Estimates of the norms of the Toeplitz operators of H∞ determined by rational inner functions

Lemma 1. If $f \in H^{\infty}$ , then $||T_f||_{H^{\infty}} \leq ||f||_{H^{\infty}} + \Lambda(f)$ . Proof. $$||T_f||_{H^{\infty}} = \sup \left\{ \lim_{r \to 1-0} \left| \int_T \frac{\overline{f}(\varsigma)h(\varsigma)}{1 - \overline{\varsigma}r\eta} dm(\varsigma) \right| : \eta \in T, \quad ||h||_{H^{\infty}} \le 1 \right\} =$$ $$= \sup \left\{ \lim_{r \to 1-0} \left| \int_T \frac{\overline{f}(\varsigma\eta)h(\varsigma\eta)}{1 - r\overline{\varsigma}} dm(\varsigma) \right| : \eta \in T, \quad ||h||_{H^{\infty}} \le 1 \right\} \le$$ $$\leq \sup \left\{ \lim_{r \to 1-0} \left| \int_T \frac{\overline{f}(\varsigma \eta) - \overline{f}(\overline{\varsigma} \eta)}{1 - r\overline{\varsigma}} h(\varsigma \eta) dm(\varsigma) \right| : \eta \in T, \quad \|h\|_{H^\infty} \leq 1 \right\} + \|f\|_{H^\infty} \leq$$ $$\leq \sup \left\{ \left| \int_{T} \frac{|f(\varsigma \eta) - f(\overline{\varsigma} \eta)|}{|1 - \varsigma|} dm(\varsigma) \right| : \eta \in T \right\} + \|f\|_{H^{\infty}} = \Lambda(f) + \|f\|_{H^{\infty}}.$$ We used, that $g(z) = \overline{f}(\overline{z}\eta) \in H^{\infty}$ and $$\left| \int_T \frac{\overline{f}(\overline{\varsigma}\eta)h(\varsigma\eta)}{1 - r\overline{\varsigma}} dm(\varsigma) \right| \leq \|f\|_{H^{\infty}} \|h\|_{H^{\infty}}.$$ 238 Peyo Stoilov

Lemma 2. If $$I(z) = \frac{z-a}{1-z\overline{a}}, \ a \in D, \ then \ \Lambda(I) \le 2.$$ Proof. $$\Lambda(I) = \sup_{\eta \in T} \int_{T} \left| \frac{\varsigma \eta - a}{1 - \varsigma \eta \overline{a}} - \frac{\overline{\varsigma} \eta - a}{1 - \overline{\varsigma} \eta \overline{a}} \right| \frac{dm(\varsigma)}{|1 - \varsigma|} =$$ $$=\sup_{\eta\in T}\int_{T}\frac{\left(1-\left|a\right|^{2}\right)\left|\varsigma-\overline{\varsigma}\right|\,dm(\varsigma)}{\left|1-\varsigma\eta\overline{a}\right|\left|1-\overline{\varsigma}\eta\overline{a}\right|\left|1-\varsigma\right|}\leq2\sup_{\eta\in T}\int_{T}\frac{1-\left|a\right|^{2}}{\left|1-\varsigma\eta\overline{a}\right|\left|1-\overline{\varsigma}\eta\overline{a}\right|}\,dm(\varsigma)\leq$$ $$\leq 2 \sup_{\eta \in T} \left( \int_{T} \frac{1 - |a|^2}{|1 - \varsigma \eta \overline{a}|^2} dm(\varsigma) \right)^{1/2} \left( \int_{T} \frac{1 - |a|^2}{|1 - \overline{\varsigma} \eta \overline{a}|^2} dm(\varsigma) \right)^{1/2} = 2.$$

Lemma 3. If $I_k(z)$ , k = 1, 2, ....., n is inner functions ( $|I_k(\varsigma)| = 1$ a.e on T), then $\Lambda(I_1I_2....I_n) \leq \Lambda(I_1) + \Lambda(I_2) + ....\Lambda(I_n)$ .

Lemma 4. If $B_n \in b_n(E)$ , then $\Lambda(B_n) \leq 2$ . Proof. Let $$B_n(z) = \prod_{k=1}^n \frac{z - a_k}{1 - z\overline{a_k}} , \quad a_k \in D.$$ From Lemmas 2 and 3 it follows that $$\Lambda(B_n) \le \sum_{k=1}^n \Lambda\left(\frac{z-a_k}{1-z\overline{a_k}}\right) \le 2n.$$

Theorem 1. Ωn(D) = Ωn(Eξ) = 1 + 2n for all ξ ∈ T, where E^ξ = {z : ε ≤ |z| < 1, z/ |z| = ξ} , ε > 0. Proof. From Lemmas 1 and 4 follows that $$\Omega_n(E_{\xi}) \le \Omega_n(D) = \sup \{ ||T_{B_n}||_{H^{\infty}} : B_n \in b_n(D) \} \le$$ $$\leq 1 + \sup \left\{ \Lambda(B_n) : B_n \in b_n(D) \right\} \leq 1 + 2n.$$ Let ξ ∈ T. We will show that $$\Omega_n(D) = \Omega_n(E_{\xi}) = 1 + 2n.$$ Let $$x_k = (1 - q^k)\xi$$ , $0 < q < 1$ , $\varepsilon < 1 - q$ , $B_n(z) = \prod_{k=1}^n \frac{z - x_k}{1 - z\overline{x_k}}$ . Let m > n and $$y_k = B'_n(x_k)(|x_k|^2 - 1)\xi, \quad k \neq m, \quad y_m = B_n(x_m).$$ Since $$|B'_n(z)| (1 - |z|^2) \le 1, |B_n(z)| \le 1, (z \in D),$$ then |yk| ≤ 1 and by a well known Carleson interpolation theorem there exists a function h⁰ ∈ H^∞ , such that $$h_0(x_k) = y_k = B'_n(x_k)(|x_k|^2 - 1)\xi, \quad k \neq m,$$ $h_0(x_m) = y_m = B_n(x_m),$ $||h_0||_{H^{\infty}} \leq A(q),$ where $$A(q) \rightarrow 1$$ as $q \rightarrow 0$ [3]. 240 Peyo Stoilov Since $B_n \in b_n(E_{\xi})$ , then $$\Omega_n(E_{\xi}) \ge \|T_{B_n}\|_{H^{\infty}} \ge \frac{1}{A(q)} \frac{1}{2\pi} \left| \int_T \frac{h_0(\varsigma)}{B_n(\varsigma)(\varsigma - x_m)} d\varsigma \right| =$$ $$= \frac{1}{A(q)} \left| \frac{h_0(x_m)}{B_n(x_m)} + \sum_{k=1}^n \frac{h_0(x_k)}{B'_n(x_k)(x_k - x_m)} \right| = \frac{1}{A(q)} \left| 1 + \sum_{k=1}^n \frac{|x_k|^2 - 1}{|x_k| - |x_m|} \right|.$$ Since m > n can be every arbitrary long positive integer and $|x_m| \to 1$ as $m \to \infty$ , then $$\Omega_n(E_{\xi}) \ge \frac{1}{A(q)} \sum_{k=1}^n (1 + |x_k|) = \frac{1}{A(q)} \left( 1 + 2n - \sum_{k=1}^n q^k \right).$$ Using the fact that $A(q) \to 1$ as $q \to 0$ , we obtain $\Omega_n(E_{\xi}) \ge 1 + 2n$ . This implies $\Omega_n(D) = \Omega_n(E_{\xi}) = 1 + 2n$ for all $\xi \in T$ .