SOME NEW RESULTS RELATED TO BESSEL AND GRÜSS INEQUALITIES FOR ORTHOGONAL FAMILIES IN INNER PRODUCT SPACES

S.S. DRAGOMIR

Abstract. Some new counterparts of Bessel’s inequality for orthornormal families in real or complex inner product spaces are pointed out. Applications for some Grüss type inequalities are also emphasized.

1. Introduction

In [1], the author has proved the following result which provides both a Grüss type inequality for orthogonal families of vectors in real or complex inner products as well as, for x = y, a counterpart of Bessel’s inequality.

Theorem 1. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in H, i.e., ⟨e_{i}, e_{j}⟩ = 0 if i ≠ j and ∥e_{i}∥ = 1, i, j ∈ I, F a finite part of I, ϕ_{i}, γ_{i}, Φ_{i}, Γ_{i} ∈ ℝ (i ∈ F), and x, y ∈ H. If either

(1.1)  Re ⟨∑_{i=1}^{n} Φ_{i}e_{i} − x, x − ∑_{i=1}^{n} ϕ_{i}e_{i}⟩ ≥ 0,

Re ⟨∑_{i=1}^{n} Γ_{i}e_{i} − y, y − ∑_{i=1}^{n} γ_{i}e_{i}⟩ ≥ 0,

or, equivalently,

(1.2)  ∥x − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,

∥y − ∑_{i∈F} \frac{Γ_{i} + γ_{i}}{2} e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} ,

Date: May 19, 2003.

2000 Mathematics Subject Classification. 26D15, 46C05.

Key words and phrases. Bessel’s inequality, Grüss’ inequality, Inner product, Lebesgue integral.

hold, then we have the inequality

(1.3)  0 ≤ |⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩|

≤ \frac{1}{4} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} · (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}}

− [Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}⟩]^{\frac{1}{2}}

× [Re ⟨∑_{i∈F} Γ_{i}e_{i} − y, y − ∑_{i∈F} γ_{i}e_{i}⟩]^{\frac{1}{2}}

≤ \frac{1}{4} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} · (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} .

The constant \frac{1}{4} is best possible in the sense that it cannot be replaced by a smaller constant.

In the follow up paper [2], and by the use of a different technique, the author has pointed out the following result as well:

Theorem 2. Let {e_{i}}_{i∈I}, F, ϕ_{i}, γ_{i}, Φ_{i}, Γ_{i} and x, y be as in Theorem 1. If either (1.1) or (1.2) holds, then we have the inequality

(1.4)  0 ≤ |⟨x, y⟩ − ∑_{i=1}^{n} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩|

≤ \frac{1}{4} (∑_{i=1}^{n} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} · (∑_{i=1}^{n} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}}

− ∑_{i∈F} |\frac{Φ_{i} + ϕ_{i}}{2} − ⟨x, e_{i}⟩| |\frac{Γ_{i} + γ_{i}}{2} − ⟨y, e_{i}⟩|

≤ \frac{1}{4} (∑_{i=1}^{n} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} · (∑_{i=1}^{n} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} .

The constant \frac{1}{4} is best possible in the sense that it cannot be replaced by a smaller constant.

It has also been shown that the bounds provided by the second inequality in (1.3) and the second inequality in (1.4) cannot be compared in general.

2. A New Counterpart of Bessel’s Inequality

The following counterpart of Bessel’s inequality holds.

Theorem 3. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in H, F a finite part of I, and ϕ_{i}, Φ_{i} (i ∈ F) , real or complex numbers such that ∑_{i∈F} Re (Φ_{i}\overline{ϕ_{i}}) > 0. If x ∈ H is such that either (i) Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}⟩ ≥ 0;

or, equivalently,

(ii) ∥x − ∑_{i∈F} \frac{ϕ_{i}+Φ_{i}}{2} e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ;

holds, then one has the inequality

(2.1)  ∥x∥^{2} ≤ \frac{1}{4} · \begin{matrix}∑ & i∈F & (|Φ_{i}| + |ϕ_{i}|) & 2 \\ ∑ & i∈F & Re & (Φ_{i}\overline{ϕ_{i}}) \end{matrix} ∑_{i∈F} |⟨x, e_{i}⟩|^{2} .

The constant \frac{1}{4} is best possible in the sense that it cannot be replaced by a smaller constant.

Proof. Firstly, we observe that for y, a, A ∈ H, the following are equivalent

(2.2)  Re ⟨A − y, y − a⟩ ≥ 0

and

(2.3)  ∥y − \frac{a + A}{2}∥ ≤ \frac{1}{2} ∥A − a∥ .

Now, for a = ∑_{i∈F} ϕ_{i}e_{i}, A = ∑_{i∈F} Φ_{i}e_{i}, we have

∥A − a∥ = ∥∑_{i∈F} (Φ_{i} − ϕ_{i}) e_{i}∥ = [∥∑_{i∈F} (Φ_{i} − ϕ_{i}) e_{i}∥^{2}]^{\frac{1}{2}}

= (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} ∥e_{i}∥^{2})^{\frac{1}{2}} = (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,

giving, for y = x, the desired equivalence.

Now, observe that

Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}⟩

= ∑_{i∈F} Re [Φ_{i}\overline{⟨x, e_{i}⟩} + \overline{ϕ_{i}} ⟨x, e_{i}⟩] − ∥x∥^{2} − ∑_{i∈F} Re (Φ_{i}\overline{ϕ_{i}}) ,

giving, from (i), that

(2.4)  ∥x∥^{2} + ∑_{i∈F} Re (Φ_{i}\overline{ϕ_{i}}) ≤ ∑_{i∈F} Re [Φ_{i}\overline{⟨x, e_{i}⟩} + \overline{ϕ_{i}} ⟨x, e_{i}⟩] .

On the other hand, by the elementary inequality

αp^{2} + \frac{1}{α}q^{2} ≥ 2pq, α > 0, p, q ≥ 0;

we deduce

(2.5)  2 ∥x∥ ≤ \begin{matrix}∥x∥ & 2 \\ [∑_{i∈F} Re (Φ_{i}\overline{ϕ_{i}})] & \frac{1}{2} \end{matrix} + [∑_{i∈F} Re (Φ_{i}\overline{ϕ_{i}})]^{\frac{1}{2}} .

Dividing (2.4) by [∑_{i∈F} Re (Φ_{i}\overline{ϕ_{i}})]^{\frac{1}{2}} > 0 and using (2.5), we obtain

(2.6)  ∥x∥ ≤ \frac{1}{2} \begin{matrix}∑ & i∈F & Re & [Φ_{i}\overline{⟨x, e_{i}⟩} + \overline{ϕ_{i}} ⟨x, e_{i}⟩] \\ [∑_{i∈F} Re (Φ_{i}\overline{ϕ_{i}})] & \frac{1}{2} \end{matrix},

which is also an interesting inequality in itself.

Using the Cauchy-Buniakowsky-Schwartz inequality for real numbers, we get

(2.7) ∑_{i∈F} Re [Φ_{i}\overline{⟨x, e_{i}⟩} + \overline{ϕ_{i}} ⟨x, e_{i}⟩] ≤ ∑_{i∈F} |Φ_{i}\overline{⟨x, e_{i}⟩} + \overline{ϕ_{i}} ⟨x, e_{i}⟩|

≤ ∑_{i∈F} (|Φ_{i}| + |ϕ_{i}|) |⟨x, e_{i}⟩|

≤ [∑_{i∈F} (|Φ_{i}| + |ϕ_{i}|)^{2}]^{\frac{1}{2}} [∑_{i∈F} |⟨x, e_{i}⟩|^{2}]^{\frac{1}{2}} .

Making use of (2.6) and (2.7), we deduce the desired result (2.1).

To prove the sharpness of the constant \frac{1}{4}, let us assume that (2.1) holds with a constant c > 0, i.e.,

(2.8)  ∥x∥^{2} ≤ c · \begin{matrix}∑ & i∈F & (|Φ_{i}| + |ϕ_{i}|) & 2 \\ ∑ & i∈F & Re & (Φ_{i}\overline{ϕ_{i}}) \end{matrix} ∑_{i∈F} |⟨x, e_{i}⟩|^{2} ,

provided x, ϕ_{i}, Φ_{i}, i ∈ F satisfies (i).

Choose F = {1} , e_{1} = e, ∥e∥ = 1, ϕ_{i} = m, Φ_{i} = M with m, M > 0, then, by (2.8) we get

(2.9)  ∥x∥^{2} ≤ c\frac{(M + m)^{2}}{mM} |⟨x, e⟩|^{2}

provided

(2.10)  Re ⟨Me − x, x − me⟩ ≥ 0.

If x = me, then obviously (2.10) holds, and by (2.9) we get

m^{2} ≤ c\frac{(M + m)^{2}}{mM} m^{2}

giving mM ≤ c (M + m)^{2} for m, M > 0. Now, if in this inequality we choose m = 1 − ε, M = 1 + ε (ε ∈ (0, 1)) , then we get 1 − ε^{2} ≤ 4c for ε ∈ (0, 1) , from where we deduce c ≥ \frac{1}{4}.

Remark 1. By the use of (2.6), the second inequality in (2.7) and the Hölder inequality, we may state the following counterparts of Bessel’s inequality as well:

(2.11) ∥x∥^{2} ≤ \frac{1}{2} · \frac{1}{[∑_{i∈F} Re (Φ_{i}\overline{ϕ_{i}})]^{\frac{1}{2}}}

× \{ \begin{matrix}max_{i∈F} & \{|Φ_{i}| + |ϕ_{i}|\} & ∑_{i∈F} & |⟨x, e_{i}⟩| \\ [ ∑_{i∈F} (|Φ_{i}| + |ϕ_{i}|)^{p}]^{\frac{1}{p}} ( ∑_{i∈F} |⟨x, e_{i}⟩|^{q})^{\frac{1}{q}} , \\ for p > 1, \frac{1}{p} + \frac{1}{q} = 1 \\ max_{i∈F} |⟨x, e_{i}⟩| ∑_{i∈F} [|Φ_{i}| + |ϕ_{i}|] . \end{matrix}

The following corollary holds.

Corollary 1. With the assumption of Theorem 3 and if either (i) or (ii) holds, then

(2.12)  0 ≤ ∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2} ≤ \frac{1}{4}M^{2} (Φ, ϕ, F) ∑_{i∈F} |⟨x, e_{i}⟩|^{2} ,

where

(2.13) M (Φ, ϕ, F) := [\begin{matrix}∑ & i∈F & \{(|Φ_{i}| + |ϕ_{i}|)^{2} + 4 [|Φ_{i}\overline{ϕ_{i}}| − Re (Φ_{i}\overline{ϕ_{i}})]\} \\ ∑ & i∈F & Re & (Φ_{i}\overline{ϕ_{i}}) \end{matrix}]^{\frac{1}{2}} .

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

Proof. The inequality (2.12) follows by (2.1) on subtracting the same quantity ∑_{i∈F} |⟨x, e_{i}⟩|^{2} from both sides.

To prove the sharpness of the constant \frac{1}{4}, assume that (2.12) holds with c > 0, i.e.,

(2.14)  0 ≤ ∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2} ≤ cM^{2} (Φ, ϕ, F) ∑_{i∈F} |⟨x, e_{i}⟩|^{2}

provided the condition (i) holds.

Choose F = {1} , e_{1} = e, ∥e∥ = 1, ϕ_{i} = ϕ, Φ_{i} = Φ, ϕ, Φ > 0 in (2.14) to get

(2.15)  0 ≤ ∥x∥^{2} − |⟨x, e⟩|^{2} ≤ c\frac{(Φ − ϕ)^{2}}{ϕΦ} |⟨x, e⟩|^{2} ,

provided

(2.16)  ⟨Φe − x, x − ϕe⟩ ≥ 0.

If H = ℝ^{2}, x = (x_{1}, x_{2}) ∈ ℝ^{2}, e = (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) then we have

∥x∥^{2} − |⟨x, e⟩|^{2} = x_{1}^{2} + x_{2}^{2} − \frac{(x_{1} + x_{2})^{2}}{2} = \frac{1}{2} (x_{1} − x_{2})^{2} ,
|⟨x, e⟩|^{2} = \frac{(x_{1} + x_{2})^{2}}{2}

and by (2.15) we get

(2.17)  \frac{(x_{1} − x_{2})^{2}}{2} ≤ c\frac{(Φ − ϕ)^{2}}{ϕΦ} · \frac{(x_{1} + x_{2})^{2}}{2} .

Now, if we let x_{1} = \frac{ϕ}{\sqrt{2}}, x_{2} = \frac{Φ}{\sqrt{2}} (ϕ, Φ > 0) then obviously

⟨Φe − x, x − ϕe⟩ = ∑_{i=1}^{2} (\frac{Φ}{\sqrt{2}} − x_{i}) (x_{i} − \frac{ϕ}{\sqrt{2}}) = 0,

which shows that (2.16) is fulfilled, and thus by (2.17) we obtain

\frac{(Φ − ϕ)^{2}}{4} ≤ c\frac{(Φ − ϕ)^{2}}{ϕΦ} · \frac{(Φ + ϕ)^{2}}{4}

for any Φ > ϕ > 0. This implies

(2.18)  c (Φ + ϕ)^{2} ≥ ϕΦ

for any Φ > ϕ > 0.

Finally, let ϕ = 1 − ε, Φ = 1 + ε, ε ∈ (0, 1). Then from (2.18) we get 4c ≥ 1 − ε^{2} for any ε ∈ (0, 1) which produces c ≥ \frac{1}{4}.

Remark 2. If {e_{i}}_{i∈I} is an orthornormal family in the real inner product (H; ⟨·, ·⟩) and M_{i}, m_{i} ∈ ℝ, i ∈ F (F is a finite part of I) and x ∈ H are such that M_{i}, m_{i} ≥ 0 for i ∈ F with ∑_{i∈F} M_{i}m_{i} ≥ 0 and

⟨∑_{i∈F} M_{i}e_{i} − x, x − ∑_{i∈F} m_{i}e_{i}⟩ ≥ 0,

then we have the inequality

(2.19)  0 ≤ ∥x∥^{2} − ∑_{i∈F} [⟨x, e_{i}⟩]^{2} ≤ \frac{1}{4} · \frac{∑_{i∈F} (M_{i} − m_{i})^{2}}{∑_{i∈F} M_{i}m_{i}} · ∑_{i∈F} [⟨x, e_{i}⟩]^{2} .

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

The following counterpart of the Schwarz’s inequality in inner product spaces holds.

Corollary 2. Let x, y ∈ H and δ, Δ ∈ 𝕂 (𝕂 = ℂ, ℝ) with the property that Re (Δ\overline{δ}) > 0. If either

(2.20)  Re ⟨Δy − x, x − δy⟩ ≥ 0

or, equivalently,

(2.21)  ∥x − \frac{δ + Δ}{2} · y∥ ≤ \frac{1}{2} |Δ − δ| ∥y∥

holds, then we have the inequalities

(2.22)  ∥x∥ ∥y∥ ≤ \frac{1}{2} · \begin{matrix}Re & [Δ\overline{⟨x, y⟩} + \overline{δ} ⟨x, y⟩] \\ √ & \overline{Δ\overline{δ}} \end{matrix}

≤ \frac{1}{2} · \frac{|Δ| + |δ|}{\sqrt{Δ\overline{δ}}} |⟨x, y⟩| ,

(2.23)  0 ≤ ∥x∥ ∥y∥ − |⟨x, y⟩|

≤ \frac{1}{2} · \frac{(\sqrt{|Δ|} − \sqrt{|δ|})^{2} + 2 (\sqrt{Δ\overline{δ}} − \sqrt{Re (Δ\overline{δ})})}{\sqrt{Δ\overline{δ}}} |⟨x, y⟩| ,

(2.24)  ∥x∥^{2} ∥y∥^{2} ≤ \frac{1}{4} · \begin{matrix}(|Δ| + |δ|) & 2 \\ Re & (Δ\overline{δ}) \end{matrix} |⟨x, y⟩|^{2} ,

and

(2.25) 0 ≤ ∥x∥^{2} ∥y∥^{2} − |⟨x, y⟩|^{2} ≤ \frac{1}{4} · \frac{(|Δ| + |δ|)^{2} + 4 (|Δ\overline{δ}| − Re (Δ\overline{δ}))}{Re (Δ\overline{δ})} |⟨x, y⟩|^{2} .

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

Proof. The inequality (2.22) follows from (2.6) on choosing F = \{1\} , e_{1} = e = \frac{y}{∥y∥}, Φ_{1} = Φ = Δ ∥y∥ , ϕ_{1} = ϕ = δ ∥y∥ (y ≠ 0) . The inequality (2.23) is equivalent with (2.22). The inequality (2.24) follows from (2.1) for F = {1} and the same choices as above. Finally, (2.25) is obviously equivalent with (2.24).

3. Some Grüss Type Inequalities

The following result holds.

Theorem 4. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in H, F a finite part of I, ϕ_{i}, Φ_{i}, γ_{i}, Γ_{i} ∈ 𝕂, i ∈ F and x, y ∈ H. If either

(3.1)  Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}⟩ ≥ 0,

Re ⟨∑_{i∈F} Γ_{i}e_{i} − y, y − ∑_{i∈F} γ_{i}e_{i}⟩ ≥ 0,

or, equivalently,

(3.2)  ∥x − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,

∥y − ∑_{i∈F} \frac{Γ_{i} + γ_{i}}{2} e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} ,

hold, then we have the inequality

(3.3)  0 ≤ |⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩|

≤ \frac{1}{4}M (Φ, ϕ, F) M (Γ, γ, F) (∑_{i∈F} |⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} (∑_{i∈F} |⟨y, e_{i}⟩|^{2})^{\frac{1}{2}} ,

where M (Φ, ϕ, F) is defined in (2.13).

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

Proof. Using Schwartz’s inequality in the inner product space (H, ⟨·, ·⟩) one has

(3.4) |⟨x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}⟩|^{2}

≤ ∥x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}∥^{2} ∥y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}∥^{2}

and since a simple calculation shows that

⟨x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}⟩ = ⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩

and

∥x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}∥^{2} ≤ ∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2}

for any x, y ∈ H, then by (3.4) and by the counterpart of Bessel’s inequality in Corollary 1, we have

|⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩|^{2} (3.5)

≤ (∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2}) (∥y∥^{2} − ∑_{i∈F} |⟨y, e_{i}⟩|^{2})

≤ \frac{1}{4}M^{2} (Φ, ϕ, F) ∑_{i∈F} |⟨x, e_{i}⟩|^{2} · \frac{1}{4}M^{2} (Γ, γ, F) ∑_{i∈F} |⟨y, e_{i}⟩|^{2} .

Taking the square root in (3.5), we deduce (3.3).

The fact that \frac{1}{4} is the best possible constant follows by Corollary 1 and we omit the details.

The following corollary for real inner product spaces holds.

Corollary 3. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in H, F a finite part of I, M_{i}, m_{i}, N_{i}, n_{i} ≥ 0, i ∈ F and x, y ∈ H such that ∑_{i∈F} M_{i}m_{i} > 0, ∑_{i∈F} N_{i}n_{i} > 0 and

(3.6) ⟨∑_{i∈F} M_{i}e_{i} − x, x − ∑_{i∈F} m_{i}e_{i}⟩ ≥ 0, ⟨∑_{i∈F} N_{i}e_{i} − y, y − ∑_{i∈F} n_{i}e_{i}⟩ ≥ 0.

Then we have the inequality

(3.7) 0 ≤ |⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨y, e_{i}⟩|^{2}

≤ \frac{1}{16} · \frac{∑_{i∈F} (M_{i} − m_{i})^{2} ∑_{i∈F} (N_{i} − n_{i})^{2} ∑_{i∈F} |⟨x, e_{i}⟩|^{2} ∑_{i∈F} |⟨y, e_{i}⟩|^{2}}{∑_{i∈F} M_{i}m_{i} ∑_{i∈F} N_{i}n_{i}} .

The constant \frac{1}{16} is best possible.

In the case where the family {e_{i}}_{i∈I} reduces to a single vector, we may deduce from Theorem 4 the following particular case first obtained in [3].

Corollary 4. Let e ∈ H, ∥e∥ = 1, ϕ, Φ, γ, Γ ∈ 𝕂 with Re (Φ\overline{ϕ}) , Re (Γ\overline{γ}) > 0 and x, y ∈ H such that either

(3.8)  Re ⟨Φe − x, x − ϕe⟩ ≥ 0, Re ⟨Γe − y, y − γe⟩ ≥ 0,

or, equivalently,

(3.9)  ∥x − \frac{ϕ + Φ}{2}e∥ ≤ \frac{1}{2} |Φ − ϕ| , ∥y − \frac{γ + Γ}{2}e∥ ≤ \frac{1}{2} |Γ − γ|

holds, then

(3.10)  0 ≤ |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| ≤ \frac{1}{4}M (Φ, ϕ) M (Γ, γ) |⟨x, e⟩ ⟨e, y⟩| ,

where

M (Φ, ϕ) := [\frac{(|Φ| − |ϕ|)^{2} + 4 [|ϕΦ| − Re (Φ\overline{ϕ})]}{Re (Φ\overline{ϕ})}]^{\frac{1}{2}} .

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

Remark 3. If H is real, e ∈ H, ∥e∥ = 1 and a, b, A, B ∈ ℝ are such that A > a > 0, B > b > 0 and

(3.11)  ∥x − \frac{a + A}{2}e∥ ≤ \frac{1}{2} (A − a) , ∥y − \frac{b + B}{2} e∥ ≤ \frac{1}{2} (B − b) ,

then

(3.12)  |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| ≤ \frac{1}{4} · \begin{matrix}(A − a) & (B − b) \\ √ & \overline{abAB} \end{matrix} |⟨x, e⟩ ⟨e, y⟩| .

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

If ⟨x, e⟩ , ⟨y, e⟩ ≠ 0, then the following equivalent form of (3.12) also holds

(3.13)  |\frac{⟨x, y⟩}{⟨x, e⟩ ⟨e, y⟩} − 1| ≤ \frac{1}{4} · \begin{matrix}(A − a) & (B − b) \\ √ & \overline{abAB} \end{matrix}.

4. Some Companion Inequalities

The following companion of the Grüss inequality also holds.

Theorem 5. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in H, F a finite part of I, ϕ_{i}, Φ_{i} ∈ 𝕂, (i ∈ F), x, y ∈ H and λ ∈ (0, 1) , such that either

(4.1)  Re ⟨∑_{i∈F} Φ_{i}e_{i} − (λx + (1 − λ) y) , λx + (1 − λ) y − ∑_{i∈F} ϕ_{i}e_{i}⟩ ≥ 0

or, equivalently,

(4.2)  ∥λx + (1 − λ) y − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} · e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,

holds. Then we have the inequality

(4.3) Re [⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩]

≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} ∑_{i∈F} M^{2} (Φ, ϕ, F) ∑_{i∈F} |⟨λx + (1 − λ) y, e_{i}⟩|^{2} .

The constant \frac{1}{16} is the best possible constant in (4.3) in the sense that it cannot be replaced by a smaller constant.

Proof. Using the known inequality

Re ⟨z, u⟩ ≤ \frac{1}{4} ∥z + u∥^{2}

we may state that for any a, b ∈ H and λ ∈ (0, 1)

(4.4)  Re ⟨a, b⟩ ≤ \frac{1}{4λ (1 − λ)} ∥λa + (1 − λ) b∥^{2} .

Since

⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩ = ⟨x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}⟩ ,

for any x, y ∈ H, then, by (4.4), we get

Re [⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩] (4.5)

= Re [⟨x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}⟩]

≤ \frac{1}{4λ (1 − λ)} ∥λ (x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}) + (1 − λ) (y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i})∥^{2}

= \frac{1}{4λ (1 − λ)} ∥λx + (1 − λ) y − ∑_{i∈F} ⟨λx + (1 − λ) y, e_{i}⟩ e_{i}∥^{2}

= \frac{1}{4λ (1 − λ)} [∥λx + (1 − λ) y∥^{2} − ∑_{i∈F} |⟨λx + (1 − λ) y, e_{i}⟩|^{2}] .

If we apply the counterpart of Bessel’s inequality from Corollary 1 for λx+(1 − λ) y, we may state that

(4.6) ∥λx + (1 − λ) y∥^{2} − ∑_{i∈F} |⟨λx + (1 − λ) y, e_{i}⟩|^{2}

≤ \frac{1}{4}M^{2} (Φ, ϕ, F) ∑_{i∈F} |⟨λx + (1 − λ) y, e_{i}⟩|^{2} .

Now, by making use of (4.5) and (4.6), we deduce (4.3).

The fact that \frac{1}{16} is the best possible constant in (4.3) follows by the fact that if in (4.1) we choose x = y, then it becomes (i) of Theorem 3, implying for λ = \frac{1}{2} (2.12), for which, we have shown that \frac{1}{4} was the best constant.

Remark 4. If in Theorem 5, we choose λ = \frac{1}{2}, then we get

(4.7)  Re [⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩] ≤ \frac{1}{4}M^{2} (Φ, ϕ, F) ∑_{i∈F} |⟨\frac{x + y}{2} , e_{i}⟩|^{2} ,

provided

Re ⟨∑_{i∈F} Φ_{i}e_{i} − \frac{x + y}{2} , \frac{x + y}{2} − ∑_{i∈F} ϕ_{i}e_{i}⟩ ≥ 0

or, equivalently,

(4.8)  ∥\frac{x + y}{2} − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} · e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} .

5. Integral Inequalities

Let (Ω, Σ, μ) be a measure space consisting of a set Ω, a σ−algebra of parts Σ and a countably additive and positive measure μ on Σ with values in ℝ ∪ {∞} . Let ρ ≥ 0 be a μ−measurable function on Ω. Denote by L_{ρ}^{2} (Ω, 𝕂) the Hilbert space of all real or complex valued functions defined on Ω and 2 − ρ−integrable on Ω, i.e.,

(5.1)  ∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) < ∞.

Consider the family {f_{i}}_{i∈I} of functions in L_{ρ}^{2} (Ω, 𝕂) with the properties that

(5.2)  ∫_{Ω} ρ (s) f_{i} (s) \overline{f_{j}} (s) dμ (s) = δ_{ij}, i, j ∈ I,

where δ_{ij} is 0 if i ≠ j and δ_{ij} = 1 if i = j. {f_{i}}_{i∈I} is an orthornormal family in L_{ρ}^{2} (Ω, 𝕂) .

The following proposition holds.

Proposition 1. Let {f_{i}}_{i∈I} be an orthornormal family of functions in L_{ρ}^{2} (Ω, 𝕂) , F a finite subset of I, ϕ_{i}, Φ_{i} ∈ 𝕂 (i ∈ F) such that ∑_{i∈F} Re (Φ_{i}\overline{ϕ_{i}}) > 0 and f ∈ L_{ρ}^{2} (Ω, 𝕂) , so that either

(5.3) ∫_{Ω} ρ (s) Re [(∑_{i∈F} Φ_{i}f_{i} (s) − f (s)) (\overline{f} (s) − ∑_{i∈F} \overline{ϕ_{i}} \overline{f_{i}} (s))] dμ (s) ≥ 0

or, equivalently,

(5.4)  ∫_{Ω} ρ (s) |f (s) − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} f_{i} (s)|^{2} dμ (s) ≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

Then we have the inequality

(5.5)  (∫_{Ω} ρ (s) |f (s)|^{2} dμ (s))^{\frac{1}{2}} ≤ \frac{1}{2} · \frac{1}{[∑_{i∈F} Re (Φ_{i}\overline{ϕ_{i}})]^{\frac{1}{2}}}

× \{ \begin{matrix}max_{i∈F} & \{|Φ_{i}| + |ϕ_{i}|\} & ∑_{i∈F} & |∫ _{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s)| \\ [∑_{i∈F} (|Φ_{i}| + |ϕ_{i}|)^{p}]^{\frac{1}{p}} (∑_{i∈F} |∫ _{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s)|^{q})^{\frac{1}{q}} , \\ for p > 1, \frac{1}{p} + \frac{1}{q} = 1 \\ max_{i∈F} |∫ _{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s)| ∑_{i∈F} [|Φ_{i}| + |ϕ_{i}|] . \end{matrix}

In particular, we have

(5.6) ∫_{Ω} ρ (s) |f (s)|^{2} dμ (s)

≤ \frac{1}{4} · \begin{matrix}∑ & i∈F & (|Φ_{i}| + |ϕ_{i}|) & 2 \\ ∑ & i∈F & Re & (Φ_{i}\overline{ϕ_{i}}) \end{matrix} ∑_{i∈F} |∫_{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s)|^{2} .

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

The proof is obvious by Theorem 3 and Remark 1. We omit the details.

The following proposition also holds. Proposition 2. Assume that f_{i}, f, ϕ_{i}, Φ_{i} and F satisfy the assumptions of Proposition 1. Then we have the following counterpart of Bessel’s inequality:

(5.7)  0 ≤ ∫_{Ω} ρ (s) f^{2} (s) dμ (s) − ∑_{i∈F} |∫_{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s)|^{2}

≤ \frac{1}{4}M^{2} (Φ, ϕ, F) · ∑_{i∈F} |∫_{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s)|^{2} ,

where, as above,

(5.8)  M (Φ, ϕ, F) := [\begin{matrix}∑_{i∈F} & \{(|Φ_{i}| − |ϕ_{i}|)^{2} + 4 [|ϕ_{i}Φ_{i}| − Re (Φ_{i}\overline{ϕ_{i}})]\} \\ Re & (Φ_{i}\overline{ϕ_{i}}) \end{matrix}]^{\frac{1}{2}} .

The constant \frac{1}{4} is the best possible.

The following Grüss type inequality also holds. Proposition 3. Let {f_{i}}_{i∈I} and F be as in Proposition 1. If ϕ_{i}, Φ_{i}, γ_{i}, Γ_{i} ∈ 𝕂 (i ∈ F) and f, g ∈ L_{ρ}^{2} (Ω, 𝕂) so that either

(5.9) ∫_{Ω} ρ (s) Re [(∑_{i∈F} Φ_{i}f_{i} (s) − f (s)) (\overline{f} (s) − ∑_{i∈F} \overline{ϕ_{i}} \overline{f_{i}} (s))] dμ (s) ≥ 0,
∫_{Ω} ρ (s) Re [(∑_{i∈F} Γ_{i}f_{i} (s) − g (s)) (\overline{g} (s) − ∑_{i∈F} \overline{γ_{i}} \overline{f_{i}} (s))] dμ (s) ≥ 0,

or, equivalently,

(5.10)  ∫_{Ω} ρ (s) |f (s) − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} · f_{i} (s)|^{2} dμ (s) ≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} ,

∫_{Ω} ρ (s) |g (s) − ∑_{i∈F} \frac{Γ_{i} + γ_{i}}{2} · f_{i} (s)|^{2} dμ (s) ≤ \frac{1}{4} ∑_{i∈F} |Γ_{i} − γ_{i}|^{2} ,

then we have the inequality

(5.11) |∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s)

− ∑_{i∈F} ∫_{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s) ∫_{Ω} ρ (s) f_{i} (s) \overline{g (s)}dμ (s)|

≤ \frac{1}{4}M (Φ, ϕ, F) M (Γ, γ, F) (∑_{i∈F} |∫_{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s)|^{2})^{\frac{1}{2}}

× (∑_{i∈F} |ρ (s) f_{i} (s) \overline{g (s)}dμ (s)|^{2})^{\frac{1}{2}} ,

where M (Φ, ϕ, F) is defined in (5.8).

The constant \frac{1}{4} is the best possible.

The proof follows by Theorem 4 and we omit the details.

In the case of real spaces, the following corollaries provide much simpler sufficient conditions for the counterpart of Bessel’s inequality (5.7) or for the Grüss type inequality (5.11) to hold.

Corollary 5. Let {f_{i}}_{i∈I} be an orthornormal family of functions in the real Hilbert space L_{ρ}^{2} (Ω) , F a finite part of I, M_{i}, m_{i} ≥ 0 (i ∈ F) , with ∑_{i∈F} M_{i}m_{i} > 0 and f ∈ L_{ρ}^{2} (Ω) so that

(5.12)  ∑_{i∈F} m_{i}f_{i} (s) ≤ f (s) ≤ ∑_{i∈F} M_{i}f_{i} (s) for μ − a.e. s ∈ Ω.

Then we have the inequality

(5.13)  0 ≤ ∫_{Ω} ρ (s) f^{2} (s) dμ (s) − ∑_{i∈F} [∫_{Ω} ρ (s) f (s) f_{i} (s) dμ (s)]^{2}

≤ \frac{1}{4} · \frac{∑_{i∈F} (M_{i} − m_{i})^{2}}{∑_{i∈F} M_{i}m_{i}} · ∑_{i∈F} [∫_{Ω} ρ (s) f (s) f_{i} (s) dμ (s)]^{2} .

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

Corollary 6. Let {f_{i}}_{i∈I} and F be as above. If M_{i}, m_{i}, N_{i}, n_{i} ≥ 0 (i ∈ F) with ∑_{i∈F} M_{i}m_{i}, ∑_{i∈F} N_{i}n_{i} > 0 and f, g ∈ L_{ρ}^{2} (Ω) such that

(5.14)  ∑_{i∈F} m_{i}f_{i} (s) ≤ f (s) ≤ ∑_{i∈F} M_{i}f_{i} (s)

and

∑_{i∈F} n_{i}f_{i} (s) ≤ g (s) ≤ ∑_{i∈F} N_{i}f_{i} (s) for μ − a.e. s ∈ Ω,

then we have the inequality

|∫_{Ω} ρ (s) f (s) g (s) dμ (s)

− ∑_{i∈F} ∫_{Ω} ρ (s) f (s) f_{i} (s) dμ (s) ∫_{Ω} ρ (s) g (s) f_{i} (s) dμ (s)|

≤ \frac{1}{4} (\frac{∑_{i∈F} (M_{i} − m_{i})^{2}}{∑_{i∈F} M_{i}m_{i}} )^{\frac{1}{2}} (\frac{∑_{i∈F} (N_{i} − n_{i})^{2}}{∑_{i∈F} N_{i}n_{i}} )^{\frac{1}{2}}

× [∑_{i∈F} (∫_{Ω} ρ (s) f (s) f_{i} (s) dμ (s))^{2} ∑_{i∈F} (∫_{Ω} ρ (s) g (s) f_{i} (s) dμ (s))^{2}]^{\frac{1}{2}} .

References

[1] S.S. DRAGOMIR, A counterpart of Bessel’s inequality in inner product spaces and some Grüss type related results, RGMIA Res. Rep. Coll., 6(2003), Supplement, Article 10. [ON

Line:  http://rgmia.vu.edu.au/v6(E).html]

[2] S.S. DRAGOMIR, On Bessel and Grüss inequalities for orthornormal families in inner product spaces, RGMIA Res. Rep. Coll., 6(2003), Supplement, [ON Line:

http://rgmia.vu.edu.au/v6(E).html]

[3] S.S. DRAGOMIR, Some Grüss’ type inequalities in inner product spaces, J. Ineq. Pure & Appl. Math., to appear. [ONLINE: http://jipam.vu.edu.au]

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

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