New Inequalities for the p-Angular Distance in Normed Spaces with Applications
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irk-123456789-1653982020-02-14T01:27:11Z New Inequalities for the p-Angular Distance in Normed Spaces with Applications Dragomir, S.S. Статті 2015 Article New Inequalities for the p-Angular Distance in Normed Spaces with Applications / S.S. Dragomir // Український математичний журнал. — 2015. — Т. 67, № 1. — С. 19–31. — Бібліогр.: 14 назв. — англ. 1027-3190 http://dspace.nbuv.gov.ua/handle/123456789/165398 517.5 en Український математичний журнал Інститут математики НАН України |
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Статті Статті Dragomir, S.S. New Inequalities for the p-Angular Distance in Normed Spaces with Applications Український математичний журнал |
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Dragomir, S.S. |
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New Inequalities for the p-Angular Distance in Normed Spaces with Applications |
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New Inequalities for the p-Angular Distance in Normed Spaces with Applications |
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New Inequalities for the p-Angular Distance in Normed Spaces with Applications |
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New Inequalities for the p-Angular Distance in Normed Spaces with Applications |
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New Inequalities for the p-Angular Distance in Normed Spaces with Applications |
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new inequalities for the p-angular distance in normed spaces with applications |
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Інститут математики НАН України |
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2015 |
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New Inequalities for the p-Angular Distance in Normed Spaces with Applications / S.S. Dragomir // Український математичний журнал. — 2015. — Т. 67, № 1. — С. 19–31. — Бібліогр.: 14 назв. — англ. |
| series |
Український математичний журнал |
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AT dragomirss newinequalitiesforthepangulardistanceinnormedspaceswithapplications |
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2025-07-14T18:24:02Z |
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2025-07-14T18:24:02Z |
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1837647730478940160 |
| fulltext |
UDC 517.5
S. S. Dragomir (College Eng. and Sci., Victoria Univ., Melbourne City, Australia;
School Comput. and Appl. Math., Univ. Witwatersrand, Johannesburg, South Africa)
NEW INEQUALITIES FOR THE p-ANGULAR DISTANCE
IN NORMED SPACES WITH APPLICATIONS
НОВI НЕРIВНОСТI ДЛЯ p-КУТОВОЇ ВIДСТАНI
В НОРМОВАНИХ ПРОСТОРАХ ТА ЇХ ЗАСТОСУВАННЯ
For nonzero vectors x and y in the normed linear space (X, ‖ · ‖) , we can define the p-angular distance by
αp[x, y] :=
∥∥‖x‖p−1x− ‖y‖p−1y
∥∥ .
We show (among other results) that, for p ≥ 2,
αp[x, y] ≤ p‖y − x‖
1∫
0
‖(1− t)x+ ty‖p−1 dt ≤
≤ p‖y − x‖
[
‖x‖p−1 + ‖y‖p−1
2
+
∥∥∥x+ y
2
∥∥∥p−1
]
≤
≤ p‖y − x‖‖x‖
p−1 + ‖y‖p−1
2
≤ p‖y − x‖ [max {‖x‖, ‖y‖}]p−1 ,
for any x, y ∈ X. This improves a result of Maligranda from [Simple norm inequalities // Amer. Math. Month. – 2006. –
113. – P. 256 – 260] who proved the inequality between the first and last terms in the estimation presented above. The
applications to functions f defined by power series in estimating a more general “distance” ‖f (‖x‖)x− f (‖y‖) y‖ for
some x, y ∈ X are also presented.
Для ненульових векторiв x та y в лiнiйному нормованому просторi (X, ‖ · ‖) можна визначити p-кутову вiдстань
таким чином:
αp[x, y] :=
∥∥‖x‖p−1x− ‖y‖p−1y
∥∥ .
У роботi, зокрема, показано, що
αp[x, y] ≤ p‖y − x‖
1∫
0
‖(1− t)x+ ty‖p−1 dt ≤
≤ p‖y − x‖
[
‖x‖p−1 + ‖y‖p−1
2
+
∥∥∥x+ y
2
∥∥∥p−1
]
≤
≤ p‖y − x‖‖x‖
p−1 + ‖y‖p−1
2
≤ p‖y − x‖ [max {‖x‖, ‖y‖}]p−1
для p ≥ 2 i будь-яких x, y ∈ X. Це покращує результат Малiгранди [Simple norm inequalities // Amer. Math. Month. –
2006. – 113. – P. 256 – 260], який встановив нерiвнiсть мiж першим та останнiм членами вказаної оцiнки. Також
наведено застосування для функцiй f, визначених степеневими рядами при оцiнюваннi бiльш загальної „вiдстанi”
‖f (‖x‖)x− f (‖y‖) y‖ для деяких x, y ∈ X.
1. Introduction. Following [3, p. 403] or [12], for nonzero vectors x and y in the normed linear
space (X, ‖ · ‖) we define the angular distance α[x, y] between x and y by
α[x, y] :=
∥∥∥∥ x
‖x‖
− y
‖y‖
∥∥∥∥ .
c© S. S. DRAGOMIR, 2015
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 1 19
20 S. S. DRAGOMIR
In 1958, Massera and Schäffer [12] (Lemma 5.1) showed that
α[x, y] ≤ 2‖x− y‖
max {‖x‖, ‖y‖}
, (1.1)
which is better than the Dunkl – Williams inequality [7]
α[x, y] ≤ 4‖x− y‖
‖x‖+ ‖y‖
. (1.2)
We notice that the Massera – Schäffer inequality was rediscovered by Gurarĭı in [8] (see also [13,
p. 516]).
In [11], Maligranda obtained the double inequality
‖x− y‖ − |‖x‖ − ‖y‖|
min {‖x‖, ‖y‖}
≤ α[x, y] ≤ ‖x− y‖+ |‖x‖ − ‖y‖|
max {‖x‖, ‖y‖}
. (1.3)
The second inequality in (1.3) is better than Massera – Schäffer’s inequality (1.1).
In the recent paper [11], L. Maligranda has also considered the p-angular distance
αp[x, y] :=
∥∥‖x‖p−1x− ‖y‖p−1y∥∥
between the vectors x and y in the normed linear space (X, ‖ · ‖) over the real or complex number
field K and showed that
αp[x, y] ≤ ‖x− y‖
(2− p) max {‖x‖p, ‖y‖p}
max {‖x‖ , ‖y‖}
if p ∈ (−∞, 0) and x, y 6= 0,
(2− p) 1
[max {‖x‖, ‖y‖}]1−p
if p ∈ [0, 1] and x, y 6= 0,
p [max {‖x‖, ‖y‖}]p−1 if p ∈ (1,∞).
(1.4)
The constants 2− p and p in (1.1) are best possible in the sense that they cannot be replaced by
smaller quantities.
As pointed out in [11], the inequality (1.1) for p ∈ [1,∞) is better than the Bourbaki inequality
obtained in 1965 [2, p. 257] (see also [13, p. 516]):
αp[x, y] ≤ 3p‖x− y‖ [‖x‖+ ‖y‖]p−1 , x, y ∈ X. (1.5)
The following results concerning upper bounds for the p-angular distance have been obtained by the
author in [5]:
αp[x, y] ≤
≤
‖x− y‖
[
max{‖x‖, ‖y‖}
]p−1
+
∣∣‖x‖p−1 − ‖y‖p−1∣∣min{‖x‖, ‖y‖} if p ∈ (1,∞),
‖x− y‖
[min {‖x‖ , ‖y‖}]1−p
+
∣∣‖x‖1−p − ‖y‖1−p∣∣min
{
‖x‖p
‖y‖1−p
,
‖y‖p
‖x‖1−p
}
if p ∈ [0, 1],
‖x− y‖
[min {‖x‖ , ‖y‖}]1−p
+
∣∣‖x‖1−p − ‖y‖1−p∣∣
max
{
‖x‖−p ‖y‖1−p , ‖y‖−p‖x‖1−p
} if p ∈ (−∞, 0),
(1.6)
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 1
NEW INEQUALITIES FOR THE p-ANGULAR DISTANCE IN NORMED SPACES WITH APPLICATIONS 21
and
αp[x, y] ≤
≤
‖x− y‖
[
min {‖x‖, ‖y‖}
]p−1
+
∣∣∣‖x‖p−1 − ‖y‖p−1∣∣∣max {‖x‖, ‖y‖} if p ∈ (1,∞),
‖x− y‖
[max {‖x‖ , ‖y‖}]1−p
+
∣∣‖x‖1−p − ‖y‖1−p∣∣max
{
‖x‖p
‖y‖1−p
,
‖y‖p
‖x‖1−p
}
if p ∈ [0, 1],
‖x− y‖
[max {‖x‖, ‖y‖}]1−p
+
∣∣‖x‖1−p − ‖y‖1−p∣∣
min {‖x‖−p‖y‖1−p, ‖y‖−p‖x‖1−p}
if p ∈ (−∞, 0),
(1.7)
for any two nonzero vectors x, y in the normed linear space (X, ‖ · ‖).
The upper bounds for αp[x, y] provided by (1.4), (1.6) and (1.7) have been compared in [5] to
conclude that some of the later ones are better in certain cases. The details are omitted here.
The following result which provides a lower bound for the p-angular distance was stated without
a proof by Gurarĭı in [8] (see also [13, p. 516]):
2−p‖x− y‖p ≤ αp[x, y], (1.8)
where p ∈ [1,∞) and x, y ∈ X. The proof of the inequality (1.8) is still an open question for the
author.
Finally, we recall the results of G. N. Hile from [4]:
αp[x, y] ≤ ‖x‖
p − ‖y‖p
‖x‖ − ‖y‖
‖x− y‖, (1.9)
for p ∈ [1,∞) and x, y ∈ X with ‖x‖ 6= ‖y‖, and
α−p−1[x, y] ≤ ‖x‖
p − ‖y‖p
‖x‖ − ‖y‖
‖x− y‖
‖x‖p ‖y‖p
, (1.10)
for p ∈ [1,∞) and x, y ∈ X \ {0} with ‖x‖ 6= ‖y‖.
2. Integral bounds for p-angular distance. The following result holds.
Theorem 2.1. Let (X; ‖ · ‖) be a normed linear space and p ≥ 1. Then for any x, y ∈ X we
have the inequality
αp[x, y] ≤ p‖y − x‖
1∫
0
‖(1− t)x+ ty‖p−1 dt. (2.1)
If the vectors x, y ∈ X are linearly independent and p < 1, then we have the inequality
αp[x, y] ≤ (2− p) ‖y − x‖
1∫
0
‖(1− t)x+ ty‖p−1 dt. (2.2)
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 1
22 S. S. DRAGOMIR
Proof. Assume that x 6= y. For p ≥ 2, consider the function fp : [0, 1] → [0,∞) given by
fp(t) = ‖(1− t)x+ ty‖p−1 . The function fp is convex on the interval [0, 1] for all p ≥ 2. Therefore
the lateral derivatives f ′p+ and f ′p− exist on each point of the interval [0, 1) and (0, 1], respectively, and
they are equal except a countably number of points in the interval (0, 1). The function fp is absolutely
continuos on [0, 1], the derivative f ′p exists almost everywhere on [0, 1] and (see, for instance, [14],
Chapter IV)
f ′p(t) = (p− 1) ‖(1− t)x+ ty‖p−2 τ+(−) ((1− t)x+ ty, y − x) (2.3)
almost everywhere on [0, 1], where the tangent functional τ+(−) is defined by
τ+(−) (u, v) :=
lims→0+(−)
‖u+ sv‖ − ‖u‖
s
if u 6= 0,
+ (−) ‖v‖ if u = 0.
(2.4)
Now, if we consider the vector valued function gp : [0, 1]→ X given by
gp(t) := fp(t) [(1− t)x+ ty]
then we observe that gp is strongly differentiable almost everywhere on [0, 1] and (see, for instance,
[1], Chapter 1)
g′p(t) = f ′p(t) [(1− t)x+ ty] + fp(t) (y − x) =
= (p− 1) ‖(1− t)x+ ty‖p−2 τ+(−) ((1− t)x+ ty, y − x)×
× [(1− t)x+ ty] + ‖(1− t)x+ ty‖p−1 (y − x)
for almost every t ∈ [0, 1].
Since for any u, v ∈ H with u 6= 0 we have∣∣τ+(−) (u, v)
∣∣ ≤ ‖v‖ ,
then ∥∥g′p(t)∥∥ ≤ (p− 1) ‖(1− t)x+ ty‖p−1
∣∣τ+(−) ((1− t)x+ ty, y − x)
∣∣+
+ ‖(1− t)x+ ty‖p−1 ‖y − x‖ ≤
≤ (p− 1) ‖(1− t)x+ ty‖p−1 ‖y − x‖+ ‖(1− t)x+ ty‖p−1 ‖y − x‖ =
= p ‖(1− t)x+ ty‖p−1 ‖y − x‖
for almost every t ∈ [0, 1].
By the norm inequality for the vector-valued integral we have (see, for instance, [1], Chapter 1)∥∥‖y‖p−1y − ‖x‖p−1x∥∥ = ‖gp (1)− gp (0)‖ =
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 1
NEW INEQUALITIES FOR THE p-ANGULAR DISTANCE IN NORMED SPACES WITH APPLICATIONS 23
=
∥∥∥∥∥∥
1∫
0
g′p(t)dt
∥∥∥∥∥∥ ≤
1∫
0
∥∥g′p(t)∥∥ dt ≤
≤ p‖y − x‖
1∫
0
‖(1− t)x+ ty‖p−1 dt
and the proof of (2.1) is complete.
Let p ∈ (1, 2). The function fp : [0, 1] → [0,∞) given by fp(t) = ‖(1− t)x+ ty‖p−1 is
absolutely continuous on [0, 1] and the equality (2.3) also holds almost everywhere on [0, 1]. The
above argument can then be extended to this case as well and the inequality (2.1) also holds.
If the vectors x, y ∈ X are linearly independent and p < 1, then ‖(1− t)x+ ty‖ > 0 for any
t ∈ [0, 1] and the function hp : [0, 1] → [0,∞) given by hp(t) = ‖(1− t)x+ ty‖p−1 is absolutely
continuous on [0, 1] and
h′p(t) = (p− 1) ‖(1− t)x+ ty‖p−2 τ+(−) ((1− t)x+ ty, y − x) (2.5)
almost everywhere on [0, 1].
If we consider the vector valued function mp : [0, 1]→ X given by
mp(t) := hp(t)
[
(1− t)x+ ty
]
,
then we observe that mp is strongly differentiable almost everywhere on [0, 1] and
m′p(t) = h′p(t) [(1− t)x+ ty] + hp(t) (y − x) =
= (p− 1) ‖(1− t)x+ ty‖p−2 τ+(−) ((1− t)x+ ty, y − x)×
× [(1− t)x+ ty] + ‖(1− t)x+ ty‖p−1 (y − x)
for almost every t ∈ [0, 1].
As above we have∥∥m′p(t)∥∥ ≤ (1− p) ‖(1− t)x+ ty‖p−1 ‖y − x‖+ ‖(1− t)x+ ty‖p−1 ‖y − x‖ =
= (2− p) ‖(1− t)x+ ty‖p−1 ‖y − x‖
for almost every t ∈ [0, 1], which implies the desired inequality (2.2).
Theorem 2.1 is proved.
Remark 2.1. If the vectors x and y are linearly dependent and y = λx with λ ∈ K, then the
p-angular distance between x and y reduces to
αp[x, y] = ‖x‖p
∣∣∣1− |λ|p−1 λ∣∣∣ = ‖x‖pβp [1, λ] .
The study of βp [1, λ] =
∣∣1− |λ∣∣p−1λ∣∣ with λ ∈ K may be done in a similar way, however the details
are omitted.
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 1
24 S. S. DRAGOMIR
Remark 2.2. If p ≥ 2, then the function fp : [0, 1]→ [0,∞) given by fp(t) = ‖(1− t)x+ ty‖p−1
is convex and by the Hermite – Hadamard type inequality for the convex function g : [a, b]→ R
1
b− a
b∫
a
g (s) ds ≤ 1
2
[
g (a) + g (b)
2
+ g
(
a+ b
2
)]
≤
≤ g (a) + g (b)
2
≤ max {g (a) , g (b)} (2.6)
we have the following chain of inequalities:
αp[x, y] ≤ p‖y − x‖
1∫
0
‖(1− t)x+ ty‖p−1 dt ≤
≤ p‖y − x‖
[
‖x‖p−1 + ‖y‖p−1
2
+
∥∥∥∥x+ y
2
∥∥∥∥p−1
]
≤
≤ p‖y − x‖‖x‖
p−1 + ‖y‖p−1
2
≤ p‖y − x‖ [max {‖x‖, ‖y‖}]p−1 , (2.7)
which provides a refinement of Maligranda’s inequality (1.4).
If p ≥ 1 and since, by the triangle inequality we have
‖(1− t)x+ ty‖ ≤ (1− t)‖x‖+ t‖y‖,
then
‖(1− t)x+ ty‖p−1 ≤ [(1− t)‖x‖+ t‖y‖]p−1
for any t ∈ [0, 1]. Integrating on [0, 1] we get
1∫
0
‖(1− t)x+ ty‖p−1 dt ≤
1∫
0
[(1− t)‖x‖+ t ‖y‖]p−1 dt =
1
p
‖y‖p − ‖x‖p
‖y‖ − ‖x‖
if ‖y‖ 6= ‖x‖, and by (2.1) we obtain the chain of inequalities
αp[x, y] ≤ p‖y − x‖
1∫
0
‖(1− t)x+ ty‖p−1 dt ≤ ‖y‖
p − ‖x‖p
‖y‖ − ‖x‖
‖y − x‖, (2.8)
which provides a refinement of Hile’s inequality (1.9).
For p ≥ 2, by the Hermite – Hadamard’s type inequalities (2.6) we also have
1
p
‖y‖p − ‖x‖p
‖y‖ − ‖x‖
=
1∫
0
[(1− t)‖x‖+ t‖y‖]p−1 dt ≤
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 1
NEW INEQUALITIES FOR THE p-ANGULAR DISTANCE IN NORMED SPACES WITH APPLICATIONS 25
≤ 1
2
[(
‖x‖+ ‖y‖
2
)p−1
+
‖x‖p−1 + ‖y‖p−1
2
]
≤
≤ ‖x‖
p−1 + ‖y‖p−1
2
≤ [max {‖x‖, ‖y‖}]p−1
which implies the following sequence of inequalities:
αp[x, y] ≤ p‖y − x‖
1∫
0
‖(1− t)x+ ty‖p−1 dt ≤
≤ ‖y‖
p − ‖x‖p
‖y‖ − ‖x‖
‖y − x‖ ≤
≤ 1
2
p‖y − x‖
[(
‖x‖+ ‖y‖
2
)p−1
+
‖x‖p−1 + ‖y‖p−1
2
]
≤
≤ p‖y − x‖‖x‖
p−1 + ‖y‖p−1
2
≤ p‖y − x‖[max {‖x‖, ‖y‖}]p−1 (2.9)
for ‖y‖ 6= ‖x‖ and p ≥ 2.
In particular, the inequality (2.9) shows that in the case p ≥ 2, Hile’s inequality (1.9) is better
than Maligranda’s inequality (1.4).
Remark 2.3. The case p = 0 is of interest, since by (2.2) we have the following upper bound
for the angular distance α[x, y]:
α[x, y] ≤ 2‖y − x‖
1∫
0
‖(1− t)x+ ty‖−1 dt, (2.10)
provided the vectors x and y are linearly independent.
Since for any t ∈ [0, 1]
‖(1− t)x+ ty‖ = ‖x− t (x− y)‖ ≥ |‖x‖ − t ‖x− y‖| ≥ ‖x‖ − t‖x− y‖ ≥ ‖x‖
and similarly
‖(1− t)x+ ty‖ ≥ ‖y‖,
then we have
‖(1− t)x+ ty‖ ≥ max {‖x‖, ‖y‖} ,
which implies that
1∫
0
‖(1− t)x+ ty‖−1 dt ≤ 1
max {‖x‖, ‖y‖}
. (2.11)
Therefore, we have the following refinement of the Massera – Schäffer’s inequality (1.1):
α[x, y] ≤ 2‖y − x‖
1∫
0
‖(1− t)x+ ty‖−1 dt ≤ 2‖y − x‖
max {‖x‖, ‖y‖}
.
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 1
26 S. S. DRAGOMIR
Remark 2.4. In [9], the authors introduced the concept of p-HH-norm on the Cartesian product
of two copies of a normed space, namely
‖(x, y)‖p−HH :=
1∫
0
‖(1− t)x+ ty‖p dt
1/p ,
where (x, y) ∈ X ×X := X2 and p ≥ 1. They showed that ‖ · ‖p−HH is a norm on X2 equivalent
with the usual p-norms
‖(x, y)‖p := (‖x‖p + ‖y‖p)1/p .
They also showed that completeness, reflexivity, smoothness, strict convexity etc. is inherited by X2
with this norm.
In [10] the authors proved the following interesting lower bound for ‖(x, y)‖p−HH :
(
‖x‖p + ‖y‖p
2(p+ 1)
)1/p
≤ ‖(x, y)‖p−HH (2.12)
for any (x, y) ∈ X2 and p ≥ 1.
Now, we observe that, by (2.1) we also have
αp+1[x, y] ≤ (p+ 1) ‖y − x‖ ‖(x, y)‖pp−HH (2.13)
for any (x, y) ∈ X2 and p ≥ 1.
For x 6= y this is equivalent with
(
‖‖x‖px− ‖y‖p y‖
(p+ 1)‖y − x‖
)1/p
≤ ‖(x, y)‖p−HH , (2.14)
where p ≥ 1.
It is natural to ask which lower bound from (2.12) and (2.14) for the p-HH-norm is better?
If we take X = C, ‖ · ‖ = |·| and p = 2, then by plotting the difference d given by
d(x, y) :=
∣∣∣|x|2 x− |y|2 y∣∣∣
3 |y − x|
1/2 −( |x|2 + |y|2
6
)1/2
for x, y ∈ R and x 6= y, we observe that d is nonnegative, showing that the new bound (2.14) is
better than (2.12). The plot is depicted in Figure as follows:
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NEW INEQUALITIES FOR THE p-ANGULAR DISTANCE IN NORMED SPACES WITH APPLICATIONS 27
8 S.S. DRAGOMIR
1;2
y
0.0-2-4
-2
-4
0 0
x
4
42
2
z 1.0
0.5
1.5
2.0
Figure 1: The variation of d in the box (x; y) 2 [4; 4] [4; 4] :
Problem 1. Is the inequality
(2.15)
kxkp + kykp
2
kkxkp x kykp yk
ky xk
true for any (x; y) 2 X2 with x 6= y and p 1?
3. Applications for Power Series
For power series f (z) =
P1
n=0 anz
n
with complex coe¢cients we can naturally
construct another power series which have as coe¢cients the absolute values of the
coe¢cient of the original series, namely, fa (z) :=
P1
n=0 janj z
n
. It is obvious that
this new power series have the same radius of convergence as the original series,
and that if all coe¢cients an 0; then fa = f .
As some natural examples that are useful for applications, we can point out that,
if
f (z) =
1X
n=1
(1)n
n
zn = ln
1
1 + z
; z 2 D (0; 1) ;(3.1)
g (z) =
1X
n=0
(1)n
(2n)!
z2n = cos z; z 2 C;
h (z) =
1X
n=0
(1)n
(2n+ 1)!
z2n+1 = sin z; z 2 C;
l (z) =
1X
n=0
(1)n zn =
1
1 + z
; z 2 D (0; 1) ;
The variation of d in the box (x, y) ∈ [−4, 4]× [−4, 4].
Problem 2.1. Is the inequality
‖x‖p + ‖y‖p
2
≤
∥∥‖x‖px− ‖y‖py∥∥
‖y − x‖
(2.15)
true for any (x, y) ∈ X2 with x 6= y and p ≥ 1?
3. Applications for power series. For power series f(z) =
∑∞
n=0
anz
n with complex
coefficients we can naturally construct another power series which have as coefficients the absolute
values of the coefficient of the original series, namely, fa(z) :=
∑∞
n=0
|an| zn. It is obvious that
this new power series have the same radius of convergence as the original series, and that if all
coefficients an ≥ 0, then fa = f .
As some natural examples that are useful for applications, we can point out that, if
f(z) =
∞∑
n=1
(−1)n
n
zn = ln
1
1 + z
, z ∈ D(0, 1),
g(z) =
∞∑
n=0
(−1)n
(2n)!
z2n = cos z, z ∈ C,
h(z) =
∞∑
n=0
(−1)n
(2n+ 1)!
z2n+1 = sin z, z ∈ C,
l(z) =
∞∑
n=0
(−1)nzn =
1
1 + z
, z ∈ D(0, 1),
(3.1)
then the corresponding functions constructed by the use of the absolute values of the coefficients are
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28 S. S. DRAGOMIR
fa(z) =
∞∑
n=1
1
n
zn = ln
1
1− z
, z ∈ D(0, 1),
ga(z) =
∞∑
n=0
1
(2n)!
z2n = cosh z, z ∈ C,
ha(z) =
∞∑
n=0
1
(2n+ 1)!
z2n+1 = sinh z, z ∈ C,
la(z) =
∞∑
n=0
zn =
1
1− z
, z ∈ D(0, 1).
(3.2)
Other important examples of functions as power series representations with nonnegative coefficients
are:
exp(z) =
∞∑
n=0
1
n!
zn, z ∈ C,
1
2
ln
(
1 + z
1− z
)
=
∞∑
n=1
1
2n− 1
z2n−1, z ∈ D(0, 1),
sin−1(z) =
∞∑
n=0
Γ
(
n+ 1
2
)
√
π(2n+ 1)n!
z2n+1, z ∈ D (0, 1) ,
tanh−1(z) =
∞∑
n=1
1
2n− 1
z2n−1, z ∈ D(0, 1),
2F1(α, β, γ, z) =
∞∑
n=0
Γ (n+ α) Γ(n+ β)Γ(γ)
n!Γ(α)Γ(β)Γ(n+ γ)
zn, α, β, γ > 0, z ∈ D(0, 1),
(3.3)
where Γ is Gamma function.
Theorem 3.1. Let f(z) =
∑∞
n=0
anz
n be a function defined by power series with complex
coefficients and convergent on the open disk D(0, R) ⊂ C, R > 0. If (X; ‖ · ‖) is a normed linear
space and x, y ∈ X with ‖x‖, ‖y‖ < R, then
‖f (‖x‖)x− f (‖y‖) y‖ ≤
≤ ‖y − x‖
1∫
0
[
fa (‖(1− t)x+ ty‖) + ‖(1− t)x+ ty‖ f ′a (‖(1− t)x+ ty‖)
]
dt. (3.4)
Proof. From the inequality (2.1) for p = n+ 1, n a natural number with n ≥ 1, we have
‖‖x‖nx− ‖y‖ny‖ ≤ (n+ 1)‖y − x‖
1∫
0
‖(1− t)x+ ty‖n dt. (3.5)
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NEW INEQUALITIES FOR THE p-ANGULAR DISTANCE IN NORMED SPACES WITH APPLICATIONS 29
We notice that the above inequality also holds for n = 0, reducing to an equality.
Let m ≥ 1. Then we have, by the generalized triangle inequality and by (3.5), that∥∥∥∥∥
(
m∑
n=0
an‖x‖n
)
x−
(
m∑
n=0
an‖y‖n
)
y
∥∥∥∥∥ ≤
≤
m∑
n=0
|an| ‖‖x‖n x− ‖y‖ny‖ ≤
≤ ‖y − x‖
m∑
n=0
(n+ 1) |an|
1∫
0
‖(1− t)x+ ty‖n dt =
= ‖y − x‖
1∫
0
(
m∑
n=0
(n+ 1) |an| ‖(1− t)x+ ty‖n
)
dt. (3.6)
Since ‖x‖, ‖y‖ < R the series
∞∑
n=0
an‖x‖n,
∞∑
n=0
an‖y‖n
and
∞∑
n=0
(n+ 1) |an| ‖(1− t)x+ ty‖n
are convergent.
Moreover, we obtain
∞∑
n=0
an‖x‖n = f (‖x‖) ,
∞∑
n=0
an‖y‖n = f (‖y‖)
and
∞∑
n=0
(n+ 1) |an| ‖(1− t)x+ ty‖n =
=
∞∑
n=0
|an| ‖(1− t)x+ ty‖n +
∞∑
n=0
n |an| ‖(1− t)x+ ty‖n =
= fa (‖(1− t)x+ ty‖) + ‖(1− t)x+ ty‖ f ′a (‖(1− t)x+ ty‖)
for any ‖x‖, ‖y‖ < R.
Taking the limit over m→∞ in (3.6) we get the desired result (3.4).
Theorem 3.1 is proved.
Remark 3.1. If we take f(z) := exp(z) =
∑∞
n=0
1
n!
zn then we have from (3.4) the following
inequality:
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 1
30 S. S. DRAGOMIR
‖exp (‖x‖)x− exp (‖y‖)y‖ ≤
≤ ‖y − x‖
1∫
0
exp (‖(1− t)x+ ty‖) (1 + ‖(1− t)x+ ty‖) dt (3.7)
for any x, y ∈ X.
If we apply the inequality (3.4) for the functions f(z) :=
1
1− z
=
∑∞
n=0
zn and f(z) :=
:=
1
1 + z
=
∑∞
n=0
(−1)nzn, then we have
∥∥∥∥ x
1± ‖x‖
− y
1± ‖y‖
∥∥∥∥ ≤ ‖y − x‖
1∫
0
dt
(1− ‖(1− t)x+ ty‖)2
(3.8)
for any x, y ∈ X with ‖x‖, ‖y‖ < 1.
Utilising the Hile’s inequality, we can also prove the following divided difference inequality:
Proposition 3.1. Let f(z) =
∑∞
n=0
anz
n be a function defined by power series with complex
coefficients and convergent on the open disk D(0, R) ⊂ C, R > 0. If (X; ‖ · ‖) is a normed linear
space and x, y ∈ X with ‖x‖, ‖y‖ < R and ‖x‖ 6= ‖y‖, then
‖f (‖x‖)x− f (‖y‖) y‖
‖y − x‖
≤ fa (‖x‖) ‖x‖ − fa (‖y‖) ‖y‖
‖x‖ − ‖y‖
. (3.9)
Proof. The proof goes along the line of the one from Theorem 3.1 by utilizing Hile’s inequal-
ity (1.9)
‖‖x‖nx− ‖y‖ny‖
‖y − x‖
≤ ‖x‖
n+1 − ‖y‖n+1
‖x‖ − ‖y‖
for any n a natural number.
Remark 3.2. If we write the inequality (3.9) for the exponential function, then we get
‖exp (‖x‖)x− exp (‖y‖) y‖
‖y − x‖
≤ exp (‖x‖) ‖x‖ − exp (‖y‖) ‖y‖
‖x‖ − ‖y‖
for any x, y ∈ X with ‖x‖ 6= ‖y‖ .
If we apply the inequality (3.9) for the functions f(z) :=
1
1− z
and f(z) :=
1
1 + z
, then we get∥∥∥∥ x
1± ‖x‖
− y
1± ‖y‖
∥∥∥∥ ≤ ‖y − x‖
(1− ‖x‖) (1− ‖y‖)
for any x, y ∈ X with ‖x‖ 6= ‖y‖ and ‖x‖, ‖y‖ < 1.
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Received 20.11.13
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