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3) is a(z) = b0 + b1 z + · · · + bs zs + b−r zn−r + · · · + b−1 zn−1 . Since a(ωnj ) = b0 + b1 ωnj + · · · + bs ωnj s + b−r ωn−j r + · · · + b−1 ωn−j = b(ωnj ), the assertions follow from the corresponding result on circ (a0 , a1 , . . , an−1 ). ✐ ✐ ✐ ✐ ✐ ✐ ✐ 34 buch7 2005/10/5 page 34 ✐ Chapter 2. 2. If n ≥ r + s + 1, then the determinant of Cn (b) is r+s det Cn (b) = bsn (−1)s(n−1) (1 − zkn ), k=1 where z1 , . . , zr+s are the zeros of the polynomial zr b(z). Proof. 1 we get n−1 det Cn (b) = ⎛ = bs ⎝ j =0 n−1 r+s n−1 bs ωn−j r b(ωnj ) = ⎞r ωn−j ⎠ j =0 j =0 (ωnj − zk ) k=1 n−1 r+s (ωnj − zk ) j =0 k=1 n−1 = bsn (−1)(n−1)r r+s (zk − ωnj ) (−1)r+s j =0 k=1 r+s n−1 = bsn (−1)(n−1)r (−1)(r+s)n (zk − ωnj ) k=1 j =0 r+s = bsn (−1)ns−r = bsn (−1) r+s (zkn − 1) = bsn (−1)ns−r (−1)r+s k=1 r+s s(n−1) (1 − zkn ) k=1 (1 − zkn ).

Consequently, there are b’s such that T (b) has a residual spectrum on 1 . 9. 9 buch7 2005/10/5 page 21 ✐ 21 Selfadjoint Operators We now consider Toeplitz operators on the space 2 . Obviously, T (b) is selfadjoint if and only if bn = b−n for all n, that is, if and only if b is real valued. Thus, let s b(eix ) = s bk eikx = b0 + k=−s (an cos nx + cn sin nx), n=1 where b0 , an , cn are real numbers. The resolution of the identity. Let A be a bounded selfadjoint operator on the space 2 . Then the operator f (A) is well defined for every bounded Borel function f on R.

26) converges absolutely. The constants G(b) and E(b) are obviously independent of the particular choice of log b. 24) holds, then k b(t) = bs 1− i=1 δi t i (t − μj )σj , j =1 where δ1 , . . , δk , μ1 , . . , μ are distinct, |δi | < 1, |μj | > 1, σ1 + · · · + σ = s. On writing k log b(t) = log bs + i i=1 log 1 − δi t 1 + ··· + σj log(−μj ) + + j =1 k = r, and σj log 1 − j =1 t μj ✐ ✐ ✐ ✐ ✐ ✐ ✐ 44 buch7 2005/10/5 page 44 ✐ Chapter 2. Determinants and using the formula log(1 − z) = −z − z2 /2 − z3 /3 − · · · (|z| < 1), we get (log b)0 = log bs + σj log(−μj ), j =1 (log b)n = − j =1 σj nμnj k (log b)−n = − i=1 (n ≥ 1), n i δi (n ≥ 1).