Step-isometries *

Peter Semrl�

Abstract

Motivated by a problem in the theory of random geometric graphs astep-isometry on a dense subset A of a real Banach space X is definedas a map f : A → X satisfying b‖x − y‖c = b‖f(x) − f(y)‖c, x, y ∈ A.Under the additional assumption that f maps A onto itself a complete de-scription of such maps on finite-dimensonal spaces has been obtained byBalister, Bollobas, Gunderson, Leader, and Walters. What happens in theabsence of this additional assumption and in the case that X is infinite-dimensional? Studying step-isometries on classical sequence spaces weshow that for certain finite-dimensional spaces we can get their result forgeneral step-isometries satisfying no surjectivity assumption. And underthe assumption of almost surjectivity, which is weaker than the assump-tion f(A) = A, this result can be extended to the infinite-dimensionalcase. By an example we show that the result of Balister, Bollobas, Gun-derson, Leader, and Walters is optimal for Rn equipped with the `∞norm and extend also this result to the infinite-dimensional setting.

AMS classification: 05C63, 05C80, 46B04.

Keywords: Banach space, isometry, step-isometry, approximate isometry, ran-dom graph.

1 Introduction and notation

Let X be a real Banach space and A ⊂ X a dense subset. A mapping f : A→ Xis said to be a step-isometry if

b‖x− y‖c = b‖f(x)− f(y)‖c (1)

for all pairs of vectors x, y ∈ A. This notion was introduced in the seriesof papers [4, 5, 6] where it was shown that the question whether a randomgraph is almost surely unique up to an isomorphism can be reformulated as the

*This research was supported by grants N1-0061, J1-8133, and P1-0288 from ARRS, Slove-nia.

�Faculty of Mathematics and Physics, University of Ljubljana, Jadranska 19, SI-1000 Ljubl-jana, Slovenia; Institute of Mathematics, Physics, and Mechanics, Jadranska 19, SI-1000Ljubljana, Slovenia, peter.semrl@fmf.uni-lj.si

1

problem of the existence of step-isometries of A onto A that are not isometries.More precisely, let A be countable and 0 < p < 1. We form a random graphon A by joining, independently and with probability p, each pair of points atdistance less than 1. The set A is called Rado if any two such random graphs arealmost surely isomorphic. Bonato and Janssen [4] observed that if X is finite-dimensional then for almost all pairs of random graphs G,G′ every functionf : A → A inducing an isomorphism of the graphs is a step-isometry on A.Therefore, a natural step towards characterizing possible graph isomorphismsis to describe the general form of bijective step-isometries.

In the finite-dimensional case the problem has been completely solved byBalister, Bollobas, Gunderson, Leader, and Walters [1] (for some first partialresults on almost sure uniqness of random graphs in the infinite-dimensionalcase we refer to [7, 8]). As pointed out by the authors the problems concerningstep-isometries are of independent interest. We are interested in such maps inthe absence of the assumption that f maps A onto itself.

When dealing with step-isometries on finite-dimensional space X satisfyingf(A) = A, Balister, Bollobas, Gunderson, Leader, and Walters first showed thatevery such map can be extended to the unique continous step-isometry on thewhole space X. For this initial step in their study of step-isometries both theasumption that f maps A bijectively onto itself and the finite-dimensionalityassumption were essential. They remarked that in the absence of the bijectivityassumption many of their results are no longer true.

Let ξ : [0, 1] → [0, 1] be a bijective strictly increasing function. Then, ofcourse, ξ(0) = 0, ξ(1) = 1, and ξ is continuous. It is trivial to verify that thefunction τ : R→ R defined by

τ(x) = ξ(x− bxc) + bxc, x ∈ R,

is a step-isometry. Each such step-isometry will be called a standard step-isometry on R.

A map f : A→ X is said to be almost surjective if there exists a positive realnumber δ such that for every y ∈ X there exists x ∈ X such that ‖f(x)−y‖ ≤ δ.

The results obtained in this paper seem to imply that if we want to get anice description of step-isometries f : A→ X defined on a dense subset A of aBanach space X we can do this

� for certain Banach spaces with no additional assumptions,

� for certain Banach spaces under the additional assumption of almost sur-jectivity,

� for certain Banach spaces under the additional assumption that the imageof f is dense in X.

Here, with a nice description we mean that either f is a restriction of a linearbijective isometry of the whole space composed with the translation, or thatf can be described with the use of linear bijective isometries, standard step-isometries of the real axis, and translations.

2

In the next section we will present our main results. These are nice descrip-tions of step-isometries of classical sequence spaces which together with coun-terexamples illustrate the three possibilities mentioned in the previous para-graph. The third section will be devoted to preliminary results which are ofindependent interest as they can be applied to Banach spaces other than classi-cal sequence spaces considered in our main theorems. Actually, a careful readerwill realize that based on these preliminary results the sequence spaces in ourmain results could have been replaced by larger classes of Banach spaces havingenough nice properties. But at present we are still very far from a completesolution of the challenging problem to identify Banach spaces that belong tothe three classes as described above.

The fourth section is devoted to the proofs of the main results. With this theproblem of describing the general form of step-isometries of classical sequencespaces is almost solved. We will finish the paper with some open questions.

2 Main results

In this section we will describe step-isometries defined on dense subsets of real`p spaces, 1 < p ≤ ∞. The symbol `np will denote the space Rn equipped withthe p-norm.

Theorem 2.1. Let n ≥ 2 be an integer and p a real number, 1 < p <∞, p 6= 2.Assume that A ⊂ `np is a dense subset and f : A → `np a step-isometry. Thenthere exist c1, . . . , cn ∈ {−1, 1}, a permutation σ ∈ Sn, and a vector a ∈ `np suchthat

f ((xk)nk=1) = (ckxσ(k))nk=1 + a

for every (xk)nk=1 ∈ A.

Theorem 2.2. Let p be a real number, 1 < p <∞, p 6= 2. Assume that A ⊂ `pis a dense subset and f : A→ `p an almost surjective step-isometry. Then thereexist a sequence c1, c2, c3, . . . ∈ {−1, 1}, a bijective map σ : N→ N, and a vectora ∈ `p such that

f ((xk)∞k=1) = (ckxσ(k))∞k=1 + a

for every (xk)∞k=1 ∈ A.

Theorem 2.3. Let n ≥ 2 be an integer. Assume that A ⊂ `n2 is a densesubset and f : A → `n2 a step-isometry. Then there exist an orthogonal lineartransformation O : `n2 → `n2 and a vector a ∈ `n2 such that

f(x) = Ox+ a

for every x ∈ A.

Theorem 2.4. Assume that A ⊂ `2 is a dense subset and f : A → `2 analmost surjective step-isometry. Then there exist an orthogonal linear operatorO : `2 → `2 and a vector a ∈ `2 such that

f(x) = Ox+ a

3

for every x ∈ A.

The first and the third theorem show that for certain finite-dimensional Ba-nach spaces we can improve the result of Balister, Bollobas, Gunderson, Leader,and Walters by completely removing any type of surjectivity condition. We donot know whether the almost surjectivity is needed for the other two results.Our method based on the stability result for approximate isometries requiresthis assumption but we do not have any counterexamples showing that it isindispensable.

When proving results on step-isometries on finite-dimensional or infinite-dimensional sequence spaces with `∞ norm the first few steps will be exactlythe same as in the case when 1 < p < ∞. However, instead of getting niceresults as in the case of `p spaces with 1 < p <∞, the analysis leads to “wild”step-isometries. We will illustrate this with an example in the two-dimensionalcase. The same idea can be applied in higher dimensions as well as in theinfinite-dimensional case.

Let c, d be any two real numbers c < d. We will need a well-known fact thatthere exists a function g : (0, 1)→ (c, d) such that

� g is strictly increasing,

� g is discontinuous at every rational number,

� g is continuous at every irational number,

� c = inf{g(x) : x ∈ (0, 1)} and d = sup{g(x) : x ∈ (0, 1)}.

Example 2.5. Let 0 < c < d < 1 and assume that the function g : (0, 1) →(c, d) is as above. Define h : R \ Z→ R by

h(x+ n) = g(x) + n, x ∈ (0, 1), n ∈ Z.

We extend h to a function h : R→ R by defining h(n) = n for every integer n.It is straightforward to verify that h is discontinuous at every rational numberand continuous at every irational number. Moreover, h is strictly increasing onR. Finally, for each real x and every integer n we have

h(x+ n) = h(x) + n.

For each rational number s we denote

s− = limx↑s

h(x) < limx↓s

h(x) = s+.

Assume that real numbers m(s, t), s, t ∈ Q, satisfy the following two conditions:for every s, t, t1 ∈ Q and every positive integer n we have

s− ≤ m(s, t) ≤ s+

andn > |t1 − t| ⇒ m(s+ n, t1)−m(s, t) ≥ n. (2)

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We define A ⊂ `2∞ to be the set of all vectors whose both coordinates are rational,A = {(s, t) : s, t ∈ Q}. We claim that the map f : A→ `2∞ defined on a densesubset A by

f((s, t)) = (m(s, t), t)

is a step-isometry.Indeed, we need to show that that for every s, t, u, v ∈ Q and n ∈ {0, 1, 2, . . .}

we have

‖(s, t)− (u, v)‖∞ ∈ [n, n+ 1)⇒ ‖(m(s, t), t)− (m(u, v), v)‖∞ ∈ [n, n+ 1).

Since ‖(s, t)− (u, v)‖∞ ∈ [n, n+ 1) is equivalent to

� |s − u| ∈ [n, n + 1) and |t − v| ∈ [k, k + 1) for some nonnegative integerk ≤ n, or

� |t − v| ∈ [n, n + 1) and |s − u| ∈ [k, k + 1) for some nonnegative integerk ≤ n,

it is enough to verify that for every s, t, u, v ∈ Q and n ∈ {0, 1, 2, . . .} we have

|s− u| < n⇒ |m(s, t)−m(u, v)| < n,

|s− u| > n⇒ |m(s, t)−m(u, v)| > n,

and|t− v| < n and |s− u| = n⇒ |m(s, t)−m(u, v)| ≥ n.

The last implication is true by (2). So, we only need to prove the first twoimplications and we will verify just one of them, say the first one. We have|s − u| < n. With no loss of generality, s > u. Because h is strictly increasingwe have

m(s, t) ≥ s− = sup{h(x) : x < s} > inf{h(x) : x > u} = u+ ≥ m(u, t),

and therefore m(s, t)−m(u, v) is positive. Hence,

|m(s, t)−m(u, v)| = m(s, t)−m(u, v) ≤ s+ − u−

= inf{h(x) : x > s} − sup{h(x) : x < u}

< h

(s+

n− (s− u)

2

)− h

(u− n− (s− u)

2

)= h

((u− n− (s− u)

2

)+ n

)− h

(u− n− (s− u)

2

)= n,

as desired.

5

The above example is based on the existence of real numbersm(s, t), s, t ∈ Q,satisfying s− ≤ m(s, t) ≤ s+ and (2). To show that such numbers exist we startwith a simple example. Set m(s, t) = s− whenever s ≤ 0 and m(s, t) = s+whenever s ≥ 1. Then clearly, we have a complete freedom when choosing

m(s, t) ∈ [s−, s+]

for rational numbers s, t, 0 < s < 1. The condition (2) will always be satisfied.This simple example gives rise to a step-isomtery with a “wild” behaviour onthe set {(s, t) ∈ Q ×Q : 0 < s < 1}. One can now easily develop this simpleidea further to get more complicated examples of step-isometries that cannotbe described in a nice way.

In our paper a step-isometry is defined as a map acting on a dense subset Aof a Banach space satisfying (1). Recall that in [1] step-isometries were definedas bijective maps from A onto itself satisfying (1). While our more generaldefinition together with the almost surjectivity assumption was good enough toobtain nice results on `p spaces, 1 < p < ∞, the above example shows that ingeneral Banach spaces we need to impose stronger assumptions if we want toget reasonable results. It turns out that we do not need to assume that f is abijection of A; it is enough to assume that f is defined on a dense subset A andits image is dense.

Theorem 2.6. Let n ≥ 2 be an integer, A ⊂ `n∞ a dense subset, and f : A→ `n∞a step-isometry. Assume that f(A) is a dense subset of `n∞. Then there exista permutation σ ∈ Sn, c1, . . . , cn ∈ {−1, 1}, a vector a ∈ `n∞ and standardstep-isometries τ1, . . . , τn : R→ R such that

f ((xk)nk=1) = (ckτk(xσ(k)))nk=1 + a

for every (xk)nk=1 ∈ A.

Theorem 2.7. Let A ⊂ `∞ be a dense subset and f : A → `∞ a step-isometry. Assume that f(A) is a dense subset of `∞. Then there exist a se-quence c1, c2, c3, . . . ∈ {−1, 1}, a bijective map σ : N→ N, a vector a ∈ `∞, anda sequence of standard step-isometries τ1, τ2, . . . : R→ R such that

f ((xk)∞k=1) = (ckτk(xσ(k)))∞k=1 + a

for every (xk)∞k=1 ∈ A.

3 Preliminary results

Throughout this section we will assume that X ia a real Banach space, dimX ≥2, K a positive real number, A ⊂ X a dense subset, 0 ∈ A, and h : A → X astep-isometry such that h(0) = 0 and

‖h(u)− u‖ ≤ K (3)

6

for every u ∈ A. We will further assume that x0 ∈ X is a vector of norm oneand Ux0

a closed subspace of X such that

X = span {x0} ⊕ Ux0.

Then the projection Px0 of X onto span {x0} along Ux0 is a bounded linearoperator. We denote ‖Px0

‖ = N .We finally assume that for every pair of positive real numbers η, L there

exists a positive Mx0,L,η ∈ R such that

|t| − η < ‖tx0 + y‖ < |t|+ η

for every real t satisfying |t| ≥Mx0,L,η and every y ∈ Ux0 satisfying ‖y‖ ≤ L.For every vector u ∈ Ux0

we define a function gu : R → R in the followingway. If tx0 + u ∈ A for some real t then h(tx0 + u) = sx0 + w for uniquelydetermined s ∈ R and w ∈ Ux0

. In this case we set gu(t) = s.If for some real t the vector tx0 + u does not belong to A then we can find

a sequence of vectors (zn) ⊂ A such that

limn→∞

zn = tx0 + u.

For every positive integer n there are uniquely determined real number sn andvector wn ∈ Ux0

such that

h(zn) = snx0 + wn.

The convergent sequence (zn) is bounded, and because h is a step-isometry, thesequence (h(zn)) is bounded, as well. But then, since Px0 is a bounded operator,the real sequence (sn) is bounded, too. We choose a real number s such thatthere exist a subsequence of (sn) converging to s, and set gu(t) = s.

Denote M = 2(1 +N)K. We start with the following simple observation.

Lemma 3.1. Let u ∈ Ux0 . Then for every ε > 0 and t ∈ R there exists z ∈ Awith h(z) = sx0 + w for some real s and some w ∈ Ux0 , such that

� ‖z − (tx0 + u)‖ < ε,

� |s− gu(t)| < ε,

� |s− t| ≤M ,

� ‖w − u‖ ≤M .

Proof. Let ε > 0 and t ∈ R. By the definition of the function gu there exists asequence (zn) ⊂ A such that (zn) converges to tx0 +u, and if we denote h(zn) =snx0 +wn, sn ∈ R, wn ∈ Ux0 , then the real sequence (sn) converges to gu(t) (iftx0 + u ∈ A then we take the constant sequence zn = tx0 + u). If we choose apositive integer n large enough then we have ‖zn − (tx0 + u)‖ < min{K, ε} and|sn− gu(t)| < ε. Hence, taking z = zn, and then, of course, s = sn and w = wn,

7

we already have the first two desired inequalities. From ‖h(z)− z‖ ≤ K we get‖(s− t)x0 + w − u‖ ≤ ‖sx0 + w − z‖+ ‖z − (tx0 + u)‖ ≤ 2K, and therefore

|s− t| = ‖Px0((s− t)x0 + w − u)‖ ≤ 2NK < M

and

‖w − u‖ = ‖(I − Px0)((s− t)x0 + w − u)‖ ≤ 2(N + 1)K = M.

Lemma 3.2. For every u ∈ Ux0the function gu is increasing.

Proof. We want to show that for every u ∈ Ux0 and every pair of real numberst1, t2 we have

t1 < t2 ⇒ gu(t1) ≤ gu(t2).

Assume on the contrary that we can find u ∈ Ux0 and t1, t2 ∈ R such that

t1 < t2 and gu(t1)− gu(t2) = ξ > 0.

We choose and fix a positive real number ε > 0 such that

ε <ξ

6. (4)

Therefore, we have|t| − ε ≤ ‖tx0 + z‖ ≤ |t|+ ε (5)

for every real t satisfying |t| ≥Mx0,2M,ε and every z ∈ Ux0 satisfying ‖z‖ ≤ 2M .Finally we choose a positive integer m such that

m > 2M + (t2 − t1) +Mx0,2M,ε. (6)

We can find a real number r such that

t1 +m < r < t2 +m. (7)

Then

m < ‖(rx0 + u)− (t1x0 + u)‖ and ‖(rx0 + u)− (t2x0 + u)‖ < m

(note that by our assumptions r > m+ t1 > t2).Using Lemma 3.1 we can find zr, zt1 , zt2 ∈ A with h(zr) = srx0+wr, h(zt1) =

st1x0 + wt1 , and h(zt2) = st2x0 + wt2 , sr, st1 , st2 ∈ R, wr, wt1 , wt2 ∈ Ux0, such

that|sr − gu(r)|, |st1 − gu(t1)|, |st2 − gu(t2)| < ε, (8)

|sr − r|, |st1 − t1|, |st2 − t2|, ‖wr − u‖, ‖wt1 − u‖, ‖wt2 − u‖ ≤M, (9)

andm < ‖zr − zt1‖ and ‖zr − zt2‖ < m. (10)

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Since h is a step-isometry we have

m ≤ ‖h(zr)− h(zt1)‖ = ‖(sr − st1)x0 + (wr − u) + (u− wt1)‖. (11)

Using (6), (7) and (9) we see that

|sr − st1 | = |(sr − r) + (t1 − st1) + (r − t1)| ≥ |r − t1| − 2M

> m− 2M > Mx0,2M,ε.

Thus, the inequality (11) together with (5) and (9) implies

m ≤ |sr − st1 |+ ε.

From (9) and r − t1 > m ≥ 2M we conclude that sr − st1 > 0 which togetherwith (8) yields

m ≤ (sr − st1) + ε ≤ gu(r)− gu(t1) + 3ε.

Hence,m− 3ε ≤ gu(r)− gu(t1) = (gu(r)− gu(t2))− ξ

< sr − st2 − ξ + 2ε. (12)

Using (6), (7), and (9) we get

sr − st2 ≥ r − t2 − 2M > m+ (t1 − t2)− 2M ≥Mx0,2M,ε

and consequently, (5) and (12) yield

m− 5ε+ ξ < ‖(sr − st2)x0 + wr − wt2‖+ ε

= ‖h(zr)− h(zt2)‖+ ε.

From (10) and the fact that h is a step-isometry we conclude that

m− 6ε+ ξ ≤ m,

contradicting (4).

For u = 0 we will write shortly g0 = g. Since g : R → R is an increasingfunction, the set S ⊂ (0, 1) of all points of the discontinuity of g on the interval(0, 1) is at most countable.

Lemma 3.3. For every integer n we have

g((n, n+ 1)) ⊂ [n, n+ 1].

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Proof. We will first prove that for every positive real number t and every positiveinteger n > t we have g(t) ≤ n. In order to do this we fix such t and n, andchoose any ε > 0. We will show that g(t) ≤ n + 5ε. Of course, since ε was anarbitrary real number this will yield the desired inequality g(t) ≤ n.

Our first step will be to prove that for every t1 > Mx0,M,ε + M and everypositive integer m > t1 we have

g(t1) ≤ m+ 2ε. (13)

By Lemma 3.1 with u = 0 we can find z ∈ A with h(z) = sx0 + w, s ∈ R,w ∈ Ux0 , such that

‖z − t1x0‖ < m− t1, |s− g(t1)| < ε, |s− t1| ≤M, and ‖w‖ ≤M. (14)

Therefore‖z‖ ≤ ‖t1x0‖+ ‖z − t1x0‖ < m,

and consequently,‖h(z)‖ = ‖sx0 + w‖ < m.

From s ≥ t1 −M > Mx0,M,ε and ‖w‖ ≤M we get

‖sx0 + w‖ ≥ |s| − ε = s− ε,

and thus, by (14) we have

g(t1) < s+ ε ≤ ‖sx0 + w‖+ 2ε < m+ 2ε,

as desired.We choose a positive integer k and a positive real number δ such that

t+ k > Mx0,M,ε +M, k > Mx0,2M,ε + 2M, and t+ δ < n. (15)

By the previous paragraph and t+ k + δ < n+ k we have

g(t+ k + δ) ≤ n+ k + 2ε. (16)

Clearly,‖tx0 − (t+ k + δ)x0‖ > k.

By Lemma 3.1 with u = 0 we can find z1, z2 ∈ A with h(z1) = s1x0 + w1,h(z2) = s2x0 + w2, s1, s2 ∈ R, w1, w2 ∈ Ux0 , such that

‖z1 − z2‖ ≥ k, (17)

|s1 − g(t)|, |s2 − g(t+ k + δ)| < ε, (18)

and|s1 − t|, |s2 − t− k − δ|, ‖w1‖, ‖w2‖ ≤M. (19)

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From (17) and the fact that h is a step-isometry we conclude that ‖h(z1) −h(z2)‖ ≥ k, and therefore

k ≤ ‖s1x0 + w1 − (s2x0 + w2)‖.

By (19) and (15) we have

|s2 − s1| ≥ |t+ k + δ − t| − 2M > k − 2M > Mx0,2M,ε.

Hence,k ≤ |s1 − s2|+ ε,

which together with (18) yields that g(t+ k+ δ)− g(t) ≥ k− 3ε. Applying (16)we finally get the desired inequality

g(t) ≤ n+ 5ε.

Hence, so far we know that if t > 0 and t < n for some positive integer n,then g(t) ≤ n. In exactly the same way we see that if t is a negative real number,n a negative integer, and n < t, then n ≤ g(t). In the next step we will showthat if t < 0, n is a nonpositive integer, and t < n, then g(t) ≤ n. A carefulreader will notice that exactly the same arguments yield that if t > 0, n is anonnegative integer, and t > n, then g(t) ≥ n. Combining all these conclusionsone can easily complete the proof.

So, assume that t < 0, n ≤ 0 is an integer, and t < n. In order to completethe proof we need to verify that then g(t) ≤ n. Assume on the contrary thatthere exist a negative real number t and a nonpositive integer n such that

t < n and g(t) = n+ δ

for some positive δ. Choose a positive η < δ3 . We further choose a positive

integer p satisfyingp > Mx0,2M,η + 2M + n. (20)

Since p− t > p− n we can find a positive real number r such that

r < p and r − t > p− n, (21)

that is, ‖rx0 − tx0‖ > p− n. By Lemma 3.1 with u = 0 we can find z1, z2 ∈ Awith h(z1) = s1x0 + w1, h(z2) = s2x0 + w2, s1, s2 ∈ R, w1, w2 ∈ Ux0

, such that

‖z1 − z2‖ > p− n, (22)

|s1 − g(r)|, |s2 − g(t)| < η, (23)

and|s1 − r|, |s2 − t|, ‖w1‖, ‖w2‖ ≤M. (24)

Therefore, by (21) and (20) we see that

|s1 − s2| ≥ r − t− 2M > p− n− 2M > Mx0,2M,η,

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and thus, (22) and (24) yield

p− n ≤ ‖h(z1)− h(z2)‖ = ‖s1x0 + w1 − (s2x0 + w2)‖ ≤ |s1 − s2|+ η.

Applying (23) we get

p− n ≤ |g(r)− g(t)|+ 3η = g(r)− g(t) + 3η.

From the conclusion formulated in the first paragraph of the proof and (21) wefurther obtain

p− n ≤ p− g(t) + 3η,

or equivalently,g(t) ≤ n+ 3η < n+ δ = g(t),

a contradiction.

Lemma 3.4. For every integer n and every t ∈ (0, 1) \ S we have

g(t+ n) = g(t) + n.

Proof. Let t be a real number, t ∈ (0, 1)\S, and n an integer. We choose ε > 0.Then, by the assumption on t we know that there exists δ > 0 such that forevery real r satisfying |r − t| < δ we have |g(r) − g(t)| < ε. Choose a positivereal number τ satisfying τ < δ.

Let m be a positive integer such that

m >τ

2+ 2M +Mx0,2M,ε + n and m >

τ

2+ 2M +Mx0,2M,ε. (25)

Clearly, ∥∥∥(t+m+τ

2

)x0 − (t+ τ)x0

∥∥∥ < m.

By Lemma 3.1 with u = 0 we can find z1, z2 ∈ A with h(z1) = s1x0 + w1,h(z2) = s2x0 + w2, s1, s2 ∈ R, w1, w2 ∈ Ux0

, such that

‖z1 − z2‖ < m, (26)∣∣∣s1 − g (t+m+τ

2

)∣∣∣ , |s2 − g(t+ τ)| < ε, (27)

and ∣∣∣t+m+τ

2− s1

∣∣∣ , |t+ τ − s2|, ‖w1‖, ‖w2‖ ≤M. (28)

Because of (26) we have

m > ‖h(z1)− h(z2)‖ = ‖s1x0 + w1 − (s2x0 + w2)‖.

Applying (25) and (28) we see that

s1 − s2 ≥(t+m+

τ

2

)−M − (t+ τ +M) > Mx0,2M,ε,

12

and therefore, the inequalities (27) yield

m > (s1 − s2)− ε > g(t+m+

τ

2

)− g(t+ τ)− 3ε,

that is,

g(t+m+

τ

2

)< g(t+ τ) +m+ 3ε < g(t) +m+ 4ε. (29)

On the other hand, since (t + m − τ/2) − (t − τ) > m, we can use Lemma3.1 with u = 0 to find z3, z4 ∈ A with h(z3) = s3x0 + w3, h(z4) = s4x0 + w4,s3, s4 ∈ R, w3, w4 ∈ Ux0

, such that

‖z3 − z4‖ > m,∣∣∣s3 − g (t+m− τ

2

)∣∣∣ , |s4 − g(t− τ)| < ε,

and ∣∣∣t+m− τ

2− s3

∣∣∣ , |t− τ − s4|, ‖w3‖, ‖w4‖ ≤M.

Therefore,

m ≤ ‖h(z3)− h(z4)‖ = ‖s3x0 + w3 − (s4x0 + w4)‖.

Since s3 − s4 ≥(t+m− τ

2

)− (t− τ)− 2M > Mx0,2M,ε we have

m ≤ (s3 − s4) + ε < g(t+m− τ

2

)− g(t− τ) + 3ε,

that is,

g(t+m− τ

2

)> g(t− τ) +m− 3ε > g(t) +m− 4ε. (30)

Clearly, ∥∥∥(t+m− τ

2

)x0 − (t+ n)x0

∥∥∥ < m− n.

By Lemma 3.1 with u = 0 we can find z5, z6 ∈ A with h(z5) = s5x0 + w5,h(z6) = s6x0 + w6, s5, s6 ∈ R, w5, w6 ∈ Ux0

, such that

‖z5 − z6‖ < m− n,∣∣∣s5 − g (t+m− τ

2

)∣∣∣ , |s6 − g(t+ n)| < ε,

and ∣∣∣t+m− τ

2− s5

∣∣∣ , |t+ n− s6|, ‖w5‖, ‖w6‖ ≤M.

Hencem− n > ‖h(z5)− h(z6)‖ = ‖s5x0 + w5 − (s6x0 + w6)‖.

Because s5−s6 ≥(t+m− τ

2

)−M − (t+n+M) > Mx0,2M,ε, we conclude that

m− n > (s5 − s6)− ε > g(t+m− τ

2

)− g(t+ n)− 3ε. (31)

13

From (30) and (31) we easily get

g(t+ n) > g(t) + n− 7ε. (32)

Since (t + m + τ/2) − (t + n) > m − n, we can use Lemma 3.1 with u = 0to find z7, z8 ∈ A with h(z7) = s7x0 + w7, h(z8) = s8x0 + w8, s7, s8 ∈ R,w7, w8 ∈ Ux0 , such that

‖z7 − z8‖ > m− n,∣∣∣s7 − g (t+m+τ

2

)∣∣∣ , |s8 − g(t+ n)| < ε,

and ∣∣∣t+m+τ

2− s7

∣∣∣ , |t+ n− s8|, ‖w7‖, ‖w8‖ ≤M.

Therefore,

m− n ≤ ‖h(z8)− h(z7)‖ = ‖s7x0 + w7 − (s8x0 + w8)‖.

Since s7 − s8 ≥(t+m+ τ

2

)− (t+ n)− 2M > Mx0,2M,ε we have

m− n ≤ (s7 − s8) + ε < g(t+m+

τ

2

)− g(t+ n) + 3ε.

From here and (29) we get

g(t+ n) < g(t) + n+ 7ε. (33)

Putting (32) and (33) together and using the fact that ε was an arbitrary positivereal number we arrive at the deisred equality

g(t+ n) = g(t) + n.

Lemma 3.5. For every integer n, every t ∈ (0, 1) \ S, and every u ∈ Ux0we

havegu(n+ t) = g(t) + n.

Proof. Let n be an integer, u ∈ Ux0 , and t ∈ (0, 1) \ S. Let ξ be any positivereal number such that t− ξ ∈ (0, 1) \ S. Let further ε be any positive number.Choose a positive integer m such that

m > ξ + 2M +Mx0,‖u‖,ξ +Mx0,2M+‖u‖,ε.

Then‖(t+ n+m− ξ)x0 − ((t+ n)x0 + u)‖ < (m− ξ) + ξ = m.

By Lemma 3.1 we can find z1, z2 ∈ A with h(z1) = s1x0+w1, h(z2) = s2x0+w2,s1, s2 ∈ R, w1, w2 ∈ Ux0

, such that

‖z1 − z2‖ < m,

14

|s1 − g(t+ n+m− ξ)|, |s2 − gu(t+ n)| < ε,

and|t+ n+m− ξ − s1|, |s2 − (t+ n)|, ‖w1‖, ‖w2 − u‖ ≤M.

It follows that ‖h(z1)− h(z2)‖ < m, that is,

m > ‖(s1 − s2)x0 + (w1 − w2)‖.

Since s1 − s2 ≥ t+ n+m− ξ − (t+ n)− 2M > Mx0,2M+‖u‖,ε we have

m > (s1−s2)−ε > g(t+n+m−ξ)−gu(t+n)−3ε = g(t−ξ)+n+m−gu(t+n)−3ε,

which further yields that gu(t+n) > g(t−ξ)+n−3ε. Because ε was arbitrarilysmall we actually have

gu(t+ n) ≥ g(t− ξ) + n.

But ξ was an arbitrary positive real number such that t− ξ belongs to (0, 1) \Sand since g is continuous at t and (0, 1) \S is dense in (0, 1) we finally concludethat

gu(t+ n) ≥ g(t) + n.

In a similar way we show that gu(t+n) ≤ g(t)+n. This completes the proof.

Lemma 3.6. Assume that for every real t and every y ∈ Ux0we have

‖tx0 + y‖ ≥ |t|.

Then for every u,w ∈ Ux0and every pair of real numbers s, t we have

t < s⇒ gu(t) < gw(s). (34)

Remark. Note the the assumption in this lemma states that x0 is Birkhoff-James orthogonal to every y ∈ Ux0

.

Proof. An arbitrary vector from A can be written in the form tx0 + u for auniquely determined real number t and vector u ∈ Ux0 . By the definition of thefunction gu we have

h(tx0 + u) = gu(t)x0 + w (35)

for some w ∈ Ux0. It follows from (3) that

‖u− w‖ = ‖(I − Px0) ((gu(t)x0 + w)− (tx0 + u))‖ ≤ (N + 1)K < M. (36)

Choose and fix u,w ∈ Ux0and assume that 0 < t < s < 1. We know

that gu and gw coincide on (0, 1) \ S. Hence, there exists r ∈ (t, s) such thatgu(r) = gw(r). Because both gu and gw are increasing, we have gu(t) ≤ gu(r) =gw(r) ≤ gw(s). Thus, we only need to show that gu(t) 6= gw(s).

Assume on the contrary that gu(t) = gw(s). We can find t1, s1 ∈ (0, 1) \ Ssuch that t < t1 < s1 < s and since gu(t) ≤ gu(t1) = g(t1) ≤ g(s1) = gw(s1) ≤gw(s) = gu(t), we have

t1 < s1 and g(t1) = g(s1).

15

Since g is an increasing function we have g(r) = g(t1) = g(s1) for every r ∈[t1, s1]. In particular, g is continuous on [t1, s1] and therefore

gv(r) = g(t1) = g(s1) = c (37)

for every r ∈ (t1, s1) and every v ∈ Ux0. Because A is dense in X we can find a

vectorrx0 + v ∈ A

(here, of course, r ∈ R and v ∈ Ux0) such that t1 < r < s1 and ‖v‖ < 1. Choose

and fix a positive real number δ such that δ < r − t1. Using the density of Aonce more we can find for every positive integer m a vector

rmx0 + vm ∈ A

such that t1 +m < rm < r +m− δ, and ‖vm‖ < 1. Using (35), (36), and (37)we see that there exists w ∈ Ux0 such that

h(rx0 + v) = cx0 + w

with ‖w‖ < M + 1. Similarly,

h(rmx0 + vm) = gvm(rm)x0 + wm

with ‖wm‖ < M + 1. By Lemma 3.5, t1 < rm −m < r < s1, and (37) we havegvm(rm) = gvm(rm −m) +m = c+m.

For m > Mx0,2,δ + 1 we have

‖(rmx0 + vm)− (rx0 + v)‖ < |rm − r|+ δ < m− δ + δ = m.

On the other hand,

‖h(rmx0+vm)−h(rx0+v)‖ = ‖(c+m)x0+wm)−(cx0+w)‖ ≥ |(c+m)−c| = m,

a contradiction. Thus, we have (34) in the special case when 0 < t < s < 1.It is now rather easy to show that for every u,w ∈ Ux0 and every pair of real

numbers s, t we havet < s⇒ gu(t) < gw(s). (38)

Clearly, all we need to check is that if t, s ∈ (n, n+ 1) for some integer n, thenfor every u,w ∈ Ux0

we have

gu(n) < gw(t) and gu(s) < gw(n+ 1).

We will prove just one of the inequalities, say the second one. Because n < s <n+ 1 we can find t1, t2 ∈ (0, 1) \ S such that

s < t1 + n < t2 + n < n+ 1.

Then

gu(s) ≤ gu(t1 + n) = gu(t1) + n < gw(t2) + n = gw(t2 + n) ≤ gw(n+ 1),

as desired.

16

We continue with an elementary observation that will be needed in the proofof one of our main results.

Lemma 3.7. Let t, δ, p be positive real numbers with 1 < p < ∞ and δ < t.Then there exist a positive integer m > t and a real number a > 32 such that

(m− t+ δ)p + (a− 32)p > mp > (m− t)p + ap.

Proof. If b is any positive real number, then obviously we have

limx→∞

(xp − (x− b)p)1/p =∞.

Further, if 0 < b < c, then

limx→∞

(xp − (x− c)p

xp − (x− b)p

)1/p

= limx→∞

(1− (1− c/x)p

1− (1− b/x)p

)1/p

=(cb

)1/p> 1,

Hence,

limx→∞

((xp − (x− c)p)1/p − (xp − (x− b)p)1/p

)= limx→∞

(xp − (x− b)p)1/p((

xp − (x− c)p

xp − (x− b)p

)1/p

− 1

)=∞.

It follows that there exists a positive integer m > t such that

(mp − (m− t)p)1/p > (mp − (m− (t− δ))p)1/p + 32.

Consequently, we can find a real number a such that

(mp − (m− (t− δ))p)1/p + 32 < a < (mp − (m− t)p)1/p .

The proof can be completed by a straightforward calculation.

We will conclude this section by formulating two known stability results forisometries that will be of great importance in our study of step-isometries. Letε be a positive real number. A map f : X → X is called an ε-isometry if

| ‖f(x)− f(y)‖ − ‖x− y‖ | ≤ ε

for every pair of vectors x, y ∈ X. Clearly, if we compose an ε-isometry witha translation then the obtained map is again an ε-isometry. Thus, when con-sidering ε-isometries we often assume that they map the origin into itself. Thisharmless normalization makes the formulation of stability results nicer.

The following result has been proved in [12].

Theorem 3.8. Let X be a real Banach space, ε > 0, and f : X → X an almostsurjective ε-isometry satisfying f(0) = 0. Then there exists a bijective linearisometry W : X → X such that

‖f(x)−Wx‖ ≤ 2ε

for every x ∈ X.

17

The assumption that f is almost surjective is indispensable for general Ba-nach spaces. However, as proved by Dilworth, it is not needed under the addi-tional assumption that X is finite-dimensional.

Theorem 3.9. [9] Let X be a finite-dimensional real normed space, ε > 0, andf : X → X an ε-isometry satisfying f(0) = 0. Then there exists a bijectivelinear isometry W : X → X such that

‖f(x)−Wx‖ ≤ 2ε

for every x ∈ X.

4 Proofs of the main results

We have six main results but since in the proofs we are using the same mainidea we will group the first four together and the last two together and thenprove them simultaneously. So let us start with the proof of Theorems 2.1, 2.2,2.3, and 2.4.

Proof. We will prove all four statements simultaneosly. In all four cases wewill denote by X the underlying real Banach space, that is, X is either `np ,1 < p < ∞, p 6= 2, or `p, 1 < p < ∞, p 6= 2, or `n2 , or `2. The general form ofbijective linear isometries of these Banach spaces is well-known. When p = 2 theset of all bijective linear isometries is the set of all orthogonal linear operators.For the other values of p we refer to [2, p.119], [10], and [11, p.112]. Hence, ourtheorems can be summarized by saying that almost surjective step-isometrieson dense subsets of `p spaces, 1 < p < ∞, are restrictions of bijective affineisometries of the whole spaces, and in the finite-dimensional case the same canbe proved without assuming almost surjectivity.

Obviously, we have

| ‖f(x)− f(y)‖ − ‖x− y‖ | ≤ 1

for all pairs of vectors x, y ∈ A.Let z ∈ A be any point. Replacing A by A − z = {x − z : x ∈ A} and f

by a map u 7→ f(u + z) − f(z), u ∈ A − z, we notice that A − z is dense in Xand the new map is a step-isometry sending 0 into itself. It is also clear that ifthis new map is a restriction of a bijective linear isometry of X to the subsetA−z, then the map f is a restriction of a bijective affine isometry. Hence, thereis no loss of generality in assuming that 0 ∈ A and f(0) = 0. Of course, afterthis harmless normalization our goal is to prove that f is a restriction of somebijective linear isometry.

We define a new map g : X → X in the following way. For x ∈ A we setg(x) = f(x). For every x ∈ X\A we can choose x1 ∈ A such that ‖x−x1‖ < 1/2and define g(x) = f(x1). Since in the case that X is infinite-dimensional the

18

map f is almost surjective, the new map g is also almost surjective. And clearly,for every pair x, y ∈ X we have

| ‖g(x)− g(y)‖ − ‖x− y‖ | ≤ 2.

Applying Theorems 3.8 and 3.9 we conclude that there exists a bijectivelinear isometry W : X → X such that

‖g(x)−Wx‖ ≤ 4

for every x ∈ X. It follows that ‖W−1g(x)−x‖ ≤ 4 for every x ∈ X. We definea new map h : A→ X by h = W−1 ◦ f . Then clearly,

‖h(x)− x‖ ≤ 4

for every x ∈ A. Obviously, h is a step-isometry.Let x be any vector in A. We will prove that the first coordinate of h(x)

coincides with the first coordinate of x. In the same way one can see that eachcoordinate of h(x) coincides with the corresponding coordinate of x. In otherwords, h(x) = x for every x ∈ A. This yields that f is the restriction of theisometry W to the subset A, as desired.

Hence, it remains to verify that the first coordinate of h(x) coincides withthe first coordinate of x for every x ∈ A. We denote by e1 the first vector inthe standard basis of X. Then

X = span {e1} ⊕ Ue1 ,

where Ue1 is the subspace of all vectors from X whose first coordinate is zero.Clearly, for every real t and every y ∈ Ue1 we have

‖te1 + y‖ = (|t|p + ‖y‖p)1p .

Let η and L be any positive real numbers. If ‖y‖ ≤ L, then

|t| ≤ (|t|p + ‖y‖p)1p ≤ (|t|p + Lp)

1p .

Sincelim|t|→∞

(|t|p + Lp)1p = |t|

we can find a positive real number Me1,L,η such that

|t| ≤ ‖te1 + y‖ < |t|+ η

for every |t| ≥Me1,L,η.Thus, if we define functions gu : R → R, u ∈ Ue1 , as at the beginning of

Section 3 and set g0 = g, then we can apply Lemmas 3.2 - 3.5 for this family ofreal functions. We need to prove that g(t) = t for every t ∈ (0, 1) \ S, where Sis defined as in Section 3. Indeed, if we assume for a moment that g(t) = t forevery t ∈ (0, 1) \ S, then by the density of the set of points of continuity of g

19

and Lemmas 3.2 and 3.5 we conclude that gu(t) = t for every real t and everyu ∈ Ue1 . An arbitrary vector from A can be written in the form te1 + u foruniquely determined real number t and vector u ∈ Ue1 . By the definition of thefunction gu we have

h(te1 + u) = gu(t)e1 + w

for some w ∈ Ue1 . Using the fact that gu(t) = t we finally conlude that the firstcoordinates of te1 + u and h(te1 + u) coincide, as desired.

So, we only need to verify that g(t) = t for every t ∈ (0, 1) \ S. Assume onthe contrary that either g(t) < t for some t ∈ (0, 1) \ S, or g(t) > t for somet ∈ (0, 1) \S. We need to check that this leads to a contradiction. We will treatjust one of the two possibilities because the other one goes through in exactlythe same way. Thus, assume that g(t) < t for some t ∈ (0, 1) \ S. Denoteδ = t − g(t) > 0. By Lemma 3.3, g(t) ≥ 0. If δ < t then choose a positiveinteger m > t and a real number a > 32 as in Lemma 3.7. Then we have

(m− t+ δ)p + (a− 32)p > mp > (m− t)p + ap.

If δ = t, then it is trivial to find a positive integer m > t and a real numbera > 32 such that the above two inequalities are fulfilled.

From here we conclude that there exists a real number s such that s > mand

((s− t)p + ap)1/p

< m.

Therefore there exists a real number ε > 0 such that 2ε < δ and

((s− t)p + ap)1/p

+ 2ε < m (39)

and(m− t+ δ − 2ε)p + (a− 32)p > mp. (40)

Choose and fix u ∈ Ue1 such that

‖u‖ = a.

We will now apply Lemma 3.1 with x0 = e1. Recall that then M = 2(1 +N)K = 8(1 + ‖Pe1‖), and since ‖Pe1‖ = 1, we have M = 16. Using Lemma 3.1we can find z1 ∈ A such that

h(z1) = s1e1 + w1

for some real s1 and some w1 ∈ Ue1 and

‖z1 − se1‖, |s1 − g(s)| < ε and |s1 − s|, ‖w1‖ ≤ 16. (41)

Applying Lemma 3.1 once more we can find z2 ∈ A such that

h(z2) = s2e1 + w2

for some real s2 and some w2 ∈ Ue1 and

‖z2 − (te1 + u)‖, |s2 − gu(t)| < ε and |s2 − t|, ‖w2 − u‖ ≤ 16. (42)

20

It follows from Lemma 3.5 that

|s2 − (t− δ)| < ε. (43)

By (41) and (42) we have

‖z1 − z2‖p ≤ (‖se1 − (te1 + u)‖+ 2ε)p =(

((s− t)p + ‖u‖p)1/p + 2ε)p.

It follows from (39) that

‖z1 − z2‖p ≤(

((s− t)p + ap)1/p

+ 2ε)p

< mp.

Applying the fact that h is a step-isometry we get

mp ≥ ‖h(z1)−h(z2)‖p = ‖(s1−s2)e1+w1−w2‖p = |s1−s2|p+‖(w1+(u−w2))−u‖p.

By (41), (42), and (43) we have

mp ≥ (|g(s)− t+ δ| − 2ε)p + (a− 32)p.

From s > m and Lemma 3.3 we infer that g(s) ≥ m. We also know that m > tand δ − 2ε > 0 and therefore, by (40) we conclude that

mp ≥ (m− t+ δ − 2ε)p + (a− 32)p > mp,

a contradiction.

And finally, we prove Theorems 2.6 and 2.7 simultaneously.

Proof. As in the previous proof we may assume with no loss of generality that0 ∈ A, f(0) = 0, and

‖f(x)− x‖ ≤ 4 (44)

for every x ∈ A, and we need to show that there exists a sequence of standardstep-isometries τ1, τ2, . . . : R→ R such that

f ((xk)∞k=1) = (τk(xk))∞k=1

for every (xk)∞k=1 ∈ A (in the finite-dimensional case we need to replace (xk)∞k=1

by (xk)nk=1).We will use the symbol X to denote either `n∞ or `∞, and e1 to denote the

first vector in the standard basis of X. Then

X = span {e1} ⊕ Ue1 ,

where Ue1 is the subspace of all vectors from X whose first coordinate is zero.Let L be any positive real number. If t is a real number, |t| ≥ L, and y ∈ Ue1

a vector satisfying ‖y‖ ≤ L, then obviously

‖te1 + y‖ = |t|.

21

Hence, if for any pair of positive real numbers η and L we set Me1,L,η = L anddefine functions gu : R → R, u ∈ Ue1 , as at the beginning of Section 3 and setg0 = g, then we can apply Lemmas 3.2 - 3.6 for this family of real functions.

An arbitrary vector from A can be written in the form te1 + u for uniquelydetermined real number t and vector u ∈ Ue1 . By the definition of the functiongu we have

f(te1 + u) = gu(t)e1 + w

for some w ∈ Ue1 . This and Lemma 3.6 yield that if x, y are two vectorsfrom A such that the first coordinates of x and y are different, then the firstcoordinates of f(x) and f(y) are different, too. The same conclusion holds forthe second coordinate, the third coordinate,... Therefore f is injective. Denotef(A) = B ⊂ X. We know that B is dense in X and f : A → B is bijective.The inverse f−1 : B → A is a step-isometry, too. For an arbitrary x ∈ B thereexists a unique y ∈ A such that f(y) = x. Consequently,

‖f−1(x)− x‖ = ‖f−1(f(y))− f(y)‖ ≤ 4.

Moreover, f−1(0) = 0. Thus, everything that has been proved so far for thestep-isometry f can be applied to its inverse. In particular, for every v ∈ Ue1there exists an increasing function hv : R→ R such that for every te1 + v ∈ Bwe have

f−1(te1 + v) = hv(t)e1 + w

for some w ∈ Ue1 . Moreover, for every u,w ∈ Ue1 and every pair of real numberss, t we have

t < s⇒ hu(t) < hw(s). (45)

Let M ⊂ R be the set of all real numbers t for which there exists u ∈ Ue1such that te1 + u ∈ B. Since B is dense in X, the set M is dense in R. Clearly,for every t ∈M there exist u,w ∈ Ue1 such that

gw(hu(t)) = t. (46)

In the next step we will prove that g is continuous on R and therefore g isa standard step-isometry on R and gu = g = τ1 for every u ∈ Ue1 . It followsthat for every x = (x1, . . .) ∈ A the first coordinate of f(x) equals τ1(x1). Asthe same must be true the second coordinate (of course, with some standardstep-isometry τ2 instead of τ1), the third coordinate,..., the proof is completed.

Hence, it remains to verify that g is continuous. Assume that this is nottrue. Then we can find t0 ∈ R such that a < b, where

a = sup{g(t) : t < t0} = sup{gu(t) : t < t0, u ∈ Ue1}

andb = inf{g(t) : t > t0} = inf{gu(t) : t > t0, u ∈ Ue1}.

Applying (46) we see that for every t ∈M , a < t < b, there exists u ∈ Ue1 suchthat hu(t) = t0, contradicting (45).

22

5 Open questions

The three most interesting open problems are the following.

Problem 5.1. Characterize Banach spaces X having the property that for eachdense subset A ⊂ X every step-isometry f : A → X is a restriction of a linearbijective isometry of the whole space composed with the translation.

We know that (R2, ‖ · ‖1) being isometric to (R2, ‖ · ‖∞) does not belong tothe above class of Banach spaces. It would be interesting to know whether thespaces (Rn, ‖ · ‖1), n ≥ 3, satisfy the above condition.

Problem 5.2. Characterize Banach spaces X satisfying the following condition:For each dense subset A ⊂ X every almost surjective step-isometry f : A→ Xis a restriction of a linear bijective isometry of the whole space composed withthe translation.

Problem 5.3. Characterize Banach spaces X having the property that for eachdense subset A ⊂ X every step-isometry f : A → X with dense image in X isa restriction of a linear bijective isometry of the whole space composed with thetranslation.

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