Canad. Math. Bull. Vol. 51 (4), 2008 pp. 604–617
The Invariant Subspace Problem for Non-Archimedean Banach Spaces ´ Wiesław Sliwa Abstract. It is proved that every infinite-dimensional non-archimedean Banach space of countable type admits a linear continuous operator without a non-trivial closed invariant subspace. This solves a problem stated by A. C. M. van Rooij and W. H. Schikhof in 1992.
1 Introduction Let K be a field with a non-trivial complete non-archimedean valuation | · | : K → [0, ∞). Every infinite-dimensional (id) Banach space E of countable type over K is isomorphic to the Banach space c0 (K ) of all sequences in K converging to 0 (with the sup-norm), [8, Theorem 3.16(ii)]. Any closed subspace F of E is (1 + ǫ)-complemented in E for every ε > 0, i.e., for every ε > 0 there exists a linear continuous projection Pε from E onto F with kPε k ≤ 1 + ε [8, Theorem 3.16(v)]. If the valuation of K is discrete, then any closed subspace of c0 (K ) is 1-complemented [8, Corollaries 2.4, 4.7]. Note that for complex Banach spaces, any closed subspace of a complex Banach space E is complemented in E if and only if E is isomorphic to a complex Hilbert space [3]. Let T be a linear operator on a linear space E. A linear subspace M of E is a non-trivial invariant subspace of T if {0} = 6 M 6= E and T(M) ⊂ M. One of the most famous problems of the operator theory is the invariant subspace problem for complex Hilbert spaces. It asks whether every linear continuous operator on an infinite-dimensional separable complex Hilbert space has a non-trivial closed invariant subspace. This problem is still open. There exists a vast literature dedicated to the invariant subspace problem for various important classes of complex Banach spaces and linear continuous operators. P. Enflo [2] and C. J. Read [5–7] negatively solved the invariant subspace problem for complex Banach spaces. Read obtained a linear continuous operator on the complex Banach space l1 without a non-trivial closed invariant subspace. This l1 -example was simplified by A. M. Davie and can be found in Beauzamy’s book [1, Ch. XIV]. Developing this example, we shall construct a linear continuous operator T on some infinite-dimensional Banach space E of countable type over K with no nontrivial closed invariant subspace. Clearly, E is isomorphic to the Banach space c0 (K ). In particular, we solve the problem stated by A. C. M. van Rooij and W. H. Schikhof Received by the editors March 14, 2006. AMS subject classification: Primary: 47S10; secondary: 46S10, 47A15. Keywords: invariant subspaces, non-archimedean Banach spaces. c
Canadian Mathematical Society 2008.
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The Invariant Subspace Problem for Non-archimedean Banach Spaces
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in [9]: if K is algebraically closed and spherically complete, does every linear continuous operator T on a Banach space E over K have a non-trivial closed subspace? Note that the Banach space E of countable type is reflexive if K is not spherically complete [8, Corollary 4.18]. Meanwhile, for reflexive complex Banach spaces, it is not known if there exists a linear continuous operator without a non-trivial closed invariant subspace on an infinite-dimensional reflexive complex Banach space. If K is algebraically closed, then it is easy to see, as in the complex case, that every linear operator T on a finite-dimensional Banach space with dim E > 1 has a nontrivial invariant subspace. If K is not algebraically closed (for example if K = Q p ), we can take an element a ∈ (K˜ \ K ), where K˜ is the algebraic closure of K . Clearly, E = K (a) is a finitedimensional linear space over K with dim E > 1 and the linear operator T : E → E, Ty = ay, has no non-trivial invariant subspace [9]. If K is not spherically complete (for example if K = C p ), then (see [8, Theorem 4.49]) there exists a spherically complete valued field Kˆ that is an immediate extension of K , i.e., K is a subfield of Kˆ and no non-zero element of Kˆ is orthogonal to K . We can take an element b ∈ (Kˆ \ K ). Clearly, the closed linear span of the set B = {bk : k ≥ 0} in Kˆ is a Banach algebra E of countable type over K with dim E > 1 (we consider Kˆ as a Banach algebra over K ). In fact E is a subfield of Kˆ : any non-zero element y ∈ E is not orthogonal to K , so there exists z ∈ K with ky −Pzk < max{kyk, kzk}. Then kyk = kzk and k1 − yz−1 k < 1. Thus ∞ −1 n −1 ∈ E. The linear continuous operator zy −1 = n=0 (1 − yz ) ∈ E, so y T : E → E, Ty = by has no non-trivial closed invariant subspace. Indeed, let M be a closed invariant subspace of T. Then V = {y ∈ E : yM ⊂ M} is a closed linear subspace of E and B ⊂ E; so V = E. Thus M is an ideal of E. Hence M = {0} or M = E [9].
2 Preliminaries Let K be a field. A function | · | : K → [0, ∞) is a valuation if:
(i) ∀α ∈ K : |α| = 0 ⇔ α = 0; (ii) ∀α, β ∈ K : |αβ| = |α||β|; (iii) ∀α, β ∈ K : |α + β| ≤ |α| + |β|. • • • • •
A valuation | · | on K is
non-trivial if |α| > 1 for some α ∈ K; complete if metric d : K × K → [0, ∞), d(x, y) = |x − y| is complete on K; archimedean if the sequence |1|, |1 + 1|, |1 + 1 + 1|, . . . is unbounded in [0, ∞); non-archimedean if it is not archimedean; discrete if the set |K| = {|α| : α ∈ K \ {0} } is discrete in (0, ∞).
Any field with a complete archimedean valuation is topologically isomorphic to (R, | · |) or (C, | · |) [8, p. 4]. A valuation | · | on K is non-archimedean if and only if ∀α, β ∈ K : |α + β| ≤ max{|α|, |β|} [8, Theorem 1.1]. A field K with a non-trivial complete non-archimedean valuation | · | is called non-archimedean. The field Q p of p-adic numbers is non-archimedean for any
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prime number p. Let K be a non-archimedean field. We say that K is spherically complete if any decreasing sequence of closed balls in K has non-empty intersection. An element y ∈ K is orthogonal to a subfield L if |y − z| = max{|y|, |z|} for any z ∈ L. By a norm on a linear space E over K we mean a function k · k : E → [0, ∞) such that (i) ∀y ∈ E : kyk = 0 ⇔ y = 0; (ii) ∀α ∈ K , x ∈ E : kαxk = |αkxk; (iii) ∀x, y ∈ E : kx + yk ≤ max{kxk, kyk}.
If k · k is a norm on a linear space E over K and x, y ∈ E with kxk = 6 kyk, then kx + yk = max{kxk, kyk}. For fundamentals of non-archimedean normed spaces we refer to [4, 8, 10]. A normed space E = (E, k · k) is of countable type if E contains a linearly dense countable set. We say that the normed spaces E and F over K are isomorphic if there −1 is a linear bijective map T : E → F such that the maps T and P∞T are continuous. A Banach space is a complete normed space. A series n=1 xn in a Banach space is convergent if and only if lim xn = 0. Every n-dimensional normed space E over K is isomorphic to the Banach space K n (with the sup-norm) and any linear operator on E is continuous. The closed graph theorem, the open mapping theorem and the Banach–Steinhaus theorem [4, Theorems 2.49, 2.73, 3.37] hold for continuous linear operators between such Banach spaces. ∞ Let E = (E, k · k) be a normed space. A sequence P∞ (y n )n=0 in E is a Schauder basis in E if each y ∈ E can be written uniquely as y = n=0 αn y n with (αn )∞ n=0 ⊂ K and the coefficient functionals gn : E → K , y → αn (n ∈ N0 ) are continuous; N0 = N ∪ {0}. A linearly dense sequence (y n )∞ n=0 in E is an orthonormal basis in E if n
X
ci y i = max |ci |
i=0
0≤i≤n
for all n ∈ N and c1 , . . . , cn ∈ K . Any orthonormal basis in E is a Schauder basis in E and every Banach space with an orthonormal basis (y n )∞ n=0 is isometrically isomorphic to c0 (K ) [8, p. 171]. We denote by lin A the linear span of a subset A of a linear space E.
3 Results Put d0 = 2. Let α ∈ K with |α| ≥ 8 and (dn ) ⊂ N with dn ≥ |α|2dn−1 for n ∈ N. It is easy to see that dn−1 ≥ 2n and dn ≥ |α|2n dn−1 ≥ 4n2 |α|dn−1 for n ∈ N. Hence we have dn ≥ 23n+1 , n ∈ N; so dn ≥ |α|2dn−1 ≥ |α|2n+2 , n ∈√ N. Since the function √ n+1 8/ d f (x) = x8/ x is decreasing in the interval (e2 , ∞), we get dn n ≤ |α|16(n+1)/|α| < √ |α|1/(n+1) for n ∈ N. Thus dn2n+2 < |α| dn /4 < dn+1 for n ∈ N. Put v0 = 0, an = d2n−1 , bn √= d2n and vn = (n − 1)(an √+ bn ) for n ∈ N. Then an /4 bn /4 4n < |α|4n ≤ an , a4n < bn , and b4n < an+1 for every n < |α| n < |α|
The Invariant Subspace Problem for Non-archimedean Banach Spaces
607 √
n ∈ N. Hence we get 8nan < bn , 8nbn < an+1 , 4(vn−1 + 1) < an , |α|a2n < |α| an /4 , √ bn /4 2 , and n2 a2n < bn for every n ∈ N. |α|bn < |α| For a, b ∈ Z we denote the set {k ∈ Z : a < k ≤ b} by (a, b]; similarly we define [a, b), [a, b] and (a, b). For nonempty sets A, B ⊂ N we write A < B if 1 + max A = min B. For n, r ∈ N with n > r we put: Jn,r = ((r − 1)an + vn−r , ran ), Ln,r = ((n − 1)an + (r − 1)bn , r(an + bn )),
In,r = [ran , ran + vn−r−1 ], Kn,r = [r(an + bn ), (n − 1)an + rbn ].
These sets are non-empty and Jn,r < In,r < Jn,r+1 , Ln,r < Kn,r < Ln,r+1 for n, r ∈ N with n > r + 1 and Jn,n−1 < In,n−1 , Ln,n−1 < Kn,n−1 for n ≥ 2. Moreover Xn := Sn−1 Sn−1 ∪ Kn,r ) = ((n − 1)an , vn ] for r=1 ( Jn,r ∪ In,r ) = (vn−1 , (n − 1)an ], Y n := r=1 (Ln,r S∞ n ∈ N with n ≥ 2; so Xn < Y n < Xn+1 for n ≥ 2 and n=2 (Xn ∪ Y n ) = N. Let (αn ), (βn ) ⊂ K with |αn | ∈ (an |α|−1 , an ] and |βn | ∈ (bn |α|−1 , bn ] for n ∈ N. Clearly 1 < |αn | < |βn | < |αn+1 | for n ∈ N. Let hai = k if k ≤ a < k + 1 and k ∈ Z. It is easy to see that hai − hbi ≤ ha − bi + 1 for a, b ∈ R. Let F = K [x] and Fn = { f ∈ F : deg( f ) ≤ n} for n ∈ N0 . Then F is a linear algebra over K and Fn is a linear subspace of F for every n ∈ N0 , where N0 = N ∪ {0}. Put xi if i = 0, √ h[(2r−1)an−2i]/ 4an i i x if i ∈ Jn,r and n, r ∈ N with n > r, α fi = αn−r (xi − xi−an ) if i ∈ In,r and n, r ∈ N with n > r, √ h[(2r−1)bn −2i]/ 4bn i i x if i ∈ Ln,r and n, r ∈ N with n > r, α xi − β xi−bn if i ∈ Kn,r and n, r ∈ N with n > r. n
Obviously, lin{ fi : 0 ≤ iP ≤ n} = Fn for n ∈ N0 . Thus ( fi )∞ i=0 is a linear base in F. m For f ∈ F of the form f = i=0 ci fi , we put k f k = max0≤i≤m |ci |; k · k is a norm on F. Clearly, ( fi )∞ basis in (F, k · k). i=0√is an orthonormal √ √ Put Am = |α| am and Bm = |α| am + bm for m ∈ N. We have the following. Lemma 1 (i) Let m ∈ N with m ≥ 2. Then max0≤i≤(m−1)am kxi k ≤ Am and max0≤i≤vm kxi k ≤ Bm . (ii) Let n, r ∈ N with n > r. Then kxi − xi−ran k ≤ |α−1 n−r | for i ∈ In,r and kxi − βnr xi−rbn k ≤ |βn |r−1 for i ∈ Kn,r . Sm Proof For m ∈ N we S have [0, mam+1 ] = [0, vm ] ∪ r=1 ( Jm+1,r ∪ Im+1,r ) and [0, vm+1 ] = [0, mam+1 ] ∪ m r=1 (Lm+1,r ∪ Km+1,r ). √ Clearly kx0 k = 1. Let n,√r ∈ N with n > r. It is easy to check that kxi k < |α| an for i ∈ Jn,r and kxi k < |α| bn for i ∈ Ln,r . Let i ∈ In,r . For j ∈ [0, r) we have i − jan ∈ In,r− j , so i− jan − xi−( j+1)an . α−1 n−r+ j f i− jan = x
´ W. Sliwa
608 Hence
Pr−1 j=0
i i−ran α−1 , so n−r+ j f i− jan = x − x −1 kxi − xi−ran k ≤ max |α−1 n−r+ j | = |αn−r | < 1. 0≤ j 2, then√j ∈ Jn−1,1 ; so f j = αh[an−1 −2 j]/ j
−h[an−1 −2(vn−2 +1)]/
4an−1 i j
x . Hence
√
4an−1 i
−h an−1 /4i
kx k = |α| ≤ |α| . If 1 < r < n−1, then j ∈ In,r−1 and using Lemma 1 we get kx√j −xvn−r−1 +1 k ≤ −1 |αn−r+1 |. Moreover we have fvn−r−1 +1 = αh[an−r −2(vn−r−1 +1)]/ because vn−r−1 + 1 ∈ Jn−r,1 . Hence
4an−r i vn−r−1 +1
x
,
√ √ kxvn−r−1 +1 k = |α|−h[an−r −2(vn−r−1 +1)]/ 4an−r i ≤ |α|−h an−r /4i . √
an−r /4i Thus kx j k ≤ max{kx j − xvn−r−1 +1 k, kxvn−r−1+1 k} ≤ |α|−h . √ h[(2n−3)an −2 j]/ 4an i j x . Hence kx j k = If r = n − 1, then j ∈ J ; so f = α n,n−1 j √ √ |α|−h[an −2]/ 4an i ≤ |α|−h an /4i . √ It follows that kT fi k ≤ |αn−r | max{kxi+1 k, kx j k} ≤ |αn−r ||α|−h an−r /4i ≤ a−1 n−r .
Case 3:
√ 4bn i i
i ∈ Ln,r . Then fi = αh[(2r−1)bn−2i]/
x.
√
3.1 i < r(an + bn ) − 1. Then i + 1 ∈ Ln,r , so fi+1 = αh[(2r−1)bn−2(i+1)]/ 4bn i xi+1 . Thus √ √ h[(2r−1)bn−2i]/ 4bn i−h[(2r−1)bn−2(i+1)]/ 4bn i f . Hence kT fi k ≤ |α|. T fi = α √ i+1 3.2 i = r(an + bn ) − 1. Then T fi = αh[−bn −2ran+2]/ 4bn i xr(an +bn ) . Put j = r(an + bn ). By Lemma 1 we have kx j − βnr x j−rbn k ≤ |βn |r−1 . In Case 1 we have shown j k ≤ max{kx j −√ βnr xran k, |βnr |kxran k} ≤ bn−1 that kxran k ≤ 1. Hence kx n . Thus √ h[−bn −2ran +2]/ 4bn i n−1 − bn /4 n−1 −1 kT fi k ≤ |α| bn ≤ |α| bn ≤ bn . Case 4:
i ∈ Kn,r . Then fi = xi − βn xi−bn and T fi = xi+1 − βn xi+1−bn .
4.1 i < (n − 1)an + rbn . Then i + 1 ∈ Kn,r , so fi+1 = xi+1 − βn xi+1−bn = T fi . Hence kT fi k = 1. 4.2 i = (n − 1)an + rbn . Put j = i + 1 − bn . Then j ∈ Ln,r , so √
f j = αh[(2r−1)bn−2 j]/ √
4bn i j
x. √
Hence kx j k = |α|−h[(2r+1)bn−2(i+1)]/ 4bn i ≤ |α|−h bn /4i . √ If r < n − 1, then i +√1 ∈ Ln,r+1 ; so fi+1 = αh[(2r+1)bn−2(i+1)]/ 4bn i xi+1 . Hence kxi+1 k = kx j k ≤ |α|−h bn /4i . √ 4an+1 i i+1 If r = n − 1, then i + 1 √∈ Jn+1,1 ; so fi+1 = αh[an+1−2(i+1)]/ x . Hence √ √ −h an+1 /4i −h bn /4i i+1 −h[an+1 −2(i+1)]/ 4an+1 i ≤ |α| ≤ |α| . Thus we have kx k = |α| √ kT fi k ≤ max{kxi+1 k, |βn |kx j k} ≤ |βn ||α|−h bn /4i ≤ |α|. From now on, by T we will denote the linear continuous operator on E such that T f = x f for all f ∈ F; clearly kTk ≤ |α|. By the proof of Lemma 2 we get the following. Remark 3. If n, r ∈ N with n > r, then kT fran−1 k ≤ an−1 , kT fran+vn−r−1 k ≤ a−1 n−r , and kT fr(an+bn )−1 k ≤ b−1 n . S∞ Let m ∈ N with m > 2. Put Sm = n=m+1 In,n−m . Let Qm : F → F(m−1)am be a
´ W. Sliwa
610 linear operator such that fi Qm fi = −αm xi−(n−m)an 0
if i ∈ [0, (m − 1)am ], if i ∈ In.n−m and n > m, if i ∈ (N \ Sm ) with i > (m − 1)am .
i−(n−m)an k≤ Clearly, kQ √ m fi k = 1 for 0 ≤ i ≤ (m − 1)am , and kQm fi k = |α√m k|x am Bm−1 < Am for i ∈ In,n−m , n > m. Thus supi∈N0 kQm fi k < Am ; so the linear √ operator Qm : (F, k · k) → (F, k · k) is continuous and kQm k < Am . From now on, by Qm we will denote its continuous extension on (E, k · k). We have the following lemma.
Lemma 4
Let m ∈ N with m > 2 and let s ∈ Km,1 . Then kT s − T s Qm k ≤ |α|.
Proof It is enough to show that kT s fi − T s Qm fi k ≤ |α| for every i ∈ N0 . If i ∈ [0, (m − 1)am ], then T s fi − T s Qm fi = 0. Let i > (m − 1)am . For some n, r ∈ N with n > r we have i ∈ Jn,r ∪ In,r ∪ Ln,r ∪ Kn,r . If i ∈ Jn,r ∪ Ln,r ∪ Kn,r , then i 6∈ Sm ; so Qm fi = 0 and T s fi − T s Qm fi = T s fi . Consider four cases. √
Case 1: i ∈ Jn,r . Then T s fi = αh[(2r−1)an−2i]/ 4an i xi+s . We have m < n, since √ (m − 1)am < i < (n − 1)an . Thus s < 2bm < an . √
1.1 i < ran − s. Then i √+ s ∈ Jn,r , so fi+s = √αh[(2r−1)an−2(i+s)]/ 4an i xi+s . Hence T s fi = αh[(2r−1)an−2i]/ 4an i−h[(2r−1)an−2(i+s)]/ 4an i fi+s. Thus kT s f√i k ≤ |α|. 1.2 i ≥ ran − s. Then T i+s−ran+1 fran−1 = αh[(2r−1)an−2(ran−1)]/ 4an i xi+s , since ran − 1 ∈ Jn,r . Thus √ √ 4an i−h[−an +2]/ 4an i
kT s fi k = |α|h[(2r−1)an−2i]/
kT i+s−ran+1 fran−1 k
≤ |α|kTki+s−ran kT fran−1 k. By Remark 3 we have kT fran−1 k ≤ a−1 n . It follows that s −1 2bm −1 am+1 ≤ 1. kT s fi k ≤ |α|1+i+s−ran a−1 n ≤ |α| an ≤ |α|
Case 2: i ∈ In,r . Then fi = αn−r (xi −xi−an ), so T s fi = αn−r (xi+s −xi+s−an ). We have n > m, since (m − 1)am < i ≤ ran + vn−r−1 ≤ (n − 1)an . Thus 4(i + s) < (4r + 1)an . 2.1 r = n − m. Then i = ran + l for some l ∈ [0, vm−1√] and T s Qm fi = −αm xl+s . We have i + s ∈ Jn,r+1 , so fi+s = αh[(2r+1)an−2(i+s)]/ 4an i xi+s . Hence kxi+sk = √ √ −h[(2r+1)an−2(i+s)]/ 4an i |α| ≤ |α|−h an /4i ≤ |α−1 n |. Using Lemma 1 we get −1 kx(r−1)an+l+s − xl+s k ≤ |α−1 n−r+1 | = |αm+1 |.
Thus kT s fi − T s Qm fi k = |αm k|xi+s − (xi−an +s − xl+s )k < 1. 2.2 r 6= n − m. Then i 6∈ Sm , so Qm fi = 0 and T s fi − T s Qm fi = T s fi .
The Invariant Subspace Problem for Non-archimedean Banach Spaces 2.2a
r > n − m. Then√for j = i + s we have j − an ∈ Jn,r , so f√j−an = n −2( j−an )]/ 4an i j−an αh[(2r−1)a . Thus kx j−an k = |α|−h[(2r+1)an−2 j]/ 4an i ≤ x √ −h an /4i . |α| √ If r + 1 < n, then j ∈ J ; so f j = αh[(2r+1)an−2 j]/ 4an i x j . Hence √ n,r+1 kx j k = kx j−an k ≤ |α|−h an /4i . √ If r + 1 =√n, then j ∈ L√n,1 ; so f j = αh[bn −2 j]/ 4bn i x j . Hence kx j k = |α|−h[bn −2 j]/ 4bn i ≤ |α|−h bn /4i . It follows that kT s fi k ≤ |αn−r | max{kx j k, kx j−an k} ≤ |αn kα|−h
2.2b
611
√ an /4i
< 1.
r < n − m. Then i = ran + l for some l ∈ [0, vn−r−1]. If l + s ≤ vn−r−1 , then i + s ∈ In,r ; so fi+s = αn−r (xi+s − xi+s−an ) = T s fi . Thus kT s fi k = 1. If l + s > vn−r−1 , then we have
T s fi = αn−r (xi+s − xi+s−an ) = T s+l−vn−r−1 [αn−r (xran+vn−r−1 − x(r−1)an+vn−r−1 )] = T s+l−vn−r−1 −1 T fran+vn−r−1 . s By Remark 3 we obtain kT fran+vn−r−1 k ≤ a−1 n−r . Thus we get kT f i k ≤ s+l−vn−r−1 −1 −1 s −1 2bm −1 kTk an−r ≤ |α| an−r ≤ |α| am+1 ≤ 1. √
Case 3: i ∈ Ln,r . Then T s fi = αh[(2r−1)bn−2i]/ 4bn i xi+s . Put j = i + s. We have n ≥ m, since vm−1 < am < i < r(an + bn ) ≤ vn . Thus 4 j < an+1 . √
3.1 n = m and j > vn . Then j ∈ Jn+1,1 ; so f j = αh[an+1−2 j]/√4an+1 i x j and √ √ kx j k√= |α|−h[an+1 −2 j]/ 4an+1 i ≤ |α|−h an+1 /4i . Thus kT s fi k ≤ |α|h bn /2i kx j k ≤ √ |α|h bn /2i−h an+1/4i ≤ 1. 3.2 n = m and j ≤ vn . Then j > nan + rbn and r < n − 1. √
n −2 j]/ 4bn i j If j < (r + 1)(an + bn ), then j ∈√Ln,r+1 ; so f j = αh[(2r+1)b x. √ s h[(2r−1)bn −2i]/ 4bn i−h[(2r+1)bn−2 j]/ 4bn i Thus kT √ fi k = |α| ≤ |α|, since s − bn < bn . 3.2b If (r + 1)(an + bn ) ≤ j ≤ (n − 1)an + (r + 1)bn , then using Lemma 1 we get kx j − βnr+1 x j−(r+1)bn k ≤ |βnr | ≤ √brn and kx j−(r+1)bn k ≤ An < b4n . Thus 4n bn /4 kx j k ≤ max{brn , br+5 . Moreover, we have n } ≤ bn < |α|
3.2a
√ 4bn i
kT s fi k = |α|h[(2r−1)bn−2i]/ 3.2c
kx j k ≤ |α|h−
√ bn /4i
kx j k,
since i = j − s ≥ [r − (1/4)]bn . It follows that kT s fi k ≤ 1. If j > (n − 1)a + (r + 1)bn , then r < n − 2 and j ∈ Ln,r+2 ; so f j = √n αh[(2r+3)bn−2 j]/ 4bn i x j . Thus √ √ 4bn i−h[(2r+3)bn−2 j]/ 4bn i
kT s fi k = |α|h[(2r−1)bn−2i]/
≤ 1.
√
h[(2r−1)bn −2 j]/ 4bn i j 3.3 n > m and j < r(an + bn ). √Then j ∈ Ln,r , so f √ x . Thus j = α s h[(2r−1)bn −2i]/ 4bn i−h[(2r−1)bn−2 j]/ 4bn i kT fi k = |α| ≤ |α|, since s < 2bm < √ bn .
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612
3.4 n > m and j ≥ r(a√n + bn ). Put k = r(an + bn ). Clearly k − 1 ∈ Ln,r , so√fk−1 = αh[(2r−1)bn−2(k−1)]/ 4bn i xk−1 . Hence√T j−k T fk−1 = αh[(2r−1)bn−2(k−1)]/ 4bn i x j . Thus kx j k ≤ |α|−h[(2r−1)bn−2(k−1)]/ 4bn i kTks−1 kT fk−1k. Using Remark 3 we obtain kT fk−1 k ≤ b−1 n . Thus √ √ 4bn i−h[(2r−1)bn−2(k−1)]/ 4bn i
kT s fi k ≤ |α|h[(2r−1)bn−2i]/
|α|s−1 b−1 n
≤ |α|s bn−1 < |α|2bm b−1 m+1 < 1, √ since k − i − 1 < s < 2bm < bn .
Case 4: i ∈ Kn,r . Then fi = xi − βn xi−bn and T s fi = xi+s − βn xi+s−bn . We have n ≥ m, since (m − 1)am < i ≤ (n − 1)an + rbn ≤ vn . Put j = i + s. 4.1 n = m and j ≤ (n − 1)an + (r + 1)bn . Then j ≥ (r + 1)(an + bn ), so r + 1 ≤ n − 1 and j ∈ Kn,r+1 . Thus f j = x j − βn x j−bn = T s fi , so kT s fi k = 1. 4.2 n = m and j > (n − 1)an + (r + 1)bn . Then 4 j < (4r + 5)bn√< an+1 . If r < n − 2, √then j ∈ Ln,r+2 and f j = αh[(2r+3)bn−2 j]/ 4bn i x j ; so kx j k = |α|−h[(2r+3)bn−2 j]/ 4bn i ≤ 1. √ If r ≥ n −√2, then j ∈ Jn+1,1 and f j = αh[an+1−2 j]/ 4an+1 i x j ; so kx j k = |α|−h[an+1 −2 j]/ 4an+1 i ≤ 1. √ If r < n − 1, then j − bn ∈ L√n,r+1 and f j−b√ = αh[(2r+1)bn−2( j−bn )]/ 4bn i x j−bn ; n so kx j−bn k = |α|−h[(2r+3)bn−2 j]/ 4bn i ≤ |α|−h bn /4i ≤ |βn−1 |. √ If r = n − 1, then j − bn ∈ √Jn+1,1 and f j−bn = αh[an+1 −2( j−bn )]/ 4an+1i x j−bn ; so √ kx j−bn k = |α|−h[an+1 −2( j−bn )]/ 4an+1 i ≤ |α|−h an+1 /4i ≤ |α−1 n+1 |. It follows that kT s fi k ≤ max{kx j k, |βn k|x j−bn k} ≤ 1. 4.3 n > m and j ≤ (n − 1)an + rbn . Then j ∈ Kn,r and f j = x j − βn x j−bn = T s fi ; so kT s fi k = 1. 4.4 n > m and j > (n − 1)an + rbn . Then 4 j < (4r + 1)bn < an+1 . √ If r < n − 1, √then j ∈ Ln,r+1 and f j = αh[(2r+1)bn−2 j]/ 4bn i x j ; so kx j k = |α|−h[(2r+1)bn−2 j]/ 4bn i ≤ 1. √ If r = n −√1, then j ∈ Jn+1,1 and f j = αh[an+1−2 j]/ 4an+1 i x j ; so kx j k = |α|−h[an+1 −2 j]/ 4an+1 i ≤ 1. √ h[(2r−1)bn−2( j−bn )]/ 4bn i j−bn ; Moreover we have j − bn ∈ L√ x n,r and f j−bn =√α so kx j−bn k = |α|−h[(2r+1)bn−2 j]/ 4bn i ≤ |α|−h bn /4i ≤ |βn−1 |. It follows that kT s fi k ≤ max{kx j k, |βn k|x j−bn k} ≤ 1. Pm For f ∈ F of the form f = i=0 ci xi we put | f | = max0≤i≤m |ci |. The functional | · | : F → [0, ∞), f → | f | is a multiplicative norm on F [8, p. 7]. It is easy to check that for m ∈ N with m√> 2 and y ∈ F(m−1)am we have kyk ≤ Am |y| and |y| ≤ max0≤i≤(m−1)am | fi |kyk ≤ Am kyk. For n ∈ N0 we denote by Pn the linear projection from F onto Fn such that Pn (xi ) = 0 for i > n. We have x(Pn v) = Pn+1 (xv) for n ∈ N and v ∈ F. We need two more lemmas to prove our theorem. Lemma 5 Let e ∈ E with e 6= 0 and k ∈ N with k > 2. Then there exists m ∈ N with m > k such that |P(m−k)am (Qm e)| ≥ a−1 m .
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613
Proof Suppose by contradiction that for every m ∈ N with m > k we have |P(m−k)am (Qm e)| < a−1 m .
(3.1)
P∞ For some (e j ) ∈ c0 (K ) we have e = Then kek = max j∈N0 |e j | > 0. Put j=0 e j f j . P Pn n cn = (n − 1)an for n ∈ N. For n ∈ N we have cj=0 e j f j = cj=0 y n, j x j for some P∞ Pvn−1 cn (y n, j ) j=0 ⊂ K . For n ∈ N with n > 2 we obtain Qn ( j=cn +1 e j f j ) = i=0 zn,i xi , where (3.2)
zn,i = −αn
∞ X
ei+(m−n)am .
m=n+1
So we get
(3.3)
Qn e =
cn X
j
y n, j x +
vn−1 X
zn, j x j .
j=0
j=0
From (3.1) and (3.3) we obtain for m ∈ N with m > k (3.4)
max
j∈(vm−1 ,(m−k)am ]
|y m, j | < a−1 m .
Let m > n > k and Mm,n = ((m − n)am + vn−2 , (m − n)am + vn−1 ]. Clearly, Mm,n ⊂ S Sm−1 i i−am ) for i ∈ Im,s , s ∈ [1, m) [am , cm ] = m−1 s=1 Im,s ∪ √ s=2 Jm,s and f i = αm−s (x − x h[(2s−1)am−2i]/ 4am i i and fi = α x for i ∈ Jm,s , s ∈ [1, m). If j ∈ Mm,n , then j ∈ Im,m−n ; if i ∈ [am , cm ] and i − am ∈ Mm,n , then i ∈ Jm,m−n+2 . Thus y m, j = αn e j for j ∈ Mm,n . Clearly, Mm,n ⊂ (vm−1 , (m − k)am ]. Using (3.2) and (3.4) we obtain for n > k (3.5) max |zn, j | ≤ max max |αn ke j+(m−n)am | = max max |y m, j | ≤ a−1 n+1 . vn−2 < j≤vn−1
m>n j∈Mm,n
m>n vn−2 < j≤vn−1
From (3.3) and (3.1) we get for n > k (3.6)
max |y n, j + zn, j | ≤ |Pan (Qn e)| < an−1 .
j∈[0,vn−1 ]
Hence, by (3.5), we have for n > k (3.7)
max
j∈(vn−2 ,vn−1 ]
|y n, j | < a−1 n .
Pcn P cn y n, j x j . Clearly, (vn−1 , cn ] = Let n > k. Put Mn = j=vn−1 +1 e j f j − Sn−1 Sn−1 Sn−1 j=vn−1 +1 ∈ In,1 , then vn−1 f i = 0; if i P s=1 ( Jn,s ∪ In,s ). If i ∈ s=1 Jn,s ∪ s=2 In,s , then PP vn−1 vn−1 j y x − Pvn−1 ( fi ) ∈ Fvn−2 . Thus Pvn−1 (Mn ) ∈ Fvn−2 ; but Mn = n, j j=0 e j f j ∈ j=0
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614
Pvn−2 Pvn−1 Pvn−1 y xj = e f − j=v Fvn−1 , so Mn ∈ Fvn−2 . Hence j=v j=0 t n, j f j for n−2 +1 n, j n−2 +1 j j vn−2 some (tn, j ) j=0 ⊂ K . By (3.7) and Lemma 1 we get max
j∈(vn−2 ,vn−1 ]
vn−2 vn−1
X X
tn, j f j ≤ ej fj − |e j | ≤ j=0
j=vn−2 +1
≤
max
j∈(vn−2 ,vn−1 ]
|y n, j |kx j k
Bn−1 1 < 1/2 . an an −1/2
−1 Hence for m ≥ k we have max j∈(vm−1 ,cm ] |e j | ≤ max j∈(vm−1 ,vm ] |e j | < am+1 < am . Using (3.2) we get for n ≥ k:
max |zn, j | ≤ |αn | max max |e j+(m−n)am | ≤ |αn | max
max
m>n j∈(vm−1 ,vm ]
m>n j∈[0,vn−1 ]
j∈[0,vn−1 ]
−1/2
|e j |
−1/2
≤ |αn | max am+1 = |αn |an+2 ≤ a−1 n+1 . m>n
Let n ∈ N with n > k. Applying (3.6) we obtain cn cn X X y n, j x j = e j f j = Pvn−1 Pvn−1 j=0
j=0
max |y n, j | < a−1 n .
j∈[0,vn−1 ]
Moreover we have cn X ej fj ≤ Pvn−1 j=vn−1 +1
max
j∈(vn−1 ,cn ]
|e j |
max
j∈(vn−1 ,cn ]
|Pvn−1 ( f j )| ≤ a−1 n an−1 .
Pvn−1 Pcn Pcn Thus | j=0 e j f j | = |Pvn−1 ( j=0 e j f j ) − Pvn−1 ( j=v e j f j )| < a−1 n an−1 . For n−1 +1 P P vn−1 vn−1 vn−1 j some (sn, j ) j=0 ⊂ K we have j=0 e j f j = j=0 sn, j x . Hence vn−1
X
sn, j x j ≤ max |e j | =
j∈[0,vn−1 ]
j=0
≤
a−1 n an−1 Bn−1
k. It follows that max j∈N0 |e j | = 0, so e = 0, which is a contradiction. Lemma 6 Let 0 < ε < 1 and 1 < M < ε−1 . Let n ∈ N and m ∈ [0, n]. Assume that y ∈ Fn with |y| ≤ M and |Pm (y)| ≥ Mε. Then there exists q ∈ Fn with |q| ≤ ε−(n+2)! such that |Pn (qy) − xm | < ε. Pn i n Proof Clearly, y = i=0 y i x for some (y i )i=0 ⊂ K . By assumptions we have max0≤i≤n |y i | ≤ M and max0≤i≤m |y i | ≥ Mε. If n = 1 and m = 0, then q = y 0−1 x0 − y 0−2 y 1 x1 satisfies our claim.
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If n = 1 and m = 1, then we can take q = y 1−1 x0 if |y 0 | < ε2 M, and q = y 0−1 x1 if |y 0 | ≥ ε2 M. Suppose that our claim is true for n = k ≥ 1. We shall prove that it is true for n = k + 1. Pi+1 Let m = 0. Then we put q0 = y 0−1 and qi+1 = −y 0−1 j=1 y j qi+1− j for 0 ≤ i ≤ k Pk+1 i and q = i=0 qi x . It is easy to see that Pk+1 (qy) − x0 = 0 and |q| = max |qi | ≤ max ε−i (Mε)−1 = ε−k−2 M −1 ≤ ε−(k+3)! . 0≤i≤k+1
0≤i≤k+1
Let 1 ≤ m ≤ k + 1. Consider two cases. Case 1: |y 0 | < ε(k+2)!+1. Then max1≤i≤k+1 |y i | ≤ M and max1≤i≤m |y i | ≥ Mε. Pk+1 Pk By the inductive assumption for y = i=1 y i xi−1 , there exists q = i=0 qi xi with |q| ≤ ε−(k+2)! and |Pk (qy) − xm−1 | < ε. Then we have |Pk+1 (qy) − xm | = |Pk+1 (q(xy + y 0 x0 )) − xxm−1 | = |x(Pk (qy) − xm−1 ) + y 0 q| < ε. Case 2: |y 0 | ≥ ε(k+2)!+1. Then we put qi = 0 for 0 ≤ i < m, qm = y 0−1 , and Pj Pk+1 qm+ j = −y 0−1 i=1 y i qm+ j−i for 1 ≤ j ≤ k + 1 − m. For q = i=0 qi xi it is easy to check that Pk+1 (qy) − xm = 0 and |q| =
max
0≤ j≤k+1−m
|qm+ j | ≤
max
0≤ j≤k+1−m
ε−(k+2)!−1(Mε−(k+2)!−1) j ≤ ε−(k+3)! .
Now we are ready to show our main result. Theorem 7 Assume that d1 ≥ |α|4 and dn+1 ≥ |α|(ndn )! for every n ∈ N. Then the linear continuous operator T on E has no non-trivial closed invariant subspace. Proof Let M be a closed subspace of E with M 6= {0} such that T(M) ⊂ M. Then g(T)(M) ⊂ M for every g ∈ F. Let e ∈ M with 0 < kek ≤ 1. We shall prove that for every δ > 0 there exists f ∈ F such that k f (T)e − x0 k < δ. Let δ > 0. Let k > 2 with ak−1 > |α|δ −1 . By Lemma 5 we have |P(m−k)am (Qm e)| ≥ a−1 m for some m > k. For Rm = (am Am )[(m−2)am+2]! we have |α|a A R < b , since am Am < |α|am and m √m m m (mam )! |α| < bm . Put y = Qm e. Then |y| ≤ Am kQm kkek ≤ Am . By Lemma 6 and its proof there exists q ∈ F(m−2)am with |q| ≤ Rm such that (3.8)
|P(m−2)am (qy) − x(m−k)am | < (am Am )−1 .
(S1) Put f = βm−1 xam +bm q and S = Km,1 . Then f = Let z = f y. Using Lemma 4 we get k f (T)e − zk = k
X s∈S
P
s∈S t s x
s
for some (ts )s∈S ⊂ K .
ts (T s − T s Qm )ek ≤ max |ts kα| = | f kα| s∈S
= |qkαkβm |−1 < Rm |αkβm |−1 < a−1 m .
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616
(S2) Let a = (m − 1)am + bm and b = 2(m − 1)am + bm . Clearly z ∈ Fb ; Pb Pb j b j so z = j=0 s j x for some (s j ) j=0 ⊂ K . Then kz − Pa zk = k j=a+1 s j x k ≤ j maxa< j≤b |s j | maxa< j≤b kx k ≤ |z|, since √ 4bm i
kx j k = |α|−h[3bm −2 j]/
√ 4bm i
≤ |α|−h[bm −4(m−1)am ]/
≤1
for j ∈ (a, b] ⊂ Lm,2 . We have |z| = | f ky| = |βm−1 kqky| ≤ |βm−1 |Rm Am < a−1 m . Thus −1 kz − Pa zk < am . P2c j (S3) Let t = xam qy and c = (m − 1)am . Clearly t ∈ F2c ; so t = j=am γ j x for 2c j −1 j+bm some (γ j ) j=am ⊂ K . For j ∈ [am , c] we have j + bm ∈ Km,1 . Thus kβm x −x k = −1 −1 |βm k| f j+bm k = |βm |. Hence we get 2c 2c
X X
kPa z − Pc tk = kPa (βm−1 xbm t) − Pc tk = Pa γ j βm−1 x j+bm − Pc γjx j j=am
j=am
c c
X X
γ j x j ≤ max |γ j |kβm−1 x j+bm − x j k γ j βm−1 x j+bm − = j=am
j=am
am ≤ j≤c
≤ |t||βm−1 |. Thus kPa z − Pc tk ≤ |βm−1 |Rm Am < a−1 m , since |t| = |q||y| ≤ Rm Am . (S4) Using (3.8) we get |Pc t − x(m−k+1)am | = |xam (Pc−am (qy) − x(m−k)am )| = |P(m−2)am (qy) − x(m−k)am | < (am Am )−1 . Hence kPc t − x(m−k+1)am k ≤ a−1 m . (S5) By Lemma 1 we have kx(m−k+1)am − x0 k < |α|a−1 k−1 . Since f (T)e−x0 = ( f (T)e−z)+(z−Pa z)+(Pa z−Pc t)+(Pc t −x(m−k+1)am )+(x(m−k+1)am −x0 ), we obtain k f (T)e − x0 k ≤ |α|a−1 k−1 < δ. We have shown that for every δ > 0 there exists f ∈ F such that k f (T)e−x0 k < δ. It follows that x0 ∈ M. Hence xn = T n x0 ∈ M for all n ∈ N. Thus F ⊂ M, so M = E.
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C. J. Read, A solution to the invariant subspace problem. Bull. London Math. Soc. 16(1984), no. 4, 337–401. , A solution to the invariant subspace problem on the space l1 . Bull. London Math. Soc. 17(1985), no. 4, 305–317. , A short proof concerning the invariant subspace problem. J. London Math. Soc. 34(1986), no. 2, 335–348. A. C. M. van Rooij, Non-Archimedean Functional Analysis. Monographs and Textbooks in Pure and Applied Math. 51, Marcel Dekker, New York, 1978. A. C. M. van Rooij and W. H. Schikhof, Open problems. In: p-Adic Functional Analysis. Lecture Notes in Pure and Appl. Math. 137, Dekker, New York, 1992, pp. 209–219. P. Schneider, Nonarchimedean Functional Analysis. Springer-Verlag, Berlin, 2002.
Faculty of Mathematics and Computer Science, A. Mickiewicz University, 61-614 Pozna´n, Poland e-mail:
[email protected] 