SOME INEQUALITIES FOR THE CHEBYSHEV FUNCTIONAL

Sever S. Dragomir School of Communications & Informatics, Victoria University, PO Box 14428, MC 8001, Melbourne, Victoria, Australia

May 17, 2002

1. Introduction

For two real n-tuples \overline{a} = (a_{1}, ..., a_{n}) , \overline{b} = (b_{1}, ..., b_{n}) and \overline{p} = (p_{1}, ..., p_{n}) with p_{i} ≥ 0, i ∈ {1, ..., n} and ∑_{i=1}^{n} p_{i} = 1, consider the Chebyshev functional:

T (\overline{p}, \overline{a}, \overline{b}) := ∑_{i=1}^{n} p_{i}a_{i}b_{i} − ∑_{i=1}^{n} p_{i}a_{i} ∑_{i=1}^{n} p_{i}b_{i}.  (1)

By Korkine’s identity (obtained in 1882, [1], see also [2] or [3, p. 242]) one has the representation

T (\overline{p}, \overline{a}, \overline{b}) = \frac{1}{2} ∑_{i,j=1}^{n} p_{i}p_{j} (a_{j} − a_{i}) (b_{j} − b_{i})  (2)

= ∑_{1≤i<j≤n}^{n} p_{i}p_{j} (a_{j} − a_{i}) (b_{j} − b_{i}) .

By Sonin’s identity (obtained in 1898, see [4], or [3, p. 246]), we also have the representation

T (\overline{p}, \overline{a}, \overline{b}) = ∑_{i=1}^{n} [p_{i}a_{i} − A_{n} (\overline{p}, \overline{a})] [b_{i} − A_{n} (\overline{p}, \overline{b})]  (3)

where

A_{n} (\overline{p}, \overline{a}) := ∑_{i=1}^{n} p_{i}a_{i}.

We note that, the above identities may be proved by direct computation. The proof is left as an exercise to the reader.

Observe that if the n-tuples \overline{a}, \overline{b} are synchronous, i.e.,

(a_{j} − a_{i}) (b_{j} − b_{i}) ≥ 0 for any i, j ∈ {1, ..., n} ,

then obviously

T (\overline{p}, \overline{a}, \overline{b}) ≥ 0  (4)

which is well known in the litarature as Chebyshev’s inequality.

Another sufficient condition for the positivity of the Chebyshev functional that easily follows from the identity (3) is

[a_{i} − A_{n} (\overline{p}, \overline{a})] [b_{i} − A_{n} (\overline{p}, \overline{b})] ≥ 0 for any i ∈ \{1, ..., n\} .

The following result for the Chebyshev functional holds:

[T (\overline{p}, \overline{a}, \overline{b})]^{2} ≤ T (\overline{p}, \overline{a}, \overline{a}) T (\overline{p}, \overline{b}, \overline{b}) .  (5)

It is the main aim of this note to provide some reverse inequalities for (5) .

2. The Results

The following lemma that is of interest in itself holds. Lemma 1. Assume that \overline{a} = (a_{1}, ..., a_{n}) , \overline{b} = (b_{1}, ..., b_{n}) are real n-tuples so that for any i, j ∈ {1, ..., n} with i < j, one has

m (b_{j} − b_{i}) ≤ a_{j} − a_{i} ≤ M (b_{j} − b_{i}) ,  (6)

where m, M are given real numbers. If \overline{p} = (p_{1}, ..., p_{n}) with p_{i} ≥ 0, i ∈ \{1, ..., n\} and ∑_{i=1}^{n} p_{i} = 1, then one has the inequality

(m + M) T (\overline{p}, \overline{a}, \overline{b}) ≥ T (\overline{p}, \overline{a}, \overline{a}) + mMT (\overline{p}, \overline{b}, \overline{b}) .  (7)

Proof. Using the assumption (6) we obviously have

[M (b_{j} − b_{i}) − (a_{j} − a_{i})] [(a_{j} − a_{i}) − m (b_{j} − b_{i})] ≥ 0  (8)

for any i, j ∈ {1, ..., n} with i < j. This relation is equivalent, after multiplication, to

(a_{j} − a_{i})^{2} + mM (b_{j} − b_{i})^{2} ≤ (m + M) (a_{j} − a_{i}) (b_{j} − b_{i}) ,  (9)

for i, j ∈ {1, ..., n} with i < j. Multiplying (9) by p_{i}p_{j} ≥ 0, summing over i, j with 1 ≤ i < j ≤ n and using Korkine’s identity we deduce the required inequality (7) . Remark 2. A sufficient condition for (6) to hold is

mΔb_{l} ≤ Δa_{l} ≤ MΔb_{l}  (10)

for any l ∈ {1, ..., n − 1} , where Δb_{l} := b_{l+1} − b_{l} denotes the forward difference.

The following result provides a reverse inequality for (5) . Theorem 3. If \overline{p}, \overline{a}, \overline{b} are as in Lemma 1 and, in addition, we assume that M ≥ m > 0, then the following inequality

[T (\overline{p}, \overline{a}, \overline{b})]^{2} ≥ \frac{4mM}{(m + M)^{2}} T (\overline{p}, \overline{a}, \overline{a}) T (\overline{p}, \overline{b}, \overline{b}) .  (11)

holds. The constant 4 in (11) is best in the sense that it cannot be replaced by a larger constant.

Proof. We use the following elementary inequality

αx^{2} + \frac{1}{α}y^{2} ≥ 2xy, x, y ≥ 0, α > 0;

to get, for the choices

α = \sqrt{mM} > 0, x = [T (\overline{p}, \overline{b}, \overline{b})]^{1/2} ≥ 0,
and y = [T (\overline{p}, \overline{a}, \overline{a})]^{1/2} ≥ 0

the inequality

\sqrt{mM}T (\overline{p}, \overline{b}, \overline{b}) + \frac{1}{\sqrt{mM}} T (\overline{p}, \overline{a}, \overline{a})  (12)
≥ 2 [T (\overline{p}, \overline{a}, \overline{a})]^{1/2} [T (\overline{p}, \overline{b}, \overline{b})]^{1/2} .

Now, making use of (7) and (12) we deduce the desired inequality (11) .

To prove the sharpness of the constant in the sense mentioned above, assume that (11) holds with a constant C > 0, i.e.,

[T (\overline{p}, \overline{a}, \overline{b})]^{2} ≥ \frac{CmM}{(m + M)^{2}} T (\overline{p}, \overline{a}, \overline{a}) T (\overline{p}, \overline{b}, \overline{b}) .  (13)

Choose n = 2, and assume that b_{2} > b_{1}, a_{2} > a_{1}, p_{1} > 0, p_{2} > 0 and p_{1} + p_{2} = 1. Also suppose that m (b_{2} − b_{1}) ≤ a_{2} − a_{1} ≤ M (b_{2} − b_{1}) with M > m > 0.

Since, in this case, we obviously have

T (\overline{p}, \overline{a}, \overline{b}) = p_{1}p_{2} (a_{2} − a_{1}) (b_{2} − b_{1})
T (\overline{p}, \overline{a}, \overline{a}) = p_{1}p_{2} (a_{2} − a_{1})^{2}
T (\overline{p}, \overline{b}, \overline{b}) = p_{1}p_{2} (b_{2} − b_{1})^{2}

then by (11) we get

[p_{1}p_{2} (a_{2} − a_{1}) (b_{2} − b_{1})]^{2}
≥ \frac{CmM}{(m + M)^{2}} p_{1}p_{2} (a_{2} − a_{1})^{2} p_{1}p_{2} (b_{2} − b_{1})^{2}

giving

(m + M)^{2} ≥ CmM.

If in this last inequality we let m := 1 − ε, M := 1 + ε, ε ∈ (0, 1) , then we obtain 4 ≥ C (1 − ε^{2}) for any ε ∈ (0, 1) showing that C ≤ 4.

The following corollary is a natural consequence of the above theorem and provides different counterparts for the inequality (5) .

Corollary 4. With the assumptions in Theorem 3 we have

0 ≤ [T (\overline{p}, \overline{a}, \overline{a})]^{1/2} [T (\overline{p}, \overline{b}, \overline{b})]^{1/2} − T (\overline{p}, \overline{a}, \overline{b})

≤ \frac{(\sqrt{M} − \sqrt{m})^{2}}{2\sqrt{mM}} T (\overline{p}, \overline{a}, \overline{b})

and

0 ≤ T (\overline{p}, \overline{a}, \overline{a}) T (\overline{p}, \overline{b}, \overline{b}) − [T (\overline{p}, \overline{a}, \overline{b})]^{2}
≤ \frac{(M − m)^{2}}{4mM} [T (\overline{p}, \overline{a}, \overline{b})]^{2} .

The following result also holds Theorem 5. Let f, g : [α, β] → R be continuous on [α, β] and differentiable on (α, β) with g′ (x) ≠ 0 for x ∈ (α, β) . Assume also that

−∞ < γ := inf_{x∈(α,β)} \frac{f′ (x)}{g′ (x)}, sup_{x∈(α,β)} \frac{f′ (x)}{g′ (x)} =: Γ < ∞.  (14)

If \overline{x} := (x_{1}, ..., x_{n}) is a real n-tuple with x_{i} ∈ [α, β] and x_{i} ≠ x_{j} for i ≠ j and if we denote by f (\overline{x}) the n-tuple (f (x_{1}) , ..., f (x_{n})) , then we have the inequality

(γ + Γ) T (\overline{p}, f (\overline{x}) , g (\overline{x})) ≥ T (\overline{p}, f (\overline{x}) , f (\overline{x})) + γΓT (\overline{p}, g (\overline{x}) , g (\overline{x}))  (15)

for any \overline{p} = (p_{1}, ..., p_{n}) with p_{i} ≥ 0, i ∈ \{1, ..., n\} and ∑_{i=1}^{n} p_{i} = 1.

Proof. Applying Cauchy Mean-Value Theorem, for any i, j ∈ {1, ..., n} with i < j there exists ξ_{ij} ∈ (α, β) so that

\frac{f (x_{j}) − f (x_{i})}{g (x_{j}) − g (x_{i})} = \frac{f′ (ξ_{ij})}{g′ (ξ_{ij})} ∈ [γ, Γ] .

Then obviously

[\frac{f (x_{j}) − f (x_{i})}{g (x_{j}) − g (x_{i})} − γ] [Γ − \frac{f (x_{j}) − f (x_{i})}{g (x_{j}) − g (x_{i})}] ≥ 0

for any i, j ∈ {1, ..., n} with i < j. Using a similar argument to the one in Lemma 1 we deduce the desired result (15) .

The following corollary is a natural consequence of the above theorem.

Corollary 6. With the assumptions in Theorem 5 and if, in addition, Γ ≥ γ > 0, then one has the inequalities

[T (\overline{p}, f (\overline{x}) , g (\overline{x}))]^{2} ≥ \frac{4γΓ}{(γ + Γ)^{2}} T (\overline{p}, f (\overline{x}) , f (\overline{x})) T (\overline{p}, g (\overline{x}) , g (\overline{x})) ,

0 ≤ [T (\overline{p}, f (\overline{x}) , f (\overline{x}))]^{1/2} [T (\overline{p}, g (\overline{x}) , g (\overline{x}))]^{1/2} − T (\overline{p}, f (\overline{x}) , g (\overline{x}))

≤ \frac{(\sqrt{Γ} − \sqrt{γ})^{2}}{2\sqrt{Γγ}} T (\overline{p}, f (\overline{x}) , g (\overline{x})) ,

and

0 ≤ T (\overline{p}, f (\overline{x}) , f (\overline{x})) T (\overline{p}, g (\overline{x}) , g (\overline{x})) − [T (\overline{p}, f (\overline{x}) , g (\overline{x}))]^{2}
≤ \frac{(Γ − γ)^{2}}{4Γγ} [T (\overline{p}, f (\overline{x}) , g (\overline{x}))]^{2} .

Now, we point out other sufficient conditions for the inequality (7) to hold.

Lemma 7. Assume that \overline{a} = (a_{1}, ..., a_{n}) , \overline{b} = (b_{1}, ..., b_{n}) are real n-tuples, \overline{p} = (p_{1}, ..., p_{n}) with p_{i} ≥ 0, i ∈ \{1, ..., n\} and ∑_{i=1}^{n} p_{i} = 1 and b_{i} ≠ A_{n} (\overline{p}, \overline{b}) for each i ∈ \{1, ..., n\} . If

−∞ < l ≤ \frac{a_{i} − A_{n} (\overline{p}, \overline{a})}{b_{i} − A_{n} (\overline{p}, \overline{b})} ≤ L < ∞;  (16)

for any i ∈ {1, ..., n} , where l, L are given real numbers, then one has the inequality

(l + L) T (\overline{p}, \overline{a}, \overline{b}) ≥ T (\overline{p}, \overline{a}, \overline{a}) + lLT (\overline{p}, \overline{b}, \overline{b}) .  (17)

Proof. Using (16) we obviously have

[L − \frac{a_{i} − A_{n} (\overline{p}, \overline{a})}{b_{i} − A_{n} (\overline{p}, \overline{b})}] [\frac{a_{i} − A_{n} (\overline{p}, \overline{a})}{b_{i} − A_{n} (\overline{p}, \overline{b})} − l] ≥ 0  (18)

for each i ∈ {1, ..., n} .

If we multiply (18) by [b_{i} − A_{n} (\overline{p}, \overline{b})]^{2} ≥ 0, we get

[a_{i} − A_{n} (\overline{p}, \overline{a})]^{2} + lL [b_{i} − A_{n} (\overline{p}, \overline{b})]^{2}  (19)
≤ (l + L) [a_{i} − A_{n} (\overline{p}, \overline{a})] [b_{i} − A_{n} (\overline{p}, \overline{b})]

for each i ∈ {1, ..., n} .

Finally, if we multiply (19) by p_{i} ≥ 0, sum over i from 1 to n and use Sonin’s identity, then we obtain the desired inequality (17) .

Using Lemma 7 and a similar argument to that in the Theorem 3 we may state the following result as well.

Theorem 8. With the assumptions of Lemma 7 and, in addition, if L ≥ l > 0, then one has the inequality

[T (\overline{p}, \overline{a}, \overline{b})]^{2} ≥ \frac{4lL}{(l + L)^{2}} T (\overline{p}, \overline{a}, \overline{a}) T (\overline{p}, \overline{b}, \overline{b}) .

The constant 4 cannot be replaced by a larger constant.

The following corollary also holds

Corollary 9. With the assumptions of Lemma 7 and, in addition, if L ≥ l > 0, then one has the inequalities

0 ≤ [T (\overline{p}, \overline{a}, \overline{a})]^{1/2} [T (\overline{p}, \overline{b}, \overline{b})]^{1/2} − T (\overline{p}, \overline{a}, \overline{b})

≤ \frac{(\sqrt{L} − \sqrt{l})^{2}}{2\sqrt{lL}} T (\overline{p}, \overline{a}, \overline{b})

and

0 ≤ T (\overline{p}, \overline{a}, \overline{a}) T (\overline{p}, \overline{b}, \overline{b}) − [T (\overline{p}, \overline{a}, \overline{b})]^{2}
≤ \frac{(L − l)^{2}}{4lL} [T (\overline{p}, \overline{a}, \overline{b})]^{2} .

References

[1] A.N. KORKINE, Ob’ odnom opredelennom’ integrale, (Russian), Math. Sbornik, 10(1882), 571-572.

[2] A.N. KORKINE, Sur un théorème de M. Tchebychef, C.R. Acad. Sci. Paris, 96(1883), 326-327.

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

[4] N.Ja. SONIN, O nekotoryh neravenstvah, otnosjscihsja k opredelennym integralam, Zap. Imp. Akad. Nauk. po Fiziko-Matem. Otd., 6(1898), 1-54.