APPROXIMATING THE STIELTJES INTEGRAL VIA THE DARST-POLLARD INEQUALITY

S.S. DRAGOMIR AND A. SOFO

Abstract. An approximation of the Stieltjes integral of bounded integrals and continuous integrators via the Darst-Pollard inequality is given. Applications for the generalised trapezoid formula and the Ostrowski inequality for functions of bounded variation are also provided.

1. Introduction

In 1970, R. Darst and H. Pollard [2] obtained by elementary arguments the following interesting inequality for the Stieltjes integral.

Theorem 1 (Darst-Pollard, 1970). If h is real and of bounded variation on the interval [a, b] and g is real and continuous there, then

(1.1)  ∫_{a}^{b} h (t) dg (t) ≤ inf_{t∈[a,b]} h (t) [g (b) − g (a)] + S (g; a, b) · ⋁_{a}^{b} (h) ,

where ⋁_{a}^{b} (h) is the total variation of h on [a, b] and

(1.2)  S (g; a, b) := sup_{a≤α<β≤b} [g (β) − g (α)] .

In the recent paper [4], in order to approximate the Stieltjes integral ∫_{a}^{b} f (t) du (t) by the quadrature rule

\frac{m + M}{2} [u (b) − u (a)] ,

where m ≤ f (t) ≤ M for each t ∈ [a, b] , the first author defined the error functional

Δ (f, u, m, M; a, b) := ∫_{a}^{b} f (t) du (t) − \frac{m + M}{2} [u (b) − u (a)]

and showed that

(1.3) |Δ (f, u, m, M; a, b)|

≤ \{ \begin{matrix}\frac{1}{2} (M − m) ⋁_{a}^{b} (u) & if u is of bounded variation, \\ \frac{1}{2} (M − m) L (b − a) & if u is L − Lipschitzian, \\ ∫_{a}^{b} |f (t) − \frac{m+M}{2} | du (t) & if u is monotonic nondecreasing. \end{matrix}

Date: 20th February, 2007.

2000 Mathematics Subject Classification. 26D15, 41A55.

Key words and phrases. Stieltjes integral, Darst-Pollard inequality, Generalised trapezoid rule, Ostrowski inequality.

The constant \frac{1}{2} is best possible in both inequalities. The last inequality in (1.3) is also sharp.

In the same paper [4], in order to approximate the Stieltjes integral ∫_{a}^{b} f (t) du (t) in terms of the generalised trapezoid type rule

[u (b) − \frac{n + N}{2} ] f (b) + [\frac{n + N}{2} − u (a)] f (a) ,

the first author also considered the error functional

∇ (f, u, n, N; a, b) := [u (b) − \frac{n + N}{2} ] f (b)+[\frac{n + N}{2} − u (a)] f (a)−∫_{a}^{b} f (t) du (t) ,

where −∞ < n ≤ u (t) ≤ N < ∞ for t ∈ [a, b] and showed that

(1.4) |∇ (f, u, n, N; a, b)|

≤ \{ \begin{matrix}\frac{1}{2} (N − n) ⋁_{a}^{b} (f) & if f is of bounded variation, \\ \frac{1}{2} (N − n) K (b − a) & if f is K − Lipschitzian, \\ ∫_{a}^{b} |u (t) − \frac{n+N}{2} | df (t) & if f is monotonic nondecreasing. \end{matrix}

The constant \frac{1}{2} is best possible in both inequalities above and the last one is sharp.

The main aim of the present paper is to provide other error estimates for the functionals Δ and ∇ defined above by utilising as a main tool the Darst-Pollard inequality stated in (1.1). Applications for the generalised trapezoid formula and the Ostrowski inequality for functions of bounded variation are also provided.

2. The Results

We can state the following result in estimating the error functional Δ (f, u, m, M; a, b) :

Theorem 2. Let u be continuous on [a, b] and f : [a, b] → ℝ of bounded variation such that

(2.1)  −∞ < m = inf_{t∈[a,b]} f (t) , sup_{t∈[a,b]} f (t) = M < ∞.

Then:

(2.2)  |Δ (f, u, m, M; a, b)| ≤ ⋁_{a}^{b} (f) · S (u; a, b) − \frac{1}{2} (M − m) [u (b) − u (a)] .

The inequality (2.2) is sharp.

Proof. Let us denote K (a, b) := ⋁_{a}^{b} (f) · S (u; a, b) .

If we apply the Darst-Pollard inequality (1.1) for h = f and g = u, we have

(2.3)  ∫_{a}^{b} f (t) du (t) ≤ m [u (b) − u (a)] + K (a, b) .

Now assume that h := M − f. Then h is of bounded variation on [a, b] ,

⋁_{a}^{b} (h) = ⋁_{a}^{b} (M − f) = ⋁_{a}^{b} (f)

and inf_{t∈[a,b]} h (t) = 0. On applying the Darst-Pollard inequality (1.1), we obtain

∫_{a}^{b} (M − f (t)) du (t) ≤ K (a, b)

which is clearly equivalent with

(2.4)  M [u (b) − u (a)] − K (a, b) ≤ ∫_{a}^{b} f (t) du (t) .

Now, subtracting the same quantity \frac{m+M}{2} [u (b) − u (a)] in both (2.3) and (2.4), we deduce

\frac{1}{2} (M − m) [u (b) − u (a)] − K (a, b) (2.5)

≤ ∫_{a}^{b} f (t) du (t) − \frac{m + M}{2} [u (b) − u (a)]

≤ −\frac{1}{2} (M − m) [u (b) − u (a)] + K (a, b)

which is clearly equivalent with the desired inequality (2.2).

For the sharpness of the inequality, let us assume that u (t) = t, t ∈ [a, b] . Then (1.2) becomes

(2.6)  |∫_{a}^{b} f (t) dt − \frac{m + M}{2} (b − a)| ≤ (b − a) [⋁_{a}^{b} (f) − \frac{1}{2} (M − m)] .

Now, if we choose the function f : [a, b] → ℝ,

f (t) = \{ \begin{matrix}0 & if t ∈ [a, b] \\ k & if t = b, k > 0, \end{matrix}

then we have m = 0, M = k, ∫_{a}^{b} f (t) dt = 0, ⋁_{a}^{b} (f) = k and in both sides of (2.6) we obtain the same quantity \frac{1}{2}k (b − a) .

The following particular cases are of interest.

Corollary 1. Let f : [a, b] → ℝ be as in Theorem 2. If u : [a, b] → ℝ is of r − H−Hölder type (L−Lipschitzian), i.e.,

(2.7)  |u (t) − u (s)| ≤ H |t − s|^{r} (≤ L |t − s|)

for any t, s ∈ [a, b] , where H > 0, r ∈ (0, 1) (L > 0) are given, then:

(2.8) |Δ (f, u, m, M; a, b)| ≤ H (b − a)^{r} ⋁_{a}^{b} (f) − \frac{1}{2} (M − m) [u (b) − u (a)]

(≤ L (b − a) ⋁_{a}^{b} (f) − \frac{1}{2} (M − m) [u (b) − u (a)]) .

The case for Lipschitzian function u provides a sharp inequality.

The proof is obvious from the above theorem since

S (u; a, b) ≤ H (b − a)^{r} (L (b − a))

and the sharpness of the inequality has been clearly proven for the function u (t) = t.

The case of absolutely continuous integrators u : [a, b] → ℝ is incorporated in the following corollary.

Corollary 2. Let f be as in Theorem 2. If u : [a, b] → ℝ is absolutely continuous on [a, b] , then

(2.9) |Δ (f, u, m, M; a, b)|

≤ \{ \begin{matrix}(b − a) ∥u′∥_{∞,[a,b]} ⋁_{a}^{b} (f) − \frac{1}{2} (M − m) [u (b) − u (a)] \\ if u′ ∈ L_{∞} [a, b] ; \\ (b − a)^{\frac{1}{q}} ∥u′∥_{p,[a,b]} ⋁_{a}^{b} (f) − \frac{1}{2} (M − m) [u (b) − u (a)] \\ if u′ ∈ L_{p} [a, b] , p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∥u′∥_{1,[a,b]} ⋁_{a}^{b} (f) − \frac{1}{2} (M − m) [u (b) − u (a)] , \end{matrix}

where ∥·∥_{p,[a,b]} are the usual Lebesgue norms, p ∈ [1, ∞] .

Proof. Since u is absolutely continuous, hence for a ≤ α < β ≤ b we have

u (β) − u (α) = ∫_{α}^{β} u′ (s) ds

≤ \{ \begin{matrix}(β − α) ∥u′∥_{∞,[a,b]} & if u′ ∈ L_{∞} [a, b] ; \\ (β − α)^{\frac{1}{q}} ∥u′∥_{p,[a,b]} & if u′ ∈ L_{p} [a, b] , p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∥u′∥_{1,[a,b]} ; \end{matrix}

which gives

S (u; a, b) ≤ \{ \begin{matrix}(b − a) ∥u′∥_{∞,[a,b]} & if u′ ∈ L_{∞} [a, b] ; \\ (b − a)^{\frac{1}{q}} ∥u′∥_{p,[a,b]} & if u′ ∈ L_{p} [a, b] , p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∥u′∥_{1,[a,b]} . \end{matrix}

These together with (2.2) produces (2.9).

The case of monotonic integrators is considered in the following.

Corollary 3. Let f : [a, b] → ℝ be as in Theorem 2 and u : [a, b] → ℝ a continuous and monotonic function on [a, b] . Then

(2.10)  |Δ (f, u, m, M; a, b)| ≤ [u (b) − u (a)] [⋁_{a}^{b} (f) − \frac{1}{2} (M − m)]

(≤ \frac{1}{2} (M − m) [u (b) − u (a)]) .

The inequality is sharp.

The proof is obvious from the fact that for monotonic functions u : [a, b] → ℝ one can derive that S (u; a, b) = u (b) − u (a) .

The following lemma may be stated (see also [4]).

Lemma 1. Let u : [a, b] → ℝ and φ, ϕ ∈ ℝ with ϕ > φ. The following statements are equivalent:

(i) The function u−\frac{φ+Φ}{2}·e, where e (t) = t, t ∈ [a, b] is \frac{1}{2} (Φ − φ) −Lipschitzian; (ii) We have the inequality:

(2.11)  φ ≤ \frac{u (t) − u (s)}{t − s} ≤ Φ for each t, s ∈ [a, b] with t > s.

(iii) We have the inequality:

(2.12) φ (t − s) ≤ u (t) − u (s) ≤ Φ (t − s) for each t, s ∈ [a, b] with t > s.

Following [5], we can introduce the concept

Definition 1. The function u : [a, b] → ℝ which satisfies one of the equivalent conditions (i) – (iii) is said to be (φ, Φ) −Lipschitzian on [a, b] .

Notice that in [5], the definition was introduced on utilising the statement (iii) and only the equivalence (i) ⟺ (iii) was considered.

Utilising Lagrange’s mean value theorem, we can state the following result that provides practical examples of (φ, Φ) −Lipschitzian functions:

Proposition 1. Let u : [a, b] → ℝ be continuous on [a, b] and differentiable on [a, b] . If

−∞ < γ = inf_{t∈(a,b)} u′ (t) , sup_{t∈(a,b)} u′ (t) = Γ < ∞,

then u is (Γ, γ) −Lipschitzian on [a, b] .

Now the following corollary can be stated as well.

Corollary 4. If f : [a, b] → ℝ is as in Theorem 2 and u : [a, b] → ℝ is (φ, Φ) −Lipschitzian with φ > 0, then

(2.13)  |Δ (f, u, m, M; a, b)| ≤ [Φ · ⋁_{a}^{b} (f) − \frac{1}{2} (M − m) · φ] (b − a) .

The following result may be stated as well.

Theorem 3. Let f : [a, b] → ℝ be continuous on [a, b] and u : [a, b] → ℝ a function of bounded variation such that

−∞ < n = inf_{t∈[a,b]} u (t) , sup_{t∈[a,b]} u (t) = N < ∞.

Then we have the inequality:

(2.14)  |∇ (f, u, n, N; a, b)| ≤ ⋁_{a}^{b} (u) S (f; a, b) − \frac{1}{2} (N − n) [f (b) − f (a)] .

The inequality is sharp.

Proof. Follows from Theorem 2 on utilising the identity

f (b) [u (b) − \frac{n + N}{2} ] + f (a) [\frac{n + N}{2} − u (a)] − ∫_{a}^{b} f (t) du (t)

= ∫_{a}^{b} [u (t) − \frac{n + N}{2} ] df (t)

= ∫_{a}^{b} u (t) df (t) − \frac{n + N}{2} [f (b) − f (a)] .

The details are omitted.

Similar corollaries to Corollary 1 – Corollary 4 may be stated. We leave them to the interested reader.

3. Applications for the Trapezoid Rule

In this section we provide some applications in connection with the generalised trapezoid rule.

In [1], in order to approximate the integral ∫_{a}^{b} f (t) dt for the function f : [a, b] → ℝ of bounded variation with the generalised trapezoid rule

f (a) (x − a) + f (b) (b − x) , x ∈ [a, b]

the authors have considered the generalised trapezoid error functional

T (f; a, b, x) := ∫_{a}^{b} f (t) dt − [f (a) (x − a) + f (b) (b − x)] ,

and obtained the following sharp bound

(3.1)  |T (f; a, b, x)| ≤ [\frac{1}{2} (b − a) + |x − \frac{a + b}{2}|] ⋁_{a}^{b} (f)

for each x ∈ [a, b] .

The best inequality we can derive from (3.1) is the following trapezoid inequality for functions of bounded variation:

(3.2)  |∫_{a}^{b} f (t) dt − \frac{f (a) + f (b)}{2} · (b − a)| ≤ \frac{1}{2} (b − a) ⋁_{a}^{b} (f) ,

where the constant \frac{1}{2} is best possible in the sense that it cannot be replaced by a smaller constant.

The version of (3.2) for continuous functions is incorporated in

Proposition 2. Let f : [a, b] → ℝ be continuous on [a, b] . Then

(3.3) |\frac{f (a) + f (b)}{2} (b − a) − ∫_{a}^{b} f (t) dt|

≤ (b − a) \{S (f; a, b) − \frac{1}{2} [f (b) − f (a)]\} ,

where, as above

S (f; a, b) = sup_{a≤α<β≤b} [f (β) − f (α)] .

Proof. We use the identity [1]

(3.4)  f (b) (b − x) + f (a) (x − a) − ∫_{a}^{b} f (t) dt = ∫_{a}^{b} (t − x) df (t)

for any x ∈ [a, b] .

On applying Theorem 2 for the Stieltjes integral ∫_{a}^{b} (t − x) df (t) we can write that

(3.5) |∫_{a}^{b} (t − x) df (t) − \frac{b − x + a − x}{2} [f (b) − f (a)]|

≤ (b − a) S (f; a, b) − \frac{1}{2} (b − a) [f (b) − f (a)] .

Finally, on utilising the identity (3.4) and the inequality (3.5), a simple calculation provides the desired inequality (3.3).

The following result may be stated as well.

Proposition 3. Let f : [a, b] → ℝ be absolutely continuous and such that the derivative f′ is of bounded variation on [a, b] . If −∞ < γ = inf_{t∈[a,b]} f′ (t) , sup_{t∈[a,b]} f′ (t) = Γ < ∞, then

(3.6) |f (b) (b − x) + f (a) (x − a) + \frac{γ + Γ}{2} (b − a) (x − \frac{a + b}{2}) − ∫_{a}^{b} f (t) dt|

≤ (b − a) [⋁_{a}^{b} (f′) [\frac{1}{2} (b − a) + |x − \frac{a + b}{2}|] − \frac{1}{2} (Γ − γ) (\frac{a + b}{2} − x)] .

Proof. If f is absolutely continuous on [a, b] , then

f (b) (b − x) + f (a) (x − a) − ∫_{a}^{b} f (t) dt = ∫_{a}^{b} f′ (t) d [\frac{1}{2} (t − x)^{2}]

for any x ∈ [a, b] .

For fixed x ∈ [a, b] , let u (t) = \frac{1}{2} (t − x)^{2} . Then u is differentiable and u′ (t) = (t − x) . Also

sup_{t∈[a,b]} |u′ (t)| = sup_{t∈[a,b]} |t − x| = max {x − a, b − x}

= \frac{1}{2} (b − a) + |x − \frac{a + b}{2}| .

Now, if we apply Corollary 1 for the Lipschitzian case, we can write:

|∫_{a}^{b} f′ (t) d [\frac{1}{2} (t − x)^{2}] − \frac{γ + Γ}{2} [\frac{(b − x)^{2} − (a − x)^{2}}{2} ]|

≤ (b − a) [\frac{1}{2} (b − a) + |x − \frac{a + b}{2}|] ⋁_{a}^{b} (f′)

− \frac{1}{2} (Γ − γ) [\frac{(b − x)^{2} − (a − x)^{2}}{2} ] ,

which is equivalent with the desired inequality (3.6).

Corollary 5. With the assumptions in Proposition 3, we have the trapezoid inequality:

(3.7)  |\frac{f (a) + f (b)}{2} (b − a) − ∫_{a}^{b} f (t) dt| ≤ \frac{1}{2} (b − a)^{2} ⋁_{a}^{b} (f′) .

4. Applications for the Ostrowski Inequality

In 1938, A. Ostrowski proved the following inequality

(4.1)  |f (x) − \frac{1}{b − a} ∫_{a}^{b} f (t) dt| ≤ [\frac{1}{4} + (\frac{x − \frac{a+b}{2}}{b − a} )^{2}] (b − a) M

for all x ∈ [a, b], provided that f is differentiable on (a, b) and |f′ (t)| ≤ M for all t ∈ (a, b).

Using the following representation, which has been obtained by Montgomery in an equivalent form [6, p. 565]

(4.2)  f (x) − \frac{1}{b − a} ∫_{a}^{b} f (t) dt = \frac{1}{b − a} ∫_{a}^{b} p (t, x) f′ (t) dt

for all x ∈ [a, b], provided that f is absolutely continuous on [a, b] and

(4.3)  p (t, x) := \{ \begin{matrix}t − a & if & t ∈ [a, x] \\ t − b & if & t ∈ (x, b\} \end{matrix} , (x, t) ∈ [a, b]^{2} ,

we can put in place of M, i.e., in (4.1), the sup norm of f′, i.e., ∥f′∥_{∞} where

∥f′∥_{∞} := ess sup_{t∈[a,b]} |f′ (t)| ,

provided that f′ ∈ L_{∞} [a, b].

The following result related to the Ostrowski inequality can be stated.

Proposition 4. Let f : [a, b] → ℝ be a continuous function on [a, b] . Then

(4.4) |f (x) (b − a) − (x − \frac{a + b}{2}) [f (b) − f (a)] − ∫_{a}^{b} f (t) dt|

≤ (b − a) [S (f; a, b) − \frac{1}{2} [f (b) − f (a)]] ,

where, as above, S (f; a, b) = sup_{a≤α<β≤b} [f (β) − f (α)] .

Proof. We use the following Montgomery type identity [3]

∫_{a}^{b} p (t, x) df (t) = f (x) (b − a) − ∫_{a}^{b} f (t) dt

for any x ∈ [a, b] , where the kernel p : [a, b]^{2} → ℝ is defined by (4.3).

For any fixed x ∈ [a, b] , the function p (·, x) is of bounded variation, and

⋁_{a}^{b} p (·, x) = ⋁_{a}^{x} p (·, x) + ⋁_{x}^{b} p (·, x) = x − a + b − x = b − a.

Also

sup_{t∈[a,b]} p (t, x) = x − a  and  inf_{t∈[a,b]} p (t, x) = x − b,

for any x ∈ [a, b] .

Now, if we apply Theorem 2 for the Stieltjes integral ∫_{a}^{b} p (t, x) df (t), then we can write:

|∫_{a}^{b} p (t, x) df (t) − \frac{1}{2} (x − a + x − b) [f (b) − f (a)]|

≤ (b − a) S (f; a, b) − \frac{1}{2} (b − a) [f (b) − f (a)] ,

for any x ∈ [a, b] , which is equivalent with the desired result (4.4).

Corollary 6. With the assumptions in Proposition 4 we have the midpoint inequality:

(4.5) |f (\frac{a + b}{2}) (b − a) − ∫_{a}^{b} f (t) dt|

≤ (b − a) [S (f; a, b) − \frac{1}{2} [f (b) − f (a)]] .

Remark 1. The interested reader can apply Theorem 2 and Theorem 3 for other quadrature rules where the remainder can be written in the form of

∫_{a}^{b} K_{n} (t, x) df^{(n)} (t)

where the n-th derivative of the integrand f is continuous and the Peano kernel K_{n} (·, x) is of bounded variation on [a, b] . The details are omitted.

References

[1] P. CERONE, S.S. DRAGOMIR and C.E.M. PEARCE, A generalised trapezoid inequality for functions of bounded variation, Turkish J. Math., 24(2) (2000), 147-163.

[2] R. DARST and H. POLLARD, An inequality for the Riemann-Stieltjes integral, Proc. Amer. Math. Soc., 25(4) (1970), 912-913.

[3] S.S. DRAGOMIR, Ostrowski’s inequality for monotonous mappings and applications, J. KSIAM (Korea), 3(1) (1999), 127-135.

[4] S.S. DRAGOMIR, Approximating the Stieltjes integral of bounded functions and applications for three point quadrature rules, Preprint RGMIA Res. Rep. Coll., 10 (2007), Supplement [ONLINE http://rgmia.vu.edu.au/v10(E).html].

[5] Z. LIU, Refinement of an inequality of Grüss type for the Riemann-Stieltjes integral, Soochow J. Math., 30(4) (2004), 483-489.

[6] D.S. MITRINOVIĆ, J.E. PEČARIĆ and A.M. FINK, Inequalities for Functions and their Integrals and Derivatives, Kluwer Academic, Dordrecht, 1994.

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

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