FATEMAH AYATOLLAH ZADEH SHIRAZI, FATEMEH EBRAHIMIFAR

Transcription

1 IS THERE ANY NONTRIVIAL COMPACT GENERALIZED SHIFT OPERATOR ON HILBERT SPACES? arxiv: v [math.fa] 2 Apr 208 FATEMAH AYATOLLAH ZADEH SHIRAZI, FATEMEH EBRAHIMIFAR Abstract. In the following text for cardinal number τ > 0, and self map ϕ : τ τ we show the generalized shift operator σ ϕ(l 2 (τ)) l 2 (τ) (where σ ϕ((x α) ) = (x ϕ(α) ) for (x α) C τ ) if and only if ϕ : τ τ is bounded and in this case σ ϕ l 2 (τ) : l2 (τ) l 2 (τ) is continuous, consequently σ ϕ l 2 (τ) : l2 (τ) l 2 (τ) is a compact operator if and only if τ is finite. 200 Mathematics Subject Classification: 46C99 Keywords: Compact operator, Generalized shift, Hilbert space.. Preliminaries The concept of generalized shifts has been introduced for the first time in [2] as a generalization of one-sided shift {,...,k} N {,...,k} N and two-sided shift (a,a 2, ) (a 2,a 3, ) {,...,k} Z {,...,k} Z [0, 9]. Suppose K is a nonempty set with at least two (a n) n Z (a n+) n Z elements, Γ is a nonempty set, and ϕ : Γ Γ is an arbitrary map, then we call σ ϕ : K Γ K Γ a generalized shift (for one-sided and two-sided shifts consider ϕ(n) = n +). It s evident that for topological space K, σ ϕ : K Γ K Γ is (x α) α Γ (x ϕ(α) ) α Γ continuous, where K Γ is equipped by product topology. For Hilbert space H there exists unique cardinal number τ such that H and l 2 (τ) are isomorphic [4, 8]. All members of the collection {l 2 (τ) : τ is a non zero cardinal number} are Hilbert spaces, moreover for cardinal number τ and (x α ) K τ (where K {R,C} depending on our choice for real Hilbert spaces or Complex Hilbertspaces)wehavex = (x α ) l 2 (τ) ifandonlyif x 2 := Σ x α 2 < +. Moreover for (x α ),(y α ) l 2 (τ) let < (x α ),(y α ) >= Σ x α y α (inner product). For ϕ : τ τ, one may consider σ ϕ : K τ K τ in particular we may study σ ϕ l 2 (τ): l 2 (τ) K τ. Convention. In the following text suppose τ > is a cardinal number and ϕ : τ τ is arbitrary, we denote σ ϕ l 2 (τ): l 2 (τ) K τ simply by σ ϕ : l 2 (τ) K τ, and equip l 2 (τ) with its usual inner product introduced in the above lines. Also for cardinal number ψ let (for properties of cardinal numbers and their arithmetic see [7]): { ψ ψ ψ isfinite, := + otherwise. Moreover for s t let δ t s = 0 and δs s =. If X,Y are normed vector spaces, we say the linear map S : X Y is an operator if it is continuous. We call (X,T) a linear dynamical system, if X is a normed

2 2 F. AYATOLLAH ZADEH SHIRAZI, F. EBRAHIMIFAR vector space and T : X X is an operator [5]. Let s recall that R is the set of real numbers, C is the set of complex numbers, and N = {,2,...} is the set of natural numbers. 2. On generalized shift operators In this section we show σ ϕ (l 2 (τ)) l 2 (τ) (and σ ϕ : l 2 (τ) l 2 (τ) is continuous) if and only if ϕ : τ τ is bounded. Moreover σ ϕ (l 2 (τ)) = l 2 (τ) if and only if ϕ : τ τ is one to one. Remark 2.. We say f : A A is bounded if there exists finite n such that for all a A we have card(ϕ (a)) n [6]. Theorem 2.2. The following statements are equivalent:. σ ϕ (l 2 (τ)) l 2 (τ), 2. ϕ : τ τ is bounded, 3. σ ϕ : l 2 (τ) l 2 (τ) is a linear continuous map. Moreover in the above case we have σ ϕ = sup{(card(ϕ (α))) : α τ}. Proof. First note that for x = (x α ) we have (*) σ ϕ (x) 2 = Σ x ϕ(α) 2 = Σ ( (card(ϕ (α))) x α 2) (where 0(+ ) = (+ )0 = 0). () (2) Suppose σ ϕ (l 2 (τ)) l 2 (τ), for θ < τ we have (δ θ α) = and: σ ϕ ((δ θ α ) ) 2 = (card(ϕ (θ))) by (*). Hence (δα) θ l 2 (τ) and σ ϕ ((δα) θ ) σ ϕ (l 2 (τ)) l 2 (τ), thus ϕ (θ) is finite. Thus {card(ϕ (α)) : α τ} is a collection of finite cardinal numbers. If sup{(card(ϕ (α))) : α τ} = +, then there exists a strictly increasing sequence {n k } k in N and sequence {α k } k in τ such that for all k we have card(ϕ (α k )) = n k. Since {n k } k is a one to one sequence, {α k } k is a one to one sequence too. Consider (x α ) with: { x α := k α = α k,k, 0 otherwise. Then Σ x α 2 = Σ x 2 α k = Σ k < + and (x 2 α ) l 2 (τ). On the other k hand by (*) we have σ ϕ ((x α ) ) 2 n = Σ k k Σ 2 k = + (note that n k k k k for all k ), in particular σ ϕ ((x α ) ) / l 2 (τ) which leads to the contradiction σ ϕ (l 2 (τ)) l 2 (τ). Therefore sup{card(ϕ (α)) : α τ} is finite and is a natural number. (2) (3) Suppose n := sup{card(ϕ (α)) : α τ} is finite. For all x = (x α ) l 2 (τ) we have: k σ ϕ (x) = Σ ((card(ϕ (α))) x α 2 ) Σ (n x α 2 ) = n x whichshowscontinuityofσ ϕ and σ ϕ n. Ontheotherhand, thereexistsθ < τ with card(ϕ (θ) = n. By (δα) θ = and (*) we have σ ϕ ((δα) θ ) = n which leads to σ ϕ n.

3 GENERALIZED SHIFTS & COMPACT OPERATORS 3 By [, Lemma 4.] and [2], ϕ : τ τ is one to one (resp. onto) if and only if σ ϕ : K τ K τ is onto (resp. one to one), however the following lemma deal with σ ϕ : l 2 (τ) l 2 (τ). Lemma 2.3. The following statements are equivalent:. σ ϕ (l 2 (τ)) = l 2 (τ), 2. σ ϕ (l 2 (τ)) is a dense subset of l 2 (τ), 3. ϕ : τ τ is one to one. In addition the following statements are equivalent too: i. σ ϕ : l 2 (τ) K τ is one to one, ii. ϕ : τ τ is onto. Proof. (2) (3) Suppose ϕ : τ τ is not one to one, then there exists θ ψ with µ := ϕ(θ) = ϕ(ψ). There exists (y α ) l 2 (τ) with σ ϕ ((y α ) ) (δα θ) < 4, thusforallα < τ wehave y ϕ(α) δα θ < 4 in particular y ϕ(ψ) δψ θ < 4 and y ϕ(θ) δθ θ < 4, thus y µ < 4 and y µ < 4, which is a contradiction, therefore ϕ : τ τ is one to one. (3) () Suppose ϕ : τ τ is one to one, then by Theorem 2.2, σ ϕ (l 2 (τ)) l 2 (τ). For y = (y α ) l 2 (τ), define x = (x α ) in the following way: { yβ α = ϕ(β),β < τ, x α = 0 α τ \ϕ(τ), then x = y and x l 2 (τ), moreover σ ϕ (x) = y, which leads to σ ϕ (l 2 (τ)) = l 2 (τ). In order to complete the proof we should prove that (i) and (ii) are equivalent however by [, Lemma 4.], (ii) implies (i), so we should just prove that (i) implies (ii). (i) (ii) Suppose ϕ : τ τ is not onto and choose θ τ \ ϕ(τ). Then (δ θ α ),(0) are two distinct elements of l 2 (τ), however and σ ϕ : l 2 (τ) K τ is not one to one. σ ϕ ((δ θ α ) ) = σ ϕ ((0) ) = (0) Corollary 2.4. The following statements are equivalent:. ϕ : τ τ is bijective, 2. σ ϕ : l 2 (τ) l 2 (τ) is bijective, 3. σ ϕ : l 2 (τ) l 2 (τ) is an isomorphism, 4. σ ϕ : l 2 (τ) l 2 (τ) is an isometry. Proof. Using Lemma 2.3, () and (2) are equivalent. It s evident that (3) implies (2), moreover if ϕ : τ τ is bijective, then by Theorem 2.2 two linear maps σ ϕ : l 2 (τ) l 2 (τ) and its inverse σ ϕ : l 2 (τ) l 2 (τ) are continuous, hence () implies (3). () implies (4), is evident by (*) in Theorem 2.2. In order to complete the proof, we should just prove that (4) implies (). (4) () Suppose σ ϕ : l 2 (τ) l 2 (τ) is an isometry, then σ ϕ : l 2 (τ) l 2 (τ) is one to one and by Lemma 2.3, ϕ : τ τ is onto. Moreover, σ ϕ = since σ ϕ : l 2 (τ) l 2 (τ) is an isometry. By Lemma 2.2 we have = σ ϕ 2 = sup{card(ϕ (α)) : α τ}, thus for all α < τ we have card(ϕ (α)) and ϕ : τ τ is one to one.

4 4 F. AYATOLLAH ZADEH SHIRAZI, F. EBRAHIMIFAR Lemma 2.5. Let D = {z l 2 (τ) : σ ϕ (z) l 2 (τ)} (consider D with induced normed and topology of l 2 (τ)), then:. D is a subspace of l 2 (τ), 2. {θ < τ : (z α ) D z θ 0} = {α < τ : ϕ (α) is finite }. Proof. Since σ ϕ : l 2 (τ) K τ is linear, we have immediately (). 2) We have { (**) σ ϕ ((δα) θ (card(ϕ (θ))) ) = ϕ (θ)isfinite, + ϕ (θ)isinfinite. Thus if ϕ (θ) is finite we have σ ϕ ((δ θ α) ) l 2 (τ) and (δ θ α) D, which shows θ {β < τ : (z α ) D z β 0}. Therefore: {α < τ : ϕ (α) is finite } {θ < τ : (z α ) D z θ 0} Now for θ < τ suppose there exists (z α ) D with z θ 0. Using the fact that D is a subspace of l 2 (τ) we may suppose z θ =, now we have (since σ ϕ (z) l 2 (τ)): σ ϕ ((δ θ α) ) = σ ϕ ((z α δ θ α) ) σ ϕ (z) < +, by (**), ϕ (θ) is finite, which completes the proof of (2). Note 2.6. For H τ the closure of subspace generated by {(δ θ α ) : θ H} (in l 2 (τ)) is {(x α ) l 2 (τ) : α / H (x α = 0)}. Theorem 2.7. For D = {z l 2 (τ) : σ ϕ (z) l 2 (τ)} as in Lemma 2.5 and M := {α < τ : ϕ (α) is finite }, the following statemnts are equivalent:. σ ϕ D : D l 2 (τ) is continuous, 2. there exists finite n with card(ϕ (α)) n for all α M, 3. D = {(x α ) l 2 (τ) : θ / M x θ = 0}, 4. D is a closed subspace of l 2 (τ), Proof. () (2) Suppose σ ϕ D : D l 2 (τ) is continuous, consider θ < τ with finite ϕ (θ). By proofofitem (2) in Lemma 2.5, (δ θ α ) D and σ ϕ ((δ θ α ) ) = (card(ϕ (θ))), thus + > σ ϕ sup{ (card(ϕ (θ))) : θ M}. (2) () For n := sup{card(ϕ (α)) : α M} < + and x = (x α ) D we have σ ϕ (x) = Σ ((card(ϕ (α))) x α 2 ) Σ (n x α 2 ) = n x which α M α M shows continuity of σ ϕ D : D l 2 (τ). (3) (4) Note that for nonempty M, {(x α ) l 2 (τ) : θ / M x θ = 0} is just l 2 (M). (4) (3) Byproofand notationsin Lemma 2.5, we have{(δα θ) : θ M} D. Use Note 2.6 to complete the proof. (2) (3) For n := sup{card(ϕ (α)) : α M} < + and x = (x α ) l 2 (τ) with x α = 0 for all α / M we have σ ϕ (x) n x which shows x D and {(x α ) l 2 (τ) : θ / M x θ = 0} D. Using Lemma 2.5 we have D {(x α ) l 2 (τ) : θ / M x θ = 0}. (3) (2) If sup{(card(ϕ (α))) : α M} = +, then there exists a strictly increasingsequence{n k } k innandsequence{α k } k inm suchthatforallk

5 GENERALIZED SHIFTS & COMPACT OPERATORS 5 we have card(ϕ (α k )) = n k. Using a similar method described in Lemma 2.2, consider (x α ) D with: { x α := k α = α k,k, 0 otherwise. Then σ ϕ ((x α ) ) 2 n = Σ k k Σ 2 k = +, in particular σ ϕ((x α ) ) / l 2 (τ) k k which is in contradiction with (x α ) D and completes the proof. 2.. Compact generalized shift operators. For normed vector spaces X, Y we say the operator T : X Y is a compact operator, if T(B X (0,)) is a compact subset of Y, where B X (0,) = {x X : x < } [3]. Theorem 2.8. The generalized shift operator σ ϕ : l 2 (τ) l 2 (τ) is compact if and only if τ is finite. Proof. If τ is infinite, then by Theorem 2.2, {ϕ (α) : α < τ}\{ } is a partition of τ to its finite subsets, thus there exists a one to one sequence {α n } n in τ such that {ϕ (α n } n is a sequence of nonempty finite and disjoint subsets of τ. For all distinct n,m we have σ ϕ ((δ αn α ) ) σ ϕ ((δ αm α ) ) = 2 so {σ ϕ ( 2 (δαn α ) )} n (is a sequence in σ ϕ (B((0),))) without any converging subsequence. Therefore σ ϕ (B((0),)) is not compact and σ ϕ : l 2 (τ) l 2 (τ) is not a compact operator. On the other hand, if τ is finite, then every linear operator on l 2 (τ) is compact, hence σ ϕ : l 2 (τ) l 2 (τ) is a compact operator. References [] F. Ayatollah Zadeh Shirazi, D. Dikranjan, Set theoretical entropy: A tool to compute topological entropy, Proceedings ICTA 20, Islamabad, Pakistan, July 4 0, 20 (Cambridge Scientific Publishers), 202, 32. [2] F. Ayatollah Zadeh Shirazi, N. Karami Kabir, F. Heidari Ardi, A Note on shift theory, Mathematica Pannonica, Proceedings of ITES 2007, 9/2 (2008), [3] J. B. Conway, A course in abstract analysis, Graduate Studies in Mathematics, 4, American Mathematical Society, Providence, RI, 202. [4] G. B. Folland, Real analysis: Modern techniques and their applications, Second edition. Pure and Applied Mathematics (New York). A Wiley-Interscience Publication, 999. [5] K. G. Grosse Erdmann, A. Peris Manguillot, Linear chaos, Universitext, Springer, 20. [6] A. Giordano Bruno, Algebraic entropy of generalized shifts on direct products, Communications in Algebra, 38, no. (200), [7] M. Holz, K. Steffens, E. Weitz, Introduction to cardinal arithmetic, Birkhuser Advanced Texts: Basler Lehrbcher, Birkhuser Verlag, Basel, 999. [8] E. Kreyszig, Introductory functional analysis with applications, Wiley Classics Library. John Wiley & Sons, Inc., 989. [9] D. Ornstein, Bernoulli shifts with the same entropy are isomorphic, Advances in Math., 4 (970), [0] P. Walters, An introduction to ergodic theory, Graduate Texts in Mathematics, 79, Springer- Verlag, 982. Fatemah Ayatollah Zadeh Shirazi, Faculty of Mathematics, Statistics and Computer Science, College of Science, University of Tehran, Enghelab Ave., Tehran, Iran ( fatemah@khayam.ut.ac.ir) Fatemeh Ebrahimifar, Faculty of Mathematics, Statistics and Computer Science, College of Science, University of Tehran, Enghelab Ave., Tehran, Iran ( ebrahimifar64@ut.ac.ir)