OLEKSIY DOVGOSHEY and OLLI MARTIO
We introduce a tangent space at an arbitrary point of a general metric space. It is proved that all tangent spaces are complete. The conditions under which these spaces have a finite cardinality are found.
AMS 2010 Subject Classification: 54E35.
Key words: metric space, finite tangent space, completeness of tangent spaces.
1. INTRODUCTION. MAIN DEFINITIONS
The recent achievements in the metric space theory are closely related to some generalizations of differentiation. The concept of upper gradient [6] and [9], Cheeger’s notion of differentiability for Rademacher’s theorem in certain metric measure spaces [4], the metric derivative in the studies of metric space valued functions of bounded variation [2], [3] and the Lipshitz type approach in [5] are interesting and important examples of such generalizations. Such generalizations lead usually to some nontrivial results only if the metric spaces have “sufficiently many” rectifiable curves.
We define “tangent” spaces at a point of a metric space as a certain quotient space of the sequences which converge to this point and after that introduce the “derivatives” of functions as corresponding quotient maps.
Let (X, d) be a metric space and let abe point of X. Fix a sequence ˜r of positive real numbers rn which tend to zero. In what follows this sequence
˜
r will be called anormalizing sequence.
Definition 1.1. Two sequences ˜x ={xn}n∈N and ˜y ={yn}n∈N,xn, yn ∈ X, are mutually stable (with respect to a normalizing sequence ˜r ={rn}n∈N) if there is a finite limit
(1.1) lim
n→∞
d(xn, yn)
rn := ˜d˜r(˜x,y) = ˜˜ d(˜x,y).˜
We shall say that a family ˜F of sequences of points fromX is maximal mutually stable(with respect to a normalizing sequence ˜r) if every two ˜x,y˜∈F˜ are mutually stable and for an arbitrary ˜z={zn}n∈Nwithzn∈Xeither ˜z∈F˜ or there is ˜x∈F˜ such that ˜x and ˜z are not mutually stable.
REV. ROUMAINE MATH. PURES APPL.,56(2011),2, 137–155
The standard application of Zorn’s Lemma leads to the following Proposition 1.2. Let (X, d) be a metric space and let a ∈ X. Then for every normalizing sequence r˜={rn}n∈N there exists a maximal mutually stable family X˜a= ˜Xa,˜r such that ˜a:={a, a, . . .} ∈X˜a.
Note that the condition ˜a∈X˜a implies the equality
n→∞lim d(xn, a) = 0 for every ˜x={xn}n∈N belonging to ˜Xa.
Consider the function ˜d: ˜Xa×X˜a→R defined by (1.1). Obviously, ˜dis symmetric and nonnegative. Moreover, the triangle inequality for dimplies
d(˜˜x,y)˜ ≤d(˜˜x,z) + ˜˜ d(˜z,y)˜
for all ˜x,y,˜ z˜from ˜Xa. Hence ( ˜Xa,d) is a pseudometric space.˜
Definition 1.3. The pretangent space to the spaceXat the pointaw.r.t.
˜
r is the metric identification of the pseudometric space ( ˜Xa,˜r,d).˜
Since the notion of pretangent space is an important step in the de- velopment of general machinery for the definition of tangent spaces and dif- ferentiability in arbitrary metric spaces, we recall this metric identification construction.
Define a relation ∼ on ˜Xa by ˜x ∼ y˜ if and only if ˜d(˜x,y) = 0. Then˜
∼ is an equivalence relation and let Ωa,˜r be the set of equivalence classes in X˜a under the equivalence relation ∼. It follows from general properties of pseudometric spaces, see for example [7, Chapter 4, Theorem 15], that if ρ is defined on Ωa by
(1.2) ρ(α, β) := ˜d(˜x,y)˜
for some ˜x∈αand ˜y∈β, thenρ is a well-defined metric on Ωa. By definition the metric space (Ωa, ρ) is the metric identification of ( ˜Xa,d).˜
Remark that Ωa,˜r 6=∅ for all ˜r because the constant sequence ˜abelongs to ˜Xa,˜r in accordance with Proposition 1.2.
Let {nk}k∈N be an increasing sequence of natural numbers. Denote by
˜
r0 the subsequence {rnk}k∈N of the normalizing sequence ˜r ={rn}n∈N. It is clear that if ˜x={xn}n∈N and ˜y={yn}n∈N are mutually stable w.r.t. ˜r, then
˜
x0 :={xnk}k∈N and ˜y0 :={ynk}k∈N are mutually stable w.r.t. ˜r0 and that (1.3) d˜˜r(˜x,y) = ˜˜ d˜r0(˜x0,y˜0).
If ˜Xa,˜r is a maximal mutually stable (w.r.t. ˜r) family, then by Zorn’s Lemma there exists a maximal mutually stable (w.r.t. ˜r0) family ˜Xa,˜r0 such that
{˜x0 : ˜x∈X˜a,˜r} ⊆X˜a,˜r0.
Denote by inr˜0 the mapping ˜Xa,˜r → X˜a,˜r0 with in˜r0(˜x) = ˜x0 for all ˜x ∈ X˜a,˜r. If follows from (1.3) that after metric identifications the mapping inr˜0 induces an isometric embedding in0: Ωa,˜r→Ωa,˜r0, i.e., the diagram
(1.4)
X˜a,˜r inr˜0
−−−−−→ X˜a,˜r0 p
y
y p0 Ωa,˜r −−−−−→in0 Ωa,˜r0
is commutative. Here p, p0 are metric identification mappings p(˜x) = {˜y ∈ X˜a,˜r : ˜dr˜(˜x,y) = 0}˜ and p0(˜x) ={˜y0∈X˜a,˜r0 : ˜d˜r0(˜x0,y˜0) = 0}.
Let X and Y be two metric spaces. Recall that a map f : X → Y is called an isometryiff is distance-preserving and onto.
Definition 1.4. A pretangent Ωa,˜r is tangent if for every ˜r0 the mapping in0: Ωa,˜r →Ωa,˜r0 is an isometry.
Remark 1.5. As has been stated above, the function in0 : Ωa,˜r → Ωa,˜r0 is an isometric embedding. Since every surjective, isometric embedding is an isometry, Ωa,˜r is tangent if and only if in0 is surjective for every ˜r0.
The following question arieses naturally.
Problem 1.6. Let (X, d) be a metric space and let a ∈ X. Find the conditions under which all pretangent spaces Ωa,˜r are tangent.
Let (X1, d1) and (X2, d2) be metric spaces, and ˜r1, ˜r2 normalizing se- quences, anda1,a2 points ofX1 andX2 respectively, and ˜Xa11,˜r1, ˜Xa22,˜r2 maxi- mal mutually stable families of sequences of points from X1 and X2 respec- tively. Let f : X1 → X2 be a function such that f(a1) = a2. For every
˜
x1:={x1n}n∈N∈X˜a11,˜r1 write
f˜(˜x1) :={f(x1n)}n∈N.
Definition 1.7. The function f is differentiable w.r.t. the pair ( ˜Xa1
1,˜r1, X˜a22,˜r2) iff(˜x1)∈X˜a21,˜r2 and ˜d1(˜x1,y˜1) = 0 implies ˜d2( ˜f(˜x1),f˜(˜y1)) = 0 for all
˜
x1,y˜1 ∈X˜a1
1,˜r1. Letpi : ˜Xai
ir˜i →Ωai,˜ri,i= 1,2, be metric identification mappings.
Definition 1.8. A functionDf : Ωa1,˜r1 →Ωa2,˜r2 is a derivative off w.r.t.
pretangent spaces Ωai,˜ri, i = 1,2, if f is differentiable w.r.t. ( ˜Xa11,˜r1,X˜a22,˜r2)
and the following diagram X˜a1
1,˜r1
f˜
−−−−−−−→ X˜a2
2,˜r2
p1
y
y
p2
Ωa1,˜r1 −−−−−−−−→Df Ωa2,˜r2 is commutative.
Definitions 1.7 and 1.8 are a main goal of all previous constructions, so we present them here, although the properties of “metric space valued derivatives”
Df are not discussed in this paper.
2. FINITE TANGENT SPACES
It is clear that Ωa is an one-point space if a is an isolated point of X.
The converse proposition is also true.
Proposition2.1. Let (X, d)be a metric space and leta∈X. Then a is an isolated point ofX if and only if every pretangent space Ωa,˜r is a singleton.
Proof. If a is not an isolated point of X, then there is a sequence ˜b = {bn}n∈N of points in X such that lim
n→∞d(a, bn) = 0 and d(a, bn) 6= 0 for all n ∈ N. Consider the normalizing sequence ˜r = {rn}n∈N with rn := d(a, bn).
It follows immediately from (1.1) that ˜dr˜(˜a,˜b) = 1 where ˜a is the constant sequence {a, a, . . .}. The application of Zorn’s Lemma shows that there is a maximal mutually stable family ˜Xa,˜r such that ˜a,˜b ∈X˜a,˜r. Then the metric identification of the pseudometric space ( ˜Xa,˜x,d) has at least two points.˜ The following proposition characterizes the points a ∈ X for which all pretangent spaces Ωa have cardinality at most two.
Define the functionF :X×X→R by the rule (2.1) F(x, y) =
d(x, y)(d(x, a)∧d(y, a))
(d(x, a)∨d(y, a))2 if (x, y)6= (a, a),
0 if (x, y) = (a, a).
Theorem 2.2. Let (X, d) be a metric space and let a ∈ X. Then the inequality
(2.2) card(Ωa,˜r)≤2
holds for every pretangent space Ωa,˜r if and only if
(2.3) lim
x→a y→a
F(x, y) = 0.
Proof. Suppose that (2.3) does not hold. Then there are some sequences {x0n}n∈N and{y0n}n∈N such that
n→∞lim x0n= lim
n→∞yn0 =a and x0n6=a6=yn0 for all n∈Nand
(2.4) lim sup
n→∞
F(x0n, y0n)>0.
Consider the normalizing sequence ˜r ={rn}n∈N withrn:=d(x0n, a)∨d(yn0, a).
Write
(xn, yn) :=
((x0n, yn0) ifd(x0n, a)≥d(yn0, a), (yn0, x0n) ifd(yn0, a)> d(x0n, a).
Then we obtain
(2.5) F(x0n, yn0) =F(xn, yn) = d(xn, yn) rn
·d(yn, a) rn
. Since
(2.6) d(yn, a)≤rn
and
(2.7) d(xn, yn)≤d(xn, a) +d(yn, a)≤2rn, we have
(2.8) F(xn, yn)≤2
for all n∈N. Let{xnk, ynk}k∈N be a subsequence of{xn, yn}N for which lim sup
n→∞ F(x0n, y0n) = lim
k→∞F(xnk, ynk).
Inequalities (2.4) and (2.8) imply that
(2.9) 0< lim
k→∞F(xnk, ynk)≤2.
We may assume, without loss of generality, that{xnk, ynk}k∈Nand{xn, yn}n∈N coincide. Conditions (2.6) and (2.7) imply the existence of finite limits
m→∞lim
d(xnm, ynm) rnm
≤2 and lim
m→∞
d(ynm, a) rnm
≤1
for some subsequence{xnm, ynm}m∈Nof{xn, yn}n∈N. Now we can, once again, take nm=n. It follows from (2.5) and (2.9) that
0< lim
n→∞F(xn, yn) = lim
n→∞
d(xn, yn) rn · lim
n→∞
d(yn, a) rn ≤2.
Consequently, we have
d(˜˜x,a) = 1,˜ 0<d(˜˜y,˜a)≤1, 0<d(˜˜x,y)˜ ≤2
for ˜x={xn}n∈N, ˜y={˜yn}n∈N. Hence, ˜a,y,˜ x˜ correspond to distinct points of the pretangent space Ωa,˜r, contrary to (2.2). The implication (2.2) ⇒(2.3) is established.
To prove that (2.3) implies (2.2), suppose card(Ωa,˜r)≥3
for some normalizing sequence ˜r. Then there exist sequences ˜x = {xn}n∈N and ˜y={yn}n∈N such that all three quantities
d(˜˜x,˜a) = lim
n→∞
d(xn, a) rn
, d(˜˜y,a) = lim˜
n→∞
d(yn, a) rn
and
d(˜˜x,y) = lim˜
n→∞
d(xn, yn) rn
are finite and positive. Consider the sequence {F(xn, yn)}n∈N where F was defined by (2.1). Since
0< lim
n→∞
d(xn, a)∨d(yn, a)
rn = ˜d(˜x,˜a)∨d(˜y,˜a)<∞ and
0< lim
n→∞
d(xn, a)∧d(yn, a) rn
= ˜d(˜x,˜a)∧d(˜y,˜a)<∞, we obtain
n→∞lim F(xn, yn) = lim
n→∞
d(xn,yn)
rn ·d(xn,a)∧d(yr n,a)
n
d(xn,a)∨d(yn,a) rn
2
= d(˜˜x,y)( ˜˜ d(˜x,˜a)∧d(˜˜y,˜a))
d(˜˜x,a)˜ ∨d(˜˜y,a)˜ 2 ∈(0,∞), contrary to the limit relation (2.3). Hence, (2.3) implies (2.2).
Theorem 2.2 can be generalized as follows.
For every natural n ≥ 2 let us denote by Xn the set of all n-tuples x = (x1, . . . , xn) with terms xk ∈ X for k = 1, . . . , n. Define the functions
Fn:Xn→Rby the rule (2.10) Fn(x1, . . . , xn) :=
:=
n
Q
k,l=1 k<l
d(xk, xl)
n
V
k=1
d(xk, a)
n W
k=1
d(xk, a)
n(n−1)2 +1 if (x1, . . . , xn)6= (a, . . . , a), 0 if (x1, . . . , xn) = (a, . . . , a), where
n
^
k=1
d(xk, a) := min
1≤k≤nd(xk, a) and
n
_
k=1
d(xk, a) := max
1≤k≤nd(xk, a).
Theorem 2.3. Let (X, d) be a metric space, a ∈X and let n≥2 be a natural number. Then the inequality
(2.11) card(Ωa,˜r)≤n
holds for every pretangent space Ωa,˜r if and only if
(2.12) lim
x1→a
···
xn→a
Fn(x1, . . . , xn) = 0.
The proof of this theorem can be obtained as a direct generalization of the proof of Theorem 2.2 so we omit it.
Remark 2.4. The value
n
Q
k,l=1 k<l
d(xk, xl) in (2.10) can be regarded as a metric-space analog of the Vandermonde Determinant.
There exist other functions Φn :Xn → Rwhich can be used instead of the function Fn in Theorem 2.3. As an example consider
(2.13) Φn(x1, . . . , xn) :=
:=
n
V
k,l=1 k<l
d(xk, xl)
n
V
k=1
d(xn, a)
n W
k=1
d(xk, a)
2 if (x1, . . . , xn)6= (a, . . . , a), 0 if (x1, . . . , xn) = (a, . . . , a).
Then (2.12) holds if and only if
(2.14) lim
x1→a
···
xn→a
Φn(x1, . . . , xn) = 0.
Indeed, it follows from the simple inequality d(xk, xl)≤2
n
_
k=1
d(xk, a), k, l∈ {1, . . . , n}
that
Fn(x1, . . . , xn)≤2n(n−1)2 −1Φn(x1, . . . , xn).
On the other hand, we have
Fn(x1, . . . , xn)≥
Vn
k,l k<l
d(xk, xl)
n
W
k=1
d(xk, a)
n(n−1) 2
·
n
V
k=1
d(xk, a)
n
W
k=1
d(xk, a)
≥
≥
n
V
k,l k<l
d(xk, xl)
n
V
k=1
d(xk, a)
n(n−1) 2
·
n
V
k=1
d(xk, a)
n
W
k=1
d(xk, a)
n(n−1) 2
= (Φn(x1, . . . , xn))
n(n−1)
2 .
Using Theorems 2.2 and 2.3 we can easily construct metric spaces with finite tangent spaces. To this end, we first establish a lemma.
Let ˜W be a family of some sequences of points fromX. Suppose that ˜a∈ W˜ and that every two ˜x,y˜∈W˜ are mutually stable w.r.t. a fixed normalizing sequence ˜r. Denote by ˜Wm a maximal mutually stable family containing ˜W, by Ωa,˜rthe pretangent space corresponding to ( ˜Wm,d) and by Ω˜ wa,˜rthe metric identification of the pseudometric space ( ˜W ,d). Clearly, there is a unique˜
“natural”, isometric embedding Em: Ωwa,˜r →Ωa,˜r for which the diagram
(2.15)
W˜ −−−−−−→in W˜m
p
y
y
pm Ωwa,˜r −−−−−−Em → Ωa,˜r
is commutative. Herepandpmare metric identification mappings and in(˜x) =
˜
x for all ˜x∈W˜.
Lemma 2.5. Let (X, d) be a metric space, a∈X and let n≥2 be a na- tural number. Suppose that the limit relation (2.12)holds. Then the inequality
(2.16) card(Ωwa,˜r)≥n
implies that the embedding Em : Ωwa,˜r → Ωa,˜r is an isometry and that a pre- tangent space Ωa,˜r is tangent.
Proof. By Theorem 2.3, the limit relation (2.12) implies that
(2.17) card(Ωa,˜r)≤n.
Since Em is an injective mapping, inequalities (2.16) and (2.17) imply card(Ωa,˜r) = card(Ωwa,˜r) =n.
These equalities show that each isometric embedding Ωwa,˜r→Ωa,˜r is an isom- etry because nis a natural number.
Consider now the commutative diagram (1.4) with ˜Xa,˜x= ˜Wm. In com- plete analogy with the above proof we can show that in0 the is an isometry for every subsequence ˜r0 of ˜r. Hence, Ωa,˜r is tangent by Definition 1.4.
Example 2.6. LetH be a Hilbert space with the norm k · kand let
(2.18) dimH=m≥1.
It follows from (2.18) that for every natural k ≤ m there are orthonormal vectors e1, . . . , ek in H. Let ˜t = {tj}j∈N be a sequence of strictly decreasing positive numbers tj for which
(2.19) lim
j→∞
tj
tj+1
=∞.
Write
(2.20) X :={tjei:i= 1, . . . , k, j ∈N} ∪ {0}, i.e., 0∈X, and for all j∈N
X∩ {x∈H :kxk=t}=∅ ift6=tj and
X∩ {x∈H:kxk=tj}={tje1, . . . , tjek}.
Consider the following sequences of elements of X
˜0 ={0, . . . ,0, . . .}, x˜1={tje1}j∈N, . . . ,x˜k={tjek}j∈N.
It is easy to see that all these sequences are pairwise mutually stable w.r.t.
the normalizing sequence ˜r= ˜t and that d(˜˜xp,x˜q) := lim
j→∞
ktjep−tjeqk rj
=
√ 2
for p6=q, and that
d(˜0,x˜p) := lim
j→∞
ktjepk rj = 1 for p∈ {1, . . . , k}.
Let ˜Wm be a maximal mutually stable family such that W˜m⊇W˜ :={˜0,x˜1, . . . ,x˜k}.
We claim the following:
(a) For every ˜y∈W˜m there is ˜x∈W˜ such that ˜d(˜y,x) = 0;˜ (b) The space Ω0,˜r, corresponding to ˜Wm, is tangent.
Since card(Ωw0,˜r) = k+ 1, for the proof of (a) and (b) it is enough to show that
(2.21) lim
x1→0
···
xk+1→0
Φk+1(x1, . . . , xk+1) = 0,
where Φn is defined by (2.13) (see Remark 2.4 and Lemma 2.5).
Let (x1, . . . , xk+1) be ak+ 1-tuple with xi ∈X, i∈ {1, . . . , k+ 1}. We may assume, without loss of generality, that xi 6= 0 for all i = 1, . . . , k+ 1 because Φk+1(x1, . . . , xk+1) = 0 in the opposite case. Using (2.20), we define the natural numbers j=j(x1, . . . , xk+1) ands=s(x1, . . . , xk+1) such that
tj =
k+1
_
l=1
kxlk, tj+s=
k+1
^
l=1
kxlk.
Note that these numbers are well defined because the sequence ˜t={tj}j∈N is strictly decreasing. It follows from the definition (2.20) that s ≥1 and that
xlim1→0 ...
xk+1→0
j(x1, . . . , xk+1) =∞. Now from (2.19) we obtain
0≤lim sup
x1→0 ...
xk+1→0
Φ(x1, . . . , xk+1)≤2 lim sup
x1→0 ...
xk+1→0
k+1
V
l=1
kxlk
k+1
W
l=1
kxlk
= 2 lim sup
x1→0 ...
xk+1→0
tj+1
tj ·tj+2
tj+1 · · ·tj+s−1
tj+s ≤2 lim
j→∞
tj+1
tj = 0.
Relation (2.21) is proved.
Let (X, ρ) and (Y, d) be metric spaces. Recall that a homeomorphism f :X→Y is asimilarityif there is a positive numberk such that
d(f(x), f(y)) :=kρ(x, y)
for all x, y ∈ X. The number k is the dilatation number of f. Two metric spaces X and Y are similarif there exists a similarity ofX onto Y.
Proposition 2.7. Let H and X be the spaces defined in Example 2.6.
Then, every tangent space at the point 0∈X, is either a singleton or similar to {0, e1, . . . , ek} ⊆H.
Proof. Let Ω0,˜r be a tangent space to X at the point 0. Equality (2.21) implies that
card(Ω0,˜r) :=m≤k+ 1.
Let ˜X0,˜rbe a maximal mutually stable family for whichp( ˜X0,˜r) = Ω0,˜r(see dia- gram (1.4)). If Ω0,˜ris not one-point, then there are sequences ˜xi ={xin}n∈N∈ X˜0,˜r such that
(2.22) lim
n→∞
kxink
rn := ˜d(˜0,x˜i)∈(0,∞) for i= 1, . . . , m−1 and
(2.23) lim
n→∞
kxin−xjnk rn
:= ˜d(˜xi,x˜j)∈(0,∞) for i, j∈ {1, . . . , m−1} ifi6=j. Relations (2.22) imply
(2.24) lim
n→∞
kxjnk
kxink = d(˜˜0,x˜j)
d(˜˜0,x˜i) ∈(0,∞).
It follows from (2.20), that for everyx∈X\ {0}there is a uniquel=l(x)∈N for which
kxk=tl. Substituting tl(x) in (2.24), we obtain
n→∞lim tl(xj
n)
tl(xi n)
=
d(˜˜0,x˜j)
d(˜0,x˜j) ∈(0,∞).
These relations and (2.19) imply the equality kxink=kxjnk,
for i, j ∈ {1, . . . , m−1}, if n is taken large enough. Hence, using (2.23) and (2.20) for all sufficiently largen we have
(2.25) kxin−xjnk=tl√
2δij, kxink=tl
where δij is Kronecker’s delta, i, j∈ {1, . . . , m−1} andl=l(xjn) =l(xin). If card(Ω0,˜r)< k+ 1,
then there exists ˜xm ={xmn}n∈N such that (2.26) kxin−xmnk=tl
√
2, kxmnk=tl
fori= 1, . . . , m−1 and for alln, lwhich satisfy (2.25). Relations (2.22), (2.23), (2.25) and (2.26) imply the existence of positive finite limits
n→∞lim kxmnk
rn
and lim
n→∞
kxin−xmnk rn
for i= 1, . . . , m−1 but it contradicts maximally of ˜X0,˜r. Consequently, card(Ω0,˜r) =k+ 1.
It also follows from (2.25) that spaces Ω0,˜rand{0, e1, . . . , ek} ⊆H are similar.
To construct an one-point tangent space Ω0,˜r consider the normalizing sequence ˜r ={rn}n∈N such that
(2.27) rn:=p
tntn+1,
where {tj}j∈N is the sequence from (2.20). For all x∈X\ {0} this definition and (2.26) give either
(2.28) kxk
rn
≥ tn
√tntn+1
= r tn
tn+1
, ifkxk ≥tn, or
(2.29) kxk
rn
≤ tn+1
√tntn+1
=
rtn+1 tn
ifkxk< tn. By (2.19),q
tn
tn+1 tends to infinity withn→ ∞. Hence, if there is a finite lim
n→∞
kxnk
rn , then ˜d(˜0,x) = 0, i.e., the pretangent space Ω˜ 0,˜ris one-point.
To complete the proof, it suffices to observe that (2.28) and (2.29) imply kxk
rnk ≥
s tnk
tnk+1 or, respectively,
kxk rnk ≤
s tnk
tnk+1
for every subsequence ˜r0 = {rnk}k∈N of ˜r. Hence, Ω0,˜r0 is also one-point.
Therefore, the one-point pretangent space Ω0,˜r is tangent.
The necessary and sufficient conditions under which there is an one-point tangent space Ωa,˜r atan accumulation pointaof (X, d) are found in [1].
Example2.8. LetCbe the complex plane with the usual distanced(z, w) =
|z−w|.Fix the following three sequences:
{zj}j∈Nwith zj ∈Cand |zj|= 1;
{αj}j∈N with αj ∈Rand lim
j→∞αj = 0;
{tj}j∈Nwith tj ∈R+, tj+1 < tj and lim
j→∞
tj+1 tj
= 0.
Let X := {tjzj : j ∈ N} ∪ {tjzjeiαj : j ∈ N} ∪ {0}. In the case under consideration function (2.1) has the form
(2.30) F(z, w) = |z−w|(|z| ∧ |w|) (|z| ∨ |w|)2 if (z, w)6= (0,0).
We claim that the equality
(2.31) lim
z→0 w→0
F(z, w) = 0
holds. Indeed, as in Example 2.7, for every (w, z) ∈(X\ {0})2 we can find j and ssuch that j≤sand
|w| ∨ |z|=tj and |w| ∧ |z|=ts. If j < s, then we have
F(z, w)≥ |tj−ts|ts t2j =
1−ts tj
ts tj
≥ ts tj
≥ tj+1 tj
but ifj =swe obtain
F(z, w) = |tjzjeiαj−tjzj|tj
t2j =|eiαj−1|.
Since
j→∞lim tj+1
tj
= lim
j→∞|1−eiαj|= 0,
(2.31) holds. Theorem 2.2 implies that each pretangent space at the point 0∈Xis either two-point or one-point. Note that two-point pretangent spaces at 0 ∈ X exist because 0 is not an isolated point of X. Moreover, all these pretangent spaces are tangent by Lemma 2.6. To construct an one-point tan- gent space it is enough to take a normalizing sequence ˜r with rj = √
tjtj+1
(see the end of the proof of Proposition 2.7).
3. COMPLETENESS OF THE TANGENT SPACES
In this section we shall prove that all tangent spaces to metric spaces are complete.
Lemma3.1. Let (X, d) be a metric space, let r˜={rn}n∈N be a normali- zing sequence and let F˜r˜1,F˜r˜2, F˜r˜3 be mutually stable families of sequences x˜= {xn}n∈N with xn∈X for n∈N. Suppose that F˜˜r1 is maximal, F˜r˜2 ⊆F˜˜r1∩F˜˜r3 and that the pseudometric space ( ˜F˜r2,d)˜ is a dense subspace of ( ˜Fr˜3,d). Then˜ the inclusion
(3.1) F˜˜r3⊆F˜r˜1
holds.
Proof. Suppose that (3.1) does not hold. Since ˜Fr˜1 is maximal mutually stable w.r.t. ˜r, there are two sequences ˜x={xn}n∈N∈F˜r˜3 and ˜y={yn}n∈N∈ F˜r˜1 such that either
(3.2) ∞ ≥lim sup
n→∞
d(xn, yn) rn
>lim inf
n→∞
d(xn, yn) rn
≥0 or
(3.3) lim
n→∞
d(xn, yn) rn
=∞.
For every ε >0 we can find ˜z={zn}n∈N∈F˜˜r2 such that d(˜˜x,y)˜ < ε
because ˜Fr˜2 is dense in ˜F˜r3 by the hypothesis of the lemma. The triangle in- equality implies
d(xn, yn) rn
− d(zn, yn) rn
≤ d(xn, zn) rn
for all n∈N. Consequently, we obtain (3.4)
lim sup
n→∞
d(xn, yn)
rn −d(˜˜z,y)˜
≤d(˜˜x,y)˜ < ε and
lim inf
n→∞
d(xn, yn)
rn −d(˜˜z,y)˜
< ε.
Therefore, the inequality
lim sup
n→∞
d(xn, yn) rn
−lim inf
n→∞
d(xn, yn) rn
<2ε
holds for all ε > 0, contrary to (3.2). To complete the proof, it suffices to observe that (3.4) contradicts (3.3).
Lemma 3.2. Let (X, d) be a metric space, let a ∈ X and let Ωa,˜r,
˜
r = {rn}n∈N, be a pretangent space with a corresponding maximal mutually stable family X˜a,˜r. Suppose thatΛ is a bounded subset ofΩa,˜r. Then there is a constant c >0such that for every α∈Λ we can findx˜={xn}n∈N∈X˜a,˜r with
(3.5) p(˜x) =α and d(xn, a)
rn ≤c for n∈N, where p is the metric identification map.
Proof. Write
A:={˜x∈X˜a,˜r :p(˜x)∈Λ}.
Since Λ is bounded in Ωa,˜r, we have
sup{d(˜˜x,˜a) : ˜x∈A}:=k <∞.
For every ˜x={xn}n∈N∈A introduce the sequence ˜x∗={x∗n}n∈N by the rule x∗n=
( xn ifd(xna)<(k+ 1)rn, a ifd(xn, a)≥(k+ 1)rn.
Note that for every ˜x ∈ A there is n(˜x) ∈N such that xn = x∗n if n ≥n(˜x) because
d(˜˜x,˜a) = lim
n→∞
d(xn, a) rn
≤k < k+ 1.
Consequently, ˜x∗∈X˜a,˜r for all ˜x∈A. Moreover, relations (3.5) evidently hold with ˜x= ˜x∗,xn=x∗n and c=k+ 1, which is what had to be proved.
Theorem 3.3. Let (X, d) be a metric space, a∈X and let r˜∈ {rn}n∈N be a normalizing sequence. Then a tangent space Ωa,˜r, if it exists, is complete.
Proof. Let{αj}j∈N be a fundamental sequence in the metric space Ωa,˜r. We must prove that {αj}j∈N is convergent. Let ˜Xa,˜r be a maximal mutually stable family with the pretangent space Ωa,˜r and let p : ˜Xa,˜r → Ωa,˜r be the metric identification mapping. Every fundamental sequence of an arbitrary metric space is bounded. Hence, by Lemma 3.5, there is c > 0 such that for every j∈N there exists ˜xj ={xjn}n∈N∈X˜a,˜r for which
(3.6) p(˜xj) =αj and d(xjn, a) rn
≤c for all n∈N.
In accordance with Definition 1.4 and Lemma 3.1 it suffices to prove that there exists a sequence ˜x = {xn}n∈N of points from X such that for some subsequence of natural numbers nk and for all j ∈ N the sequences
˜
x0 ={xnk}k∈N and ˜x0j = {xjnk}k∈N are mutually stable w.r.t. ˜r0 ={rnk}k∈N and
(3.7) lim
j→∞ lim
k→∞
d(xnk, xjnk) rnk
= 0.
Suppose now that (3.7) holds and, in addition, we have the following condition: for every k∈Nthere is j(k)∈Nsuch that
(3.8) xnk =xj(k)nk ,
where xj(k)nk is the nkth element of ˜xj(k). Then it follows from (3.6) that the inequality
d(xjnk, xnk) rnk
≤2c holds for all j, k∈N. In particular, we obtain
d(x1nk, xnk) rnk ≤2c
for allk∈N. Since every bounded infinite sequence of real numbers has a con- vergent subsequence, there is an increasing infinite subsequence {n(1)k }k∈N of the sequence{nk}k∈Nsuch that lim
k→∞
d(xn(1) k
,x1
n(1) k
)
rn(1) k
is finite. Hence, the sequences {xn(1)k }k∈N and{x1
n(1)k }k∈Nare mutually stable w.r.t. {r
n(1)k }k∈N. Analogously, by induction, we can prove that, for every integeri≥2, there is a subsequence {n(i)k }k∈N of the sequence {n(i−1)k }k∈N such that {x
n(i)k }k∈N and {xi
n(i)k }k∈N are mutually stable w.r.t. {r
n(i)k }k∈N. Consider now the diagonal sequence {n(k)k }k∈N. For everyi∈Nthe sequences{xi
n(k)k }k∈Nand{x
n(k)k }k∈Nare mutu- ally stable w.r.t. {r
n(k)k }k∈N. Moreover, (3.7) evidently holds withnk=j(k) = n(k)k . Consequently, for the proof of the theorem it suffices to construct ˜xsuch that (3.8) and (3.7) are satisfied. In order to construct this ˜x, we estimate
d(xjn,xin)
rn .
Lemma 3.4. Let {˜xj}j∈N (˜xj ={xjn}n∈N) be a fundamental sequence in a pseudometric space ( ˜Xa,˜r,d). Then there is a sequence˜ {˜x∗j}j∈N∈X˜a,˜r with the following properties:
(i) The equality
(3.9) d(˜˜xj,x˜∗j) = 0 holds for all j∈N.
(ii)For every ε >0 there existsj0(ε)∈N such that the inequality
(3.10) sup
n∈N
d(x∗jn, x∗in) rn
≤ε holds if i∧j≥j0(ε).
Proof. Let {˜xjk}k∈N be a subsequence of {˜xj}j∈N such that j1 < j2 <
· · ·< jk· · · and
(3.11) d(˜˜xj,x˜jk)≤ 1
2 k+1
whenever j ≥jk. For all j, n∈Nwrite
(3.12) (1)xjn=
xjn ifj≤j1,
xjn ifj > j1 and d(xjn1, xjn) rn
≤ 12, xjn1 ifj > j1 and d(xjn1, xjn)
rn > 12, and put
(1)x˜j :={(1)xjn}n∈N.
Inequality (3.11) implies that, for every j ∈N, there is n0(j) ∈N such that
(1)xjn=xjn for all n≥n0(j). Hence, we have the equality (3.13) d(˜˜xj,(1)x˜j) = 0
for all j∈N. Note that (3.13) implies the inequality d(˜(1)x˜j,x˜jk)≤
1 2
k+1
whenever j ≥jk. Moreover, it follows from (3.12) that sup
n∈N
d((1)xjn,(1)xin) rn
≤sup
n∈N
d((1)xjn, xjn1) rn
+ sup
n∈N
d((1)xin, xjn1) rn
≤1 whenever i∧j > j1. Now we define (k)xjn by induction ink.
Ifk≥2 andj, n∈N write
(k)xjn=
(k−1)xjn ifj≤jk,
(k−1)xjn ifj > jk and d(xjnk,(k−1)xjn)
rn ≤
1 2
k
, xjnk ifj > jk and d(xjnk,(k−1)xjn)
rn >
1 2
k
,
and put
(k)x˜j :={(k)xjn}n∈N.
In the same manner as in the case k= 1 we obtain the equality (3.14) d(˜xj,(k)x˜j) = 0
for all j, k∈N and the inequality
(3.15) sup
n∈N
d((k)xjn,(k)xin)
rn ≤
1 2
k
whenewer i∧j > jk. Now set for all n∈N
x∗jn =(1)xjn if 1≤j≤j1, x∗jn =(2)xjn ifj1 < j≤j2,
...
x∗jn =(k)xjn ifjk−1 < j≤jk,
and so on. The sequence ˜x∗j :={x∗jn}n∈Nhas the properties (i) and (ii) because (3.14) implies (3.9) and, moreover, (3.10) follows from (3.15).
Continuation of the proof of Theorem 3.3. Using Lemma 3.4, we may assume that, for every ε >0, there isj0 =j0(ε)∈Nsuch that
(3.16) sup
n∈N
d(xin, xjn) rn ≤ε
ifi∧j ≥j0(ε). Setxn=xnn for all n∈N. It follows from (3.16) that
(3.17) lim sup
n→∞
d(xn, xjn) rn
≤ε.
As was shown in the first part of the proof there is a subsequence ˜r0 ={rnk}k∈N of the sequence ˜r such that the sequences ˜x0j ={xjnk}k∈N and ˜x0 ={xnk}k∈N are mutually stable w.r.t. ˜r0 for all j∈N. Hence, by (3.17), we obtain
n→∞lim
d(xnk, xjnk) rnk
≤ε
ifj≥j0(ε). Lettingj→ ∞ we obtain (3.7). To complete the proof, it suffices to observe that (3.8) holds withj(k) =nk.
Recall that a map f : X → Y is called closed if the image of each set closed in X is closed in Y.
Corollary 3.5. Let (X, d) be a metric space, let Y be a subspace of X and let a ∈ Y. If a pretangent space Ωya,˜r is tangent, then iny : Ωya,˜r → Ωa,˜r is closed.
Proof. The map iny is an isometric embedding. Hence, iny is closed if and only if the set iny(Ωya,˜r) is a closed subset of Ωa,˜r. The space Ωya,˜r is complete by Theorem 3.6. Since a metric space is complete if and only if this space is closed in every its superspace (see, for example, [8, Theorem 10.2.1]), iny(Ωya,˜r) is closed in Ωa,˜r.
Acknowledgements.The first author is thankful to the Finnish Academy of Science and Letters for the support.
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Received 29 February 2009 Institute of Applied Mathematics and Mechanics of NASU
R. Luxemburg str. 74, Donetsk 83114, Ukraine dovgoshey@iamm.ac.donetsk.ua
and
University of Helsinki
Department of Mathematics and Statistics P.O. Box 68, FI-00014, Finland
olli.martio@helsinki.fi
