On a multiplier operator induced by the Schwarzian derivative of univalent functions

Theorem 1.1. Let $\alpha > 1$ . Let f be a univalent function in $\Delta$ admitting a quasiconformal extension to $\widehat{\mathbb{C}}$ with $\|\mu_f\|_{\infty} = k \in [0,1)$ . Then, for any $\phi \in \mathcal{H}_{\alpha}(\Delta)$ , we have $$||M_f(\phi)||_{\alpha+4}^2 = ||S_f(z)\phi(z)||_{\alpha+4}^2 \le \frac{36(\alpha+1)k^2}{(\alpha-1)}||\phi(z)||_{\alpha}^2.$$ Remark 1.2. Note that, for a univalent function f in $\Delta$ admitting a quasiconformal extension to $\widehat{\mathbb{C}}$ with $\|\mu_f\|_{\infty} = k \in [0,1)$ , we have $$\sum_{m=1}^{\infty} m \left| \sum_{n=1}^{\infty} \gamma_{mn} \lambda_n \right|^2 \le k^2 \sum_{n=1}^{\infty} \frac{|\lambda_n|^2}{n}, \ \lambda_n \in \mathbb{C},$$ where $\gamma_{mn}$ are the Grunsky's coefficients of f, see [22, Chapter 9]. Then Theorem 1.1 follows from the arguments in [26]. We use Theorem 1.1 to show that

Theorem 1.3. Let f be a univalent function in $\Delta$ admitting a quasiconformal extension to $\widehat{\mathbb{C}}$ . If $\|\mu_f\|_{\infty} \leq \sqrt{\frac{5}{8}} \approx 0.79056$ , then $\beta_f(-2) \leq 1$ and the Brennan conjecture is satisfied for the domain $f(\Delta)$ . Remark 1.4. Let $t \in \mathbb{R}$ . In [15], Hedenmalm proved that, for a univalent function f admitting a quasiconformal extension to $\widehat{\mathbb{C}}$ with $\|\mu_f\|_{\infty} = k \in (0,1)$ , one has $$\beta_f(t) \le \frac{1}{4}k^2|t|^2(1+7k)^2$$ , when $|t| \le \frac{2}{k(1+7k)^2}$ , and $$\beta_f(t) \le k|t| - \frac{1}{(1+7k)^2}$$ , when $|t| \ge \frac{2}{k(1+7k)^2}$ . We consider t = -2. - (1) When $k(1+7k)^2 \le 1$ , i.e., $k \in (0, k_0]$ , here $k_0 \approx 0.18726$ is the real root of the equation $k(1+7k)^2 = 1$ . We see that $\beta_f(-2) \le 1$ in this case. - (2) When $k(1+7k)^2 \ge 1$ , i.e., $k \in [k_0, 1)$ , we have $$\beta_f(-2) \le 2k - \frac{1}{(1+7k)^2} = \frac{2k(1+7k)^2 - 1}{(1+7k)^2}.$$ Thus, if $k_0 \le k < 1$ and $$\frac{2k(1+7k)^2 - 1}{(1+7k)^2} \le 1,$$ i.e., $k_0 \le k \le k_1 \approx 0.52301$ , here $k_1$ is the real root of the equation $$\frac{2k(1+7k)^2 - 1}{(1+7k)^2} = 1,$$ then we have $\beta_f(-2) \leq 1$ . Consequently, we see that, if $0 \le k \le k_1 \approx 0.52301$ , then we have $\beta_f(-2) \le 1$ . Hence Theorem 1.3 provides an improvement of the results in [15]. The paper is organized as follows. We will give the proof of Theorem 1.3 in the next section. By refining results in [26], we will show in Section 3 that the Brennan conjecture is satisfied for another class of quasidisks. In Section 4, we establish a new characterization of asymptotically conformal curves and of the Weil-Petersson curves in terms of the multiplier operator. We will present some remarks in Section 5.

Proposition 2.1. If f is a univalent function in $\Delta$ and $$||S_f(z)\phi(z)||_{\alpha+4}^2 \le \frac{36(\alpha+1)(\alpha+3)}{\alpha(\alpha+2)} ||\phi(z)||_{\alpha}^2$$ holds for any $\alpha > 2$ and $\phi \in \mathcal{H}_{\alpha}(\Delta)$ , then $\beta_f(-2) \leq 1$ . Remark 2.2. Proposition 2.1 is Proposition 8 of [26], where it is shown that under the assumption, $(f')^{-1} \in \mathcal{H}_{\alpha}(\Delta)$ for any $\alpha > 2$ . <span id="page-3-0"></span>The following lemma will be used later.

Lemma 2.3. Let f be a univalent function in $\Delta$ admitting a quasiconformal extension to $\widehat{\mathbb{C}}$ . If g is another univalent function from $\Delta$ to $f(\Delta)$ , then g admits a quasiconformal extension to $\widehat{\mathbb{C}}$ with $\|\mu_g\|_{\infty} = \|\mu_f\|_{\infty}$ .

Theorem 3.1. Let f be a univalent function in $\Delta$ admitting a quasiconformal extension to $\widehat{\mathbb{C}}$ . Let h(f) be the boundary dilatation of f. If $h(f) \leq \sqrt{\frac{3}{11}} \approx 0.52223$ , then $\beta_f(-2) \leq 1$ and the Brennan conjecture is satisfied for the domain $f(\Delta)$ . For the proof of Theorem 3.1, we need the following refinement of Proposition 2.1.

Proposition 3.2. Let f be a univalent function in $\Delta$ , $r \in (0,1)$ and let $f_r(z) = f(rz)$ . If, for any $\alpha > 2$ , there is a constant $R = R(f, \alpha) \in (0,1)$ such that for any $r \in (0,1)$ , $$\iint_{A_R} |S_{f_r}(z)\phi(z)|^2 (1-|z|^2)^{\alpha+2} dx dy \le \frac{36(\alpha+1)}{\alpha(\alpha+2)} \|\phi(z)\|_{\alpha}^2$$ holds for all $\phi \in \mathcal{H}_{\alpha}(\Delta)$ , then $\beta_f(-2) \leq 1$ . Here, $A_R := \Delta \setminus \Delta(0,R)$ is an annulus.

Lemma 3.3. A function $\phi \in \mathcal{H}_{\alpha}(\Delta)$ if and only if $\phi' \in \mathcal{H}_{\alpha+2}(\Delta)$ . Moreover, for any $\varepsilon$ such that $0 < \varepsilon < \alpha(\alpha+1)$ , <span id="page-4-3"></span>(3.2) $$\|\phi'\|_{\alpha+2}^2 \le [\alpha(\alpha+1)+\varepsilon] \|\phi\|_{\alpha}^2 + C_1(\phi,\varepsilon);$$ <span id="page-4-1"></span>(3.3) $$\|\phi\|_{\alpha}^{2} \leq \frac{1}{[\alpha(\alpha+1)-\varepsilon]} \|\phi'\|_{\alpha+2}^{2} + C_{2}(\phi,\varepsilon),$$ where the constants $C_1(\phi, \varepsilon)$ and $C_2(\phi, \varepsilon)$ depend only on finitely many first Taylor coefficients of the function $\phi$ . We begin the proof of Proposition 3.2. In view of (1.2), we see that it is enough to show that $(f')^{-1} \in \mathcal{H}_{\alpha}(\Delta)$ for any fixed $\alpha > 2$ . First, we have (3.4) $$-\frac{d^3}{dz^3} \left[ (f')^{-1} \right] = \frac{d}{dz} \left[ S_f(z)(f')^{-1} \right] + S_f(z) \frac{d}{dz} \left[ (f')^{-1} \right].$$ Also, for $r \in (0,1)$ , $\phi \in \mathcal{A}(\Delta)$ , it follows from the assumption that there is a constant R > 0 such that <span id="page-4-2"></span> $$\iint_{A_R} |S_{f_r}(z)\phi(rz)|^2 (1-|z|^2)^{\alpha+2} dx dy \le \frac{36(\alpha+1)}{\alpha(\alpha+2)} \|\phi(rz)\|_{\alpha}^2.$$ Hence <span id="page-4-4"></span> $$||S_{f_r}(z)\phi(rz)||_{\alpha+4}^2 = (\alpha+3) \iint_{\Delta} |S_{f_r}(z)\phi(rz)|^2 (1-|z|^2)^{\alpha+2} dx dy$$ $$\leq \frac{36(\alpha+1)(\alpha+3)}{\alpha(\alpha+2)} ||\phi(rz)||_{\alpha}^2 + C_3(\phi,\alpha,R).$$ We let $$\mathbf{A}(\alpha) := \frac{36(\alpha+1)(\alpha+3)}{\alpha(\alpha+2)},$$ and we use $(f'_r)^{-1}$ to denote $[f'_r(z)]^{-1}$ for simplicity. Note that $(f'_r)^{-1} = [rf'(rz)]^{-1}$ . Then, by using (3.3) three times for the function $[f'(rz)]^{-1}$ , we see from (3.4) that <span id="page-4-5"></span> $$[\alpha(\alpha+1)(\alpha+2)(\alpha+3)(\alpha+4)(\alpha+5) - \varepsilon] \|(f_r')^{-1}\|_{\alpha}^{2}$$ $$\leq \|\frac{d^{3}}{dz^{3}} [(f_r')^{-1}] \|_{\alpha+6}^{2} + \frac{1}{r^{2}} C_{4}(f,\varepsilon)$$ $$(3.6) \leq \left( \left\| \frac{d}{dz} \left[ S_{f_r}(z) (f_r')^{-1} \right] \right\|_{\alpha+6} + \left\| S_{f_r}(z) \frac{d}{dz} \left[ (f_r')^{-1} \right] \right\|_{\alpha+6} \right)^2 + \frac{1}{r^2} C_4(f, \varepsilon).$$ Here $\varepsilon > 0$ is a small number. On the other hand, since $S_{f_r}(z)(f'_r)^{-1} = rS_f(rz)[f'(rz)]^{-1}$ , we see from (3.2) and (3.5) that <span id="page-5-0"></span> $$\left\| \frac{d}{dz} \left[ S_{f_r}(z)(f_r')^{-1} \right] \right\|_{\alpha+6}$$ $$\leq \sqrt{(\alpha+4)(\alpha+5) + \varepsilon} \left\| S_{f_r}(z)(f_r')^{-1} \right\|_{\alpha+4} + C_5(f,\varepsilon)$$ $$\leq \sqrt{(\alpha+4)(\alpha+5) + \varepsilon} \cdot \sqrt{\mathbf{A}(\alpha) \| (f_r')^{-1} \|_{\alpha}^2 + \frac{1}{r^2} C_6(f,\alpha)} + C_5(f,\varepsilon)$$ $$\leq \sqrt{\mathbf{A}(\alpha) [(\alpha+4)(\alpha+5) + \varepsilon]} \| (f_r')^{-1} \|_{\alpha} + \frac{1}{r} C_7(f,\alpha,\varepsilon).$$ $$(3.7)$$ In the second inequality of (3.7) we have used $[f'(rz)]^{-1}$ instead of $\phi(rz)$ . Since $[(f'_r)^{-1}]' = f''(rz)[f'(rz)]^{-2}$ , then, from (3.5) and (3.2) again, we obtain <span id="page-5-1"></span> $$\left\| S_{f_r}(z) \frac{d}{dz} \left[ (f'_r)^{-1} \right] \right\|_{\alpha+6} \leq \sqrt{\mathbf{A}(\alpha+2)} \left\| \frac{d}{dz} \left[ (f'_r)^{-1} \right] \right\|_{\alpha+2}^2 + C_8(f,\alpha) \\ \leq \sqrt{\mathbf{A}(\alpha+2)} \left\{ \left[ \alpha(\alpha+1) + \varepsilon \right] \| (f'_r)^{-1} \|_{\alpha}^2 + \frac{1}{r^2} C_9(f,\varepsilon) \right\} + C_8(f,\alpha) \\ \leq \sqrt{\mathbf{A}(\alpha+2)} \left[ \alpha(\alpha+1) + \varepsilon \right] \| (f'_r)^{-1} \|_{\alpha} + \frac{1}{r} C_{10}(f,\alpha,\varepsilon). \\$$ (3.8) In (3.7) and (3.8) we have used that $$\sqrt{A+B} < \sqrt{A} + \sqrt{B}, A > 0, B > 0.$$ Thus, combining (3.6), (3.7), (3.8), we obtain $$[\alpha(\alpha+1)(\alpha+2)(\alpha+3)(\alpha+4)(\alpha+5) - \varepsilon] \| (f'_r)^{-1} \|_{\alpha}^{2}$$ $$\leq \left[ \sqrt{\mathbf{A}(\alpha)[(\alpha+4)(\alpha+5) + \varepsilon]} + \sqrt{\mathbf{A}(\alpha+2)[\alpha(\alpha+1) + \varepsilon]} \right]^{2} \| (f'_r)^{-1} \|_{\alpha}^{2}$$ $$+ \frac{1}{r} C_{11}(f, \alpha, \varepsilon) \| (f'_r)^{-1} \|_{\alpha} + \frac{1}{r^{2}} C_{12}(f, \alpha, \varepsilon).$$ Let $$\mathbf{B}(\alpha,\varepsilon) := \alpha(\alpha+1)(\alpha+2)(\alpha+3)(\alpha+4)(\alpha+5) - \varepsilon;$$ $$\mathbf{C}(\alpha,\varepsilon) := \left[ \sqrt{\mathbf{A}(\alpha)[(\alpha+4)(\alpha+5) + \varepsilon]} + \sqrt{\mathbf{A}(\alpha+2)[\alpha(\alpha+1) + \varepsilon]} \right]^2.$$ It is not difficult to see that for $\alpha > 2$ and $\varepsilon$ small enough. $$\mathbf{D}(\alpha, \varepsilon) = \mathbf{B}(\alpha, \varepsilon) - \mathbf{C}(\alpha, \varepsilon) > 0.$$ Hence <span id="page-5-2"></span> $$(3.9) \quad \mathbf{D}(\alpha, \varepsilon) \| (f_r')^{-1} \|_{\alpha}^2 \le \frac{1}{r} C_{11}(f, \alpha, \varepsilon) \| (f_r')^{-1} \|_{\alpha} + \frac{1}{r^2} C_{12}(f, \alpha, \varepsilon).$$ We conclude that there exist $\mathcal{R} \in (0,1)$ , $\mathcal{M} > 0$ , such that $\|(f'_r)^{-1}\|_{\alpha}^2 \leq \mathcal{M}$ when $r > \mathcal{R}$ . Otherwise, there is an increasing sequence $\{r_n\}$ with $r_n < 1$ and $r_n \to 1$ as $n \to \infty$ , such that $\|(f'_{r_n})^{-1}\|_{\alpha}^2 \to \infty$ . This contradicts the inequality (3.9). On the other hand, by Fatou's lemma, we have $\|(f')^{-1}\|_{\alpha}^2 \leq \underline{\lim}_{r \to 1^-} \|(f'_r)^{-1}\|_{\alpha}^2$ . Consequently, we see that $(f')^{-1} \in \mathcal{H}_{\alpha}(\Delta)$ for any fixed $\alpha > 2$ , which finishes the proof. 3.3. Proof of Theorem 3.1. To prove Theorem 3.1, we will use the following key lemma.

Lemma 3.4. Let f be a univalent function in $\Delta$ , which has a quasiconformal extension to $\widehat{\mathbb{C}}$ . For $r \in (0,1)$ , let $f_r(z) = f(rz)$ and h(f) be the boundary dilatation of f. Then, for any $\varepsilon \in (0,1-h(f))$ there is a constant $R \in (0,1)$ such that for any $r \in (0,1)$ , $$\iint_{A_R} |S_{f_r}(z)\phi(z)|^2 (1-|z|^2)^{\alpha+2} dx dy \le \frac{36(h(f)+\varepsilon)^2}{(\alpha-1)[1-(h(f)+\varepsilon)^2]} \|\phi(z)\|_{\alpha}^2$$ holds for any $\alpha > 1$ and $\phi \in \mathcal{H}_{\alpha}(\Delta)$ . Here, $A_R = \Delta \setminus \Delta(0,R)$ is an annulus. Proof of Lemma 3.4. We need an integral expression of the Schwarzian derivative of a univalent function which can be extended to a quasiconformal mapping in $\widehat{\mathbb{C}}$ . This integral expression has appeared in [1]. For the completeness, we will give a detailed derivation of this integral expression and clarify some arguments presented in [1]. Let $\bar{f} = \tau \circ f$ , here $\tau(z) = \frac{1}{f'(0)}[z - f(0)]$ . Then we have $\bar{f}(0) = 0$ , $\bar{f}'(0) = 1$ . We assume that $\bar{f}(z)$ have the series expansion at origin as $$\bar{f}(z) = z + a_2 z^2 + a_3 z^3 + \cdots$$ The mapping $\hat{f} = \varsigma \circ \bar{f} \circ \varsigma$ , $\varsigma(z) = \frac{1}{z}$ is univalent (conformal) in $\Delta^* \setminus \{\infty\}$ , and has the series expansion at infinity $$\widehat{f}(z) = z + b_0 + \frac{b_1}{z} + \cdots.$$ It is easy to see that $b_0 = -a_2$ , $b_1 = a_2^2 - a_3$ . For any $z \in \Delta^* \setminus \{\infty\}$ , let $$\phi_z(w) = \frac{w+z}{1+\bar{z}w}.$$ The Koebe transformation $\mathcal{K}_{\widehat{f}}(w)$ of $\widehat{f}$ (see [22, page 21]) is defined as $$\mathcal{K}_{\widehat{f}}(w) = \frac{\widehat{f}(\phi_z(w)) - \widehat{f}(z)}{(1 - |z|^2)\widehat{f}'(z)}.$$ It follows that $F(w) = \varsigma \circ \mathcal{K}_{\widehat{f}} \circ \varsigma$ is univalent in $\Delta^* \setminus \{\infty\}$ and has a series expansion at infinity $$F(w) = w + c_0 + \frac{c_1}{w} + \cdots.$$ Then, by Pompieu's formula, for any $z \in \Delta^* \setminus \{\infty\}$ , we have $$F(z) = \frac{1}{2\pi i} \oint_{\Gamma} \frac{f(w)}{w-z} dw - \frac{1}{\pi} \iint_{\Delta(z,|z|+2)} \frac{\bar{\partial} F(w)}{w-z} du dv.$$ Here, $\Gamma = \partial \Delta(z, |z| + 2)$ is a circle. Since F is univalent in $\Delta^ \setminus \{\infty\}$ , it follows that $\bar{\partial} F(w) = 0$ when $w \in \Delta^ \setminus \{\infty\}$ . Then, by Laurent's theorem, we have $$F(z) = z + c_0 - \frac{1}{\pi} \iint_{\Delta} \frac{\bar{\partial} F(w)}{w - z} du dv, \ z \in \Delta^* \setminus \{\infty\}.$$ Consequently, <span id="page-7-0"></span>(3.10) $$\iint_{\Delta} \bar{\partial}F(w)dudv = \lim_{z \to \infty} z^2 \iint_{\Delta} \frac{\bar{\partial}F(w)}{(z-w)^2} dudv$$ $$= -\pi \lim_{z \to \infty} z^2 (F'(z) - 1)$$ $$= \pi c_1 = -\frac{\pi}{6} \lim_{z \to \infty} z^4 S_F(z).$$ Note that $F = \varsigma \circ \chi \circ \widehat{f} \circ \phi_z \circ \varsigma$ , where $$\chi(w) = \frac{w - \hat{f}(z)}{(1 - |z|^2)\hat{f}'(z)}.$$ Let $$\rho_z(w) := \phi_z \circ \varsigma(w) = \frac{1 + wz}{w + \bar{z}}.$$ From the transformation rule (1.1), it follows that $$S_F(w) = S_{\widehat{f}} \circ \rho_z(w) [\rho_z'(w)]^2.$$ Consequently, we see from $$\rho_z'(w) = \frac{|z|^2 - 1}{(w + \bar{z})^2},$$ and $\rho_z(w) \to z$ as $w \to \infty$ that $$S_{\widehat{f}}(z) = \lim_{w \to \infty} S_F(w) [\rho_z'(w)]^{-2} = (|z|^2 - 1)^{-2} \lim_{w \to \infty} w^4 S_F(w),$$ and then, in view of (3.10), we obtain $$S_{\widehat{f}}(z) = -\frac{6}{\pi} (|z|^2 - 1)^{-2} \iint_{\Delta} \bar{\partial} F(w) \, du dv$$ $$= -\frac{6}{\pi} (|z|^2 - 1)^{-2} \iint_{\Delta} \mu_F(w) \partial F(w) \, du dv.$$ It follows that $$\begin{split} |S_{\widehat{f}}(z)| & \leq & \frac{6}{\pi} (|z|^2 - 1)^{-2} \iint_{\Delta} |\mu_F(w)| |\partial F(w)| \, du dv \\ & = & \frac{6}{\pi} (|z|^2 - 1)^{-2} \iint_{\Delta} |\mu_F(w)| \left[ \frac{J_F(w)}{1 - |\mu_F(w)|^2} \right]^{\frac{1}{2}} \, du dv, \end{split}$$ where $J_F$ is the Jacobian of F. Hence, by Cauchy's inequality, we have <span id="page-7-1"></span> $$(3.11) |S_{\widehat{f}}(z)|^2 \le \frac{36}{\pi^2} (|z|^2 - 1)^{-4} \iint_{\Lambda} \frac{|\mu_F(w)|^2}{1 - |\mu_F(w)|^2} du dv \iint_{\Lambda} J_F(w) du dv.$$ By the well-known area theorem, we have $$\iint_{\Lambda} J_F(w) \, du dv \le \pi.$$ On the other hand, since $\mu_F(w) = \mu_{\widehat{f} \circ \rho_z}(w)$ , a change of variables in the first integral of inequality (3.11) gives $$|S_{\widehat{f}}(z)|^{2}(|z|^{2}-1)^{2} \leq \frac{36}{\pi} \iint_{\Delta} \frac{|\mu_{\widehat{f}}(\zeta)|^{2}}{1-|\mu_{\widehat{f}}(\zeta)|^{2}} \cdot \frac{d\xi d\eta}{|\zeta-z|^{4}}, \ z \in \Delta^{*} \setminus \{\infty\}.$$ From (1.1), we have <span id="page-8-0"></span> $$|S_{\widehat{f}}(z)| = |S_f(\frac{1}{z})| \frac{1}{|z|^4}, z \in \Delta^* \setminus \{\infty\}.$$ Thus, for any $z \in \Delta \setminus \{0\}$ , we have $$(3.12) |S_f(z)|^2 (1 - |z|^2)^2 \le \frac{36}{\pi} \iint_{\Lambda} \frac{|\mu_{\widehat{f}}(\zeta)|^2}{1 - |\mu_{\widehat{f}}(\zeta)|^2} \cdot \frac{d\xi d\eta}{|1 - \zeta z|^4}.$$ It is easy to see that (3.12) still holds for z=0. Hence (3.12) holds for all $z \in \Delta$ . For any $r \in (0,1)$ , we see from $S_{f_r}(z) = r^2 S_f(rz)$ that <span id="page-8-2"></span>(3.13) $$|S_{f_r}(z)|^2 (1 - |z|^2)^4 \le \frac{36r^4}{\pi} (1 - |z|^2)^2 \iint_{\Delta} \frac{|\mu_{\widehat{f}}(\zeta)|^2}{1 - |\mu_{\widehat{c}}(\zeta)|^2} \cdot \frac{d\xi d\eta}{|1 - r\zeta z|^4}, \ z \in \Delta.$$ Since $|\mu_f(z^{-1})| = |\mu_{\widehat{f}}(z)|$ , $z \in \Delta \setminus \{0\}$ , we see that for any $\varepsilon \in (0, 1 - h(f))$ there is a constant $r_0 \in (0, 1)$ such that $$|\mu_{\widehat{f}}(\zeta)| \le h(f) + \frac{\varepsilon}{2}$$ for all $r_0 \leq |\zeta| < 1$ . Let $$\mathbf{E}(r,z) := \iint_{\Delta} \frac{|\mu_{\widehat{f}}(\zeta)|^2}{1 - |\mu_{\widehat{f}}(\zeta)|^2} \cdot \frac{d\xi d\eta}{|1 - r\zeta z|^4}.$$ From $$|1 - r\zeta z|^4 > (1 - r_0)^4$$ , for any $|\zeta| < r_0, z \in \Delta$ , we see that $$\begin{split} \mathbf{E}(r,z) & \leq \frac{[h(f) + \frac{1}{2}\varepsilon]^2}{1 - [h(f) + \frac{1}{2}\varepsilon]^2} \iint_{A_{r_0}} \frac{d\xi d\eta}{|1 - r\zeta z|^4} \\ & + \frac{\|\mu_{\widehat{f}}\|_{\infty}^2}{1 - \|\mu_{\widehat{f}}\|_{\infty}^2} \cdot \iint_{\Delta(0,r_0)} \frac{d\xi d\eta}{|1 - r\zeta z|^4} \\ & \leq \frac{[h(f) + \frac{1}{2}\varepsilon]^2}{1 - [h(f) + \frac{1}{2}\varepsilon]^2} \iint_{\Delta} \frac{d\xi d\eta}{|1 - r\zeta z|^4} + \frac{\|\mu_{\widehat{f}}\|_{\infty}^2}{1 - \|\mu_{\widehat{f}}\|_{\infty}^2} \cdot \frac{\pi r_0^2}{(1 - r_0)^4} \\ & = \frac{[h(f) + \frac{1}{2}\varepsilon]^2}{1 - [h(f) + \frac{1}{2}\varepsilon]^2} \cdot \frac{\pi}{(1 - |rz|^2)^2} + \frac{\|\mu_{\widehat{f}}\|_{\infty}^2}{1 - \|\mu_{\widehat{f}}\|_{\infty}^2} \cdot \frac{\pi r_0^2}{(1 - r_0)^4}, \end{split}$$ where $A_{r_0} = \Delta \setminus \Delta(0, r_0)$ . It follows that <span id="page-8-1"></span> $$\frac{36r^{4}}{\pi}(1-|z|^{2})^{2} \iint_{\Delta} \frac{|\mu_{\widehat{f}}(\zeta)|^{2}}{1-|\mu_{\widehat{f}}(\zeta)|^{2}} \cdot \frac{d\xi d\eta}{|1-r\zeta z|^{4}}$$ $$\leq \frac{36r^{4}[h(f)+\frac{1}{2}\varepsilon]^{2}}{1-[h(f)+\frac{1}{2}\varepsilon]^{2}} \cdot \frac{(1-|z|^{2})^{2}}{(1-|rz|^{2})^{2}} + \frac{\|\mu_{\widehat{f}}\|_{\infty}^{2}}{1-\|\mu_{\widehat{f}}\|_{\infty}^{2}} \cdot \frac{36r^{4}r_{0}^{2}(1-|z|^{2})^{2}}{(1-r_{0})^{4}}$$ $$\leq \frac{36[h(f)+\frac{1}{2}\varepsilon]^{2}}{1-[h(f)+\frac{1}{2}\varepsilon]^{2}} + \frac{\|\mu_{\widehat{f}}\|_{\infty}^{2}}{1-\|\mu_{\widehat{f}}\|_{\infty}^{2}} \cdot \frac{36(1-|z|^{2})^{2}}{(1-r_{0})^{4}}.$$ (3.14) We see from (3.14) that there is a constant $R \in (0,1)$ such that <span id="page-9-0"></span> $$(3.15) \ \frac{36r^4}{\pi} (1 - |z|^2)^2 \iint_{\Delta} \frac{|\mu_{\widehat{f}}(\zeta)|^2}{1 - |\mu_{\widehat{f}}(\zeta)|^2} \cdot \frac{d\xi d\eta}{|1 - r\zeta z|^4} \le \frac{36(h(f) + \varepsilon)^2}{1 - (h(f) + \varepsilon)^2}$$ holds for $R \leq |z| < 1$ . Here we have used that the function $x^2(1-x^2)^{-1}$ is increasing in [0,1) and $h(f) + \varepsilon < 1$ . Consequently, from (3.13) and (3.15), we find that $$\iint_{A_R} |S_{f_r}(z)\phi(z)|^2 (1-|z|^2)^{\alpha+2} dx dy \le \frac{36(h(f)+\varepsilon)^2}{(\alpha-1)[1-(h(f)+\varepsilon)^2]} \|\phi(z)\|_{\alpha}^2$$ holds for any $r \in (0,1)$ and $\phi \in \mathcal{H}_{\alpha}(\Delta)$ . The lemma is proved. The following variant of Lemma 2.3 can be established in analogous form, and will be stated without

Lemma 3.5. Let f be a univalent function in $\Delta$ admitting a quasiconformal extension to $\widehat{\mathbb{C}}$ . If g is another univalent function from $\Delta$ to $f(\Delta)$ , then g admits a quasiconformal extension to $\widehat{\mathbb{C}}$ with h(g) = h(f). We can now proceed with the proof of Theorem 3.1. We see from Proposition 3.2, Lemma 3.4 and Lemma 3.5 that, if the inequality (3.16) $$\frac{36h^2(f)}{(\alpha - 1)(1 - h^2(f))} \le \frac{36(\alpha + 1)}{\alpha(\alpha + 2)}$$ holds for any $\alpha > 2$ , then $\beta_f(-2) \leq 1$ . Meanwhile, it follows from (3.16) that <span id="page-9-2"></span> $$\frac{h^2(f)}{1-h^2(f)} \leq \frac{\alpha^2-1}{\alpha^2+2\alpha} = 1 - \frac{2\alpha+1}{\alpha^2+2\alpha}, \text{ and } \inf_{\alpha>2} \left[1 - \frac{2\alpha+1}{\alpha^2+2\alpha}\right] = \frac{3}{8}.$$ Consequently, if $$\frac{h^2(f)}{1 - h^2(f)} \le \frac{3}{8}, \text{ i.e., } 0 \le h(f) \le \sqrt{\frac{3}{11}} \approx 0.52223,$$ then the inequality (3.16) holds for any $\alpha > 2$ and $\beta_f(-2) \le 1$ . This finishes the

Theorem 4.2. Let $\alpha > 1$ . Let f be a univalent function from $\Delta$ to a bounded Jordan domain in $\mathbb{C}$ . - (I) $f(S^1)$ is an asymptotically conformal curve if and only if the multiplier operator $M_f$ , acting from $\mathcal{H}_{\alpha}(\Delta)$ to $\mathcal{H}_{\alpha+4}(\Delta)$ , is compact. Moreover, - (II) $f(S^1)$ is a Weil-Petersson curve if and only if the multiplier operator $M_f$ belongs to the Hilbert-Schmidt class. Proof of the sufficiency of (I) of Theorem 4.2. Suppose that $f(S^1)$ is an asymptotically conformal curve. To show that $M_f$ is compact operator, it is sufficient to show that $M_f(\psi_n) \to 0$ for each sequence $(\psi_n)$ which converges to zero weakly. It is easy to check that $(\psi_n)$ converges to zero weakly if and only if $(\psi_n)$ is bounded and $(\psi_n)$ converges to zero locally. On the other hand, we recall that $f(S^1)$ is an asymptotically conformal curve if and only if $S_f(z)(1-|z|^2)^2 \to 0$ , $|z| \to 1^-$ . Thus, for any $\varepsilon > 0$ , there exists some $r \in (0,1)$ such that $|S_f(z)(1-|z|^2)^2| < \varepsilon$ , when |z| > r. It follows that, for $\psi \in \mathcal{H}_{\alpha}(\Delta)$ , we have $$||M_f(\psi)||_{\alpha+4}^2 = (\alpha+3) \iint_{\Delta} |S_f(z)\psi(z)|^2 (1-|z|^2)^{\alpha+2} dx dy$$ $$= (\alpha+3) \iint_{\Delta} |S_f(z)|^2 (1-|z|^2)^4 |\psi(z)|^2 (1-|z|^2)^{\alpha-2} dx dy$$ $$\leq 36(\alpha+3) \iint_{|z|< r} |\psi(z)|^2 (1-|z|^2)^{\alpha-2} dx dy + (\alpha+3)\varepsilon^2 ||\psi||_{\alpha}^2.$$ Consequently, we see that $M_f(\psi_n) \to 0$ for each sequence $(\psi_n)$ which converges to zero weakly. The sufficiency of (I) is proved. Proof of the necessity of (I) of Theorem 4.2. If $M_f$ is a compact, we consider the function $$\psi_a(z) = \frac{(1-|a|^2)^{\frac{\alpha}{2}}}{(1-az)^{\alpha}}, \ a \in \Delta.$$ From [28, Lemma 4.2.2], we see that $\psi_a(z) \in \mathcal{H}_{\alpha}(\Delta)$ and that $\psi_a(z)$ tends to zero locally uniformly in $\Delta$ when $|a| \to 1^-$ . We conclude that $\psi_a$ converges to zero weakly, hence $M_f(\psi_a) \to 0$ as $|a| \to 1^-$ , i.e., <span id="page-10-3"></span>(4.1) $$\lim_{|a| \to 1^{-}} \iint_{\Lambda} |S_f(z)|^2 \frac{(1 - |a|^2)^{\alpha} (1 - |z|^2)^{\alpha+2}}{|1 - az|^{2\alpha}} dx dy = 0.$$ For $a \in \Delta$ , let $l \in (0,1)$ be such that the disk $\Delta(a, l(1-|a|)) = \{|z-a| \le l(1-|a|)\}$ is contained in $\Delta$ . Hence, for any $z \in \Delta(a, l(1-|a|))$ , <span id="page-10-1"></span> $$(4.2) (1-l)(1-|a|) \le 1-|z| \le (1+l)(1-|a|)$$ and <span id="page-10-2"></span> $$(4.3) (1-|a|) \le |1-az| \le (2+l)(1-|a|).$$ It follows from (4.2) and (4.3) that (4.4) $$\frac{(1-|a|^2)^{\alpha}(1-|z|^2)^{\alpha}}{|1-az|^{2\alpha}} \ge \frac{(1-l)^{\alpha}}{(2+l)^{2\alpha}}$$ holds for any $z \in \Delta(a, l(1-|a|))$ . On the other hand, since $|S_f(z)|^2$ is a subharmonic function <span id="page-11-0"></span> $$|S_f(a)|^2 (1-|a|^2)^2 \le \frac{4}{\pi l^2} \iint_{|z-a|< l(1-|a|)} |S_f(z)|^2 dx dy.$$ It follows from (4.2) that <span id="page-11-1"></span> $$(4.5) \quad |S_f(a)|^2 (1-|a|^2)^4 \le \frac{16}{\pi l^2 (1-l)^2} \iint_{|z-a| < l(1-|a|)} |S_f(z)|^2 (1-|z|^2)^2 dx dy.$$ Combining (4.4), (4.5), we see that there is a constant $C(l, \alpha) > 0$ such that $$(4.6) |S_f(a)|^2 (1-|a|^2)^4 \le C(l,\alpha) \iint_{\Lambda} |S_f(z)|^2 \frac{(1-|a|^2)^{\alpha} (1-|z|^2)^{\alpha+2}}{|1-az|^{2\alpha}} dx dy.$$ Thus, from (4.1), $S_f(a)(1-|a|^2)^2 \to 0$ , $|a| \to 1^-$ . Hence $f(S^1)$ is an asymptotically conformal curve. This finishes the proof of (I). For part (II), let $n \in \mathbb{N} \cup \{0\}$ and let $$e_n(z) = \sqrt{\frac{\Gamma(n+\alpha)}{n!\Gamma(\alpha)}} z^n, z \in \Delta.$$ Here, $\Gamma(s)$ stands for the usual Gamma function. It is easy to see that $\{e_n\}$ is an orthonormal set in $\mathcal{H}_{\alpha}(\Delta)$ . It is known that $M_f$ belongs to the Hilbert-Schmidt class if and only if $$\sum_{n=0}^{\infty} ||M_f(e_n)||_{\alpha+4}^2 < \infty.$$ Since $$\frac{1}{(1-|z|^2)^{\tau}} = \sum_{r=0}^{\infty} \frac{\Gamma(\tau+n)}{n!\Gamma(\tau)} |z|^{2n}, \ \tau > 0, \ z \in \Delta,$$ we have $$\sum_{n=0}^{\infty} ||M_f(e_n)||_{\alpha+4}^2$$ $$= (\alpha+3) \sum_{n=0}^{\infty} \iint_{\Delta} |S_f(z)|^2 \frac{\Gamma(n+\alpha)}{n!\Gamma(\alpha)} |z|^{2n} (1-|z|^2)^{\alpha+2} dx dy$$ $$= (\alpha+3) \iint_{\Delta} |S_f(z)|^2 \sum_{n=0}^{\infty} \frac{\Gamma(n+\alpha)}{n!\Gamma(\alpha)} |z|^{2n} (1-|z|^2)^{\alpha+2} dx dy$$ $$= (\alpha+3) \iint_{\Delta} |S_f(z)|^2 (1-|z|^2)^2 dx dy,$$ which shows that $$\iint_{\Lambda} |S_f(z)|^2 (1-|z|^2)^2 dx dy < \infty.$$ Thus, $f(S^1)$ is a Weil-Petersson curve if and only if the multiplier operator $M_f$ belongs to the Hilbert-Schmidt class. This finishes the proof of Theorem 4.2.

Proposition 5.1. Let f be a univalent function in $\Delta$ admitting a quasiconformal extension to $\widehat{\mathbb{C}}$ . If $\|\mu_f\|_{\infty} \leq \frac{\sqrt{1105}}{48} \approx 0.69253$ , then $\beta_f(-1) \leq \frac{1}{4}$ . We also have

Proposition 5.2. Let f be a univalent function in $\Delta$ admitting a quasiconformal extension to $\widehat{\mathbb{C}}$ . Let h(f) be the boundary dilatation of f. If $h(f) \leq \sqrt{\frac{65}{321}} \approx 0.44999$ , then $\beta_f(-1) \leq \frac{1}{4}$ . Proof of Proposition 5.2. By Lemma 3.4, we have, for any $\varepsilon_1 \in (0, 1 - h(f))$ , there is a constant $R \in (0, 1)$ such that for any $r \in (0, 1)$ , $$\iint_{A_R} |S_{f_r}(z)\phi(z)|^2 (1-|z|^2)^{\alpha+2} dx dy \le \frac{36(h(f)+\varepsilon_1)^2}{(\alpha-1)[1-(h(f)+\varepsilon_1)^2]} \|\phi(z)\|_{\alpha}^2$$ holds for any $\alpha > 1$ and $\phi \in \mathcal{H}_{\alpha}(\Delta)$ . Thus, for any $\phi \in \mathcal{A}(\Delta)$ , we have <span id="page-13-0"></span> $$||S_{f_r}(z)\phi(rz)||_{\alpha+4}^2 = (\alpha+3) \iint_{\Delta} |S_{f_r}(z)\phi(rz)|^2 (1-|z|^2)^{\alpha+2} dx dy$$ $$(5.2) \qquad \leq \frac{36(\alpha+3)(h(f)+\varepsilon_1)^2}{(\alpha-1)[1-(h(f)+\varepsilon_1)^2]} ||\phi(rz)||_{\alpha}^2 + C_{14}(\phi,\alpha,h(f),\varepsilon_1).$$ By letting $[f'(rz)]^{-1/2}$ to be instead of $\phi(rz)$ in (5.2), we have $$||S_{f_r}(z)(f_r')^{-1/2}||_{\alpha+4}^2 \le \frac{36(\alpha+3)(h(f)+\varepsilon_1)^2}{(\alpha-1)[1-(h(f)+\varepsilon_1)^2]}||(f_r')^{-1/2}||_{\alpha}^2 + \frac{1}{r}C_{15}(f,\alpha,\varepsilon_1).$$ It follows from (5.1) that, for small number $\varepsilon_2 > 0$ , $$\|(f_r')^{-1/2}\|_{\alpha}^2 \leq \mathbf{F}(h(f), \varepsilon_1)\mathbf{G}(\alpha, \varepsilon_2)\|(f_r')^{-1/2}\|_{\alpha}^2 + \frac{1}{r}C_{16}(f, \alpha, \varepsilon_1, \varepsilon_2),$$ where $$\mathbf{F}(h(f), \varepsilon_1) := \frac{[h(f) + \varepsilon_1]^2}{1 - [(h(f) + \varepsilon_1]^2},$$ $$\mathbf{G}(\alpha, \varepsilon_2) := \frac{9(\alpha + 3)}{(\alpha - 1)[\alpha(\alpha + 1)(\alpha + 2)(\alpha + 3) - \varepsilon_2]}.$$ On one hand, since $\mathbf{G}(\alpha, \varepsilon_2)$ is decreasing with respect to $\alpha$ when $\alpha > 1$ , hence for fixed $\alpha > \frac{5}{4}$ there is a constant $\theta > 0$ such that when $\varepsilon_2$ sufficiently small we have (5.3) $$\mathbf{G}(\alpha, \varepsilon_2) < \mathbf{G}(\frac{5}{4}, 0) - \theta = \frac{256}{65} - \theta.$$ On the other hand, since $x^2(1-x^2)^{-1}$ is increasing in [0,1) and $h(f) + \varepsilon_1 < 1$ , then when <span id="page-13-1"></span> $$h(f) \le \sqrt{\frac{65}{321}} \approx 0.44999$$ , i.e., $\frac{[h(f)]^2}{1 - [(h(f)]^2} \le \frac{65}{256}$ , and $\varepsilon_1$ sufficiently small we have (5.4) $$\mathbf{F}(h(f), \varepsilon_1) < \frac{65}{256} + (\frac{65}{256})^2 \theta.$$ Consequently, for $\alpha > \frac{5}{4}$ and $\varepsilon_1, \varepsilon_2$ sufficiently small, we see from (5.3) and (5.4) that <span id="page-13-2"></span> $$\mathbf{F}(h(f), \varepsilon_1)\mathbf{G}(\alpha, \varepsilon_2) < 1.$$ It follows that, for fixed $\alpha > \frac{5}{4}$ , $\|(f'_r)^{-1/2}\|_{\alpha}^2 \leq \mathcal{M}_2$ for some $\mathcal{M}_2 > 0$ when r is close enough to 1. Then, by Fatou's lemma, we see that $(f')^{-1/2} \in \mathcal{H}_{\alpha}(\Delta)$ for any fixed $\alpha > \frac{5}{4}$ . This implies that $\beta_f(-1) \leq \frac{1}{4}$ , The proof of Theorem 5.2 is finished. $\square$ Theorem 1.3 and 3.1 can be restated in the language of Teichmüller theory. We recall the definition of the universal Teichmüller space and the universal asymptotic Teichmüller space. For primary references, see [11, 8, 9, 10]. Let $M(\Delta^)$ denote the open unit ball of the Banach space $L^{\infty}(\Delta^)$ of essentially bounded measurable functions in $\Delta$ . For $\mu \in M(\Delta^)$ , let $f_{\mu}$ be the quasiconformal mapping in the extended complex plane $\widehat{\mathbb{C}}$ with complex dilatation equal to $\mu$ in $\Delta$ , equal to 0 in $\Delta$ , normalized $f_{\mu}(0) = 0$ , $f'_{\mu}(0) = 1$ , $f_{\mu}(\infty) = \infty$ . We say two elements $\mu$ and $\nu$ in $M(\Delta^)$ are equivalent, denoted by $\mu \sim \nu$ , if $f_{\mu}|_{\Delta} = f_{\nu}|_{\Delta}$ . The equivalence class of $\mu$ is denoted by $[\mu]_T$ . Then $T = M(\Delta^*)/\sim$ is one model of the universal Teichmüller space. The Teichmüller distance $d([\mu]_T, [\nu]_T)$ of two points $[\mu]$ , $[\nu]$ in T is defined as $$d([\mu]_T, [\nu]_T) = \frac{1}{2} \inf \left\{ \log \frac{1 + \|(\mu_1 - \nu_1)/(1 - \overline{\nu_1}\mu_1)\|_{\infty}}{1 - \|(\mu_1 - \nu_1)/(1 - \overline{\nu_1}\mu_1)\|_{\infty}}, |\mu_1|_T = [\mu]_T, [\nu_1]_T = [\nu]_T \right\}.$$ In particular, the distance between $[\mu]_T$ and the basepoint $[0]_T$ is $$d([\mu]_T, [0]_T) = \frac{1}{2} \log \frac{1 + k_0([\mu]_T)}{1 - k_0([\mu]_T)}, \ k_0([\mu]_T) = \inf\{\|\nu\|_{\infty}, \ \nu \sim \mu\}.$$ We say $\mu$ and $\nu$ in $M(\Delta^)$ are asymptotically equivalent if there exists some $\tilde{\nu}$ such that $\tilde{\nu}$ and $\nu$ are equivalent and $\tilde{\nu}(z) - \mu(z) \to 0$ as $|z| \to 1^+$ . The asymptotic equivalence of $\mu$ will be denoted by $[\mu]_{AT}$ . The universal asymptotic Teichmüller space AT is the set of all the asymptotic equivalence classes $[\mu]_{AT}$ of elements $\mu$ in $M(\Delta^)$ . The Teichmüller distance $d([\mu]_{AT}, [\nu]_{AT})$ of two points $[\mu]_{AT}, [\nu]_{AT}$ in AT is defined as $$d([\mu]_{AT}, [\nu]_{AT}) = \frac{1}{2} \inf \left\{ \log \frac{1 + \|(\mu_1 - \nu_1)/(1 - \overline{\nu_1}\mu_1)\|_{\infty}}{1 - \|(\mu_1 - \nu_1)/(1 - \overline{\nu_1}\mu_1)\|_{\infty}}, [\mu_1]_{AT} = [\mu]_{AT}, [\nu_1]_{AT} = [\nu]_{AT} \right\}.$$ In particular, the distance between $[\mu]_{AT}$ and the basepoint $[0]_{AT}$ is $$d([\mu]_{AT}, [0]_{AT}) = \frac{1}{2} \log \frac{1 + h_0([\mu]_{AT})}{1 - h_0([\mu]_{AT})}, \ h_0([\mu]_{AT}) = \inf\{h([\nu]_T), \ [\nu]_{AT} = [\mu]_{AT}\}.$$ Here. $$h([v]_T) := \inf\{h^*(\mu) : \mu \sim \nu\},\$$ and $$h^(\mu) = \inf\{\|\mu|_{\Delta^ \setminus E}\|_{\infty} : E \text{ is a compact set in } \Delta^*\}.$$ Remark 5.3. It is known that $h_0([\mu]_{AT}) = h([\mu]_T)$ and it is easy to see that $h([\mu]_T) = \inf\{h(f_\nu), \ \nu \sim \mu\}$ , here $h(f_\nu)$ is defined as in (3.1). Then, we can restate Theorem 1.3 and Theorem 3.1 as

Theorem 5.4. Let $\mu \in M(\Delta^)$ . Let $f_{\mu}$ be a quasiconformal mapping in the extended complex plane $\widehat{\mathbb{C}}$ with complex dilatation equal to $\mu$ in $\Delta$ , equal to 0 in $\Delta$ , normalized $f_{\mu}(0) = 0$ , $f'_{\mu}(0) = 1$ , $f_{\mu}(\infty) = \infty$ . If $$d([\mu]_T, [0]_T) \le \frac{1}{2} \log \frac{1 + \sqrt{5/8}}{1 - \sqrt{5/8}}, \ or \ d([\mu]_{AT}, [0]_{AT}) \le \frac{1}{2} \log \frac{1 + \sqrt{3/11}}{1 - \sqrt{3/11}}$$ then $\beta_{f_{\mu}}(-2) \leq 1$ and the Brennan conjecture is satisfied for the domain $f_{\mu}(\Delta)$ .