JOURNAL Series A - Pure and Applied Mathematics

Construction of analytic functions, which determine Toeplitz operators 73S n (f) =n∑k=0Applying the Abel’s formula we obtainˆf(k) z k ; σ n (f) = 1n + 1n∑S k (f).k=0f ∗ α = ∑ n≥0ˆf(n) α n z n = ∑ n≥0( α n − α n+1 ) S n (f) =Since by Lemma 1.= ∑ n≥0(α n − 2α n+1 + α n+2 )(n + 1) σ n (f).then‖σ n (f)‖ N≤ 3 ‖σ n (f)‖ H ∞ log(n + 2) ≤ 3 ‖f‖ H ∞ log(n + 2),‖f ∗ α‖ N ≤ ∑ n≥0(α n − 2α n+1 + α n+2 )(n + 1) ‖σ n (f)‖ N≤∑≤ 3 ‖f‖ H ∞ (α n − 2α n+1 + α n+2 )(n + 1) log(n + 2) ≤n≥0≤ 12 ‖f‖ H ∞∑n≥0α nn + 1 = 12 ‖f‖ H ∞ ‖α‖ < ∞.□The following proposition follows at once from Theorem 2.Theorem 3. Let α denote one of the sequences (ε > 0) :(1;(n + 1))n≥0ε(1)log 1+ε (n + 2)n≥0(1)log(n + 2) log 1+ε log(n + 3);n≥0, ..............................Then f ∗ α ∈ N , Toeplitz operator T f∗α is bounded on H 1 and H ∞ for allf ∈ H ∞ .