Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 5, Article 193, 2006
A GENERALIZATION OF HÖLDER AND MINKOWSKI INEQUALITIES
YILMAZ YILMAZ, M. KEMAL ÖZDEMİR, AND İHSAN SOLAK
D EPARTMENT OF M ATHMEMATICS
İ NÖNÜ U NIVERSITY
44280 M ALATYA / TURKEY
yyilmaz@inonu.edu.tr
kozdemir@inonu.edu.tr
isolak@inonu.edu.tr
Received 29 June, 2006; accepted 09 November, 2006
Communicated by K. Nikodem
A BSTRACT. In this work, we give a generalization of Hölder and Minkowski inequalities to
normal sequence algebras with absolutely monotone seminorm. Our main result is Theorem 2.1
and Theorem 2.2 which state these extensions. Taking F = `1 and k·kF = k·k1 in both these
theorems, we obtain classical versions of these inequalities. Also, using these generalizations we
construct the vector-valued sequence space F (X, λ, p) as a paranormed space which is a most
general form of the space c0 (X, λ, p) investigated in [6].
Key words and phrases: Inequalities, Hölder, Minkowski, Sequence algebra, Vector-valued sequence space.
2000 Mathematics Subject Classification. Primary 26D15, 47A30; Secondary 46A45.
1. I NTRODUCTION
Hölder and Minkowski inequalities have been used in several areas of mathematics, especially in functional analysis. These inequalities have been generalized in various directions.
The purpose of this paper is to give some extensions of the classical Hölder and Minkowski
inequalities. We discovered that the classical versions are only a type of these extensions in `1
which is a normal sequence algebra with absolutely monotone seminorm k·k1 .
We now recall some definitions and facts.
A Frechet space is a complete total paranormed space. If H is an Hausdorff space then an
FH-space is a vector subspace X of H which is a Frechet space and is continuously embedded
in H, that is, the topology of X is larger than the relative topology of H. Moreover if X is
a normed FH-space then it is called a BH-space. An FH-space with H = w, the space of all
complex sequences, is called an FK-space, so a BK-space is a normed FK-space. We know
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
177-06
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Y ILMAZ Y ILMAZ , M. K EMAL Ö ZDEMIR , AND İ HSAN S OLAK
that `∞ , c, c0 and `p (1 ≤ p < ∞) are BK-spaces. The following relation exists among these
sequence spaces:
`p ⊂ c0 ⊂ c ⊂ `∞ .
A basis for a topological P
vector space X is a sequence (bn ) such that every
P x ∈ X has a
unique representation x =
tn bn . This is equivalent to the fact that x − m
n=1 tn bn → 0
(m → ∞) in the vector topology of X. For example, c0 and `p have (en ) as a basis (en is a
sequence x where xn =
P1, xk = 0 for n 6= k). If X has a basis (bn ) the functionals ln , given by
ln (x) = tn when x =
tn bn , are linear. They are called the coordinate functionals and (bn ) is
called a Schauder basis if each ln ∈ X 0 , the continuous dual of X. A basis of a Frechet space
must be a Schauder basis [7]. An FK-space X is said to have AK, or be an AK-space, if X ⊃ φ
(the space of all finite sequences)
and (en ) is a basis for X, i.e. forPeach x, x[n] → x, where
P
n
x[n] , the nth section of x is k=1 xk ek ; otherwise expressed, x =
xk ek for all x ∈ X [8].
The spaces c0 and `p are AK-spaces but c and `∞ are not. We say that a sequence space F is an
AK-BK space if it is both a BK and an AK-space.
An algebra A over a field K is a vector space A over K such that for each ordered pair of
elements x, y ∈ A a unique product xy ∈ A is defined with the properties
(1) (xy)z = x(yz)
(2a) x(y + z) = xy + xz
(2b) (x + y)z = xz + yz
(3) α(xy) = (αx)y = x(αy)
for all x, y, z ∈ A and scalars α [4].
If K = R (real field) or C (complex field) then A is said to be a real or complex algebra,
respectively.
Let F be a sequence space and x, y be arbitrary members of F . F is called a sequence algebra
if it is closed under the multiplication defined by xy = (xi yi ), i ≥ 1, and is called normal or
solid if y ∈ F whenever |yi | ≤ |xi |, for some x ∈ F . If F is both a normal and sequence
algebra then it is called a normal sequence algebra. For example, c is a sequence algebra but
not normal. w, `∞ , c0 and `p (0 < p < ∞) are normal sequence algebras.
A paranorm p on a normal sequence space F is said to be absolutely monotone if p(x) ≤ p(y)
for x, y ∈ F with |xi | ≤ |yi | for each i [3].
The norm kxk∞ = sup |xk | which makes the spaces `∞ , c, c0 a BK-space, is absolutely
P∞
p 1/p
over `p is absolutely monotone. Also,
monotone. For p ≥ 1, the norm kxk
=
(
k=1 |xk | )
P∞
p
for 0 < p < 1, the p-norm kxkp = k=1 |xk | over `p is absolutely monotone.
An Orlicz function is a function M : [0, ∞) −→ [0, ∞) which is continuous, non-decreasing
and convex with M (0) = 0, M (x) > 0 for x > 0 and M (x) → ∞ as x → ∞. We say
that the Orlicz function M satisfies the ∆´-condition if there exist positive constants a and u
such that M (xy) ≤ aM (x) M (y) (x, y ≥ u). By means of M , Lindenstrauss and Tzafriri [2]
constructed the sequence space
`M =
x∈w:
X
M
|xk |
ρ
< ∞ for some ρ > 0
n
o
P |xk | with the norm kxkM = inf ρ > 0 :
M ρ ≤ 1 . This norm is absolutely monotone
and `M is normal since M is non-decreasing. Also if M satisfies the ∆´-condition then `M is a
sequence algebra.
Now we give a useful inequality from classical analysis.
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Lemma 1.1. Let f be a function such that f 00 (x) ≥ 0 for x > 0. Then for 0 < a < x < b
Z x
f (x) − f (a)
1
=
f 0 (t)dt
x−a
x−a a
≤ f 0 (x)
Z b
1
≤
f 0 (t)dt
b−x x
f (b) − f (x)
=
.
b−x
Hence
b−x
x−a
f (x) ≤
f (a) +
f (b)
b−a
b−a
[7].
Apply this to the function f (x) = − ln x with θ = (b − x)/(b − a). Then for all a, b positive
numbers and 0 ≤ θ ≤ 1, we have
aθ b1−θ ≤ aθ + (1 − θ)b.
(1.1)
Next, we give a lemma associated with the theorems in Section 2.
Lemma 1.2.
a) Let F be a normal sequence algebra, u = (un ) ∈ F and p ≥ 1. Then up = (upn ) ∈ F .
b) If F is a normal sequence space, k·kF is an absolutely monotone seminorm on F and
u = (un ) ∈ F then |u| = (|un |) ∈ F and k|u|kF = kukF .
Proof. a) We define two sequences a = (an ) and b = (bn ) such that
(
(
un if |un | ≥ 1
0 if |un | ≥ 1
an =
and bn =
.
0 if |un | < 1
un if |un | < 1
So un = an + bn and upn = apn + bpn . Obviously, a, b ∈ F . Since p < [p] + 1, we have
|an |p ≤ |an |[p]+1 ,
where [p] denotes the integer part of p. Since F is a sequence algebra, the sequence a[p]+1 is a
member of F by induction, and so ap ∈ F . Furthermore, since F is normal and |bn |p ≤ |bn |,
we have bp ∈ F . Hence up ∈ F .
b) It is a direct consequence of normality and absolute monotonicity.
2. G ENERALIZATIONS
Our main results are the following theorems which state the extensions of Hölder and Minkowski
inequalities. Taking F = `1 and k·kF = k·k1 in both Theorem 2.1 and Theorem 2.2, we get
classical versions of these inequalities. Moreover, if we change the choices of F and k·kF then
we can obtain many different inequalities corresponding to these generalizations. Therefore,
the following results are quite productive.
Theorem 2.1. Let F be a sequence algebra and k·kF be an absolutely monotone seminorm on
F . Suppose u = (un ) , v = (vn ) ∈ F . Then
1/p
1/p
kuvkF ≤ kup kF kv q kF ,
where p > 1 and
1
p
+
1
q
= 1.
J. Inequal. Pure and Appl. Math., 7(5) Art. 193, 2006
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Y ILMAZ Y ILMAZ , M. K EMAL Ö ZDEMIR , AND İ HSAN S OLAK
Proof. Assume that xn = |un |p and yn = |vn |q . It is immediate from Lemma 1.2(a) that
x = (xn ) and y = (yn ) are members of F . Let M = kxkF and N = kykF . Then it follows
from inequality (1.1) that for each n,
x θ y 1−θ
xn
yn
n
n
≤θ
+ (1 − θ)
M
N
M
N
as 0 ≤ θ ≤ 1. Because k·kF is an absolutely monotone seminorm we write
x
xn θ yn 1−θ yn n
≤
θ
+
(1
−
θ)
.
M
N
M
N F
F
Hence
1
M θ N 1−θ
θ 1−θ xn yn ≤ 1,
F
so that
θ 1−θ xn yn ≤ k(xn )kθ k(yn )k1−θ .
F
F
F
Setting θ = 1/p, we get
1/p 1/q xn yn ≤ k(xn )k1/p k(yn )k1/q ,
F
F
F
and putting xn = |un |p and yn = |vn |q , we obtain
1/p
1/q
k(|un vn |)kF ≤ k(|un |p )kF k(|vn |q )kF .
So, it follows from Lemma 1.2(b) that
kuvkF ≤ kup k1/p
kv q k1/q
.
F
F
Theorem 2.2. Let F be a normal sequence algebra and k·kF be an absolutely monotone seminorm on F . Then for every u = (un ) , v = (vn ) ∈ F and p ≥ 1,
1/p
1/p
1/p
k(u + v)p kF ≤ kup kF + kv p kF ,
where (u + v)p = ((un + vn )p ).
Proof. For p = 1, it is obvious.
Let p > 1. Proceeding with the manner of the proof in the classical version, we write
(u + v)p = u (u + v)p−1 + v (u + v)p−1 .
It follows from Theorem 2.1 that
1/q
1/q
1/p 1/p k(u + v)p kF ≤ kup kF (u + v)(p−1)q + kv p kF (u + v)(p−1)q F
F
1/q
1/p
1/p = kup kF + kv p kF (u + v)(p−1)q ,
F
1/q
(p−1)q p 1/q
1
1
where p + q = 1. Hence, dividing the first and last terms by (u + v)
= k(u + v) kF ,
F
we obtain the inequality.
Example 2.1. Taking F = `∞ and k·kF = k·k∞ in both Theorem 2.1 and Theorem 2.2, we
obtain the inequalities
1q
p1 p
q
· sup|vn |
sup|un vn | ≤ sup|un |
n
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and
p
sup|un + vn |
n
p1
p1 p1
p
p
≤ sup|un |
+ sup|vn |
,
n
n
where u, v ∈ `∞ and p1 + 1q = 1 as p > 1. Hence, in fact, these elementary inequalities are
extended Hölder and Minkowski inequalities respectively.
Example 2.2. Now put F = `M and k·kF = k·kM in Theorem 2.1 and Theorem 2.2, where M
satisfies the ∆´-condition. In this case, we write the inequalities
X |xk yk | inf ρ > 0 :
M
≤1
ρ
p1
X |xk |p ≤ inf ρ > 0 :
M
≤1
ρ
1q
X |yk |q · inf ρ > 0 :
M
≤1
ρ
and
p1
X |xk + yk |p inf ρ > 0 :
M
≤1
ρ
p1
X |xk |p ≤ inf ρ > 0 :
M
≤1
ρ
p1
X |yk |p + inf ρ > 0 :
M
≤1
ρ
as Hölder and Minkowski inequalities respectively.
3. A N A PPLICATION
Now let us introduce the class F (X, λ, p) of vector-valued sequence spaces which includes
the space c0 (X, λ, p) investigated in [6] with some linear topological properties. Theorem 2.2
makes it possible to improve some topological properties of the space F (X, λ, p).
Let F be an AK-BK normal sequence algebra such that the norm k·kF of F is absolutely
monotone and X be a seminormed space. Also suppose that λ = (λk ) is a non-zero complex
sequence and p = (pk ) is a sequence of strictly positive real numbers. Define the vector-valued
sequence class
F (X, λ, p) = {x ∈ s(X) : ([q (λk xk )]pk ) ∈ F } ,
where q is the seminorm of X and s(X) is the most general X-termed sequence space. F (X, λ, p)
becomes a linear space under natural co-ordinatewise vector operations if and only if p ∈ `∞
(see Lascarides [1]). Taking F = c0 and X as a Banach space we get the space c0 (X, λ, p) in
[6].
Lemma 3.1. Let 0 < tk ≤ 1. If ak and bk are complex numbers then we have
|ak + bk |tk ≤ |ak |tk + |bk |tk
[5, p.5].
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Y ILMAZ Y ILMAZ , M. K EMAL Ö ZDEMIR , AND İ HSAN S OLAK
Lemma 3.2. Let (X, q) be a seminormed space, and F a normal AK-BK space with an absolutely monotone norm k·kF . Suppose p = (pk ) is a bounded sequence of positive real numbers.
Then the map
n
X
x
en : [0, ∞) → [0, ∞) ; x
en (u) = [uq (λk xk )]pk ek k=1
F
defined by means of x = (xk ) ∈ F (X, λ, p) and a positive integer n, is continuous, where (ek )
is a unit vector basis of F .
Proof. Since the norm function is continuous it is sufficient to show that the mappings defined
by
gk : [0, ∞) → F, gk (u) = [uqk (λk xk )]pk ek
are continuous. Let ui → 0 (i → ∞), then
gk (ui ) → (0, 0, . . .)
(i → ∞)
for each k. Hence, each gk is sequential continuous (it is equivalent to continuity here).
Theorem 3.3. Define the function g : F (X, λ, p) −→ R by
1/M
g (x) = k([q (λk xk )]pk )kF
,
where M = max (1, sup pn ). Then g is a paranorm on F (X, λ, p).
Proof. It is obvious that g(θ) = 0 and g(−x) = g(x). From the absolute monotonicity of k·kF ,
Lemma 3.1 and Theorem 2.2, we get
M 1/M
pk /M
g(x + y) = [q (λk xk + λk yk )]
F
M 1/M
pk /M
pk /M
≤ [q (λk xk )]
+ [q (λk yk )]
F
pk
≤ k([q (λk xk )]
1/M
)kF
pk
+ k([q (λk yk )]
1/M
)kF
= g(x) + g(y)
for x, y ∈ F (X, λ, p).
To show the continuity of scalar multiplication assume that (µn ) is a sequence of scalars
such that |µn − µ| → 0 (n → ∞) and g (xn − x) → 0 (n → ∞) for an arbitrary sequence
(xn ) ⊂ F (X, λ, p). We shall show that
g (µn xn − µx) → 0 (n → ∞) .
Say τn = |µn − µ| and we get
1/M
g (µn xn − µx) = k([q (λk (µn xnk − µxk ))]pk )kF
1/M
= k([q (λk (µn xnk − µn xk + µn xk − µxk ))]pk )kF
1/M
≤ k([|µn | q (λk (xnk − xk ))) + τn q (λk xk )]pk )kF
n
oM 1/M
pk /M
pk /M
≤ [A (k, n)]
,
+ [B (k, n)]
F
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where A (k, n) = Rq (λk (xnk − xk )), B (k, n) = τn q (λk xk ) and R = max {1, sup |µn |}. Again
by Theorem 2.2 we can write
1/M
1/M
g (µn xn − µx) ≤ k(A (k, n))kF + k(B (k, n))kF
pk 1/M
A
1/M
≤ R
+ k(B (k, n))kF
R
F
1/M
= Rg(xn − x) + k(B (k, n))kF
.
1/M
Since g (xn − x) → 0 (n → ∞) we must show that k(B (k, n))kF → 0 (n → ∞). We can
find a positive integer n0 such that 0 ≤ τn ≤ 1 for n ≥ n0 . Say tk = [q (λk xk )]pk . Since
t = (tk ) ∈ F and F is an AK-space, we get
m
∞
X
X
tk ek = [q (λk xk )]pk ek → 0
t −
k=1
F
k=m+1
(m → ∞) ,
F
where (ek ) is a unit vector basis of F . Therefore, for every ε > 0 there exists a positive integer
m0 such that
1
∞
X
M
ε
pk
[q (λk xk )] ek < .
2
k=m +1
0
F
For n ≥ n0 write [(τn q (λk xk ))]pk ≤ [f (q (λk xk ))]pk for each k. On the other hand, we can
write
1
1
∞
∞
X
M
M
X
ε
[τn q (λk xk )]pk ek ≤ [q (λk xk )]pk ek < .
2
k=m +1
k=m +1
0
0
F
F
Now, from Lemma 3.2, the function
m
0
X
x
em0 (u) = [(uq (λk xk ))]pk ek k=1
F
is continuous. Hence, there exists a δ (0 < δ < 1) such that
x
em0 (u) ≤
ε M
2
,
for 0 < u < δ. Also we can find a number ∆ such that τn < δ for n > ∆. So for n > ∆ we
have
m
0
X
ε
1/M
pk
(e
xm0 (τn ))
=
[τn q (λk xk )] ek < ,
2
k=1
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Y ILMAZ Y ILMAZ , M. K EMAL Ö ZDEMIR , AND İ HSAN S OLAK
and eventually we get
1/M
k([τn q (λk xk )]pk )kF
1
∞
X
M
pk
=
[τn q (λk xk )] ek k=1
F
1
m
∞
0
M
X
X
pk
pk
=
[τn q (λk xk )] ek +
[τn q (λk xk )] ek k=1
k=m0 +1
F
m
1
∞
1
0
X
M X
M
≤
[τn q (λk xk )]pk ek + [τn q (λk xk )]pk ek k=1
k=m0 +1
F
F
ε ε
< + = ε.
2 2
1/M
This shows that k(B (k, n))kF
→ 0 (n → ∞).
Theorem 3.4. Let (X, q) be a complete seminormed space. Then F (X, λ, p) is complete with
the paranorm g. If X is a Banach space then F (X, λ, p) is an FK-space, in particular, an
AK-space.
Proof. Let (xn ) be a Cauchy sequence in F (X, λ, p). Therefore
1/M
pk
g(xn − xm ) = k([q (λk (xnk − xm
k ))] )kF
→0
(m, n → ∞) ,
also, since F is an FK-space, for each k
pk
→0
[q (λk (xnk − xm
k ))]
(m, n → ∞)
and so |λk | q (xnk − xm
k ) → 0 (m, n → ∞). Because of the completeness of X, there exists an
xk ∈ X such that q (xnk − xk ) → 0 (n → ∞) for each k. Define the sequence x = (xk ) with
these points. Now we can determine a sequence ηk ∈ c0 (0 < ηkn ≤ 1) such that
(3.1)
[|λk | q (xnk − xk )]pk ≤ ηkn [q (λk xnk )]pk
since q (xnk − xk ) → 0. On the other hand,
[q (λk xk )]pk ≤ D {[q (λk (xnk − xk ))]pk + [q (λk xnk )]pk } ,
where D = max 1, 2H−1 ; H = sup pk . From (3.1) we have
[q (λk xk )]pk ≤ D (1 + ηkn ) [q (λk xnk )]pk
≤ 2D [q (λk xnk )]pk .
So we get x ∈ F (X, λ, p). Now, for each ε > 0 there exist n0 (ε) such that
[g(xn − xm )]M < εM
for n, m > n0 .
Also, we may write from the AK-property of F that
m
∞
0
X
X
p
p
n
m
n
m
k
k
[q (λk (xk − xk ))] ek ≤ [q (xk − xk )] ek k=1
k=1
F
F
n
m M
= g(x − x ) .
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Letting m → ∞ we have
m
m
0
0
X
X
pk
pk
n
m
n
[q
(λ
(x
−
x
))]
e
→
[q
(λ
(x
−
x
))]
e
k
k
k
k
k
k
k
k
k=1
F
<ε
k=1
M
F
for n > n0 .
Since (ek ) is a Schauder basis for F ,
m
0
X
pk
pk
n
n
−
x
))]
e
[q
(λ
(x
k
k → k([q (λk (xk − xk ))] )kF
k
k
k=1
F
< εM
as m0 → ∞.
Then we get g(xn − x) < ε for n > n0 so g(xn − x) → 0 (n → ∞).
For the rest of the theorem; we can say immediately that F (X, λ, p) is a Frechet space,
because X is a Banach space. Also, the projections
P̂k : F (X, λ, p) −→ X;
P̂k (x) = xk
are continuous since Pk = |λk | q ◦ P̂k for each k. Where Pk ’s are coordinate mappings on F
and they are continuous since F is an FK-space.
Let x[n] be the nth section of an element x of F (X, λ, p). We must prove that x[n] → x in
F (X, λ, p) for each x ∈ F (X, λ, p). Indeed,
g x − x[n] = g (0, 0, . . . , 0, xn+1 , xn+2 , . . .)
∞
X
=
[q (λk xk )]pk ek → 0
k=n+1
since F is an AK-space. Hence F (X, λ, p) is an AK-space.
F
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AND
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J. Inequal. Pure and Appl. Math., 7(5) Art. 193, 2006
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