QUADRATIC REVERSES OF THE CONTINUOUS TRIANGLE

INEQUALITY FOR BOCHNER INTEGRAL OF VECTOR-VALUED FUNCTIONS IN HILBERT SPACES

SEVER S. DRAGOMIR

Abstract. Some quadratic reverses of the continuous triangle inequality for Bochner integral of vector-valued functions in Hilbert spaces are given. Applications for complex-valued functions are provided as well.

1. Introduction

Let f : [a, b] → 𝕂, 𝕂 = ℂ or ℝ be a Lebesgue integrable function. The following inequality is the continuous version of the triangle inequality

(1.1)  |∫_{a}^{b} f (x) dx| ≤ ∫_{a}^{b} |f (x)| dx

and plays a fundamental role in Mathematical Analysis and its applications.

It seems, see [6, p. 492], that the first reverse inequality for (1.1) was obtained by J. Karamata in his book from 1949, [4]:

(1.2)  cos θ ∫_{a}^{b} |f (x)| dx ≤ |∫_{a}^{b} f (x) dx|

provided

|arg f (x)| ≤ θ, x ∈ [a, b] ,

where θ is a given angle in (0, \frac{π}{2} ) .

This integral inequality is the continuous version of a reverse inequality for the generalised triangle inequality

(1.3)  cos θ ∑_{i=1}^{n} |z_{i}| ≤ |∑_{i=1}^{n} z_{i}| ,

provided

a − θ ≤ arg (z_{i}) ≤ a + θ, for i ∈ {1, …, n} ,

where a ∈ ℝ and θ ∈ (0, \frac{π}{2} ) , which, as pointed out in [6, p. 492], was first discovered by M. Petrovich in 1917, [7], and, subsequently rediscovered by other authors, including J. Karamata [4, p. 300 – 301], H.S. Wilf [8], and in an equivalent form by M. Marden [5].

The first to consider the problem for sums in the more general case of Hilbert and Banach spaces, were J.B. Diaz and F.T. Metcalf [1].

Date: April 16, 2004.

2000 Mathematics Subject Classification. 46C05, 26D15, 26D10.

Key words and phrases. Triangle inequality, Reverse inequality, Hilbert spaces, Bochner integral.

In our previous work [2], we pointed out some continuous versions of Diaz and Metcalf results providing reverses of the generalised triangle inequality in Hilbert spaces.

We mention here some results from [2] which may be compared with the new ones obtained in Sections 2 and 3 below.

Theorem 1. If f ∈ L ([a, b] ; H) , the space of Bochner integral functions defined on [a, b] and with values in the Hilbert space H, and there exists a constant K ≥ 1 and a vector e ∈ H, ∥e∥ = 1 such that

(1.4)  ∥f (t)∥ ≤ K Re ⟨f (t) , e⟩ for a.e. t ∈ [a, b] ,

then we have the inequality:

(1.5)  ∫_{a}^{b} ∥f (t)∥ dt ≤ K ∥∫_{a}^{b} f (t) dt∥ .

The case of equality holds in (1.5) if and only if

(1.6)  ∫_{a}^{b} f (t) dt = \frac{1}{K} (∫_{a}^{b} ∥f (t)∥ dt) e.

As particular cases of interest that may be applied in practice, we note the following corollaries established in [2].

Corollary 1. Let e be a unit vector in the Hilbert space (H; ⟨·, ·⟩) , ρ ∈ (0, 1) and f ∈ L ([a, b] ; H) so that

(1.7)  ∥f (t) − e∥ ≤ ρ for a.e. t ∈ [a, b] .

Then we have the inequality

(1.8)  \sqrt{1 − ρ^{2}} ∫_{a}^{b} ∥f (t)∥ dt ≤ ∥∫_{a}^{b} f (t) dt∥ ,

with equality if and only if

(1.9)  ∫_{a}^{b} f (t) dt = \sqrt{1 − ρ^{2}} (∫_{a}^{b} ∥f (t)∥ dt) · e.

Corollary 2. Let e be a unit vector in H and M ≥ m > 0. If f ∈ L ([a, b] ; H) is such that

(1.10)  Re ⟨Me − f (t) , f (t) − me⟩ ≥ 0

or, equivalently,

(1.11)  ∥f (t) − \frac{M + m}{2}e∥ ≤ \frac{1}{2} (M − m)

for a.e. t ∈ [a, b] , then we have the inequality

(1.12)  \frac{2\sqrt{mM}}{M + m} ∫_{a}^{b} ∥f (t)∥ dt ≤ ∥∫_{a}^{b} f (t) dt∥ ,

or, equivalently

(1.13)  0 ≤ ∫_{a}^{b} ∥f (t)∥ dt − ∥∫_{a}^{b} f (t) dt∥ ≤ \frac{(\sqrt{M} − \sqrt{m})^{2}}{M + m} ∥∫_{a}^{b} f (t) dt∥ .

The equality holds in (1.12) (or in the second part of (1.13)) if and only if

∫_{a}^{b} f (t) dt = \frac{2\sqrt{mM}}{M + m} (∫_{a}^{b} ∥f (t)∥ dt) e.

The case of additive reverse inequalities for the continuous triangle inequality has been considered in [3].

We recall here the following general result.

Theorem 2. If f ∈ L ([a, b] ; H) is such that there exists a vector e ∈ H, ∥e∥ = 1 and k : [a, b] → [0, ∞) a Lebesgue integrable function such that

(1.14)  ∥f (t)∥ − Re ⟨f (t) , e⟩ ≤ k (t) for a.e. t ∈ [a, b] ,

then we have the inequality:

(1.15)  (0 ≤) ∫_{a}^{b} ∥f (t)∥ dt − ∥∫_{a}^{b} f (t) dt∥ ≤ ∫_{a}^{b} k (t) dt.

The equality holds in (1.15) if and only if

(1.16)  ∫_{a}^{b} ∥f (t)∥ dt ≥ ∫_{a}^{b} k (t) dt

and

(1.17)  ∫_{a}^{b} f (t) dt = (∫_{a}^{b} ∥f (t)∥ dt − ∫_{a}^{b} k (t) dt) e.

This general result has some particular cases of interest that may be easily applied [3].

Corollary 3. If f ∈ L ([a, b] ; H) is such that there exists a vector e ∈ H, ∥e∥ = 1 and ρ ∈ (0, 1) such that

(1.18)  ∥f (t) − e∥ ≤ ρ for a.e. t ∈ [a, b] ,

then

(1.19)  0 ≤ ∫_{a}^{b} ∥f (t)∥ dt − ∥∫_{a}^{b} f (t) dt∥

≤ \begin{matrix}ρ & 2 \\ \sqrt{1 − ρ^{2}} & (1 + \sqrt{1 − ρ^{2}}) \end{matrix} Re ⟨∫_{a}^{b} f (t) dt, e⟩ .

Corollary 4. If f ∈ L ([a, b] ; H) is such that there exists a vector e ∈ H, ∥e∥ = 1 and M ≥ m > 0 such that either

(1.20)  Re ⟨Me − f (t) , f (t) − me⟩ ≥ 0

or, equivalently,

(1.21)  ∥f (t) − \frac{M + m}{2}e∥ ≤ \frac{1}{2} (M − m)

for a.e. t ∈ [a, b] , then

(1.22)  0 ≤ ∫_{a}^{b} ∥f (t)∥ dt − ∥∫_{a}^{b} f (t) dt∥

≤ \frac{(\sqrt{M} − \sqrt{m})^{2}}{2\sqrt{mM}} Re ⟨∫_{a}^{b} f (t) dt, e⟩;

and finally,

Corollary 5. If f ∈ L ([a, b] ; H) and r ∈ L_{2} ([a, b] ; H) , e ∈ H, ∥e∥ = 1 are such that

(1.23)  ∥f (t) − e∥ ≤ r (t) for a.e. t ∈ [a, b] ,

then

(1.24)  (0 ≤) ∫_{a}^{b} ∥f (t)∥ dt − ∥∫_{a}^{b} f (t) dt∥ ≤ \frac{1}{2} ∫_{a}^{b} r^{2} (t) dt.

The main aim of this paper is to point out some quadratic reverses for the continuous triangle inequality, namely, upper bounds for the nonnegative difference

(∫_{a}^{b} ∥f (t)∥ dt)^{2} − ∥∫_{a}^{b} f (t) dt∥^{2}

under various assumptions on the functions f ∈ L ([a, b] ; H) .Some related results are also pointed out. Applications for complex-valued functions are provided as well.

2. Quadratic Reverses of the Triangle Inequality

The following lemma holds.

Lemma 1. Let f ∈ L ([a, b] ; H) be such that there exists a functions k : Δ ⊂ ℝ^{2} → ℝ, Δ := {(t, s) |a ≤ t ≤ s ≤ b} with the property that k ∈ L (Δ) and

(2.1)  (0 ≤) ∥f (t)∥ ∥f (s)∥ − Re ⟨f (t) , f (s)⟩ ≤ k (t, s) ,

for a.e. (t, s) ∈ Δ. Then we have the following quadratic reverse of the continuous triangle inequality:

(2.2)  (∫_{a}^{b} ∥f (t)∥ dt)^{2} ≤ ∥∫_{a}^{b} f (t) dt∥^{2} + 2 ∫∫_{Δ} k (t, s) dtds.

The case of equality holds in (2.2) if and only if it holds in (2.1) for a.e. (t, s) ∈ Δ.

Proof. We observe that the following identity holds

(∫_{a}^{b} ∥f (t)∥ dt)^{2} − ∥∫_{a}^{b} f (t) dt∥^{2} (2.3)

= ∫_{a}^{b} ∫_{a}^{b} ∥f (t)∥ ∥f (s)∥ dtds − ⟨∫_{a}^{b} f (t) dt, ∫_{a}^{b} f (s) ds⟩

= ∫_{a}^{b} ∫_{a}^{b} ∥f (t)∥ ∥f (s)∥ dtds − ∫_{a}^{b} ∫_{a}^{b} Re ⟨f (t) , f (s)⟩ dtds
= ∫_{a}^{b} ∫_{a}^{b} [∥f (t)∥ ∥f (s)∥ − Re ⟨f (t) , f (s)⟩] dtds := I.

Now, observe that for any (t, s) ∈ [a, b] × [a, b] , we have

∥f (t)∥ ∥f (s)∥ − Re ⟨f (t) , f (s)⟩ = ∥f (s)∥ ∥f (t)∥ − Re ⟨f (s) , f (t)⟩

and thus

(2.4)  I = 2 ∫∫_{Δ} [∥f (t)∥ ∥f (s)∥ − Re ⟨f (t) , f (s)⟩] dtds.

Using the assumption (2.1), we deduce

∫∫_{Δ} [∥f (t)∥ ∥f (s)∥ − Re ⟨f (t) , f (s)⟩] dtds ≤ ∫∫_{Δ} k (t, s) dtds,

and, by the identities (2.3) and (2.4), we deduce the desired inequality (2.2). The case of equality is obvious and we omit the details.

Remark 1. From (2.2) one may deduce a coarser inequality that can be useful in some applications. It is as follows:

(0 ≤) ∫_{a}^{b} ∥f (t)∥ dt − ∥∫_{a}^{b} f (t) dt∥ ≤ \sqrt{2} (∫∫_{Δ} k (t, s) dtds)^{\frac{1}{2}} .

Remark 2. If the condition (2.1) is replaced with the following refinement of the Schwarz inequality

(2.5)  (0 ≤) k (t, s) ≤ ∥f (t)∥ ∥f (s)∥ − Re ⟨f (t) , f (s)⟩

for a.e. (t, s) ∈ Δ, then the following refinement of the quadratic triangle inequality is valid

(2.6)  (∫_{a}^{b} ∥f (t)∥ dt)^{2} ≥ ∥∫_{a}^{b} f (t) dt∥^{2} + 2 ∫∫_{Δ} k (t, s) dtds

(≥ ∥∫_{a}^{b} f (t) dt∥^{2}) .

The equality holds in (2.6) iff the case of equality holds in (2.5) for a.e. (t, s) ∈ Δ.

The following result holds.

Theorem 3. Let f ∈ L ([a, b] ; H) be such that there exists M ≥ 1 ≥ m ≥ 0 such that either

(2.7)  Re ⟨Mf (s) − f (t) , f (t) − mf (s)⟩ ≥ 0 for a.e. (t, s) ∈ Δ,

or, equivalently,

(2.8)  ∥f (t) − \frac{M + m}{2}f (s)∥ ≤ \frac{1}{2} (M − m) ∥f (s)∥ for a.e. (t, s) ∈ Δ.

Then we have the inequality:

(2.9) (∫_{a}^{b} ∥f (t)∥ dt)^{2} ≤ ∥∫_{a}^{b} f (t) dt∥^{2} + \frac{1}{2} · \frac{(M − m)^{2}}{M + m} ∫_{a}^{b} (s − a) ∥f (s)∥^{2} ds.

The case of equality holds in (2.9) if and only if

(2.10)  ∥f (t)∥ ∥f (s)∥ − Re ⟨f (t) , f (s)⟩ = \frac{1}{4} · \frac{(M − m)^{2}}{M + m} ∥f (s)∥^{2}

for a.e. (t, s) ∈ Δ.

Proof. Firstly, observe that, in an inner product space (H; ⟨·, ·⟩) and for x, z, Z ∈ H, the following statements are equivalent

(i) Re ⟨Z − x, x − z⟩ ≥ 0 and

(ii) ∥x − \frac{Z+z}{2} ∥ ≤ \frac{1}{2} ∥Z − z∥ .

This shows that (2.7) and (2.8) are obviously equivalent.

Now, taking the square in (2.8), we get

∥f (t)∥^{2} + (\frac{M + m}{2})^{2} ∥f (s)∥^{2}

≤ 2 Re ⟨f (t) , \frac{M + m}{2}f (s)⟩ + \frac{1}{4} (M − m)^{2} ∥f (s)∥^{2} ,

for a.e. (t, s) ∈ Δ, and obviously, since

2 (\frac{M + m}{2}) ∥f (t)∥ ∥f (s)∥ ≤ ∥f (t)∥^{2} + (\frac{M + m}{2})^{2} ∥f (s)∥^{2} ,

we deduce that

2 (\frac{M + m}{2}) ∥f (t)∥ ∥f (s)∥

≤ 2 Re ⟨f (t) , \frac{M + m}{2}f (s)⟩ + \frac{1}{4} (M − m)^{2} ∥f (s)∥^{2} ,

giving the much simpler inequality:

(2.11)  ∥f (t)∥ ∥f (s)∥ − Re ⟨f (t) , f (s)⟩ ≤ \frac{1}{4} · \frac{(M − m)^{2}}{M + m} ∥f (s)∥^{2}

for a.e. (t, s) ∈ Δ.

Applying Lemma 1 for k (t, s) := \frac{1}{4} · \frac{(M−m)^{2}}{M+m} ∥f (s)∥^{2} , we deduce

(2.12)  (∫_{a}^{b} ∥f (t)∥ dt)^{2} ≤ ∥∫_{a}^{b} f (t) dt∥^{2} + \frac{1}{2} · \frac{(M − m)^{2}}{M + m} ∫∫_{Δ} ∥f (s)∥^{2} ds

with equality if and only if (2.11) holds for a.e. (t, s) ∈ Δ.

QUADRATIC REVERSES OF THE CONTINUOUS TRIANGLE INEQUALITY 7

Since

∫∫_{Δ} ∥f (s)∥^{2} ds = ∫_{a}^{b} (∫_{a}^{s} ∥f (s)∥^{2} dt) ds = ∫_{a}^{b} (s − a) ∥f (s)∥^{2} ds,

then by (2.12) we deduce the desired result (2.9).

Another result which is similar to the one above is incorporated in the following theorem. Theorem 4. With the assumptions of Theorem 3, we have

(2.13)  (∫_{a}^{b} ∥f (t)∥ dt)^{2} − ∥∫_{a}^{b} f (t) dt∥^{2} ≤ \frac{(\sqrt{M} − \sqrt{m})^{2}}{2\sqrt{Mm}} ∥∫_{a}^{b} f (t) dt∥^{2}

or, equivalently,

(2.14)  ∫_{a}^{b} ∥f (t)∥ dt ≤ (\frac{M + m}{2\sqrt{Mm}})^{\frac{1}{2}} ∥∫_{a}^{b} f (t) dt∥ .

The case of equality holds in (2.13) or (2.14) if and only if

(2.15)  ∥f (t)∥ ∥f (s)∥ = \frac{M + m}{2\sqrt{Mm}} Re ⟨f (t) , f (s)⟩ ,

for a.e. (t, s) ∈ Δ. Proof. From (2.7), we deduce

∥f (t)∥^{2} + Mm ∥f (s)∥^{2} ≤ (M + m) Re ⟨f (t) , f (s)⟩

for a.e. (t, s) ∈ Δ. Dividing by \sqrt{Mm} > 0, we deduce

\begin{matrix}∥f (t)∥ & 2 \\ √ & \overline{Mm} \end{matrix} + \sqrt{Mm} ∥f (s)∥^{2} ≤ \frac{M + m}{\sqrt{Mm}} Re ⟨f (t) , f (s)⟩

and, obviously, since

2 ∥f (t)∥ ∥f (s)∥ ≤ \begin{matrix}∥f (t)∥ & 2 \\ √ & \overline{Mm} \end{matrix} + \sqrt{Mm} ∥f (s)∥^{2} ,

hence

∥f (t)∥ ∥f (s)∥ ≤ \frac{M + m}{\sqrt{Mm}} Re ⟨f (t) , f (s)⟩

for a.e. (t, s) ∈ Δ, giving

∥f (t)∥ ∥f (s)∥ − Re ⟨f (t) , f (s)⟩ ≤ \frac{(\sqrt{M} − \sqrt{m})^{2}}{2\sqrt{Mm}} Re ⟨f (t) , f (s)⟩ .

Applying Lemma 1 for k (t, s) := ^{\begin{matrix}(\sqrt{M}−\sqrt{m}) & ^{2} \\ √ & \overline{Mm} \end{matrix}} Re ⟨f (t) , f (s)⟩ , we deduce

(2.16) (∫_{a}^{b} ∥f (t)∥ dt)^{2} ≤ ∥∫_{a}^{b} f (t) dt∥^{2} + \frac{(\sqrt{M} − \sqrt{m})^{2}}{2\sqrt{Mm}} Re ⟨f (t) , f (s)⟩ .

On the other hand, since Re ⟨f (t) , f (s)⟩ = Re ⟨f (s) , f (t)⟩ for any (t, s) ∈ [a, b]^{2} , hence

∫∫_{Δ} Re ⟨f (t) , f (s)⟩ dtds = \frac{1}{2} ∫_{a}^{b} ∫_{a}^{b} Re ⟨f (t) , f (s)⟩ dtds

= \frac{1}{2} Re ⟨∫_{a}^{b} f (t) dt, ∫_{a}^{b} f (s) ds⟩

= \frac{1}{2} ∥∫_{a}^{b} f (t) dt∥^{2}

and thus, from (2.16), we get (2.13).

The equivalence between (2.13) and (2.14) is obvious and we omit the details.

3. Related Results

The following result also holds.

Theorem 5. Let f ∈ L ([a, b] ; H) and γ, Γ ∈ ℝ be such that either

(3.1)  Re ⟨Γf (s) − f (t) , f (t) − γf (s)⟩ ≥ 0 for a.e. (t, s) ∈ Δ,

or, equivalently,

(3.2)  ∥f (t) − \frac{Γ + γ}{2} f (s)∥ ≤ \frac{1}{2} |Γ − γ| ∥f (s)∥ for a.e. (t, s) ∈ Δ.

Then we have the inequality:

(3.3)  ∫_{a}^{b} [(b − s) + γΓ (s − a)] ∥f (s)∥^{2} ds ≤ \frac{Γ + γ}{2} ∥∫_{a}^{b} f (s) ds∥^{2} .

The case of equality holds in (3.3) if and only if the case of equality holds in either (3.1) or (3.2) for a.e. (t, s) ∈ Δ.

Proof. The inequality (3.1) is obviously equivalent to

(3.4)  ∥f (t)∥^{2} + γΓ ∥f (s)∥^{2} ≤ (Γ + γ) Re ⟨f (t) , f (s)⟩

for a.e. (t, s) ∈ Δ.

Integrating (3.4) on Δ, we deduce

(3.5) ∫_{a}^{b} (∫_{a}^{s} ∥f (t)∥^{2} dt) ds + γΓ ∫_{a}^{b} (∥f (s)∥^{2} ∫_{a}^{s} dt) ds

= (Γ + γ) ∫_{a}^{b} (∫_{a}^{s} Re ⟨f (t) , f (s)⟩ dt) ds.

It is easy to see, on integrating by parts, that

∫_{a}^{b} (∫_{a}^{s} ∥f (t)∥^{2} dt) ds = s ∫_{a}^{s} ∥f (t)∥^{2} dt|_{a}^{b} − ∫_{a}^{b} s ∥f (s)∥^{2} ds

= b ∫_{a}^{s} ∥f (s)∥^{2} ds − ∫_{a}^{b} s ∥f (s)∥^{2} ds
= ∫_{a}^{b} (b − s) ∥f (s)∥^{2} ds

and

∫_{a}^{b} (∥f (s)∥^{2} ∫_{a}^{s} dt) ds = ∫_{a}^{b} (s − a) ∥f (s)∥^{2} ds.

Since

\frac{d}{ds} (∥∫_{a}^{b} f (t) dt∥^{2}) = \frac{d}{ds} ⟨∫_{a}^{s} f (t) dt, ∫_{a}^{s} f (t) dt⟩

= ⟨f (s) , ∫_{a}^{s} f (t) dt⟩ + ⟨∫_{a}^{s} f (t) dt, f (s)⟩
= 2 Re ⟨∫_{a}^{s} f (t) dt, f (s)⟩ ,

hence

∫_{a}^{b} (∫_{a}^{s} Re ⟨f (t) , f (s)⟩ dt) ds = ∫_{a}^{b} Re ⟨∫_{a}^{s} f (t) dt, f (s)⟩ ds

= \frac{1}{2} ∫_{a}^{b} \frac{d}{ds} (∥∫_{a}^{s} f (t) dt∥^{2}) ds

= \frac{1}{2} ∥∫_{a}^{s} f (t) dt∥^{2}|_{a}^{b}

= \frac{1}{2} ∥∫_{a}^{b} f (t) dt∥^{2} .

Utilising (3.5), we deduce the desired inequality (3.3).

The case of equality is obvious and we omit the details.

Remark 3. Consider the function φ (s) := (b − s) + γΓ (s − a) , s ∈ [a, b] . Obviously,

φ (s) = (Γγ − 1) s + b − γΓa.

Observe that, if Γγ ≥ 1, then

b − a = φ (a) ≤ φ (s) ≤ φ (b) = γΓ (b − a) , s ∈ [a, b]

and, if Γγ < 1, then

γΓ (b − a) ≤ φ (s) ≤ b − a, s ∈ [a, b] .

Taking into account the above remark, we may state the following corollary.

Corollary 6. Assume that f, γ, Γ are as in Theorem 5.

a) If Γγ ≥ 1, then we have the inequality

(b − a) ∫_{a}^{b} ∥f (s)∥^{2} ds ≤ \frac{Γ + γ}{2} ∥∫_{a}^{b} f (s) ds∥^{2} .

b) If 0 < Γγ < 1, then we have the inequality

γΓ (b − a) ∫_{a}^{b} ∥f (s)∥^{2} ds ≤ \frac{Γ + γ}{2} ∥∫_{a}^{b} f (s) ds∥^{2} .

4. Applications for Complex-Valued Functions

Let f : [a, b] → ℂ be a Lebesgue integrable function and M ≥ 1 ≥ m ≥ 0. The condition (2.7) from Theorem 3, which plays a fundamental role in the results obtained above, can be translated in this case as

(4.1)  Re [(Mf (s) − f (t)) (\overline{f (t)} − m\overline{f (s)})] ≥ 0

for a.e. a ≤ t ≤ s ≤ b.

Since, obviously

Re [(Mf (s) − f (t)) (\overline{f (t)} − m\overline{f (s)})]
= [(M Re f (s) − Re f (t)) (Re f (t) − m Re f (s))]
+ [(M Im f (s) − Im f (t)) (Im f (t) − m Im f (s))]

hence a sufficient condition for the inequality in (4.1) to hold is

(4.2) m Re f (s) ≤ Re f (t) ≤ M Re f (s) and m Im f (s) ≤ Im f (t) ≤ M Im f (s)

for a.e. a ≤ t ≤ s ≤ b.

Utilising Theorems 3,4 and 5 we may state the following results incorporating quadratic reverses of the continuous triangle inequality:

Proposition 1. With the above assumptions for f, M and m, and if (4.1) holds true, then we have the inequalities

(∫_{a}^{b} |f (t)| dt)^{2} ≤ |∫_{a}^{b} f (t) dt|^{2} + \frac{1}{2} · \frac{(M − m)^{2}}{M + m} ∫_{a}^{b} (s − a) |f (s)|^{2} ds,

∫_{a}^{b} |f (t)| dt ≤ (\frac{M + m}{2\sqrt{Mm}})^{\frac{1}{2}} |∫_{a}^{b} f (t) dt| ,

and

∫_{a}^{b} [(b − s) + mM (s − a)] |f (s)|^{2} ds ≤ \frac{M + m}{2} |∫_{a}^{b} f (s) ds|^{2} .

Remark 4. One may wonder if there are functions satisfying the condition (4.2) above. It suffices to find examples of real functions φ : [a, b] → ℝ verifying the following double inequality

(4.3)  γφ (s) ≤ φ (t) ≤ Γφ (s)

for some given γ, Γ with 0 ≤ γ ≤ 1 ≤ Γ < ∞ for a.e. a ≤ t ≤ s ≤ b.

For this purpose, consider ψ : [a, b] → ℝ a differentiable function on (a, b), continuous on [a, b] and with the property that there exists Θ ≥ 0 ≥ θ such that

(4.4)  θ ≤ ψ′ (u) ≤ Θ for any u ∈ (a, b) .

By Lagrange’s mean value theorem, we have, for any a ≤ t ≤ s ≤ b

ψ (s) − ψ (t) = ψ′ (ξ) (s − t)

with t ≤ ξ ≤ s. Therefore, for a ≤ t ≤ s ≤ b, by (4.4), we have the inequality

θ (b − a) ≤ θ (s − t) ≤ ψ (s) − ψ (t) ≤ Θ (s − t) ≤ Θ (b − a) .

If we choose the function φ : [a, b] → ℝ given by

φ (t) := exp [−ψ (t)] , t ∈ [a, b] ,

and γ := exp [θ (b − a)] ≤ 1, Γ := exp [Θ (b − a)] , then (4.3) holds true for any a ≤ t ≤ s ≤ b.

References

[1] J.B. DIAZ and F.T. METCALF, A complementary triangle inequality in Hilbert and Banach spaces, Proceedings Amer. Math. Soc., 17(1) (1966), 88-97.

[2] S.S. DRAGOMIR, Reverses of the continuous triangle inequality for Bochner integral of vector valued function in Hilbert spaces. Preprint, RGMIA Res. Rep. Coll., to appear.

[3] S.S. DRAGOMIR, Additive reverses of the continuous triangle inequality for Bochner integral of vector valued functions in Hilbert spaces. Preprint, RGMIA Res. Rep. Coll., to appear.

[4] J. KARAMATA, Teorija i Praksa Stieltjesova Integrala (Serbo-Coratian) (Stieltjes Integral, Theory and Practice), SANU, Posebna izdanja, 154, Beograd, 1949.

[5] M. MARDEN, The Geometry of the Zeros of a Polynomial in a Complex Variable, Amer. Math. Soc. Math. Surveys, 3, New York, 1949.

[6] D.S. MITRINOVIĆ, J.E. PEČARIĆ and A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993.

[7] M. PETROVICH, Module d’une somme, L’ Ensignement Mathématique, 19 (1917), 53-56.

[8] H.S. WILF, Some applications of the inequality of arithmetic and geometric means to polynomial equations, Proceedings Amer. Math. Soc., 14 (1963), 263-265.

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

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