http://jipam.vu.edu.au/
Volume 6, Issue 3, Article 79, 2005
ON AN ε-BIRKHOFF ORTHOGONALITY
JACEK CHMIELI ´NSKI
INSTYTUTMATEMATYKI, AKADEMIAPEDAGOGICZNA WKRAKOWIE
PODCHOR ¸A ˙ZYCH2, 30-084 KRAKÓW, POLAND
jacek@ap.krakow.pl
Received 18 February, 2005; accepted 28 July, 2005 Communicated by S.S. Dragomir
ABSTRACT. We define an approximate Birkhoff orthogonality relation in a normed space. We compare it with the one given by S.S. Dragomir and establish some properties of it. In particular, we show that in smooth spaces it is equivalent to the approximate orthogonality stemming from the semi-inner-product.
Key words and phrases: Birkhoff (Birkhoff-James) orthogonality, Approximate orthogonality, Semi-inner-product.
2000 Mathematics Subject Classification. 46B20, 46C50.
1. INTRODUCTION
In an inner product space, with the standard orthogonality relation⊥, one can consider the approximate orthogonality defined by:
x⊥εy ⇔ | hx|yi | ≤εkxk kyk. (|cos(x, y)| ≤εforx, y 6= 0).
The notion of orthogonality in an arbitrary normed space, with the norm not necessarily coming from an inner product, may be introduced in various ways. One of the possibilities is the following definition introduced by Birkhoff [1] (cf. also James [6]). Let X be a normed space over the fieldK∈ {R,C}; then forx, y ∈X
x⊥By⇐⇒ ∀λ∈K:kx+λyk ≥ kxk.
We call the relation⊥B, a Birkhoff orthogonality (often called a Birkhoff-James orthogonality).
Our aim is to define an approximate Birkhoff orthogonality generalizing the⊥ε one. Such a definition was given in [3]:
(1.1) x⊥
εBy⇐⇒ ∀λ∈K:kx+λyk ≥(1−ε)kxk. We are going to give another definition of this concept.
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
The paper has been completed during author’s stay at the Silesian University in Katowice.
043-05
2. BIRKHOFF APPROXIMATEORTHOGONALITY
Let us define an approximate Birkhoff orthogonality. Forε ∈[0,1):
(2.1) x⊥εBy ⇐⇒ ∀λ ∈K:kx+λyk2 ≥ kxk2−2εkxk kλyk. If the above holds, we say thatxisε-Birkhoff orthogonal toy.
Note, that the relation⊥εBis homogeneous, i.e.,x⊥εByimpliesαx⊥εBβy (for arbitraryα, β ∈ K). Indeed, for anyλ∈Kwe have (excluding the obvious caseα= 0)
kαx+λβyk2 =|α|2
x+λβ αy
2
≥ |α|2
kxk2−2εkxk
λβ αy
=kαxk2−2εkαxkkλβyk.
Proposition 2.1. IfXis an inner product space then, for arbitraryε ∈[0,1), x⊥εy ⇐⇒ x⊥εBy.
We omit the proof – a more general result will be proved later (Theorem 3.3). As a corollary, forε = 0, we obtain the well known fact: x⊥By ⇔ x⊥y(in an inner product space).
Let us modify slightly the definition of Dragomir (1.1). Replacing 1 −ε by √
1−ε2 we obtain:
x⊥εDy ⇐⇒ ∀λ ∈K: kx+λyk ≥√
1−ε2kxk. Thusx⊥εDy ⇔ x⊥ρBywithρ=ρ(ε) = 1−√
1−ε2. Then, for inner product spaces we have:
x⊥εDy ⇐⇒ x⊥εy (see [3, Proposition 1]).
T. Szostok [10], considering a generalization of the sine function introduced, for a real normed spaceX, the mapping:
s(x, y) =
infλ∈R kx+λyk
kxk , forx∈X\ {0};
1, forx= 0.
It is easily seen that x⊥By ⇔ s(x, y) = 1. It is also apparent that x⊥εDy ⇔ s(x, y) ≥
√1−ε2. Definingc(x, y) := ±p
1−s2(x, y)(generalized cosine) one getsx⊥εDy ⇔ |c(x, y)|
≤ε.
Let us compare the approximate orthogonalities⊥εDand⊥εB. In an inner product space both of them are equal to ε-orthogonality⊥ε. Thus one may ask if they are equal in an arbitrary normed space. This is not true. Moreover, neither⊥εB ⊂ ⊥εDnor⊥εD⊂ ⊥εBholds generally (i.e., for an arbitrary normed space and allε ∈[0,1)). For, considerX =R2(overR) equipped with the maximum normk(x1, x2)k:= max{|x1|,|x2|}. Now, letx= (1,0),y= 12,1
,ε= 12. One can verify thatx⊥εBy(i.e., that max
1 + λ2
,|λ| 2 ≥1− |λ|holds for eachλ ∈R) but not x⊥εDy(takeλ=−23). Thus⊥εB 6⊂ ⊥εD.
On the other hand, for x = 1,12
, y = (1,0), ε =
√ 3
2 we have max{|1 +λ|,12}2
≥ 1−√
3 2
2
, i.e.,x⊥εDybut notx⊥εBy(consider, for example,λ=
√3
2 −1). Thus⊥εD6⊂ ⊥εB. See also Remark 4.1 for further comparison of⊥εBand⊥εD.
3. SEMI–INNER–PRODUCT(APPROXIMATE) ORTHOGONALITY
LetX be a normed space overK ∈ {R,C}. The norm inX need not come from an inner product. However, (cf. G. Lumer [7] and J.R. Giles [5]) there exists a mapping[·|·] :X×X → Ksatisfying the following properties:
(s1) [λx+µy|z] =λ[x|z] +µ[y|z], x, y, z ∈X, λ, µ∈K; (s2) [x|λy] =λ[x|y], x, y ∈X, λ∈K;
(s3) [x|x] =kxk2, x∈X;
(s4) |[x|y]| ≤ kxk · kyk, x, y ∈X.
(Cf. also [4].) We will call each mapping[·|·]satisfying (s1)–(s4) a semi-inner-product (s.i.p.) in a normed spaceX. Let us stress that we assume that a s.i.p. generates the given norm inX (i.e., (s3) is satisfied). Note, that there may exist infinitely many different semi-inner-products inX. There is a unique s.i.p. inX if and only ifX is smooth (i.e., there is a unique supporting hyperplane at each point of the unit sphereSor, equivalently, the norm is Gâteaux differentiable onS – cf. [2, 4]). IfXis an inner product space, the only s.i.p. onX is the inner-product itself ([7, Theorem 3]).
We say that s.i.p. is continuous iffRe [y|x+λy]→ Re [y|x]asR3 λ →0for allx, y ∈S.
The continuity of s.i.p is equivalent to the smoothness of X (cf. [5, Theorem 3] or [4]). It follows also in that case (see the proof of Theorem 3 in [5]):
(3.1) lim
λ→0λ∈R
kx+λyk −1
λ = Re [y|x], x, y ∈S.
Extending previous notations we define semi-orthogonality and approximate semi-orthogo- nality:
x⊥sy ⇔ [y|x] = 0;
x⊥εsy ⇔ |[y|x]| ≤εkxk · kyk, forx, y ∈X and0≤ε <1.
Obviously, for an inner–product space:⊥s=⊥and⊥εs=⊥ε. Proposition 3.1. Forx, y ∈X, ifx⊥εsy, thenx⊥εBy(i.e.,⊥εs⊂ ⊥εB).
Proof. Suppose that x⊥εsy, i.e.,|[y|x]| ≤ εkxk · kyk. Then, for some θ ∈ [0,1]and for some ϕ∈[−π, π]we have:
[y|x] =θεkxk · kyk ·eiϕ. For arbitraryλ ∈Kwe have:
kx+λyk · kxk ≥ |[x+λy|x]|
=
kxk2+λ[y|x]
=
kxk2+θεkxk · kyk ·λ·eiϕ whence
kx+λyk ≥
kxk+θεkyk ·λ·eiϕ
=
kxk+θεkykRe λeiϕ
+iθεkykIm λeiϕ .
Therefore
kx+λyk2 ≥ kxk+θεkykRe λeiϕ2
+ θεkykIm λeiϕ2
=kxk2+ 2θεkxk kykRe λeiϕ +θ2ε2kyk2
Re λeiϕ2
+ Im λeiϕ2
=kxk2+ 2θεkxk kykRe λeiϕ
+θ2ε2kλyk2
≥ kxk2+ 2θεkxk kykRe λeiϕ
≥ kxk2+ 2θεkxk kyk − λeiϕ
=kxk2−2θεkxk kλyk
≥ kxk2−2εkxk kλyk,
i.e.,x⊥εBy.
Since|[y|x]| ≤ kxk kyk, i.e., x⊥1syfor arbitrary x, y, the above result gives alsox⊥1Byfor allx, y. That is the reason we restrictεto the interval[0,1).
Proposition 3.2. IfXis a continuous s.i.p. space andε∈[0,1), then⊥εB ⊂ ⊥εs.
Proof. Suppose thatx⊥εBy. Because of the homogeneity of relations⊥εBand⊥εswe may assume, without loss of generality, thatx, y ∈S. Then, for arbitraryλ∈Kwe have:
0≤ kx+λyk2−1 + 2ε|λ|= [x|x+λy] + [λy|x+λy]−1 + 2ε|λ|. Therefore
0≤Re [x|x+λy] + Re [λy|x+λy]−1 + 2ε|λ|
≤ |[x|x+λy]|+ Re [λy|x+λy]−1 + 2ε|λ|
≤ kx+λyk+ Re [λy|x+λy]−1 + 2ε|λ|
whence
(3.2) Re [λy|x+λy] +kx+λyk −1≥ −2ε|λ|, for allλ∈K. Letλ0 ∈K\ {0},n ∈Nandλ = λn0. Then from (3.2) we have
Re λ0
ny|x+ λ0 n y
+
x+ λ0 n y
−1≥ −2ε|λ0| n ;
Re λ0
|λ0|y|x+|λ0| n
λ0
|λ0|y
+
x+ |λn0||λλ0
0|y −1
|λ0| n
≥ −2ε.
Putting y0 := |λλ0
0|y ∈ S, ξn := |λn0| ∈ R (ξn → 0 as n → ∞) we obtain from the above inequality
Re [y0|x+ξny0] + kx+ξny0k −1
ξn ≥ −2ε.
Lettingn → ∞, using continuity of the s.i.p. and (3.1) Re [y0|x] + Re [y0|x]≥ −2ε whence
Re [λ0y|x]≥ −ε|λ0|.
Putting−λ0 in the place ofλ0we obtainRe [λ0y|x]≤ε|λ0|whence
|Re [λ0y|x]| ≤ε|λ0| for arbitraryλ0 ∈K.
Now, takingλ0 = [y|x]we get Reh
[y|x]y|xi
≤ε|[y|x]|
whence|[y|x]|2 ≤ε|[y|x]|and finally|[y|x]| ≤ε, i.e.,x⊥εsy.
Without the additional continuity assumption, the inclusion⊥εB⊂ ⊥εs need not hold.
Example 3.1. Consider the spacel1(with the normkxk=P∞
i=1|xi|forx= (x1, x2, . . .)∈l1).
Define
[x|y] :=kyk
∞
X
i=1 yi6=0
xiyi
|yi|, x, y ∈l1
— a semi-inner-product inl1. Letε∈[0,√
2−1)and letx= (1,0,0, . . .),y= (1,1, ε,0, . . .).
Then, for an arbitraryλ∈K:
kx+λyk2− kxk2+ 2εkxk kλyk= (|1 +λ|+|λ|+|λε|)2−1 + 2ε(2 +ε)|λ|
≥(1 +|λ|ε)2−1 + 2ε(2 +ε)|λ|
= 2ε(3 +ε)|λ|+|λ|2ε2
≥0, i.e.,x⊥εBy(in fact,x⊥By). On the other hand,
[y|x] = 1 = 1
2 +εkxk kyk> εkxk kyk
whence¬(x⊥εsy). In particular, forε= 0, this shows that⊥B 6⊂ ⊥s(cf. [4, 8, 9]).
From the last two propositions we have:
Theorem 3.3. IfXis a continuous s.i.p. space, then
⊥εB =⊥εs. Moreover we obtain, forε= 0, (cf. [5, Theorem 2]) Corollary 3.4. IfXis a continuous s.i.p. space, then
⊥B =⊥s.
Conversely,⊥B ⊂ ⊥simplies continuity of s.i.p. (smoothness) – cf. [4] and [8].
4. SOME REMARKS
Remark 4.1. Dragomir [3, Definition 5] introduces the following concept: The s.i.p.[·|·]is of (APP)-type if there exists a mapping η : [0,1) → [0,1) such thatη(ε) = 0 ⇔ ε = 0and x⊥η(ε)D yimpliesx⊥εsyfor allε∈[0,1). It follows from Proposition 3.1 that in that case we have also
(4.1) x⊥η(ε)D y ⇒ x⊥εBy
for allε∈[0,1).
It follows from [3, Lemma 1] that for a closed, proper linear subspaceGof a normed space X and for an arbitrary ε ∈ (0,1), the setG⊥εD of all vectors ⊥εD-orthogonal toG is nonzero.
Using (4.1) we get
(4.2) G⊥η(ε)D ⊂G⊥εB.
Therefore, we have
Lemma 4.2. IfXis a normed space with the s.i.p. [·|·]of the (APP)-type, then for an arbitrary proper and closed linear subspace G and an arbitrary ε ∈ [0,1) the setG⊥εB of all vectors ε-Birkhoff orthogonal toGis nonzero.
We have also
Theorem 4.3. IfXis a normed space with the s.i.p. [·|·]of the (APP)-type, then for an arbitrary closed linear subspaceGand an arbitraryε∈[0,1)the following decomposition holds:
X =G+G⊥εB.
Proof. FixGandε∈[0,1). It follows from [3, Theorem 3] that X =G+G⊥η(ε)D .
Using (4.2) we get the assertion.
The final example shows that the set of allε-orthogonal vectors may be equal to the set of all orthogonal ones.
Example 4.1. Consider again the space l1 with the s.i.p. defined above. Let e = (1,0, . . .).
Observe that vectorsε-orthogonal toeare, in fact, orthogonal toe:
(4.3) x⊥εBe ⇒ x⊥Be.
Indeed, let ε ∈ [0,1) be fixed and let x = (x1, x2, . . .) ∈ l1 satisfy x⊥εBe. Because of the homogeneity of⊥εBwe may assume, without loss of generality, thatkxk= 1andx1 ≥0. Thus we have
∀λ∈K:kx+λek2 ≥1−2ε|λ|. Therefore
∀λ ∈K: (|x1+λ|+ 1−x1)2 ≥1−2ε|λ|.
Suppose thatx1 >0. Takeλ∈Rsuch thatλ <0,λ >−x1 andλ >−2(1−ε). Then we have (x1+λ+ 1−x1)2 ≥1 + 2ελ,
which leads to λ ≤ −2(1−ε)– a contradiction. Thus x1 = 0, i.e, x = (0, x2, x3, . . .)and
|x2|+|x3|+· · ·= 1. This yields, for arbitraryλ ∈K,
kx+λek=|λ|+ 1≥1 =kxk, i.e.,x⊥Be. It follows from (4.3) that forG:= linewe have
G⊥εB =G⊥B.
Note, that the implicatione⊥εBx ⇒ e⊥Bxis not true. Take for examplex = 34,14,0, . . . . Then [x|e] = 34kek kxk, i.e, e⊥
3
4sx, whence (Proposition 3.1) e⊥
3 4
Bx. On the other hand, for λ=−53 one has
ke+λxk= 2
3 <1 = kek, i.e.,¬(e⊥Bx).
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