Isometries and Approximate Isometries

Some properties of isometric mappings as well as approximate isometries are studied. 2000 Mathematics Subject Classification. Primary 46B04. 1. Isometry and linearity. Mazur and Ulam [17] proved the following well-known result concerning isometries, that is, transformations which preserve distances. Theorem 1.1. Given two real normed vector spaces X and Y , let U be a surjective mapping from X onto Y such that ‖U(x)−U(y)‖ = ‖x−y‖ for all x and y in X. Then the mapping x U(x)−U(0) is linear. Since continuity is implied by isometry, the proof of this theorem consists of showing that U(x)−U(0) is additive, and the additivity of this mapping will follow if we can prove that U satisfies Jensen’s equation: U ( x 2 + y 2 ) = U(x) 2 + U(y) 2 ∀x,y ∈X. (1.1) In general this is not easy. However, for the special case in which the space Y is strictly convex, the proof is simple (see Baker [1]). A (real) normed space is called strictly convex if, for each pair of its nonzero elements y , z such that ‖y+z‖ = ‖y‖+‖z‖, it follows that y = cz for some real number c > 0. When Y is strictly convex, it is easy to show that the unique solution to the two equations ∥∥m−y1∥∥= ∥∥y2−m∥∥, ∥∥m−y1∥∥+∥∥y2−m∥∥= ∥∥y2−y1∥∥ (1.2) is m = (y1/2)+(y2/2). Following Baker [1], we find that, from the second equation, since Y is strictly convex, m−y1 = c(y2−m) with c > 0. From the first equation of (1.2), it follows that c = 1 and m= (y1/2)+(y2/2). Now, for any given pair x1, x2 in X, let h= (x1+x2)/2 and put y1 =U(x1), y2 =U(x2). Clearly, we have ∥∥h−x1∥∥= ∥∥x2−h∥∥, ∥∥h−x1∥∥+∥∥x2−h∥∥= ∥∥x2−x1∥∥. (1.3) Since U is an isometry, it follows that ∥∥U(h)−y1∥∥= ∥∥y2−U(h)∥∥, ∥∥U(h)−y1∥∥+∥∥y2−U(h)∥∥= ∥∥y2−y1∥∥. (1.4) Hence, m = U(h) is the unique solution of (1.2), so that U is a solution of Jensen’s equation (1.1). It follows that U(x)−U(0) is additive (cf. Aczél [9, page 43]). 74 THEMISTOCLES M. RASSIAS In this strictly convex case, it was not necessary to assume that the mapping U was surjective. However, in general, the Mazur-Ulam Theorem 1.1 fails without this assumption. Among counterexamples in the literature which show this, we cite the following from Figiel [10]. Let Y denote the space of real number pairs y = (y1,y2) with the norm ‖y‖ = max[|y1|,|y2|], and consider the mapping T : R → Y given by T(s)= (s,sin(s)), s in R. Then, for s1 and s2 in R, we have ∥∥T(s2)−T(s1)∥∥=max ∣∣s2−s1∣∣,∣∣sins2−sins1∣∣ } . (1.5) Now ∣∣sin(s2)−sin(s1)∣∣= ∣∣∣ ∫ s2 s1 (cosx)dx ∣∣∣≤ ∣∣s2−s1∣∣, (1.6) so that T is an isometry but clearly is not additive. Of course, this space Y is far from being strictly convex. Indeed, if we take y1 = (0,0) and y2 = (2,0) in (1.2), we see that the solutions m to these equations form an infinite set, namely the line segment joining the points (1,−1) and (1,1). In their proof of Theorem 1.1, Mazur and Ulam found a way to produce a metric characterization of the midpoint of the segment joining two points y1 and y2 of an arbitrary normed vector space Y . They did this by constructing a sequence of sets Hn in Y , n= 1,2,3, . . . , defined recursively by H1 = { y ∈ Y : ∥∥y−y1∥∥= ∥∥y−y2∥∥= ‖y1−y2‖ 2 } , Hn = { y ∈Hn−1 : ‖y−z‖ ≤ δ ( Hn−1 ) 2 ∀z ∈Hn−1 } for n≥ 2, (1.7) where δ(Hn−1) denotes the diameter of the set Hn−1. The intersection of all the sets Hn is called the metric center of y1 and y2. They proved that it consists of a single point (y1/2)+(y2/2). Vogt [24] generalized Theorem 1.1 by considering mappings which preserve “equality of distance.” These are mappings f : X → Y between normed vector spaces such that there exists a function p : [0,∞)→ [0,∞) with ‖f(x)−f(y)‖ = p(‖x−y‖). Such mappings were studied by Schoenberg [20] and by von Neumann and Schoenberg [25]. In Vogt’s approach, a basic result was the following theorem concerning isometries in metric spaces. Theorem 1.2 (Vogt [24]). Let (M,d) be a bounded metric space. Suppose that there is an element m ∈ M , a surjective isometry g :M →M and a constant k > 1 such that d(gx,x)≥ kd(m,x) for all x inM . Thenm is a fixed point for every surjective isometry h :M →M . Proof. Let h : M → M be any surjective isometry. Since surjective isometries are injective, g−1 and h−1 exist and g, h, g−1, h−1 are bijective isometries of M together with finite compositions of them. We define a sequence of isometries gn :M →M and a sequence of elements mn (n∈N) defined recursively by g1 = g, m1 =m, g2 = hgh−1, m2 = hm, (1.8) gn+1 = gn−1gn(gn−1)−1, mn+1 = gn−1mn, n≥ 2. ISOMETRIES AND APPROXIMATE ISOMETRIES 75 Clearly, each gn is invertible on M . We now prove by induction that d ( gnx,x )≥ kd(mn,x) ∀x ∈M. (1.9) By hypothesis, (1.9) is true for n= 1. When n= 2, d(m2,x)= d(hm,x)= d(m,h−1x) since h−1 is an isometry. Also, d ( g2x,x )= d(hgh−1x,x)= d(gh−1x,h−1x)≥ kd(m,h−1x)= kd(m2,x), (1.10) which demonstrates (1.9) for n= 2. Make the induction assumption that d(gnx,x)≥ kd(mn,x) when n≥ 2. Then d ( gn+1x,x )= d(gn−1gng−1 n−1x,x)= d(gng−1 n−1x,g n−1x) ≥ kd(mn,g−1 n−1x)= kd(gn−1mn,x)= kd(mn+1,x), (1.11) so the induction proof is complete and (1.9) is true for all n∈N . If we put x =mn+1 in (1.9), we find that d ( mn+2,mn+1 )= d(gnmn+1,mn+1)≥ kd(mn+1,mn). (1.12) By another induction, we obtain d(mn+2,mn+1)≥ kd(m2,m1) for all n≥ 1. Since the metric space M is bounded there exists a positive number B such that B > d(mn+2,mn+1) for all n ≥ 1. Hence (B/kn) ≥ d(m2,m1) for all n ∈ N . By hypothesis, k > 1, and it follows that d(m2,m1) = d(hm,m) = 0, so that m is a fixed point of h. Proof of Theorem 1.1. Instead of using the original proof of Theorem 1.1 due to Mazur and Ulam, we shall rely on Vogt’s method. This will “set the stage” for later developments when we deal with approximate isometries. Put f(x) = U(x)−U(0), so that f : X → Y is a surjective isometry with f(0) = 0. Given a fixed element x in X, define M as the set M = {y ∈ Y : ‖y‖ = ‖2f(x)−y‖ ≤ 2‖f(x)‖}. (1.13) Put m= f(x) and let g :M →M be given by g(y)= 2f(x)−y . Then we see that (i) M is a bounded metric space, with d(·,·)= ‖·−·‖, (ii) m= f(x) belongs to M , (iii) g is an isometry from M onto M , and g(M)=M since g is its own inverse, (iv) d(gy,y)≥ kd(m,y) for all y in M with k= 2. Property (iv) is true because d(gy,y)= ‖g(y)−y‖ = ‖2f(x)−2y‖ = 2‖f(x)−y‖ = 2d(m,y). (1.14) By Theorem 1.2 and properties (i)–(iv), we conclude that m is a fixed point of every surjective isometryh ofM . In particular, we putx0 = f−1(2f(x)) and defineh :M →M by h(y)= f(x0−f−1(y)). (v) h is a surjective isometry from M onto M . 76 THEMISTOCLES M. RASSIAS With f(x1)=y1 in M , we have ∥∥2f(x)−h(y1)∥∥= ∥∥2f(x)−f (x0−x1)∥∥= ∥∥f (x0)−f (x0−x1)∥∥= ‖x1‖ = ∥∥f (x1)∥∥= ‖y1‖ = ∥∥2f(x)−y1∥∥= ∥∥f (x0)−f (x1)∥∥ = ∥∥f (x0−x1)∥∥= ‖f (x0−f−1(y1))∥∥= ∥∥h(y1)∥∥. (1.15) Thus, h(y1) is in M . But h is its own inverse, so h(M)=M . Now, by Theorem 1.2, m is a fixed point of h. Thus, we have f(x)=m= h(m)= f (x0−f−1(m))= f (x0−x), 0= ∥∥f(x)−f (x0−x)∥∥= ∥∥x−(x0−x)∥∥= ∥∥2x−x0∥∥= ∥∥f(2x)−f (x0)∥∥ = ∥∥f(2x)−2f(x)∥∥. (1.16) Thus, we have proved that (vi) f(2x)= 2f(x) for all x ∈ X. For any given y in X, put fy(x) = f(x+y)−f(y) for all x ∈ X. Then fy : X → Y is a surjective isometry and fy(0) = 0, so that fy satisfies (vi). For any pair x,y in X, we have f(x)−f(y)= f [(x−y)+y]−f(y)= fy(x−y) = 2fy ( x−y 2 ) = 2 ( f ( x−y 2 +y ) −f(y) ) . (1.17) Thus, f(x)+f(y)= 2f ( x−y 2 +y ) = 2f ( x+y 2 ) = f(x+y). (1.18) Comments: generalizations of the Mazur-Ulam theorem. As indicated above, Vogt used his method, that was used above to prove Theorem 1.1, together with other considerations in order to prove the following generalization. Theorem 1.3. Given real normed spaces X and Y , let f : X → Y with f(0)= 0 be a surjective mapping which preserves equality of distance. Then f is linear and f = βU , where β = 0 is a real number and U :X → Y is a surjective isometry, providing that the dimension of X is greater than one. Vogt also gave an example to show that this theorem fails when X is one-dimensional. We have seen that, in general, the isometry must be surjective in order to be linear. However, Figiel [10] proved the following result which amounts to another generalization of the Mazur-Ulam theorem. Theorem 1.4. Given real normed vector spaces X and Y , let φ : X → Y be an isometric embedding of X into Y . Then there exists a linear (not necessarily continuous) mapping F : Y → X such that F ◦φ is the identity on X and the restriction of F to the linear span of φ(X) is of norm one. Various other generalizations or variations of the Mazur-Ulam theorem may be found in the literature. Charzyński [7] proved the theorem for metric vector spaces ISOMETRIES AND APPROXIMATE ISOMETRIES 77 of the same finite dimension. Day [8, pages 110–111], demonstrated a version of the theorem involving semi-norms (instead of norms or metrics) in locally convex topological vector spaces. Rolewicz [19] proved the theorem for a class of metric vector spaces which are locally bounded and such that the function φ(t) = d(tx,0) is concave in the real t for each fixed x in the space. An example of such a space is the set of sequences x = (x1,x2, . . .) such that d(x,0)= ∑∞ k=1 |xk|p converges, with 0 1. (1.19) Using Baker’s lemma, which states that ‖a+b‖ = ‖a‖+‖b‖ implies that ‖sa+tb‖ = s‖a‖+t‖b‖ when s ≥ 0 and t ≥ 0, it is easy to see that f is an isometry. Also, f(0)= 0 and f is nonlinear. The second example was constructed by Baker in answer to a problem of Chernoff (Advanced Problem 5688, Amer. Math. Monthly 76 (1969), 835). It shows that an isometry can be not only nonlinear but also homogeneous of degree one, and is defined as follows: let Y be the set R3 but with the metric given by the norm ‖(x,y,z)‖ = max [√ x2+y2,|z|]. Let f :R2 → Y be given by f(x,y)= (x,y,g(x,y)), with g(x,y)=   y if 0≤y ≤ x or x ≤y ≤ 0, x if 0≤ x ≤y or y ≤ x ≤ 0, 0 otherwise. (1.20) Clearly, g satisfies g(tx,ty)= tg(x,y) for real t, x, y , and it is easily seen that g is not linear. Also it can be shown that |g(x,y)−g(u,ν)| ≤ √ (x−u)2+(y−v)2. Thus, f :R2 → Y is an isometry which is nonlinear and homogeneous. 2. Approximate isometries: special cases. By an ε-isometry of one metric space E into another E′, we mean a mapping T : E → E′ which changes distances between elements of E by at most ε for some fixed ε > 0, that is, ∣∣d′(T(x),T(y))−d(x,y) ∣∣≤ ε ∀x, y ∈ E, (2.1) where d and d′ are the metrics for E and E′, respectively. The stability question here is: given η > 0, does there exist ε > 0 and an isometry U : E → E′ such that d′(T(x),U(x)) < η for all x in E? In the case, where E is a real Hilbert space and E = E′, an answer was given in the affirmative by Hyers and Ulam [14], providing that T was surjective. Using the geometry of Hilbert space, they showed that, if T : E → E is an ε-isometry with T(0)= 0, then ∥∥∥T ( x 2 ) − T(x) 2 ∥∥∥≤ √ 2ε‖x‖+2ε, x ∈ E. (2.2a) 78 THEMISTOCLES M. RASSIAS They proved that the limit U(x)= lim n→∞ T ( 2nx ) 2n (2.2b) exists for all x in E and that U is an isometry. Finally, it was shown that, if the ε-isometry T is surjective, then U is also surjective and the inequality ∥∥T(x)−U(x)∥∥< kε (2.3) is satisfied for all x in E, with k= 10. This theorem was generalized by Bourgin [3], who obtained the results (2.2b) and (2.3) with k = 12 for an ε-isometry T of a Banach space E1 onto a Banach space E2, where E2 belongs to a class of uniformly convex spaces which includes the spaces Lp(0,1) for 1 0, define the sets M1(p,b,h)= { φ ∈ C(S1) : |φ(p)−b| ≤ h}, M2(q,c,k)= { ψ∈ C(S2) : |ψ(q)−c| ≤ k}. (2.4) Then, for each point p of S1, each b in R and each a ≥ 0, there exists a point q in S2 and c in R such that T ( M1(p,b,a) ) ⊂M2 ( q,c,a+ 3ε 2 ) . (2.5) This lemma was the first step in establishing a correspondence between points p of S1 and q of S2. A corollary was proved to show that, when T is a surjective homeomorphism as well as an ε-isometry of C(S1) onto C(S2), then the point q of C(S2) is uniquely determined by the point p of S1 and is independent of the choice of the parameters b and a. Finally, the following stability result was proved. Theorem 2.2. Let S1 and S2 be compact metric spaces. If T is a homeomorphic mapping of C(S1) onto C(S2) which is also an ε-isometry, then there exists an isometry U : C(S1)→ C(S2) such that ‖U(f)−T(f)‖ ≤ 21ε for all f in C(S1). ISOMETRIES AND APPROXIMATE ISOMETRIES 79 Corollary 2.3. Under the hypotheses of Theorem 2.2, the underlying spaces S1 and S2 are homeomorphic. Remark 2.4. Incidentally, Corollary 2.3 shows that we have a generalization of Banach’s theorem, where isometry is replaced by ε-isometry. Stone [22] generalized Banach’s theorem by replacing compact metric spaces with completely regular topological spaces, and this result is now known as the BanachStone theorem (cf. [8, page 86]). Bourgin [4] obtained a significant generalization of Theorem 2.2 by adapting the basic idea of the proof outlined above to the more general case, with the following principal result. Here C(Sj) denotes the space of bounded continuous functions on Sj , j = 1,2. Theorem 2.5. Let S1 and S2 be completely regular Hausdorff spaces and let T : C(S1) → C(S2) be a surjective ε-isometry with T(0) = 0. Then there exists a linear isometry U : C(S1)→ C(S2) for which ‖T(f)−U(f)‖ ≤ 10ε, for all f ∈ C(S1). Note. No continuity conditionswere needed in Bourgin’s theoremon the ε-isometry T , in contrast to the conditions of Theorem 2.2. His theorem also results in a generalization of the Banach-Stone theorem, again by replacing isometries by ε-isometries. More explicitly, he defined a μ-ideal as the subset of C(S1) given by J(p,μ) = {φ ∈ C(S1) : |φ(p)|< μ}, and similarly for C(S2), and proved the following theorem. Theorem 2.6. Let S1 and S2 be compact Hausdorff spaces. IfH is a possibly multiplevalued mapping of C(S1) onto C(S2) and yields a correspondence of μ-ideals of either algebra to (μ+p)-ideals of the other, then S1 and S2 are homeomorphic. 3. Approximate isometries: the general case. The problem of stability of isometry for mappings between arbitrary real normed vector spaces remained open for many years after the work on special cases discussed in the previous section. In the present section, all normed spaces are assumed to be real. The study of ε-isometries of Banach spaces was revived by Bourgin [6] and Bourgin [5], Omladǐc and Šemrl [18]. Gruber [13] demonstrated the stability of surjective isometries between all finite dimensional normed vector spaces. In addition, these authors obtained information concerning ε-isometries between arbitrary normed vector spaces which will be indispensible in the proof of the general case to be given below. In particular, the following lemma is a slight modification of Bourgin [6, Lemma 2.8]. Lemma 3.1 (R. D. Bourgin and P. L. Renz). Let X and Y be normed vector spaces, η and δ ∈ R+, and let T : X → Y be an η-isometry with T(0) = 0. Then there exists a continuous (2δ+3η)-isometry T ′ : X → Y such that ‖T ′(x)−T(x)‖ < η+δ for each x ∈X. Proof. Consider all subsets of X such that the distance between each pair of distinct elements is at least δ/2. By Zorn’s lemma, there exists a maximal such collection which will be denoted by {xγ : γ ∈ Γ}. Then, for any x ∈ X, there is a γ ∈ Γ with ‖x−xγ‖ ≤ δ/2. Let Bγ denote the ball {x ∈X : ‖x−xγ‖< δ}. 80 THEMISTOCLES M. RASSIAS Then {Bγ : γ ∈ Γ} is an open cover of X. Hence, there is a partition of unity {fξ : ξ ∈ Ξ} subordinate to {Bγ : γ ∈ Γ}. For each ξ ∈ Ξ, pick any γ ∈ Γ such that {y ∈ X : fξ(y) = 0} ⊂ Bγ and denote this γ by γ(ξ). Thus, {y ∈X : fξ(y) = 0} ⊂ Bγ(ξ) for each ξ ∈ Ξ. Define T ′ : X → Y by the formula T ′(x) = ∑ξ∈Ξfξ(x)T(xγ(ξ)) for each x ∈ X. Thus the function T ′ is well defined and continuous. Also, for x ∈X, ∥∥T ′(x)−T(x)∥∥= ∥∥∥∥ ∑ ξ∈Ξ fξ(x) ( T ( xγ(ξ) )−T(x)) ∥∥∥∥ ≤ ∑ ξ∈Ξ fξ(x) ∥∥T(xγ(ξ))−T(x)∥∥ ≤ ∑ ξ∈Ξ fξ(x) (∥∥xγ(ξ)−x∥∥+η). (3.1) Now, if fξ(x) = 0, then ‖x−xγ(ξ)‖< δ, so that ∥∥T ′(x)−T(x)∥∥≤ ∑ ξ∈Ξ fξ(x)(δ+η)= δ+η ∀x ∈X. (3.2) It follows that T ′ is a continuous (2δ+3η)-isometry. Bourgin [6] also introduced the following concept which also will be useful later. Definition 3.2. Given δ > 0, a function f : X → Y between normed vector spaces is said to be δ-onto if, for each y ∈ Y , there is a point x ∈X for which ‖f(x)−Y‖ ≤ δ. Gruber [13] obtained an elegant and definitive result, as follows. Theorem 3.3. Let X and Y be normed spaces. Given ε > 0, suppose that T : X → Y is a surjective ε-isometry while U : X → Y is an isometry such that T(0) = U(0) = 0. If ‖T(x)−U(x)‖/‖x‖→ 0 uniformly as ‖x‖→∞ uniformly, then U is a surjective linear isometry and ∥∥T(x)−U(x)∥∥≤ 5ε ∀x ∈X. (3.3) The proof requires several lemmas subject to the hypotheses of Theorem 3.3. Lemma 3.4. U is a surjective isometry. Proof. From the hypotheses of the theorem, we have ‖x‖/‖Tx‖ ≤ 1+ε for large ‖x‖, and ‖T(x)−U(x)‖ ‖T(x)‖ = (‖T(x)−U(x)‖ ‖x‖ )( ‖x‖ ‖T(x)‖ ) ≤ (‖T(x)−U(x)‖ ‖x‖ ) (1+ε), (3.4) so that (‖T(x)−U(x)‖ ‖T(x)‖ ) → 0 (3.5) uniformly as ‖T(x)‖→∞ uniformly. Hence, Y is the closure of the linear hull of U(X). Now we use Theorem 1.4 above to deduce the existence of a map J : Y → X, where J is linear and such that J ◦U = idX = identity on X, and ‖J‖ = 1. We must show that J is bijective. Assume that J is not injective. Then, because J is linear, there exists y = 0 in Y with J(y) = 0, so that J(βy) = 0 for all β ∈ R+. As was shown above, ISOMETRIES AND APPROXIMATE ISOMETRIES 81 ‖T(x)−U(x)‖/‖T(x)‖ → 0 uniformly as ‖T(x)‖ → ∞ uniformly. Therefore, we may choose for each β ∈ R+ a point xβ ∈ X such that ‖βy −U(xβ)‖/‖βy‖ and so also ‖βy−U(xβ)‖/β→ 0 as β→∞. Recalling that J ◦U = idX and ‖J‖ = 1, we have 0= ∥∥J(βy)∥∥≥ ∥∥J(U(xβ))∥∥−∥∥J(βy)−J(U(xβ))∥∥≥ ∥∥xβ∥∥−∥∥βy−U(xβ)∥∥ = ∥∥U(xβ)∥∥−∥∥βy−U(xβ)∥∥≥ ‖βy‖−2∥∥βy−U(xβ)∥∥, (3.6) so that 0≥ ‖y‖− 2 ∥∥βy−U(xβ)∥∥ β →‖y‖ as β →∞. (3.7) This contradiction proves that J is injective. Since J ◦U = idX , J is also surjective and thus bijective. Lemma 3.5. The isometry U :X → Y is given by the formula U(x)= lim2−nT(2nx), x ∈X. (3.8) Proof. By the hypotheses of the theorem, given δ > 0, there exists Mδ > 0 such that ‖T(z)−U(z)‖ ≤ δ‖z‖ for all z ∈X with ‖z‖>Mδ. For x = 0, equation (3.8) holds. For a given x = 0 in X, choose m so large that ‖2mx‖ ≥Mδ. Now U is linear, so, when n≥m, we have ∥∥T(2nx)−2nU(x)∥∥≤ δ∥∥2nx∥∥, ∥∥2−nT(2nx)−U(x)∥∥≤ δ‖x‖. (3.9) Given η > 0, choose δ < η/‖x‖ and we have ∥∥∥T ( 2nx ) 2n −U(x) ∥∥∥≤ η for n≥m. (3.10) By Lemmas 3.4, 3.5, and the fact that T(0)= 0, we find that: V :=U−1T :X →X is a surjective ε-isometry with V(0)= 0, x = lim n→∞ −nV ( 2nx ) ∀x ∈X. (3.11) Now choose a mapping V−1 : X → X as follows. Put V−1(0) = 0, and for x ∈ X with x = 0, let V−1(x) be any pointy ∈X with V(y)= x. Then clearly V−1 has the following properties: V−1 :X →X is an ε-isometry with V−1(0)= 0, (3.12) V ( V−1(x) )= x ∀x ∈X. (3.13) For x ∈X, by (3.11), (3.12), and (3.13), we have ∥∥2−nV−1(2nx)−x∥∥= 2−n∥∥V−1(2nx)−2nx∥∥ ≤ 2−n(∥∥2nx−V(2nx)∥∥+ε) = ∥∥x−2−nV(2nx)∥∥+2−nε, (3.14) so that x = lim n→∞ −nV−1 ( 2nx ) ∀x ∈X. (3.15) 82 THEMISTOCLES M. RASSIAS We now use Lemma 3.1 to approximate the ε-isometry V as follows. For each integer k > 0 let a continuous mapping Vk :X →X be chosen so that ∥∥Vk(x)−V(x)∥∥< (1+k−1)ε. (3.16) Lemma 3.6. Let y ∈ X and h in the dual space X∗ be chosen so that ‖y‖ = 1 and ‖h‖ = h(y)= 1. Then h(V(βy))≥ β−4ε for each β > 0. Proof. Since ‖h‖ = 1= ‖y‖, we use (3.11), (3.12), (3.13), and (3.16) to obtain h ( −Vk ( 2−nV−1 ( 2nβy) )+2nβy)≤ ∥∥Vk(2−nV−1(2nβy))−2nβy∥∥ ≤ ∥∥V(2−n(V−1(2nβy)))−2nβy∥∥+(1+k−1)ε = ∥∥V(2−n(V−1(2nβy)))−V(V−1(2nβy))∥∥+(1+k−1)ε ≤ ∥∥2−nV−1(2nβy)−V−1(2nβy)∥∥+(2+k−1)ε = (1−2−n)∥∥V−1(2nβy)∥∥+(2+k−1)ε ≤ (1−2−n)(∥∥2nβy∥∥+ε)+(2+k−1)ε = (2n−1)β+(3+k−1−2−n)ε. (3.17) Since h is linear and h(y) = 1 we have h(Vk(2−nV−1(2nβy))) ≥ β−(3+k−1−2−n)ε. Now we use (3.15) and the fact that h and Vk are both continuous, so that whenn→∞, we obtain h ( Vk(βy) )≥ β−(3+k−1)ε. (3.18) Therefore, by (3.16) and the fact that ‖h‖ = 1, we have h(V(βy)) ≥ h(Vk(βy))− (1+k−1)ε, so that h(V(βy))≥ β−(4+2k−1)ε. Since this is true for each integer k > 0, Lemma 3.6 follows. Lemma 3.7. For each x ∈X, ‖x−V(x)‖ ≤ 5ε. Proof. Denote the closed ball in X with center at zero and radius 1 by B. Let x ∈X be chosen. Take y ∈X with ‖y‖ = 1 and such that x−V(x) is an element of the halfray starting from the origin and containing y . Thus, x−V(x) = ωy , where ω ≥ 0. Next, choose u∈X such that y−u is a half-tangent to B at y , that is, ‖y−δu‖−1 δ → 0 as δ → 0+, (3.19) and also that x = μy+θu (3.20) for suitable real numbers μ and θ, where θ ≥ 0. Thus, V(x) has the form V(x)= νy+θu (3.21) with ν = μ−ω∈R. We note that x−V(x)= (μ−ν)y with μ−ν =ω≥ 0. (3.22) ISOMETRIES AND APPROXIMATE ISOMETRIES 83 Denote by L the linear subspace generated by u and y . The line through y and y−u is a tac-line of B. Thus, the linear functional k : L→R defined by k(ξy+ηu)= ξ for ξ and η ∈ R, is of norm one. By the Hahn-Banach theorem, we can extend k to a linear functional h : X → R which also has norm one. Now ‖h‖ = h(y) = k(y) = ‖y‖ = 1, so by Lemma 3.6 we have h(V(βy)) ≥ β−4ε with β > 0. Since V is an ε-isometry, we have ‖βy−x‖ ≥ ∥∥V(βy)−V(x)∥∥−ε ≥ h(V(βy)−V(x))−ε = h(V(βy))−h(V(x))−ε. (3.23) Using (3.21) and noting that h(u)= k(u)= 0, we obtain the inequality ‖βy−x‖ ≥ β−5ε−h(νy+θu)= β−ν−5ε for each β > 0. (3.24) On the other hand, by (3.20) for β > μ, we have ‖βy−x‖ = ‖βy−μy−θu‖ = (β−μ) ∥∥∥y− ( θ β−μ ) u ∥∥∥. (3.25) By (3.19), given any η > 0, we may choose β= β(η) so large that ‖βy−x‖ ≤ β−μ+η. Using (3.24), we obtain β−μ ≥ ‖βy −x‖−η ≥ β−ν −5ε−η, or 0 ≥ μ−ν −5ε−η. Since η may be chosen arbitrarily small, we have μ− ν ≤ 5ε, so, from (3.22), since ‖y‖ = 1, we see that ‖x−V(x)‖ ≤ 5ε. Now, by definition, V = U−1T , and, since U is an isometry, we conclude that ‖U(x)−T(x)‖ ≤ 5ε. Corollary 3.8. Under the hypotheses of Theorem 3.3, suppose that in addition T is continuous. Then the above inequality may be improved as ‖U(x)−T(x)‖ ≤ 3ε. Proof. When T is continuous, we can eliminate the approximations Vk and use V directly instead. Then the conclusion of Lemma 3.6 can be changed to h(V(βy)) ≥ β−2ε, and the corollary follows. It remained for Gevirtz [12] to at last prove the following result which establishes the stability of isometries between arbitrary Banach spaces. Theorem 3.9. Given real Banach spaces X and Y , let f : X → Y be a surjective εisometry. Then there exists a surjective isometryU :X→Y for which ‖f(x)−U(x)‖≤5ε. The main part of the proof is to demonstrate the following result. Theorem 3.10 (Gevirtz [12]). If f : X → Y is a surjective ε-isometry, then, for any pair x0,x1 ∈X, we have ∥∥∥f ( x0+x1 2 ) − f ( x0 )+f (x1) 2 ∥∥∥≤ 10 √ ε‖x0−x1‖+20ε. (3.26) Proof. The idea of the proof was to “epsilonize” the method of Vogt which was used above in demonstrating Theorems 1.1 and 1.2. From Bourgin’s δ-onto idea, Gevirtz developed the following definition. Definition 3.11. Given f :X → Y , a mapping F : Y →X for which ‖fF(y)−y‖ ≤ δ (3.27) 84 THEMISTOCLES M. RASSIAS for all y ∈ Y is called a δ-inverse of f and f is called δ-onto if it has a δ-inverse. Also, the term (δ,ε)-isometry is an abbreviation for “δ-onto ε-isometry.” We will need the following lemmas. Lemma 3.12. If f : X → Y is a (δ,ε)-isometry and F is a δ-inverse of f , then F is a (δ+ε,2δ+ε)-isometry. Proof. ‖Ff(x)−x‖ ≤ ‖f(Ff(x))−f(x)‖+ε ≤ δ+ε by (3.27), so F is (δ+ε)-onto. For y0 and y1 ∈ Y we have by (3.27) that ‖fF(yj)−yj‖ ≤ δ, j = 0,1. Since f is an ε-isometry, |‖fF(y0)−fF(y1)‖−‖F(y0)−F(y1)‖| ≤ ε. Hence, ∣∣∥∥F(y0)−F(y1)∥∥−∥∥y0−y1∥∥∣∣ ≤ ∣∣∥∥fF(y0)−fF(y1)∥∥−∥∥y0−y1∥∥∣∣+ε ≤ ∣∣∥∥fF(y0)−y0∥∥+∥∥y1−fF(y1)∥∥+∥∥y0−y1∥∥−∥∥y0−y1∥∥∣∣+ε ≤ 2δ+ε, (3.28) so that F is a (2δ+ε)-isometry. Lemma 3.13. Let f1 :X → Y be a (δ1,ε1)-isometry, let Z be another Banach space and let f2 : Y → Z be a (δ2,ε2)-isometry. Then f2f1 :X → Z is a (δ1+δ2+ε2,ε1+ε2)-isometry. Proof. By hypotheses, we have ∣∣∥∥f2f1(x0)−f2f1(x1)∥∥−∥∥f1(x0)−f1(x1)∥∥∣∣≤ ε2, ∣∣∥∥f1(x0)−f1(x1)∥∥−∥∥x0−x1∥∥∣∣≤ ε1, (3.29) so that ∣∣∥∥f2f1(x0)−f2f1(x1)∥∥−∥∥x0−x1∥∥∣∣≤ ε2+ε1. (3.30) Thus, f2f1 is an (ε1+ε2)-isometry. Now let Fj be a δj-inverse of fj , j = 1,2, and let z be any element of Z . Since f2 is an ε2-isometry, we can apply (3.27), to f2 and f1, to obtain ∥∥f2f1F1F2(z)−z∥∥≤ ∥∥f2f1F1F2(z)−f2F2(z)∥∥+∥∥f2F2(z)−z∥∥ ≤ ∥∥f1F1F2(z)−F2(z)∥∥+ε2+δ2 ≤ δ1+δ2+ε2. (3.31) With the given x0 and x1 ∈X, let y0 = f(x0), y1 = f(x1) and put p = x0+x1 2 , q = y0+y1 2 . (3.32) For the present, we assume that y0 = y1. Since f is surjective it has an inverse, so it is a (0,ε)-isometry. By Lemma 3.12, its inverse F is an (ε,ε)-isometry for which F ( y0 )= x0, F(y1)= x1. (3.33) We now define sequences (gk)k≥0 and (Gk)k≥0 as follows: g0(y)= f ( 2p−F(y)) for y ∈ Y . (3.34) ISOMETRIES AND APPROXIMATE ISOMETRIES 85 By Lemma 3.13, g0 : Y → Y is a (2ε,2ε)-isometry, and clearly g0(yj)=y1−j for j = 0,1. We define G0 : Y → Y to be any mapping such that: G0 is a (4ε)-inverse of g0, G0(yj)=y1−j , j = 0,1. (3.35) For y ∈ Y , let g1(y)=G1(y)= 2q−y. (3.36) Finally, if we are given g0, . . . ,gn and G0, . . . ,Gn, we define Gn+1 as any mapping which is a 4n+2 ε-inverse of gn+1 and Gn+1(yj)=y1−j , j = 0,1, where gn+1 = gn−1gnGn−1 (n≥ 1). (3.37) Lemma 3.14. The sequences (gk) and (Gk) defined by (3.34), (3.35), (3.36), and (3.37) have the following properties: gk is a ( 4k+1ε,4k+1ε ) -isometry, gk(yj)=y1−j , j = 0,1, (3.38) Gk is a ( 4k+1ε ) -inverse of gk, Gk ( yj )=y1−j , j = 0,1. (3.39) Proof. The properties (3.38) and (3.39) are true for k = 0 and k = 1, as we have seen. Make the induction assumption that both are true when 0 ≤ k ≤ n. By (3.37) we have gn+1 = gn−1gnGn−1 and we know that gn−1 is a (4nε,4nε)-isometry and Gn−1 is a (4nε)-inverse of gn−1. Hence, by Lemmas 3.12 and 3.13, it follows that gn+1 is a (4n+2ε,4n+2ε)-isometry. Clearly, gn+1(yj) = gn−1gnGn−1(yj) = y1−j , j = 0,1 so gn+1 satisfies (3.38) with k = n+1. By definition, Gn+1 satisfies (3.39) with k = n+1, and the induction is complete. Proof of Theorem 3.10. We define a sequence of points an ∈ Y recursively by a1 = q = y0+y1 2 , an+1 = gn−1 ( an ) , n≥ 1. (3.40) Put d = ‖y0 −y1‖/2. Let B(y,r) denote the closed ball in Y with center y and radius r > 0. Since gk is a (4k+1ε)-isometry for k ≥ 0, when y ∈ B(y0,r ), we have |‖gk(y)−gk(y0)‖−‖y−y0‖| ≤ 4k+1ε and therefore ‖gk(y)−y1‖ ≤ r +4k+1ε, where we have used gk(y0)=y1. Thus, gk(B(y0,r ))⊂ B(y1,r+4k+1ε), and clearly the same is true with y0 and y1 interchanged. Therefore, we have gk ( B ( yj,r ))⊂ B(y1−j ,r +4k+1ε), j = 0,1. (3.41) Evidently, a1 ∈ B(y0,d)∩B(y1,d). Also an = gn−2(an−1) = gn−2gn−3 ···g0(a1), so an induction based on successive applications of (3.41) with k = 0,1, . . . ,n−2 gives an ∈ B(y0,d+4nε)∩B(y1,d+4nε)⊂ B(q,d+4nε) for n≥ 1. Now the diameter of this last ball is 2(d+4nε), so that ∥∥an−an−1∥∥≤ 2(d+4nε), n≥ 2. (3.42) Next we prove by induction that ∥∥gn(y)−y∥∥≥ 2∥∥an−y‖−2(4nε), n≥ 1. (3.43) 86 THEMISTOCLES M. RASSIAS For n = 1, we have by (3.36) and (3.40) that g1(y) = 2q − y = 2a1 − y , so that, ‖g1(y)−y‖ = ‖2a1 − 2y‖ = 2‖a1 −y‖ ≥ 2‖a1 −y‖ − 2(4ε), which verifies (3.43) for n= 1. Assuming that (3.43) is true, we shall show that it holds when n is replaced by n+1. Using the recursion formula (3.37) and Lemma 3.14, we obtain ∥∥gn+1(y)−y∥∥= ∥∥gn−1gnGn−1(y)−y∥∥ = ∥∥gn−1gnGn−1(y)−gn−1Gn−1(y)+gn−1Gn−1(y)−y∥∥ ≥ ∥∥gn−1gnGn−1(y)−gn−1Gn−1(y)∥∥−∥∥gn−1Gn−1(y)−y∥∥ ≥ ∥∥gn−1gnGn−1(y)−gn−1Gn−1(y)∥∥−4nε ≥ ∥∥gnGn−1(y)−Gn−1(y)∥∥−2(4nε). (3.44) Now use the induction hypothesis with y replaced by Gn−1(y) to find that ∥∥gn+1(y)−y∥∥≥ 2∥∥an−Gn−1(y)∥∥−4n+1ε ≥ 2(∥∥gn−1(an)−gn−1Gn−1(y)∥∥−4nε)−4n+1ε ≥ 2(∥∥an+1−y∥∥−2(4nε))−4n+1ε = 2∥∥an+1−y∥∥−2(4n+1ε), (3.45) so the induction is complete for (3.43). By (3.40) and (3.43), we have ‖an+1 − an‖ = ‖gn−1(an)− an‖ ≥ 2‖an − an−1‖ − 2(4n−1ε), for n ≥ 1. We replace n by n− 1 to get ‖an −an−1‖ ≥ 2‖an−1 −an−2‖− 2(4n−2ε) and again ‖an−1−an−2‖ ≥ 2‖an−2−an−3‖−2(4n−3ε). Substituting the last inequality into the preceding one, we have ∥∥an−an−1∥∥≥ 22∥∥an−2−an−3∥∥−4n−1ε. (3.46) By induction, we arrive at the inequality ∥∥an−an−1∥∥≥ 2n−2∥∥a2−a1∥∥−4n−1ε, for n≥ 2. (3.47) Now, by (3.42), ∥∥an−an−1∥∥≤ 2(d+4nε), (3.48) and we have ∥∥a2−a1∥∥≤ 22−n(2d+2(4nε)+4n−1ε) for n≥ 2. (3.49) Changing n to n+2, now for n≥ 0, we find that ∥∥a2−a1∥∥≤ 2(d2−n+18ε2n). (3.50) By (3.34) and (3.40), a1 = q and a2 = g0(q)= f(2p−F(q)). Hence, ∥∥a2−a1∥∥= ∥∥f (2p−F(q))−q∥∥= ∥∥f (2p−F(q))−fF(q)∥∥≥ ∥∥2p−2F(q)∥∥−ε = 2∥∥p−F(q)∥∥−ε ≥ 2(∥∥f(p)−fF(q)∥∥−ε)−ε = 2∥∥f(p)−q∥∥−3ε. (3.51) Thus, by (3.50), for n≥ 0 we have ∥∥f(p)−q∥∥≤ d2−n+18ε2n+2ε. (3.52) We consider two cases according to whether or not d> 18ε. ISOMETRIES AND APPROXIMATE ISOMETRIES 87 Case 1 (d > 18ε). Let the real number t be defined by d2−t = 18ε2t , that is, t = log4 (d/18ε) > 0. Let n be the greatest integer in t. Then 2−n < 2·2−t so by (3.52) ∥∥f(p)−q∥∥≤ 2d·2−t+18ε2t+2ε = 3d·2−t+2ε, (3.53) by the definition of t. Note also from this definition that d ·2−2t = 18εd, d·2−t = √ 18εd. (3.54) Hence, ∥∥f(p)−q∥∥≤ 3√18εd+2ε. (3.55) Since f is an ε-isometry and d> 18ε, we have ∥∥x0−x1∥∥≥ ∥∥y1−y0∥∥−ε = 2d−ε > 2d− d 18 = 35d 18 , (3.56) and so 18d≤ ( 182 35 ∥∥x0−x1∥∥. (3.57) Thus, in Case 1, we find that ∥∥f(p)−q∥∥≤ 10 √ ε‖x0−x1‖+2ε. (3.58) Case 2 (d≤ 18ε). This case covers the situation, where y0 =y1 which was excluded earlier. Here ∥∥y0−y1∥∥= 2d≤ 36ε, (3.59) so that ∥∥x0−x1∥∥≤ 37ε. (3.60) Thus, ‖xj−p‖ ≤ 19ε and ‖yj−f(p)‖ ≤ 20ε, j = 0,1. Since q = y0+y1 2 , ‖f(p)−q‖ = ∥∥∥f(p) 2 − y0 2 + f(p) 2 − y 2 ∥∥∥≤ ∥∥∥f(p)−y0 2 ∥∥∥+ ∥∥∥f(p)−y1 2 ∥∥∥≤ 20ε. (3.61) Therefore, in either Case 1 or Case 2, we have demonstrated the inequality ∥∥f(p)−q∥∥≤ 10 √ ε ∥∥x0−x1∥∥+20ε, (3.62) which by (3.32) is the required inequality (3.26) of Theorem 3.10. Proof of Theorem 3.9. We may assume, without loss of generality, that f(0)= 0. Put x0 = 2x, x1 = 0 in (3.26) to get ∥∥∥f(2x) 2 −f(x) ∥∥∥≤ 10 √ 2 ( ε‖x‖)1/2+20ε. (3.63) We prove by induction that ∥∥∥f ( 2nx ) 2n −f(x) ∥∥∥≤ 10 √ 2   n−1 ∑ k=0 2−k/2  (ε‖x‖)1/2+20ε n−1 ∑