Volume 8 (2007), Issue 3, Article 99, 9 pp.
ON THE B-ANGLE AND g-ANGLE IN NORMED SPACES
PAVLE M. MILIČIĆ
FACULTY OF M ATHEMATICS
U NIVERSITY OF B ELGRADE
S ERBIA .
pmilicic@hotmail.com
Received 15 February, 2007; accepted 16 July, 2007
Communicated by S.S. Dragomir
A BSTRACT. It is known that in a strictly convex normed space, the B−orthogonality (Birkhoff
orthogonality) has the property, “B−orthogonality is unique to the left“. Using this property,
we introduce the definition of the so-called B−angle between two vectors, in a smooth and
uniformly convex space. Also, we define the so-called g−angle between two vectors. It is
demonstrated that the g−angle in a unilateral triangle, in a quasi-inner product space, is π/3.
The g−angle between a side and a diagonal, in a so-called g−quandrangle, is π/4.
Key words and phrases: Smooth normed spaces, quasi-inner product spaces, oriented (non-oriented) B−angle between two
vectors, oriented (non-oriented) g−angle between two vectors.
2000 Mathematics Subject Classification. 46B20, 46C15, 51K05.
Let X be a real smooth normed space of dimension greater than 1. It is well known that the
functional
kx + tyk − kxk
(1)
g(x, y) := kxk lim
(x, y ∈ X)
t→0
t
always exists (see [5]).
This functional is linear in the second argument and it has the following properties:
(2)
g(αx, y) = αg(x, y) (α ∈ R),
g(x, x) = kxk2 ,
|g(x, y)| ≤ kxk kyk .
Definition 1 ([10]). A normed space X is a quasi-inner product space (q.i.p. space) if the
equality
(3)
kx + yk4 − kx − yk4 = 8 kxk2 g(x, y) + kyk2 g(y, x)
holds for all x, y ∈ X.
The space of sequences l4 is a q.i.p. space, but l1 is not a q.i.p. space.
It is proved in [10] and [11] that a q.i.p. space X is very smooth, uniformly smooth, strictly
convex and, in the case of Banach spaces, reflexive.
The orthogonality of the vector x 6= 0 to the vector y 6= 0 in a normed space X may be
defined in several ways. We mention some kinds of orthogonality and their notations:
• x⊥B y ⇔ (∀λ ∈ R) kxk ≤ kx + λyk (Birkhoff orthogonality),
053-07
2
PAVLE M. M ILI ČI Ć
• x⊥J y ⇔ kx
− yk = kx + yk (Jamesorthogonality),
x
x
y y • x⊥S y ⇔ kxk
− kyk
+ kyk
= kxk
(Singer orthogonality).
In the papers [8], [6] and [9], by using the functional g, the following orthogonal relations
were introduced:
x⊥g y ⇔ g(x, y) = 0,
g
x ⊥ y ⇔ g(x, y) + g(y, x) = 0,
x ⊥ y ⇔ kxk2 g(x, y) + kyk2 g(y, x) = 0.
g
In [6, Theorem 2] the following assertion is proved: If X is smooth, then x⊥g y ⇔ x⊥B y.
In [11] we have proved the following assertion: If X is a q.i.p. space, then
x ⊥ y ⇔ x⊥J y
g
g
and x ⊥ y ⇔ x⊥S y.
If there exists an inner product h·, ·i in X, (i.p.), then it is easy to see that xρy ⇔ hx, yi = 0
holds for every
ρ∈
g
⊥B , ⊥J , ⊥S , ⊥g , ⊥, ⊥ .
g
For more on B−orthogonality and g−orthogonality, see the papers [1], [2], [13] and [14].
Some additional properties of this orthogonality are quoted below. Denote by P[x] y the set of
the best approximations of y with vectors from [x].
Theorem 1. Let X be a smooth and uniformly convex normed space, and let x, y ∈ X − {0}
be fixed linearly independent vectors. The following assertions are valid.
(1) There exists a unique a ∈ R such that
P[x] y = ax ⇔ g(y − ax, x) = 0 ⇔ ky − axk2 = g(y − ax, y),
sgn a = sgn g(y, x).
(2) If z ∈ span {x, y} and y⊥B x ∧ z⊥B x, then there exists λ ∈ R such that z = λy.
(3) If x⊥B y − αx ∧ x⊥B y − βx then α = β.
Proof.
(1) The proof can be found in [14].
(2) Since X is smooth, the equivalence
y⊥B x ∧ z⊥B x ⇔ g(y, x) = 0 ∧ g(z, x) = 0
holds.
Hence
x = αy + βz ⇒ g(y, αx + βz) = 0 ∧ g(z, αx + βz) = 0.
We get the system of equations
α kyk2 + βg(y, z) = 0
αg(z, x) + β kzk2 = 0.
This system has a non-trivial solution for α and β iff
g(y, z)g(z, y) = kyk2 kzk2 ⇔ |g(y, z)| |g (z, y)| = kyk2 kzk2 .
The last equation is not correct if |g(y, z)| < kyk kzk . So, |g(y, z)| = kyk kzk.Then by
Lemma 5 of [3], there exists λ ∈ R such that z = λy.
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 99, 9 pp.
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O N THE B- ANGLE AND g- ANGLE IN NORMED SPACES
3
(3) In accordance with 1) we have
g(x, y − αx) = 0 ∧ g(x, y − βx) = 0
⇔ g(x, y) − α kxk2 = 0 ∧ g(x, y) − β kxk2 = 0 ⇒ α = β.
From now on we assume that points 0, x, y are the vertices of the triangle (0, x, y) and points
0, x, y, x+y are the vertices of the parallelogram (0, x, y, x+y). The numbers kx − yk , kx + yk
are the lengths of diagonal of this parallelogram. If kxk = kyk, we say thatthis parallelogram
is a rhomb, and if x⊥ρ y, we say that this parallelogram is a ρ-rectangle, ρ ∈
⊥B , ⊥J , ⊥S , ⊥ .
g
From the next theorem, we see the similarity of q.i.p. spaces to inner-product spaces ( i.p.
spaces).
Theorem 2. Let X be a q.i.p. space. The following assertions are valid.
(1) The lengths of the diagonals in parallelogram (0, x, y, x + y) are equal if and only if the
parallelogram is a g−rectangle, i.e., x ⊥ y.
g
(2) The diagonals of the rhomb (0, x, y, x + y) are g−orthogonal, i.e., (x − y) ⊥(x + y).
g
(3) The parallelogram (0, x, y, x + y) is a g−quadrangle if and only if the lengths of its
diagonals are equal and the diagonals are g−orthogonal.
The proof of Theorem 2 can be found in [11].
The angle between two vectors x and y in a real normed space was introduced in [7] as
∠(x, y) := arccos
g(x, y) + g(y, x)
2 kxk kyk
(x, y ∈ X − {0}).
g
So, x ⊥ y ⇔ cos ∠(x, y) = 0.
In this paper we introduce several definitions of angles in a smooth normed space X.
Let us begin with the following observations. By (2), it is easily seen that we have
−1 ≤
(4)
kxk2 g(x, y) + kyk2 g(y, x)
≤1
kxk kyk (kxk2 + kyk2 )
(x, y ∈ X − {0}).
Hence we define new angle between the vectors x and y ,denoted as ∠(x, y).
g
Definition 2. The number
kxk2 g(x, y) + kyk2 g(y, x)
∠(x, y) := arccos
g
kxk kyk (kxk2 + kyk2 )
is called the g−angle between the vector x and the vector y.
It is very easy to see that :
∠(x, y) = ∠(y, x),
g
g
∠(λx, λy) = ∠(x, y),
g
g
x ⊥ y ⇔ cos ∠(x, y) = 0.
g
g
Theorem 3. Let X be a q.i.p. space. Then the following assertions hold.
(1) The g−angle over the diameter of a circle is g−right, i.e., if c is the circle in span {x, y},
centered at x+y
of radius kx−yk
, then z ∈ c ⇒ (x − z) ⊥(y − z), Figure 1.
2
2
g
(2) The centre of the circumscribed circumference about the g−right triangle is the centre
of the g−hypotenuse.
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 99, 9 pp.
http://jipam.vu.edu.au/
4
PAVLE M. M ILI ČI Ć
Proof.
(1) If z ∈ c, then z −
x+y 2
=
kx−yk
,
2
i.e. k2z − (x + y)k = kx − yk. Hence
(x − z)⊥J (y − z) ⇔ (x − z) ⊥(y − z),
g
because X is a q.i.p. space.
(2) Let c be the circle defined by the equation z −
x+y 2
, where x ⊥ y i.e.
= x−y
2
g
kx − yk = kx + yk. Then 0 ∈ c.
z
c
x
y
o
Figure 1:
In accordance with B−orthogonality, now we define the oriented B−angle between vectors
x and y.
Firstly, we have the following observation. Let P[x] y = ax, (a = a(x, y)).If kaxk ≤ kyk
for every x, y ∈ X − {0}, then X is an i.p. space (see (18.1) in [4]). So, in a normed (non
trivial) space, a B−catheti may be greater than the hypotenuse.
Lemma 4. Let X be a smooth and uniformly convex space and x, y ∈ X − {0} linearly
independent. Then there exists a unique τ = τ (x, y) such that kyk = ky − τ xk. If X is a q.i.p.
space and y is not B−orthogonal to x, then there exist unique p ∈ R such that (y − px) ⊥ px.
g
Proof. We consider the function
f (t) = ky − txk
(x, y ∈ X − {0} ,
t ∈ R).
Since X is smooth and uniformly convex, there exists a unique a = a(x, y) ∈ R such that
(5)
min f (t) = f (a) = ky − axk ,
t∈R
g(y − ax, x) = 0,
sgn a = sgn g(y, x).
(The vector ax is the best approximation of vector y with vectors of [x], i.e., P[x] y = ax (see
[14]).
On the other hand, the function f is continuous and convex on R and therefore there exists a
unique τ = τ (x, y) ∈ R (see Figure 2) such that
f (a) < kyk = ky − τ xk .
If X is a q.i.p. space, we get p = τ2 . In this case, we have kyk = ky − 2pxk, hence
k(y − px) + pxk = k(y − px) − pxk ,
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 99, 9 pp.
http://jipam.vu.edu.au/
O N THE B- ANGLE AND g- ANGLE IN NORMED SPACES
5
i.e.
(y − px)⊥J px ⇔ (y − px) ⊥ px.
g
In this case we shall write Pxg y = px. Clearly kyk = ky − 2pxk ⇒ kpxk ≤ kyk .
In (5) we have:
0 < a < τ ⇔ g(y, x) > 0,
τ < a < 0 ⇔ g(y, x) < 0 (Figure 2).
Hence, by kyk = ky − τ xk we get kτ xk − kyk ≤ kyk, i.e.
kτ xk
(6)
≤ kyk .
2
f
y
0
a
?
t
Figure 2:
Assume that g(y, x) > 0. If a < τ2 , then by (5) we have kaxk ≤ kτ2xk ≤ kyk . If a ≥ τ2 , then
τ − a ≤ τ2 and we have k(τ − a)xk ≤ kτ2xk ≤ kyk. Hence we get min {a, τ − a} ≤ τ2 .
Of course, if g(y, x) < 0, we get min {|a| , |τ − a|} ≤ |τ2| . Thus, we conclude that
kkxk
sgn g(y, x) ≤ 1
kyk
where k = min {|a| , |τ − a|} (k = k(x, y)).
(7)
−1 ≤
(x, y ∈ X − {0}),
Keeping in mind (7) and the characteristics of B− orthogonality, we introduce the following
definitions of the oriented B−angle between the vector x and the vector y.
Definition 3. Let X be smooth and uniformly convex. The number
kkxk
→
(8)
cosB (x, y) :=
sgn g(y, x),
kyk
k = min {|a| , |τ − a|} , (x, y ∈ X − {0})
is called the B−cosine of the oriented angle between x and y.
The number
∠B (−
x,→
y) := arccosB (−
x,→
y)
is the oriented B−angle between the vector x and the vector y.
Definition 4.
cosB (x, y) :=
q
|cosB (−
x,→
y) cosB (−
y,→
x)| sgn g(x, y) sgn g(y, x).
The number ∠B (x, y) := arccosB (x, y) is called the B−angle between the vector x and the
vector y.
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 99, 9 pp.
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6
PAVLE M. M ILI ČI Ć
If X is an i.p. space with i.p. h·, ·i , we have a =
→
→
hx,yi
.
kxkkyk
g(x,y)
kxk2
=
hx,yi
kxk2
=
g(y,x)
kxk2
(see [14]). So, in
this case cosB (x, y) =
Observe that cosB (x, y) is not symmetric in x and y, so, in the
triangle (0, x, y) we have 6 oriented B−angles.
|g(x,y)|
Since inequalities −1 ≤ kxk
≤ 1 are valid for every x, y ∈ X − {0} and y⊥B x ⇔
kyk
g(y, x) = 0 in a smooth space, we may ask whether cos (−
x,→
y) = g(y,x) for every x, y ∈ X.
B
The answer is no. Namely, in this case we have a(x, y) =
X − {0}, we get kaxk =
|g(y,x)|
kxk
kxk kyk
g(y,x)
kxk2
and hence, for every x, y ∈
≤ kyk. It follows from 18.1 of [4] that X is an i.p. space.
Theorem 5. Let X be a smooth and strictly convex space. Then,
−−→
(1) cosB (λx, y) = cosB (−
x,→
y) sgn λ (λ ∈ R − {0}),
−−→
−
→
(2) cosB (x, λy) = cosB (x, y) sgn λ (λ ∈ R − {0}).
Proof.
(1) Assume that P[x] y = ax, k = {|a| , |τ − a|} , kyk = ky − τ xk , P[λx] y = bλx.
Then bλ = a and min {|bλ| , |τ − bλ|} = min {|a| , |τ − a|} = k. Hence, by Definition
3, we have
−−→
min {|λb| , |τ − λb|} kxk
cosB (λx, y) =
sgn g(y, λx)
kyk
kkxk
=
sgn λg(y, x) = cosB (−
x,→
y) sgn λ.
kyk
(2) Let be P[x] y = ax kyk = ky − τ xk and kλyk = kλy − τλ xk. Then P[x] λy = λax
and by kλyk = kλy − λτ xk we get τλ = λτ and kλ = min {|λa| , |λτ − λa|} = |λ| k.
Thus
−−→
kkλ xk
cosB (x, λy) =
sgn g(λy, x)
kλyk
kkxk
=
sgn λg(y, x)
kyk
= cosB (−
x,→
y) sgn λ.
Theorem 6.
Let X be smooth, x, y ∈ X − {0} linearly independent, ky − xk = kyk. Then
∠B (x, y) = ∠B (−x, y − x), (Figure 3).
y
o
?
?
x
Figure 3:
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 99, 9 pp.
http://jipam.vu.edu.au/
O N THE B- ANGLE AND g- ANGLE IN NORMED SPACES
7
Proof. In a smooth space X (see [12]), for x, y ∈ X, we have
kxk (kxk − kx − yk) ≤ g(x, y) ≤ kxk (kx + yk − kxk).
(9)
Since ky − xk = kyk , we get g(y, x) > 0 and g(y − x, −x) > 0. Let P[x] y = ax and P[x] (y −
x) = b. Then: a > 0, b > 0 (see [14]), g(y − ax, x) = 0 and
g(y − x − bx, x) = 0 ⇔ g(y − (1 + b)x, x) = 0.
By virtue of 2) in Theorem 1, we get 1 + b = a such that P[x] (y − x) = (a − 1)x. From this and
Definition 3, we have
kkxk
−−−−−−→
sgn g(y − x, −x)
cosB (−x, y − x) =
ky − xk
min {a, 1 − a} kxk
=
kyk
−
→
= cosB (x, y).
We now assume that X is a s.i.p. space.
Analogous to Definition 3 and Definition 4, in a q.i.p. space, we will introduce a new definition of an oriented g−angle and the corresponding non oriented g−angle.
Definition 5. Let x 6= 0, y ∈ X and p =
τ
2
(see Lemma 4). Then
kpxk
cosg (−
x,→
y) :=
sgn(kxk2 g(x, y) + kyk2 g(y, x)).
kyk
−
→
−
→
The number ∠g (x, y) := arccosg (x, y) is the oriented g−angle between vector x and vector y.
We observe that, for all λ 6= 0,
y − px ⊥ px ⇒ λy − λpx ⊥ λpx,
g
i.e.,
Pxg y
=a⇒
g
Pλx
λy
(10)
g
= ax. Hence we have
−−−→
cosg (λx, λy) = cosg (−
x,→
y) sgn λ
(λ 6= 0).
Definition 6.
q
−−→
cosg (x, y) := cosg (x, y) cosg (−
y,→
x) sgn(kxk2 g(x, y) + kyk2 g(y, x)).
The number ∠g (x, y) := arccosg (x, y) is the non-oriented g−angle between x and y.
Clearly, in a q.i.p. space we have cosg (x, y) = cosg (y, x).
If X is an i.p. space with i.p. h·, ·i we have
(y − px) ⊥ px ⇔ ky − pxk2 g(y − px, px) + kpxk2 g(px, y − px) = 0
g
⇔ (ky − pxk2 + kpxk2 ) hx, y − pxi = 0
hx, yi
⇔ p=
kxk2
|hx, yi|
⇒ kpxk =
kxk
kpxk
hx, yi
2
2
⇒ cosg (−
x,→
y) =
.
2 sgn((kxk + kyk ) hx, yi) =
kxk kyk
kyk
Thus, Definition 5 and Definition 6 are correct.
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 99, 9 pp.
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8
PAVLE M. M ILI ČI Ć
Theorem 7. Let X be a q.i.p. space and kxk = kyk = kx − yk, i.e., let triangle (0, x, y) be
equilateral. Then
π
∠g (−
x,→
y) = ∠g (x, y) = ∠g (y, x) = .
3
Proof. At first, from equations kxk = kyk = ky − xk and inequalities (9) we get inequalities
0 < g(x, y) and 0 < g(y, x). By this sgn(kxk2 g(x, y) + kyk2 g(y, x)) = 1.
Let c be the circle centred at x2 with diameter kxk, (see Figure 4). Then y2 , x+y
∈ c. Ac2
Figure 4:
x−y
cording to 1), Theorem 3, we have (x − y2 ) ⊥ y2 and x+y
. That is, we have Pxg y = x2 and
2 ⊥ 2
g
g
P g x = y . By Definition 5 we get cos (−
x,→
y) = cos (y, x) = 1 . Hence, by Definition 6, we have
y
g
2
g
∠g (x, y) = π3 .
2
Theorem 8. Let (0, x, y, x + y) be a g−quadrangle, i.e. let kxk = kyk ∧ x ⊥ y. Then ∠g (x, x +
g
y) = π4 , i.e., the non-oriented g−angle between a diagonal and a side is π4 .
Figure 5:
Proof. We observe that in a q.i.p. space
sgn(kxk2 g(x, y) + kyk2 g(y, x)) = sgn(kx + yk − kx − yk)
and that
k2x + yk − kxk ≥ k2xk − kyk − kyk = 0.
Now consider Figure 5. Since Pxg (x + y) = x, we have
−−−−−→
cosg (x, x + y) =
kxk
kxk
sgn(k2x + yk − kyk) =
.
kx + yk
kx + yk
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 99, 9 pp.
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O N THE B- ANGLE AND g- ANGLE IN NORMED SPACES
9
Let s be the crossing point of the diagonal [0, x + y] and the diagonal [x, y]. Then, by Theorem
g
3, Px+y
x = s. It follows, by Definition 5, that
ksk
−−−−−→
cosg (x + y, x) =
sgn(ks + xk − ks − xk)
kxk
kx + yk
sgn(k2xk − kxk)
=
2 kxk
kx + yk
=
.
2 kxk
So, by Definition 6, we have
q
−−−−−→
−−−−−→
cosg (x, x + y) = cosg (x, x + y) cosg (x + y, x) sgn(k2x + yk − kyk)
r
√
1
2
=
=
.
2
2
Hence ∠g (x, x + y) = π4 .
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J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 99, 9 pp.
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