A COUNTERPART OF BESSEL’S INEQUALITY IN INNER PRODUCT SPACES AND SOME GRÜSS TYPE RELATED RESULTS

S.S. DRAGOMIR

Abstract. A counterpart of the famous Bessel’s inequality for orthornormal families in real or complex inner product spaces is given. Applications for some Grüss type inequalities are also provided.

1. Introduction

In [1], the author has proved the following Grüss type inequality in real or complex inner product spaces.

Theorem 1. Let (H, ⟨·, ·⟩) be an inner product space over 𝕂 (𝕂 = ℝ, ℂ) and e ∈ H, ∥e∥ = 1. If ϕ, Φ, γ, Γ are real or complex numbers and x, y are vectors in H such that the conditions

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

hold, then we have the inequality

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

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

In [2], the following refinement of (1.2) has been pointed out.

Theorem 2. Let H, 𝕂 and e be as in Theorem 1. If ϕ, Φ, γ, Γ, x, y satisfy (1.1) or, equivalently (see [2, Lemma 1])

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

then

(1.4) |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩|

≤ \frac{1}{4} |Φ − ϕ| |Γ − γ| − [Re ⟨Φe − x, x − ϕe⟩]^{\frac{1}{2}} [Re ⟨Γe − y, y − γe⟩]^{\frac{1}{2}} .

In [3], N. Ujević has generalised Theorem 1 for the case of real inner product spaces as follows.

Date: May 16, 2003.

2000 Mathematics Subject Classification. 26D15, 46C05.

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

Theorem 3. Let (H, ⟨·, ·⟩) be an inner product space over the real numbers field ℝ, and {e_{i}}_{i∈{1,...,n}} an orthornormal family in H, i.e., we recall that ⟨e_{i}, e_{j}⟩ = 0 if i ≠ j and ∥e_{i}∥ = 1, i, j ∈ {1, …, n} . If ϕ_{i}, γ_{i}, Φ_{i}, Γ_{i} ∈ ℝ, i ∈ {1, …, n} satisfy the condition

(1.5) ⟨∑_{i=1}^{n} Φ_{i}e_{i} − x, x − ∑_{i=1}^{n} ϕ_{i}e_{i}⟩ ≥ 0,  ⟨∑_{i=1}^{n} Γ_{i}e_{i} − y, y − ∑_{i=1}^{n} γ_{i}e_{i}⟩ ≥ 0,

then one has the inequality:

(1.6)  |⟨x, y⟩ − ∑_{i=1}^{n} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩| ≤ \frac{1}{4} [∑_{i=1}^{n} (Φ_{i} − ϕ_{i})^{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.

We note that the key point in his proof is the following identity:

(1.7) ∑_{i=1}^{n} (⟨x, e_{i}⟩ − ϕ_{i}) (Φ_{i} − ⟨x, e_{i}⟩) − ⟨x − ∑_{i=1}^{n} ϕ_{i}e_{i}, ∑_{i=1}^{n} Φ_{i}e_{i} − x⟩

= ∥x∥^{2} − ∑_{i=1}^{n} ⟨x, e_{i}⟩^{2} ,

holding for x ∈ H, ϕ_{i}, Φ_{i} ∈ ℝ, i ∈ {1, …, n} and {e_{i}}_{i∈{1,...,n}} an orthornormal family of vectors in the real inner product space H.

In this paper we point out a counterpart of Bessel’s inequality in both real and complex inner product spaces. This result will then be employed to provide a refinement of the Grüss type inequality (1.6) for real or complex inner products. Related results as well as integral inequalities for general measure spaces are also given.

2. A Counterpart of Bessel’s Inequality

Let (H, ⟨·, ·⟩) be an inner product over the real or complex number field 𝕂 and {e_{i}}_{i∈I} a finite or infinite family of orthornormal vectors in H, i.e.,

(2.1)  ⟨e_{i}, e_{j}⟩ = \{ \begin{matrix}0 & if i ≠ j \\ 1 & if i = j \end{matrix} , i, j ∈ I.

It is well known that, the following inequality due to Bessel holds

(2.2)  ∑_{i∈I} |⟨x, e_{i}⟩|^{2} ≤ ∥x∥^{2} for any x ∈ H,

where the meaning of the sum is:

(2.3)  ∑_{i∈I} |⟨x, e_{i}⟩|^{2} := sup_{F ⊂I} {∑_{i∈F} |⟨x, e_{i}⟩|^{2} , F is a finite part of I} .

The following lemma holds.

Lemma 1. 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. The following statements are equivalent for x ∈ H

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

Proof. It is easy to see that for y, a, A ∈ H, the following are equivalent (see [2, Lemma 1])

(a) Re ⟨A − y, y − a⟩ ≥ 0 and (aa) ∥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.

The following counterpart of Bessel’s inequality holds.

Theorem 4. Let {e_{i}}_{i∈I} , F, ϕ_{i}, Φ_{i}, i ∈ F and x ∈ H so that either (i) or (ii) of Lemma 1 holds. Then we have the inequality:

(2.4)  0 ≤ ∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2}

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} − Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}⟩

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

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

Proof. Define

I_{1} := ∑_{i∈H} Re [(Φ_{i} − ⟨x, e_{i}⟩) (\overline{⟨x, e_{i}⟩} − \overline{ϕ_{i}})]

and

I_{2} := Re [⟨∑_{i∈H} Φ_{i}e_{i} − x, x − ∑_{i∈H} ϕ_{i}e_{i}⟩] .

Observe that

I_{1} = ∑_{i∈H} Re [Φ_{i}\overline{⟨x, e_{i}⟩}] + ∑_{i∈H} Re [\overline{ϕ_{i}} ⟨x, e_{i}⟩] − ∑_{i∈H} Re [Φ_{i}\overline{ϕ_{i}}] − ∑_{i∈H} |⟨x, e_{i}⟩|^{2}

and

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

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

Consequently, subtracting I_{2} from I_{1}, we deduce the following equality that is useful in its turn

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

− Re [⟨∑_{i∈H} Φ_{i}e_{i} − x, x − ∑_{i∈H} ϕ_{i}e_{i}⟩] .

Using the following elementary inequality for complex numbers

Re [a\overline{b}] ≤ \frac{1}{4} |a + b|^{2} , a, b ∈ 𝕂,

for the choices a = Φ_{i} − ⟨x, e_{i}⟩ , b = ⟨x, e_{i}⟩ − ϕ_{i} (i ∈ F) , we deduce

(2.6)  ∑_{i∈H} Re [(Φ_{i} − ⟨x, e_{i}⟩) (\overline{⟨x, e_{i}⟩} − \overline{ϕ_{i}})] ≤ \frac{1}{4} ∑_{i∈H} |Φ_{i} − ϕ_{i}|^{2} .

Making use of (2.5), (2.6) and the assumption (i), we deduce (2.4).

The sharpness of the constant \frac{1}{4} was proved for a single element e, ∥e∥ = 1 in [1], or for the real case in [3].

We can give here a simple proof as follows.

Assume that there is a c > 0 such that

(2.7)  0 ≤ ∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2}

≤ c ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} − Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}⟩ ,

provided ϕ_{i}, Φ_{i}, x and F satisfy (i) or (ii).

We choose F = \{1\} , e_{1} = e_{2} = (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) ∈ ℝ^{2}, x = (x_{1}, x_{2}) ∈ ℝ^{2}, Φ_{1} = Φ =

m > 0, ϕ_{1} = ϕ = −m, H = ℝ^{2} to get from (2.7) that

(2.8)  0 ≤ x_{1}^{2} + x_{2}^{2} − \frac{(x_{1} + x_{2})^{2}}{2}

≤ 4cm^{2} − (\frac{m}{\sqrt{2}} − x_{1}) (x_{1} + \frac{m}{\sqrt{2}}) − (\frac{m}{\sqrt{2}} − x_{2}) (x_{2} + \frac{m}{\sqrt{2}}) ,

provided

(2.9)  0 ≤ ⟨me − x, x + me⟩

= (\frac{m}{\sqrt{2}} − x_{1}) (x_{1} + \frac{m}{\sqrt{2}}) + (\frac{m}{\sqrt{2}} − x_{2}) (x_{2} + \frac{m}{\sqrt{2}}) .

If we choose x_{1} = \frac{m}{\sqrt{2}}, x_{2} = −\frac{m}{\sqrt{2}}, then (2.9) is fulfilled and by (2.8) we get m^{2} ≤ 4cm^{2}, giving c ≥ \frac{1}{4}.

3. A Refinement of the Grüss Inequality

The following result holds.

BESSEL’S INEQUALITY IN INNER PRODUCT SPACES 5

Theorem 5. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in H, F a finite part of I and 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 inequalities

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

≤ \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. 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 Theorem 4, 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} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} − Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}⟩]

× [\frac{1}{4} ∑_{i∈F} |Γ_{i} − γ_{i}|^{2} − Re ⟨∑_{i∈F} Γ_{i}e_{i} − y, y − ∑_{i∈F} γ_{i}e_{i}⟩]

≤ [\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}} ]

where, for the last inequality, we have made use of the inequality

(m^{2} − n^{2}) (p^{2} − q^{2}) ≤ (mp − nq)^{2} ,

where m, n, p, q > 0.

Taking the square root in (3.5) and observing that the quantity in the last square bracket is nonnegative (see for example (2.4)), we deduce the desired result (3.3).

The best constant has been proved in [1] for one element and we omit the details.

4. Some Companion Inequalities

The following companion of the Grüss inequality also holds.

Theorem 6. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in H, F a finite part of I and ϕ_{i}, Φ_{i} ∈ 𝕂, i ∈ F and x, y ∈ H such that

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

or, equivalently,

(4.2)  ∥\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}} ,

then we have the inequality

(4.3)  Re [⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩] ≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

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

Proof. Start with the well known inequality

(4.4)  Re ⟨z, u⟩ ≤ \frac{1}{4} ∥z + u∥^{2} , z, u ∈ H.

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

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

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

= ∥\frac{x + y}{2} − ∑_{i∈F} ⟨\frac{x + y}{2} , e_{i}⟩ e_{i}∥^{2}

= ∥\frac{x + y}{2} ∥^{2} − ∑_{i∈F} |⟨\frac{x + y}{2} , e_{i}⟩|^{2} .

If we apply the counterpart of Bessel’s inequality in Theorem 4 for \frac{x+y}{2} , we may state that

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

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

The fact that \frac{1}{4} is the best constant in (4.3) follows by the fact that if in (4.1) we choose x = y, then it becomes (i) of Lemma 1, implying (2.4), for which, we have shown that \frac{1}{4} was the best constant.

The following corollary may be of interest if we wish to evaluate the absolute value of

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

Corollary 1. With the assumptions of Theorem 6 and if

(4.7)  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}} ,

then we have the inequality

(4.9)  |Re [⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩]| ≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

Proof. We only remark that, if

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

holds, then by Theorem 6 for (−y) instead of y, we have

Re [− ⟨x, y⟩ + ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩] ≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} ,

showing that

(4.10)  Re [⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩] ≥ −\frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

Making use of (4.3) and (4.10), we deduce the desired inequality (4.9).

Remark 1. If H is a real inner product space and m_{i}, M_{i} ∈ ℝ with the property that

(4.11)  ⟨∑_{i∈F} M_{i}e_{i} − \frac{x ± y}{2} , \frac{x ± y}{2} − ∑_{i∈F} m_{i}e_{i}⟩ ≥ 0

or, equivalently,

(4.12)  ∥\frac{x ± y}{2} − ∑_{i∈F} \frac{M_{i} + m_{i}}{2} · e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} (M_{i} − m_{i})^{2})^{\frac{1}{2}} ,

then we have the Grüss type inequality

(4.13)  |⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩| ≤ \frac{1}{4} ∑_{i∈F} (M_{i} − m_{i})^{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) 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

BESSEL’S INEQUALITY IN INNER PRODUCT SPACES 9

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

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2}

− ∫_{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)

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

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

The proof follows by Theorem 4 applied for the Hilbert space L_{ρ}^{2} (Ω, 𝕂) and the orthornormal family {f_{i}}_{i∈I} .

The following Grüss type inequality also holds. Proposition 2. 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

(5.6) ∫_{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,

(5.7)  ∫_{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 inequalities

(5.8) |∫_{Ω} ρ (s) f (s) 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} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}}

− [∫_{Ω} ρ (s) Re [(∑_{i∈F} Φ_{i}f_{i} (s) − f (s)) (\overline{f} (s) − ∑_{i∈F} ϕ\overline{i}\overline{f_{i}} (s))] dμ (s)]^{\frac{1}{2}}

× [∫_{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)]^{\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 the best possible.

The proof follows by Theorem 5 and we omit the details. Remark 2. Similar results may be stated if we apply the inequalities in Section 4. 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 (5.5) or for the Grüss type inequality (5.8) to hold. Corollary 2. Let {f_{i}}_{i∈I} be an orthornormal family of functions in the real Hilbert space L_{ρ}^{2} (Ω, 𝕂) , F a finite part of I, M_{i}, m_{i} ∈ ℝ (i ∈ F) and f ∈ L_{ρ}^{2} (Ω) so that

(5.9)  ∑_{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 inequalities

(5.10) 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} ∑_{i∈F} (M_{i} − m_{i})^{2}

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

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

The constant \frac{1}{4} is best possible. Corollary 3. Let {f_{i}}_{i∈I} and F be as in Corollary 2. If M_{i}, m_{i}, N_{i}, n_{i} ∈ ℝ (i ∈ F) and f, g ∈ L_{ρ}^{2} (Ω) such that

(5.11)  ∑_{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 inequalities

|∫_{Ω} ρ (s) f (s) g (s) dμ (s) (5.12)

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

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

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

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

References

[1] S.S. DRAGOMIR, A generalisation of Grüss’ inequality in inner product spaces and applications, J. Math. Anal. Appl., 237 (1999), 74-82.

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

[3] N. UJEVIĆ, A generalisation of Grüss inequality in prehilbertian spaces, Math. Ineq. & Appl., to appear.

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