# Orthogonal Symmetries and Reflections in Banach Spaces

Authors: Ali Jaballah; Fathi B. Saidi
Journal: Journal of Mathematics
 (2017)

---

## Abstract

LetX be a Banach space. We introduce a concept of orthogonal symmetry and reflection in X. We then establish its relation with the concept of best approximation and investigate its implication on the shape of the unit ball of the Banach space X by considering sections over subspaces. The results are then applied to the space C(I) of continuous functions on a compact set I. We obtain some nontrivial symmetries of the unit ball of C(I). We also show that, under natural symmetry conditions, every odd function is orthogonal to every even function in X. We conclude with some suggestions for further investigations.


---

## Body

## 1. Introduction

When we try to imagine or picture a reflection of a pointx in a Banach space X with respect to, say, a line L passing through the origin, we tend to put ourselves in the context of a Euclidean space and think of a “mirror reflection” of the point x, that is, of a point xr that satisfies xr=x; x and xr are equidistant from a point m∈L (m=x+xr/2), such that, in addition, if we move x and xr “away from L” an equal distance, the two new points x1 and x1r still satisfy the same conditions; namely, x1r=x1 and x1 and x1r are equidistant from m; see Definition 6. It would be indeed nice and convenient if such a symmetry always existed, given its implications on the geometry of the unit ball of X. This would be a valuable asset that could help in establishing results in X that otherwise may prove to be difficult. Moreover, there are many results in the literature that rely either directly or indirectly on the geometry of the unit ball of a Banach space. It may be interesting to revisit these and investigate the presence of such symmetries and their consequences. For some recent results in this direction, we refer the readers to [1, 2]. We also note that the investigation of many concepts within Banach spaces is highly active. For some recent results along these lines we refer the readers for examples to [3, 4].However, we all know that, in general, these “mirror symmetries” do not hold when we work inside a Banach space. So naturally, one should wonder about the types of spaces that do possess such symmetries. Our aim in this paper is to give necessary and sufficient conditions for this to be true and to investigate the consequences of these symmetries on the geometry of the unit ball of the Banach space; see Section3. Before doing so, we investigate in Section 2 a weaker version of symmetry, which we term as “weak symmetry”; see Definition 1. We will establish a link between weak symmetry and the concept of best approximation (see Definition 2) and establish some of its properties and characteristics. In Section 4, we apply our results to the space C(I) of continuous functions on a compact subset I of R. We are able to obtain some nontrivial geometric regularity for the unit ball of C(I). We also show that every odd function is orthogonal to every even function (Theorem 23). We conclude at the end of the paper by suggesting some directions for further investigations.


## 2. Weak Symmetry in Banach Spaces

Throughout this section,X is a Banach space and H is a closed subspace of X.Definition 1.
LetH be a closed subspace of a Banach space X. An element xr∈X is called a weak-reflection point with respect to H of the element x∈X if (1)x+xr2∈H,x-xr2=dx,H. If every element x in a nonempty subset A of X admits a weak-reflection point xr∈A with respect to H, then we say that A is weakly symmetric with respect to H.We also need the following definition.Definition 2.
LetH be a closed subspace of a Banach space X. An element ho∈H is called a best approximation from H of an element x∈X if it satisfies (2)x-ho=dx,H≔inf⁡x-h:h∈H. If every element in X admits a best approximation from H then we say that H is proximinal in X. The set of all best approximations of an element x in X is denoted by PH(x). The set-valued map PH:X→2H is called the metric projection of X onto H. If PH(x) is a singleton for every x∈X, then H is said to be Chebyshev in X. If PH(x) is a singleton for some x∈X, then we denote the best approximation of x also by PH(x).It is clear that, in general, an elementx∈X may have more than one weak-reflection point with respect to H. In fact, we have the following characterization of uniqueness.Lemma 3.
LetH be a closed subspace of a Banach space X. Then we have the following.
(i) An elementx∈X admits a weak-reflection point with respect to H if and only if PH(x)≠∅, in which case the set of weak-reflection points of x is given by (3)xr:xris  a  weak-refection  point  ofx=2PHx-x.
(ii) An elementx∈X admits a unique weak-reflection point with respect to H if and only if PH(x) is a singleton.
(iii) Every element inX admits a unique weak-reflection point with respect to H if and only if H is Chebyshev in X.Proof.
It follows directly from Definitions1 and 2 that if an element x∈X admits a weak-reflection point xr with respect to H then (4)hx≔x+xr2∈PHx; hence PH(x)≠∅. Conversely, suppose that PH(x)≠∅ for some x∈X. Let hx∈PH(x) and let (5)xr≔2hx-x. Then one easily checks that xr is a weak-reflection point of x. This completes the proof of Part (i). Parts (ii) and (iii) follow directly from Part (i). This ends the proof.To explain the use of the term “weak” symmetry in Definition1, we observe the following.Lemma 4.
LetH be a closed subspace of a Banach space X. The following statements are equivalent.
(i) The Banach spaceX is weakly symmetric with respect to H.
(ii) The subspaceH is proximinal in X.
(iii) For everyx∈X, there exists an element xr∈X and a linear functional fx∈X∗, where X∗ is the continuous dual space of X, such that (6)f=1,H⊂ker⁡f,x+xr2∈H,fx-xr=x-xr, where ker⁡f denotes the kernel of the linear functional f.Proof.
(i)⇔ (ii): this follows directly from Lemma 3.
(ii)⇔ (iii): this follows from Lemma 3 and Theorem1.1 in [5].It follows from the previous lemma that one can find a Banach spaceX and a closed subspace H of X such that X is not weakly symmetric with respect to H. In fact, it is well known that every nonreflexive Banach space X admits a nonproximinal hyperplane H; see Corollary2.4 in [5]; hence X is not weakly symmetric with respect to H. An example of a nonreflexive Banach space is the Banach space BV[0,1] of functions of bounded variation on the closed interval [0,1] [6]. On the other hand, every reflexive subspace H of a Banach space X is proximinal; see Corollary2.5 in [5]; hence a Banach space is always weakly symmetric with respect to its reflexive subspaces. In particular, every Banach space is weakly symmetric with respect to its finite dimensional subspaces. The following follows from the previous observations.Remark 5.
A Banach spaceX is weakly symmetric with respect to all of its closed subspaces if and only if it is reflexive.


## 3. Orthogonal Symmetry in Banach Spaces

The notions of weak-reflection and weak symmetry that were discussed above are quite different from the “usual” notions of reflection and symmetry. As mentioned above, when we think about a reflection pointxr of a point x with respect to a subspace H of a Banach space X, we tend to visualize (and would like it to be true) points xr and x that are “mirror images” of each other with respect to H, that is, that are equidistant from the origin and from H, such that if we move them “away from H” an equal distance, the two new points x1 and x1r still satisfy the same conditions. With this in mind, we introduce the following definition of symmetry.Definition 6.
LetH be a closed subspace of a Banach space X. An element xr∈X is called a reflection point with respect to H of the element x∈X if (7)hx≔x+xr2∈H,hx+tx-hx=hx+txr-hx,∀t∈R. If every element x in a nonempty subset A of X admits a reflection point xr∈A with respect to H, then we say that A is symmetric with respect to H.Settingt=1 in the second condition in Definition 6, we obtain immediately the following.Remark 7.
Ifxr is a reflection point with respect to H of an element x∈X, then (8)xr=x.We recall the following notion of orthogonality, which was first introduced by Roberts [7].Definition 8.
Two elementsx and y of a Banach space X are said to be orthogonal (we write x⊥y) if, for every λ∈R, (9)x+λy=x-λy.
Two nonempty subsetsF and G of X are said to be orthogonal, and we write F⊥G, if every element of F is orthogonal to every element of G. Given x∈X, we write x⊥G if x⊥g for every g∈G.
IfH is a closed subspace of a Banach space X and x∈X, then we denote the set of orthogonal projections of x onto H by PH⊥(x): (10)PH⊥x≔h∈H:x-h⊥H. If PH⊥(x)=ho is a singleton, then we denote ho also by PH⊥(x).Note that this definition of orthogonality is symmetric in the sense that(11)x⊥yiffy⊥x. For a generalization of this notion of orthogonality to the case of orthogonal sequences in complex Banach spaces, we refer the reader to [8].The following observation gives an alternative way of thinking about the notion of symmetry introduced in Definition6 in terms of orthogonality.Remark 9.
LetH be a closed subspace of a Banach space X. An element xr∈X is a reflection point with respect to H of the element x∈X if and only if hx≔x+xr/2 is an orthogonal projection of the point x onto H, that is, if and only if (12)hx≔x+xr2∈H,x-hx⊥H.To address the question of uniqueness of reflection points (equivalently of orthogonal projections), we recall the following definition.Definition 10.
LetA be a nonempty subset of a Banach space X. An element c∈X is called a center of A if A is symmetric with respect to c, that is, if (13)c-a-c=2c-a∈A,for  everya∈A. Note that A is symmetric with respect to c if and only if A-c is symmetric with respect to the origin.First we take a look at the uniqueness of centers. We have the following.Lemma 11.
IfA is a bounded nonempty subset of a Banach space X, then A admits at most one center.Proof.
SupposeA admits two distinct centers c1 and c2. Let Ri denote the reflection with respect to ci, i=1,2. For each a∈A, we have Ri(a)-ci=ci-a; hence (14)Ria=2ci-a,i=1,2. It follows that (15)R1R2a=R12c2-a=2c1-2c2-a=a-2c2-c1. By induction we obtain that(16)R1R2na=a-2nc2-c1,a∈A,n≥1. Similarly we show that(17)R2R1na=a+2nc2-c1,a∈A,n≥1.
Now leta∈A be fixed and consider the two sequences of points in A defined by (18)xn≔R1R2na,yn≔R2R1na,n=1,2,…. It follows from (16) and (17) that (19)yn-xn=4nc2-c1⟶∞,asn⟶∞. This implies that A is unbounded, which contradicts the assumption in the lemma. This ends the proof.The following follows from the proof of the previous lemma.Corollary 12.
LetA be a nonempty subset of a Banach space X. If A admits two distinct centers, then A is unbounded.We note that there are various notions of centers in the literature and that they all play an important role in approximation theory and in geometric functional analysis in general.We have seen above in Section2 that, in general, weak-reflection points are not unique and that a weak-reflection point of an element x∈X with respect to a closed subspace H is unique if and only if PH(x) is a singleton. The situation is quite different in the case of symmetry. The following theorem shows, in particular, that the concept of symmetry is stronger than the concept of weak symmetry and that reflection points and orthogonal projections, when they exist, are unique.Theorem 13.
LetH be a closed subspace of a Banach space X. Then we have the following.
(i) IfX is symmetric with respect to H, then X is weakly symmetric with respect to H.
(ii) Every elementx∈X admits at most one orthogonal projection PH⊥(x) onto H. Moreover, if an element x∈X admits an orthogonal projection PH⊥(x) onto H then the set of best approximations PH(x) admits a unique center cx, x admits a unique reflection point xr with respect to H, and the reflection point xr of x is given by (20)xr=2cx-x=2PH⊥x-x.
(iii) Every elementx in X admits at most one reflection point xr with respect to H. Moreover, if an element x∈X admits a reflection point xr with respect to H, then set of best approximations PH(x) admits a unique center cx, x admits a unique orthogonal projection PH⊥(x) onto H, and the orthogonal projection of x onto H is given by (21)PH⊥x=cx=x+xr2.Proof.
(i) Suppose thatX is symmetric with respect to H. Let x∈X be given and let xr be a reflection point of x. We will show that xr is a weak-reflection point of x. Indeed we have, by Remark 9, (22)hx≔x+xr2∈H,x-hx⊥H. Hence it suffices to show that x-hx=d(x,H). For every h∈H we have, since x-hx⊥H, (23)x-h=z1≔x-hx-h-hx=z2≔x-hx+h-hx. It follows, since x is a convex function of x, that (24)x-hx=z1+z22≤z1+z22=x-h,for  everyh∈H. Hence x-hx=d(x,H) and the proof of Part (i) is complete.
(ii) Letx∈X be fixed and suppose that x admits an orthogonal projection hx onto H; hence (25)x-hx⊥H. Then, by Remark 9, x admits a reflection point and hence, by Part (i) and Lemma 3, PH(x)≠⌀. We now show that hx is a center of PH(x). We need to show that (2hx-h)∈PH(x), for every h∈PH(x). Let h∈PH(x) be given. It follows, since (hx-h)∈H and since (x-hx)⊥H, that (26)x-2hx-h=x-hx-hx-h=x-hx+hx-h=x-h=dx,H. Hence (2hx-h)∈PH(x) for every h∈PH(x). This implies that hx is indeed a center of PH(x). It follows, by Lemma 11 and since PH(x) is bounded, that hx is the unique center cx of PH(x). Hence the orthogonal projection of x onto H is unique and is given by (27)PH⊥x=cx. It follows from Remark 9 that xr≔2PH⊥(x)-x is a reflection point of x and that it is unique since, again by Remark 9, (xr+x)/2 must be an orthogonal projection of x onto H. This completes the proof of Part (ii).
(iii) This follows directly from Part (ii) and Remark9. This completes the proof of the theorem.The following follows immediately from Theorem13.Corollary 14.
A Banach spaceX is symmetric with respect to a closed subspace H if and only if every element x∈X admits an orthogonal projection onto H.Remark 15.
For every elementx∈X, PH(x) is always a convex subset of the subspace H of X; hence the center cx and the orthogonal projection PH⊥(x) always belong to PH(x) whenever they exist.It is important to note thatH does not have to be Chebyshev in X in order for X to be symmetric with respect to H. Indeed we have the following.Example 16.
Consider the Banach space(28)X,·≔R2,·∞, where (a,b)∞≔max⁡a,b, and let (29)H≔0,b:b∈R. Then we have, for every (a,b)∈X, (30)PHa,b=0,b+λa:λ∈-1,1; hence H is proximinal but not Chebyshev in X. Also, one can easily verify that X is symmetric with respect to H and that, for every x≔(a,b)∈X, the unique center cx of PH(x) and the unique reflection point xr of x are given by (31)cx=0,b,xr=-a,b.We also point out that the existence of a center ofPH(x) does not guarantee the existence of a reflection point (or of an orthogonal projection) of x. Indeed we have the following.Example 17.
LetX and H be as in Example 16 but, instead of ·∞, consider the norm (32)a,b≔a2+b2ifab≥0a+bifab<0. Note that · is the Minkowski functional associated with the closed convex and symmetric subset K of X given by (33)K≔a,b∈X:a,b≤1; hence indeed it is a norm on X. To verify this, all we need to check is the triangle inequality as the other properties are trivial: First, recall that we always have, for all s,t∈R, (34)s-t≤s2+t2≤s+t; hence, for all s,t∈R,(35)s-t≤s2+t2≤s,t≤s+t. Also, for all s1, s2, t1, and t2 in R, we have(36)s1+s22+t1+t22≤s12+t12+s22+t22. Now, let x=(a,b) and y=(c,d) be two arbitrary elements in X.If(a+c)(b+d)≥0, then we have, by (35) and (36), (37)x+y=a+c2+b+d2≤a2+b2+c2+d2≤x+y.If(a+c)(b+d)<0, then one of the factors is positive and one is negative, say (38)a+c>0,b+d<0. It follows, by (35), that (39)x+y=a+c+b+d=a+c-b-d=a-b+c-d≤x+y, where the last inequality follows from the fact that (40)a-b≤max⁡a,b≤a2+b2≤a,bifab≥0a+b=a,bifab<0, and we can also easily obtain a similar inequality for (c-d). Going back to our example, it is clear that, for every (a,b)∈X, (41)PHa,b=0,b. Hence H is Chebyshev in X and, for each x≔(a,b)∈X, PH(x) admits a unique center, namely, cx=(0,b). It follows from Theorem 13 that if an element x≔(a,b) in X admitted a reflection point xr with respect to H, then we would have xr=2cx-x=(-a,b) and, by Remark 9, we would also have x-cx⊥H or, in other words,(42)a,0⊥H. But clearly this is not possible if a≠0. Indeed, in this case h≔(0,a)∈H and (43)a,0+h=a,a=a2, while (44)a,0-h=a,-a=2a≠a2. Hence (42) does not hold and consequently (a,b) does not have a reflection point if a≠0.We now highlight the effect of symmetry on the geometry of the unit ballBX of the Banach space X by looking at its 1-dimensional sections over the subspace H, that is, by looking at sections of BX by subspaces of X in which H is of codimension 1.Theorem 18.
If a Banach spaceX is symmetric with respect to a proper closed subspace H, then its unit ball BX is symmetric with respect to H. Moreover, given any point y∈X∖H, the subspace H divides the unit ball BF of F≔span⁡y,H into two halves which are identical modulo reflection with respect to H. The two halves of BF are given by (45)BF±≔BF∩h±αy-cy:h∈H,α∈R=BF∩h±αy:h∈H,α∈R, where cy is the unique center of PH(y).Proof.
SinceX is symmetric with respect to H, it follows immediately from Remark 7 that BX is symmetric with respect to H. Now let x∈BX be given. It follows from Remark 9 and Theorem 13 that (46)cx=x+xr2∈H,x-cx⊥H, where xr is the reflection point of x. If x∈BF⊂BX, then we have (47)xr=2cx-x∈span⁡x,H≔F. Hence xr∈BX∩F=BF. Therefore BF is symmetric with respect to H. Since H is a hyperplane in F, H divides BF into two halves, BF+ and BF-. Clearly, if x≔h+α(y-cy)∈BF+ for some h∈H and α∈R, then the reflection point of x is given by (48)xr≔h-αy-cy∈BF-. This follows from Remark 9 and from the fact that (y-cy)⊥H. Hence BF- is the reflection of BF+. Similarly we show that BF+ is the reflection of BF-. This ends the proof.One should note that the orthogonal symmetry of the unit ball given by Theorem18 is not always true in Banach spaces. It follows from Definition 6 and Remark 9 that it holds if and only if the unit ball BX of X is symmetric with respect to H in the sense of Definition 6. In particular, it is possible for a Banach space X to be weakly symmetric with respect to a closed subspace H, while the unit ball BX is neither symmetric nor weakly symmetric with respect to H. Indeed we have the following.Example 19.
Consider again the example given in Example17 above. Then X is weakly symmetric with respect to H, since H is Chebyshev in X. It is easy to verify that (49)x≔22,22∈BXbut the weak reflection point(50)xr=-22,22∉BX.Hence BX is neither weakly symmetric nor symmetric with respect to H.We now consider the case whereX is the direct sum of two mutually orthogonal subspaces.Theorem 20.
Suppose that a Banach spaceX is the direct sum of two mutually orthogonal closed subspaces H and G: (51)H⊥G,H⊕G=X. Then X is symmetric with respect to both H and G and we have, for each x∈X, (52)x=PH⊥x+PG⊥x,xHr=PH⊥x-PG⊥x,xGr=PG⊥x-PH⊥x, where xHr and xGr are the reflections of x with respect to H and G, respectively.Proof.
Letx∈X. Then x=h+g, for some h∈H and g∈G. Now let xHr≔h-g. Then we have, since G⊥H and x-h=g∈G, (53)x+xHr2=h∈H,x-h⊥H. Hence, by Remark 9, xHr is a reflection point of x with respect to H and, by Theorem 13, (54)PH⊥x=x+xHr2=h. Since x was arbitrary in X, X is symmetric with respect to H. Similarly we show that xGr≔g-h is a reflection point of x with respect to G, that X is symmetric with respect to G, and that (55)PG⊥x=x+xGr2=g. Hence (56)PH⊥x+PG⊥x=x+xHr2+x+xGr2=x, since xHr+xGr=0. This ends the proof.The reader may be wondering about the situation when a Banach spaceX is symmetric with respect to all of its closed subspaces. For one thing, it follows from Remark 5 and Theorem 13 that X must be reflexive. Also it is clear that this condition cannot be sufficient, as symmetry and weak symmetry are not equivalent. It turns out that the situation is possible only if X is isometric to a Hilbert space. We have the following stronger result.Theorem 21.
LetX be a Banach space satisfying dim⁡X≥2. Then X is symmetric with respect to all of its 1-dimensional closed subspaces if and only if it is isometric to a Hilbert space.Proof.
SupposeX is symmetric with respect to all of its 1-dimensional closed subspaces. Let F be an arbitrary 2-dimensional subspace of X, let x be a nonzero element in F, and let (57)H≔span⁡x≔rx:r∈R. We first show that x admits a nonzero orthogonal element in F. Let y∈F be linearly independent of x. Since H is 1-dimensional, X is symmetric with respect to H by assumption. Therefore there exists hy∈H such that (y-hy)⊥H. It follows, since x∈H and y∉H, that (58)y-hy≠0,y-hy⊥x,y-hy∈span⁡x,y=F. Therefore we have proved that in every 2-dimensional subspace of X every nonzero element admits a nonzero orthogonal element. It follows by a theorem of James [9] that X is isometric to a Hilbert space. This ends the proof.A Final RemarkRemark 22.
Since, by Remark7, the concept of symmetry given by Definition 6 is norm-preserving, it follows immediately that in Theorem 18 we may replace the unit balls by balls or spheres of radius ρ.


## 4. Symmetries of the SpaceC(I)

LetI be a nonempty compact subset of the set of real numbers R, I containing at least two elements, and let C(I) be the Banach space of real-valued continuous functions defined on I, where the norm of an element f∈C(I) is given by (59)f≔sup⁡fx:x∈I≔supI⁡f. Given f∈C(I) and c∈R, we let fc be the element of C(Ic) defined by (60)fcx≔fx+c,x∈Ic≔I-c. Note that fc∈C(Ic) whenever f∈C(I). If I is symmetric with respect to some c∈R, we denote by Ec(I) and Oc(I) the closed subspaces of C(I) given by (61)EcI≔f∈CI:fcis  an  even  function,OcI≔f∈CI:fcis  an  odd  function. When c=0, we let (62)OI≔O0I,EI≔E0I.With these definitions in mind, we have the following.Theorem 23.
IfI is symmetric with respect to some c∈R, then Ec(I)⊥Oc(I) and C(I) is the direct sum of Ec(I) and Oc(I): (63)CI=EcI⊕OcI; hence C(I) is symmetric with respect to both Ec(I) and Oc(I). Moreover, for each f≔fe⊕fo∈Ec(I)⊕Oc(I)=C(I), we have (64)fEcIr=fe-fo,fOcIr=fo-fe,hencefEcIr=-fOcIr, where fEc(I)r and fOc(I)r are the reflections of f with respect to Ec(I) and Oc(I), respectively.Proof.
First we prove thatEc(I) is orthogonal to Oc(I): Let h∈Ec(I) and g∈Oc(I) be given. Since hc is an even function, gc is an odd function, and -Ic=Ic; we have (65)h+tg≔supx∈I⁡hx+tgx=supx∈Ic⁡hcx+tgcx=supx∈-Ic⁡hc-x+tgc-x=supx∈Ic⁡hcx-tgcx=supx∈I⁡hx-tgx=h-tg, for every t∈R. Hence (66)EcI⊥OcI. Now, let f∈C(I) be given. Then (67)fx=fx+f2c-x2+fx-f2c-x2≔fex+fox,∀x∈I. Note that (68)fe∈EcI,fo∈OcI. It follows that C(I)=Ec(I)+Oc(I) and, since Ec(I)∩Oc(I)=0, that (69)CI=EcI⊕OcI. The theorem now follows from Theorem 20. This completes the proof.Note that Theorems23 and 18 shed some light on the shape and geometry of the unit ball of the Banach space C(I), which otherwise are not easy to visualize. In general the only guaranteed symmetry in a Banach space is symmetry with respect to the origin. But here, with the help of Theorems 23 and 18, one can actually visualize sections of the unit ball of C(I). Indeed, if we consider, for example, a 2-dimensional section of the unit ball, say of C(-1,1), by a 2-dimensional subspace F≔span⁡f,g, where f(≠0) ∈O(-1,1) and g(≠0)∈E(-1,1), then Theorem 23 tells us that the two lines Lf≔span⁡f and Lg≔span⁡g divide the 2-dimensional unit sphere of F into four identical quarters modulo reflections with respect to Lf and Lg. Note that this characteristic is not always true in Banach spaces. More generally, the following follows immediately from Theorems 23 and 18.Corollary 24.
LetI be symmetric with respect to some c∈R and let H represent one of the two subspaces Oc(I) and Ec(I). Then, for every g∈C(I)∖H, H divides each of the unit sphere SF and the unit ball BF of the subspace F≔span⁡g,H of C(I) into two identical halves modulo reflection with respect to H. The two halves of SF and BF are given by, respectively, (70)SF±≔SF∩αh±βg:h∈H,α,β∈R,BF±≔BF∩αh±βg:h∈H,α,β∈R.We note that there are symmetries inC(I) other than those mentioned above. Indeed, let I denote the cardinality of I and, for each f∈C(I), let supp⁡(f) be defined by (71)supp⁡f≔t∈I:ft≠0.Then we have the following.Proposition 25.
LetI be symmetric with respect to some c∈R, I≥4, and let H,K≔Oc(I),Ec(I). Choose any two elements k1 and k2 from the unit sphere SK of K such that (72)supp⁡k1∩supp⁡k2=∅. Now let F=span⁡H,k1,k2, F1≔span⁡H,k1, and F2=span⁡H,k2. Then (73)Fi⊥kj,F=Fi⊕span⁡kj,i,j=1,2. Moreover, the hyperplanes F1 and F2 of F together divide each of the unit sphere SF and the unit ball BF of F into four identical quarters modulo reflections with respect to F1 and F2. The four quarters of the unit sphere SF and the unit ball BF of F are given by, respectively, (74)SF±±≔SF∩αh±βk1±γk2:h∈H,α,β,γ∈R,(75)BF±±≔BF∩αh±βk1±γk2:h∈H,α,β,γ∈R.Proof.
First note that, sinceI≥4, one can indeed choose two elements k1 and k2 from the unit sphere SK of K such that (76)supp⁡k1∩supp⁡k2=∅. Let Ij≔supp⁡(kj), j=1,2, and let I3≔I∖I1∪I2. Then I is the disjoint union of I1, I2, and I3. Note that, since k1=k2=1, I1 and I2 are nonempty. Since I is symmetric with respect to c and k1,k2∈K, it follows that (77)Ijis  symmetric  with  respect  toc,j=1,2,3. First, we show that Fi⊥kj, i,j=1,2. Let h+αki∈Fi be fixed. Then we have, since I1, I2, and I3 have disjoint supports, (78)h+αki+λkj≔supI⁡h+αki+λkj=max⁡supIi⁡h+αki,supIj⁡h+λkj,supI3⁡h. Since Ij is nonempty and symmetric with respect to c, so is the closure cl(Ij) of Ij. It follows from Theorem 23 that Oc(cl(Ij))⊥Ec(cl(Ij)) and hence the restrictions of h and kj to cl(Ij) are orthogonal in CclIj. This implies that (79)supIj⁡h+λkj=supclIj⁡h+λkj=supclIj⁡h-λkj=supIj⁡h-λkj. It follows that (80)h+αki+λkj=max⁡supIi⁡h+αki,supIj⁡h-λkj,supI3h=supIh+αki-λkj=h+αki-λkj; hence h+αki⊥kj and consequently Fi⊥kj, as required.To prove thatF=Fi⊕span⁡kj,i,j=1,2, it suffices to show that kj∉Fi. But this follows directly from the fact that kj≠0 and kj⊥Fi.Note thatk1 and k2 are linearly independent. Therefore, since Ec(I)⊥Oc(I), (81)F1≔span⁡k1,H⊥span⁡k2,F=F1⊕span⁡k2. Also (82)F2:span⁡k2,H⊥span⁡k1,F=F2⊕span⁡k1. It follows from Theorem 18 that the hyperplanes F1 and F2 of F together divide the unit sphere and the unit ball of F into four identical quarters modulo reflections with respect to F1 and F2. The four quarters of the unit sphere BF of F are (83)BF±±≔BF∩αh±βk1±γk2:h∈H,α,β,γ∈R. Similarly we obtain the four quarters of the unit sphere SF of F.We conclude with some suggestions for further investigations:(1)
With Theorem18 in mind, one may look at n-dimensional or infinite-dimensional sections of the unit ball of the Banach space X over the subspace H, n≥2. For this one may need to consider the extension of the notion of orthogonality introduced in [8].(2)
Instead of a general Banach space, we may consider specific Banach spacesX and identify the subspaces H such that X is symmetric with respect to H. For instance, it is not difficult to show that if X is the usual lp sequence-space, 1≤p≤∞, then X is orthogonally symmetric with respect to all subspaces H spanned by the coordinate-vectors, H≔span⁡ei:i∈J, where J is a nonempty finite or infinite subset of positive integers and ei is the usual coordinate vector with 1 in the ith-place and zeros elsewhere. Still, are there other subspaces with respect to which X is symmetric?(3)
As mentioned above in Introduction, there are many results in the literature that rely either directly or indirectly on the geometry of the unit ball of a Banach space. It may be interesting to revisit these and investigate the presence of orthogonal symmetries and their consequences.



---

*Source: 1073589-2017-06-14.xml*