Volume 9 (2008), Issue 1, Article 27, 8 pp.
ON SOME INEQUALITIES FOR p-NORMS
U.S. KIRMACI, M. KLARIČIĆ BAKULA, M. E. ÖZDEMIR, AND J. PEČARIĆ
ATATÜRK U NIVERSITY
K. K. E DUCATION FACULTY
D EPARTMENT OF M ATHEMATICS
25240 K AMPÜS , E RZURUM
T URKEY
kirmaci@atauni.edu.tr
D EPARTMENT OF M ATHEMATICS
FACULTY OF NATURAL S CIENCES , M ATHEMATICS AND E DUCATION
U NIVERSITY OF S PLIT
T ESLINA 12, 21000 S PLIT
C ROATIA
milica@pmfst.hr
ATATÜRK U NIVERSITY
K. K. E DUCATION FACULTY
D EPARTMENT OF M ATHEMATICS
25240 K AMPÜS , E RZURUM
T URKEY
emos@atauni.edu.tr
FACULTY OF T EXTILE T ECHNOLOGY
U NIVERSITY OF Z AGREB
P RILAZ BARUNA F ILIPOVI ĆA 30, 10000 Z AGREB
C ROATIA
pecaric@hazu.hr
Received 01 August, 2007; accepted 18 March, 2008
Communicated by S.S. Dragomir
A BSTRACT. In this paper we establish several new inequalities including p-norms for functions
whose absolute values aroused to the p-th power are convex functions.
Key words and phrases: Convex functions, p-norm, Power means, Hölder’s integral inequality.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
Integral inequalities have become a major tool in the analysis of integral equations, so it is
not surprising that many of them appear in the literature (see for example [2], [5], [3] and [1]).
251-07
2
U.S. K IRMACI , M. K LARI ČI Ć BAKULA , M. E. Ö ZDEMIR , AND J. P E ČARI Ć
One of the most important inequalities in analysis is the integral Hölder’s inequality which is
stated as follows (for this variant see [3, p. 106]).
Theorem A. Let p, q ∈ R {0} be such that p1 + 1q = 1 and let f, g : [a, b] → R, a < b, be
such that |f (x)|p and |g (x)|q are integrable on [a, b] . If p, q > 0, then
Z
b
Z
|f (x) g (x)| dx ≤
(1.1)
a
b
p
p1 Z
1q
b
|f (x)| dx
q
|g (x)| dx
.
a
a
If p < 0 and additionally f ([a, b]) ⊆ R {0} , or q < 0 and g ([a, b]) ⊆ R {0} , then the
inequality in (1.1) is reversed.
The Hermite-Hadamard inequalities for convex functions is also well known. This double
inequality is stated as follows (see for example [3, p. 10]): Let f be a convex function on
[a, b] ⊂ R, where a 6= b. Then
Z b
a+b
1
f (a) + f (b)
(1.2)
f
≤
f (x)dx ≤
.
2
b−a a
2
To prove our main result we need comparison inequalities between the power means defined
by
 1
Pn

1
r r

p
x
, r 6= −∞, 0, ∞;

i=1 i i

Pn




 Qn pi P1n
[r]
,
r = 0;
i=1 xi
Mn (x; p) =




min (x1 , . . . , xn ) , r = −∞;





max (x1 , . . . , xn ) , r = ∞,
P
where x, p are positive n-tuples and Pn = ni=1 pi . It is well known that for such means the
following inequality holds:
Mn[r] (x; p) ≤ Mn[s] (x; p)
(1.3)
whenever r < s (see for example [3, p. 15]).
In this paper we also use the following result (see [5, p. 152]):
Theorem B. Let ξ ∈ [a, b]n , 0 < a < b, and p ∈ [0, ∞)n be two n-tuples such that
n
X
n
X
pi ξi ∈ [a, b] ,
i=1
pi ξi ≥ ξj ,
j = 1, 2, . . . , n.
i=1
If f : [a, b] → R is such that the function f (x) /x is decreasing, then
!
n
n
X
X
(1.4)
f
pi ξi ≤
pi f (ξi ) .
i=1
i=1
If f (x) /x is increasing, then the inequality in (1.4) is reversed.
Our goal is to establish several new inequalities for functions whose absolute values raised to
some real powers are convex functions.
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 27, 8 pp.
http://jipam.vu.edu.au/
O N S OME I NEQUALITIES FOR p-N ORMS
3
2. R ESULTS
In the literature, the following definition is well known.
Let f : [a, b] → R and p ∈ R+ . The p-norm of the function f on [a, b] is defined by
 1

 R b |f (x)|p dx p , 0 < p < ∞;
a
kf kp =


sup |f (x)| ,
p = ∞,
and Lp ([a, b]) is the set of all functions f : [a, b] → R such that kf kp < ∞.
Observe that if |f |p is convex (or concave) on [a, b] it is also integrable on [a, b] , hence
0 ≤ kf kp < ∞, that is, f belongs to Lp ([a, b]) .
Although p-norms are not defined for p < 0, for the sake of the simplicity we will use the
same notation kf kp when p ∈ R {0} .
In order to prove our results we need the following two lemmas.
Lemma 2.1. Let x and p be two n-tuples such that
(2.1)
xi > 0, pi ≥ 1,
i = 1, 2, . . . , n.
If r < s < 0 or 0 < r < s, then
n
X
(2.2)
! 1s
pi xsi
n
X
≤
i=1
! r1
pi xri
,
i=1
and if r < 0 < s, then
n
X
! r1
pi xri
n
X
≤
i=1
! 1s
pi xsi
.
i=1
If the n-tuple x is only nonnegative, then (2.2) holds whenever 0 < r < s.
Proof. Suppose that x and p are such that the inequalities in (2.1) hold. It can be easily seen
that in this case for any q ∈ R
n
X
pi xqi ≥ xqj > 0,
j = 1, 2, . . . , n.
i=1
To prove the lemma we must consider three cases: (i) r < s < 0, (ii) 0 < r < s and
s
(iii) r < 0 < s. In case (i) we define the function f : R+ → R+ by f (x) = x r . Since in this
case we have (s − r) /r < 0, the function
s
f (x) /x = x r −1 = x
s−r
r
is decreasing. Applying Theorem B on f, ξ = (xr1 , . . . , xrn ) and p we obtain
! rs
n
n
n
X
X
X
s
≤
pi (xri ) r =
pi xri
pi xsi ,
i=1
i=1
i=1
i.e.,
n
X
! r1
pi xri
i=1
≥
n
X
! 1s
pi xsi
i=1
since s is negative.
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 27, 8 pp.
http://jipam.vu.edu.au/
4
U.S. K IRMACI , M. K LARI ČI Ć BAKULA , M. E. Ö ZDEMIR , AND J. P E ČARI Ć
In case (ii) for the same f as in (i) we have (s − r) /r > 0, so similarly as before from
Theorem B we obtain
! rs
n
n
n
X
X
X
s
pi xri
≥
pi (xri ) r =
pi xsi ,
i=1
i=1
i=1
and since s is positive, (2.2) immediately follows.
And in the end, in case (iii) we have (s − r) /r < 0, so using again Theorem B we obtain
(2.2) reversed.
Remark 2.2. In this paper we will use Lemma 2.1 only in a special case when all weights are
equal to 1. Then for r < s < 0 or 0 < r < s, (2.2) becomes
! r1
! 1s
n
n
X
X
(2.3)
xsi
≤
xri
i=1
i=1
and for r < 0 < s,
n
X
! 1s
xsi
≥
i=1
n
X
! r1
xri
.
i=1
In the rest of the paper we denote
 −1

2 p , p ≤ −1 or p ≥ 1;


Cp =
2,
−1 < p < 0;


 −1
2 , 0 < p < 1;
ep =
C

2,



p ≤ −1;
1
2− p , −1 < p < 1, p 6= 0; .



2−1 , p ≥ 1.
Lemma 2.3. Let f : [a, b] → R, a < b. If |f |p is convex on [a, b] for some p > 0, then
p
p p1
1
a
+
b
|f
(a)|
+
|f
(b)|
−
≤ (b − a) p kf k ≤
f
≤ Cp (|f (a)| + |f (b)|) ,
p
2
2
and if |f |p is concave on [a, b] , then
p
p p1
a + b |f
(a)|
+
|f
(b)|
− p1
e
Cp (|f (a)| + |f (b)|) ≤
≤ (b − a) kf kp ≤ f
.
2
2
Proof. Suppose first that |f |p is convex on [a, b] for some p > 0. We have
Z b
p1
p1
Z b
1
1
p
p
kf kp =
|f (x)| dx
= (b − a) p
|f (x)| dx .
b−a a
a
From (1.2) we obtain
p
Z b
a + b 1
|f (a)|p + |f (b)|p
p
(2.4)
f
≤
|f
(x)|
dx
≤
,
2
b−a a
2
hence
p
p p1
1
f a + b ≤ (b − a)− p kf k ≤ |f (a)| + |f (b)|
.
p
2
2
Now we must consider two cases. If p ≥ 1 we can use (2.3) to obtain
1
(|f (a)|p + |f (b)|p ) p ≤ |f (a)| + |f (b)| ,
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 27, 8 pp.
http://jipam.vu.edu.au/
O N S OME I NEQUALITIES FOR p-N ORMS
5
hence
(2.5)
|f (a)|p + |f (b)|p
2
p1
≤ Cp (|f (a)| + |f (b)|) ,
1
where Cp = 2− p .
In the other case, when 0 < p < 1, from (1.3) we have
1
|f (a)|p + |f (b)|p p
|f (a)| + |f (b)|
≤
,
2
2
so again we obtain (2.5) , where Cp = 2−1 . This completes the proof for |f |p convex.
Suppose now that |f |p is concave on [a, b] for some p > 0. In that case − |f |p is convex on
[a, b] , hence (1.2) implies
p
Z b
|f (a)|p + |f (b)|p
1
a + b p
≤
|f (x)| dx ≤ f
.
2
b−a
2
a
If p ≥ 1 from (1.3) we obtain
1
|f (a)|p + |f (b)|p p
|f (a)| + |f (b)|
≥
,
2
2
hence
1
|f (a)|p + |f (b)|p p
ep (|f (a)| + |f (b)|) ,
≥C
2
ep = 2−1 .
where C
In the other case, when 0 < p < 1, from (2.3) we have
1
(|f (a)|p + |f (b)|p ) p ≥ |f (a)| + |f (b)| ,
hence
|f (a)|p + |f (b)|p
2
p1
ep (|f (a)| + |f (b)|) ,
≥C
1
ep = 2− p . This completes the proof.
where C
Lemma 2.4. Let f : [a, b] → R {0} , a < b. If |f |p is convex on [a, b] for some p < 0, then
1
a + b |f (a) f (b)|
|f (a)|p + |f (b)|p p
− p1
Cp
≤
≤ (b − a) kf kp ≤ f
|f (a)| + |f (b)|
2
2
and if |f |p is concave on [a, b] , then
p
p p1
1
|f
(a)|
+
|f
(b)|
a
+
b
−
ep |f (a) f (b)| .
f
≤ (b − a) p kf k ≤
≤C
p
|f (a)| + |f (b)|
2
2
Proof. Suppose that |f |p is convex on [a, b] for some p < 0. From (2.4) , using the fact that
p < 0, we obtain
1
|f (a)|p + |f (b)|p p
a
+
b
− p1
.
≤ (b − a) kf kp ≤ f
2
2
Again we consider two cases. If −1 < p < 0, then from (1.3) we have
!−1 1
|f (a)|−1 + |f (b)|−1
|f (a)|p + |f (b)|p p
≤
,
2
2
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 27, 8 pp.
http://jipam.vu.edu.au/
6
U.S. K IRMACI , M. K LARI ČI Ć BAKULA , M. E. Ö ZDEMIR , AND J. P E ČARI Ć
hence
|f (a) f (b)|
Cp
≤
|f (a)| + |f (b)|
|f (a)|p + |f (b)|p
2
p1
,
where Cp = 2.
In the other case, when p ≤ −1, from (2.3) we have
−1
1
|f (a)|−1 + |f (b)|−1
≤ (|f (a)|p + |f (b)|p ) p ,
hence
|f (a) f (b)|
Cp
≤
|f (a)| + |f (b)|
|f (a)|p + |f (b)|p
2
p1
,
1
where Cp = 2− p .
In the other case, when |f |p is concave on [a, b] for some p < 0, the proof is similar.
Theorem 2.5. Let p, q > 0 and let f, g : [a, b] → R, a < b, be such that
m (|g (a)| + |g (b)|) ≤ |f (a)| + |f (b)| ≤ M (|g (a)| + |g (b)|)
(2.6)
for some 0 < m ≤ M.
If |f |p and |g|q are convex on [a, b] , then
1
1
1
M
Cp (b − a) p +
Cq (b − a) q K (f, g) ,
(2.7)
kf kp + kgkq ≤
M +1
m+1
where
K (f, g) = |f (a)| + |f (b)| + |g (a)| + |g (b)| .
p
q
If |f | and |g| are concave on [a, b] , then
1
1
m e
1 e
p
q
(2.8)
kf kp + kgkq ≥
Cp (b − a) +
Cq (b − a) K (f, g) .
m+1
M +1
Proof. Suppose that |f |p and |g|q are convex on [a, b] for some fixed p, q > 0. From Lemma 2.3
we have that
kf kp + kgkq
1
1
1
b−a p
b−a q
p
p p1
≤
(|f (a)| + |f (b)| ) +
(|g (a)|q + |g (b)|q ) q
2
2
1
1
≤ Cp (b − a) p (|f (a)| + |f (b)|) + Cq (b − a) q (|g (a)| + |g (b)|) .
(2.9)
Using (2.6) we can write
|f (a)| + |f (b)| ≤ M (|f (a)| + |f (b)| + |g (a)| + |g (b)|) − M (|f (a)| + |f (b)|) ,
i.e.,
(2.10)
|f (a)| + |f (b)| ≤
M
M
(|f (a)| + |f (b)| + |g (a)| + |g (b)|) =
K (f, g) ,
M +1
M +1
and analogously
1
K (f, g) .
m+1
Combining (2.10) and (2.11) with (2.9) we obtain (2.7) .
Suppose now that |f |p and |g|q are concave on [a, b] for some fixed p, q > 0. From Lemma
2.3 we have that
1
1
ep (b − a) p (|f (a)| + |f (b)|) + C
eq (b − a) q (|g (a)| + |g (b)|) .
kf k + kgk ≥ C
|g (a)| + |g (b)| ≤
(2.11)
p
q
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 27, 8 pp.
http://jipam.vu.edu.au/
O N S OME I NEQUALITIES FOR p-N ORMS
7
Using again (2.6) we can write
|f (a)| + |f (b)| ≥ m (|f (a)| + |f (b)| + |g (a)| + |g (b)|) − m (|f (a)| + |f (b)|) ,
i.e.,
|f (a)| + |f (b)| ≥
m
K (f, g) ,
m+1
|g (a)| + |g (b)| ≥
1
K (f, g) ,
M +1
and analogously
from which (2.8) easily follows.
Remark 2.6. A similar type of condition as in (2.6) was used in [1, Theorem 1.1] where a
variant of the reversed Minkowski’s integral inequality for p > 1 was proved.
Theorem 2.7. Let p, q < 0 and let f, g : [a, b] → R {0} , a < b, be such that
|g (a) g (b)|
|f (a) f (b)|
|g (a) g (b)|
≤
≤M
|g (a)| + |g (b)|
|f (a)| + |f (b)|
|g (a)| + |g (b)|
for some 0 < m ≤ M.
If |f |p and |g|q are concave on [a, b] , then
1
1
M e
1 e
p
q
kf kp + kgkq ≤
Cp (b − a) +
Cq (b − a) H (f, g) ,
M +1
m+1
where
|f (a) f (b)|
|g (a) g (b)|
H (f, g) =
+
.
|f (a)| + |f (b)| |g (a)| + |g (b)|
If |f |p and |g|q are convex on [a, b] , then
1
1
m
1
p
q
kf kp + kgkq ≥
Cp (b − a) +
Cq (b − a) H (f, g) .
m+1
M +1
m
Proof. Similar to that of Theorem 2.5.
Theorem 2.8. Let f, g : [a, b] → R, a < b, be such that |f |p and |g|q are convex on [a, b] for
some fixed p, q > 1, where p1 + 1q = 1. Then
Z b
b−a
1
1
f (x) g (x) dx ≤
(|f (a)|p + |f (b)|p ) p (|g (a)|q + |g (b)|q ) q
2
a
≤
b−a
[M (f, g) + N (f, g)] ,
2
where
M (f, g) = |f (a)| |g (a)| + |f (b)| |g (b)| ,
N (f, g) = |f (a)| |g (b)| + |f (b)| |g (a)| .
Proof. First note that since |f |p and |g|q are convex on [a, b] we have f ∈ Lp ([a, b]) and g ∈
Lq ([a, b]) , and since p1 + 1q = 1 we know that f g ∈ L1 ([a, b]) , that is, f g is integrable on [a, b].
Using Hölder’s integral inequality (1.1) we obtain
Z b
Z b
≤
f
(x)
g
(x)
dx
|f (x) g (x)| dx ≤ kf kp kgkq .
a
a
From Lemma 2.4 we have that
1
1
b−a p
b−a p
p
p p1
(|f (a)| + |f (b)| ) ≤
(|f (a)| + |f (b)|)
kf kp ≤
2
2
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 27, 8 pp.
http://jipam.vu.edu.au/
8
U.S. K IRMACI , M. K LARI ČI Ć BAKULA , M. E. Ö ZDEMIR , AND J. P E ČARI Ć
and
kgkq ≤
b−a
2
1q
q
q
1
q
(|g (a)| + |g (b)| ) ≤
b−a
2
1q
(|g (a)| + |g (b)|) ,
hence
Z b
b−a
p
p p1
q
q 1q
≤
(|g
(a)|
+
|g
(b)|
)
(|f
(a)|
+
|f
(b)|
)
f
(x)
g
(x)
dx
2
a
b−a
≤
(|f (a)| + |f (b)|) (|g (a)| + |g (b)|)
2
b−a
[M (f, g) + N (f, g)] .
=
2
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J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 27, 8 pp.
http://jipam.vu.edu.au/