aa r X i v : . [ m a t h . F A ] O c t ON EXPANSIVE MAPPINGS
MARAT V. MARKIN AND EDWARD S. SICHEL
Abstract.
When finding an original proof to a known result describing ex-pansive mappings on compact metric spaces as surjective isometries, we revealthat relaxing the condition of compactness to total boundedness preserves theisometry property and nearly that of surjectivity.While a counterexample is found showing that the converse to the abovedescriptions do not hold, we are able to characterize boundedness in terms ofspecific expansions we call anticontractions . O God, I could be bounded in anutshell, and count myself a king ofinfinite space - were it not that Ihave bad dreams.
William Shakespeare(Hamlet, Act 2, Scene 2) Introduction
We take a close look at the nature of expansive mappings on certain metric spaces(compact, totally bounded, and bounded), provide a finer classification for suchmappings, and use them to characterize boundedness.When finding an original proof to a known result describing all expansive mappingson compact metric spaces as surjective isometries [1, Problem X . . ∗ ], we revealthat relaxing the condition of compactness to total boundedness still preserves theisometry property and nearly that of surjectivity.We provide a counterexample of a not totally bounded metric space, on which theonly expansion is the identity mapping, demonstrating that the converse to theabove descriptions do not hold.Various examples for different types of expansions are furnished, in particular theone of a nonsurjective expansion on a totally bounded “dial set” in the complexplane which allows to better understand the essence of the latter.2. Preliminaries
Here, we outline certain preliminaries essential for the subsequent discourse (formore, see, e.g., [2–6]).
Mathematics Subject Classification.
Primary 54E40, 54E45; Secondary 46T99.
Key words and phrases.
Metric space, expansion, compactness, total boundednessF.
Definition 2.1 (Sequential Compactness) . A set A in a metric space ( X, d ) is called sequentially compact , or compact in the
Bolzano-Weierstrass sense , if every sequence ( x n ) n ∈ N of its elements contains asubsequence convergent to an element of A .A metric space ( X, d ) is said to be sequentially compact if sequentially compact isthe set X . Remark 2.1.
In a metric space setting, the above definition of compactness isequivalent to compactness in the
Heine-Borel sense defined via open covers (see,e.g., [3, 5]).It is convenient for us to use a sequential definition for total boundedness as well(see e.g., [3, 4]).
Definition 2.2 (Total Boundedness) . A set A in a metric space ( X, d ) is called totally bounded if every sequence of itselements contains a fundamental (Cauchy) subsequence.A metric space (
X, d ) is said to be totally bounded if totally bounded is the set X . Definition 2.3 (Boundedness) . A set A in a metric space ( X, d ) is said to be bounded ifdiam( A ) := sup x,y ∈ X d ( x, y ) < ∞ , the number diam( A ) being called the diameter of A A metric space (
X, d ) is said to be bounded if bounded is the set X . Remark 2.2.
In a metric space, a (sequentially) compact set is totally boundedand a totally bounded set is bounded but not vice versa (see, e.g., [3]).3.
Expansive Mappings
Now, we introduce and further classify the focal subject of our study, expansivemappings (or expansions ). Definition 3.1 (Expansive Mapping) . Let (
X, d ) be a metric space. A mapping T : X → X on ( X, d ) such that ∀ x, y ∈ X : d ( T x, T y ) ≥ d ( x, y )is called an expansive mapping (or expansion ).It is important for our discourse to introduce a finer classification of expansions. Definition 3.2 (Types of Expansions) . Let (
X, d ) be a metric space.(1) An expansion T : X → X such that ∀ x, y ∈ X : d ( T x, T y ) = d ( x, y )is called an isometry , which is the weakest form of expansive mappings. N EXPANSIVE MAPPINGS 3 (2) An expansion T : X → X such that ∃ x, y ∈ X, x = y : d ( T x, T y ) > d ( x, y )we call a proper expansion .(3) An expansion T : X → X such that ∀ x, y ∈ X, x = y : d ( T x, T y ) > d ( x, y )we call a strict expansion .(4) Finally, an expansion T : X → X such that ∃ E > ∀ x, y ∈ X : d ( T x, T y ) ≥ Ed ( x, y )we call an anticontraction with expansion constant E . Remark 3.1.
Clearly, any anticontraction is necessarily a strict expansion , whichin turn is also a proper expansion . However, as the following examples demonstrate,the converse statements are not true.
Examples 3.1.
1. On C with the standard metric, the mapping g ( z ) := e i z, i.e., the counterclockwise rotation by one radian, is an isometry which isnot a proper expansion.2. On the space ℓ ∞ of all real- or complex-termed bounded sequences with itsstandard supremum metric ℓ ∞ ∋ x := ( x k ) k ∈ N , y := ( y k ) k ∈ N d ∞ ( x, y ) := sup k ∈ N | x k − y k | , the right shift mapping ℓ ∞ ∋ ( x , x , x . . . ) T ( x , x , x . . . ) := (0 , x , x , x . . . ) ∈ ℓ ∞ is also an isometry which is not a proper expansion.3. On ℓ ∞ , the mapping ℓ ∞ ∋ ( x , x , x . . . ) T ( x , x , x . . . ) := ( x , x , x , x , . . . ) ∈ ℓ ∞ is a proper expansion that is not strict, since, for x := (1 , , , . . . ) , y :=(1 / , , , . . . ) ∈ ℓ ∞ , d ∞ ( T x, T y ) = 3 / > / d ∞ ( x, y ) , but, for x := (1 , , , . . . ) , y := (0 , , , . . . ) ∈ ℓ ∞ , d ∞ ( T x, T y ) = 1 = d ∞ ( x, y ) .
4. In the space L (0 , ∞ ), consider the set of the equivalence classes { f n } n ∈ N represented by the functions f n ( x ) := √ nχ [0 , /n ] ( x ) , n ∈ N , x ∈ (0 , ∞ ) , ( χ · ( · ) is the characteristic function of a set), which is a subset of the unitsphere S (0 ,
1) := { f ∈ L (0 , ∞ ) | d ( f,
0) = k f k = 1 } . MARAT V. MARKIN AND EDWARD S. SICHEL
For any m, n ∈ N with n > m , we have: d ( f n , f m ) = k f n − f m k = (cid:20)Z ∞ | f n ( x ) − f m ( x ) | dx (cid:21) / = (cid:20)Z ∞ (cid:12)(cid:12) ( √ n − √ m ) χ [0 , /n ] ( x ) − √ mχ (1 /n, /m ] ( x ) (cid:12)(cid:12) dx (cid:21) / = "Z /n ( √ n − √ m ) dx + Z /m /n √ m dx / = (cid:20) m − √ m √ n + nn + m (cid:18) m − n (cid:19)(cid:21) / = (cid:20) − r mn (cid:21) / . The map
T f n := f kn , n ∈ N , with an arbitrary fixed k ∈ N is an isometry on { f n } n ∈ N since, for any m, n ∈ N with n > m , d ( T f n , T f m ) = k T f n − T f m k = k f kn − f km k = " − r kmkn / = (cid:20) − r mn (cid:21) / = k f m − f n k = d ( f n , f m ) . On the other hand, the map Sf n := f n , n ∈ N , is a strict expansion on { f n } n ∈ N since, for any m, n ∈ N with n > m , d ( Sf n , Sf m ) = k Sf n − Sf m k = k f n − f m k = " − r m n / = h − mn i / > (cid:20) − r mn (cid:21) / = k f n − f m k = d ( f n , f m ) , which is not an anticontraction since d ( Sf n , Sf n ) d ( f n , f n ) = (cid:2) − n (cid:3) / h − √ n i / → , n → ∞ .
5. On R with the standard metric, the mapping f ( x ) = 2 x is an anticontraction with expansion constant E = 2. However, the samemapping, when considered on R equipped with the metric R ∋ x, y ρ ( x, y ) := | x − y || x − y | + 1 , turning R into a bounded space (see, e.g., [3]), is merely a strict expansion,which is not an anticontraction since ρ ( f ( x ) , f (0)) ρ ( x,
0) = ρ (2 x, ρ ( x,
0) = | x || x | +1 | x || x | +1 → , x → ∞ . N EXPANSIVE MAPPINGS 5 Expansions on Compact Metric Spaces
Theorem 4.1 (Expansions on Compact Metric Spaces [1, Problem X . . ∗ ]) . An expansive mapping T on a compact metric space ( X, d ) is a surjection, i.e., T ( X ) = X, and an isometry, i.e., ∀ x, y ∈ X : d ( T x, T y ) = d ( x, y ) . Proof.
For an arbitrary point x ∈ X , and an increasing sequence ( n ( k )) k ∈ N ofnatural numbers, consider the sequence (cid:16) x n ( k ) := T n ( k ) x (cid:17) k ∈ N in ( X, d ).Since the space (
X, d ) is compact , there exists a convergent subsequence (cid:0) x n ( k ( j )) (cid:1) j ∈ N ,which is necessarily fundamental . Remark 4.1.
Subsequently, we use only the fundamentality , and not the con-vergence of the subsequence, and hence, only the total boundedness and not thecompactness of the underlying space (Remark 2.2).By the fundamentality of (cid:0) x n ( k ( j )) (cid:1) j ∈ N , without loss of generality, we can regardthe indices n ( k ( j )), j ∈ N , chosen sparsely enough so that d ( x n ( k ( j )) , x n ( k ( j )) ) ≤ j , j ∈ N . Since T is an expansion, d ( x, x n ( k ( j )) ) ≤ d ( T n ( k ( j )) x, T n ( k ( j )) x n ( k ( j )) ) = d ( x n ( k ( j )) , x n ( k ( j )) ) ≤ j , j ∈ N . We thus conclude that x n ( k ( j )) = T n ( k ( j )) x → x, j → ∞ , which implies that the range T ( X ) is dense in ( X, d ), i.e., T ( X ) = X. Now, let x, y ∈ X be arbitrary. Then, for the sequence ( x n := T n x ) n ∈ N , we can, bythe above argument, select a subsequence (cid:0) x n ( k ) (cid:1) k ∈ N such that x n ( k ) → x, k → ∞ , and then, in turn, for the sequence (cid:0) y n ( k ) := T n ( k ) y (cid:1) k ∈ N , we choose a subsequence (cid:0) y n ( k ( j )) (cid:1) j ∈ N for which y n ( k ( j )) → y, j → ∞ . Since (cid:0) x n ( k ( j )) (cid:1) j ∈ N is a subsequence of (cid:0) x n ( k ) (cid:1) k ∈ N , we also have:lim j →∞ x n ( k ( j )) = lim k →∞ x n ( k ) = x. MARAT V. MARKIN AND EDWARD S. SICHEL
Then, in view of the expansiveness of T , for any j ∈ N , d ( x, y ) ≤ d ( T x, T y ) ≤ d ( T n ( k ( j )) x, T n ( k ( j )) y ) = d ( x n ( k ( j )) , y n ( k ( j )) ) . Whence, passing to the limit as j → ∞ , by joint continuity of metric, we arrive at d ( x, y ) ≤ d ( T x, T y ) ≤ d ( x, y ) , which implies that ∀ x, y ∈ X : d ( T x, T y ) = d ( x, y ) , i.e., T is an isometry . Remark 4.2.
Thus far, only the total boundedness and not the compactness ofthe underlying space has been utilized (Remark 2.2).Being an isometry, the mapping T is continuous , whence, since X is compact , weinfer that the image T ( X ) is compact as well, and therefore closed in ( X, d ) (see,e.g., [3]).In view of the denseness and the closedness of T ( X ), we conclude that T ( X ) = T ( X ) = X, i.e., T is also a surjection , as desired, which completes the proof. Remark 4.3.
For the surjectivity of T , the requirement of the compactness ofthe underlying space is essential, as we rely on the fact the continuous image of acompact set is compact. Example 5.1 demonstrates that this requirement cannotbe relaxed even to total boundedness. (cid:3) Expansions on Totally Bounded Metric Spaces
We proceed now to demonstrate that relaxing the condition of the compactness ofthe underlying space to total boundedness yields a slightly weaker result, in whichexpansions emerge as “presurjective” isometries.
Theorem 5.1 (Expansions on Totally Bounded Metric Spaces) . An expansive mapping T on a totally bounded metric space ( X, d ) has a dense range,i.e., T ( X ) = X (“presurjection”), and is an isometry, i.e., ∀ x, y ∈ X : d ( T x, T y ) = d ( x, y ) . Proof.
As is shown in the corresponding part of the proof of Theorem 4.1 (seeRemarks 4.1 and 4.2), the image T ( X ) is dense in ( X, d ), i.e., T ( X ) = X, and T is an isometry . (cid:3) N EXPANSIVE MAPPINGS 7
As is mentioned in Remark 4.3, the compactness of the underlying space is essentialfor the surjectivity of expansions, the following example demonstrating that, whencompactness is relaxed to total boundedness, surjectivity is not guaranteed.
Example 5.1 (Dial Set) . Let D := (cid:8) e in (cid:9) n ∈ Z + ⊂ { z ∈ C | | z | = 1 } ( Z + is the set of nonnegative integers ) be a dial set in the complex plane C withthe usual distance, which is bounded in C , and hence, totally bounded (see, e.g.,[3]), and D ∋ e in T e in := e i ( n +1) ∈ D, n ∈ Z + , be the counterclockwise rotation by one radian, which is, clearly, an isometry (seeExamples 3.1) but not a surjection on D since, as is easily seen, D ∋ e i / ∈ T ( D ) . Remarks 5.1. • This, in particular, implies that, by Theorem 4.1, the dial set D is notcompact , and hence, not closed , in C (see, e.g., [3]). • Thus, on a totally bounded, in particular compact, metric space, any ex-pansion is not proper but is an isometry which may fall a little short ofbeing surjective.By Theorem 5.1, the range T ( D ) is dense in the dial set D , which is not closed,relative to the usual distance. This allows us to “turn the tables” on the dial setand derive the following rather interesting immediate corollary. Corollary 5.1.
Let D := (cid:8) e in (cid:9) n ∈ Z + . Then,(1) for an arbitrary n ∈ Z + , there exists an increasing sequence ( n ( k )) k ∈ N ofnatural numbers such that e in ( k ) → e in , k → ∞ ; (2) there exists a θ ∈ R \ Z + for which there is an increasing sequence ( n ( k )) k ∈ N of natural numbers such that e in ( k ) → e iθ , k → ∞ . Proof. (1) Part (1) immediately follows from the fact that, by Theorem 5.1, the range T ( D ) = (cid:8) e in (cid:9) n ∈ N is dense in D .(2) Part (2) follows from the fact that the set D , being not closed (see Remarks5.1), has at least one limit point not belonging to D , which, by continuityof metric, is located on the unit circle { z ∈ C | | z | = 1 } , i.e., is of the form e iθ with some θ ∈ R \ Z + . (cid:3) MARAT V. MARKIN AND EDWARD S. SICHEL
Remark 5.2.
If posed as a problem, the prior statement, although simply stated,might be quite challenging to be proved exclusively via the techniques of classicalanalysis. 6.
Are the Converse Statements True?
Now, there are two natural questions to ask. • If every expansive map T on a metric space ( X, d ) is a surjective isometry,is the space compact? • If every expansive map T on a metric space ( X, d ) is a presurjective isometry(see Theorem 5.1), is the space totally bounded?In other words, do the converse statements to Theorems 4.1 and 5.1 hold?The following example answers both questions in the negative.
Example 6.1.
In the space ℓ ∞ , consider the bounded set { x n } n ∈ N defined by x n := , . . . , , n | {z } n th term , , . . . , n ∈ N , and let T be an arbitrary expansion on { x n } n ∈ N . First, we note that, for anyexpansion, if ∃ m, n ∈ N , m = n : T x m = T x n , then 0 = d ( T x m , T x n ) < d ( x m , x n )contradicting the expansiveness of T . Thus, the mapping T is injective .Observe that ∀ m, n = 2 , , . . . : d ( x m , x n ) < d ( x , x n ) . Assume(6.1)
T x = x . Then
T x = x k with some k ∈ N , k ≥
2. Let n ∈ N , n ≥
2, be arbitrary.There are two possibilities: either
T x n = x or T x n = x . In the first case, we have: d ( T x , T x n ) = d ( x k , T x n ) < d ( x , x n ) . contradicting the expansiveness of T . N EXPANSIVE MAPPINGS 9
In the second case, for any m ∈ N , m = n , by the injectivity of T , T x m = x , and hence, d ( T x , T x m ) = d ( x k , T x m ) < d ( x , x m ) , which again contradicts the expansiveness of T .The obtained contradictions making assumption (6.1) false, we conclude that T x = x . Therefore, by the injectivity of T , we can restrict the expansion T to the subset { x n } n ≥ . Applying the same argument, one can show that T x = x . Continuing inductively, we see that ∀ n ∈ N : T x n = x n , i.e. T is the identity map , which is both a surjection and an isometry , even thoughthe set { x n } n ∈ N is not totally bounded, let alone compact (see Remark 2.2), as ∀ m, n ∈ N , m = n : d ∞ ( x m , x n ) > . Remark 6.1.
Thus, a metric space with the property that every expansion onit is a presurjective isometry need not be totally bounded. Such spaces, which,by Theorems 4.1 and 5.1, encompass compact and totally bounded, can be called nonexpansive . 7.
A Characterization of Boundedness
Although bounded sets support strict expansions (see Examples 3.1 4, 5). Any at-tempt to produce an anticontraction on a bounded set would be futile, the followingcharacterization explaining why.
Theorem 7.1 (Anticontraction Characterization of Boundedness) . A metric space ( X, d ) is bounded iff no subset of X supports an anticontraction.Proof. The case of a singleton being trivial, suppose that X consists of at least twodistinct elements. “Only if” part. We proceed by contradiction , assuming that X is bounded and thereexists a subset A ⊆ X supporting an anticontraction T : A → A with expansionconstant E . Then ∀ x, y ∈ A, x = y ∀ n ∈ N : T n x, T n y ∈ A, which implies diam( A ) ≥ d ( T n x, T n y ) ≥ E n d ( x, y ) → ∞ , n → ∞ . Hence, A is unbounded, and since A ⊆ X , this contradicts the boundedness of X ,the obtained contradiction proving the “only if ” part. “If” part. Here, we proceed by contrapositive assuming X to be unbounded andshowing that there exists a subset of X which supports an anti-contraction.Since X is unbounded, we can select two distinct points x , x ∈ X , and subse-quently pick x so that min ≤ i ≤ d ( x , x i ) > ≤ i,j ≤ d ( x i , x j )Continuing inductively in this fashion, we construct a countably infinite subset S := { x n } n ∈ N of X such thatmin ≤ i ≤ n d ( x n +1 , x i ) > ≤ i,j ≤ n d ( x i , x j ) . Let If we then define T : { x n } n ∈ N → { x n } n ∈ N by: S ∋ x n T x n := x n +1 ∈ S, n ∈ N . Then, for any m, n ∈ N with n > m , d ( T x n , T x m ) = d ( x n +1 , x m +1 ) ≥ min ≤ i ≤ n d ( x n +1 , x i ) > ≤ i,j ≤ n d ( x i , x j ) ≥ d ( x n , x m ) , which implies that T is an anti-contraction with expansion constant E = 2 on S ⊆ X completing the proof of the “if ” part and the entire statement. (cid:3) Reformulating equivalently, we arrive at
Theorem 7.2 (Anticontraction Characterization of Unboundedness) . A metric space ( X, d ) is unbounded iff there exists a subset of X which supports ananticontraction. Acknowledgments
The authors would like to express their appreciation to Drs. Michael Bishop, Prze-myslaw Kajetanowicz, and other members of the Functional Analysis and Math-ematical Physics Interdepartmental Research Group (FAMP) of California StateUniversity, Fresno for insightful questions and stimulating discussions.
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Department of MathematicsCalifornia State University, Fresno5245 N. Backer Avenue, M/S PB 108Fresno, CA 93740-8001
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