Construction of analytic functions, which determine bounded Toeplitz operators

Theorem 1. If $f \in \mathfrak{N}$ , then Toeplitz operator $T_f$ is bounded on $H^p$ $(p = 1, \infty)$ and $$||T_f||_{H^p} \le ||f||_{\mathfrak{N}}.$$

Theorem 2. Let α = (αn)n≥⁰ be a monotone decreasing, concave sequence of positive numbers and $$\|\alpha\| \stackrel{def}{=} \sum_{n \ge 0} \frac{\alpha_n}{n+1} < \infty.$$ Then f ∗α ∈ N , Toeplitz operator Tf∗^α is bounded on H¹ and H^∞ for all f ∈ H^∞ and $$||T_{f\alpha}||_{H^p} \le ||f\alpha||_{\mathfrak{N}} \le 12 ||f||_{H^\infty} ||\alpha||, \quad p = 1, \infty.$$ Proof. Using Abel's formula two times we obtain $$\sum_{n\geq 0} \frac{\alpha_n}{n+1} = \sum_{n\geq 0} (\alpha_n - \alpha_{n+1}) \sum_{k=0}^n \frac{1}{k+1} \ge$$ $$\geq \sum_{n\geq 0} (\alpha_n - \alpha_{n+1}) \log(n+2) =$$ $$= \sum_{n\geq 0} (\alpha_n - 2\alpha_{n+1} + \alpha_{n+2}) \sum_{k=0}^n \log(k+2).$$ Since $$\sum_{k=0}^{n} \log(k+2) \ge \sum_{k=\lfloor n/2 \rfloor}^{n} \log(k+2) \ge (n/2+1) \log(\lfloor n/2 \rfloor + 2) \ge \frac{1}{4}(n+1) \log(n+2),$$ then $$4\sum_{n\geq 0} \frac{\alpha_n}{n+1} \geq \sum_{n\geq 0} (\alpha_n - 2\alpha_{n+1} + \alpha_{n+2})(n+1)\log(n+2).$$ Further let f ∈ H^∞ and $$S_n(f) = \sum_{k=0}^n \hat{f}(k) z^k; \qquad \sigma_n(f) = \frac{1}{n+1} \sum_{k=0}^n S_k(f).$$ Applying the Abel's formula we obtain $$f * \alpha = \sum_{n \ge 0} \hat{f}(n) \alpha_n z^n = \sum_{n \ge 0} (\alpha_n - \alpha_{n+1}) S_n(f) =$$ $$= \sum_{n \ge 0} (\alpha_n - 2\alpha_{n+1} + \alpha_{n+2}) (n+1) \sigma_n(f).$$ Since by Lemma 1. $$\|\sigma_n(f)\|_{\mathfrak{N}} \le 3 \|\sigma_n(f)\|_{H^{\infty}} \log(n+2) \le 3 \|f\|_{H^{\infty}} \log(n+2),$$ then $$||f * \alpha||_{\mathfrak{N}} \leq \sum_{n \geq 0} (\alpha_n - 2\alpha_{n+1} + \alpha_{n+2})(n+1) ||\sigma_n(f)||_{\mathfrak{N}} \leq$$ $$\leq 3 ||f||_{H^{\infty}} \sum_{n \geq 0} (\alpha_n - 2\alpha_{n+1} + \alpha_{n+2})(n+1) \log(n+2) \leq$$ $$\leq 12 ||f||_{H^{\infty}} \sum_{n \geq 0} \frac{\alpha_n}{n+1} = 12 ||f||_{H^{\infty}} ||\alpha|| < \infty. \square$$ The following proposition follows at once from Theorem 2.

Theorem 3. Let $\alpha$ denote one of the sequences $(\varepsilon > 0)$ : $$\left(\frac{1}{(n+1)^{\varepsilon}}\right)_{n\geq 0};$$ $$\left(\frac{1}{\log^{1+\varepsilon}(n+2)}\right)_{n\geq 0};$$ $$\left(\frac{1}{\log(n+2)\log^{1+\varepsilon}\log(n+3)}\right)_{n\geq 0},\dots$$ Then $f \alpha \in \mathfrak{N}$ , Toeplitz operator $T_{f\alpha}$ is bounded on $H^1$ and $H^{\infty}$ for all $f \in H^{\infty}$ . Remark. Theorem 3 was proved by another method in [3] ( Theorem 7. ) for the bounded Toeplitz operators $T_{f*\alpha}$ on $H^{\infty}$ .

Theorem 4. Let the sequence $\alpha = (\alpha_n)_{n\geq 0}$ satisfy the conditions of Theorem 3. If the sequence $a = (a_n)_{n\geq 0} \in \ell^2$ , then there exists a function $f \in \mathfrak{N}$ , satisfying $$|\hat{f}(n)| \ge \alpha_n |a_n|, \qquad ||f||_{\mathfrak{N}} \le c_0 ||\alpha|| ||a||_{\ell^2},$$ where $c_0$ is an absolute constant.