LETTER TO THE EDITOR 321 Now Theorem 2. 1 is an easy consequence of Lemmas 2.1-2.3 Thus, the problem of constructing tight frames, generated by a refinable function, can be reduced to finding mk, that satisfy (7). Now we shall describe all possible solutions to(7) Let the symbol mo satisfy (10). Unit eigenvectors of the matrix M(o) can be represented in the form v1(a)= v2(a)= B(a)≠0, where B(o) is an arbitrary -periodic measurable functions, satisfying B(o)l= Imo(o)12+ mo(o+I)12 a.e. For definiteness, we can take here the positive root the right-hand expression. For those o when mo()= mo(o+I)=0 the matrix M( becomes the identity matrix. So any non-zero vector is its eigenvector. In this case we pu v1(a)=(1,0),v2(a)=(0 Thus, we have M(O)=P(o)A(o)P"(o), e mofc A(a)= (a)2-|mo(a+x)2 We note that eigenvectors are determined up to multiplication by a scalar function of absolute value 1 ae we have chosen the normalization convenient for further consideration THEOREM 2.2. Let a 2T-Periodic function mo(o)satisfy (10). Then there exists a pair of 2IT-periodic measurable functions mI, m2 which satisfy(7)for n= 2. Any solution of (7)can be represented in the form of the first row of the matrix 1(a)=P(a)√A(o)Q(a), where Q(o) is an arbitrary unitary (a.e. ) matrix with T-periodic measurable components Proof. The matrix My can be represented in the form of its singular decomposition My(o)=P(o)D(o)(o) where P, 2 are unitary matrices, D(a) is a nonnegative diagonal matrix. These epresentations may differ by multiplication of columns of the matrix p by functions a1(o),a2(o), laI(o)I=la2()I= I and simultaneous multiplication of rows of the matrix@ by ai(o)and a2(o).Thus, in view of(9)and(14)without loss of generality we can suppose=P,D=√ALETTER TO THE EDITOR 321 Now Theorem 2.1 is an easy consequence of Lemmas 2.1–2.3. Thus, the problem of constructing tight frames, generated by a refinable function, can be reduced to finding mk, that satisfy (7). Now we shall describe all possible solutions to (7). Let the symbol m0 satisfy (10). Unit eigenvectors of the matrix M(ω) can be represented in the form v1(ω) =    e iωm0(ω+π ) B(ω)  −  e iωm0(ω) B(ω)   , v2(ω) =   m0(ω) B(ω) m0(ω+π ) B(ω)  , B(ω) = 0, where B(ω) is an arbitrary π-periodic measurable functions, satisfying |B(ω)| 2 = |m0(ω)| 2 + |m0(ω + π )| 2 a.e. For definiteness, we can take here the positive root of the right-hand expression. For those ω when m0(ω) = m0(ω + π ) = 0 the matrix M(ω) becomes the identity matrix. So any non-zero vector is its eigenvector. In this case we put v1(ω) = (1, 0) T , v2(ω) = (0, 1) T . Thus, we have M(ω) = P(ω)0(ω)P ∗ (ω), (14) where P (ω) =    e iωm0(ω+π ) B(ω)  m0(ω) B(ω) −  e iωm0(ω) B(ω)  m0(ω+π ) B(ω)  , 0(ω) = 1 0 0 1 − |m0(ω)| 2 − |m0(ω + π )| 2  . We note that eigenvectors are determined up to multiplication by a scalar function of absolute value 1 a.e. We have chosen the normalization convenient for further consideration. THEOREM 2.2. Let a 2π-periodic function m0(ω) satisfy (10). Then there exists a pair of 2π-periodic measurable functions m1, m2 which satisfy (7) for n = 2. Any solution of (7) can be represented in the form of the first row of the matrix M(ω) = P (ω) 0(ω)Q(ω), where Q(ω) is an arbitrary unitary (a.e.) matrix with π-periodic measurable components. Proof. The matrix Mψ can be represented in the form of its singular decomposition Mψ (ω) = P(ω)D(ω)Q(ω), where P, Q are unitary matrices, D(ω) is a nonnegative diagonal matrix. These representations may differ by multiplication of columns of the matrix P by functions α1(ω), α2(ω), |α1(ω)| = |α2(ω)| ≡ 1 and simultaneous multiplication of rows of the matrix Q by α −1 1 (ω) and α −1 2 (ω). Thus, in view of (9) and (14) without loss of generality we can suppose P ≡ P, D ≡ √ 0.