YOUNG’S INTEGRAL INEQUALITY ON TIME
SCALES REVISITED
Young’s Inequality Revisited
DOUGLAS R. ANDERSON
Concordia College
Department of Mathematics and Computer Science
Moorhead, MN 56562 USA
EMail: andersod@cord.edu
Douglas R. Anderson
vol. 8, iss. 3, art. 64, 2007
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Received:
09 June, 2007
Accepted:
22 August, 2007
Communicated by:
D. Hinton
2000 AMS Sub. Class.:
26D15, 39A12.
Key words:
Dynamic equations, Integral inequalities.
Abstract:
A more complete Young’s integral inequality on arbitrary time scales (unbounded
above) is presented.
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Contents
1
Introduction
3
2
Revisiting Young’s Inequality on Time Scales
4
Young’s Inequality Revisited
Douglas R. Anderson
vol. 8, iss. 3, art. 64, 2007
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1.
Introduction
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].
Young’s Inequality Revisited
Douglas R. Anderson
vol. 8, iss. 3, art. 64, 2007
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2.
Revisiting Young’s Inequality on Time Scales
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].
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
Z b
Z b
−1
(2.1)
f (t)∆t +
f (t)∇t +
f (y)∆y +
f −1 (y)∇y ≥ 2ab,
0
0
with equality if and only if b = f (a).
0
0
Young’s Inequality Revisited
Douglas R. Anderson
vol. 8, iss. 3, art. 64, 2007
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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 ,
Young’s Inequality Revisited
b ∈ [0, ∞) ∩ f (T),
with equality if and only if b = f (a). For any such a and b, we have
Z b
Z b
−1
−1
F (a, b) − F (a, f (a)) =
f (y) − a ∆y +
f (y) − a ∇y
f (a)
Z
a − f −1 (y) ∆y +
b
Z
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f (a)
a − f −1 (y) ∇y.
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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
Z b
−1
−1
F (a, b) − F (a, f (a)) =
f (y) − a ∆y +
f (y) − a ∇y ≥ 0.
f (a)
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f (a)
f (a)
=
Douglas R. Anderson
vol. 8, iss. 3, art. 64, 2007
f (a)
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−1
Since f (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)
Z f (a)
−1
F (a, b) − F (a, f (a)) =
a − f (y) ∆y +
a − f −1 (y) ∇y ≥ 0.
b
b
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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 a
Z a
Z f (a)
Z f (a)
−1
δ(a) :=
f (t)∆t +
f (t)∇t +
f (y)∆y +
f −1 (y)∇y − 2af (a).
0
0
0
0
Young’s Inequality Revisited
First, assume a is a right-scattered point. Then
Douglas R. Anderson
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 rightdense 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
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f (an )
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
Z f (an )
Z f (an )
−1
−1
+
f (y) − a ∆y +
f (y) − a ∇y.
+
f (a)
vol. 8, iss. 3, art. 64, 2007
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Since the functions f and f −1 are strictly increasing,
Z an
Z an
δ(an ) − δ(a) ≥
[f (a) − f (an )] ∆t +
[f (a) − f (an )] ∇t
a
a
Z f (an )
Z f (an )
−1
−1
+
f (f (a)) − a ∆y +
f (f (a)) − a ∇y
f (a)
f (a)
= 2(an − a) [f (a) − f (an )] .
Young’s Inequality Revisited
Douglas R. Anderson
Similarly,
vol. 8, iss. 3, art. 64, 2007
Z
δ(an ) − δ(a) ≤
an
Z
an
[f (an ) − f (an )] ∆t +
[f (an ) − f (an )] ∇t
a
a
Z f (an )
Z f (an )
−1
−1
+
f (f (an )) − a ∆y +
f (f (an )) − a ∇y
f (a)
f (a)
= 2 [f (an ) − f (a)] (an − a).
Therefore,
δ(an ) − δ(a)
≤ lim 2 [f (an ) − f (a)] = 0.
0 = lim 2 [f (a) − f (an )] ≤ lim
n→∞
n→∞
n→∞
an − a
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.
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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
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 (y) + f −1 (σ(y)) ∆y ≥ 2ab,
[f (t) + f (σ(t))] ∆t +
0
0
Young’s Inequality Revisited
Douglas R. Anderson
vol. 8, iss. 3, art. 64, 2007
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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
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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 1. 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
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If T = Z, then σ(t) = t + 1 and the theorem yields Young’s discrete inequality,
a−1
X
X
[f (t) + f (t + 1)] +
t=0
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
Young’s Inequality Revisited
Douglas R. Anderson
vol. 8, iss. 3, art. 64, 2007
rτ f (rτ ) + f (rτ +1 ) + (r + 1)
τ =−∞
X
µ(y)f −1 (y) ≥ 2ab,
y∈[0,b)∩f (Tr )
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where a = rα and t = rτ for α, τ ∈ Z.
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Corollary 2.3. Assume T is any time scale with 0 ∈ T. Let p, q > 1 be real numbers
with p1 + 1q = 1. Then for any a ∈ [0, ∞)T and b ∈ [0, ∞)T∗ , where T∗ := {tp−1 :
t ∈ T}, we have
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
0
0
with equality if and only if b = ap−1 .
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.
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References
[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
1934.
AND
G. PÓLYA, Inequalities, Cambridge,
[4] S. HILGER, Ein Maßkettenkalkül mit Anwendung auf Zentrumsmannigfaltigkeiten, PhD. thesis, Universität Würzburg (1988).
Young’s Inequality Revisited
Douglas R. Anderson
vol. 8, iss. 3, art. 64, 2007
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[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.
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