e., $\mathbb{K=R}$, then a sufficient condition for (\[7.4\]) to hold is:$$f\left( s\right) \geq 0,\ \ g\left( s\right) \geq 0\text{ \ for }\mu -\text{%
a.e. }s\in \Omega .  \label{7.5}$$

We provide now, by the use of certain result obtained in Section \[s2\], some integral inequalities that may be used in practical applications.

\[p7.1\]Let $f,g\in L_{\rho }^{2}\left( \Omega ,\mathbb{K}\right) $ and $%
r>0$ with the properties that$$\left\vert f\left( s\right) -g\left( s\right) \right\vert \leq r\leq
\left\vert g\left( s\right) \right\vert \ \text{\ for }\mu -\text{a.e. }s\in
\Omega ,  \label{7.6}$$and $\int_{\Omega }\rho \left( s\right) \left\vert g\left( s\right)
\right\vert ^{2}d\mu \left( s\right) \neq r.$ Then we have the inequalities$$\begin{aligned}
0& \leq \int_{\Omega }\rho \left( s\right) \left\vert f\left( s\right)
\right\vert ^{2}d\mu \left( s\right) \int_{\Omega }\rho \left( s\right)
\left\vert g\left( s\right) \right\vert ^{2}d\mu \left( s\right) -\left\vert
\int_{\Omega }\rho \left( s\right) f\left( s\right) \overline{g\left(
s\right) }d\mu \left( s\right) \right\vert ^{2}  \label{7.7} \\
& \leq \int_{\Omega }\rho \left( s\right) \left\vert f\left( s\right)
\right\vert ^{2}d\mu \left( s\right) \int_{\Omega }\rho \left( s\right)
\left\vert g\left( s\right) \right\vert ^{2}d\mu \left( s\right)  \notag \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\left[ \int_{\Omega
}\rho \left( s\right) \func{Re}\left( f\left( s\right) \overline{g\left(
s\right) }\right) d\mu \left( s\right) \right] ^{2}  \notag \\
& \leq r^{2}\int_{\Omega }\rho \left( s\right) \left\vert g\left( s\right)
\right\vert ^{2}d\mu \left( s\right) .  \notag\end{aligned}$$The constant $c=1$ in front of $r^{2}$ is best possible.

The proof follows by Theorem \[t2.1\] and we omit the details.

\[p7.2\]Let $f,g\in L_{\rho }^{2}\left( \Omega ,\mathbb{K}\right) $ and $%
\gamma ,\Gamma \in \mathbb{K}$ such that $\func{Re}\left( \Gamma \overline{%
\gamma }\right) >0$ and$$\func{Re}\left[ \left( \Gamma g\left( s\right) -f\left( s\right) \right)
\left( \overline{f\left( s\right) }-\overline{\gamma }\overline{g\left(
s\right) }\right) \right] \geq 0\text{ \ for }\mu -\text{a.e. }s\in \Omega .
\label{7.8}$$Then we have the inequalities$$\begin{aligned}
& \int_{\Omega }\rho \left( s\right) \left\vert f\left( s\right) \right\vert
^{2}d\mu \left( s\right) \int_{\Omega }\rho \left( s\right) \left\vert
g\left( s\right) \right\vert ^{2}d\mu \left( s\right)   \label{7.9} \\
& \leq \frac{1}{4}\cdot \frac{\left\{ \func{Re}\left[ \left( \overline{%
\Gamma }+\overline{\gamma }\right) \int_{\Omega }\rho \left( s\right)
f\left( s\right) \overline{g\left( s\right) }d\mu \left( s\right) \right]
\right\} ^{2}}{\func{Re}\left( \Gamma \overline{\gamma }\right) }  \notag \\
& \leq \frac{1}{4}\cdot \frac{\left\vert \Gamma +\gamma \right\vert ^{2}}{%
\func{Re}\left( \Gamma \overline{\gamma }\right) }\left\vert \int_{\Omega
}\rho \left( s\right) f\left( s\right) \overline{g\left( s\right) }d\mu
\left( s\right) \right\vert ^{2}.  \notag\end{aligned}$$The constant $\frac{1}{4}$ is best possible in both inequalities.

The proof follows by Theorem \[t2.2\] and we omit the details.

\[c7.3\]With the assumptions of Proposition \[p7.2\], we have the inequality$$\begin{aligned}
0& \leq \int_{\Omega }\rho \left( s\right) \left\vert f\left( s\right)
\right\vert ^{2}d\mu \left( s\right) \int_{\Omega }\rho \left( s\right)
\left\vert g\left( s\right) \right\vert ^{2}d\mu \left( s\right) 
\label{7.10} \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\left\vert \int_{\Omega
}\rho \left( s\right) f\left( s\right) \overline{g\left( s\right) }d\mu
\left( s\right) \right\vert ^{2}  \notag \\
& \leq \frac{1}{4}\cdot \frac{\left\vert \Gamma -\gamma \right\vert ^{2}}{%
\func{Re}\left( \Gamma \overline{\gamma }\right) }\left\vert \int_{\Omega
}\rho \left( s\right) f\left( s\right) \overline{g\left( s\right) }d\mu
\left( s\right) \right\vert ^{2}.  \notag\end{aligned}$$The constant $\frac{1}{4}$ is best possible.

If the space is real and we assume, for $M>m>0,$ that$$mg\left( s\right) \leq f\left( s\right) \leq Mg\left( s\right) \text{ \ for }%
\mu -\text{a.e. }s\in \Omega ,  \label{7.11}$$then, by (\[7.9\]) and (\[7.10\]), we deduce the inequalities$$\begin{gathered}
\int_{\Omega }\rho \left( s\right) \left[ f\left( s\right) \right] ^{2}d\mu
\left( s\right) \int_{\Omega }\rho \left( s\right) \left[ g\left( s\right) %
\right] ^{2}d\mu \left( s\right)   \label{7.12} \\
\leq \frac{1}{4}\cdot \frac{\left( M+m\right) ^{2}}{mM}\left[ \int_{\Omega
}\rho \left( s\right) f\left( s\right) g\left( s\right) d\mu \left( s\right) %
\right] ^{2}.\end{gathered}$$and $$\begin{aligned}
0& \leq \int_{\Omega }\rho \left( s\right) \left[ f\left( s\right) \right]
^{2}d\mu \left( s\right) \int_{\Omega }\rho \left( s\right) \left[ g\left(
s\right) \right] ^{2}d\mu \left( s\right)   \label{7.13} \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\left[ \int_{\Omega }\rho \left(
s\right) f\left( s\right) g\left( s\right) d\mu \left( s\right) \right] ^{2}
\notag \\
& \leq \frac{1}{4}\cdot \frac{\left( M-m\right) ^{2}}{mM}\left[ \int_{\Omega
}\rho \left( s\right) f\left( s\right) g\left( s\right) d\mu \left( s\right) %
\right] ^{2}.  \notag\end{aligned}$$The inequality (\[7.12\]) is known in the literature as Cassel’s inequality.

The following Grüss type integral inequality for real or complex-valued functions also holds.

\[p.7.3\]Let $f,g,h\in L_{\rho }^{2}\left( \Omega ,\mathbb{K}\right) $ with $\int_{\Omega }\rho \left( s\right) \left\vert h\left( s\right)
\right\vert ^{2}d\mu \left( s\right) =1$ and $a,A,b,B\in \mathbb{K}$ such that $\func{Re}\left( A\overline{a}\right) ,\func{Re}\left( B\overline{b}%
\right) >0$ and$$\begin{aligned}
\func{Re}\left[ \left( Ah\left( s\right) -f\left( s\right) \right) \left( 
\overline{f\left( s\right) }-\overline{a}\overline{h\left( s\right) }\right) %
\right]  &\geq &0, \\
\func{Re}\left[ \left( Ah\left( s\right) -g\left( s\right) \right) \left( 
\overline{g\left( s\right) }-\overline{b}\overline{h\left( s\right) }\right) %
\right]  &\geq &0\text{,}\end{aligned}$$for $\mu -$a.e. $s\in \Omega .$ Then we have the inequalities$$\begin{aligned}
& \left\vert \int_{\Omega }\rho \left( s\right) f\left( s\right) \overline{%
g\left( s\right) }d\mu \left( s\right) -\int_{\Omega }\rho \left( s\right)
f\left( s\right) \overline{h\left( s\right) }d\mu \left( s\right)
\int_{\Omega }\rho \left( s\right) h\left( s\right) \overline{g\left(
s\right) }d\mu \left( s\right) \right\vert  \\
& \leq \frac{1}{4}\cdot \frac{\left\vert A-a\right\vert \left\vert
B-b\right\vert }{\sqrt{\func{Re}\left( A\overline{a}\right) \func{Re}\left( B%
\overline{b}\right) }}\left\vert \int_{\Omega }\rho \left( s\right) f\left(
s\right) \overline{h\left( s\right) }d\mu \left( s\right) \int_{\Omega }\rho
\left( s\right) h\left( s\right) \overline{g\left( s\right) }d\mu \left(
s\right) \right\vert   \notag\end{aligned}$$The constant $\frac{1}{4}$ is best possible.

The proof follows by Theorem \[t4.2\].

All the other inequalities in Sections \[s3\] – \[s6\] may be used in a similar way to obtain the corresponding integral inequalities. We omit the details.

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