REVERSING THE CBS-INEQUALITY FOR SEQUENCES OF VECTORS IN HILBERT SPACES WITH APPLICATIONS (I)

S.S. DRAGOMIR

Abstract. Several reverses for the Cauchy-Bunyakovsky-Schwarz (CBS) inequality for sequences of vectors in Hilbert spaces are obtained. Applications for bounding the distance to a finite-dimensional subspace, in reversing the generalised triangle inequality and for Fourier coefficients are also given.

1. Introduction

Let (H; ⟨·, ·⟩) be an inner product space over the real or complex number field 𝕂. One of the most important inequalities in inner product spaces with numerous applications, is the Schwarz inequality

(1.1)  |⟨x, y⟩|^{2} ≤ ∥x∥^{2} ∥y∥^{2} , x, y ∈ H

or, equivalently,

(1.2)  |⟨x, y⟩| ≤ ∥x∥ ∥y∥ , x, y ∈ H.

The case of equality holds iff there exists a scalar α ∈ 𝕂 such that x = αy.

By a multiplicative reverse of the Schwarz inequality we understand an inequality of the form

(1.3)  (1 ≤) \begin{matrix}∥x∥ & ∥y∥ \\ | & ⟨x, y⟩ & | \end{matrix} ≤ k_{1} or (1 ≤) \begin{matrix}∥x∥ & 2 & ∥y∥ & 2 \\ |⟨x, y⟩| & 2 \end{matrix} ≤ k_{2}

with appropriate k_{1} and k_{2} and under various assumptions for the vectors x and y, while by an additive reverse we understand an inequality of the form

(1.4)  (0 ≤) ∥x∥ ∥y∥ − |⟨x, y⟩| ≤ h_{1} or (0 ≤) ∥x∥^{2} ∥y∥^{2} − |⟨x, y⟩|^{2} ≤ h_{2}.

Similar definition apply when |⟨x, y⟩| is replaced by Re ⟨x, y⟩ or |Re ⟨x, y⟩| .

The following recent reverses for the Schwarz inequality hold (see for instance the monograph on line [3, p. 20]):

Theorem 1. Let (H; ⟨·, ·⟩) be an inner product space over the real or complex number field 𝕂. If x, y ∈ H and r > 0 are such that

(1.5)  ∥x − y∥ ≤ r < ∥y∥ ,

then we have the following multiplicative reverse of the Schwarz inequality

(1.6)  (1 ≤) \begin{matrix}∥x∥ & ∥y∥ \\ | & ⟨x, y⟩ & | \end{matrix} ≤ \begin{matrix}∥x∥ & ∥y∥ \\ Re & ⟨x, y⟩ \end{matrix} ≤ \frac{∥y∥}{\sqrt{∥y∥^{2} − r^{2}}}

Date: 07 October, 2004.

2000 Mathematics Subject Classification. 46C05, 26D15.

Key words and phrases. CSB-inequality, Gram determinant, Gram inequality, Fourier coefficients, Triangle inequality.

2 S.S. DRAGOMIR

and the subsequent additive reverses

(1.7)  (0 ≤) ∥x∥ ∥y∥ − |⟨x, y⟩| ≤ ∥x∥ ∥y∥ − Re ⟨x, y⟩

≤ \frac{r^{2}}{\sqrt{∥y∥^{2} − r^{2}} (∥y∥ + \sqrt{∥y∥^{2} − r^{2}})} Re ⟨x, y⟩

and

(1.8)  (0 ≤) ∥x∥^{2} ∥y∥^{2} − |⟨x, y⟩|^{2} ≤ ∥x∥^{2} ∥y∥^{2} − [Re ⟨x, y⟩]^{2}
≤ r^{2} ∥x∥^{2} .

All the above inequalities are sharp.

Other additive reverses of the quadratic Schwarz’s inequality are incorporated in the following result [3, p. 18-19]: Theorem 2. Let x, y ∈ H and a, A ∈ 𝕂. If

(1.9)  Re ⟨Ay − x, x − ay⟩ ≥ 0

or, equivalently,

(1.10) ∥x − \frac{a + A}{2} · y∥ ≤ \frac{1}{2} |A − a| ∥y∥ ,

then

(1.11)  (0 ≤) ∥x∥^{2} ∥y∥^{2} − |⟨x, y⟩|^{2}

≤ \frac{1}{4} |A − a|^{2} ∥y∥^{4} − \{ \begin{matrix}|\frac{A+a}{2} ∥y∥^{2} − ⟨x, y⟩| & 2 \\ ∥y∥ & 2 & Re & ⟨Ay − x, x − ay⟩ \end{matrix}

≤ \frac{1}{4} |A − a|^{2} ∥y∥^{4} .

The constant \frac{1}{4} is best possible in all inequalities.

If one were to assume more about the complex numbers A and a, then one may state the following result as well [3, p. 21-23]. Theorem 3. With the assumptions of Theorem 2 and, if in addition, Re (Aā) > 0, then

(1.12)  ∥x∥ ∥y∥ ≤ \frac{1}{2} · \begin{matrix}Re & [(Ā + ā) ⟨x, y⟩] \\ √ & \overline{Re (Aā)} \end{matrix} ≤ \frac{1}{2} · \frac{|A + a|}{\sqrt{Re (Aā)}} |⟨x, y⟩| ,
(1.13)  (0 ≤) ∥x∥ ∥y∥ − Re ⟨x, y⟩ ≤ \frac{1}{2} · \begin{matrix}Re & [(Ā + ā − 2\sqrt{Re (Aā)}) ⟨x, y⟩] \\ √ & \overline{Re (Aā)} \end{matrix}

and

(1.14)  (0 ≤) ∥x∥^{2} ∥y∥^{2} − |⟨x, y⟩|^{2} ≤ \frac{1}{4} · \begin{matrix}|A − a| & 2 \\ Re & (Aā) \end{matrix} |⟨x, y⟩|^{2} .

The constants \frac{1}{2} and \frac{1}{4} are best possible.

Remark 1. If A = M, a = m and M ≥ m > 0, then (1.12) and (1.13) may be written in a more convenient form as

(1.15)  ∥x∥ ∥y∥ ≤ \frac{M + m}{2\sqrt{mM}} Re ⟨x, y⟩

and

(1.16)  (0 ≤) ∥x∥ ∥y∥ − Re ⟨x, y⟩ ≤ \frac{(\sqrt{M} − \sqrt{m})^{2}}{2\sqrt{mM}} Re ⟨x, y⟩ .

Here the constant \frac{1}{2} is sharp in both inequalities.

In this paper several reverses for the Cauchy-Bunyakovsky-Schwarz (CBS) inequality for sequences of vectors in Hilbert spaces are obtained. Applications for bounding the distance to a finite-dimensional subspace and in reversing the generalised triangle inequality are also given.

A continuation of this work for different classes of reverse inequalities is planed to be considered in the subsequent paper [5].

2. Reverses of the (CBS) −Inequality for Two Sequences in ℓ_{p}^{2} (K)

Let (K, ⟨·, ·⟩) be a Hilbert space over 𝕂, p_{i} ≥ 0, i ∈ ℕ with ∑_{i=1}^{∞} p_{i} = 1. Consider ℓ_{p}^{2} (K) as the space

ℓ_{p}^{2} (K) := {x = (x_{i})_{i∈ℕ} |x_{i} ∈ K, i ∈ ℕ and ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} < ∞} .

It is well known that ℓ_{p}^{2} (K) endowed with the inner product

⟨x, y⟩_{p} := ∑_{i=1}^{∞} p_{i} ⟨x_{i}, y_{i}⟩

is a Hilbert space over 𝕂. The norm ∥·∥_{p} of ℓ_{p}^{2} (K) is given by

∥x∥_{p} := (∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2})^{\frac{1}{2}} .

If x, y ∈ ℓ_{p}^{2} (K) , then the following Cauchy-Bunyakovsky-Schwarz (CBS) inequality holds true

(2.1)  ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} ∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2} ≥ |∑_{i=1}^{∞} p_{i} ⟨x_{i}, y_{i}⟩|^{2}

with equality iff there exists a λ ∈ 𝕂 such that x_{i} = λy_{i} for each i ∈ ℕ.

This is an obvious consequence of the Schwarz inequality (1.1) written for the inner product ⟨·, ·⟩_{p} defined on ℓ_{p}^{2} (K) .

The following proposition may be stated.

Proposition 1. Let x, y ∈ ℓ_{p}^{2} (K) and r > 0. Assume that

(2.2)  ∥x_{i} − y_{i}∥ ≤ r < ∥y_{i}∥ for each i ∈ ℕ.

4 S.S. DRAGOMIR

Then we have the inequality

(2.3)  (1 ≤)\frac{(∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} ∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2})^{\frac{1}{2}}}{|∑_{i=1}^{∞} p_{i} ⟨x_{i}, y_{i}⟩|}

≤ \frac{(∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} ∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2})^{\frac{1}{2}}}{∑_{i=1}^{∞} p_{i} Re ⟨x_{i}, y_{i}⟩}

≤ \begin{matrix}(∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2}) & \frac{1}{2} \\ √ & \overline{∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2} − r^{2}} \end{matrix} ,

(2.4) (0 ≤) (∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} ∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2})^{\frac{1}{2}} − |∑_{i=1}^{∞} p_{i} ⟨x_{i}, y_{i}⟩|

≤ (∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} ∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2})^{\frac{1}{2}} − ∑_{i=1}^{∞} p_{i} Re ⟨x_{i}, y_{i}⟩

≤ \frac{r^{2} · ∑_{i=1}^{∞} p_{i} Re ⟨x_{i}, y_{i}⟩}{\sqrt{∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2} − r^{2}} [(∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2})^{\frac{1}{2}} + \sqrt{∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2} − r^{2}}]}

and

(2.5)  (0 ≤) ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} ∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2} − |∑_{i=1}^{∞} p_{i} ⟨x_{i}, y_{i}⟩|^{2}

≤ ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} ∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2} − [∑_{i=1}^{∞} p_{i} Re ⟨x_{i}, y_{i}⟩]^{2}

≤ r^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} .

Proof. From (2.2), we have

∥x − y∥_{p}^{2} = ∑_{i=1}^{∞} p_{i} ∥x_{i} − y_{i}∥^{2} ≤ r^{2} ∑_{i=1}^{∞} p_{i} ≤ ∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2} = ∥y∥_{p}^{2} ,

giving ∥x − y∥_{p} ≤ r ≤ ∥y∥_{p} . Applying Theorem 1 for ℓ_{p}^{2} (K) and ⟨·, ·⟩_{p} , we deduce the desired inequality.

The following proposition holds. Proposition 2. Let x, y ∈ ℓ_{p}^{2} (K) and a, A ∈ 𝕂. If

(2.6)  Re ⟨Ay_{i} − x_{i}, x_{i} − ay_{i}⟩ ≥ 0 for each i ∈ ℕ

or, equivalently,

(2.7)  ∥x_{i} − \frac{a + A}{2}y_{i}∥ ≤ \frac{1}{2} |A − a| ∥y_{i}∥ for each i ∈ ℕ

then

(2.8)  (0 ≤ ) ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} ∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2} − |∑_{i=1}^{∞} p_{i} ⟨x_{i}, y_{i}⟩|^{2}

≤ \frac{1}{4} |A − a|^{2} (∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2})^{2}

− \{ \begin{matrix}|\frac{A+a}{2} ∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2} − ∑_{i=1}^{∞} p_{i} ⟨x_{i}, y_{i}⟩| & 2 \\ ∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2} ∑_{i=1}^{∞} p_{i} Re ⟨Ay_{i} − x_{i}, x_{i} − ay_{i}⟩ \end{matrix}

≤ \frac{1}{4} |A − a|^{2} (∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2})^{2} .

The proof follows by Theorem 2, we omit the details.

Finally, on using Theorem 3, we may state:

Proposition 3. Assume that x, y, a and A are as in Proposition 2. Moreover, if Re (Aā) > 0, then we have the inequality:

(2.9)  (∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} ∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2})^{\frac{1}{2}} ≤ \frac{1}{2} · \begin{matrix}Re & [(Ā + ā) ∑_{i=1}^{∞} p_{i} ⟨x_{i}, y_{i}⟩] \\ √ & \overline{Re (Aā)} \end{matrix}

≤ \frac{1}{2} · \frac{|A − a|}{\sqrt{Re (Aā)}} |∑_{i=1}^{∞} p_{i} ⟨x_{i}, y_{i}⟩| ,

(2.10)  (0 ≤) (∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} ∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2})^{\frac{1}{2}} − ∑_{i=1}^{∞} p_{i} Re ⟨x_{i}, y_{i}⟩

≤ \frac{1}{2} · \begin{matrix}Re & [(Ā + ā − 2\sqrt{Re (Aā)}) ∑_{i=1}^{∞} p_{i} ⟨x_{i}, y_{i}⟩] \\ √ & \overline{Re (Aā)} \end{matrix}

and

(2.11)  (0 ≤) ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} ∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2} − |∑_{i=1}^{∞} p_{i} ⟨x_{i}, y_{i}⟩|^{2}

≤ \frac{1}{4} · \begin{matrix}|A − a| & 2 \\ Re & (Aā) \end{matrix} |∑_{i=1}^{∞} p_{i} ⟨x_{i}, y_{i}⟩|^{2} .

3. Reverses of the (CBS) −Inequality for Mixed Sequences

Let (K, ⟨·, ·⟩) be a Hilbert space over 𝕂 and for p_{i} ≥ 0, i ∈ ℕ with ∑_{i=1}^{∞} p_{i} = 1, and ℓ_{p}^{2} (K) the Hilbert space defined in the previous section.

If

α ∈ ℓ_{p}^{2} (𝕂) := {α = (α_{i})_{i∈ℕ} |α_{i} ∈ 𝕂, i ∈ ℕ and ∑_{i=1}^{∞} p_{i} |α_{i}|^{2} < ∞}

and x ∈ ℓ_{p}^{2} (K) , then the following Cauchy-Bunyakovsky-Schwarz (CBS) inequality holds true:

(3.1)  ∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} ≥ ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥^{2} ,

with equality if and only if there exists a vector v ∈ K such that x_{i} = \overline{α_{i}}v for any i ∈ ℕ.

The inequality (3.1) follows by the obvious identity

∑_{i=1}^{n} p_{i} |α_{i}|^{2} ∑_{i=1}^{n} p_{i} ∥x_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}α_{i}x_{i}∥^{2}

= \frac{1}{2} ∑_{i=1}^{n} ∑_{j=1}^{n} p_{i}p_{j} ∥\overline{α_{i}}x_{j} − \overline{α_{j}}x_{i}∥^{2} ,

for any n ∈ ℕ, n ≥ 1.

In the following we establish some reverses of the (CBS) −inequality in some of its various equivalent forms that will be specified where they occur.

Theorem 4. Let α ∈ ℓ_{p}^{2} (𝕂) , x ∈ ℓ_{p}^{2} (K) and a ∈ K, r > 0 such that ∥a∥ > r. If the following condition holds

(3.2)  ∥x_{i} − \overline{α_{i}}a∥ ≤ r |α_{i}| for each i ∈ ℕ,

(note that if α_{i} ≠ 0 for any i ∈ ℕ, then the condition (3.2) is equivalent to

(3.3)  ∥\frac{x_{i}}{\overline{α_{i}}} − a∥ ≤ r for each i ∈ ℕ),

then we have the following inequalities

(3.4)  (∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2})^{\frac{1}{2}} ≤ \frac{1}{\sqrt{∥a∥^{2} − r^{2}}} Re ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, a⟩

≤ \frac{∥a∥}{\sqrt{∥a∥^{2} − r^{2}}} ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥;

(3.5)  0 ≤ (∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2})^{\frac{1}{2}} − ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥

≤ (∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2})^{\frac{1}{2}} − Re ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, \frac{a}{∥a∥}⟩

≤ \frac{r^{2}}{\sqrt{∥a∥^{2} − r^{2}} (∥a∥ + \sqrt{∥a∥^{2} − r^{2}})} Re ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, \frac{a}{∥a∥}⟩

≤ \frac{r^{2}}{\sqrt{∥a∥^{2} − r^{2}} (∥a∥ + \sqrt{∥a∥^{2} − r^{2}})} ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥;

(3.6)  ∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} ≤ \frac{1}{∥a∥^{2} − r^{2}} [Re ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, a⟩]^{2}

≤ \frac{∥a∥^{2}}{∥a∥^{2} − r^{2}} ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥^{2}

and

(3.7)  0 ≤ ∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} − ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥^{2}

≤ ∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} − [Re ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, \frac{a}{∥a∥}⟩]^{2}

≤ \frac{r^{2}}{∥a∥^{2} (∥a∥^{2} − r^{2})} [Re ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, a⟩]^{2}

≤ \frac{r^{2}}{∥a∥^{2} − r^{2}} ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥^{2} .

All the inequalities in (3.4) – (3.7) are sharp. Proof. From (3.2) we deduce

∥x_{i}∥^{2} − 2 Re ⟨x_{i}, \overline{α_{i}}a⟩ + |α_{i}|^{2} ∥a∥^{2} ≤ |α_{i}|^{2} r^{2}

for any i ∈ ℕ, which is clearly equivalent to

(3.8)  ∥x_{i}∥^{2} + (∥a∥^{2} − r^{2}) |α_{i}|^{2} ≤ 2 Re ⟨α_{i}x_{i}, a⟩

for each i ∈ ℕ.

If we multiply (3.8) by p_{i} ≥ 0 and sum over i ∈ ℕ, then we deduce

(3.9)  ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} + (∥a∥^{2} − r^{2}) ∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ≤ 2 Re ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, a⟩ .

Now, dividing (3.9) by \sqrt{∥a∥^{2} − r^{2}} > 0 we get

(3.10) \frac{1}{\sqrt{∥a∥^{2} − r^{2}}} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} + \sqrt{∥a∥^{2} − r^{2}} ∑_{i=1}^{∞} p_{i} |α_{i}|^{2}

≤ \frac{2}{\sqrt{∥a∥^{2} − r^{2}}} Re ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, a⟩ .

On the other hand, by the elementary inequality

\frac{1}{α}p + αq ≥ 2\sqrt{pq}, α > 0, p, q ≥ 0,

we can state that:

(3.11) 2 [∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2}]^{\frac{1}{2}}

≤ \frac{1}{\sqrt{∥a∥^{2} − r^{2}}} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} + \sqrt{∥a∥^{2} − r^{2}} ∑_{i=1}^{∞} p_{i} |α_{i}|^{2} .

Making use of (3.10) and (3.11), we deduce the first part of (3.4).

The second part is obvious by Schwarz’s inequality

Re ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, a⟩ ≤ ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥ ∥a∥ .

If p_{1} = 1, x_{1} = x, α_{1} = 1 and p_{i} = 0, α_{i} = 0, x_{i} = 0 for i ≥ 2, then from (3.4) we deduce the inequality

∥x∥ ≤ \frac{1}{\sqrt{∥a∥^{2} − r^{2}}} Re ⟨x, a⟩ ≤ \begin{matrix}∥x∥ & ∥a∥ \\ √ & \overline{∥a∥^{2} − r^{2}} \end{matrix}

provided ∥x − a∥ ≤ r < ∥a∥ , x, a ∈ K. The sharpness of this inequality has been shown in [3, p. 20], and we omit the details.

The other inequalities are obvious consequences of (3.4) and we omit the details.

The following corollary may be stated.

Corollary 1. Let α ∈ ℓ_{p}^{2} (𝕂) , x ∈ ℓ_{p}^{2} (K) , e ∈ H, ∥e∥ = 1 and φ, ϕ ∈ 𝕂 with Re (ϕφ̄) > 0. If

(3.12)  ∥x_{i} − \overline{α_{i}} · \frac{φ + ϕ}{2} · e∥ ≤ \frac{1}{2} |ϕ − φ| |α_{i}|

for each i ∈ ℕ, or, equivalently

(3.13)  Re ⟨ϕ\overline{α_{i}}e − x_{i}, x_{i} − φ\overline{α_{i}}e⟩ ≥ 0

for each i ∈ ℕ, (note that, if α_{i} ≠ 0 for any i ∈ ℕ, then (3.12) is equivalent to

(3.14)  ∥\frac{x_{i}}{\overline{α_{i}}} − \frac{φ + ϕ}{2} · e∥ ≤ \frac{1}{2} |ϕ − φ|

for each i ∈ ℕ and (3.13) is equivalent to

Re ⟨ϕe − \frac{x_{i}}{\overline{α_{i}}} , \frac{x_{i}}{\overline{α_{i}}} − φe⟩ ≥ 0

for each i ∈ ℕ), then the following reverses of the (CBS) −inequality are valid:

(3.15)  (∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2})^{\frac{1}{2}} ≤ \frac{Re [(ϕ̄ + φ̄) ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, e⟩]}{2 [Re (ϕ\overline{φ})]^{\frac{1}{2}}}

≤ \frac{1}{2} · \frac{|φ + ϕ|}{[Re (ϕ\overline{φ})]^{\frac{1}{2}}} ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥;

REVERSING THE CBS-INEQUALITY 9

(3.16) 0 ≤ (∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2})^{\frac{1}{2}} − ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥

≤ (∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2})^{\frac{1}{2}} − Re [\frac{ϕ̄ + φ̄}{|φ + ϕ|} ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, e⟩]

≤ \frac{|ϕ − φ|^{2}}{2\sqrt{Re (ϕφ)} (|φ + ϕ| + 2\sqrt{Re (ϕ\overline{φ})})} Re [\frac{ϕ̄ + φ̄}{|φ + ϕ|} ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, e⟩]

≤ \frac{|ϕ − φ|^{2}}{2\sqrt{Re (ϕφ)} (|φ + ϕ| + 2\sqrt{Re (ϕ\overline{φ})})} ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥;

(3.17) ∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} ≤ \frac{1}{4 Re (ϕφ̄)} [Re \{(ϕ̄ + φ̄) ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, e⟩\}]^{2}

≤ \frac{1}{4} · \begin{matrix}|φ + ϕ| & 2 \\ Re & (ϕφ̄) \end{matrix} ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥^{2}

and

(3.18)  0 ≤ ∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} − ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥^{2}

≤ ∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} − [Re \{\frac{ϕ̄ + φ̄}{|φ + ϕ|} ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, e⟩\}]^{2}

≤ \frac{|ϕ − φ|^{2}}{4 |ϕ + φ|^{2} Re (ϕφ̄)} \{Re [(ϕ̄ + φ̄) ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, e⟩]\}^{2}

≤ \frac{|ϕ − φ|^{2}}{4 Re (ϕφ̄)} ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥^{2} .

All the inequalities in (3.15) – (3.18) are sharp. Remark 2. We remark that if M ≥ m > 0 and for α ∈ ℓ_{p}^{2} (𝕂) , x ∈ ℓ_{p}^{2} (K) , e ∈ H with ∥e∥ = 1, one would assume that either

(3.19)  ∥\frac{x_{i}}{\overline{α_{i}}} − \frac{M + m}{2} · e∥ ≤ \frac{1}{2} (M − m)

for each i ∈ ℕ, or, equivalently

(3.20)  Re ⟨Me − \frac{x_{i}}{\overline{α_{i}}} , \frac{x_{i}}{\overline{α_{i}}} − me⟩ ≥ 0

for each i ∈ ℕ, then the following, much simpler reverses of the (CBS) −inequality may be stated:

(3.21)  (∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2})^{\frac{1}{2}} ≤ \frac{M + m}{2\sqrt{mM}} Re ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, e⟩

≤ \frac{M + m}{2\sqrt{mM}} ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥;

(3.22)  0 ≤ (∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2})^{\frac{1}{2}} − ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥

≤ (∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2})^{\frac{1}{2}} − Re ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, e⟩

≤ \frac{(M − m)^{2}}{2 (\sqrt{M} + \sqrt{m})^{2} \sqrt{mM}} Re ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, e⟩

≤ \frac{(M − m)^{2}}{2 (\sqrt{M} + \sqrt{m})^{2} \sqrt{mM}} ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥;

∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} − ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥^{2} (3.23)

≤ \frac{(M + m)^{2}}{4mM} [Re ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, e⟩]^{2}

≤ \frac{(M + m)^{2}}{4mM} ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥^{2}

and

(3.24)  0 ≤ ∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} − ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥^{2}

≤ ∑_{i=1}^{∞} p_{i} |α_{i}|^{2} ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} − [Re ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, e⟩]^{2}

≤ \frac{(M − m)^{2}}{4mM} [Re ⟨∑_{i=1}^{∞} p_{i}α_{i}x_{i}, e⟩]^{2}

≤ \frac{(M − m)^{2}}{4mM} ∥∑_{i=1}^{∞} p_{i}α_{i}x_{i}∥^{2} .

4. Reverses for the Generalised Triangle Inequality

In 1966, J.B. Diaz and F.T. Metcalf [2] proved the following reverse of the generalised triangle inequality holding in an inner product space (H; ⟨·, ·⟩) over the real or complex number field 𝕂:

(4.1)  r ∑_{i=1}^{n} ∥x_{i}∥ ≤ ∥∑_{i=1}^{n} x_{i}∥

provided the vectors x_{1}, …, x_{n} ∈ H\ {0} satisfy the assumption

(4.2)  0 ≤ r ≤ \frac{Re ⟨x_{i}, a⟩}{∥x_{i}∥},

where a ∈ H and ∥a∥ = 1.

In an attempt to diversify the assumptions for which such reverse results hold, the author pointed out in [4] that

(4.3)  \sqrt{1 − ρ^{2}} ∑_{i=1}^{n} ∥x_{i}∥ ≤ ∥∑_{i=1}^{n} x_{i}∥ ,

where the vectors x_{i}, i ∈ {1, …, n} satisfy the condition

(4.4)  ∥x_{i} − a∥ ≤ ρ,  i ∈ {1, …, n}

where a ∈ H, ∥a∥ = 1 and ρ ∈ (0, 1) .

If, for M ≥ m > 0, the vectors x_{i} ∈ H, i ∈ {1, …, n} verify either

(4.5)  Re ⟨Ma − x_{i}, x_{i} − ma⟩ ≥ 0,  i ∈ {1, …, n} ,

or, equivalently,

(4.6)  ∥x_{i} − \frac{M + m}{2} · a∥ ≤ \frac{1}{2} (M − m) ,  i ∈ \{1, …, n\} ,

where a ∈ H, ∥a∥ = 1, then the following reverse of the generalised triangle inequality may be stated as well [4]

(4.7)  \frac{2\sqrt{mM}}{M + m} ∑_{i=1}^{n} ∥x_{i}∥ ≤ ∥∑_{i=1}^{n} x_{i}∥ .

Note that the inequalities (4.1), (4.3), and (4.7) are sharp; necessary and sufficient equality conditions were provided (see [2] and [4]).

It is obvious, from Theorem 4, that, if

(4.8)  ∥x_{i} − a∥ ≤ r, for i ∈ {1, …, n} ,

where ∥a∥ > r, a ∈ H and x_{i} ∈ H, i ∈ {1, …, n} , then one can state the inequalities

(4.9)  ∑_{i=1}^{n} ∥x_{i}∥ ≤ \sqrt{n} (∑_{i=1}^{n} ∥x_{i}∥^{2})^{\frac{1}{2}}

≤ \frac{1}{\sqrt{∥a∥^{2} − r^{2}}} Re ⟨∑_{i=1}^{n} x_{i}, a⟩

≤ \frac{∥a∥}{\sqrt{∥a∥^{2} − r^{2}}} ∥∑_{i=1}^{n} x_{i}∥;

and

(4.10)  0 ≤ ∑_{i=1}^{n} ∥x_{i}∥ − ∥∑_{i=1}^{n} x_{i}∥

≤ \sqrt{n} (∑_{i=1}^{n} ∥x_{i}∥^{2})^{\frac{1}{2}} − ∥∑_{i=1}^{n} x_{i}∥

≤ \sqrt{n} (∑_{i=1}^{n} ∥x_{i}∥^{2})^{\frac{1}{2}} − Re ⟨∑_{i=1}^{n} x_{i}, \frac{a}{∥a∥}⟩

≤ \frac{r^{2}}{\sqrt{∥a∥^{2} − r^{2}} (∥a∥ + \sqrt{∥a∥^{2} − r^{2}})} Re ⟨∑_{i=1}^{n} x_{i}, \frac{a}{∥a∥}⟩

≤ \frac{r^{2}}{\sqrt{∥a∥^{2} − r^{2}} (∥a∥ + \sqrt{∥a∥^{2} − r^{2}})} ∥∑_{i=1}^{n} x_{i}∥ .

We note that for ∥a∥ = 1 and r ∈ (0, 1) , the inequality (3.9) becomes

(4.11)  \sqrt{1 − r^{2}} ∑_{i=1}^{n} ∥x_{i}∥ ≤ \sqrt{(1 − r^{2}) n} (∑_{i=1}^{n} ∥x_{i}∥^{2})^{\frac{1}{2}}

≤ Re ⟨∑_{i=1}^{n} x_{i}, a⟩ ≤ ∥∑_{i=1}^{n} x_{i}∥ ,

which is a refinement of (4.3).

With the same assumptions for a and r, we have from (4.10) the following additive reverse of the generalised triangle inequality:

(4.12)  0 ≤ ∑_{i=1}^{n} ∥x_{i}∥ − ∥∑_{i=1}^{n} x_{i}∥

≤ \frac{r^{2}}{\sqrt{1 − r^{2}} (1 + \sqrt{1 − r^{2}})} Re ⟨∑_{i=1}^{n} x_{i}, a⟩

≤ \frac{r^{2}}{\sqrt{1 − r^{2}} (1 + \sqrt{1 − r^{2}})} ∥∑_{i=1}^{n} x_{i}∥ .

We can obtain the following reverses of the generalised triangle inequality from Corollary 1 when the assumptions are in terms of complex numbers ϕ and φ :

If φ, ϕ ∈ 𝕂 with Re (ϕφ̄) > 0 and x_{i} ∈ H, i ∈ {1, …, n} , e ∈ H, ∥e∥ = 1 are such that

(4.13)  ∥x_{i} − \frac{φ + ϕ}{2}e∥ ≤ \frac{1}{2} |ϕ − φ| for each i ∈ \{1, …, n\} ,

or, equivalently,

Re ⟨ϕe − x_{i}, x_{i} − φe⟩ ≥ 0 for each i ∈ {1, …, n} ,

then we have the following reverses of the generalised triangle inequality:

(4.14)  ∑_{i=1}^{n} ∥x_{i}∥ ≤ \sqrt{n} (∑_{i=1}^{n} ∥x_{i}∥^{2})^{\frac{1}{2}}

≤ \frac{Re [(ϕ̄ + φ̄) ⟨∑_{i=1}^{n} x_{i}, e⟩]}{2\sqrt{Re (ϕφ̄)}}

≤ \frac{1}{2} · \frac{|ϕ̄ + φ̄|}{\sqrt{Re (ϕφ̄)}} ∥∑_{i=1}^{n} x_{i}∥

and

(4.15)  0 ≤ ∑_{i=1}^{n} ∥x_{i}∥ − ∥∑_{i=1}^{n} x_{i}∥

≤ \sqrt{n} (∑_{i=1}^{n} ∥x_{i}∥^{2})^{\frac{1}{2}} − ∥∑_{i=1}^{n} x_{i}∥

≤ \sqrt{n} (∑_{i=1}^{n} ∥x_{i}∥^{2})^{\frac{1}{2}} − Re [\frac{|ϕ̄ + φ̄|}{\sqrt{Re (ϕ̄φ̄)}} ⟨∑_{i=1}^{n} x_{i}, e⟩]

≤ \frac{|ϕ − φ|^{2}}{2\sqrt{Re (ϕφ̄)} (|ϕ + φ| + 2\sqrt{Re (ϕφ̄)})} Re [\frac{ϕ̄ + φ̄}{|ϕ̄ + φ̄|} ⟨∑_{i=1}^{n} x_{i}, e⟩]

≤ \frac{|ϕ − φ|^{2}}{2\sqrt{Re (ϕφ̄)} (|ϕ + φ| + 2\sqrt{Re (ϕφ̄)})} ∥∑_{i=1}^{n} x_{i}∥ .

Obviously (4.14) for ϕ = M, φ = m, M ≥ m > 0 provides a refinement for (4.7).

5. Lower Bounds for the Distance to Finite-Dimensional Subspaces

Let (H; ⟨·, ·⟩) be an inner product space over the real or complex number field 𝕂, {y_{1}, …, y_{n}} a subset of H and G (y_{1}, …, y_{n}) the gram matrix of {y_{1}, …, y_{n}} where (i, j) −entry is ⟨y_{i}, y_{j}⟩ . The determinant of G (y_{1}, …, y_{n}) is called the Gram determinant of {y_{1}, …, y_{n}} and is denoted by Γ (y_{1}, …, y_{n}) . Thus,

(5.1)  Γ (y_{1}, …, y_{n}) = | \begin{matrix}⟨y_{1}, y_{1}⟩ & ⟨y_{1}, y_{2}⟩ & ⋯ & ⟨y_{1}, y_{n}⟩ \\ ⟨y_{2}, y_{1}⟩ & ⟨y_{2}, y_{2}⟩ & ⋯ & ⟨y_{2}, y_{n}⟩ \\ ⋯ & ⋯ & ⋯ & ⋯ & ⋯ \\ ⟨y_{n}, y_{1}⟩ & ⟨y_{n}, y_{2}⟩ & ⋯ & ⟨y_{n}, y_{n}⟩ \end{matrix} | .

Following [1, p. 129 – 133], we state here some general results for the Gram determinant that will be used in the sequel.

(1) Let {x_{1}, …, x_{n}} ⊂ H. Then Γ (x_{1}, …, x_{n}) ≠ 0 if and only if {x_{1}, …, x_{n}} is linearly independent;

(2) Let M = span {x_{1}, …, x_{n}} be n−dimensional in H, i.e., {x_{1}, …, x_{n}} is linearly independent. Then for each x ∈ H, the distance d (x, M) from x to the linear subspace H has the representations

(5.2)  d^{2} (x, M) = \frac{Γ (x_{1}, …, x_{n}, x)}{Γ (x_{1}, …, x_{n})}

and

(5.3)  d^{2} (x, M) = \{ \begin{matrix}∥x∥^{2} − \begin{matrix}(∑_{i=1}^{n}|⟨x,x_{i}⟩|^{2}) & ^{2} \\ ∥∑_{i=1}^{n}⟨x,x_{i}⟩x_{i}∥ & ^{2} \end{matrix} & if x ∉ M^{⊥}, \\ ∥x∥^{2} & if x ∈ M^{⊥}, \end{matrix}

where M^{⊥} denotes the orthogonal complement of M.

(3) If {x_{1}, …, x_{n}} is an orthornormal set in H, i.e., ⟨x_{i}, x_{j}⟩ = δ_{ij}, i, j ∈ {1, …, n} , where δ_{ij} is Kronecker’s delta, then

(5.4)  d^{2} (x, M) = ∥x∥^{2} − ∑_{i=1}^{n} |⟨x, x_{i}⟩|^{2} .
(4) Let {x_{1}, …, x_{n}} be a set of nonzero vectors in H. Then
(5.5)  0 ≤ Γ (x_{1}, …, x_{n}) ≤ ∥x_{1}∥^{2} ∥x_{2}∥^{2} ⋯ ∥x_{n}∥^{2} .

The equality holds on the left (respectively right) side of (5.5) if and only if {x_{1}, …, x_{n}} is linearly dependent (respectively orthogonal). The first inequality in (5.5) is known in the literature as Gram’s inequality while the second one is known as Hadamard’s inequality.

The following result may be stated.

Proposition 4. Let {x_{1}, …, x_{n}} be a system of linearly independent vectors, M = span {x_{1}, …, x_{n}} , x ∈ H\M^{⊥}, a ∈ H, r > 0 and ∥a∥ > r. If

(5.6)  ∥x_{i} − \overline{⟨x, x_{i}⟩}a∥ ≤ |⟨x, x_{i}⟩| r for each i ∈ \{1, …, n\} ,

(note that if ⟨x, x_{i}⟩ ≠ 0 for each i ∈ {1, …, n} , then (5.6) can be written as

(5.7)  ∥\frac{x_{i}}{\overline{⟨x, x_{i}⟩}} − a∥ ≤ r for each i ∈ \{1, …, n\}),

then we have the inequality

(5.8)  d^{2} (x, M) ≥ ∥x∥^{2} − \frac{∥a∥^{2}}{∥a∥^{2} − r^{2}} · \begin{matrix}∑ & {n \atop i=1} & |⟨x, x_{i}⟩| & 2 \\ ∑ & {n \atop i=1} & ∥x_{i}∥ & 2 \end{matrix}
≥ 0.

Proof. Utilising (5.3) we can state that

(5.9)  d^{2} (x, M) = ∥x∥^{2} − \begin{matrix}∑ & {n \atop i=1} & |⟨x, x_{i}⟩| & 2 \\ ∥∑_{i=1}^{n} ⟨x, x_{i}⟩ x_{i}∥ & 2 \end{matrix} · ∑_{i=1}^{n} |⟨x, x_{i}⟩|^{2} .

Also, by the inequality (3.6) applied for α_{i} = ⟨x, x_{i}⟩ , p_{i} = \frac{1}{n}, i ∈ \{1, …, n\} , we can state that

(5.10)  \begin{matrix}∑ & {n \atop i=1} & |⟨x, x_{i}⟩| & 2 \\ ∥∑_{i=1}^{n} ⟨x, x_{i}⟩ x_{i}∥ & 2 \end{matrix} ≤ \frac{∥a∥^{2}}{∥a∥^{2} − r^{2}} · \frac{1}{∑_{i=1}^{n} ∥x_{i}∥^{2}}

provided the condition (5.7) holds true.

Combining (5.9) with (5.10) we deduce the first inequality in (5.8).

The last inequality is obvious since, by Schwarz’s inequality

∥x∥^{2} ∑_{i=1}^{n} ∥x_{i}∥^{2} ≥ ∑_{i=1}^{n} |⟨x, x_{i}⟩|^{2} ≥ \frac{∥a∥^{2}}{∥a∥^{2} − r^{2}} ∑_{i=1}^{n} |⟨x, x_{i}⟩|^{2} .

Remark 3. Utilising (5.2), we can state the following result for Gram determinants

(5.11) Γ (x_{1}, …, x_{n}, x) ≥ [∥x∥^{2} − \frac{∥a∥^{2}}{∥a∥^{2} − r^{2}} · \begin{matrix}∑ & {n \atop i=1} & |⟨x, x_{i}⟩| & 2 \\ ∑ & {n \atop i=1} & ∥x_{i}∥ & 2 \end{matrix} ] Γ (x_{1}, …, x_{n}) ≥ 0

for x ∉ M^{⊥} and x, x_{i}, a and r are as in Proposition 4.

The following corollary of Proposition 4 may be stated as well.

Corollary 2. Let {x_{1}, …, x_{n}} be a system of linearly independent vectors, M = span {x_{1}, …, x_{n}} , x ∈ H\M^{⊥} and ϕ, φ ∈ K with Re (ϕφ̄) > 0. If e ∈ H, ∥e∥ = 1 and

(5.12)  ∥x_{i} − \overline{⟨x, x_{i}⟩} · \frac{φ + ϕ}{2}e∥ ≤ \frac{1}{2} |ϕ − φ| |⟨x, x_{i}⟩|

or, equivalently,

Re ⟨ϕ\overline{· ⟨x, x_{i}⟩}e − x_{i}, x_{i} − φ · \overline{⟨x, x_{i}⟩}e⟩ ≥ 0,

for each i ∈ {1, …, n} , then

(5.13)  d^{2} (x, M) ≥ ∥x∥^{2} − \frac{1}{4} · \begin{matrix}|φ + ϕ| & 2 \\ Re & (ϕφ̄) \end{matrix} · \begin{matrix}∑ & {n \atop i=1} & |⟨x, x_{i}⟩| & 2 \\ ∑ & {n \atop i=1} & ∥x_{i}∥ & 2 \end{matrix} ≥ 0,

or, equivalently,

(5.14) Γ (x_{1}, …, x_{n}, x)

≥ [∥x∥^{2} − \frac{1}{4} · \begin{matrix}|φ + ϕ| & 2 \\ Re & (ϕφ̄) \end{matrix} · \begin{matrix}∑ & {n \atop i=1} & |⟨x, x_{i}⟩| & 2 \\ ∑ & {n \atop i=1} & ∥x_{i}∥ & 2 \end{matrix} ] Γ (x_{1}, …, x_{n}) ≥ 0.

6. Applications for Fourier Coefficients

Let (H; ⟨·, ·⟩) be a Hilbert space over the real or complex number field 𝕂 and {e_{i}}_{i∈I} an orthornormal basis for H. Then (see for instance [1, p. 54 – 61]):

(i) Every element x ∈ H can be expanded in a Fourier series, i.e.,

x = ∑_{i∈I} ⟨x, e_{i}⟩ e_{i},

where ⟨x, e_{i}⟩ , i ∈ I are the Fourier coefficients of x; (ii) (Parseval identity)

∥x∥^{2} = ∑_{i∈I} ⟨x, e_{i}⟩ e_{i}, x ∈ H;

(iii) (Extended Parseval identity)

⟨x, y⟩ = ∑_{i∈I} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩ , x, y ∈ H;

(iv) (Elements are uniquely determined by their Fourier coefficients)

⟨x, e_{i}⟩ = ⟨y, e_{i}⟩ for every i ∈ I implies that x = y.

Now, we must remark that all the results from the second and third sections can be stated for K = 𝕂 where 𝕂 is the Hilbert space of complex (real) numbers endowed with the usual norm and inner product.

Therefore, we can state the following proposition.

Proposition 5. Let (H; ⟨·, ·⟩) be a Hilbert space over 𝕂 and {e_{i}}_{i∈I} an orthornormal base for H. If x, y ∈ H (y ≠ 0) , a ∈ 𝕂 (ℂ, ℝ) and r > 0 such that |a| > r and

(6.1)  |\frac{⟨x, e_{i}⟩}{⟨y, e_{i}⟩} − a| ≤ r for each i ∈ I,

then we have the following reverse of the Schwarz inequality

(6.2)  ∥x∥ ∥y∥ ≤ \frac{1}{\sqrt{|a|^{2} − r^{2}}} Re [ā · ⟨x, y⟩]

≤ \frac{|a|}{\sqrt{|a|^{2} − r^{2}}} |⟨x, y⟩|;

(6.3)  (0 ≤) ∥x∥ ∥y∥ − |⟨x, y⟩|

≤ ∥x∥ ∥y∥ − Re [\frac{ā}{|a|} · ⟨x, y⟩]

≤ \frac{r^{2}}{\sqrt{|a|^{2} − r^{2}} (|a| + \sqrt{|a|^{2} − r^{2}})} Re [\frac{ā}{|a|} · ⟨x, y⟩]

≤ \frac{r^{2}}{\sqrt{|a|^{2} − r^{2}} (|a| + \sqrt{|a|^{2} − r^{2}})} |⟨x, y⟩|;

(6.4)  ∥x∥^{2} ∥y∥^{2} ≤ \frac{1}{|a|^{2} − r^{2}} (Re [ā · ⟨x, y⟩])^{2}

≤ \frac{|a|^{2}}{|a|^{2} − r^{2}} |⟨x, y⟩|^{2}

and

(6.5)  (0 ≤) ∥x∥^{2} ∥y∥^{2} − |⟨x, y⟩|^{2}

≤ ∥x∥^{2} ∥y∥^{2} − (Re [\frac{ā}{|a|} · ⟨x, y⟩])^{2}

≤ \frac{r^{2}}{|a|^{2} (|a|^{2} − r^{2})} − (Re [\frac{ā}{|a|} · ⟨x, y⟩])^{2}

≤ \frac{r^{2}}{|a|^{2} − r^{2}} |⟨x, y⟩| .

The proof is similar to the one in Theorem 4, where instead of x_{i} we take ⟨x, e_{i}⟩ , instead of α_{i} we take ⟨e_{i}, y⟩ , ∥·∥ = |·| , p_{i} = 1, and we use the Parseval identities mentioned above in (ii) and (iii). We omit the details.

The following result may be stated as well.

Proposition 6. Let (H; ⟨·, ·⟩) be a Hilbert space over 𝕂 and {e_{i}}_{i∈I} an orthornormal base for H. If x, y ∈ H (y ≠ 0) , e, φ, ϕ ∈ 𝕂 with Re (ϕφ̄) > 0, |e| = 1 and, either

(6.6)  |\frac{⟨x, e_{i}⟩}{⟨y, e_{i}⟩} − \frac{φ + ϕ}{2} · e| ≤ \frac{1}{2} |ϕ − φ|

or, equivalently,

(6.7)  Re [(ϕe − \frac{⟨x, e_{i}⟩}{⟨y, e_{i}⟩}) (\frac{⟨e_{i}, x⟩}{⟨e_{i}, y⟩} − φ̄ē)] ≥ 0

for each i ∈ I, then the following reverses of the Schwarz inequality hold:

(6.8)  ∥x∥ ∥y∥ ≤ \frac{Re [(ϕ̄ + φ̄) ē⟨x, y⟩]}{2\sqrt{Re (ϕφ̄)}} ≤ \frac{1}{2} · \frac{|φ + ϕ|}{\sqrt{Re (ϕφ̄)}} |⟨x, y⟩| .

(6.9)  (0 ≤) ∥x∥ ∥y∥ − |⟨x, y⟩|

≤ ∥x∥ ∥y∥ − Re [\frac{(ϕ̄ + φ̄) ē}{|φ + ϕ|} ⟨x, y⟩]

≤ \frac{|ϕ − φ|^{2}}{2\sqrt{Re (ϕφ̄)} (|φ + ϕ| + 2\sqrt{Re (ϕφ̄)})} Re [\frac{(ϕ̄ + φ̄) ē}{|φ + ϕ|} ⟨x, y⟩]

≤ \frac{|ϕ − φ|^{2}}{2\sqrt{Re (ϕφ̄)} (|φ + ϕ| + 2\sqrt{Re (ϕφ̄)})} |⟨x, y⟩|

and

(6.10)  (0 ≤) ∥x∥^{2} ∥y∥^{2} − |⟨x, y⟩|^{2}

≤ ∥x∥^{2} ∥y∥^{2} − [Re [\frac{(ϕ̄ + φ̄) ē}{|φ + ϕ|} ⟨x, y⟩]]^{2}

≤ \frac{|ϕ − φ|^{2}}{4 |ϕ + φ|^{2} Re (ϕφ̄)} \{Re [(ϕ̄ + φ̄) ē⟨x, y⟩]\}^{2}

≤ \frac{|ϕ − φ|^{2}}{4 Re (ϕφ̄)} |⟨x, y⟩|^{2} .

Remark 4. If ϕ = M ≥ m = φ > 0, then one may state simpler inequalities from (6.8) – (6.10). We omit the details.

References

[1] F. DEUTSCH, Best Approximation in Inner Product Spaces, CMS Books in Mathematics, Springer-Verlag, New York, Berlin, Heidelberg, 2001.

[2] J.B. DIAZ and F.T. METCALF, A complementary triangle inequality in Hilbert and Banach spaces, Proc. Amer. Math. Soc., 17(1) (1966), 88-99.

[3] S.S. DRAGOMIR, Advances in Inequalities of the Schwarz, Grüss and Bessel Type in Inner Product Spaces, RGMIA Monographs, Victoria University, 2004. [ONLINE

http://rgmia.vu.edu.au].

[4] S.S. DRAGOMIR, Reverses of the triangle inequality in inner product spaces, RGMIA Res. Rep. Coll., 7 (2004), Supplement, Article 7. [ONLINE: http://rgmia.vu.edu.au/v7(E).html].

[5] S.S. DRAGOMIR, Reversing the CBS-inequality for sequences of vectors in Hilbert spaces with applications (II), RGMIA Res. Rep. Coll., 8(2005), Supplement, Article 3 [ONLINE

http://rgmia.vu.edu.au/v8(E).html].

School of Computer Science and Mathematics, Victoria University of Technology, PO Box 14428, MCMC 8001, VIC, Australia.

E-mail address: sever@matilda.vu.edu.au
URL: http://rgmia.vu.edu.au/SSDragomirWeb.html