On linear isometries on nonarchimedean power series spaces


 Erick Lane
 2 years ago
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1 On linear isometries on nonarchimedean power series spaces Wies law Śliwa and Agnieszka Ziemkowska Abstract. The nonarchimedean power series spaces A p (a, t) are the most known and important examples of nonarchimedean nuclear Fréchet spaces. We study when the spaces A p (a, t) and A q (b, s) are isometrically isomorphic. Next we determine all linear isometries on the space A p (a, t) and show that all these maps are surjective. 1 Introduction In this paper all linear spaces are over a nonarchimedean nontrivially valued field K which is complete under the metric induced by the valuation : K [0, ). For fundamentals of locally convex Hausdorff spaces (lcs) and normed spaces we refer to [2], [4] and [6]. Let Γ be the family of all nondecreasing unbounded sequences of positive real numbers. Let a = (a n ), b = (b n ) Γ. The power series spaces of finite type A 1 (a) and infinite type A (b) were studied in [1] and [7] [9]. In [7] it has been proved that A p (a) has the quasiequivalence property i.e. any two Schauder bases in A p (a) are quasiequivalent ([7], Corollary 6). The problem when A p (a) has a subspace (or quotient) isomorphic to A q (b) was studied in [8]. In particular, the spaces A p (a) and A q (b) are isomorphic if and only if p = q and the sequences a, b are equivalent i.e. 0 < inf n (a n /b n ) sup n (a n /b n ) < ([8], Corollary 6) Mathematics Subject Classification: 46S10, 47S10, 46A45. Key words: Nonarchimedean power series space, linear isometry, Schauder basis. The first author was supported by the National Center of Science, Poland, grant no. N N
2 For p (0, ] we denote by Λ p the family of all strictly increasing sequences t = (t k ) of real numbers such that lim k t k = ln p (if p =, then ln p := ). Let p (0, ], a = (a n ) Γ and t = (t k ) Λ p. Then the following linear space A p (a, t) = {(x n ) K : lim n x n e t ka n = 0 for all k N} with the base ( k ) of the norms (x n ) k = max n x n e t ka n, k N, is a Fréchet space with a Schauder basis. Clearly, A 1 (a) = A 1 (a, t) for a = (a n ) Γ, t = (t k ) = (ln k k+1 ), and A (b) = A (b, s) for b = (b n ) Γ, s = (s k ) = (ln k). Let q(p) = 1 for p (0, ) and q( ) =. It is not hard to show that for every p (0, ], a = (a n ) Γ and t = (t k ) Λ p the space A p (a, t) is isomorphic to A q(p) (b) for some b Γ. Thus we can consider the spaces A p (a, t) as power series spaces. In this paper we study linear isometries on power series spaces. First we show that the spaces A p (a, t) and A q (b, s), for p, q (0, ], t = (t k ) Λ p, s = (s k ) Λ q and a = (a n ), b = (b n ) Γ, are isometrically isomorphic if and only if there exist C, D R such that s k = Ct k + D and a k = Cb k for all k N, and for every k N there is ψ k K with ψ k = e (D/C)a k (Theorem 1). Let p (0, ], t = (t k ) Λ p and a = (a n ) Γ. Let (N s ) be a partition of N into nonempty finite subsets such that (1) a i = a j for all i, j N s, s N; (2) a i < a j for all i N s, j N s+1, s N. We prove that a linear map T : A p (a, t) A p (a, t) with T e j = t i,je i, j N, is an isometry if and only if (1) t i,j e (a j a i )t 1 when a i < a j ; (2) t i,j e (a j a i ) ln p when a i > a j (e := 0); (3) max (i,j) Ns N s t i,j = 1 and det[t i,j ] (i,j) Ns N s = 1 for s N; (Theorem 5 and Proposition 7). In particular, if the sequence (a n ) is strictly increasing, then a linear map T : A p (a, t) A p (a, t) with T e j = t i,je i, j N, is an isometry if and only if (1) t i,j e (a j a i )t 1 when i < j; (2) t i,j e (a j a i ) ln p when i > j; (3) t i,i = 1 for i N. Finally we show that every linear isometry on A p (a, t) is surjective (Corollary 10 and Theorem 12). Thus the family I p (a, t) of all linear isometries on A p (a, t) forms a group by composition of maps. 2 Preliminaries The linear span of a subset A of a linear space E is denoted by lin A. By a seminorm on a linear space E we mean a function p : E [0, ) such that p(αx) = α p(x) for all α K, x E and p(x + y) max{p(x), p(y)} for all x, y E. A seminorm p on E is a norm if {x E : p(x) = 0} = {0}. 2
3 If p is a seminorm on a linear space E and x, y E with p(x) p(y), then p(x + y) = max{p(x), p(y)}. The set of all continuous seminorms on a lcs E is denoted by P(E). A nondecreasing sequence (p k ) of continuous seminorms on a metrizable lcs E is a base in P(E) if for any p P(E) there are C > 0 and k N such that p Cp k. A complete metrizable lcs is called a Fréchet space. Let E and F be locally convex spaces. A map T : E F is called an isomorphism if it is linear, injective, surjective and the maps T, T 1 are continuous. If there exists an isomorphism T : E F, then we say that E is isomorphic to F. The family of all continuous linear maps from E to F we denote by L(E, F ). Let E and F be Fréchet spaces with fixed bases ( k ) and ( k ) in P(E) and P(F ), respectively. A map T : E F is an isometry if T x T y k = x y k for all x, y E and k N; clearly, a linear map T : E F is an isometry if and only if T x k = x k for all x E and k N. A linear map T : E F is a contraction if T x k x k for all x E and k N. A rotation on E is a surjective isometry T : E E with T (0) = 0. By [3], Corollary 1.7, we have the following Proposition A. Let m N. Equip the linear space K m with the maximum norm. Let T : K m K m be a linear map with T e j = m t i,je i for 1 j m. Then T is an isometry if and only if max i,j t i,j = 1 and det[t i,j ] = 1. A sequence (x n ) in a lcs E is a Schauder basis in E if each x E can be written uniquely as x = n=1 α nx n with (α n ) K, and the coefficient functionals f n : E K, x α n (n N) are continuous. The coordinate sequence (e n ) is an unconditional Schauder basis in A p (a, t). 3 Results First we show when the power series spaces A p (a, t) and A q (b, s) are isometrically isomorphic. Theorem 1. Let p, q (0, ], t = (t k ) Λ p, s = (s k ) Λ q and a = (a n ), b = (b n ) Γ. Then the spaces A p (a, t) and A q (b, s) are isometrically isomorphic if and only if (1) there exist C, D R such that s k = Ct k + D and a k = Cb k for all k N; 3
4 (2) for every k N there is ψ k K with ψ k = e (D/C)a k. In this case the linear map P : A p (a, t) A q (b, s), (x n ) (ψ n x n ) is an isometric isomorphism. Proof. Let T : A p (a, t) A q (b, s) be an isometric isomorphism and let T e j = t i,je i for j N. Then max i t i,j e s kb i = e t ka j for all j, k N; so max i t i,j e s kb i t k a j = 1 for j, k N. Let j, k N with k > 1. Then for some i N we have t i,j = e t ka j s k b i, t i,j e t k+1a j s k+1 b i and t i,j e t k 1a j s k 1 b i. Hence we get (s k+1 s k )b i (t k+1 t k )a j and (t k t k 1 )a j (s k s k 1 )b i ; so s k+1 s k t k+1 t k a j b i s k s k 1 t k t k 1. Thus the sequence ( s k+1 s k t k+1 t k ) is nonincreasing. Similarly we infer that the sequence ( t k+1 t k s k+1 s k ) is nonincreasing, since the map T 1 : A q (b, s) A p (a, t) is an isometric isomorphism, too. It follows that the sequence ( s k+1 s k t k+1 t k ) is constant, so there is C > 0 such that s k+1 s k t k+1 t k = C for all k N. Moreover, for every j N there is i N with a j /b i = C and for every i N there is j N with b i /a j = 1/C. Thus {a j : j N} = {Cb i : i N}. For l > 1 we have s l Ct l = s 1 Ct 1, since l 1 l 1 s l s 1 = (s k+1 s k ) = C (t k+1 t k ) = C(t l t 1 ). k=1 k=1 Put D = s 1 Ct 1, then s k = Ct k + D for k N. Let (j k ) N, (i k ) N be strictly increasing sequences such that {a jk : k N} = {a j : j N}, {b ik : k N} = {b i : i N} and a jk < a jk +1, b ik < b ik +1 for k N. Hence we get a jk = Cb ik for k N, since {a j : j N} = {Cb i : i N}. Put j 0 = i 0 = 0 and M r = {j N : j r 1 < j j r }, W r = {i N : i r 1 < i i r } for r N; clearly W r = {i N : Cb i = a jr }. Let r N and (φ j ) j Mr max φ j e t ka jr K with max j Mr φ j > 0. Then we have = max φ j e t ka j = φ j e j k = T ( φ j e j ) k = Thus φ j t i,j e i k = ( t i,j φ j )e i k = max t i,j φ j e skbi. i max t i,j φ j e skbi tkajr i 4 = max φ j.
5 and Let k > 1. For some i N we have t i,j φ j = max φ j e t ka jr s k b i, t i,j φ j max φ j e t k+1a jr s k+1 b i t i,j φ j max φ j e t k 1a jr s k 1 b i. Hence we get (s k+1 s k )b i (t k+1 t k )a jr and (t k t k 1 )a jr (s k s k 1 )b i ; so Cb i a jr and a jr Cb i. Thus a jr = Cb i, so i W r. It follows that max t i,j φ j e skbi tkajr i W r = max φ j. We have s k b i t k a jr = (Ct k + D)a jr /C t k a jr = (D/C)a jr for i W r ; so max t i,j φ j e (D/C)ajr i W r = max φ j. Thus e (D/C)a jr = γ r for some γ r K. Put ψ j = γ r for every j M r. Then ψ j = e (D/C)a j for j M r. Since max i Wr t i,j φ j ψ 1 j = max j Mr φ j, the linear map U : K Mr K Wr, (φ j ) j Mr ( t i,j ψ 1 j φ j ) i Wr is an isometry, so M r W r. We have shown that j r j r 1 i r i r 1 for every r N. Similarly we get i r i r 1 j r j r 1 for every r N, since T 1 is an isometric isomorphism. Thus j r j r 1 = i r i r 1 for every r N; so j r = i r for r N. It follows that a j = Cb j for j N. Now we assume that (1) and (2) hold. Then the linear map P : A p (a, t) A q (b, s), (x j ) (ψ j x j ) is an isometric isomorphism. Indeed, P is surjective since for any y = (y j ) A q (b, s) we have x = (ψ 1 j y j ) A p (a, t) and P x = y. For x A p (a, t) and k N we have P x k = max j ψ j x j e s kb j = max j x j e (D/C)a j+s k b j = max x j e t ka j = x k. j By obvious modifications of the proof of Theorem 1 we get the following two propositions. 5
6 Proposition 2. Let p (0, ], t Λ p and a = (a n ), b = (b n ) Γ. Then A p (b, t) contains a linear isometric copy of A p (a, t) if and only if a is a subsequence of b. If (n j ) N is a strictly increasing sequence with a j = b nj, j N, then the map T : A p (a, t) A p (b, t), (x j ) (y j ), where y j = x k if j = n k for some k N, and y j = 0 for all other j N, is a linear isometry. Proposition 3. Let p, q (0, ], t Λ p, s Λ q and a, b Γ. If there exist linear isometries T : A p (a, t) A q (b, s) and S : A q (b, s) A p (a, t), then A p (a, t) and A q (b, s) are isometrically isomorphic. Remark 4. Let p, q (0, ], t Λ p, s Λ q and a, b Γ. If P : A p (a, t) A q (b, s) is an isometric isomorphism, then every isometric isomorphism T : A p (a, t) A q (b, s) is of the form P S where S is an isometric automorphism of A p (a, t). Now we determine all linear isometries on the space A p (a, t). Recall that (N s ) is a partition of N into nonempty finite subsets such that (1) a i = a j for all i, j N s, s N; (2) a i < a j for all i N s, j N s+1, s N. Theorem 5. Let p (0, ], t = (t k ) Λ p and a = (a n ) Γ. Let T : A p (a, t) A p (a, t) be a continuous linear map and let T e j = t i,je i for j N. Then T is an isometry if and only if (1) t i,j e (a j a i )t 1 when a i < a j, and t i,j e (a j a i ) ln p when a i > a j ; (2) max (i,j) Ns N s t i,j = 1 and det[t i,j ] (i,j) Ns N s = 1 for all s N. Proof. ( ) For k, j N we have T e j k = max i t i,j e t ka i and e j k = e t ka j. Thus max i t i,j e t k(a i a j ) = 1 for all j, k N. Hence t i,j e t k(a j a i ) for all i, j, k N; so t i,j inf k e t k(a j a i ) for all i, j N. It follows (1); moreover t i,j 1 when a i = a j. Let s N, j s = min N s and (β j ) j Ns K with max j Ns β j > 0. Then we have T ( j N s β j e j k = j N s β j t i,j e i k = ( j N s β j t i,j )e i k = max i j N s β j t i,j e t ka i and j N s β j e j k = max j Ns β j e t ka j = (max j Ns β j )e t ka js for all k N. Thus max β j t i,j e t k(a i a js ) = max β j, k N; i j N s j N s hence max i Ns j N s β j t i,j max j Ns β j. 6
7 Let k > 1. For some i k N we have If a ik < a js, then j N s β j t ik,j e t k(a ik a js ) = max j N s β j. max β j β j t ik,j e t k 1(a ik a js ) > β j t ik,j e t k(a ik a js ) = max β j ; j N s j N s j N s j N s if a ik > a js, then max β j β j t ik,j e t k+1(a ik a js ) > β j t ik,j e t k(a ik a js ) = max β j. j N s j N s j N s j N s It follows that a ik = a js, so i k N s and j N s β j t ik,j = max j Ns β j. Thus the following linear map is an isometry S : K Ns K Ns, (β j ) j Ns ( j N s β j t i,j ) i Ns. By Proposition A we get max (i,j) Ns N s t i,j = 1 and det[t i,j ] (i,j) Ns N s = 1. ( ) Let x = (β j ) A p (a, t) and k N. Clearly, T x k = lim m T ( m j=1 β je j ) k and x k = lim m m j=1 β je j k. Thus to prove that T x k = x k it is enough to show that T ( m j=1 β je j ) k = m j=1 β je j k for all m N. Let m N. We have T ( m β j e j ) = j=1 m j=1 β j t i,j e i = m ( β j t i,j )e i, so L := T ( m j=1 β je j ) k = max i m j=1 β jt i,j e t ka i ; clearly P := m j=1 β je j k = max 1 j m β j e t ka j. We shall prove that L = P. By (1) and (2) we have t i,j e t k(a j a i ) for all i, j N. Hence for i N we get m j=1 j=1 β j t i,j e t ka i max 1 j m β j e t ka j = P ; so L P. If P = 0, then L = P. Assume that P > 0. Put j 0 = max{1 j m : β j e t ka j = P } and β j = 0 for j > m. Let q, s N with m N q, j 0 N s. Put W s = {N l : 1 l < s} and M s = {N l : s < l q}. Then β j e t ka j β j0 e t ka j0 for j W s, β j e t ka j < β j0 e t ka j0 for j M s and max j Ns β j = β j0 > 0. By (2) and Proposition A, the linear map S : K Ns K Ns, (x j ) j Ns ( j N s t i,j x j ) i Ns 7
8 is an isometry, so max i Ns j N s t i,j β j = max j Ns β j = β j0. i 0 N s, we have j N s t i0,jβ j = β j0 ; clearly a i0 = a j0. If j W s, then β j t i0,j β j0 e t k(a j0 a j ) e (a j a j0 ) ln p = β j0 e (a j a j0 )(ln p t k ) < β j0, so j W s β j t i0,j < β j0. If j M s, then β j t i0,j < β j0 e t k(a j0 a j ) e t 1(a j a j0 ) = β j0 e (a j a j0 )(t 1 t k ) β j0, so j M s β j t i0,j < β j0. Thus m β j t i0,j = β j t i0,j + β j t i0,j + β j t i0,j = β j0, j=1 j W s j N s j M s so m j=1 β jt i0,j e t ka i0 = β j0 e t ka j0 = P. Hence P L. Thus L = P. By the proof of Theorem 5 we get the following. Thus for some Corollary 6. Let p (0, ], t = (t k ) Λ p and a = (a n ) Γ. Let T L(A p (a, t)) and T e j = t i,je i for j N. Then T is a contraction if and only if t i,j e (a j a i )t 1 when a i a j and t i,j e (a j a i ) ln p when a i > a j. Proposition 7. Let p (0, ], t = (t k ) Λ p and a = (a n ) Γ. Let (t i,j ) K with (1) t i,j e (a j a i )t 1 when a i < a j, and t i,j e (a j a i ) ln p when a i > a j ; (2) max (i,j) Ns N s t i,j = 1 and det[t i,j ] (i,j) Ns N s = 1 for all s N. Then there exists a linear isometry T on A p (a, t) such that T e j = t i,je i, j N. Proof. Let j N and k N. For i N with a i > a j we have t i,j e t ka i e (a j a i ) ln p+t k a i = e a j ln p+a i (t k ln p) if p (0, ), and t i,j e t ka i = 0, if p =. Thus lim i t i,j e i k = 0 for k N; so lim i t i,j e i = 0. Therefore the series t i,je i is convergent in A p (a, t) to some element T e j. Let x = (x j ) A p (a, t). We shall prove that lim j x j T e j = 0 in A p (a, t). By (1) and (2) we have t i,j e t k(a j a i ) for all i, j, k N. Let k N and j N. Then x j e t k+1a j x k+1 ; moreover T e j k = max i t i,j e t ka i j, k N; so lim j x j T e j = 0. e t ka j. Hence x j T e j k e (t k t k+1 )a j x k+1 for Thus the series j=1 x jt e j is convergent in A p (a, t) to some T x for every x A p (a, t). Clearly T x = lim n T n x, where T n : A p (a, t) A p (a, t), T n x = n j=1 x jt e j. The linear operators T n, n N, are continuous, so using the BanachSteinhaus theorem we infer that the operator T : A p (a, t) A p (a, t), x T x is linear and continuous. By Theorem 5, T is an isometry. By Proposition 7 and the proof of Theorem 5 we get the following. 8
9 Corollary 8. Let p (0, ], t Λ p and a Γ. Then a linear map T : A p (a, t) A p (a, t) is an isometry if and only if T e j k = e j k for all j, k N. Finally we shall show that every linear isometry on the space A p (a, t) is a surjection. For p = it follows from Theorem 5 and our next proposition. For p (0, ) the proof is much more complicated. Let p (0, ], t = (t k ) Λ p and a = (a n ) Γ. Put W k = k N i, M k = i=k N i for k N and N k,m = N k N m for all k, m N. For every m N there is v(m) N with m N v(m). Proposition 9. Let p (0, ], t = (t k ) Λ p and a = (a n ) Γ. Let D L(A p (a, t)) with De j = d i,je i for j N. Assume that (1) d i,j e t 1(a j a i ) when a i < a j, and d i,j = 0 when a i > a j ; (2) max (i,j) Ns,s d i,j = 1 and det[d i,j ] (i,j) Ns,s = 1 for all s N. Then D is surjective. Proof. We have lin{de j : j W k } lin{e i : i W k } for k N, since De j = i W k d i,j e i for j N k, k N. By Theorem 5 the operator D is a linear isometry, so D(A p (a, t)) is a closed subspace of A p (a, t) and the sequence (De j ) j Wk is linearly independent for every k N. Thus lin{de j : j W k } = lin{e i : i W k }, k N; so D(A p (a, t)) lin{e i : i N}. It follows that D is surjective. Corollary 10. Let t = (t k ) Λ and a = (a n ) Γ. Every linear isometry on A (a, t) is surjective. Proposition 11. Let p (0, ], t = (t k ) Λ p and a = (a n ) Γ. Let S L(A p (a, t)) with Se j = s i,je i for j N. Assume that (1) s i,j = 0 when a i < a j, and s i,j e (a j a i ) ln p when a i > a j ; (2) max (i,j) Nk,k s i,j = 1 and det[s i,j ] (i,j) Nk,k = 1 for k N. Then S is surjective. Proof. For x = (x j ) A p (a, t) we have Sx = x j Se j = x j s i,j e i = j=1 j=1 j=1 ( s i,j x j )e i = ( j W v(i) s i,j x j )e i. Let y = (y i ) A p (a, t). By (2) and Proposition A, there exists (x i ) i N1 K with max i N1 x i = max i N1 y i such that j N 1 s i,j x j = y i for i N 1. Assume that for some l N with l > 1 we have chosen (x j ) j Ns K for 1 s < l. 9
10 By (2) and Proposition A, there exists (x j ) j Nl K with max x i = max y i s i,j x j i N l i N l j W l 1 such that j N l s i,j x j = y i j W l 1 s i,j x j for i N l. Thus by induction we get x = (x j ) K N such that j W l s i,j x j = y i for all i N l, l N and max x i = max y i, and max x i = max y i s i,j x j for l > 1. i N 1 i N 1 i N l i N l j W l 1 Let k N. Clearly, max i W1 x i e t ka i = max i W1 y i e t ka i. For l > 1, i N l, j W l 1 we have s i,j x j e t ka i e (a j a i ) ln p+t k a i x j = e (a j a i )(ln p t k ) x j e t ka j x j e t ka j. Thus by induction we get max i Wl x i e t ka i max i Wl y i e t ka i for all l N. It follows that x A p (a, t). We have Sx = ( s i,j x j )e i = i N l j W l l=1 l=1 i N l y i e i = y i e i = y. Thus S is a surjection. Theorem 12. Let p (0, ), t = (t k ) Λ p and a = (a n ) Γ. Every linear isometry T on A p (a, t) is surjective. Proof. Let k, m N. Denote by M k,m the family of all matrixes B = [β i,j ] (i,j) Nk,m with (β i,j ) K such that a) β i,j e (a j a i ) ln p for (i, j) N k,m, if k > m; b) β i,j e t 1(a j a i ) for (i, j) N k,m, if k < m; c) β i,j 1 for (i, j) N k,m and det[β i,j ] (i,j) Nk,m = 1, if k = m. By Proposition A, for every k N and B M k,k we have B 1 M k,k. Let T e j = t i,je i for j N. Put T k,m = [t i,j ] (i,j) Nk,m and I k,m = [δ i,j ] (i,j) Nk,m for all k, m N. We define matrixes D k,m, S k,m M k,m for k N and m = 1, 2, 3,.... Put D k,1 = I k,1 and S k,1 = T k,1 for k N; clearly D k,1, S k,1 M k,1 for k N. Assume that for some m N with m > 1 we have D k,j, S k,j M k,j for k N and 1 j < m. Let D 1,m = S 1 1,1T 1,m. It is easy to see that D 1,m M 1,m, since S 1 1,1 M 1,1 and T 1,m M 1,m. 10
11 Let C k,m = k 1 v=1 S k,vd v,m and D k,m = S 1 k,k [T k,m C k,m ] for k = 2, 3,..., m 1. Let 1 < k < m. Let [s i,n ] (i,n) Nk,v = S k,v and [d n,j ] (n,j) Nv,m = D v,m for 1 v < k. Put [c i,j ] (i,j) Nk,m = C k,m. Then k 1 c i,j = s i,n d n,j max s i,n d n,j n W k 1 v=1 n N v for (i, j) N k,m. For i N k, j N m and n W k 1 we have s i,n d n,j e (an a i) ln p+t 1 (a j a n) = e (an a i)(ln p t 1 )+t 1 (a j a i ) e t 1(a j a i ) ; hence C k,m M k,m. Since S 1 k,k M k,k and T k,m M k,m, we infer that D k,m M k,m for k = 2,..., m 1. Let D k,m = I k,m for k m; clearly D k,m M k,m. Let S k,m = I k,m for 1 k < m; then S k,m M k,m. Let C k,m = m 1 v=1 S k,vd v,m and S k,m = T k,m C k,m for k m. Let k m. Let [s i,n ] (i,n) Nk,v = S k,v and [d n,j ] (n,j) Nv,m = D v,m for 1 v < m. Put [c i,j ] (i,j) Nk,m = C k,m. Then m 1 c i,j = v=1 n N v s i,n d n,j max s i,n d n,j n W m 1 for (i, j) N k,m. For i N k, j N m and n W m 1 we have s i,n d n,j e (an a i) ln p+t 1 (a j a n) = e (an a j)(ln p t 1 )+ln p(a j a i ) < e (a j a i ) ln p ; hence C k,m M k,m. Since t i,j e (a j a i ) ln p for (i, j) N k,m, we get S k,m M k,m for k > m and c i,j < 1 for all (i, j) N m,m. Thus for some (ϕ σ ) σ S(Nm) {α K : α < 1} we have det S m,m = σ S(N m) sgn σ det(t m,m ) It follows that S m,m M m,m. i N m (t i,σ(i) c i,σ(i) ) = σ S(N m) By definition of D k,m and S k,m we get σ S(N m) sgn σ[( sgn σ ϕ σ = det(t m,m ) = 1. i N m t i,σ(i) ) ϕ σ ] = a) T k,1 = S k,1 = k v=1 S k,vd v,1 for k N; b) S 1,1 D 1,m = T 1,m for m 2 and S k,k D k,m = T k,m k 1 v=1 S k,vd v,m for 2 k < m, so T k,m = k v=1 S k,vd v,m for 1 k < m; c) S k,m D m,m = S k,m = T k,m m 1 v=1 S k,vd v,m for k m > 1, 11
12 so T k,m = m v=1 S k,vd v,m = k v=1 S k,vd v,m for k m > 1. Thus ( ) T k,m = k v=1 S k,vd v,m = v=1 S k,vd v,m for all k, m N. Let [s i,j ] (i,j) N N and [d i,j ] (i,j) N N be matrixes such that [s i,j ] (i,j) Nk,m = S k,m and [d i,j ] (i,j) Nk,m = D k,m for all k, m N. By Theorem 5 and Proposition 7, there exist linear isometries S and D on A p (a, t) such that Se j = s i,je i and De j = d i,je i for all j N; by Propositions 9 and 11, these isometries are surjective. Using ( ) we get t i,j = k v=1 n N v s i,n d n,j = s i,n d n,j v=1 n=1 n N v s i,n d n,j = for (i, j) N k,m and k, m N. Hence for j N we get SDe j = S( d n,j e n ) = d n,j ( s i,n e i ) = n=1 n=1 so T = SD. Thus T is surjective. ( s i,n d n,j )e i = n=1 t i,j e i = T e j ; Let p (0, ], t = (t k ) Λ p and a = (a n ) Γ. For every m N there is v(m) with m N v(m). Denote by D p (a, t), K p (a, t) and S p (a, t) the families of all linear isometries on A p (a, t) such that T e j = i W v(j) t i,j e i, T e j = i N v(j) t i,j e i and T e j = i M v(j) t i,j e i for j N, respectively. We have the following two propositions. Proposition 13. D p (a, t), K p (a, t) and S p (a, t) are subgroups of the group I p (a, t) of all linear isometries on A p (a, t). Moreover D (a, t) = I (a, t) and S (a, t) = K (a, t). For every T I p (a, t) there exist D D p (a, t) and S S p (a, t) such that T = S D. Proof. The last part of the proposition follows by the proof of Theorem 12. Clearly, I p (a, t) is a subgroup of the group of all automorphisms of A p (a, t); moreover D (a, t) = I (a, t) and S (a, t) = K (a, t). Let S, T S p (a, t). Let j N. We have ST e j = S( t i,j e i ) = t i,j ( s k,i e k ) = k=1 ( s k,i t i,j )e k. k=1 12
13 If a k < a j, then for every i N we have a k < a i or a i < a j ; so s k,i = 0 or t i,j = 0 for i N. Thus s k,it i,j = 0 for k N with a k < a j ; so ST S p (a, t). Let k N. For some x k = (x j,k ) A p (a, t) we have Sx k = e k. By the proof of Proposition 11 we have max{ x j,k e t 1a j : a j < a k } = 0, so x j,k = 0 for j N with a j < a k. Hence S 1 (e k ) = j M v(k) x j,k e j, so S 1 S p (a, t). We have shown that S p (a, t) is a subgroup of I p (a, t). Let D, T D p (a, t). Let j N. We have DT e j = k=1 ( d k,it i,j )e k. If a k > a j, then for every i N we have a k > a i or a i > a j ; so d k,i = 0 or t i,j = 0 for i N. Thus d k,it i,j = 0 for every k N with a k > a j, so DT D p (a, t). Let k N. Put F k = lin{e i : a i a k }. We know that D(F k ) = F k. Thus there exists x k = (x j,k ) F k such that Dx k = e k. Then x j,k = 0 for j N with a j > a k and D 1 (e k ) = x k = j W v(k) x j,k e j, so D 1 D p (a, t). Thus D p (a, t) is a subgroup of I p (a, t). Clearly, K p (a, t) = S p (a, t) D p (a, t), so K p (a, t) is subgroup of I p (a, t). Proposition 14. I p (a, t) I p (a, s) if and only if t 1 s 1. In particular, I p (a, t) = I p (a, s) if and only if t 1 = s 1. Proof. If t 1 s 1, then using Theorem 5 we get I p (a, t) I p (a, s). Assume that t 1 > s 1. Then lim j e (t 1 s 1 )(a j a 1 ) =, so there exists j 0 > 1 and β 0 K such that e s 1(a j0 a 1 ) < β 0 e t 1(a j0 a 1 ). Let T L(A p (a, t)) with T e j j N. By Theorem 5, we have T I p (a, t) and T / I p (a, s). = e j + β 0 δ j0,je 1 for In relation with Corollary 10 and Theorem 12 we give the following two examples and state one open problem. Let p (0, ], t = (t k ) Λ p and a = (a n ) Γ. For every isometry F on K the map T F : A p (a, t) A p (a, t), (x n ) (F x n ) is an isometry on A p (a, t). Example 1. Assume that the field K is not spherically complete or the residue class field of K is infinite. Then there exists an isometry on A p (a, t) which is not a surjection. Indeed, by [5], Theorem 2, there is an isometry F on K which is not surjective. Then the map T F is an isometry on A p (a, t) which is not a surjection. Problem. Assume that K is spherically complete with finite residue class. Does every isometry on A p (a, t) is surjective? 13
14 Example 2. On A p (a, t) there exists a nonlinear rotation. Indeed, put S K = {β K : β = 1} and let f : [0, ) S K be a function which is not constant on the set { α : α K with α > 0}. Then the map F : K K, F (x) = f( x )x is a nonlinear surjective isometry with F (0) = 0. In fact, let x, y K. If x = y, then F (x) F (y) = f( x )x f( y )y = f( x ) x y = x y. If x y, then F (x) = x y = F (y), so F (x) F (y) = max{ F (x), F (y) } = max{ x, y } = x y. If α S K, then F (αx) = αf (x), so F (x/f( x )) = (1/f( x ))f( x )x = x for every x K. Let α (K \ {0}) with f( α ) f(1), then F (α1) αf (1). Then T F is an nonlinear surjective isometry on A p (a, t) with T F (0) = 0. References [1] De GrandeDe Kimpe, N. Nonarchimedean Fréchet spaces generalizing spaces of analytic functions, Indag. Mathem., 44 (1982), [2] PerezGarcia, C. and Schikhof, W.H. Locally Convex Spaces over Non Archimedean Valued Fields, Cambridge studies in advanced mathematics, 119, Cambridge University Press, [3] Rooij, A.C.M. van Notes on padic Banach spaces, Report 7633, Department of Mathematics, University of Nijmegen, [4] Rooij, A.C.M. van Nonarchimedean functional analysis, Marcel Dekker, New York (1978). [5] Schikhof, W.H. Isometrical embeddings of ultrametric spaces into nonarchimedean valued fields, Indag. Mathem., 87 (1984), [6] Schikhof, W.H. Locally convex spaces over nonspherically complete valued fields, Bull. Soc. Math. Belg., 38 (1986), [7] Śliwa, W. On the quasi equivalence of orthogonal bases in nonarchimedean metrizable locally convex spaces, Bull. Belg. Math. Soc., 9 (2002),
15 [8] Śliwa, W. On relations between nonarchimedean power series spaces, Indag. Mathem., N.S., 17 (2006), [9] Śliwa, W. and Ziemkowska, A. On complemented subspaces of nonarchimedean power series spaces, Canad. J. Math., to appear. Authors Address: Faculty of Mathematics and Computer Science A. Mickiewicz University ul. Umultowska 87, Poznań, POLAND 15
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