302

Xionghui He and Hans Volkmer

for all m, n = 0, 1,...,/: (if 5 = 0 then /oo = 2.) If 5 > 0, the matrix M whose elements are defined by

m , n = 0, 1,...,^ (3.5)

m - { - n - \ - 28

IS the Gram matrix of {t^^^ г} in L^[0, 1]. Since Gram matrices are positive semidefimte, we have {Mc, c) > 0 for any с € C^'^^. From

sin ( 257r ) F = I ^-------------M

and sm(257r) > 0, we get (Fc, c) > (c, c) for all с e C^+^ If 5 = 0, then F = diag(2, 1,..., 1) is a diagonal matrix and clearly (Fc, c) > (c, c) for all с e C^'^^. This completes the proof of part (a), (b) Let F' be the Gram matrix of {/i(x),..., Л (л:)}, and let M' be the matrix with ments (3 5) for m, n = 1,..., k. The Hubert matrix L whose elements are defined by (m-\-n —1)~^ m, « = 1,..., /:, satisfies (Lc, c) < n(c, c) for all с e C^, due to Hubert's Inequality [4]. Therefore we have

{ M'c , c) < 7t{c, c) for all с G C^ .

Hence ,

. „ , Ч sin(2^7r) . , .

( Гс , с) = (с, с) + —!^------ (M'с, с) < (1 + sin(257r))(c, с) .

This completes the proof of part (b).

( c ) Let gn{x) = (—1)"л/2/7Г sin((n + 5)x). The elements g;„„ of the Gram matrix G of the system {^1 (jf),..., ^fc(x)} are given by

Г f л ^ л^ я sm(2(57r)

Jo (m 4- /1 + 25)7Г

for m, л = 1,..., /: Since M' is positive semidefimte, we have (Gc, c) < (c, c), which shows that {^Ai(-^)} has upper bound I in L^[0, n] D

Lemma 6.

Let 0 < 5 < 2- TTien {cos(n -f 8)x], n e No, w a Riesz basis in L^[0, тг].

Proof . By Lemma 5, the sequence has lower bound V^/2 and a finite upper bound. So it is enough to show that the sequence IS complete. Let/n(x) = ^'('^+^)'^ and^„(jc) = e''^^"''^^'^^,n e Щ. If 5 € (-|, |), Kadec's 1/4-Theorem shows that {/„} U [gn], n e No, is complete m L^[—тг, тг]. Hence,

{ cos ( n -f 5)jc} и {sin(n + 8)x], П 6 No (3.6)

IS complete m L^[—я, тг]. It follows that {cos(n + 8)x], n e No, is complete in L^[0, тг]

Let Un{x) = e^^^+^^-l)^ = e-^ifn{x) and u„(jc) = ^^(-«-'5+î)-^ = e'ign{x), n e No-

If 8 e i\, |), Kadec's 1/4-Theorem shows that [un] U {un},n € No, is complete m L^[—7г,7г].

Hence , {/„}U{^„},n e No, is complete m L^[—тг, тг]. It follows that (3.6) is complete in L^[—7r,7r].

Therefore , {cos(n + 8)x}, n e No, is complete in L^[0, тг]. Thus, {cos(n-f 5)jc}, n e No, is complete

mL2 [ 0 , 7r ] for ( 5e ( - | , i ) U ( i , | ) .

If 5 = ^, then {cos(n -h 8)x], n e No, is also complete in L^[0, тг]. This follows from [12,

Theorem 5, p. 122]. D

Now we shall introduce the mam theorem in this section.