Volume 8 (2007), Issue 3, Article 64, 5 pp.
YOUNG’S INTEGRAL INEQUALITY ON TIME SCALES REVISITED
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 09 June, 2007; accepted 22 August, 2007
Communicated by D. Hinton
A BSTRACT. A more complete Young’s integral inequality on arbitrary time scales (unbounded
above) is presented.
Key words and phrases: Dynamic equations, Integral inequalities.
2000 Mathematics Subject Classification. 26D15, 39A12.
1. I NTRODUCTION
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 [4]. Since then, time-scale calculus has made steady inroads in explaining the
interconnections that exist among the various calculuses, and in extending our understanding to
a new, more general and overarching theory.
The purpose of this note is to illustrate this new understanding by extending a continuous
result, Young’s inequality [3, 6], to arbitrary time scales. Throughout this note a knowledge and
understanding of time scales and time-scale notation is assumed; for an excellent introduction
to calculus on time scales, see Bohner and Peterson [1].
2. R EVISITING YOUNG ’ S I NEQUALITY ON T IME S CALES
Recently Wong, Yeh, Yu, and Hong [5] presented a version of Young’s inequality on time
scales. An important subplot in the story of Young’s inequality includes an if and only if clause
concerning an actual equality; this is missing in the formulation in [5]. Moreover, in [5] the
authors implicitly assume that the integrand function in the proposed integral inequality is delta
differentiable, an unnecessarily strong assumption. In this note we will rectify these shortcomings by presenting a more complete version of Young’s inequality on time scales with standard
assumptions on the integrand function. To this end, let T be any time scale (unbounded above)
that contains 0. Then we have the following extension of Young’s inequality to arbitrary time
scales, whose statement and proof are quite different from that found in [5].
196-07
2
D. A NDERSON
Theorem 2.1 (Young’s Inequality I). Let T be any time scale (unbounded above) with 0 ∈ T.
Further, suppose that f : [0, ∞)T → R is a real-valued function satisfying
(1) f (0) = 0;
(2) f is continuous on [0, ∞)T , right-dense continuous at 0;
(3) f is strictly increasing on [0, ∞)T such that f (T) is also a time scale.
Then for any a ∈ [0, ∞)T and b ∈ [0, ∞) ∩ f (T), we have
Z
a
Z
a
f (t)∆t +
(2.1)
Z
f (t)∇t +
0
b
f
0
−1
b
Z
f −1 (y)∇y ≥ 2ab,
(y)∆y +
0
0
with equality if and only if b = f (a).
Proof. The proof is modelled after the one given on R in [2]. Note that f is both delta and nabla
integrable by the continuity assumption in (ii). For simplicity, define
Z a
Z a
Z b
Z b
−1
F (a, b) :=
f (t)∆t +
f (t)∇t +
f (y)∆y +
f −1 (y)∇y − 2ab.
0
0
0
0
Then, the inequality to be shown is just F (a, b) ≥ 0.
(I). We will first show that
F (a, b) ≥ F (a, f (a)),
a ∈ [0, ∞)T ,
b ∈ [0, ∞) ∩ f (T),
with equality if and only if b = f (a). For any such a and b, we have
Z
b
−1
f (y) − a ∆y +
F (a, b) − F (a, f (a)) =
Z
f (a)
Z
−1
f (y) − a ∇y
f (a)
f (a)
a − f −1 (y) ∆y +
=
b
f (a)
Z
b
a − f −1 (y) ∇y.
b
There are two cases to consider. The first case is b ≥ f (a). Here, whenever y ∈ [f (a), b]∩f (T),
we have f −1 (b) ≥ f −1 (y) ≥ f −1 (f (a)) = a. Consequently,
Z
b
F (a, b) − F (a, f (a)) =
−1
f (y) − a ∆y +
f (a)
Z
b
−1
f (y) − a ∇y ≥ 0.
f (a)
Since f −1 (y) − a is continuous and strictly increasing for y ∈ [f (a), b] ∩ f (T), equality will
hold if and only if b = f (a). The second case is b ≤ f (a). Here, whenever y ∈ [b, f (a)] ∩ f (T),
we have f −1 (b) ≤ f −1 (y) ≤ f −1 (f (a)) = a. Consequently,
Z
F (a, b) − F (a, f (a)) =
f (a)
a − f −1 (y) ∆y +
b
Z
f (a)
a − f −1 (y) ∇y ≥ 0.
b
Since a − f −1 (y) is continuous and strictly decreasing for y ∈ [b, f (a)] ∩ f (T), equality will
hold if and only if b = f (a).
(II). We will next show that F (a, f (a)) = 0. For brevity, put δ(a) = F (a, f (a)), that is
Z
δ(a) :=
a
Z
f (t)∆t +
0
a
Z
f (t)∇t +
0
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 64, 5 pp.
f (a)
f
0
−1
Z
(y)∆y +
f (a)
f −1 (y)∇y − 2af (a).
0
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YOUNG ’ S I NEQUALITY R EVISITED
3
First, assume a is a right-scattered point. Then
δ σ (a) − δ(a) = [σ(a) − a]f (a) + [σ(a) − a]f σ (a) + [f σ (a) − f (a)] f −1 (f (a))
+ [f σ (a) − f (a)] f −1 (f σ (a)) − 2 [σ(a)f σ (a) − af (a)]
= [σ(a) − a] [f (a) + f σ (a)] + [f σ (a) − f (a)] [a + σ(a)]
− 2 [σ(a)f σ (a) − af (a)]
= 0.
Therefore, if a is a right-scattered point, then δ ∆ (a) = 0. Next, assume a is a right-dense point.
Let {an }n∈N ⊂ [a, ∞)T be a decreasing sequence converging to a. Then
Z an
Z an
Z f (an )
δ(an ) − δ(a) =
f (t)∆t +
f (t)∇t +
f −1 (y)∆y
a
a
Z
f (an )
Z
f (an )
f (a)
f −1 (y)∇y − 2an f (an ) + 2af (a)
f (a)
Z an
Z an
=
[f (t) − f (an )] ∆t +
[f (t) − f (an )] ∇t
+
a
a
−1
f (y) − a ∆y +
+
f (a)
f (an )
Z
−1
f (y) − a ∇y.
f (a)
Since the functions f and f −1 are strictly increasing,
Z an
Z an
[f (a) − f (an )] ∇t
[f (a) − f (an )] ∆t +
δ(an ) − δ(a) ≥
a
a
Z
f (an )
−1
f (f (a)) − a ∆y +
+
Z
f (a)
f (an )
−1
f (f (a)) − a ∇y
f (a)
= 2(an − a) [f (a) − f (an )] .
Similarly,
an
Z
δ(an ) − δ(a) ≤
Z
[f (an ) − f (an )] ∆t +
a
an
[f (an ) − f (an )] ∇t
a
Z
f (an )
+
−1
f (f (an )) − a ∆y +
Z
f (a)
f (an )
−1
f (f (an )) − a ∇y
f (a)
= 2 [f (an ) − f (a)] (an − a).
Therefore,
δ(an ) − δ(a)
≤ lim 2 [f (an ) − f (a)] = 0.
n→∞
n→∞
an − a
0 = lim 2 [f (a) − f (an )] ≤ lim
n→∞
It follows that δ ∆ (a) exists, and δ ∆ (a) = 0 for right-dense a as well. In other words, in either
case, δ ∆ (a) = 0 for a ∈ (0, ∞)T . As δ(0) = 0, by the uniqueness theorem for initial value
problems, we have that δ(a) = 0 for all a ∈ [0, ∞)T .
As an overall result, we have that
F (a, b) ≥ F (a, f (a)) = 0,
with equality if and only if b = f (a), as claimed.
Theorem 2.2 (Young’s Inequality II). Let T be any time scale (unbounded above) with 0 ∈ T.
Further, suppose that f : [0, ∞)T → R is a real-valued function satisfying
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 64, 5 pp.
http://jipam.vu.edu.au/
4
D. A NDERSON
(1) f (0) = 0;
(2) f is continuous on [0, ∞)T , right-dense continuous at 0;
(3) f is strictly increasing on [0, ∞)T such that f (T) is also a time scale.
Then for any a ∈ [0, ∞)T and b ∈ [0, ∞) ∩ f (T), we have
Z a
Z b
−1
(2.2)
[f (t) + f (σ(t))] ∆t +
f (y) + f −1 (σ(y)) ∆y ≥ 2ab,
0
0
with equality if and only if b = f (a).
Proof. For a continuous function g and a ∈ [0, ∞)T , define the function
Z a
Z a
Z a
G(a) :=
g(t)∆t +
g(t)∇t −
[g(t) + g(σ(t))] ∆t.
0
0
0
Then G(0) = 0, and
G∆ (a) := g(a) + g(σ(a)) − [g(a) + g(σ(a))] = 0.
Therefore G ≡ 0, and Theorem 2.2 follows from Theorem 2.1.
Remark 2.3. Consider (2.2). If T = R, then σ(t) = t and the theorem yields the classic Young
inequality,
Z a
Z b
f (t)dt +
f −1 (y)dy ≥ ab.
0
0
If T = Z, then σ(t) = t + 1 and the theorem yields Young’s discrete inequality,
a−1
X
[f (t) + f (t + 1)] +
t=0
X
µ(y) 2f −1 (y) + 1 ≥ 2ab,
y∈[0,b)∩f (Z)
since by the discrete nature of the expression, f −1 (σ(y)) = σ (f −1 (y)) = f −1 (y) + 1. If
T = Tr , where r > 1 and Tr := {0} ∪ {rz }z∈Z , then we have Young’s quantum inequality
(r − 1)
α−1
X
rτ f (rτ ) + f (rτ +1 ) + (r + 1)
τ =−∞
α
X
µ(y)f −1 (y) ≥ 2ab,
y∈[0,b)∩f (Tr )
τ
where a = r and t = r for α, τ ∈ Z.
Corollary 2.4. Assume T is any time scale with 0 ∈ T. Let p, q > 1 be real numbers with
1
+ 1q = 1. Then for any a ∈ [0, ∞)T and b ∈ [0, ∞)T∗ , where T∗ := {tp−1 : t ∈ T}, we have
p
Z a
Z a
Z b
Z b
p−1
p−1
q−1
t ∆t +
t ∇t +
y ∆y +
y q−1 ∇y ≥ 2ab,
0
0
p−1
with equality if and only if b = a
0
0
.
Proof. Let f (t) := tp−1 for t ∈ [0, ∞)T . Then f −1 (y) = y q−1 for y ∈ T∗ , and all the hypotheses
of Theorem 2.1 are satisfied; the result follows.
R EFERENCES
[1] M. BOHNER AND A. PETERSON, Dynamic Equations on Time Scales: An Introduction with Applications, Birkhäuser, Boston (2001).
[2] J.B. DIAZ AND F.T. METCALF, An analytic proof of Young’s inequality, American Math. Monthly,
77 (1970), 603–609.
[3] G.H. HARDY, J.E. LITTLEWOOD AND G. PÓLYA, Inequalities, Cambridge, 1934.
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 64, 5 pp.
http://jipam.vu.edu.au/
YOUNG ’ S I NEQUALITY R EVISITED
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[4] S. HILGER, Ein Maßkettenkalkül mit Anwendung auf Zentrumsmannigfaltigkeiten, PhD. thesis,
Universität Würzburg (1988).
[5] F.H. WONG, C.C. YEH, S.L. YU AND C.H. HONG, Young’s inequality and related results on time
scales, Appl. Math. Letters, 18 (2005), 983–988.
[6] W.H. YOUNG, On classes of summable functions and their Fourier series, Proc. Royal Soc., Series
(A), 87 (1912), 225–229.
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 64, 5 pp.
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