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Article

A New Semi-Inner Product and pn-Angle in the Space of p-Summable Sequences

1
Department of Mathematics, Hasanuddin University, Makassar 90245, Indonesia
2
Department of Statistics, Hasanuddin University, Makassar 90245, Indonesia
3
Department of Mathematics, Pattimura University, Ambon 97233, Indonesia
*
Author to whom correspondence should be addressed.
Mathematics 2023, 11(14), 3139; https://doi.org/10.3390/math11143139
Submission received: 8 May 2023 / Revised: 16 June 2023 / Accepted: 14 July 2023 / Published: 16 July 2023

Abstract

:
In this paper, we propose a definition for a semi-inner product in the space of p-summable sequences equipped with an n-norm. Using this definition, we introduce the concepts of  p n -orthogonality and the  p n -angle between two vectors in the space of p-summable sequences. For the special case n = 1, these concepts are identical to previous studies. We also introduce the notion of the  p n -angle between one-dimensional subspaces and arbitrary-dimensional subspaces. The authors believe that the results obtained in this paper are very significant, especially in the theory of n-normed space in functional analysis.
MSC:
15A03; 46B20; 46B45; 46C50

1. Introduction

Let X be the vector space. A semi-inner product on X is a mapping  [ · , · ] : X 2 R , which satisfies the following properties:
(1)
[ x , x ] 0  for every  x X  and  [ x , x ] = 0  if and only if  x = 0 ;
(2)
[ α x , β y ] = α β · [ x , y ]  for every  x , y X  and  α , β R ;
(3)
[ x + y , z ] = [ x , z ] + [ y , z ]  for every  x , y , z X ;
(4)
| [ x , y ] | [ x , x ] 1 2 [ y , y ] 1 2  for every  x , y X .
The pair  ( X , [ · , · ] )  is called a semi-inner product space. In this space, we can define a norm, that is,  · = [ · , · ] 1 2  [1].
Let  ( X , · )  be a normed space. As it is known, not all normed spaces are inner product spaces, but we can define the semi-inner product. For instance, the space  p  for  1 p < , which is the space of all the p-summable sequences with norm  x p = n = 1 x n p 1 p , is not an inner product space, except for  p = 2 . Konca et al. [2] define a (weighted) inner product  · , · v  and a weighted norm on  p . In this space with usual norm, we may check that
[ x , y ] = y p 2 p j y j p 1 sgn ( y j ) x j , x : = x j , y : = y j l p
is a semi-inner product on  ( p , · p )  for  1 p <  [3].
Next, the mapping  g : X 2 R  defined by the formula
g ( x , y ) : = 1 2 x τ + ( x , y ) + τ ( x , y ) ,
with
τ ± ( x , y ) : = lim t ± 0 x + t y x t
is the semi-inner product on X if  g ( x , y )  is linear in y.
Using the concept of semi-inner product g, Miličić [4] introduced the notions of g-orthogonality, namely  x y y  if and only if  g ( x , y ) = 0  and the g-angle, namely  A g ( x , y ) : = arccos g ( x , y ) x y . Many researchers have studied the g-orthogonal and g-angle between two vectors and two subspaces in X; see, for example, [5,6,7,8]. In 2018, Nur et al. [9] developed the notion of the g-angle between two subspaces. If  V = span { v }  and  W = span { w 1 , , w m }  of X with  m 1 , then the g-angle between V and W is defined by  A g ( V , W )  with  cos 2 A g ( V , W ) = v W 2 v 2 . Here,  v W  denotes the g-orthogonal projection of v on W. Recently, Nur et al. [10] defined the standard n-norm using the (weighted) inner product and discussed the angle between two subspaces in the space of the p-summable.
In general, an n-norm on a real vector space X is a mapping  · , , · : X × × X R , which satisfies the following four conditions:
(1)
x 1 , , x n = 0  if and only if  x 1 , , x n  are linearly dependent;
(2)
x 1 , , x n  is invariant under permutation;
(3)
α x 1 , , x n = | α | x 1 , , x n  for every  x 1 , , x n X  and for every  α R ;
(4)
x 1 , , x n 1 , y + z x 1 , , x n 1 , y + x 1 , , x n 1 , z  for every  x , y , z X .
The pair  ( X , · , , · )  is called an n-normed space.
Geometrically,  x 1 , , x n  may be interpreted as the volume of the n-dimensional parallelepiped spanned by  x 1 , , x n . The theory of n-normed spaces for  n 2  was developed in the late 1960s [11,12,13]. Recent results can be found, for example, in [14,15,16]. On the space  p  for  1 p < , the following n-norm was defined by Gunawan in [17]:
x 1 , , x n p : = 1 n ! k 1 k n abs x 1 k 1 x 1 k n x n k 1 x n k n p 1 p .
The aim of this paper is to define a semi-inner product in an n-normed space  ( p , · , , · p )  with  1 p < . Using this result, we can introduce the  p n -orthogonal and the  p n -angle between two vectors. We also will discuss their properties. Moreover, we will define the  p n -angle between a one-dimensional subspace and an m-dimensional subspace in the n-normed space  ( p , · , , · p ) .

2. Main Results

2.1. Semi-Inner Product and  p n -Angle between Two Vectors

In this subsection, we shall begin with the new semi-inner product on  p  spaces equipped with an n-norm. Let  ( p , · , , · )  be an n-normed space and  { a 1 , , a n }  be a linearly independent set on  p . Now, we define the following mapping.
x p n : = { i 2 , , i n } { 1 , , n } x , a i 2 , , a i n p p 1 p ,
for every  x p . Next, we have the following proposition.
Proposition 1
([17]). The mapping  · p n  defines a norm on  p .
Example 1.
For  p = 1 , n = 3 . Let  { a 1 , a 2 , a 3 }  be the Schauder basis on  1 , that is,  a i = δ i j , i = 1 , 2 , 3  and  x = ( 2 , 1 , 3 , 0 , ) . We observe that
x 1 3 = x , a 1 , a 2 1 + x , a 1 , a 3 1 + x , a 2 , a 3 1 = 1 6 k 1 k 2 k 3 a b s x k 1 x k 2 x k 3 a 1 k 1 a 1 k 2 a 1 k 3 a 2 k 1 a 2 k 2 a 2 k 3 + 1 6 k 1 k 2 k 3 a b s x k 1 x k 2 x k 3 a 1 k 1 a 1 k 2 a 1 k 3 a 3 k 1 a 3 k 2 a 3 k 3 + 1 6 k 1 k 2 k 3 a b s y k 1 y k 2 y k 3 a 2 k 1 a 2 k 2 a 2 k 3 a 3 k 1 a 3 k 2 a 3 k 3 = 3 + 1 + 2 = 6 .
Using the norm  · p n  with  a i = ( a i j )  for  i = 1 , , n , we define a mapping  [ · , · ] p n  on the n-normed space  ( l p , · , , · p )  with  1 p <  by
[ x , y ] p n = x p n 2 p n ! { i 2 , , i n } { 1 , , n } k 1 k n abs x 1 k 1 x 1 k n a i 2 k 1 a i 2 k n a i n k 1 a i n k n p 1 × × sgn x k 1 x k n a i 2 k 1 a i 2 k n a i n k 1 a i n k n y k 1 y k n a i 2 k 1 a i 2 k n a i n k 1 a i n k n .
for every  x = ( x j ) , y = ( y j ) p .
Then we have the following result.
Theorem 1.
The mapping  [ x , y ] p n  in (4) defines a semi-inner product on  ( p , · , · p ) .
Proof. 
We will verify that  [ x , y ] p n  satisfies the properties (1–4) of the semi-inner product.
  • Observe that
    [ x , x ] p n = x p n 2 p { i 2 , , i n } { 1 , , n } x , a i 2 , , a i n p p = ( x p n ) 2 .
  • Observe that
    [ α x , β y ] p n = α x p n 2 p n ! { i 2 , , i n } { 1 , , n } k 1 k n abs α x 1 k 1 α x 1 k n a i 2 k 1 a i 2 k n a i n k 1 a i n k n p 1 × × sgn α x k 1 α x k n a i 2 k 1 a i 2 k n a i n k 1 a i n k n β y k 1 β y k n a i 2 k 1 a i 2 k n a i n k 1 a i n k n = α β [ x , y ] p n .
  • Using the properties of the determinant, we have
    [ x , y + y ] p n = x p n 2 p n ! { i 2 , , i n } { 1 , , n } k 1 k n abs x k 1 x k n a i 2 k 1 a i 2 k n a i n k 1 a i n k n p 1 × × sgn x k 1 x k n a i 2 k 1 a i 2 k n a i n k 1 a i n k n y k 1 + y k 1 y k n + y k n a i 2 k 1 a i 2 k n a i n k 1 a i n k n = [ x , y ] p n + [ x , y ] p n .
  • Observe that
    [ x , y ] p n x p n 2 p n ! { i 2 , , i n } { 1 , , n } k 1 k n abs x k 1 x k n a i 2 k 1 a i 2 k n a i n k 1 a i n k n p 1 × × abs y k 1 y k n a i 2 k 1 a i 2 k n a i n k 1 a i n k n . ( x p n ) 2 p 1 n ! { i 2 , , i n } { 1 , , n } k 1 k n abs x k 1 x k n a i 2 k 1 a i 2 k n a i n k 1 a i n k n p p 1 p × × 1 n ! { i 2 , , i n } { 1 , , n } k 1 k n abs y k 1 y k n a i 2 k 1 a i 2 k n a i n k 1 a i n k n p 1 p . = x p n y p n .
Therefore,  [ x , y ] p n  defines a semi-inner product on  p . □
Example 2.
For  p = 1 , n = 3 . Let  { a 1 , a 2 , a 3 }  be the Schauder basis on  1 , that is,  a i = δ i j , i = 1 , 2 , 3  and  · 1 3  in Proposition 1. If  x = ( 2 , 1 , 3 , 0 , )  and  y = ( 1 , 2 , 1 , 0 , )  in  1 , we have
x 1 3 = x , a 1 , a 2 1 + x , a 1 , a 3 1 + x , a 2 , a 3 1 = 1 6 k 1 k 2 k 3 a b s x k 1 x k 2 x k 3 a 1 k 1 a 1 k 2 a 1 k 3 a 2 k 1 a 2 k 2 a 2 k 3 + 1 6 k 1 k 2 k 3 a b s x k 1 x k 2 x k 3 a 1 k 1 a 1 k 2 a 1 k 3 a 3 k 1 a 3 k 2 a 3 k 3 + 1 6 k 1 k 2 k 3 a b s y k 1 y k 2 y k 3 a 2 k 1 a 2 k 2 a 2 k 3 a 3 k 1 a 3 k 2 a 3 k 3 = 3 + 1 + 2 = 6 .
and
y 1 3 = y , a 1 , a 2 1 + y , a 1 , a 3 1 + y , a 2 , a 3 1 = 1 6 k 1 k 2 k 3 a b s y k 1 y k 2 y k 3 a 1 k 1 a 1 k 2 a 1 k 3 a 2 k 1 a 2 k 2 a 2 k 3 + 1 6 k 1 k 2 k 3 a b s y k 1 y k 2 y k 3 a 1 k 1 a 1 k 2 a 1 k 3 a 3 k 1 a 3 k 2 a 3 k 3 + 1 6 k 1 k 2 k 3 a b s y k 1 y k 2 y k 3 a 2 k 1 a 2 k 2 a 2 k 3 a 3 k 1 a 3 k 2 a 3 k 3 = 1 + 2 + 1 = 4
As a consequence,
[ x , y ] 1 3 = x 1 3 3 ! { i 2 , i 3 } { 1 , 2 , 3 } k 1 k 2 k 3 s g n x k 1 x k 2 x k 3 a i 2 k 1 a i 2 k 2 a i 2 k 3 a i 3 k 1 a i 3 k 2 a i 3 k 3 y k 1 y k 2 y k 3 a i 2 k 1 a i 2 k 2 a i 2 k 3 a i 3 k 1 a i 3 k 2 a i 3 k 3 = 6 + 12 + 6 = 12
and
[ y , x ] 1 3 = y 1 3 3 ! { i 2 , i 3 } { 1 , 2 , 3 } k 1 k 2 k 3 s g n y k 1 y k 2 y k 3 a i 2 k 1 a i 2 k 2 a i 2 k 3 a i 3 k 1 a i 3 k 2 a i 3 k 3 x k 1 x k 2 x k 3 a i 2 k 1 a i 2 k 2 a i 2 k 3 a i 3 k 1 a i 3 k 2 a i 3 k 3 = 18 + 6 + 12 = 0 .
Remark 1.
The function  [ x , y ] 1 3  does not satisfy commutative property. In the example above, we observe that  [ x , y ] 1 3 = 12 [ y , x ] 1 3 = 0 .
Specifically for  p = 2 , we observe that
[ x , y ] 2 n = 1 n ! { i 2 , , i n } { 1 , , n } k 1 k n x k 1 x k n a i 2 k 1 a i 2 k 1 a i n k 1 a i n k 1 y k 1 y k n a i 2 k 1 a i 2 k 1 a i n k 1 a i n k 1 = [ y , x ] 2 n .
As consequence, we have following corollary:
Corollary 1.
The mapping  [ x , y ] 2 n  in (4) defines an inner product on  ( 2 , · , · 2 ) .
Example 3.
For  n = 2 . Let  { a 1 , a 2 }  be the Schauder basis on  2 , that is,  a i = δ i j , i = 1 , 2 . If  x = ( 2 , 1 , 0 , 0 , )  and  y = ( 1 , 2 , 0 , 0 , )  in  2 , we have
[ x , y ] 2 2 = 1 2 { i 2 } { 1 , 2 } k 1 k 2 x k 1 x k 2 a i 2 k 1 a i 2 k 2 y k 1 y k 2 a i 2 k 1 a i 2 k 2 = 1 2 k 1 k 2 x k 1 x k 2 a 1 k 1 a 1 k 2 y k 1 y k 2 a 1 k 1 a 1 k 2 + 1 2 k 1 k 2 x k 1 x k 2 a 2 k 1 a 2 k 2 y k 1 y k 2 a 2 k 1 a 2 k 2 = 4 .
By using the semi-inner product  [ x , y ] p n , we define orthogonality in  ( p , · , · n )  as follows:
Definition 1
( p n -orthogonality). Let  ( p , · , · n )  be the n-normed space. Vector x is  p n -orthogonal to vector y, and we write  x p n y  if and only if  [ x , y ] p n = 0 .
Example 4.
For  p = 1 , n = 3 . Let  { a 1 , a 2 , a 3 }  be the Schauder basis on  1 , that is,  a i = δ i j , i = 1 , 2 , 3 . If  x = ( 1 , 2 , 1 , 0 , )  and  y = ( 2 , 1 , 3 , 0 , )  in  1 , using Example 2, we have  [ x , y ] p n = 0 . Hence, x is  1 3 -orthogonal to y.
Next,  p n -orthogonality has the following properties.
Proposition 2.
The  p n -orthogonality satisfies the following properties:
(a) 
Nondegeneracy property: If  x p n x , then  x = 0 .
(b) 
Homogeneity property: If  x p n y , then  α x p n β y  for every  α , β R .
(c) 
Right additive property: If  x p n y  and  x p n z , then  x g ( y + z ) .
(d) 
Resolvability property: For every  x , y X  there is  γ R  such that  x p n ( γ x + y ) .
Proof. 
By using Theorem 1, the properties (a)−(c) are obviously true.
(d)
Let  x , y p . For case  x = 0 , the resolvability property is fulfilled. Next, for case  x 0 , choose  γ = [ x , y ] p n ( x p n ) 2 . Using Theorem 1, we have
[ x , γ x + y ] p n = 1 γ [ γ x , γ x + y ] p n = 1 γ ( γ x p n 2 + [ γ x , y ] p n ) = γ x p n 2 + [ x , y ] p n = 0 ,
as desired.
 □
Remark 2.
Note that for  n = 1 , the  p 1 -orthogonality coincides with the g-orthogonality in  p . Specifically for  p = 2 , the  2 n -orthogonality satisfies the symmetry and continuity property. Next, by using Remark 1, the  p n -orthogonality does not satisfy the symmetry property. The  p n -orthogonality also does not satisfy continuity property.
Example 5.
For  p = 1 . Take  x k = ( 1 k , 1 , 0 , ) x = ( 0 , 1 , 0 , ) , and  y = ( 1 , 1 , 0 , . . )  in  1 . Using inequality  x 1 x 1 n n x 1  in [17], we have  x k x  (in norm  · 1 n ). Next, we observe that  [ x k , y ] 1 n = 0  for every  k N , but  [ x , y ] 1 n 0 .
Using a semi-inner product in (4), we define the angle between two nonzero vectors x and y on  ( p , · , , · p )  as follows:
A p n ( x , y ) : = arccos [ y , x ] p n x p n y p n .
Note that  y p n x  if and only if  A p n ( x , y ) = 1 2 π .  We can observe that the angle  A p n ( x , y )  for  n = 1  is identical with the g-angle  A g ( x , y )  in [9].
Example 6.
Let  ( 1 , · , · , · 1 )  be 3 normed space and  { a 1 , a 2 , a 3 }  be the Schauder basis on  1 , that is,  a i = δ i j , i = 1 , 2 , 3  and  · 1 3  in Proposition 1. If  x = ( 1 , 2 , 1 , 0 , )  and  y = ( 2 , 1 , 3 , 0 , )  in  1 , we observe that
x 1 3 = x , a 1 , a 2 1 + x , a 1 , a 3 1 + x , a 2 , a 3 1 = 1 6 k 1 k 2 k 3 a b s x k 1 x k 2 x k 3 a 1 k 1 a 1 k 2 a 1 k 3 a 2 k 1 a 2 k 2 a 2 k 3 + 1 6 k 1 k 2 k 3 a b s x k 1 x k 2 x k 3 a 1 k 1 a 1 k 2 a 1 k 3 a 3 k 1 a 3 k 2 a 3 k 3 + 1 6 k 1 k 2 k 3 a b s x k 1 x k 2 x k 3 a 2 k 1 a 2 k 2 a 2 k 3 a 3 k 1 a 3 k 2 a 3 k 3 = 1 + 2 + 1 = 4
and
y 1 3 = y , a 1 , a 2 1 + y , a 1 , a 3 1 + y , a 2 , a 3 1 = 1 6 k 1 k 2 k 3 a b s y k 1 y k 2 y k 3 a 1 k 1 a 1 k 2 a 1 k 3 a 2 k 1 a 2 k 2 a 2 k 3 + 1 6 k 1 k 2 k 3 a b s y k 1 y k 2 y k 3 a 1 k 1 a 1 k 2 a 1 k 3 a 3 k 1 a 3 k 2 a 3 k 3 + 1 6 k 1 k 2 k 3 a b s y k 1 y k 2 y k 3 a 2 k 1 a 2 k 2 a 2 k 3 a 3 k 1 a 3 k 2 a 3 k 3 = 3 + 1 + 2 = 6 .
As a consequence,
[ y , x ] 1 3 = y 1 3 3 ! { i 2 , i 3 } { 1 , 2 , 3 } k 1 k 2 k 3 s g n y k 1 y k 2 y k 3 a i 2 k 1 a i 2 k 2 a i 2 k 3 a i 3 k 1 a i 3 k 2 a i 3 k 3 x k 1 x k 2 x k 3 a i 2 k 1 a i 2 k 2 a i 2 k 3 a i 3 k 1 a i 3 k 2 a i 3 k 3 = 6 + 12 + 6 = 12
Hence,  A 1 3 ( x , y ) = arccos ( 1 2 ) .
The angle  A p n ( · , · )  has the following properties.
Proposition 3.
Let  ( p , · , · n )  be the n-normed space. The angle  A p n  between two nonzero vectors x and y on  p  satisfies the following properties:
(a) 
If x and y are of the same direction, then  A p n ( x , y ) = 0 ; if x and y are of the opposite direction, then  A p n ( x , y ) = π  (part of parallelism property).
(b) 
A p n ( a x , b y ) = A p n ( x , y )  if  a b > 0 A p n ( a x , b y ) = π A p n ( x , y )  if  a b < 0  (homogeneity property).
(c) 
If  x n x  (in norm  · p n ), then  A p n ( x n , y ) A p n ( x , y )  (part of continuity property).
Proof. 
(a)
Let  y = k x  for an arbitrary nonzero vector x in X and  k R { 0 } .  We have
A p n ( x , y ) = arccos [ k x , x ] p n x p n k x p n = arccos k · [ x , x ] p n | k | x p n 2 = arccos k x p n 2 | k | x p n 2 .
Hence,  A p n ( x , y ) = arccos ( 1 ) = 0  for  k > 0  and  A p n ( x , y ) = arccos ( 1 ) = π  for  k < 0 .
(b)
Let  α  and  β R { 0 } .  Observe that
A p n ( α x , β y ) = arccos α β · [ y , x ] p n | α β | ( x p n · y p n ) .
If  α β > 0 , then  A p n ( α x , β y ) = arccos [ y , x ] p n x p n · y p n = A p n ( x , y ) .  Likewise, if  α β < 0 , then  A p n ( α x β b y ) = arccos [ y , x ] p n x p n · y p n = π A p n ( x , y ) .
(c)
If  x k x  (in norm  · p n ), then
| [ y , x k x ] p n | y p n x n x p n 0 .
Observe that  [ y , x k x ] p n = [ y , x k ] p n [ y , x ] p n . We have  [ y , x k ] p n [ y , x ] p n . Hence,
A p n ( x n , y ) A p n ( x , y ) ,
as desired.
 □
Remark 3.
Since the mapping  [ · , · ] p n  in general is not commutative, the angle  A p n ( · , · )  does not satisfy the symmetry property. For example, we can see Remark 1. Likewise, the g-angle does not satisfy the continuity property.
Example 7.
For  n = 1  and  p = 1 . Take  x k = ( 1 k , 1 , 0 , ) x = ( 0 , 1 , 0 , ) , and  y = ( 1 , 1 , 0 , )  in  1 . We observe that  cos A 1 1 ( x k , y ) = 0  for every  k N , but  cos A 1 1 ( x , y ) 0 .

2.2.  p n -Angle between Two Subspaces

In this section, we will discuss the  p n -angle between two subspaces in  ( p , · , , · p ) . In particular, we define the  p n -angle between a one-dimensional subspace and an m-dimensional subspace for  m 1 . Using  [ · , · ] p n  in (4), we have the Gram determinant  Γ p n  as follows.
Definition 2.
Let  ( p , · , , · p )  be an n-normed space and  W = s p a n { w 1 , , w m }  fo  1 m n  is subspace of  p . The Gram determinant  p n  of  { w 1 , , w m } , denoted by  Γ p n { w 1 , , w m }  is defined by
Γ p n { w 1 , , w m } : = [ w 1 , w 1 ] p n [ w 1 , w m ] p n [ w m , w 1 ] p n [ w m , w m ] p n .
Example 8.
For  p = 1  and  n = 2 . Let  { a 1 , a 2 }  be the Schauder basis on  1 , that is,  a i = δ i j , i = 1 , 2 , w 1 = ( 1 , 3 , 0 , )  and  w 2 = ( 3 , 1 , 0 , )  in  1 . We observe that  w 1 1 2 = 4  and  w 2 1 2 = 4 . Using the semi-inner product in (4), we have
[ w 1 , w 2 ] 1 2 = w 1 1 2 2 i = 1 2 k 1 k 2 s g n w 1 k 1 w 1 k 2 a i k 1 a i k 2 w 2 k 1 w 2 k 2 a i k 1 a i k 2 = 2 1 + 1 + 3 + 3 = 16
and
[ w 2 , w 1 ] 1 2 = w 2 1 2 2 i = 1 2 k 1 k 2 s g n w 2 k 1 w 2 k 2 a i k 1 a i k 2 w 1 k 1 w 1 k 2 a i k 1 a i k 2 = 2 3 + 3 + 1 + 1 = 16 .
Hence,  Γ 1 2 ( w 1 , w 2 ) = 0 .
We have the connection between the Gram determinant  p n  of  { w 1 , , w m }  for  1 m n  and the linearly independence set of  w 1 , , w m  as follows.
Theorem 2.
If  Γ p n ( w 1 , , w m ) 0 ,  then  w 1 , , w m  is a linearly independent set.
Proof. 
Suppose by contradiction that  w 1 , w m  is linearly dependent. Therefore, there is a i with  1 i m  so that  w i  is a linear combination of  w 1 , , w i 1 , w i + 1 , , w m . Using the properties of the determinant, we observe that the i-th column of  Γ p n  is a linear combination of the other columns. This implies  Γ p n ( w 1 , , w m ) = 0 , which is a contradiction. Hence,  w 1 , , w m  is a linearly independent set. □
Example 9.
Suppose that  p = 1 n = 2 . Let  { a 1 , a 2 }  be the Schauder basis on  1 , that is,  a i = δ i j , i = 1 , 2  and  · 1 3  in Proposition 1. If  w 1 = ( 1 , 0 , 0 , 0 , )  and  w 2 = ( 0 , 1 , 0 , 0 , )  in  1 . We observe that  w 1 1 2 = 1  and  w 2 1 2 = 1 .  As a consequence,
[ w 1 , w 2 ] 1 2 = w 1 1 2 2 i = 1 2 j k s g n w 1 j w 1 k a i j a i k w 2 j w 2 k a i j a i k = 0 .
In a similar way is obtained  [ w 2 , w 1 ] 1 2 = 0 . As a consequence,  Γ 1 2 ( w 1 , w 2 ) = 1 .  Hence, if  Γ p n ( w 1 , w 2 ) = 1 0 ,  then  w 1 , w 2  is a linearly independent set.
Remark 4.
The converse of the above theorem is not true. For instance, let  ( 1 , · , · 1 )  be a 2-normed space with  a 1 = ( 1 , 0 , 0 , 0 , )  and  a 2 = ( 0 , 1 , 0 , 0 , ) . Take  w 1 = ( 1 , 3 , 0 , )  and  w 2 = ( 3 , 1 , 0 , )  in  1 . We observe that  w 1 1 2 = 4  and  w 2 1 2 = 4 . Using the semi-inner product in (4), we have
[ w 1 , w 2 ] 1 2 = w 1 1 2 2 i = 1 2 k 1 k 2 s g n w 1 k 1 w 1 k 2 a i k 1 a i k 2 w 2 k 1 w 2 k 2 a i k 1 a i k 2 = 2 1 + 1 + 3 + 3 = 16
and
[ w 2 , w 1 ] 1 2 = w 2 1 2 2 i = 1 2 k 1 k 2 s g n w 2 k 1 w 2 k 2 a i k 1 a i k 2 w 1 k 1 w 1 k 2 a i k 1 a i k 2 = 2 3 + 3 + 1 + 1 = 16 .
Hence,  Γ 1 2 ( w 1 , w 2 ) = 0  but  w 1  and  w 2  are linearly independent.
Next, we can discuss the  p n -orthogonal projection and the  p n - orthogonal complement as follows:
Definition 3.
Let v be a vector of  p  and  W = s p a n { w 1 , , w m }  be a subspace of  p  with  Γ p n ( w 1 , , w m ) 0 . The  p n -orthogonal projection of v on W, denoted by  v W p n , is defined by
v W p n : = 1 Γ p n ( w 1 , , w m ) 0 w 1 w m [ w 1 , v ] p n [ w 1 , w 1 ] p n [ w 1 , w m ] p n [ w m , v ] p n [ w m , w 1 ] p n [ w m , w m ] p n ,
and its  p n -orthogonal complement  v W p n  is given by
v W p n : = 1 Γ p n ( w 1 , , w m ) v w 1 w m [ w 1 , v ] p n [ w 1 , w 1 ] p n [ w 1 , w m ] p n [ w m , v ] p n [ w m , w 1 ] p n [ w m , w m ] p n .
Example 10.
Suppose that  p = 1 n = 2  and  m = 2 . Let  { a 1 , a 2 }  be the Schauder basis on  1 , that is,  a i = δ i j , i = 1 , 2  and  · 1 3  in Proposition 1. If  v = ( 1 , 1 , 3 , 0 , ) , w 1 = ( 1 , 0 , 0 , 0 , )  and  w 2 = ( 0 , 1 , 0 , 0 , ) , then  Γ 1 2 ( w 1 , w 2 ) = 1  by Example 9. Therefore,
v W 1 2 = 1 Γ 1 2 ( w 1 , w 2 ) 0 w 1 w 2 [ w 1 , v ] 1 2 [ w 1 , w 1 ] 1 2 [ w 1 , w 2 ] 1 2 [ w 2 , v ] 1 2 [ w 2 , w 1 ] 1 2 [ w 2 , w 2 ] 1 2 = 0 w 1 w 2 1 1 0 1 0 1 = w 1 + w 2 = 1 , 1 , 0 ,
and
v W 1 2 = 1 Γ 1 2 ( w 1 , w 2 ) v w 1 w 2 [ w 1 , v ] 1 2 [ w 1 , w 1 ] 1 2 [ w 1 , w 2 ] 1 2 [ w 2 , v ] 1 2 [ w 2 , w 1 ] 1 2 [ w 2 , w 2 ] 1 2 = v w 1 w 2 1 1 0 1 0 1 = v w 1 w 2 = 0 , 0 , 3 , 0 , .
Using the properties of the determinant and the semi-inner product in (4), we obtain
[ w i , v W p n ] p n = 1 Γ p n ( w 1 , , w m ) [ w i , v ] p n [ w i , w 1 ] p n [ w i , w m ] p n [ w 1 , v ] p n [ w 1 , w 1 ] p n [ w 1 , w m ] p n [ w m , v ] p n [ w m , w 1 ] p n [ w m , w m ] p n = 0 .
Hence,  w i p n v W p n  for every  i = 1 , , n .
The definition of the  p n -angle between  V = span { v }  and  W = span { w 1 , , w m }  of  ( p , · , , · p )  is as follows.
Definition 4.
Let  ( p , · , , · p )  be an n-normed space for  1 p < . If  V = s p a n { v }  and  W = s p a n { w 1 , , w m }  are subspaces of  p  with  Γ p n ( w 1 , , w m ) 0  dan  m 1  then the  p n -angle between subspaces V and W is denoted by  A p n ( V , W )  with
cos 2 A p n ( V , W ) = [ v W p n , v ] p n 2 ( v p n ) 2 ( v W p n p n ) 2 .
where  v W p n  denotes the  p n -orthogonal projection of v on W.
Example 11.
Suppose that  p = 1 n = 2  and  m = 2  in Definition 4 Let  { a 1 , a 2 }  be the Schauder basis on  1 , that is,  a i = δ i j , i = 1 , 2  and  · 1 3  in Proposition 1. If  V = s p a n { v }  and  W = s p a n { w 1 , w 2 }  with  v = ( 1 , 1 , 3 , 0 , ) , w 1 = ( 1 , 0 , 0 , 0 , )  and  w 2 = ( 0 , 1 , 0 , 0 , )  in  1 , we observe that
v 1 2 = v , a 1 , a 2 1 + v , a 1 , a 2 1 = 1 2 k 1 k 2 a b s v k 1 v k 2 a 1 k 1 a 1 k 2 + 1 2 k 1 k 2 a b s v k 1 v k 2 a 2 k 1 a 2 k 2 = 4 + 4 = 8 .
By using Example 9, we have  Γ 1 2 ( w 1 , w 2 ) = 1 .  Therefore,
v W 1 2 = 1 Γ 1 2 ( w 1 , w 2 ) 0 w 1 w 2 [ w 1 , v ] 1 2 [ w 1 , w 1 ] 1 2 [ w 1 , w 2 ] 1 2 [ w 2 , v ] 1 2 [ w 2 , w 1 ] 1 2 [ w 2 , w 2 ] 1 2 = 0 w 1 w 2 1 1 0 1 0 1 = w 1 + w 2 = 1 , 1 , 0 , .
Moreover, we observe that  v W 1 2 1 2 = 2  and  [ v W 1 2 , v ] 1 2 = 4 . Hence,
cos 2 A 1 2 ( V , W ) = [ v W 1 2 , v ] 1 2 2 ( v 1 2 ) 2 ( v W 1 2 1 2 ) 2 = 1 16 .
Remark 5.
If  W = s p a n w , then the  p n -orthogonal projection of v on W is
v W p n = [ w , v ] p n w p n 2 w .
Using this  p n -orthogonal projection, the properties of the semi-inner product  p n  and Definition 4, we have the  p n -angle between subspaces V and W, that is,
cos 2 A p n ( V , W ) = [ w , v ] p n 2 ( v p n ) 2 ( w p n ) 2 .
Hence, we can see that the  p n -angle between the two vectors is also the  p n -angle between these two vectors in the subspace spanned by them. Next, by using Definition 3, we have  v = v W p n + v W p n . Therefore, Definition 4 may be rewritten as
cos 2 A p n ( V , W ) = [ v W p n , v W p n + v W p n ] p n 2 ( v p n ) 2 ( v W p n p n ) 2 = [ v W p n , v W p n ] p n + [ v W p n , v W p n + v W p n ] p n 2 ( v p n ) 2 ( v W p n p n ) 2 = [ v W p n , v W p n ] p n 2 ( v p n ) 2 ( v W p n p n ) 2 = ( v W p n p n ) 2 ( v p n ) 2 .
This tells us that the value of  cos A p n ( V , W )  is equal to the ratio between the norm  · p n  of the  p n -orthogonal projection of v on W and the norm  · p n  of v.

3. Further Results

Let X be a measured space with at least n disjoint subsets of positive measure. Our results also extend the n-normed space  ( L p ( X ) , · , , · L p )  for  1 p <  with a n-norm that was defined by Gunawan in [17]
f 1 , , f n L p = 1 n ! X X abs f 1 ( x 1 ) f n ( x n ) f n ( x 1 ) f n ( x n ) p d x 1 d x 2 .
Next, Ekariani et al. [18] defines a norm on  L p ( X )  by
f L p n : = { i 2 , , i n } { 1 , , n } f , a i 2 , , a i n L p 1 p ,
where  { a 1 , , a n }  is a linearly independent set in  L p ( X ) . Using the norm  · L p n  with  a i = ( a i j )  for  i = 1 , , n , we define a mapping  [ · , · ] L p n  with  1 p <  by
[ f , g ] L p n = f L p n 2 p n ! { i 2 , , i n } { 1 , , n } X X abs f ( x 1 ) f ( x n ) a i 2 ( x 1 ) a i 2 ( x n ) a i n ( x 1 ) a i n ( x n ) p 1 × × sgn f ( x 1 ) f ( x n ) a i 2 ( x 1 ) a i 2 ( x n ) a i n ( x 1 ) a i n ( x n ) g ( x 1 ) g ( x n ) a i 2 ( x 1 ) a i 2 ( x n ) a i n ( x 1 ) a i n ( x n ) d x 1 d x 2 .
and check  [ f , g ] L p n  defines a semi-inner product on  L p ( X ) . Next, the results are analogous to section main results.

4. Conclusions

In this article, we have introduced the semi-inner product in an n-normed space  ( p , · , , · p )  with  1 p < . We have introduced the  p n -orthogonal and the  p n -angle between two vectors. We have proven their properties. Moreover, we have introduced the notion of  p n -angle between one-dimensional subspaces and arbitrary-dimensional subspaces in the n-normed space  ( p , · , , · p ) .

Author Contributions

Conceptualization, M.N.; formal analysis, M.N.; funding acquisition, M.B. and M.N.; investigation and methodology, H.B. and A.I.; resources, M.B.; validation, A.I. and H.B.; writing review and editing, M.N. All authors have read and agreed to the published version of the manuscript.

Funding

This research is supported by PDUPT Program 2022 No. 956/UN4.22/PT.01.03/2022.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Conflicts of Interest

The authors declare no conflict of interest.

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Nur, M.; Bahri, M.; Islamiyati, A.; Batkunde, H. A New Semi-Inner Product and pn-Angle in the Space of p-Summable Sequences. Mathematics 2023, 11, 3139. https://doi.org/10.3390/math11143139

AMA Style

Nur M, Bahri M, Islamiyati A, Batkunde H. A New Semi-Inner Product and pn-Angle in the Space of p-Summable Sequences. Mathematics. 2023; 11(14):3139. https://doi.org/10.3390/math11143139

Chicago/Turabian Style

Nur, Muh, Mawardi Bahri, Anna Islamiyati, and Harmanus Batkunde. 2023. "A New Semi-Inner Product and pn-Angle in the Space of p-Summable Sequences" Mathematics 11, no. 14: 3139. https://doi.org/10.3390/math11143139

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