Vector Inequalities for a Projection in Hilbert Spaces and Applications

Abstract

In this paper we establish some vector inequalities related to Schwarz and Buzano results. Applications for norm and numerical radius inequalities of two bounded operators are given as well.

1 Introduction

Let $\left (H,\left \langle \cdot,\cdot \right \rangle \right )$ be an inner product space over the real or complex numbers field $\mathbb{K}$. The following inequality is well known in literature as the Schwarz’s inequality

$$\displaystyle{ \left \Vert x\right \Vert \left \Vert y\right \Vert \geq \left \vert \left \langle x,y\right \rangle \right \vert \text{ for any }x,y \in H. }$$

(1)

The equality case holds in (1) if and only if there exists a constant $\lambda \in \mathbb{K}$ such that x = λ y. 

In 1985 the author [5] (see also [23]) established the following refinement of (1):

$$\displaystyle{ \left \Vert x\right \Vert \left \Vert y\right \Vert \geq \left \vert \left \langle x,y\right \rangle -\left \langle x,e\right \rangle \left \langle e,y\right \rangle \right \vert + \left \vert \left \langle x,e\right \rangle \left \langle e,y\right \rangle \right \vert \geq \left \vert \left \langle x,y\right \rangle \right \vert }$$

(2)

for any x, y, e ∈ H with $\left \Vert e\right \Vert = 1.$

Using the triangle inequality for modulus we have

$$\displaystyle{ \left \vert \left \langle x,y\right \rangle -\left \langle x,e\right \rangle \left \langle e,y\right \rangle \right \vert \geq \left \vert \left \langle x,e\right \rangle \left \langle e,y\right \rangle \right \vert -\left \vert \left \langle x,y\right \rangle \right \vert }$$

and by (2) we get

$$\displaystyle\begin{array}{rcl} \left \Vert x\right \Vert \left \Vert y\right \Vert & \geq & \left \vert \left \langle x,y\right \rangle -\left \langle x,e\right \rangle \left \langle e,y\right \rangle \right \vert + \left \vert \left \langle x,e\right \rangle \left \langle e,y\right \rangle \right \vert {}\\ & \geq & 2\left \vert \left \langle x,e\right \rangle \left \langle e,y\right \rangle \right \vert -\left \vert \left \langle x,y\right \rangle \right \vert, {}\\ \end{array}$$

which implies the Buzano’s inequality [2]

$$\displaystyle{ \frac{1} {2}\left [\left \Vert x\right \Vert \left \Vert y\right \Vert + \left \vert \left \langle x,y\right \rangle \right \vert \right ] \geq \left \vert \left \langle x,e\right \rangle \left \langle e,y\right \rangle \right \vert }$$

(3)

that holds for any x, y, e ∈ H with $\left \Vert e\right \Vert = 1.$

For other Schwarz and Buzano related inequalities in inner product spaces, see [1–10, 12–15, 17, 19–25, 27–36], and the monographs [11, 16] and [18].

Now, let us recall some basic facts on orthogonal projection that will be used in the sequel.

If K is a subset of a Hilbert space $\left (H,\left \langle \cdot,\cdot \right \rangle \right )$, the set of vectors orthogonal to K is defined by

$$\displaystyle{ K^{\perp }:= \left \{x \in H: \left \langle x,k\right \rangle = 0\text{ for all }k \in K\right \}. }$$

We observe that K ^⊥ is a closed subspace of H and so forms itself a Hilbert space. If V is a closed subspace of H, then V ^⊥ is called the orthogonal complement of V. In fact, every x in H can then be written uniquely as x = v + w, with v in V and w in K ^⊥. Therefore, H is the internal Hilbert direct sum of V and V ^⊥, and we denote that as H = V ⊕ V ^⊥. 

The linear operator P _V : H → H that maps x to v is called the orthogonal projection onto V. There is a natural one-to-one correspondence between the set of all closed subspaces of H and the set of all bounded self-adjoint operators P such that P ² = P. Specifically, the orthogonal projection P _V is a self-adjoint linear operator on H of norm ≤ 1 with the property P _V ² = P _V . Moreover, any self-adjoint linear operator E such that E ² = E is of the form P _V , where V is the range of E. For every x in H, P _V (x) is the unique element v of V, which minimizes the distance $\left \Vert x - v\right \Vert$. This provides the geometrical interpretation of P _V (x): it is the best approximation to x by elements of V.

Projections P _U and P _V are called mutually orthogonal if P _U P _V  = 0. This is equivalent to U and V being orthogonal as subspaces of H. The sum of the two projections P _U and P _V is a projection only if U and V are orthogonal to each other, and in that case P _U + P _V  = P _U+V . The composite P _U P _V is generally not a projection; in fact, the composite is a projection if and only if the two projections commute, and in that case P _U P _V  = P _U∩V .

A family $\left \{e_{j}\right \}_{j\in J}$ of vectors in H is called orthonormal if

$$\displaystyle{ e_{j} \perp e_{k}\text{ for any }j,k \in J\text{ with }j\neq k\text{ and }\left \Vert e_{j}\right \Vert = 1\text{ for any }j,k \in J. }$$

If the linear span of the family $\left \{e_{j}\right \}_{j\in J}$ is dense in H, then we call it an orthonormal basis in H. 

It is well known that for any orthonormal family $\left \{e_{j}\right \}_{j\in J}$ we have Bessel’s inequality

$$\displaystyle{ \sum _{j\in J}\left \vert \left \langle x,e_{j}\right \rangle \right \vert ^{2} \leq \left \Vert x\right \Vert ^{2}\text{ for any }x \in H. }$$

This becomes Parseval’s identity

$$\displaystyle{ \sum _{j\in J}\left \vert \left \langle x,e_{j}\right \rangle \right \vert ^{2} = \left \Vert x\right \Vert ^{2}\text{ for any }x \in H, }$$

when $\left \{e_{j}\right \}_{j\in J}$ an othonormal basis in H. 

For an othonormal family $\mathcal{E} = \left \{e_{j}\right \}_{j\in J}$ we define the operator $P_{\mathcal{E}}: H \rightarrow H$ by

$$\displaystyle{ P_{\mathcal{E}}x:=\sum _{j\in J}\left \langle x,e_{j}\right \rangle e_{j\text{ }},\text{ }x \in H. }$$

(4)

We know that $P_{\mathcal{E}}$ is an orthogonal projection and

$$\displaystyle{ \left \langle P_{\mathcal{E}}x,y\right \rangle =\sum _{j\in J}\left \langle x,e_{j}\right \rangle \left \langle e_{j},y\right \rangle,\text{ }x,y \in H\text{ and }\left \langle P_{\mathcal{E}}x,x\right \rangle =\sum _{j\in J}\left \vert \left \langle x,e_{j}\right \rangle \right \vert ^{2},\text{ }x \in H. }$$

The particular case when the family reduces to one vector, namely $\mathcal{E} = \left \{e\right \},\ \left \Vert e\right \Vert = 1,$ is of interest since in this case $P_{e}x:= \left \langle x,e\right \rangle e,\ x \in H,$

$$\displaystyle{ \left \langle P_{e}x,y\right \rangle = \left \langle x,e\right \rangle \left \langle e,y\right \rangle,\text{ }x,y \in H\text{ } }$$

(5)

and Buzano’s inequality can be written as

$$\displaystyle{ \frac{1} {2}\left [\left \Vert x\right \Vert \left \Vert y\right \Vert + \left \vert \left \langle x,y\right \rangle \right \vert \right ] \geq \left \vert \left \langle P_{e}x,y\right \rangle \right \vert }$$

(6)

that holds for any x, y, e ∈ H with $\left \Vert e\right \Vert = 1.$

Motivated by the above results we establish in this paper some vector inequalities for an orthogonal projection P that generalizes amongst others the Buzano’s inequality (6). Applications for norm and numerical radius inequalities are provided as well.

2 Vector Inequalities for a Projection

Assume that P: H → H is an orthogonal projection on H, namely it satisfies the condition P ² = P = P ^∗. We obviously have in the operator order of $\mathcal{B}\left (H\right )$ that 0 ≤ P ≤ 1 _H .

The following result holds:

Theorem 1.

Let P: H → H is an orthogonal projection on H. Then for any x,y ∈ H we have the inequalities

$$\displaystyle{ \left \Vert x\right \Vert \left \Vert y\right \Vert -\left \langle Px,x\right \rangle ^{1/2}\left \langle Py,y\right \rangle ^{1/2} \geq \left \vert \left \langle x,y\right \rangle -\left \langle Px,y\right \rangle \right \vert. }$$

(7)

and

$$\displaystyle{ \left \Vert x\right \Vert \left \Vert y\right \Vert -\left (\left \Vert x\right \Vert ^{2} -\left \langle Px,x\right \rangle \right )^{1/2}\left (\left \Vert y\right \Vert ^{2} -\left \langle Py,y\right \rangle \right )^{1/2} \geq \left \vert \left \langle Px,y\right \rangle \right \vert. }$$

(8)

Proof.

Using the properties of projection, we have

$$\displaystyle\begin{array}{rcl} \left \langle x - Px,y - Py\right \rangle & =& \left \langle x,y\right \rangle -\left \langle Px,y\right \rangle -\left \langle x,Py\right \rangle + \left \langle Px,Py\right \rangle \\ & =& \left \langle x,y\right \rangle - 2\left \langle Px,y\right \rangle + \left \langle P^{2}x,y\right \rangle \\ & =& \left \langle x,y\right \rangle -\left \langle Px,y\right \rangle {}\end{array}$$

(9)

for any x, y ∈ H. 

By Schwarz’s inequality we have

$$\displaystyle{ \left \Vert x - Px\right \Vert ^{2}\left \Vert y - Py\right \Vert ^{2} \geq \left \vert \left \langle x - Px,y - Py\right \rangle \right \vert ^{2} }$$

(10)

for any x, y ∈ H. 

Since, by (7), we have

$$\displaystyle{ \left \Vert x - Px\right \Vert ^{2} = \left \Vert x\right \Vert ^{2} -\left \langle Px,x\right \rangle,\text{ }\left \Vert y - Py\right \Vert ^{2} = \left \Vert y\right \Vert ^{2} -\left \langle Py,y\right \rangle, }$$

then by (10) we have

$$\displaystyle{ \left (\left \Vert x\right \Vert ^{2} -\left \langle Px,x\right \rangle \right )\left (\left \Vert y\right \Vert ^{2} -\left \langle Py,y\right \rangle \right ) \geq \left \vert \left \langle x,y\right \rangle -\left \langle Px,y\right \rangle \right \vert ^{2} }$$

(11)

for any x, y ∈ H. 

Using the elementary inequality that holds for any real numbers a, b, c, d

$$\displaystyle{ \left (ac - bd\right )^{2} \geq \left (a^{2} - b^{2}\right )\left (c^{2} - d^{2}\right ), }$$

we have

$$\displaystyle{ \left (\left \Vert x\right \Vert \left \Vert y\right \Vert -\left \langle Px,x\right \rangle ^{1/2}\left \langle Py,y\right \rangle ^{1/2}\right )^{2} \geq \left (\left \Vert x\right \Vert ^{2} -\left \langle Px,x\right \rangle \right )\left (\left \Vert y\right \Vert ^{2} -\left \langle Py,y\right \rangle \right ) }$$

(12)

for any x, y ∈ H. 

Since

$$\displaystyle{ \left \Vert x\right \Vert \geq \left \langle Px,x\right \rangle ^{1/2},\text{ }\left \Vert y\right \Vert \geq \left \langle Py,y\right \rangle ^{1/2}, }$$

then

$$\displaystyle{ \left \Vert x\right \Vert \left \Vert y\right \Vert -\left \langle Px,x\right \rangle ^{1/2}\left \langle Py,y\right \rangle ^{1/2} \geq 0, }$$

for any x, y ∈ H. 

By (11) and (12) we get

$$\displaystyle{ \left (\left \Vert x\right \Vert \left \Vert y\right \Vert -\left \langle Px,x\right \rangle ^{1/2}\left \langle Py,y\right \rangle ^{1/2}\right )^{2} \geq \left \vert \left \langle x,y\right \rangle -\left \langle Px,y\right \rangle \right \vert ^{2} }$$

for any x, y ∈ H, which, by taking the square root, is equivalent to the desired inequality (7).

Observe that, if P is an orthogonal projection, then Q: = 1 _H − P is also a projection. Indeed we have

$$\displaystyle{ Q^{2} = \left (1_{ H} - P\right )^{2} = 1_{ H} - 2P + P^{2} = 1_{ H} - P = Q. }$$

Now, if we write the inequality (7) for the projection Q we get the desired inequality (8).

Corollary 1.

With the assumptions of Theorem 1, we have the following refinements of Schwarz inequality:

$$\displaystyle\begin{array}{rcl} \left \Vert x\right \Vert \left \Vert y\right \Vert & \geq & \left \langle Px,x\right \rangle ^{1/2}\left \langle Py,y\right \rangle ^{1/2} + \left \vert \left \langle x,y\right \rangle -\left \langle Px,y\right \rangle \right \vert \\ & \geq & \left \vert \left \langle Px,y\right \rangle \right \vert + \left \vert \left \langle x,y\right \rangle -\left \langle Px,y\right \rangle \right \vert \geq \left \vert \left \langle x,y\right \rangle \right \vert {}\end{array}$$

(13)

and

$$\displaystyle\begin{array}{rcl} \left \Vert x\right \Vert \left \Vert y\right \Vert & \geq & \left (\left \Vert x\right \Vert ^{2} -\left \langle Px,x\right \rangle \right )^{1/2}\left (\left \Vert y\right \Vert ^{2} -\left \langle Py,y\right \rangle \right )^{1/2} + \left \vert \left \langle Px,y\right \rangle \right \vert \\ & \geq & \left \vert \left \langle x,y\right \rangle -\left \langle Px,y\right \rangle \right \vert + \left \vert \left \langle Px,y\right \rangle \right \vert \geq \left \vert \left \langle x,y\right \rangle \right \vert {}\end{array}$$

(14)

for any x,y ∈ H.

Remark 1.

Since

$$\displaystyle{ \left \vert \left \langle x,y\right \rangle -\left \langle Px,y\right \rangle \right \vert \geq \left \vert \left \langle x,y\right \rangle \right \vert -\left \vert \left \langle Px,y\right \rangle \right \vert }$$

then by the first inequality in (13) we have

$$\displaystyle{ \left \Vert x\right \Vert \left \Vert y\right \Vert \geq \left \langle Px,x\right \rangle ^{1/2}\left \langle Py,y\right \rangle ^{1/2} + \left \vert \left \langle x,y\right \rangle \right \vert -\left \vert \left \langle Px,y\right \rangle \right \vert }$$

that produces the inequality

$$\displaystyle{ \left \Vert x\right \Vert \left \Vert y\right \Vert -\left \vert \left \langle x,y\right \rangle \right \vert \geq \left \langle Px,x\right \rangle ^{1/2}\left \langle Py,y\right \rangle ^{1/2} -\left \vert \left \langle Px,y\right \rangle \right \vert \geq 0 }$$

(15)

for any x, y ∈ H. 

We notice that the second inequality follows by Schwarz’s inequality for the nonnegative self-adjoint operator P. 

Since

$$\displaystyle{ \left \vert \left \langle x,y\right \rangle -\left \langle Px,y\right \rangle \right \vert \geq \left \vert \left \langle Px,y\right \rangle \right \vert -\left \vert \left \langle x,y\right \rangle \right \vert }$$

then by (13) we have

$$\displaystyle\begin{array}{rcl} \left \Vert x\right \Vert \left \Vert y\right \Vert & \geq & \left \langle Px,x\right \rangle ^{1/2}\left \langle Py,y\right \rangle ^{1/2} + \left \vert \left \langle x,y\right \rangle -\left \langle Px,y\right \rangle \right \vert {}\\ & \geq & \left \langle Px,x\right \rangle ^{1/2}\left \langle Py,y\right \rangle ^{1/2} + \left \vert \left \langle Px,y\right \rangle \right \vert -\left \vert \left \langle x,y\right \rangle \right \vert, {}\\ \end{array}$$

which implies that

$$\displaystyle\begin{array}{rcl} \left \Vert x\right \Vert \left \Vert y\right \Vert + \left \vert \left \langle x,y\right \rangle \right \vert & \geq & \left \langle Px,x\right \rangle ^{1/2}\left \langle Py,y\right \rangle ^{1/2} + \left \vert \left \langle Px,y\right \rangle \right \vert {}\\ & \geq & 2\left \vert \left \langle Px,y\right \rangle \right \vert {}\\ \end{array}$$

and is equivalent to

$$\displaystyle\begin{array}{rcl} \frac{1} {2}\left [\left \Vert x\right \Vert \left \Vert y\right \Vert + \left \vert \left \langle x,y\right \rangle \right \vert \right ]& \geq & \frac{1} {2}\left [\left \langle Px,x\right \rangle ^{1/2}\left \langle Py,y\right \rangle ^{1/2} + \left \vert \left \langle Px,y\right \rangle \right \vert \right ] \\ & \geq & \left \vert \left \langle Px,y\right \rangle \right \vert {}\end{array}$$

(16)

for any x, y ∈ H. 

The inequality between the first and last term in (16), namely

$$\displaystyle{ \frac{1} {2}\left [\left \Vert x\right \Vert \left \Vert y\right \Vert + \left \vert \left \langle x,y\right \rangle \right \vert \right ] \geq \left \vert \left \langle Px,y\right \rangle \right \vert }$$

(17)

for any x, y ∈ H is a generalization of Buzano’s inequality (3).

From the inequality (14) we can state that

$$\displaystyle\begin{array}{rcl} \left \Vert x\right \Vert \left \Vert y\right \Vert -\left \vert \left \langle Px,y\right \rangle \right \vert & \geq & \left (\left \Vert x\right \Vert ^{2} -\left \langle Px,x\right \rangle \right )^{1/2}\left (\left \Vert y\right \Vert ^{2} -\left \langle Py,y\right \rangle \right )^{1/2} \\ & \geq & \left \vert \left \langle x,y\right \rangle -\left \langle Px,y\right \rangle \right \vert {}\end{array}$$

(18)

for any x, y ∈ H. 

From the inequality (14) we also have

$$\displaystyle\begin{array}{rcl} \left \Vert x\right \Vert \left \Vert y\right \Vert & \geq & \left (\left \Vert x\right \Vert ^{2} -\left \langle Px,x\right \rangle \right )^{1/2}\left (\left \Vert y\right \Vert ^{2} -\left \langle Py,y\right \rangle \right )^{1/2} + \left \vert \left \langle Px,y\right \rangle \right \vert {}\\ & \geq & \left \vert \left \langle x,y\right \rangle -\left \langle Px,y\right \rangle \right \vert + \left \vert \left \langle Px,y\right \rangle \right \vert \geq \left \vert \left \langle Px,y\right \rangle \right \vert -\left \vert \left \langle x,y\right \rangle \right \vert + \left \vert \left \langle Px,y\right \rangle \right \vert {}\\ & =& 2\left \vert \left \langle Px,y\right \rangle \right \vert -\left \vert \left \langle x,y\right \rangle \right \vert, {}\\ \end{array}$$

which implies that

$$\displaystyle\begin{array}{rcl} \frac{1} {2}\left [\left \Vert x\right \Vert \left \Vert y\right \Vert + \left \vert \left \langle x,y\right \rangle \right \vert \right ]& \geq & \frac{1} {2}\left [\left (\left \Vert x\right \Vert ^{2} -\left \langle Px,x\right \rangle \right )^{1/2}\left (\left \Vert y\right \Vert ^{2} -\left \langle Py,y\right \rangle \right )^{1/2}\right ] \\ +\frac{1} {2}\left [\left \vert \left \langle Px,y\right \rangle \right \vert + \left \vert \left \langle x,y\right \rangle \right \vert \right ]& \geq & \left \vert \left \langle Px,y\right \rangle \right \vert {}\end{array}$$

(19)

for any x, y ∈ H. 

The case of orthonormal families which is related to Bessel’s inequality is of interest.

Let $\mathcal{E} = \left \{e_{j}\right \}_{j\in J}$ be an othonormal family in H. Then for any x, y ∈ H we have from (13) and (14) the inequalities

$$\displaystyle\begin{array}{rcl} \left \Vert x\right \Vert \left \Vert y\right \Vert & \geq & \left (\sum _{j\in J}\left \vert \left \langle x,e_{j}\right \rangle \right \vert ^{2}\right )^{1/2}\left (\sum _{ j\in J}\left \vert \left \langle y,e_{j}\right \rangle \right \vert ^{2}\right )^{1/2} \\ & +& \left \vert \left \langle x,y\right \rangle -\sum _{j\in J}\left \langle x,e_{j}\right \rangle \left \langle e_{j},y\right \rangle \right \vert \\ & \geq & \left \vert \sum _{j\in J}\left \langle x,e_{j}\right \rangle \left \langle e_{j},y\right \rangle \right \vert + \left \vert \left \langle x,y\right \rangle -\sum _{j\in J}\left \langle x,e_{j}\right \rangle \left \langle e_{j},y\right \rangle \right \vert \geq \left \vert \left \langle x,y\right \rangle \right \vert {}\end{array}$$

(20)

and

$$\displaystyle\begin{array}{rcl} \left \Vert x\right \Vert \left \Vert y\right \Vert & \geq & \left (\left \Vert x\right \Vert ^{2} -\sum _{ j\in J}\left \vert \left \langle x,e_{j}\right \rangle \right \vert ^{2}\right )^{1/2}\left (\left \Vert y\right \Vert ^{2} -\sum _{ j\in J}\left \vert \left \langle y,e_{j}\right \rangle \right \vert ^{2}\right )^{1/2} \\ & +& \left \vert \left \langle \sum _{j\in J}\left \langle x,e_{j}\right \rangle \left \langle e_{j},y\right \rangle \right \rangle \right \vert \\ & \geq & \left \vert \left \langle x,y\right \rangle -\sum _{j\in J}\left \langle x,e_{j}\right \rangle \left \langle e_{j},y\right \rangle \right \vert + \left \vert \sum _{j\in J}\left \langle x,e_{j}\right \rangle \left \langle e_{j},y\right \rangle \right \vert \geq \left \vert \left \langle x,y\right \rangle \right \vert.{}\end{array}$$

(21)

By (15) and (16) we have

$$\displaystyle\begin{array}{rcl} & & \left \Vert x\right \Vert \left \Vert y\right \Vert -\left \vert \left \langle x,y\right \rangle \right \vert \\ & & \geq \left (\sum _{j\in J}\left \vert \left \langle x,e_{j}\right \rangle \right \vert ^{2}\right )^{1/2}\left (\sum _{ j\in J}\left \vert \left \langle y,e_{j}\right \rangle \right \vert ^{2}\right )^{1/2} -\left \vert \left \langle \sum _{ j\in J}\left \langle x,e_{j}\right \rangle \left \langle e_{j},y\right \rangle \right \rangle \right \vert \geq 0{}\end{array}$$

(22)

and

$$\displaystyle\begin{array}{rcl} \frac{1} {2}\left [\left \Vert x\right \Vert \left \Vert y\right \Vert + \left \vert \left \langle x,y\right \rangle \right \vert \right ]& \geq & \frac{1} {2}\left (\sum _{j\in J}\left \vert \left \langle x,e_{j}\right \rangle \right \vert ^{2}\right )^{1/2}\left (\sum _{ j\in J}\left \vert \left \langle y,e_{j}\right \rangle \right \vert ^{2}\right )^{1/2} \\ & & +\frac{1} {2}\left \vert \left \langle \left \langle \sum _{j\in J}\left \langle x,e_{j}\right \rangle \left \langle e_{j},y\right \rangle \right \rangle \right \rangle \right \vert \\ & & \qquad \qquad \qquad \geq \left \vert \left \langle \sum _{j\in J}\left \langle x,e_{j}\right \rangle \left \langle e_{j},y\right \rangle \right \rangle \right \vert {}\end{array}$$

(23)

for any x, y ∈ H. 

The inequality between the first and last term in (23) provides a generalization of Buzano’s inequality for orthonormal families $\mathcal{E} = \left \{e_{j}\right \}_{j\in J}$.

The following result holds:

Theorem 2.

Let P: H → H is an orthogonal projection on H. Then for any x,y ∈ H we have the inequalities

$$\displaystyle{ \left \vert \left \langle x,y\right \rangle - 2\left \langle Px,y\right \rangle \right \vert \leq \left \Vert x\right \Vert \left \Vert y\right \Vert, }$$

(24)

$$\displaystyle\begin{array}{rcl} & & \left \vert \left \langle x,y\right \rangle -\left \langle Px,y\right \rangle \right \vert \\ & & \leq \min \left \{\left \Vert x\right \Vert \left (\left \Vert y\right \Vert ^{2} -\left \langle Py,y\right \rangle \right )^{1/2},\left \Vert y\right \Vert \left (\left \Vert x\right \Vert ^{2} -\left \langle Px,x\right \rangle \right )^{1/2}\right \} \\ & & \leq \frac{1} {2}\left [\left \Vert x\right \Vert \left (\left \Vert y\right \Vert ^{2} -\left \langle Py,y\right \rangle \right )^{1/2} + \left \Vert y\right \Vert \left (\left \Vert x\right \Vert ^{2} -\left \langle Px,x\right \rangle \right )^{1/2}\right ] \\ & & \leq \frac{1} {2}\left (\left \Vert x\right \Vert ^{2} + \left \Vert y\right \Vert ^{2}\right )^{1/2}\left (\left \Vert x\right \Vert ^{2} + \left \Vert y\right \Vert ^{2} -\left \langle Py,y\right \rangle -\left \langle Px,x\right \rangle \right )^{1/2}{}\end{array}$$

(25)

and

$$\displaystyle\begin{array}{rcl} \left \vert \left \langle Px,y\right \rangle \right \vert & \leq & \min \left \{\left \Vert x\right \Vert \left \langle Py,y\right \rangle ^{1/2},\left \Vert y\right \Vert \left \langle Px,x\right \rangle ^{1/2}\right \} \\ & \leq & \frac{1} {2}\left [\left \Vert x\right \Vert \left \langle Py,y\right \rangle ^{1/2} + \left \Vert y\right \Vert \left \langle Px,x\right \rangle ^{1/2}\right ] \\ & \leq & \frac{1} {2}\left (\left \Vert x\right \Vert ^{2} + \left \Vert y\right \Vert ^{2}\right )^{1/2}\left (\left \langle Px,x\right \rangle + \left \langle Py,y\right \rangle \right )^{1/2}.{}\end{array}$$

(26)

Proof.

Observe that

$$\displaystyle\begin{array}{rcl} \left \Vert x - 2Px\right \Vert ^{2}& =& \left \Vert x\right \Vert ^{2} - 4\mathop{\mathrm{Re}}\left \langle x,Px\right \rangle + 4\left \langle Px,Px\right \rangle {}\\ & =& \left \Vert x\right \Vert ^{2} - 4\left \langle x,Px\right \rangle + 4\left \langle P^{2}x,x\right \rangle {}\\ & =& \left \Vert x\right \Vert ^{2} - 4\left \langle x,Px\right \rangle + 4\left \langle Px,x\right \rangle = \left \Vert x\right \Vert ^{2} {}\\ \end{array}$$

for any x ∈ H. 

Using Schwarz’s inequality we have

$$\displaystyle{ \left \Vert x\right \Vert \left \Vert y\right \Vert = \left \Vert x - 2Px\right \Vert \left \Vert y\right \Vert \geq \left \vert \left \langle x - 2Px,y\right \rangle \right \vert = \left \vert \left \langle x,y\right \rangle - 2\left \langle Px,y\right \rangle \right \vert }$$

for any x, y ∈ H and the inequality (24) is proved.

By Schwarz’s inequality we also have

$$\displaystyle{ \left \Vert x - Px\right \Vert \left \Vert y\right \Vert \geq \left \vert \left \langle x - Px,y\right \rangle \right \vert = \left \vert \left \langle x,y\right \rangle -\left \langle Px,y\right \rangle \right \vert }$$

and

$$\displaystyle{ \left \Vert x\right \Vert \left \Vert y - Py\right \Vert \geq \left \vert \left \langle x,y - Py\right \rangle \right \vert = \left \vert \left \langle x,y\right \rangle -\left \langle x,Py\right \rangle \right \vert = \left \vert \left \langle x,y\right \rangle -\left \langle Px,y\right \rangle \right \vert }$$

for any x, y ∈ H, which implies the first inequality in (25).

The second and the third inequalities are obvious by the elementary inequalities

$$\displaystyle{ \min \left \{a,b\right \} \leq \frac{1} {2}\left (a + b\right ),\text{ }a,b \in \mathbb{R}_{+} }$$

and

$$\displaystyle{ ac + bd \leq \left (a^{2} + b^{2}\right )^{1/2}\left (c^{2} + d^{2}\right )^{1/2},\text{ }a,b,c,d \in \mathbb{R}_{ +}. }$$

The inequality (26) follows from (25) by replacing P with 1 _H − P. 

Remark 2.

By the triangle inequality we have

$$\displaystyle{ \left \Vert x\right \Vert \left \Vert y\right \Vert + \left \vert \left \langle x,y\right \rangle \right \vert \geq \left \vert \left \langle x,y\right \rangle - 2\left \langle Px,y\right \rangle \right \vert + \left \vert \left \langle x,y\right \rangle \right \vert \geq 2\left \vert \left \langle Px,y\right \rangle \right \vert, }$$

which implies that [see also (16) and (19)]

$$\displaystyle{ \frac{1} {2}\left [\left \Vert x\right \Vert \left \Vert y\right \Vert + \left \vert \left \langle x,y\right \rangle \right \vert \right ] \geq \left \vert \left \langle Px,y\right \rangle \right \vert }$$

(27)

for any x, y ∈ H. 

From (25) we also have

$$\displaystyle\begin{array}{rcl} & & \left \vert \left \langle Px,y\right \rangle \right \vert \\ & & \leq \left \vert \left \langle x,y\right \rangle \right \vert +\min \left \{\left \Vert x\right \Vert \left (\left \Vert y\right \Vert ^{2} -\left \langle Py,y\right \rangle \right )^{1/2},\left \Vert y\right \Vert \left (\left \Vert x\right \Vert ^{2} -\left \langle Px,x\right \rangle \right )^{1/2}\right \}{}\end{array}$$

(28)

and

$$\displaystyle\begin{array}{rcl} & & \left \vert \left \langle x,y\right \rangle \right \vert \\ & & \leq \left \vert \left \langle Px,y\right \rangle \right \vert +\min \left \{\left \Vert x\right \Vert \left (\left \Vert y\right \Vert ^{2} -\left \langle Py,y\right \rangle \right )^{1/2},\left \Vert y\right \Vert \left (\left \Vert x\right \Vert ^{2} -\left \langle Px,x\right \rangle \right )^{1/2}\right \}{}\end{array}$$

(29)

for any x, y ∈ H. 

Now, if $\mathcal{E} = \left \{e_{j}\right \}_{j\in J}$ is an orthonormal family, then by the inequalities (24) and (25) we have

$$\displaystyle{ \left \vert \left \langle x,y\right \rangle - 2\sum _{j\in J}\left \langle x,e_{j}\right \rangle \left \langle e_{j},y\right \rangle \right \vert \leq \left \Vert x\right \Vert \left \Vert y\right \Vert, }$$

(30)

and

$$\displaystyle\begin{array}{rcl} & & \left \vert \left \langle x,y\right \rangle -\sum _{j\in J}\left \langle x,e_{j}\right \rangle \left \langle e_{j},y\right \rangle \right \vert \\ & & \leq \min \left \{\left \Vert x\right \Vert \left (\left \Vert y\right \Vert ^{2} -\sum _{ j\in J}\left \vert \left \langle y,e_{j}\right \rangle \right \vert ^{2}\right )^{1/2},\left \Vert y\right \Vert \left (\left \Vert x\right \Vert ^{2} -\sum _{ j\in J}\left \vert \left \langle x,e_{j}\right \rangle \right \vert ^{2}\right )^{1/2}\right \} \\ & & \leq \frac{1} {2}\left [\left \Vert x\right \Vert \left (\left \Vert y\right \Vert ^{2} -\sum _{ j\in J}\left \vert \left \langle y,e_{j}\right \rangle \right \vert ^{2}\right )^{1/2} + \left \Vert y\right \Vert \left (\left \Vert x\right \Vert ^{2} -\sum _{ j\in J}\left \vert \left \langle x,e_{j}\right \rangle \right \vert ^{2}\right )^{1/2}\right ] \\ & & \leq \frac{1} {2}\left (\left \Vert x\right \Vert ^{2} + \left \Vert y\right \Vert ^{2}\right )^{1/2}\left (\left \Vert x\right \Vert ^{2} + \left \Vert y\right \Vert ^{2} -\sum _{ j\in J}\left \vert \left \langle y,e_{j}\right \rangle \right \vert ^{2} -\sum _{ j\in J}\left \vert \left \langle x,e_{j}\right \rangle \right \vert ^{2}\right )^{1/2}{}\end{array}$$

(31)

for any x, y ∈ H. 

From (28) we also have

$$\displaystyle\begin{array}{rcl} & & \left \vert \sum _{j\in J}\left \langle x,e_{j}\right \rangle \left \langle e_{j},y\right \rangle \right \vert \\ & \leq & \left \vert \left \langle x,y\right \rangle \right \vert +\min \left \{\left \Vert x\right \Vert \left (\left \Vert y\right \Vert ^{2} -\sum _{ j\in J}\left \vert \left \langle y,e_{j}\right \rangle \right \vert ^{2}\right )^{1/2},\left \Vert y\right \Vert \left (\left \Vert x\right \Vert ^{2} -\sum _{ j\in J}\left \vert \left \langle x,e_{j}\right \rangle \right \vert ^{2}\right )^{1/2}\right \}{}\end{array}$$

(32)

for any x, y ∈ H. 

3 Inequalities for Norm and Numerical Radius

Let $\left (H;\left \langle \cdot,\cdot \right \rangle \right )$ be a complex Hilbert space. The numerical range of an operator T is the subset of the complex numbers $\mathbb{C}$ given by Gustafson and Rao [26, p. 1]:

$$\displaystyle{ W\left (T\right ) = \left \{\left \langle Tx,x\right \rangle,\ x \in H,\ \left \Vert x\right \Vert = 1\right \}. }$$

The numerical radius $w\left (T\right )$ of an operator T on H is defined by Gustafson and Rao [26, p. 8]:

$$\displaystyle{ w\left (T\right ) =\sup \left \{\left \vert \lambda \right \vert,\lambda \in W\left (T\right )\right \} =\sup \left \{\left \vert \left \langle Tx,x\right \rangle \right \vert,\left \Vert x\right \Vert = 1\right \}. }$$

It is well known that $w\left (\cdot \right )$ is a norm on the Banach algebra $B\left (H\right )$ and the following inequality holds true:

$$\displaystyle{ w\left (T\right ) \leq \left \Vert T\right \Vert \leq 2w\left (T\right ),\text{ for any }T \in B\left (H\right ). }$$

Utilizing Buzano’s inequality (3) we obtained the following inequality for the numerical radius [13] or [15]:

Theorem 3.

Let $\left (H;\left \langle \cdot,\cdot \right \rangle \right )$ be a Hilbert space and T: H → H a bounded linear operator on H. Then

$$\displaystyle{ w^{2}\left (T\right ) \leq \frac{1} {2}\left [w\left (T^{2}\right ) + \left \Vert T\right \Vert ^{2}\right ]. }$$

(33)

The constant $\frac{1} {2}$ is best possible in (33).

The following general result for the product of two operators holds [26, p. 37]:

Theorem 4.

If A,B are two bounded linear operators on the Hilbert space $\left (H,\left \langle \cdot,\cdot \right \rangle \right ),$ then $w\left (AB\right ) \leq 4w\left (A\right )w\left (B\right ).$ In the case that AB = BA, then $w\left (AB\right ) \leq 2w\left (A\right )w\left (B\right ).$ The constant 2 is best possible here.

The following results are also well known [26, p. 38].

Theorem 5.

If A is a unitary operator that commutes with another operator B, then

$$\displaystyle{ w\left (AB\right ) \leq w\left (B\right ). }$$

(34)

If A is an isometry and AB = BA, then (34) also holds true.

We say that A and B double commute if AB = BA and AB ^∗ = B ^∗ A. The following result holds [26, p. 38].

Theorem 6.

If the operators A and B double commute, then

$$\displaystyle{ w\left (AB\right ) \leq w\left (B\right )\left \Vert A\right \Vert. }$$

(35)

As a consequence of the above, we have [26, p. 39]:

Corollary 2.

Let A be a normal operator commuting with B. Then

$$\displaystyle{ w\left (AB\right ) \leq w\left (A\right )w\left (B\right ). }$$

(36)

A related problem with the inequality (35) is to find the best constant c for which the inequality

$$\displaystyle{ w\left (AB\right ) \leq cw\left (A\right )\left \Vert B\right \Vert }$$

holds for any two commuting operators $A,B \in B\left (H\right ).$ It is known that 1. 064 < c < 1. 169, see [3, 32] and [33].

In relation to this problem, it has been shown in [24] that

Theorem 7.

For any $A,B \in B\left (H\right )$ we have

$$\displaystyle{ w\left (\frac{AB + BA} {2} \right ) \leq \sqrt{2}w\left (A\right )\left \Vert B\right \Vert. }$$

(37)

For other numerical radius inequalities see the recent monograph [18] and the references therein.

The following result holds.

Theorem 8.

Let P: H → H be an orthogonal projection on the Hilbert space $\left (H,\left \langle \cdot,\cdot \right \rangle \right ).$ If A,B are two bounded linear operators on H, then

$$\displaystyle{ \left \vert \left \langle BPAx,x\right \rangle \right \vert \leq \frac{1} {2}\left [\left \Vert Ax\right \Vert \left \Vert B^{{\ast}}x\right \Vert + \left \vert \left \langle BAx,x\right \rangle \right \vert \right ] }$$

(38)

and

$$\displaystyle{ \left \Vert BPAx\right \Vert \leq \frac{1} {2}\left [\left \Vert Ax\right \Vert \left \Vert B\right \Vert + \left \Vert BAx\right \Vert \right ] }$$

(39)

for any x ∈ H.

Moreover, we have

$$\displaystyle{ w\left (BPA\right ) \leq \frac{1} {2}\left [\left \Vert A\right \Vert \left \Vert B\right \Vert + w\left (BA\right )\right ] }$$

(40)

and

$$\displaystyle{ \left \Vert BPA\right \Vert \leq \frac{1} {2}\left [\left \Vert A\right \Vert \left \Vert B\right \Vert + \left \Vert BA\right \Vert \right ]. }$$

(41)

Proof.

From the inequality (17) we have

$$\displaystyle{ \left \vert \left \langle PAx,B^{{\ast}}y\right \rangle \right \vert \leq \frac{1} {2}\left [\left \Vert Ax\right \Vert \left \Vert B^{{\ast}}y\right \Vert + \left \vert \left \langle Ax,B^{{\ast}}y\right \rangle \right \vert \right ] }$$

that is equivalent to

$$\displaystyle{ \left \vert \left \langle BPAx,y\right \rangle \right \vert \leq \frac{1} {2}\left [\left \Vert Ax\right \Vert \left \Vert B^{{\ast}}y\right \Vert + \left \vert \left \langle BAx,y\right \rangle \right \vert \right ] }$$

(42)

for any x, y ∈ H. 

If we take y = x in (42), then we get (38).

Taking the supremum over y ∈ H with $\left \Vert y\right \Vert = 1$ in (42) we have

$$\displaystyle\begin{array}{rcl} \left \Vert BPAx\right \Vert & =& \sup _{\left \Vert y\right \Vert =1}\left \vert \left \langle BPAx,y\right \rangle \right \vert \leq \frac{1} {2}\sup _{\left \Vert y\right \Vert =1}\left [\left \Vert Ax\right \Vert \left \Vert B^{{\ast}}y\right \Vert + \left \vert \left \langle BAx,y\right \rangle \right \vert \right ] {}\\ & \leq & \frac{1} {2}\left [\left \Vert Ax\right \Vert \sup _{\left \Vert y\right \Vert =1}\left \Vert B^{{\ast}}y\right \Vert +\sup _{\left \Vert y\right \Vert =1}\left \vert \left \langle BAx,y\right \rangle \right \vert \right ] {}\\ & =& \frac{1} {2}\left [\left \Vert Ax\right \Vert \left \Vert B\right \Vert + \left \Vert BAx\right \Vert \right ] {}\\ \end{array}$$

for any x ∈ H.

The inequalities (40) and (41) follow from (38) and (39) by taking the supremum over x ∈ H with $\left \Vert x\right \Vert = 1.$

Corollary 3.

Let P: H → H be an orthogonal projection on the Hilbert space $\left (H,\left \langle \cdot,\cdot \right \rangle \right ).$ If A,B are two bounded linear operators on H, then

$$\displaystyle{ \left \vert \left \langle APAx,x\right \rangle \right \vert \leq \frac{1} {2}\left [\left \Vert Ax\right \Vert \left \Vert A^{{\ast}}x\right \Vert + \left \vert \left \langle A^{2}x,x\right \rangle \right \vert \right ] }$$

(43)

and

$$\displaystyle{ \left \Vert APAx\right \Vert \leq \frac{1} {2}\left [\left \Vert Ax\right \Vert \left \Vert A\right \Vert + \left \Vert A^{2}x\right \Vert \right ] }$$

(44)

for any x ∈ H.

Moreover, we have

$$\displaystyle{ w\left (APA\right ) \leq \frac{1} {2}\left [\left \Vert A\right \Vert ^{2} + w\left (A^{2}\right )\right ] }$$

(45)

and

$$\displaystyle{ \left \Vert APA\right \Vert \leq \frac{1} {2}\left [\left \Vert A\right \Vert ^{2} + \left \Vert A^{2}\right \Vert \right ]. }$$

(46)

Remark 3.

Let $e \in H,\ \left \Vert e\right \Vert = 1.$ If we write the inequalities (38) and (39) for the projector P _e defined by $P_{e}x = \left \langle x,e\right \rangle e,\ x \in H,$ we have

$$\displaystyle{ \left \vert \left \langle Ax,e\right \rangle \right \vert \left \vert \left \langle Be,x\right \rangle \right \vert \leq \frac{1} {2}\left [\left \Vert Ax\right \Vert \left \Vert B^{{\ast}}x\right \Vert + \left \vert \left \langle BAx,x\right \rangle \right \vert \right ] }$$

(47)

and

$$\displaystyle{ \left \vert \left \langle Ax,e\right \rangle \right \vert \left \Vert Be\right \Vert \leq \frac{1} {2}\left [\left \Vert Ax\right \Vert \left \Vert B\right \Vert + \left \Vert BAx\right \Vert \right ] }$$

(48)

for any x ∈ H. 

Now, if we take the supremum over $x \in H,\ \left \Vert x\right \Vert = 1$ in (48), then we get

$$\displaystyle{ \left \Vert A^{{\ast}}e\right \Vert \left \Vert Be\right \Vert \leq \frac{1} {2}\left [\left \Vert A\right \Vert \left \Vert B\right \Vert + \left \Vert BA\right \Vert \right ] }$$

(49)

for any $e \in H,\ \left \Vert e\right \Vert = 1.$

If in (49) we take B = A, we have

$$\displaystyle{ \left \Vert A^{{\ast}}e\right \Vert \left \Vert Ae\right \Vert \leq \frac{1} {2}\left [\left \Vert A\right \Vert ^{2} + \left \Vert A^{2}\right \Vert \right ] }$$

(50)

for any $e \in H,\ \left \Vert e\right \Vert = 1.$

If in (47) we take B = A, then we get

$$\displaystyle{ \left \vert \left \langle Ax,e\right \rangle \right \vert \left \vert \left \langle e,A^{{\ast}}x\right \rangle \right \vert \leq \frac{1} {2}\left [\left \Vert Ax\right \Vert \left \Vert A^{{\ast}}x\right \Vert + \left \vert \left \langle A^{2}x,x\right \rangle \right \vert \right ] }$$

(51)

for any x ∈ H and $e \in H,\ \left \Vert e\right \Vert = 1,$ and in particular

$$\displaystyle{ \left \vert \left \langle Ae,e\right \rangle \right \vert ^{2} \leq \frac{1} {2}\left [\left \Vert Ae\right \Vert \left \Vert A^{{\ast}}e\right \Vert + \left \vert \left \langle A^{2}e,e\right \rangle \right \vert \right ] }$$

(52)

for any $e \in H,\ \left \Vert e\right \Vert = 1.$

Taking the supremum over $e \in H,\ \left \Vert e\right \Vert = 1$ in (52) we recapture the result in Theorem 3.

For a given operator T we consider the modulus of T defined as $\left \vert T\right \vert:= \left (T^{{\ast}}T\right )^{1/2}.$

Corollary 4.

Let P: H → H be an orthogonal projection on the Hilbert space $\left (H,\left \langle \cdot,\cdot \right \rangle \right ).$ If A,B are two bounded linear operators on H, then

$$\displaystyle{ w\left (BPA\right ) \leq \frac{1} {2}w\left (BA\right ) + \frac{1} {4}\left \Vert \left \vert A\right \vert ^{2} + \left \vert B^{{\ast}}\right \vert ^{2}\right \Vert. }$$

(53)

In particular, we have

$$\displaystyle{ w\left (APA\right ) \leq \frac{1} {2}w\left (A^{2}\right ) + \frac{1} {4}\left \Vert \left \vert A\right \vert ^{2} + \left \vert A^{{\ast}}\right \vert ^{2}\right \Vert. }$$

(54)

Proof.

From the inequality (38) we have

$$\displaystyle\begin{array}{rcl} \left \vert \left \langle BPAx,x\right \rangle \right \vert & \leq & \frac{1} {2}\left [\left \Vert Ax\right \Vert \left \Vert B^{{\ast}}x\right \Vert + \left \vert \left \langle BAx,x\right \rangle \right \vert \right ] \\ & \leq & \frac{1} {2}\left \vert \left \langle BAx,x\right \rangle \right \vert + \frac{1} {4}\left [\left \Vert Ax\right \Vert ^{2} + \left \Vert B^{{\ast}}x\right \Vert ^{2}\right ]{}\end{array}$$

(55)

for any x ∈ H, where for the second inequality we used the elementary inequality

$$\displaystyle{ ab \leq \frac{1} {2}\left (a^{2} + b^{2}\right ),\text{ }a,b \in \mathbb{R}. }$$

(56)

Since

$$\displaystyle\begin{array}{rcl} \left \Vert Ax\right \Vert ^{2} + \left \Vert B^{{\ast}}x\right \Vert ^{2}& =& \left \langle Ax,Ax\right \rangle + \left \langle B^{{\ast}}x,B^{{\ast}}x\right \rangle = \left \langle A^{{\ast}}Ax,x\right \rangle + \left \langle BB^{{\ast}}x,x\right \rangle {}\\ & =& \left \langle \left (\left \vert A\right \vert ^{2} + \left \vert B^{{\ast}}\right \vert ^{2}\right )x,x\right \rangle {}\\ \end{array}$$

for any x ∈ H, then from (55) we have

$$\displaystyle{ \left \vert \left \langle BPAx,x\right \rangle \right \vert \leq \frac{1} {2}\left \vert \left \langle BAx,x\right \rangle \right \vert + \frac{1} {4}\left \langle \left (\left \vert A\right \vert ^{2} + \left \vert B^{{\ast}}\right \vert ^{2}\right )x,x\right \rangle }$$

(57)

for any x ∈ H. 

Taking the supremum over $x \in H,\ \left \Vert x\right \Vert = 1$ in (57) we get the desired result (53).

Remark 4.

We observe that by (52) we have

$$\displaystyle\begin{array}{rcl} \left \vert \left \langle Ae,e\right \rangle \right \vert ^{2}& \leq & \frac{1} {2}\left [\left \Vert Ae\right \Vert \left \Vert A^{{\ast}}e\right \Vert + \left \vert \left \langle A^{2}e,e\right \rangle \right \vert \right ] \\ & \leq & \frac{1} {2}\left \vert \left \langle A^{2}e,e\right \rangle \right \vert + \frac{1} {4}\left [\left \Vert Ae\right \Vert ^{2} + \left \Vert A^{{\ast}}e\right \Vert ^{2}\right ] \\ & =& \frac{1} {2}\left \vert \left \langle A^{2}e,e\right \rangle \right \vert + \frac{1} {4}\left \langle \left (\left \vert A\right \vert ^{2} + \left \vert A^{{\ast}}\right \vert ^{2}\right )e,e\right \rangle {}\end{array}$$

(58)

for any $e \in H,\ \left \Vert e\right \Vert = 1.$

Taking the supremum over $e \in H,\ \left \Vert e\right \Vert = 1$ in (58) we get

$$\displaystyle{ w^{2}\left (A\right ) \leq \frac{1} {2}w\left (A^{2}\right ) + \frac{1} {4}\left \Vert \left \vert A\right \vert ^{2} + \left \vert A^{{\ast}}\right \vert ^{2}\right \Vert, }$$

(59)

for any bounded linear operator A. 

Since

$$\displaystyle{ \left \Vert \left \vert A\right \vert ^{2} + \left \vert A^{{\ast}}\right \vert ^{2}\right \Vert \leq \left \Vert \left \vert A\right \vert ^{2}\right \Vert + \left \Vert \left \vert A^{{\ast}}\right \vert ^{2}\right \Vert = 2\left \Vert A\right \Vert ^{2}, }$$

then the inequality (59) is better than the inequality in Theorem 3.

The following result also holds:

Theorem 9.

Let P: H → H be an orthogonal projection on the Hilbert space $\left (H,\left \langle \cdot,\cdot \right \rangle \right ).$ If A,B are two bounded linear operators on H, then

$$\displaystyle{ w\left (B\left (\frac{1} {2}1_{H} - P\right )A\right ) \leq \frac{1} {4}\left \Vert \left \vert A\right \vert ^{2} + \left \vert B^{{\ast}}\right \vert ^{2}\right \Vert. }$$

(60)

In particular, we have

$$\displaystyle{ w\left (A\left (\frac{1} {2}1_{H} - P\right )A\right ) \leq \frac{1} {4}\left \Vert \left \vert A\right \vert ^{2} + \left \vert A^{{\ast}}\right \vert ^{2}\right \Vert. }$$

(61)

Proof.

From the inequality (24) we have

$$\displaystyle{ \left \vert \left \langle \left (1_{H} - 2P\right )Ax,B^{{\ast}}x\right \rangle \right \vert \leq \left \Vert Ax\right \Vert \left \Vert B^{{\ast}}x\right \Vert, }$$

that is equivalent to

$$\displaystyle{ \left \vert \left \langle B\left (\frac{1} {2}1_{H} - P\right )Ax,x\right \rangle \right \vert \leq \frac{1} {2}\left \Vert Ax\right \Vert \left \Vert B^{{\ast}}x\right \Vert }$$

(62)

for any x ∈ H. 

Using the elementary inequality (56) we have

$$\displaystyle{ \frac{1} {2}\left \Vert Ax\right \Vert \left \Vert B^{{\ast}}x\right \Vert \leq \frac{1} {4}\left (\left \Vert Ax\right \Vert ^{2} + \left \Vert B^{{\ast}}x\right \Vert ^{2}\right ) = \frac{1} {4}\left \langle \left (\left \vert A\right \vert ^{2} + \left \vert B^{{\ast}}\right \vert ^{2}\right )x,x\right \rangle }$$

and by (62) we get

$$\displaystyle{ \left \vert \left \langle B\left (\frac{1} {2}1_{H} - P\right )Ax,x\right \rangle \right \vert \leq \frac{1} {4}\left \langle \left (\left \vert A\right \vert ^{2} + \left \vert B^{{\ast}}\right \vert ^{2}\right )x,x\right \rangle }$$

(63)

for any x ∈ H. 

Taking the supremum over $x \in H,\ \left \Vert x\right \Vert = 1$ in (63) we get the desired result (60).

Remark 5.

If we take in (60) P = 1 _H , then we get [18, p. 6]

$$\displaystyle{ w\left (BA\right ) \leq \frac{1} {2}\left \Vert \left \vert A\right \vert ^{2} + \left \vert B^{{\ast}}\right \vert ^{2}\right \Vert }$$

(64)

for any A, B bounded linear operators on H.