Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 3, Article 66, 2005
TIME-SCALE INTEGRAL INEQUALITIES
DOUGLAS R. ANDERSON
C ONCORDIA C OLLEGE
D EPARTMENT OF M ATHEMATICS AND C OMPUTER S CIENCE
M OORHEAD , MN 56562 USA
andersod@cord.edu
Received 16 March, 2005; accepted 02 June, 2005
Communicated by H. Gauchman
A BSTRACT. Some recent and classical integral inequalities are extended to the general timescale calculus, including the inequalities of Steffensen, Iyengar, Čebyšev, and Hermite-Hadamard.
Key words and phrases: Taylor’s Formula, Nabla integral, Delta integral.
2000 Mathematics Subject Classification. 34B10, 39A10.
1. P RELIMINARIES ON T IME S CALES
The unification and extension of continuous calculus, discrete calculus, q-calculus, and indeed arbitrary real-number calculus to time-scale calculus was first accomplished by Hilger in
his Ph.D. thesis [8]. Since then, time-scale calculus has made steady inroads in explaining the
interconnections that exist among the various calculi, and in extending our understanding to a
new, more general and overarching theory. The purpose of this work is to illustrate this new
understanding by extending some continuous and q-calculus inequalities and some of their applications, such as those by Steffensen, Hermite-Hadamard, Iyengar, and Čebyšev, to arbitrary
time scales.
The following definitions will serve as a short primer on the time-scale calculus; they can
be found in Agarwal and Bohner [1], Atici and Guseinov [3], and Bohner and Peterson [4]. A
time scale T is any nonempty closed subset of R. Within that set, define the jump operators
ρ, σ : T → T by
ρ(t) = sup{s ∈ T : s < t}
and
σ(t) = inf{s ∈ T : s > t},
where inf ∅ := sup T and sup ∅ := inf T. The point t ∈ T is left-dense, left-scattered, rightdense, right-scattered if ρ(t) = t, ρ(t) < t, σ(t) = t, σ(t) > t, respectively. If T has a
right-scattered minimum m, define Tκ := T − {m}; otherwise, set Tκ = T. If T has a leftscattered maximum M , define Tκ := T−{M }; otherwise, set Tκ = T. The so-called graininess
functions are µ(t) := σ(t) − t and ν(t) := t − ρ(t).
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
081-05
2
D. A NDERSON
For f : T → R and t ∈ Tκ , the nabla derivative [3] of f at t, denoted f ∇ (t), is the number
(provided it exists) with the property that given any ε > 0, there is a neighborhood U of t such
that
|f (ρ(t)) − f (s) − f ∇ (t)[ρ(t) − s]| ≤ ε|ρ(t) − s|
for all s ∈ U . Common special cases again include T = R, where f ∇ = f 0 , the usual derivative;
T = Z, where the nabla derivative is the backward difference operator, f ∇ (t) = f (t)−f (t−1);
q-difference equations with 0 < q < 1 and t > 0,
f (t) − f (qt)
f ∇ (t) =
.
(1 − q)t
For f : T → R and t ∈ Tκ , the delta derivative [4] of f at t, denoted f ∆ (t), is the number
(provided it exists) with the property that given any ε > 0, there is a neighborhood U of t such
that
|f (σ(t)) − f (s) − f ∆ (t)[σ(t) − s]| ≤ ε|σ(t) − s|
for all s ∈ U . For T = R, f ∆ = f 0 , the usual derivative; for T = Z the delta derivative is the
forward difference operator, f ∆ (t) = f (t + 1) − f (t); in the case of q-difference equations with
q > 1,
f (qt) − f (t)
f (s) − f (0)
f ∆ (t) =
,
f ∆ (0) = lim
.
s→0
(q − 1)t
s
A function f : T → R is left-dense continuous or ld-continuous provided it is continuous at
left-dense points in T and its right-sided limits exist (finite) at right-dense points in T. If T = R,
then f is ld-continuous if and only if f is continuous. It is known from [3] or Theorem 8.45 in
[4] that if f is ld-continuous, then there is a function F such that F ∇ (t) = f (t). In this case,
we define
Z b
f (t)∇t = F (b) − F (a).
a
In the same way, from Theorem 1.74 in [4] we have that if g is right-dense continuous, there is
a function G such that G∆ (t) = g(t) and
Z b
g(t)∆t = G(b) − G(a).
a
The following theorem is part of Theorem 2.7 in [3] and Theorem 8.47 in [4].
Theorem 1.1 (Integration by parts). If a, b ∈ T and f ∇ , g ∇ are left-dense continuous, then
Z b
Z b
∇
f (t)g (t)∇t = (f g)(b) − (f g)(a) −
f ∇ (t)g(ρ(t))∇t
a
and
Z
a
b
b
Z
∇
f ∇ (t)g(t)∇t.
f (ρ(t))g (t)∇t = (f g)(b) − (f g)(a) −
a
a
2. TAYLOR ’ S T HEOREM U SING N ABLA P OLYNOMIALS
The generalized polynomials for nabla equations [2] are the functions ĥk : T2 → R, k ∈ N0 ,
defined recursively as follows: The function ĥ0 is
(2.1)
ĥ0 (t, s) ≡ 1
for all
and, given ĥk for k ∈ N0 , the function ĥk+1 is
Z t
(2.2)
ĥk+1 (t, s) =
ĥk (τ, s)∇τ
s, t ∈ T,
for all
s, t ∈ T.
s
J. Inequal. Pure and Appl. Math., 6(3) Art. 66, 2005
http://jipam.vu.edu.au/
T IME -S CALE I NTEGRAL I NEQUALITIES
3
Note that the functions ĥk are all well defined, since each is ld-continuous. If for each fixed s
we let ĥ∇
k (t, s) denote the nabla derivative of ĥk (t, s) with respect to t, then
(2.3)
ĥ∇
k (t, s) = ĥk−1 (t, s)
k ∈ N, t ∈ Tκ .
for
The above definition implies
ĥ1 (t, s) = t − s
for all
s, t ∈ T.
Obtaining an expression for ĥk for k > 1 is not easy in general, but for a particular given time
scale it might be easy to find these functions; see [2] for some examples.
Theorem 2.1 (Taylor’s Formula [2]). Let n ∈ N. Suppose f is n + 1 times nabla differentiable
on Tκn+1 . Let s ∈ Tκn , t ∈ T, and define the functions ĥk by (2.1) and (2.2), i.e.,
Z t
ĥ0 (t, s) ≡ 1 and ĥk+1 (t, s) =
ĥk (τ, s)∇τ for k ∈ N0 .
s
Then we have
f (t) =
n
X
ĥk (t, s)f
∇k
Z
(s) +
t
ĥn (t, ρ(τ ))f ∇
n+1
(τ )∇τ.
s
k=0
We may also relate the functions ĥk as introduced in (2.1) and (2.2) (which we repeat below)
to the functions hk and gk in the delta case [1, 4], and the functions ĝk in the nabla case, defined
below.
Definition 2.1. For t, s ∈ T define the functions
h0 (t, s) = g0 (t, s) = ĥ0 (t, s) = ĝ0 (t, s) ≡ 1,
and given hn , gn , ĥn , ĝn for n ∈ N0 ,
Z t
Z t
hn+1 (t, s) =
hn (τ, s)∆τ, gn+1 (t, s) =
gn (σ(τ ), s)∆τ,
s
s
Z t
Z t
ĥn+1 (t, s) =
ĥn (τ, s)∇τ, ĝn+1 (t, s) =
ĝn (ρ(τ ), s)∇τ.
s
s
The following theorem combines Theorem 9 of [2] and Theorem 1.112 of [4].
n
Theorem 2.2. Let t ∈ Tκκ and s ∈ Tκ . Then
ĥn (t, s) = gn (t, s) = (−1)n hn (s, t) = (−1)n ĝn (s, t)
for all n ≥ 0.
3. S TEFFENSEN ’ S INEQUALITY
For a q-difference equation version of the following result and most results in this paper,
including proof techniques, see [7]. In fact, the presentation of the results to follow largely
mirrors the organisation of [7].
Theorem 3.1 (Steffensen’s Inequality (nabla)). Let a, b ∈ Tκκ with a < b and f, g : [a, b] → R
be nabla-integrable functions, with f of one sign and decreasing and 0 ≤ g ≤ 1 on [a, b].
Assume `, γ ∈ [a, b] such that
Z b
b−`≤
g(t)∇t ≤ γ − a if f ≥ 0, t ∈ [a, b],
a
J. Inequal. Pure and Appl. Math., 6(3) Art. 66, 2005
http://jipam.vu.edu.au/
4
D. A NDERSON
b
Z
γ−a≤
g(t)∇t ≤ b − ` if f ≤ 0,
t ∈ [a, b].
a
Then
Z
b
b
Z
f (t)∇t ≤
(3.1)
γ
Z
f (t)g(t)∇t ≤
`
f (t)∇t.
a
a
Proof. The proof given in the q-difference case [7] can be extended to general time scales. As
in [7], we prove only the case in (3.1) where f ≥ 0 for the left inequality; the proofs of the
other cases are similar. After subtracting within the left inequality,
Z
b
b
Z
f (t)g(t)∇t −
a
f (t)∇t
`
Z
`
=
b
Z
f (t)g(t)∇t −
f (t)g(t)∇t +
a
Z
Z
`
`
Z
b
f (t)∇t
`
b
f (t)(1 − g(t))∇t
Z `
Z b
≥
f (t)g(t)∇t − f (`)
(1 − g(t))∇t
a
`
Z b
Z `
g(t)∇t
f (t)g(t)∇t − (b − `)f (`) + f (`)
=
`
a
Z `
Z b
Z b
≥
f (t)g(t)∇t − f (`)
g(t)∇t + f (`)
g(t)∇t
a
a
`
Z b
Z b
Z `
g(t)∇t −
g(t)∇t
=
f (t)g(t)∇t − f (`)
f (t)g(t)∇t −
=
`
a
a
a
Z
`
Z
f (t)g(t)∇t − f (`)
=
a
Z
`
`
g(t)∇t
a
`
(f (t) − f (`)) g(t)∇t ≥ 0
=
a
since f is decreasing and g is nonnegative.
Note that in the theorem above, we could easily replace the nabla integrals with delta integrals
under the same hypotheses and get a completely analogous result. The following theorem more
closely resembles the theorem in the continuous case; the proof is identical to that above and is
omitted.
Theorem 3.2 (Steffensen’s Inequality II). Let a, b ∈ Tκκ and f, g : [a, b] → R be nablaRb
integrable functions, with f decreasing and 0 ≤ g ≤ 1 on [a, b]. Assume λ := a g(t)∇t
such that b − λ, a + λ ∈ T. Then
Z
b
Z
f (t)∇t ≤
(3.2)
b−λ
J. Inequal. Pure and Appl. Math., 6(3) Art. 66, 2005
b
Z
f (t)g(t)∇t ≤
a
a+λ
f (t)∇t.
a
http://jipam.vu.edu.au/
T IME -S CALE I NTEGRAL I NEQUALITIES
5
4. TAYLOR ’ S R EMAINDER
Suppose f is n + 1 times nabla differentiable on Tκn+1 . Using Taylor’s Theorem, Theorem
2.1, we define the remainder function by R̂−1,f (·, s) := f (s), and for n > −1,
Z s
n
X
n+1
∇j
(4.1)
R̂n,f (t, s) := f (s) −
ĥj (s, t)f (t) =
ĥn (s, ρ(τ ))f ∇ (τ )∇τ.
t
j=0
Lemma 4.1. The following identity involving nabla Taylor’s remainder holds:
Z b
Z t
Z b
∇n+1
ĥn+1 (t, ρ(s))f
(s)∇s =
R̂n,f (a, s)∇s +
R̂n,f (b, s)∇s.
a
a
t
Proof. Proceed by mathematical induction on n. For n = −1,
Z b
Z b
Z t
Z b
∇0
ĥ0 (t, ρ(s))f (s)∇s =
f (s)∇s =
f (s)∇s +
f (s)∇s.
a
a
a
t
Assume the result holds for n = k − 1:
Z b
Z t
Z b
∇k
ĥk (t, ρ(s))f (s)∇s =
R̂k−1,f (a, s)∇s +
R̂k−1,f (b, s)∇s.
a
a
t
Let n = k. By Corollary 11 in [2], for fixed t ∈ T we have
s
ĥ∇
k+1 (t, s) = −ĥk (t, ρ(s)).
(4.2)
Thus using the nabla integration by parts rule, Theorem 1.1, we have
Z b
k+1
ĥk+1 (t, ρ(s))f ∇ (s)∇s
a
Z b
k
k
k
ĥk (t, ρ(s))f ∇ (s)∇s + ĥk+1 (t, b)f ∇ (b) − ĥk+1 (t, a)f ∇ (a).
=
a
By the induction assumption and the definition of ĥk+1 ,
Z b
Z b
Z t
∇k+1
ĥk+1 (t, ρ(s))f
(s)∇s =
R̂k−1,f (a, s)∇s +
R̂k−1,f (b, s)∇s
a
t
a
∇k
k
(b) − ĥk+1 (t, a)f ∇ (a)
Z b
Z t
R̂k−1,f (b, s)∇s
R̂k−1,f (a, s)∇s +
=
t
a
Z t
Z t
k
∇k
ĥk (s, a)f ∇ (a)∇s
ĥk (s, b)f (b)∇s −
+
a
b
Z th
i
k
=
R̂k−1,f (a, s) − ĥk (s, a)f ∇ (a) ∇s
a
Z bh
i
k
+
R̂k−1,f (b, s) − ĥk (s, b)f ∇ (b) ∇s
t
Z b
Z t
R̂k,f (b, s)∇s.
R̂k,f (a, s)∇s +
=
+ ĥk+1 (t, b)f
a
t
J. Inequal. Pure and Appl. Math., 6(3) Art. 66, 2005
http://jipam.vu.edu.au/
6
D. A NDERSON
Corollary 4.2. For n ≥ −1,
Z b
Z b
∇n+1
(s)∇s =
ĥn+1 (a, ρ(s))f
R̂n,f (b, s)∇s,
a
a
Z b
Z b
∇n+1
(s)∇s =
ĥn+1 (b, ρ(s))f
R̂n,f (a, s)∇s.
a
a
Lemma 4.3. The following identity involving delta Taylor’s remainder holds:
Z b
Z t
Z b
∆n+1
hn+1 (t, σ(s))f
(s)∆s =
Rn,f (a, s)∆s +
Rn,f (b, s)∆s,
a
a
where
Rn,f (t, s) := f (s) −
t
n
X
j
hj (s, t)f ∆ (t).
j=0
5. A PPLICATIONS OF S TEFFENSEN ’ S I NEQUALITY
In the following we generalize to arbitrary time scales some results from [7] by applying
Steffensen’s inequality, Theorem 3.1.
Theorem 5.1. Let f : [a, b] → R be an n + 1 times nabla differentiable function such that
n+1
n
f∇
is increasing and f ∇ is monontonic (either increasing or decreasing) on [a, b]. Assume
`, γ ∈ [a, b] such that
b−`≤
γ−a≤
ĥn+2 (b, a)
ĥn+1 (b, ρ(a))
ĥn+2 (b, a)
≤γ−a
n
if f ∇ is decreasing,
n
ĥn+1 (b, ρ(a))
≤ b − ` if f ∇ is increasing.
Then
f
∇n
(γ) − f
∇n
(a) ≤
Z
1
ĥn+1 (b, ρ(a))
b
n
n
R̂n,f (a, s)∇s ≤ f ∇ (b) − f ∇ (`).
a
n
n
Proof. Assume f ∇ is decreasing; the case where f ∇ is increasing is similar and is omitted.
n+1
n
n+1
Let F := −f ∇ . Because f ∇ is decreasing, f ∇
≤ 0, so that F ≥ 0 and decreasing on
[a, b]. Define
ĥn+1 (b, ρ(t))
g(t) :=
∈ [0, 1], t ∈ [a, b], n ≥ −1.
ĥn+1 (b, ρ(a))
Note that F, g satisfy the assumptions of Steffensen’s inequality, Theorem 3.1; using (4.2),
Z b
Z b
1
ĥn+2 (b, a)
g(t)∇t =
ĥn+1 (b, ρ(t))∇t =
.
ĥn+1 (b, ρ(a)) a
ĥn+1 (b, ρ(a))
a
Thus if
b−`≤
ĥn+2 (b, a)
ĥn+1 (b, ρ(a))
≤ γ − a,
then
Z
b
Z
F (t)∇t ≤
`
J. Inequal. Pure and Appl. Math., 6(3) Art. 66, 2005
b
Z
F (t)g(t)∇t ≤
a
γ
F (t)∇t.
a
http://jipam.vu.edu.au/
T IME -S CALE I NTEGRAL I NEQUALITIES
7
By Corollary 4.2 and the fundamental theorem of nabla calculus, this simplifies to
Z b
1
n
γ
∇n
f (t)|t=a ≤
R̂n,f (a, s)∇s ≤ f ∇ (t)|bt=` .
ĥn+1 (b, ρ(a)) a
It is evident that an analogous result can be found for the delta integral case using the delta
equivalent of Theorem 3.1.
Definition 5.1. A twice nabla-differentiable function f : [a, b] → R is convex on [a, b] if and
2
only if f ∇ ≥ 0 on [a, b].
The following corollary is the first Hermite-Hadamard inequality, derived from Theorem 5.1
with n = 0.
Corollary 5.2. Let f : [a, b] → R be convex and monotonic. Assume `, γ ∈ [a, b] such that
`≥b−
ĥ2 (b, a)
,
b − ρ(a)
γ≥
ĥ2 (b, a)
+a
b − ρ(a)
if f is decreasing,
`≤b−
ĥ2 (b, a)
,
b − ρ(a)
γ≤
ĥ2 (b, a)
+a
b − ρ(a)
if f is increasing.
Then
ρ(a) − a
1
f (a) ≤
f (γ) +
b − ρ(a)
b − ρ(a)
Z
b
f (t)∇t ≤
a
b−a
f (a) + f (b) − f (`).
b − ρ(a)
Another, slightly different, form of the first Hermite-Hadamard inequality is the following;
this implies that for time scales with left-scattered points there are at least two inequalities of
this type.
Theorem 5.3. Let f : [a, b] → R be convex and monotonic. Assume `, γ ∈ [a, b] such that
ĥ2 (b, a)
,
b−a
ĥ2 (b, a)
`≤a+
,
b−a
ĥ2 (b, a)
b−a
ĥ2 (b, a)
γ ≤b−
b−a
`≥a+
γ ≥b−
Then
1
f (γ) ≤
b−a
Z
if f is decreasing,
if f is increasing.
b
f (ρ(t))∇t ≤ f (a) + f (b) − f (`).
a
2
Proof. Assume f is decreasing and convex. Then f ∇ ≥ 0, f ∇ ≤ 0, and f ∇ is increasing. Then
b−t
F := −f ∇ is decreasing and satisfies F ≥ 0. For G := b−a
, 0 ≤ G ≤ 1 and F, G satisfy the
hypotheses of Theorem 3.1. Now the inequality expression
Z b
b−`≤
G(t)∇t ≤ γ − a
a
takes the form
Z b
1
b−`≤
(b − t)∇t ≤ γ − a.
b−a a
Concentrating on the left inequality,
Z b
Z b
1
1
`≥b−
(b − t)∇t = b −
(b − a + a − t)∇t,
b−a a
b−a a
J. Inequal. Pure and Appl. Math., 6(3) Art. 66, 2005
http://jipam.vu.edu.au/
8
D. A NDERSON
which simplifies to
`≥a+
ĥ2 (b, a)
;
b−a
similarly,
ĥ2 (b, a)
.
b−a
Rs
Furthermore, note that r F (t)∇t = f (r) − f (s), and integration by parts yields
Z b
Z b
Z b
1
1
∇
F (t)G(t)∇t =
(t − b)f (t)∇t = f (a) −
f (ρ(t))∇t.
b−a a
b−a a
a
γ ≥b−
It follows that Steffensen’s inequality takes the form
Z b
1
f (`) − f (b) ≤ f (a) −
f (ρ(t))∇t ≤ f (a) − f (γ),
b−a a
which can be rearranged to match the theorem’s stated conclusion.
Theorem 5.4. Let f : [a, b] → R be an n + 1 times nabla differentiable function such that
m ≤ f∇
n+1
≤M
on [a, b] for some real numbers m < M . Also, let `, γ ∈ [a, b] such that
b−`≤
∇n
1
n
f (b) − f ∇ (a) − m(b − a) ≤ γ − a.
M −m
Then
Z
mĥn+2 (b, a) + (M − m)ĥn+2 (b, `) ≤
b
R̂n,f (a, t)∇t
a
≤ M ĥn+2 (b, a) + (m − M )ĥn+2 (b, γ).
Proof. Let
h
i
1
k(t) :=
f (t) − mĥn+1 (t, a) , F (t) := ĥn+1 (b, ρ(t)),
M −m
h n+1
i
1
n+1
G(t) := k ∇ (t) =
f ∇ (t) − m ∈ [0, 1].
M −m
Observe that F is nonnegative and decreasing, and
Z b
∇n
1
n
G(t)∇t =
f (b) − f ∇ (a) − m(b − a) .
M −m
a
Since F, G satisfy the hypotheses of Theorem 3.1, we compute the various integrals given in
(3.1). First, by (4.2),
Z b
Z b
b
F (t)∇t =
ĥn+1 (b, ρ(t))∇t = −ĥn+2 (b, t)t=` = ĥn+2 (b, `),
`
`
and
Z
a
γ
γ
F (t)∇t = −ĥn+2 (b, t)a = ĥn+2 (b, a) − ĥn+2 (b, γ).
J. Inequal. Pure and Appl. Math., 6(3) Art. 66, 2005
http://jipam.vu.edu.au/
T IME -S CALE I NTEGRAL I NEQUALITIES
9
Moreover, using Corollary 4.2, we have
Z b
Z b
n+1
1
F (t)G(t)∇t =
ĥn+1 (b, ρ(t)) f ∇ (t) − m ∇t
M −m a
a
Z b
b
1
m
=
R̂n,f (a, t)∇t +
ĥn+2 (b, t)a
M −m a
M −m
Z b
1
m
=
R̂n,f (a, t)∇t −
ĥn+2 (b, a).
M −m a
M −m
Using Steffensen’s inequality (3.1), we obtain
Z b
1
ĥn+2 (b, `) ≤
R̂n,f (a, t)∇t − mĥn+2 (b, a) ≤ ĥn+2 (b, a) − ĥn+2 (b, γ),
M −m a
which yields the conclusion of the theorem.
Theorem 5.5. Let f : [a, b] → R be a nabla and delta differentiable function such that
m ≤ f ∇, f ∆ ≤ M
on [a, b] for some real numbers m < M .
(i) If there exist `, γ ∈ [a, b] such that
1
b−`≤
[f (b) − f (a) − m(b − a)] ≤ γ − a,
M −m
then
Z b
mĥ2 (b, a) + (M − m)ĥ2 (b, `) ≤
f (t)∇t − (b − a)f (a)
a
≤ M ĥ2 (b, a) + (m − M )ĥ2 (b, γ).
(ii) If there exist `, γ ∈ [a, b] such that
1
γ−a≤
[f (b) − f (a) − m(b − a)] ≤ b − `,
M −m
then
Z b
mh2 (a, b) + (M − m)h2 (a, γ) ≤ (b − a)f (b) −
f (t)∆t
a
≤ M h2 (a, b) + (m − M )h2 (a, `).
Proof. The first part is just Theorem 5.4 with n = 0. For the second part, let
1
k(t) :=
[f (t) − m(t − b)] , F (t) := h1 (a, σ(t)),
M −m
∆
1
G(t) := k ∆ (t) =
f (t) − m ∈ [0, 1].
M −m
Clearly F is decreasing and nonpositive, and
Z b
1
G(t)∆t =
[f (b) − f (a) − m(b − a)] ∈ [γ − a, b − `].
M −m
a
Since F, G satisfy the hypotheses of Steffensen’s inequality for delta integrals, we determine
the corresponding integrals. First,
Z b
Z b
b
F (t)∆t =
h1 (a, σ(t))∆t = −h2 (a, t)t=` = −h2 (a, b) + h2 (a, `),
`
`
J. Inequal. Pure and Appl. Math., 6(3) Art. 66, 2005
http://jipam.vu.edu.au/
10
D. A NDERSON
and
Z
a
γ
γ
F (t)∆t = −h2 (a, t)a = −h2 (a, γ).
Moreover, using the formula for integration by parts for delta integrals,
Z b
Z b
F (t)G(t)∆t =
h1 (a, σ(t))k ∆ (t)∆t
a
a
Z b
b
= h1 (a, t)k(t) a −
h∆
1 (a, t)k(t)∆t
a
Z b
1
=
−(b − a)f (b) +
f (t)∆t + mh2 (a, b) .
M −m
a
Using Steffensen’s inequality for delta integrals, we obtain
Z b
1
−h2 (a, b) + h2 (a, `) ≤
−(b − a)f (b) +
f (t)∆t + mh2 (a, b)
M −m
a
≤ −h2 (a, γ),
which yields the conclusion of (ii).
In [7], part (ii) of the above theorem also involved the equivalent of nabla derivatives for
q-difference equations with 0 < q < 1. However, the function used there, F (t) = a − qt =
a − ρ(t), is not of one sign on [a, b], since F (a) = a(1 − q) > 0, F (a/q) = 0, and F (a/q 2 ) =
a(1 − 1/q) < 0. For this reason we introduced a delta-derivative perspective in (ii) above and
in the following.
Corollary 5.6. Let f : [a, b] → R be a nabla and delta differentiable function such that
m ≤ f ∇, f ∆ ≤ M
on [a, b] for some real numbers m < M . Assume there exist `, γ ∈ [a, ρ(b)] such that
(5.1)
(5.2)
1
[f (b) − f (a) − m(b − a)] ≤ γ − a,
M −m
1
b−`≤
[f (b) − f (a) − m(b − a)] ≤ b − ρ(`).
M −m
ρ(γ) − a ≤
Then
2mh2 (a, b) + (M − m) [h2 (`, b) + h2 (a, ρ(γ))]
Z b
Z b
≤
f (t)∇t −
f (t)∆t + (b − a)(f (b) − f (a))
a
a
≤ 2M h2 (a, b) − (M − m) [h2 (γ, b) + h2 (a, ρ(`))] .
Proof. By the previous theorem, Theorem 5.5,
Z
mĥ2 (b, a) + (M − m)ĥ2 (b, `) ≤
b
f (t)∇t − (b − a)f (a)
a
≤ M ĥ2 (b, a) + (m − M )ĥ2 (b, γ)
(5.3)
using (i) and the fact that
b−`≤
1
[f (b) − f (a) − m(b − a)] ≤ γ − a;
M −m
J. Inequal. Pure and Appl. Math., 6(3) Art. 66, 2005
http://jipam.vu.edu.au/
T IME -S CALE I NTEGRAL I NEQUALITIES
11
in like manner
Z
b
mh2 (a, b) + (M − m)h2 (a, ρ(γ)) ≤ (b − a)f (b) −
f (t)∆t
a
≤ M h2 (a, b) + (m − M )h2 (a, ρ(`))
(5.4)
using (ii) and the fact that
ρ(γ) − a ≤
1
[f (b) − f (a) − m(b − a)] ≤ b − ρ(`).
M −m
Add (5.3) to (5.4) and use Theorem 2.2 to arrive at the conclusion.
Remark 5.7. If T = R, set λ := b − ` = γ − a, so that b − γ = ` − a = b − a − λ. Here the
nabla and delta integrals of f on [a, b] are identical, and h2 (s, t) = (t − s)2 /2, so the conclusion
of the previous corollary, Corollary 5.6, is the known [7] inequality
m+
(M − m)λ2
f (b) − f (a)
(M − m)(b − a − λ)2
≤
≤
M
−
.
(b − a)2
b−a
(b − a)2
If T = Z, then h2 (s, t) = (t − s)(t − s + 1)/2 = (t − s)2 /2 and
Z b
Z b
b
b−1
X
X
f (t)∇t −
f (t)∆t =
f (t) −
f (t) = f (b) − f (a).
a
a
t=a+1
t=a
This time take λ = b − ` = γ − 1 − a. The discrete conclusion of Corollary 5.6 is thus
m+
(M − m)λ2
f (b) − f (a)
(M − m)(b − a − λ − 1)2
≤
≤
M
−
.
b−a
(b − a)2
(b − a)2
Corollary 5.8 (Iyengar’s Inequality). Let f : [a, b] → R be a nabla and delta differentiable
function such that
m ≤ f ∇, f ∆ ≤ M
on [a, b] for some real numbers m < M . Assume there exist `, γ ∈ [a, ρ(b)] such that (5.1), (5.2)
are satisfied. Then
(M − m) [h2 (`, b) + h2 (a, ρ(`)) − h2 (a, b)]
Z b
Z b
≤
f (t)∇t +
f (t)∆t − (b − a)(f (b) + f (a))
a
a
≤ (M − m) [h2 (a, b) − h2 (γ, b) − h2 (a, ρ(γ))] .
Proof. Subtract (5.4) from (5.3) and use Theorem 2.2 to arrive at the conclusion.
Rb
Rb
Rb
Remark 5.9. Again if T = R, then a f (t)∇t = a f (t)∆t = a f (t)dt and h2 (t, s) =
(t − s)2 /2. Moreover, ρ(`) = ` and ρ(γ) = γ; set
λ=b−`=γ−a=
1
[f (b) − f (a) − m(b − a)] .
M −m
This transforms the conclusion of Corollary 5.8 into a continuous calculus version,
Z b
[f (b)−f (a)−m(b−a)] [M (b−a)+f (a)−f (b)]
f (a)+f (b)
≤
f
(t)dt−
(b−a)
.
2
2(M − m)
a
J. Inequal. Pure and Appl. Math., 6(3) Art. 66, 2005
http://jipam.vu.edu.au/
12
D. A NDERSON
6. A PPLICATIONS OF Č EBYŠEV ’ S I NEQUALITY
Recently, Čebyšev’s inequality on time scales for delta integrals was proven [9]. We repeat
the statement of it here in the case of nabla integrals for completeness.
Theorem 6.1 (Čebyšev’s inequality). Let f and g be both increasing or both decreasing in
[a, b]. Then
Z b
Z b
Z b
1
f (t)g(t)∇t ≥
f (t)∇t
g(t)∇t.
b−a a
a
a
If one of the functions is increasing and the other is decreasing, then the above inequality is
reversed.
The following is an application of Čebyšev’s inequality, which extends a similar result in [7]
to general time scales.
Theorem 6.2. Assume that f ∇
(i) If f
∇n+1
(ii) If f ∇
n+1
n+1
is monotonic on [a, b].
is increasing, then
∇n
Z b
n
f (b) − f ∇ (a)
0≥
R̂n,f (a, t)∇t −
ĥn+2 (b, a)
b−a
a
h n+1
i
n+1
≥ f ∇ (a) − f ∇ (b) ĥn+2 (b, a).
is decreasing, then
∇n
Z b
n
f (b) − f ∇ (a)
0≤
ĥn+2 (b, a)
R̂n,f (a, t)∇t −
b−a
a
h n+1
i
∇
∇n+1
≤ f
(a) − f
(b) ĥn+2 (b, a).
n+1
Proof. The situation for (ii) is analogous to that of (i). Assume (i), and set F (t) := f ∇ (t),
G(t) := ĥn+1 (b, ρ(t)). Then F is increasing by assumption, and G is decreasing, so that by
Čebyšev’s nabla inequality,
Z b
Z b
Z b
1
F (t)G(t)∇t ≤
F (t)∇t
G(t)∇t.
b−a a
a
a
By Corollary 4.2,
Z b
Z b
Z b
∇n+1
F (t)G(t)∇t =
f
(t)ĥn+1 (b, ρ(t))∇t =
R̂n,f (a, t)∇t.
a
a
a
We also have
Z b
n
n
F (t)∇t = f ∇ (b) − f ∇ (a),
Z
Thus Čebyšev’s inequality implies
Z b
R̂n,f (a, t)∇t ≤
a
Z
b
G(t)∇t =
a
a
b
ĥn+1 (b, ρ(t)) = ĥn+2 (b, a).
a
1 ∇n
n
f (b) − f ∇ (a) ĥn+2 (b, a),
b−a
n+1
which subtracts to the left side of the inequality. Since f ∇ is increasing on [a, b],
∇n
n
f (b) − f ∇ (a)
n+1
∇n+1
f
(a)ĥn+2 (b, a) ≤
ĥn+2 (b, a) ≤ f ∇ (b)ĥn+2 (b, a),
b−a
J. Inequal. Pure and Appl. Math., 6(3) Art. 66, 2005
http://jipam.vu.edu.au/
T IME -S CALE I NTEGRAL I NEQUALITIES
13
and we have
∇n
Z b
Z b
n
f (b)−f ∇ (a)
n+1
R̂n,f (a, t)∇t−
ĥn+2 (b, a) ≥
R̂n,f (a, t)∇t − f ∇ (b)ĥn+2 (b, a).
b−a
a
a
n+1
Now Corollary 4.2 and f ∇ is increasing imply that
Z b
Z b
Z b
∇n+1
∇n+1
f
(b)
ĥn+1 (b, ρ(t))∇t ≥
R̂n,f (a, t)∇t ≥ f
(a)
ĥn+1 (b, ρ(t))∇t,
a
a
a
which simplifies to
f
∇n+1
Z
b
(b)ĥn+2 (b, a) ≥
R̂n,f (a, t)∇t ≥ f ∇
n+1
(a)ĥn+2 (b, a).
a
This, together with the earlier lines give the right side of the inequality.
n+1
Theorem 6.3. Assume that f ∆
is monotonic on [a, b] and the function gk is as defined in
Definition 2.1.
n+1
(i) If f ∆ is increasing, then
∆n
Z b
n
f (b) − f ∆ (a)
n+1
Rn,f (b, t)∆t −
gn+2 (b, a)
0 ≤ (−1)
b−a
a
h n+1
i
n+1
≤ f ∆ (b) − f ∆ (a) gn+2 (b, a).
(ii) If f ∆
n+1
is decreasing, then
∆n
Z b
n
f (b) − f ∆ (a)
n+1
Rn,f (b, t)∆t −
0 ≥ (−1)
gn+2 (b, a)
b−a
a
h n+1
i
n+1
≥ f ∆ (b) − f ∆ (a) gn+2 (b, a).
n+1
Proof. The situation for (ii) is analogous to that of (i). Assume (i), and set F (t) := f ∆ (t),
G(t) := (−1)n+1 hn+1 (a, σ(t)). Then F and G are increasing, so that by Čebyšev’s delta inequality,
Z b
Z b
Z b
1
F (t)G(t)∆t ≥
F (t)∆t
G(t)∆t.
b−a a
a
a
By Lemma 4.3 with t = a,
Z b
Z b
Z b
n+1
∆n+1
n+1
F (t)G(t)∆t = (−1)
f
(t)hn+1 (a, σ(t))∆t = (−1)
Rn,f (b, t)∆t.
a
We also have
a
Rb
a
∆n
a
∆n
F (t)∆t = f (b) − f (a), and, using Theorem 2.2,
Z b
Z b
n+1
G(t)∆t = (−1)
hn+1 (a, σ(t))∆t = gn+2 (b, a).
a
a
Thus Čebyšev’s inequality implies
Z b
n+1
(−1)
Rn,f (b, t)∆t ≥
a
1 ∆n
n
f (b) − f ∆ (a) gn+2 (b, a),
b−a
n+1
which subtracts to the left side of the inequality. Since f ∆ is increasing on [a, b],
∆n
n
f (b) − f ∆ (a)
n+1
∆n+1
f
(a)gn+2 (b, a) ≤
gn+2 (b, a) ≤ f ∆ (b)gn+2 (b, a),
b−a
J. Inequal. Pure and Appl. Math., 6(3) Art. 66, 2005
http://jipam.vu.edu.au/
14
D. A NDERSON
and we have
Z b
n+1
n+1
(−1)
Rn,f (b, t)∆t − f ∆ (a)gn+2 (b, a)
a
∆n
Z b
n
f (b) − f ∆ (a)
n+1
Rn,f (b, t)∆t −
≥ (−1)
gn+2 (b, a).
b−a
a
Now Theorem 2.2 and Lemma 4.3 again with t = a yield
Z b
Z b
n+1
n+1
(−1)
Rn,f (b, t)∆t =
gn+1 (σ(t), a)f ∆ (t)∆t.
a
Since f
f
∆n+1
∆n+1
a
is increasing,
Z
b
n+1
Z
b
gn+1 (σ(t), a)∆t ≥ (−1)
(b)
a
Rn,f (b, t)∆t ≥ f
∆n+1
Z
(a)
a
b
gn+1 (σ(t), a)∆t,
a
which simplifies to
f
∆n+1
n+1
Z
(b)gn+2 (b, a) ≥ (−1)
b
Rn,f (b, t)∆t ≥ f ∆
n+1
(a)gn+2 (b, a).
a
This, together with the earlier lines give the right side of the inequality.
Remark 6.4. If T = R, then combining Theorem 6.2 and Theorem 6.3 yields Theorem 3.1 in
[6].
Remark 6.5. In Theorem 6.2 (i), if n = 0, we obtain
Z b
ĥ2 (b, a)
(f (b) − f (a)).
f (t)∇t ≤ (b − a)f (a) +
(6.1)
b−a
a
Compare that with the following result.
2
Theorem 6.6. Assume that f is nabla convex on [a, b]; that is, f ∇ ≥ 0 on [a, b]. Then
Z b
ĥ2 (b, a)
(6.2)
f (ρ(t))∇t ≤ (b − a)f (b) −
(f (b) − f (a)).
b−a
a
Proof. If F := f ∇ and G(t) := t − a = ĥ1 (t, a), then both F and G are increasing functions.
By Čebyšev’s inequality on time scales, and the definition of ĥ in (2.2),
Z b
Z b
Z b
1
∇
∇
f (t)∇t
ĥ1 (t, a)∇t.
f (t)(t − a)∇t ≥
b−a a
a
a
Using nabla integration by parts on the left, and calculating the right yields the result.
The following result is a Hermite-Hadamard-type inequality for time scales; compare with
Corollary 5.2.
Corollary 6.7. Let f be nabla convex on [a, b]. Then
Z b
1
f (ρ(t)) + f (t)
f (b) + f (a)
∇t ≤
.
b−a a
2
2
Proof. Use (6.1), (6.2) and rearrange accordingly.
J. Inequal. Pure and Appl. Math., 6(3) Art. 66, 2005
http://jipam.vu.edu.au/
T IME -S CALE I NTEGRAL I NEQUALITIES
15
R EFERENCES
[1] R.P. AGARWAL AND M. BOHNER, Basic calculus on time scales and some of its applications,
Results Math., 35(1-2) (1999), 3–22.
[2] D.R. ANDERSON, Taylor polynomials for nabla dynamic equations on time scales, PanAmerican
Math. J., 12(4) (2002), 17–27.
[3] F.M. ATICI AND G.Sh. GUSEINOV, On Green’s functions and positive solutions for boundary value
problems on time scales, J. Comput. Appl. Math., 141 (2002) 75–99.
[4] M. BOHNER AND A. PETERSON, Dynamic Equations on Time Scales: An Introduction with Applications, Birkhäuser, Boston (2001).
[5] M. BOHNER AND A. PETERSON (Eds.), Advances in Dynamic Equations on Time Scales,
Birkhäuser, Boston (2003).
[6] H. GAUCHMAN, Some Integral Inequalities Involving Taylor’s Remainder II, J. Inequal. in Pure
& Appl. Math., 4(1) (2003), Art. 1. [ONLINE: http://jipam.vu.edu.au/article.php?
sid=237]
[7] H. GAUCHMAN, Integral Inequalities in q-Calculus, Comp. & Math. with Applics., 47 (2004), 281–
300.
[8] S. HILGER, Ein Maßkettenkalkül mit Anwendung auf Zentrumsmannigfaltigkeiten, PhD thesis, Universität Würzburg (1988).
[9] C.C. YEH, F.H. WONG AND H.J. LI, Čebyšev’s inequality on time scales, J. Inequal. in Pure &
Appl. Math., 6(1) (2005), Art. 7. [ONLINE: http://jipam.vu.edu.au/article.php?
sid=476]
J. Inequal. Pure and Appl. Math., 6(3) Art. 66, 2005
http://jipam.vu.edu.au/