LETTER TO THE EDITOR We split the proof of Theorem 2.1 into several lemmas. LEMMA 2.1. Let the symbols (mk k-o satisfy(7). Then for any o m(o)2+lm(a+x)2≤1,=0,1,,n. Proof. Obviously, without lose of generality it suffices to prove inequality(8)only for 1=0. Let us rewrite relation (7)in the form Imo(o)12 mo(a)mo(a+丌) M(o): =My(o)My(o) o(a)mo(a+丌) m0(0+丌 m1(a)m2(a) My(o) m1(ox)m2(a+丌) The Hermitian matrix M(o) has eigenvalues 入1()=1,A2(a)=1-|m0(a)2-|mo(a+r)2 By definition(9), M(o) is a positive definite matrix. Hence, 12(0)>0, which is(8)for 0 LEMMA 2.2. If peLZ(R)is a refinable function with a symbol m(o) that satisfies the m(a)2+m(a+丌)≤1ae then Sj:=∑ezl(f,中,k)2<∞ for any function f∈L2(R)and (i)imS;=‖f‖ (ii lim Si=0 where ;. k=2/4(2Jx-k Proof. First, we prove that ∑|中(x+2x)2≤2 We note that due to (10)and the continuity (o)at a=0 we have |d(o)l <( 2x)-/2a.e Thus, for any positive /EZ we obtain ∑(+2xmk)2=∑∏m(2-"(o+2mk)21(2--1(o+2k)2 k==21 k==2ln=1 2-11+1 ∑∏m(2-"(o+2xk)2LETTER TO THE EDITOR 317 We split the proof of Theorem 2.1 into several lemmas. LEMMA 2.1. Let the symbols {mk} n k=0 satisfy (7). Then for any ω |ml(ω)| 2 + |ml(ω + π )| 2 ≤ 1, l = 0, 1,..., n. (8) Proof. Obviously, without lose of generality it suffices to prove inequality (8) only for l = 0. Let us rewrite relation (7) in the form M(ω) := Mψ (ω)M∗ ψ (ω) = 1 − |m0(ω)| 2 −m0(ω)m0(ω + π ) −m0(ω)m0(ω + π ) 1 − |m0(ω + π )| 2  , (9) where Mψ (ω) = m1(ω) m2(ω) ... mn(ω) m1(ωπ ) m2(ω + π) ... mn(ωπ )  . The Hermitian matrix M(ω) has eigenvalues λ1(ω) ≡ 1, λ2(ω) = 1 − |m0(ω)| 2 − |m0(ω + π )| 2 . By definition (9), M(ω) is a positive definite matrix. Hence, λ2(ω) ≥ 0, which is (8) for l = 0. LEMMA 2.2. If " ∈ L 2 (R) is a refinable function with a symbol m(ω) that satisfies the condition |m(ω)| 2 + |m(ω + π )| 2 ≤ 1 a.e., (10) then Sj :=  k∈Z |
f,"j,k|2 < ∞ for any function f ∈ L 2 (R) and (i) lim j→∞ Sj = f  2 ; (ii) lim j→−∞ Sj = 0, where "j,k = 2 j/2"(2 j x − k). Proof. First, we prove that  k∈Z |"(x ˆ + 2πk)| 2 ≤ 1 2π . (11) We note that due to (10) and the continuity "(ω ˆ ) at ω = 0 we have |"(ˆ ω)| ≤ (2π )−1/2 a.e. Thus, for any positive l ∈ Z we obtain 2 l−1 k=−2 l |"( ˆ ω + 2πk)| 2 = 2 l−1 k=−2 l  l+1 n=1 |m(2 −n (ω + 2πk))| 2 |"( ˆ 2 −l−1 (ω + 2πk))| 2 ≤ 1 2π 2 l−1 k=−2 l  l+1 n=1 |m(2 −n (ω + 2πk))| 2