Normal bases for the space of continuous functions defined on a subset of Zp

Publicacions Matemátiques, Vol 38 (1994), 371-380 . NORMAL BASES FOR THE SPAC E OF CONTINUOUS FUNCTION S DEFINED ON A SUBSET OF z p ANN VERDOOD T A bstract Let K be a non -archimedean valued field which contains Q p an d suppose that K is complete for the valuation • 1, which extends th e p-adic valuation . Vq is the closure of the set {agn in = 0, 1, 2, . . . } where a and q are two units of Z p, q not a root of unity . C(Vq -- > K) is the Banach space of continuous functions from Vq to K , equipped with the supremum norm . Our aim is to find norma l bases (rn (x)} for C(Vq --~ K), where rn (x) does not have to be a polynomial . 1 . Introductio n The main aim of this paper is to find normal bases (r(x)) for th e space of continuous functions on V q , where r n (x) does not have to be a polynomial . Therefore we start by recalling some definitions and some previou s results . Let E be a nonarchimedean Banach space over a non-archimedea n valued field L . Let f 1 , f 2 , . . . be a finite or infinite sequence of elements of E . We sa y that this sequence is orthogonal if IIaifi + • • • +  f k II = max { i i acz f i l j i = 1, . . . , k} for all k in N (or for all k that do not exceed the length o f the sequence) and for all a l , . . . , ak in L . If the sequence is infinite, i t follows that E a i f i = max {ii a i fi : i = 1, 2, . . . } for all a l , a2, . . . i n i= z L for which i hm ai f i = O . An orthogonal sequence f 1 , f 2 , . . . is calle d orthonormal if 11 f i II = 1 for all ~ . This leads us to the following definition : 372  A . VERDOOD T If E is a nonarchimedean Banach space over a nonarchimedean valued field L, then a family (fi) of elements of E is a (ortho)normal basi s of E if the family (fi) is orthonormal and also a basis . An equivalent formulation is (see [1, Propositions 50 .4 and 50 .61 ) If E is a nonarchimedean Banach space over a nonarchimedean valued fieid L, then a family (fi) of elements of E is a (ortho)normal basi s of E if each element x of E has a unique representation x = E xi f i wher e x i E L and x i--> 0 if i  oo, and if the norm of x is the supremum o f the norms of x i . Let 7L p be the ring of p -adic integers, Qp the fieid of p -adic numbers , and K is a nonarchimedean valued fieid, K containing Qp, and we suppose that K is complete for the valuation ~ • ~ , which extends the p -adi c valuation . Let a and q be two units of Z p , q not a root of unity . We defin e Vq to be the closure of the set {aqnln = O, 1, 2, . . . } . The set Vq has bee n described in [3] . Let C(V q - -} K) (resp . C(Lp ---~ K)) be the Banac h space of continous functions from Vg to K (resp . Z p to K) equippe d with the supremum norm . N denotes the set of natural numbers, and N o is the set of natural numbers without zero . We introduce the following : If x is an element of Q p , x can be written in the following way : +o o x = E a i p j where  is zero for i sufficiently large (i E N) (se e i_—o c [1, section 3 and section 4]) . This is called the Henseldevelopment o f the p -adic integer x . We then define the p -adic entire part [x] 1 , of x b y 1  n- 1 [x] p = E a i p a and we put xn — p n [p —n x] p —E a j p i (n E N) . .7=-00  i=-o o We write m q x, if m is one of the numbers xo, x 1 , . . . . We then sa y that "m is an initial part of x" or "x starts with m" (see [1, section 62D . . s  If n belongs to No, n = E a i p 2 where a s  0, then we put n_ = j= o s- 1 E cl i p . ) . We remark that n_ q n . j= o In [1, Theorem 62 .2], we find the following result which is due to va n der Put : Theorem . The functions go, gla . . . defined b y gn(x) =1 ifn q x , = 0 otherwise, NORMAL BASES  373 foren a normal basis for C(7L P —> K) . If f is an element of C(7G P —> K) , 0 0 then f can be written as a uniformly convergent series f (x) = E ry k g k(x ) k= 0 where 'yo = f (0) and ryn = f (n) — f (n_) if n E Np . We now survey the content of this paper : In Theorem 1 of section 2, our aim is to find a basis (e(x)) analogou s to van der Put's basis, but with the space C (7L p -- -+ K) replaced b y C(V q —} K) . If f is an element of C(V q ~ K), then there exist element s a k of K such that f (x) = Eakek(x) where the series on the right-hand- k= o side is uniformly convergent . We are able to give an expression for th e coefficients a k . In Theorem 2 of section 3, we prove the following result : n Define r n (x) = E c n ; j e j (x), c n ; j E K, Cn ;n  0 ((e(x)) as in Theoj= o rem 1 below) . Then (r(x)) forms a normal basis for C(V q ~ K) if and only if fo r all n f ~rnll = 1 and Icn ;n1 = 1 . In Theorem 3 of section 3, we give an extension of Theorem 2 : Let (r n (x)) be such a sequence which forms a normal basis fo r n C( V q - --} K), and let (sn(x)) be a sequence such that s n (x) = E d n ;j r j (x) , j= o dn ; j E K, d n ;n O . Then (s(x)) forms a normal basis for C( V g --} K) <= > II Snll = 1, k1 n ;n I = 1 <=> I d n ;j 1 ç 1, Id n ;n I = 1 . Acknowledgement . I thank professor Van Hamme for the advice h e gave me during the preparation of this paper . 2 . Proof of the first theore m We start with some lemmas and some definitions . Definition . If b and x are elements of Z, b 1 (mod p), then we put b~ = lim b n . n--} x The :mapping : 7L p ---} 7L p : x ~ b x is continuous . For more details, we refer the reader to [1, section 32] . Not at ion . Take m ~ 1, m the smallest integer such that q m 1 (rnodp) . We have 1 ç m ç p -- 1 and (qm)X is defined for all x in 7L . 374  A . VERDOOD T Definition . Let k be a natural number prime to p . We denote by 7L P (k) the projective limit 7L p (k) = lim(7L/kp j 7L) (Z/kZ) x Z . In the following lemma we use the fact that 7L p (m) = (Z/mZ)x 7L p t o denote an element of 7% p ( rn ) as x = (r, y) . Also, if n E N, n = r + m k (o Ç r C rri) then the map n —+ (r, k) imbeds N in 7Lp (m) . Lemma 1 . T he mapping c,o : 7L p (m) - -} V q : (r, y) —> ag r ( g m ) y is a homeomor - phisrn . The proof of this lemma can be found in [2, p . 377] . Corollary . If g 1 (mod p ) , i . e . m = 1, then the mapping : 7L p ~Vg : x  aq ' is a homeomorphism . Let ~ be an element of 7L p \{0} . We want to know the valuation of th e p -adic integer (g m ) O — 1 . Therefore we need two lemmas : The following lemmas (2 and 3) are proved in [3] : Lemma 2 . Let a be an element of 7L p , a 1 (moda 1 (modr 1 . If (p, r)  (2,1), ,C3 E 71 p \{ o} then a @  1 (modp r + ordp,Q ),  1 (mod pr+ l +vrdp ,Q ) . Corollary . Let q m  1 (modpko), q m  1 (modpko+l) . lf (p, k 0 )  ( 2,1 ) , ,C3 E 71 p \ {o} then (g m )) 3 1 (modp" +ord P Q ), (g m ) P 1 (modpk°+1+°rdp Q) . In Lemma 2 we excluded the case where (p, r) = (2, 1) . This case wil l be handled in the following lemma : Lemma 3 . Let a be an element of Z 2 , a 3 (mod 4) . Define a natural number n by a = 1+2+2 2 E, E = eo + ~12 + E222 + . . ., Eo =El = ... — En_l = 1 , En = O . If ,3 E Z 2 \{O}, ord 2 ~ = 0 then a 0 T 1 (mod 2), a P # 1 (mod 4) . NORMAL BASES  37 5 If i 5' E Z 2 \{O}, ord 2 fi = k ~ 1 then cx~ - 1 (mod2 2+02 ~), a 0 # 1 (mod 2n+3 +ord2 i 5') . Corollary . If q 3 (mod 4), we define a natural number N by q = 1 + 2 + 2 2 e , e = ea + e 1 2 + e 2 2 2 + . . . = e i = . . . = EN—1 = 1, E N = 0. If fl E Z 2 \{O}, orde fi = 0 then q Q - 1 (mod 2), q a # 1 (mod 4) . If Q E Z 2 \{O}, orde fi = k > 1 then q Q 1 (mod2 2+01 ' 2 ~), qp 1 (mod 2N+3+ord e We remark that is possible to write each x and element of V q in th e following way : x = agi x (gm )° e x where i x is a natural number, 0 < i x C m, and where a s is an element of Z p . This immediately follows fro m Lemma 1 . This leads us to the following definition : Definition . We now define a sequence of functions ek in the following way . Writ e k( E N) in the form k = i + m j, 0 Ç i C m (i, j E N) . The functions e k are defined b y e k( x ) = e i -I-mj( x ) = 1 if x = agzx (gm )' x where i x = 2, q (x x . = = 0 otherwise . Let us use the notation B (b, r - ) for the 'open' disc with radius r and wit h center b, i .e . B(b, r - ) = {x E V q l Ix -bi C r}, and B(b,r) far the 'closed ' disc with radius r and with center b, Le . B (b, r) = {x E V g ! i x— b l Ç r} . Inthe following lemmas we will show that the functions e k (x) ar e characteristic functions of discs . There exists a ko such that q m EE . 1 (modp c 0) , q m # 1 (modp c o) . We distinguish two cases : (p, ko ) (2,1) ( :Lemma 4), and(p, ko ) = (2,1) Le . q - 3 (mod4) (Lemma 5) . I f we use the same notation in Lemmas 4 and 5 as in the definition, w e hav e Lemma 4 . Let q m . - 1 (modp'0), q m  1 (modp'0+l) and suppose (p, kp ) (2,1) . If 0 < i < m then e i (x) is the characteristic function of the close d disc B (aq i , p -k® ) , and if 0 Ç i C m, j > 1 then e k (x) = e i + j m(x) is th e -,~ - characteristic function o f the open disc B (aq(qm)2, (P0)  . 376  A . VERDOOD T Pr oof : s Let j = E  be the Henseldevelopment of j E No, with a s differen t i= o from zero . If we use the notation x = ag ix (g m }°` x (O Ç i s C m) far an element x of Vq , we will show the following : a) if o Ç i C m : 1 x— ag i 1 Ç p - " if and only if i s = i . b) if o Ç i c m, j  1 : Ix - agi (gm )j 1 C P  j ° if and only if = j q a x . We first prove a) . If ix = i, then Ix — ag i = 1  — ag i = 1x — 11 Ç pk ° by the corollary to Lemma 2 . If  the n Ix - agz I= agZx ( gm } ax _ aq z = max{ I ag zx (gm rx — aq ix 1, Iaq zx — agz 1 } = 1 , since laqix (qm)x — agzx C p - k0 , Iaqix — ag i = 1 . This proves a) . Now we prove b) .  Suppose i s — i, j  a s . Then 1x - - ag i ( g m ) j 1= I(g m )" x-j - - 11 < p - k ° - (s+ 1 ) by the corollary following Lemma 2, since j is an initia l part of a~ . Since j is strictly smaller than p (8+1) , we conclude tha t I  ¡  11  ° ix— agzlgmliI C  k . k For the converse, suppose Ix — agi (qm)i < P j ° . Then we must hav e that  equals í, since otherwise 1x -- ag i (g m ) j I = 1 : Ix — agz ( gm ) a 1 = lagzx ( qm ) "x -- aqi ( gm ) a } 1 = max{ ag zx ( gm )" x — aq zx 1, I aqix _ aqz 1, laqi — agz l gm l i 1 } = 1 since laqix (qm)ESx — aq zx < p-'0, 1aq i — ag z (g m ) 3 I Ç p - k ° (corollary t o Lemma 2) and lag ix — aq i = 1 if is different from i . So we have 1  ax -j — 11 <  and from this it follows tha t 1 — 11 Ç p - k° - ( s + 1 ) since j is at least p s . This means tha t ord P (a s — j) is at least s + 1 (again by the corollary to Lemma 2) an d so we conclude that j is an initialpart of a~ . n Lemma 5 . lf q 3 (mod 4), with q = 1 + 2 + 2 2 e, where s = so + e 1 2 + s222 + . . . , Eo = E1 = . . . = E N -1 = 1, EN = O, then eo(x) is the characteristic NORMAL BASES  37 7 function of V 9 , and e i (x) is the characteristic function of the open dis c B (aq j , (2-'2)) J Proof : In this case m equals one and we use the notation x = ag ax far a n element x of Vq . It is clear that eo(x) is the characteristic function of Vq . If j is at least one, we prove : ix — ag j ~ < 2   ( N . +2 ) if and only if j q ax . 3 Suppose j q cx x . Then l x — aq j I _ le x- i — 11 < 2-(N+2)-(s+1 ) (corollary following Lemma 3), and since j is strictly smaller than 2 s+1 , we conclude — aq j < 2 -~N+ 2 ) For the converse, suppose l x— aq j < 2  7 2 ) . Then q~` x Tj — C 2 -(N+2) and so  -1 i C 2 -(N +2 ) -(s +1) since j is at least 2 S . By th e 3 corollary to Lemma 3, we have that or d 2 (c xx — j) is at least s + 1 and s o j is an initialpart of cx x . ■ Corollary . The functions (e k (x)) are continuous functions on V q . In the following theorem we prove that the sequence (e k (x)} forms a normal basis for C(V q -~ K) . This implies that if f is an element o f c o C(V q -~ K), there exists elements a k of K such that f (x) = > l a k e k( x ) k= o where the righthand -side is uniformly convergent . We are able to give a n expression for the coefficients a k . The proof of this theorem is analogou s to the proof of Theorem 62 .2 in [1] . Theorem 1 . The functions (e k (x)) form a normal basis for C(V q ~ K) . lf f is a n element of C(V q ---~ K) then f can be written as a uniformly convergen t c c series f(x) = Eakek(x) wher e k= o a k = f (aq k )  if 0 < k < m a k = a i+im = f( ag2 (4 m ) j ) — f(a4 2 ((I m ) J -) if 0 < i c m, j > 0 . Proof : Let f be an element of C (I/ q ~ K), and let a k be defined as a k = f (aq k ) if 6 Ç k < m, ak = ai+ jm — f (ag z (q m ) j ) — f (ag z (g m ) j -) i f 0 ÇiCm, j > O . 378  A . VERDOOD T We first prove that a k tends to zero if k tends to infinity : for all e > 0 , there exists a J such that k ~ J implies ia k l < E . To prove this, w e distinguish two cases : i) Let q m  1 (modp k °), q m  1 (modp k ° +1 ), with (p, ko)  (2,1) . Since the function f is continuous on V q , it is uniformly continuou s on V q , and so there exist an S, such that 1 x - y i < p- (ko +s ) implie s f (x) - f (y) ~ C E . We then put J = p s m . If k ~ J, and k equals i+ j m with 0 Ç i C m, then we have that j ~ p s and so (corollary to Lemma 2) I  - agi ( gm ) i- ! = (qm)J_a__1 Ç p -(kQ+s) and this implies that kkI = f ( agi (gm ) 3 ) - f (aq(q m ) 3 ) I C E . ii} Let g 3 (mod 4 ) , q-- 1+2+2 2 e, e= eo+e 1 2 + E22 2 + . . . , EO = El = • • • = e N -1 = 1, EN = O . We remark that m equals one in thi s case . Since the function f is continous on V q , it is uniformly continuou s on V q , and so there exist an S, such that ix - y i < 2- ( N + 2 + s ) implie s ~ - f( y ) ~ C E . We then put J = 2 s . If k ~ J, then (corollary to Lemma 3) I qk - qk- I = I qk-k- - 1I Ç 2( 2 + 8 ) and this implies that la k = f (q k ) - f (q k - } 1 C E . We conclude that a k tends to zero if k tends to infinity . If f is an element of C (1/ g -4 K), we introduce a function g(x) define d ~ by g(x) = > akek(x) with a k as in (*) . Since lia k e k ~I Ç ia k i —> 0 , k= 0 the series onthe righthand -sideconverges uniformly, so the function g is continous as a uniformly limit of continuous functions . We can prov e that g(aq k ) = f (aq k ) if O . < k< m and that g(ag i (g m ) j )—g(aq i (q m } j - ) _ f (agZ(gm ) 3 )- f (ag i (g m )j- ) for 0 Ç i C m, j > 0 . Then we have g (aq k ) = f(aq k ) for all natural numbers k and by continuity, we conclude tha t f (x) = g(x) . 0 0 So we have f (x) = E a k e k (x), with a k as in (*) . k- o It is clear that iif < ó á~x{ia k but we also have f (aq k ) ~ IfM an d (ag i (q m )i ) - f(ag i (q m ) i_ )i < IIfM so we conclude f ii = óax{la k i} . Finally we prove the uniqueness of the coefficients . 00  00  0 0 If f(x) = E a k e k (x) = E b k e k (x), then E (a k - b k )e k (x) = 0 . S o k= a  k=o  k= o o max{ I a k - b i d} = 0, from which it follows . that a k = b k for all k . Thi s proves the t heorem . ■ NORMAL BASES  37 9 3 . More bases for C(V q K ) We can make more normal bases, using the basis (ek(x)) of Theorem 1 : Theorem 2 . n Let (e(x)) be as above, and define r n (x) = E c n , j e j (x), cn, j E K , j= o c n ; n0 . Then (r(x)) forms a normalbasis f or C (Vq - --} K) if and onl y i f  = 1 and iCn ;n! = 1 for all n . The proof of this theorem will not be given here, since it is analogou s to the proof of Theorem 2 in [3] . Reraark . An analogous result can be found on the space C(7Lp —> K), if w e replace the sequence (e(x)) by the van der Put basis (g(x)) from th e introduction . We can extend Theorem 2 to the following : Theorem 3 . Let (r n (x)) be a sequence as found in Theorem 2, which forms a norma l basis for C(V q —> K), and let (s n (x)) be a sequence such, that sn(x) = n Edm ; j n j (3), d n ;j E K, d n ;n D . J=o Then the following are equivalent : i) (s n (x)} forms a normal basis for C (V q —> K) . ii}  =1, I d n ;nl = 1 . iii) Idn ;j Ç 1, k41 ;nI = 1 . Proo f : i)  ii} follows from Theorem 2, using the expressio n n E c n ; j e j (x), and ii} <=> iii) follows from the fact that (r(x)) forms a j= o normal basis . ■ Exa :mples . 1) If a sequence (r(x)), as found in Theorem 2, forms a norma l basis of C(V q --> K), then so does (s(x)), where s n (x) = ro (x) + r l (x) + . . . + r n (x) : apply iii) . 2) If we put for O Ç i C m , r i (x) = 1 if x = ag zx (g m )" where i x = = 0 otherwise, r n (x)