On approximate Birkhoff orthogonality in normed spaceskafedra.schoolsite.org.ua/BOOK/CHMIELINSKI.pdf · 2019. 7. 4. · On approximate Birkho orthogonality in normed spaces Jacek

On approximate Birkhoff orthogonality in normed spaces

Jacek Chmielinski

Instytut MatematykiUniwersytet Pedagogiczny w Krakowie

Banach Spaces and their ApplicationsLviv (Ukraine), June 26-29, 2019

J. Chmielinski (Krakow, Poland) Approximate Birkhoff orthogonality Lviv 2019 1 / 28

Introduction — inner product space

(X , 〈·|·〉) — inner product space; x⊥y ⇔ 〈x |y〉 = 0.

Approximate orthogonality (ε-orthogonality with ε ∈ [0, 1)):

x⊥ε y ⇔ | 〈x |y〉 | ≤ ε ‖x‖ ‖y‖, x , y ∈ X .

Observation

x⊥ε y ⇔ ∃ z ∈ X : x⊥z , ‖z − y‖ ≤ ε‖y‖.

Indeed, if x⊥ε y take z = − 〈x |y〉‖x‖2 x + y (z = y in case x = 0).

Conversely, assuming x⊥z and ‖z − y‖ ≤ ε‖y‖,

| 〈x |y〉 | = | 〈x |y − z〉 | ≤ ‖x‖ ‖y − z‖ ≤ ε‖x‖ ‖y‖.





x⊥ε y ⇔ | 〈x |y〉 | ≤ ε ‖x‖ ‖y‖, x , y ∈ X .

Observation

x⊥ε y ⇔ ∃ z ∈ X : x⊥z , ‖z − y‖ ≤ ε‖y‖.



| 〈x |y〉 | = | 〈x |y − z〉 | ≤ ‖x‖ ‖y − z‖ ≤ ε‖x‖ ‖y‖.





x⊥ε y ⇔ | 〈x |y〉 | ≤ ε ‖x‖ ‖y‖, x , y ∈ X .

Observation

x⊥ε y ⇔ ∃ z ∈ X : x⊥z , ‖z − y‖ ≤ ε‖y‖.



| 〈x |y〉 | = | 〈x |y − z〉 | ≤ ‖x‖ ‖y − z‖ ≤ ε‖x‖ ‖y‖.





x⊥ε y ⇔ | 〈x |y〉 | ≤ ε ‖x‖ ‖y‖, x , y ∈ X .

Observation

x⊥ε y ⇔ ∃ z ∈ X : x⊥z , ‖z − y‖ ≤ ε‖y‖.



| 〈x |y〉 | = | 〈x |y − z〉 | ≤ ‖x‖ ‖y − z‖ ≤ ε‖x‖ ‖y‖.





x⊥ε y ⇔ | 〈x |y〉 | ≤ ε ‖x‖ ‖y‖, x , y ∈ X .

Observation

x⊥ε y ⇔ ∃ z ∈ X : x⊥z , ‖z − y‖ ≤ ε‖y‖.



| 〈x |y〉 | = | 〈x |y − z〉 | ≤ ‖x‖ ‖y − z‖ ≤ ε‖x‖ ‖y‖.





x⊥ε y ⇔ | 〈x |y〉 | ≤ ε ‖x‖ ‖y‖, x , y ∈ X .

Observation

x⊥ε y ⇔ ∃ z ∈ X : x⊥z , ‖z − y‖ ≤ ε‖y‖.



| 〈x |y〉 | = | 〈x |y − z〉 | ≤ ‖x‖ ‖y − z‖ ≤ ε‖x‖ ‖y‖.


Birkhoff orthogonality


Birkhoff orthogonalityG. Birkhoff, Orthogonality in linear metric spaces. Duke Math. J., 1 (1935), 169–172.

(X , ‖ · ‖) a real normed space.

x⊥By :⇐⇒ ∀λ ∈ R : ‖x + λy‖ ≥ ‖x‖.

xy

x+λy

Figure: R2 with the maximum norm; x⊥By




x⊥By :⇐⇒ ∀λ ∈ R : ‖x + λy‖ ≥ ‖x‖.

xy

x+λy





x⊥By :⇐⇒ ∀λ ∈ R : ‖x + λy‖ ≥ ‖x‖.

xy

x+λy





x⊥By :⇐⇒ ∀λ ∈ R : ‖x + λy‖ ≥ ‖x‖.

xy

x+λy



Approximate Birkhoff orthogonality



For ε ∈ [0, 1) we consider an ε-Birkhoff orthogonality ⊥εB.

J. Chmielinski, On an ε-Birkhoff orthogonality, J. Inequal. Pure andAppl. Math. 6 (2005), Art. 79.

x⊥εBy :⇐⇒ ∀λ ∈ K : ‖x + λy‖2 ≥ ‖x‖2 − 2ε‖x‖ ‖λy‖.

J. Chmielinski, T. Stypu la, P. Wojcik, Approximate orthogonality innormed spaces and its applications, Linear Algebra and itsApplications 531 (2017), 305–317.

x⊥εBy ⇐⇒ ∃z ∈ Lin{x , y} : x⊥Bz , ‖z − y‖ ≤ ε‖y‖.





x⊥εBy :⇐⇒ ∀λ ∈ K : ‖x + λy‖2 ≥ ‖x‖2 − 2ε‖x‖ ‖λy‖.







x⊥εBy :⇐⇒ ∀λ ∈ K : ‖x + λy‖2 ≥ ‖x‖2 − 2ε‖x‖ ‖λy‖.







x⊥εBy :⇐⇒ ∀λ ∈ K : ‖x + λy‖2 ≥ ‖x‖2 − 2ε‖x‖ ‖λy‖.







x⊥εBy :⇐⇒ ∀λ ∈ K : ‖x + λy‖2 ≥ ‖x‖2 − 2ε‖x‖ ‖λy‖.





1−1

1

−1

x

y

z

Figure: R2 with l∞-l1-norm

x 6⊥By , x⊥z , ‖z − y‖ ≤ ε⇒ x⊥εBy .



1−1

1

−1

x

y

z


x 6⊥By , x⊥z , ‖z − y‖ ≤ ε⇒ x⊥εBy .



1−1

1

−1

x

y

z


x 6⊥By

, x⊥z , ‖z − y‖ ≤ ε⇒ x⊥εBy .



1−1

1

−1

x

y

z


x 6⊥By , x⊥z

, ‖z − y‖ ≤ ε⇒ x⊥εBy .



1−1

1

−1

x

y

z


x 6⊥By , x⊥z , ‖z − y‖ ≤ ε

⇒ x⊥εBy .



1−1

1

−1

x

y

z


x 6⊥By , x⊥z , ‖z − y‖ ≤ ε⇒ x⊥εBy .


For 0 6= x ∈ X we consider the class of its supporting functionals:

J(x) = {ϕ ∈ X ∗ : ‖ϕ‖ = 1, ϕ(x) = ‖x‖ }.

Theorem

Let X be a real normed space, x , y ∈ X and ε ∈ [0, 1). Then

x⊥εBy ⇔ ∃ϕ ∈ J(x) : |ϕ(y)| ≤ ε‖y‖.

In particular (James),

x⊥By ⇔ ∃ϕ ∈ J(x) : ϕ(y) = 0.



J(x) = {ϕ ∈ X ∗ : ‖ϕ‖ = 1, ϕ(x) = ‖x‖ }.

Theorem


x⊥εBy ⇔ ∃ϕ ∈ J(x) : |ϕ(y)| ≤ ε‖y‖.


x⊥By ⇔ ∃ϕ ∈ J(x) : ϕ(y) = 0.



J(x) = {ϕ ∈ X ∗ : ‖ϕ‖ = 1, ϕ(x) = ‖x‖ }.

Theorem


x⊥εBy ⇔ ∃ϕ ∈ J(x) : |ϕ(y)| ≤ ε‖y‖.


x⊥By ⇔ ∃ϕ ∈ J(x) : ϕ(y) = 0.


Applications


Orthogonality of operators on a Hilbert space

H – Hilbert space; L(H) – the space of linear bounded operators on H.For T ∈ L(H):

MT := {x ∈ SH : ‖Tx‖ = ‖T‖}.

R. Bhatia, P. Semrl, Orthogonality of matrices and some distanceproblems, Linear Algebra Appl. 287 (1999), 77-85.

Theorem (Bhatia-Semrl)

Let H be a Hilbert space and let T ,S ∈ L(H). Then, the followingconditions are equivalent:

(1) T⊥BS ;

(2) ∃ (xn)∞n=1 ⊂ SH : ‖Txn‖ → ‖T‖, 〈Txn|Sxn〉 → 0 (n→∞).

Moreover, if dimH <∞ and T ,S ∈ L(H), then each of the aboveconditions is equivalent to:

(3) ∃ x0 ∈ SH : ‖Tx0‖ = ‖T‖, Tx0⊥Sx0.




MT := {x ∈ SH : ‖Tx‖ = ‖T‖}.




(1) T⊥BS ;



(3) ∃ x0 ∈ SH : ‖Tx0‖ = ‖T‖, Tx0⊥Sx0.




MT := {x ∈ SH : ‖Tx‖ = ‖T‖}.




(1) T⊥BS ;



(3) ∃ x0 ∈ SH : ‖Tx0‖ = ‖T‖, Tx0⊥Sx0.




MT := {x ∈ SH : ‖Tx‖ = ‖T‖}.



Let H be a Hilbert space and let T , S ∈ L(H). Then, the followingconditions are equivalent:

(1) T⊥BS ;



(3) ∃ x0 ∈ SH : ‖Tx0‖ = ‖T‖, Tx0⊥Sx0.


Approximate orthogonality in L(H)

H – Hilbert space; T ∈ L(H); MT := {x ∈ SH : ‖Tx‖ = ‖T‖}.

Theorem

For T ,S ∈ L(H) the following conditions are equivalent:

(1) T⊥εBS ;

(2) ∃ (xn)∞n=1 ⊂ SH : ‖Txn‖ → ‖T‖, limn→∞ | 〈Txn|Sxn〉 | ≤ ε‖T‖ ‖S‖.

Moreover, if dimH <∞, then the above conditions are equivalent to

(3) ∃ x0 ∈ SH : ‖Tx0‖ = ‖T‖, | 〈Tx0|Sx0〉 | ≤ ε‖T‖ ‖S‖.

If dimH <∞ or if T is compact and if additionally MT ⊂ MS , the abovethree conditions are equivalent also to

(4) ∃ x0 ∈ SH : ‖Tx0‖ = ‖T‖, Tx0⊥εSx0.




Theorem

For T ,S ∈ L(H) the following conditions are equivalent:

(1) T⊥εBS ;



(3) ∃ x0 ∈ SH : ‖Tx0‖ = ‖T‖, | 〈Tx0|Sx0〉 | ≤ ε‖T‖ ‖S‖.


(4) ∃ x0 ∈ SH : ‖Tx0‖ = ‖T‖, Tx0⊥εSx0.




Theorem

For T , S ∈ L(H) the following conditions are equivalent:

(1) T⊥εBS ;



(3) ∃ x0 ∈ SH : ‖Tx0‖ = ‖T‖, | 〈Tx0|Sx0〉 | ≤ ε‖T‖ ‖S‖.


(4) ∃ x0 ∈ SH : ‖Tx0‖ = ‖T‖, Tx0⊥εSx0.




Theorem


(1) T⊥εBS ;



(3) ∃ x0 ∈ SH : ‖Tx0‖ = ‖T‖, | 〈Tx0|Sx0〉 | ≤ ε‖T‖ ‖S‖.


(4) ∃ x0 ∈ SH : ‖Tx0‖ = ‖T‖, Tx0⊥εSx0.




Theorem


(1) T⊥εBS ;



(3) ∃ x0 ∈ SH : ‖Tx0‖ = ‖T‖, | 〈Tx0|Sx0〉 | ≤ ε‖T‖ ‖S‖.


(4) ∃ x0 ∈ SH : ‖Tx0‖ = ‖T‖, Tx0⊥εSx0.




Theorem


(1) T⊥εBS ;



(3) ∃ x0 ∈ SH : ‖Tx0‖ = ‖T‖, | 〈Tx0|Sx0〉 | ≤ ε‖T‖ ‖S‖.


(4) ∃ x0 ∈ SH : ‖Tx0‖ = ‖T‖, Tx0⊥εSx0.




Theorem


(1) T⊥εBS ;



(3) ∃ x0 ∈ SH : ‖Tx0‖ = ‖T‖, | 〈Tx0|Sx0〉 | ≤ ε‖T‖ ‖S‖.


(4) ∃ x0 ∈ SH : ‖Tx0‖ = ‖T‖, Tx0⊥εSx0.


Approximate orthogonality in C0(K )

L(H) Let K be a locally compact topological space.

C0(K ) := {f : K → R cont. : ∀ ε > 0, {t ∈ K : |f (t)| ≥ ε} compact}

– with the supremum norm. For f ∈ C0(K ), Mf := {t ∈ K : |f (t)| = ‖f ‖}(nonempty and compact).

Theorem

Let f , g ∈ C0(K ), f 6= 0 6= g . Assume that Mf is connected. Then, thefollowing conditions are equivalent:

(a) f⊥εBg ,

(b) ∃ t1 ∈ Mf : |g(t1)| ≤ ε‖g‖.






Theorem


(a) f⊥εBg ,

(b) ∃ t1 ∈ Mf : |g(t1)| ≤ ε‖g‖.






Theorem


(a) f⊥εBg ,

(b) ∃ t1 ∈ Mf : |g(t1)| ≤ ε‖g‖.






Theorem


(a) f⊥εBg ,

(b) ∃ t1 ∈ Mf : |g(t1)| ≤ ε‖g‖.






Theorem


(a) f⊥εBg ,

(b) ∃ t1 ∈ Mf : |g(t1)| ≤ ε‖g‖.






Theorem


(a) f⊥εBg ,

(b) ∃ t1 ∈ Mf : |g(t1)| ≤ ε‖g‖.


Approximate symmetry of B-orthogonality


Symmetry of ⊥B

Birkhoff orthogonality ⊥B is generally not symmetric.

x

y

x+λy

y+λx

Figure: R2 with the maximum norm; x⊥By , y 6⊥Bx

If dimX ≥ 3 and ⊥B – symmetric, then X is an inner product space.If dimX = 2, then the symmetry of ⊥B is possible even if the norm doesnot come from an inner product (Radon plane).


Symmetry of ⊥B


x

y

x+λy

y+λx




Symmetry of ⊥B


x

y

x+λy

y+λx




Symmetry of ⊥B


x

y

x+λy

y+λx


If dimX ≥ 3 and ⊥B – symmetric, then X is an inner product space.

If dimX = 2, then the symmetry of ⊥B is possible even if the norm doesnot come from an inner product (Radon plane).


Symmetry of ⊥B


x

y

x+λy

y+λx




Approximate symmetry of ⊥B

J. Chmielinski, P. Wojcik, Approximate symmetry of the Birkhofforthogonality, J. Math. Anal. Appl. 461 (2018), 625–640.

Definition

The Birkhoff orthogonality relation in a normed space X is calledε-symmetric (for some ε ∈ [0, 1)), if for any x , y ∈ X :

x⊥By =⇒ y⊥εBx .

⊥B is ε-symmetric for some ε ∈ [0, 1) if and only if:

x⊥By =⇒ ∃ z ∈ Lin {x , y} : y⊥Bz , ‖z − x‖ ≤ ε‖x‖.

The approximate symmetry of ⊥B does not imply that the norm comesfrom an inner product (even if dimX ≥ 3).




Definition









Definition









Definition









Definition







Sufficient conditions for approximate symmetry of ⊥B

A modulus of convexity of a normed space X , δX : [0, 2]→ [0, 1]:

δX (ε) := inf{

1−∥∥∥x + y

2

∥∥∥ : ‖x‖ ≤ 1, ‖y‖ ≤ 1, ‖x − y‖ ≥ ε}.

Theorem

If δX (1) > 0 and 1− 2δX (1) ≤ ε < 1, relation ⊥B is ε-symmetric.

Corollary

Suppose that for any ε ∈ [0, 1) the relation ⊥B is not ε-symmetric. Then

ε0(X ) := sup{ε ∈ [0, 2] : δX (ε) = 0} ≥ 1.

Moreover, if X is finite-dimensional, then R(X ) ≥ 1.




δX (ε) := inf{

1−∥∥∥x + y

2

∥∥∥ : ‖x‖ ≤ 1, ‖y‖ ≤ 1, ‖x − y‖ ≥ ε}.

Theorem


Corollary


ε0(X ) := sup{ε ∈ [0, 2] : δX (ε) = 0} ≥ 1.





δX (ε) := inf{

1−∥∥∥x + y

2

∥∥∥ : ‖x‖ ≤ 1, ‖y‖ ≤ 1, ‖x − y‖ ≥ ε}.

Theorem


Corollary


ε0(X ) := sup{ε ∈ [0, 2] : δX (ε) = 0} ≥ 1.





δX (ε) := inf{

1−∥∥∥x + y

2

∥∥∥ : ‖x‖ ≤ 1, ‖y‖ ≤ 1, ‖x − y‖ ≥ ε}.

Theorem


Corollary


ε0(X ) := sup{ε ∈ [0, 2] : δX (ε) = 0} ≥ 1.




Theorem

Let X be a real, uniformly convex normed space.Then, ⊥B is approximately-symmetric.

Theorem

Let X be a finite-dimensional real smooth normed space.Then, ⊥B is approximately-symmetric.

Theorem

Let X be a real uniformly convex and smooth Banach space. Then, theBirkhoff orthogonality relations in X , X ∗ and X ∗∗ are approximatelysymmetric (actually, with the same ε).



Theorem


Theorem


Theorem




Theorem


Theorem


Theorem



There are spaces in which the Birkhoff orthogonality is not approximatelysymmetric, i.e., for any ε ∈ [0, 1), ⊥B is not ε-symmetric.

Example

X = R2 with the maximum norm.

x

yz

y+λx

y + λz

Figure: x⊥By , y 6⊥Bx , y 6⊥Bz , y 6⊥εBx



Example


x

yz

y+λx

y + λz




Example


x

yz

y+λx

y + λz



Geometrical properties connected with approximatesymmetry of B-orthogonality


R(X ) := sup{‖x − y‖ : conv {x , y} ⊂ SX}.

We consider the following property of X :

x , y ∈ X , x 6= y , conv {x , y} ⊂ SX =⇒ X is smooth at x − y . (∗)

Examples

Each smooth or strictly convex space satisfies (∗).R2 with the supremum norm (which is neither strictly convex nor smooth)also satisfies (∗).X = R2 with the norm for which the unit ball is a symmetric polygon suchthat sides are not parallel to diagonals, the condition (∗) is satisfied.

Theorem

Let X be a two-dimensional strictly convex normed space and let Y be astrictly convex and smooth normed space. Then the space L (X ,Y )satisfies (∗).





Examples


Theorem






Examples

Each smooth or strictly convex space satisfies (∗).

R2 with the supremum norm (which is neither strictly convex nor smooth)also satisfies (∗).X = R2 with the norm for which the unit ball is a symmetric polygon suchthat sides are not parallel to diagonals, the condition (∗) is satisfied.

Theorem






Examples

Each smooth or strictly convex space satisfies (∗).R2 with the supremum norm (which is neither strictly convex nor smooth)also satisfies (∗).

X = R2 with the norm for which the unit ball is a symmetric polygon suchthat sides are not parallel to diagonals, the condition (∗) is satisfied.

Theorem






Examples


Theorem






Examples


Theorem



Theorem

Let X be a real normed space satisfying (∗) and let ε ∈ (0, 1). If theorthogonality relation ⊥B in X is ε-symmetric, then R(X ) ≤ 2ε.

Corollary

If X is a real normed space satisfying (∗) and R(X ) = 2, then the Birkhofforthogonality in X is not approximately symmetric.


Theorem

Let X be a real normed space satisfying (∗) and let ε ∈ (0, 1). If theorthogonality relation ⊥B in X is ε-symmetric, then R(X ) ≤ 2ε.

Corollary

If X is a real normed space satisfying (∗) and R(X ) = 2, then the Birkhofforthogonality in X is not approximately symmetric.


Symmetry constant S(X )

Definition

S(X ) := inf{ε ∈ [0, 1] : ∀ x , y ∈ X x⊥By ⇒ y⊥εBx}.

S(X ) ∈ [0, 1]S(X ) = 0 means that ⊥B is symmetric.S(X ) = 1 means that ⊥B is not approximately symmetric.

S(X ) = sup{S(X0) : X0 is a two-dimensional subspace of X}.



Definition






Definition


S(X ) ∈ [0, 1]

S(X ) = 0 means that ⊥B is symmetric.S(X ) = 1 means that ⊥B is not approximately symmetric.




Definition


S(X ) ∈ [0, 1]S(X ) = 0 means that ⊥B is symmetric.

S(X ) = 1 means that ⊥B is not approximately symmetric.




Definition






Definition





If X is uniformly convex, then S(X ) < 1 (the reverse is not true).

If X satisfies (∗), then1

2R(X ) ≤ S(X ).

If X is a real uniformly convex and smooth Banach space, then

S(X ) = S(X ∗) = S(X ∗∗) < 1.




2R(X ) ≤ S(X ).


S(X ) = S(X ∗) = S(X ∗∗) < 1.




2R(X ) ≤ S(X ).


S(X ) = S(X ∗) = S(X ∗∗) < 1.


Example

Let X = R2 with an l1-l∞ norm.

For δ > 0 let Y = Yδ = R2 with the norm such that the unit sphere is ahexagon with vertices at

y1 = (1, 0), y2 = (0, 1), y3 = (−1− δ, 1− δ), y4 = (−1, 0),

y5 = (0,−1), y6 = (1 + δ,−1 + δ).

x1 = y1

x2 = y2

x3

y3

x4 = y4

x5 = y5 x6

y6

Figure: Unit spheres in X and Y


Example

Let X = R2 with an l1-l∞ norm.For δ > 0 let Y = Yδ = R2 with the norm such that the unit sphere is ahexagon with vertices at

y1 = (1, 0), y2 = (0, 1), y3 = (−1− δ, 1− δ), y4 = (−1, 0),

y5 = (0,−1), y6 = (1 + δ,−1 + δ).

x1 = y1

x2 = y2

x3

y3

x4 = y4

x5 = y5 x6

y6



Example

Let X = R2 with an l1-l∞ norm.For δ > 0 let Y = Yδ = R2 with the norm such that the unit sphere is ahexagon with vertices at

y1 = (1, 0), y2 = (0, 1), y3 = (−1− δ, 1− δ), y4 = (−1, 0),

y5 = (0,−1), y6 = (1 + δ,−1 + δ).

x1 = y1

x2 = y2

x3

y3

x4 = y4

x5 = y5 x6

y6



It can be checked that the Banach-Mazur distance d(X ,Y ) can bearbitrarily close to 1; namely:

d(X ,Y ) ≤ 1 + δ

1− δ.

The space X is a Radon plane, therefore S(X ) = 0.

No matter how small is δ > 0, the space Y satisfies (∗), whence

S(Y ) ≥ 1

2R(Y ) >

1

2.



d(X ,Y ) ≤ 1 + δ

1− δ.



S(Y ) ≥ 1

2R(Y ) >

1

2.



d(X ,Y ) ≤ 1 + δ

1− δ.


No matter how small is δ > 0, the space Y satisfies (∗)

, whence

S(Y ) ≥ 1

2R(Y ) >

1

2.



d(X ,Y ) ≤ 1 + δ

1− δ.



S(Y ) ≥ 1

2R(Y ) >

1

2.


S(lpn ) (p > 1)

The Banach-Mazur distance between lpn and l2n is equal to:

d := d(lpn , l2n ) = n

∣∣∣ 1p− 1

2

∣∣∣.

We were able to estimate that for p > 1, sufficiently close to 2, we have

S(lpn ) ≤ max

{(2p −

(1 +

1

d2

)p) 1p

,

(2q −

(1 +

1

d2

)q) 1q

}

(with q such that 1p + 1

q = 1).If p → 2, then q → 2 and d → 1.Therefore

limp→2

S(lpn ) = 0 = S(l2n ).


S(lpn ) (p > 1)


d := d(lpn , l2n ) = n

∣∣∣ 1p− 1

2

∣∣∣.


S(lpn ) ≤ max

{(2p −

(1 +

1

d2

)p) 1p

,

(2q −

(1 +

1

d2

)q) 1q

}



limp→2

S(lpn ) = 0 = S(l2n ).


S(lpn ) (p > 1)


d := d(lpn , l2n ) = n

∣∣∣ 1p− 1

2

∣∣∣.


S(lpn ) ≤ max

{(2p −

(1 +

1

d2

)p) 1p

,

(2q −

(1 +

1

d2

)q) 1q

}



limp→2

S(lpn ) = 0 = S(l2n ).


S(lpn ) (p > 1)


d := d(lpn , l2n ) = n

∣∣∣ 1p− 1

2

∣∣∣.


S(lpn ) ≤ max

{(2p −

(1 +

1

d2

)p) 1p

,

(2q −

(1 +

1

d2

)q) 1q

}


q = 1).

If p → 2, then q → 2 and d → 1.Therefore

limp→2

S(lpn ) = 0 = S(l2n ).


S(lpn ) (p > 1)


d := d(lpn , l2n ) = n

∣∣∣ 1p− 1

2

∣∣∣.


S(lpn ) ≤ max

{(2p −

(1 +

1

d2

)p) 1p

,

(2q −

(1 +

1

d2

)q) 1q

}


q = 1).If p → 2, then q → 2 and d → 1.

Thereforelimp→2

S(lpn ) = 0 = S(l2n ).


S(lpn ) (p > 1)


d := d(lpn , l2n ) = n

∣∣∣ 1p− 1

2

∣∣∣.


S(lpn ) ≤ max

{(2p −

(1 +

1

d2

)p) 1p

,

(2q −

(1 +

1

d2

)q) 1q

}



limp→2

S(lpn ) = 0 = S(l2n ).


Thank you!


Bibliography

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625–640.

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Documents

On approximate Birkhoff orthogonality in normed spaceskafedra.schoolsite.org.ua/BOOK/CHMIELINSKI.pdf · 2019. 7. 4. · On approximate Birkho orthogonality in normed spaces Jacek