Univalent approximation by Fourier series of step functions

Theorem 1. If $F \in S_H^o$ , then F can be approximated uniformly on compact subsets by functions in $B_n^o$ . It may seem that Theorem 1 should be proved simply by interpolating F(tz)/t by Poisson integrals of step functions. However, the task of making such approximations univalent in U seems daunting. Therefore, we shall use the tools involving first order elliptic systems as developed in [1] and [10]. Consider a first order system $$(2.1) u_x = a_{11}(x,y)v_x + a_{12}(x,y)v_y, -u_y = a_{21}(x,y)v_x + a_{22}(x,y)v_y.$$ By a theorem of Bers and Nirenberg [1, p.132] we have Theorem A. If the coefficients of the system (2.1) are defined and continuous in a Jordan domain D and if the coefficients satisfy the ellipticity conditions $$(2.2) 4a_{12}a_{21} - (a_{11} + a_{22})^2 > 0 and a_{12} > 0,$$ then, given a Jordan domain $\Lambda$ , there exists a solution w = u + iv which is a homeomorphism of the closure of D onto the closure of $\Lambda$ and takes three given boundary points of D onto three assigned boundary points on the boundary of $\Lambda$ . This will be supplemented with a result of Mcleod, Gergen, and Dressel [10, p. 174]. Theorem B. Let D be bounded by a continuously differentiable Jordan curve and suppose that the coefficients $a_{ij}$ in (2.1) are continuously differentiable in D and continuous on $\overline{D}$ . Assume the system satisfies the ellipticity conditions (2.2), and let $u_1$ , $v_1$ and $u_2$ , $v_2$ be solution pairs of the system. If $F_1 = u_1 + iv_1$ and $F_2 = u_2 + iv_2$ are both continuously differentiable on D and continuous on $\overline{D}$ , mapping D homeomorphically onto a set T such that three distinct points on $\partial D$ correspond to the same three points on $\partial T$ , then $F_1 \equiv F_2$ in D. We shall apply Theorem A and Theorem B to univalent harmonic mappings in U which satisfy (2.3) $$\overline{f_{\overline{z}}(z)} = a(z)f_z(z),$$ where the dilatation a(z) is analytic in U and satisfies the condition $|a(z)| \le k < 1$ . Writing f = u + iv and $a = a_1 + ia_2$ with $|a(z)| \le k < 1$ in U, (2.3) becomes $$u_x - iv_x - i(u_y - iv_y) = (a_1 + ia_2)(u_x + iv_x - i(u_y + iv_y))$$ so that $$u_x - v_y = a_1(u_x + v_y) - a_2(v_x - u_y)$$ - $u_y - v_x = a_1(v_x - u_y) + a_2(u_x + v_y)$ . Solving these equations we obtain $$u_x = \frac{2a_2}{-a_2^2 - (1 - a_1)^2} v_x + \frac{a_1^2 - 1 + a_2^2}{-a_2^2 - (1 - a_1)^2} v_y$$ $$-u_y = \frac{-a_2^2 + 1 - a_1^2}{a_2^2 + (1 - a_1)^2} v_x + \frac{2a_2}{a_2^2 + (1 - a_1)^2} v_y,$$ and thus the conditions (2.2) are satisfied. Our procedure will also rely on the work of Hengartner and Schober; in particular [7, Theorem 4.3]. Theorem C. Let D be a bounded simply connected domain whose boundary is locally connected. Suppose that f is a univalent harmonic orientation preserving mapping from U into D for which the radial limits $\hat{f}(e^{i\theta}) = \lim_{r\to 1} f(re^{i\theta})$ belong to $\partial D$ for almost every $\theta$ . Then there exists a countable set $E \subseteq \partial U$ such that - (a) the unrestricted limit $\hat{f}(e^{i\theta}) = \lim_{z \to e^{i\theta}} \int_{z \in U} f(z) exists$ , is continuous, and belongs to $\partial D$ for $e^{i\theta} \in \partial U \setminus E$ , - (b) $\lim_{\theta \uparrow \theta_0} \hat{f}(e^{i\theta})$ and $\lim_{\theta \downarrow \theta_0} \hat{f}(e^{i\theta})$ exist, are different, and belong to $\partial D$ for $\varepsilon^{i\theta} \in E$ , (c) the cluster set of f at $e^{i\theta} \in E$ is a straight line segment joining $\lim_{\theta \uparrow \theta_0} \hat{f}(e^{i\theta})$ and $\lim_{\theta \downarrow \theta_0} \hat{f}(e^{i\theta})$ . Proof of Theorem 1. With F as in Theorem 1 having dilatation A(z) and 0 < t < 1, let $f(z) = f_t(z) = F(tz)/t$ so that $f(z) \in S_H^o$ having dilatation $a(z) = a_t(z) = A(tz) \le k$ for some k < 1. As such, the image $\Omega = \Omega_t = f(U)$ is a bounded domain with real analytic boundary. It suffices to approximate a fixed such f(z) by taking t close to 1. We fix three points $z_1$ , $z_2$ , $z_3$ positively oriented on $\partial U$ , and the corresponding points $w_i = f(z_i)$ j = 1, 2, 3 on $\partial \Omega$ . Since $\partial\Omega$ is smooth, we may take a sequence of Jordan polygons $P_n$ interior to $\Omega$ having vertices $W_n = \{w_{n,1}, ... w_{n,k_n}\}$ such that $P_1 \subset P_2 \subset ...$ , and having vertices with $w_1, w_2, w_3$ in such a way that $P_n \to \Omega$ in the sense that $\operatorname{dist}(\partial\Omega, \partial P_n) \to 0$ , $\operatorname{dist}(w_{n,j}, w_{n,j+1}) \to 0$ $(j = 1, ..., k_n, w_{n,k_n+1} = w_{n,1})$ uniformly as $n \to \infty$ , and the lengths of the $\partial P_n$ are uniformly bounded. We note that in the continuation, the term "vertices' can also refer to interior points of the line segments of the polygon. We next take a sequence of Blaschke products $a_n(z)$ converging uniformly on compact subsets of U (in fact pointwise in U) to a(z) [6, p. 7]. If $a_{n,\rho}(z) = a_n(\rho z)$ (0 < $\rho$ < 1), then by Theorem A, there exists a homeomorphism $f_{n,\rho}(z)$ of U onto $P_n$ with $f_{n,\rho}(z_j) = w_j$ , j = 1, 2, 3 and satisfying (2.3) with dilatation $a_{n,\rho}$ . In fact, since each $a_{n,\rho}$ satisfies $|a_{n,\rho}(z)| < k_{n,\rho} < 1$ for some constants $k_{n,\rho}$ and all $z \in U$ , the $f_{n,\rho}$ in (2.3) are univalent harmonic mappings (cf. [4, p.6]). Letting $\rho \to 1$ , we can take subsequence which converges to a univalent harmonic mapping $f_n$ of U into $P_n$ , having dilatation $a_n(z)$ . In fact the functions $f_{n,\rho}(z)$ are Poisson integrals of some boundary functions $\varphi_{n,\rho}(e^{i\theta})$ of uniformly bounded variation so that a subsequence converges to a function $\varphi_n(e^{i\theta})$ a.e., [5, p.3] and the limit function $f_n$ is also a Poisson integral of a radial limit function $\psi_n(e^{i\theta})$ . Thus $\psi_n(e^{i\theta}) = \varphi_n(e^{i\theta})$ a.e., and consequently $f_n(z)$ has radial limits on $\partial P_n$ , a.e., and Theorem C applies. It is important to emphasize that the functions $f_n$ are in B(n) since the dilatations are Blaschke products and there can be no nonconstant intervals of continuity on $\partial U$ , since otherwise there would be an interval which is mapped onto a line segment which is not possible since the image of such an interval has to be strictly concave with respect to the interior (cf. [4, p. 116]. The 3 specified boundary points need not correspond. They either reside on the images of arcs where $f_n$ is constant, or points of discontinuity which create the "collapsing line segments." Again, the functions $f_n(z)$ are Poisson integrals of sense preserving step functions $\varphi_n$ that have their values in the $P_n$ respectively. We now take a subsequence of the $f_n$ converging to a function $\tilde{f}$ thus having dilatation a(z) Let $\tilde{f}_0$ denote the a.e. radial limit function for $\tilde{f}$ . Then, (2.4) $$\tilde{f}(z) = \frac{1}{2\pi} \int_{U} P(r, \theta - t) \tilde{f}_0(e^{it}) dt.$$ Since the functions $\{\varphi_n\}$ are of uniformly bounded variation, as before there exists a function $\varphi$ on $\partial U$ and a subsequence $\{\varphi_{n_k}\}$ converging a.e. to $\varphi$ . Therefore, $\varphi(e^{i\theta}) = \tilde{f}_0(e^{i\theta})$ a.e., so that in particular, the values taken by $\tilde{f}_0(e^{i\theta})$ are a.e. in $\partial\Omega$ . Now $\tilde{f}(z)$ in (2.4) satisfies the conditions of Theorem C. Since $\tilde{f}$ has dilatation bounded strictly less than 1 it is quasiconformal in U and hence a homeomorphism on $\overline{U}$ [9, p.98]. This implies that the set E in Theorem C is empty, and thus it follows that $\hat{f}$ must map $\partial U$ homeomorphically onto $\partial D$ . From our construction it follows further that $\hat{f}(z_j) = w_j$ j = 1, 2, 3. Thus, from Theorem B we conclude that $\tilde{f}(z) \equiv f(z)$ . The functions $\{f_n\}$ that approximate f are in B(n). Set $f_n(z) = \sum_{k=0}^{\infty} a_{nk} z^k + \sum_{k=1}^{\infty} \overline{b_{nk} z^k}$ . Since the $f_n$ converge locally uniformly to f(z), we infer that $a_{n0}, a_{n1}, b_{n1}$ converge to 0, 1, 0 respectively. Accordingly the functions $$g_n(z) = \frac{\overline{a}_{n1}(f_n(z) - a_{n0}) - \overline{b}_{n1}(\overline{f_n(z) - a_0^n})}{|a_n^n|^2 - |b_n^n|^2} \in B_n^0$$ are univalent harmonic mappings converging locally uniformly to f(z). This completes the

Theorem 2. If $f = h + g \in \overline{B_n^o}$ and h has a pole at $\zeta \in \partial U$ , then the order of the pole is at most 3. Proof of Theorem 2. Arguing by contradiction, we assume that h has a pole of order k at least 4, and we consider only even k. The odd case is similar. We may assume that $\zeta = 1$ . As described above, we then have $$\begin{split} w &= f(z) = \frac{e^{i\alpha}}{(z-1)^k} + \frac{e^{i\beta}}{(\overline{z}-1)^k} + \text{lower order terms} \\ &= e^{i(\alpha+\beta)/2} \left( \frac{e^{i(\alpha-\beta)/2}}{(z-1)^k} + \frac{e^{i(-\alpha+\beta)/2}}{(\overline{z}-1)^k} \right) + \text{lower order terms} \\ &= 2e^{i(\alpha+\beta)/2} \Re e^{\frac{e^{i(\alpha-\beta)/2}}{(z-1)^k}} + \text{lower order terms}. \end{split}$$ Writing $z-1=re^{i\varphi}$ and $(\alpha-\beta)/2=\varphi_0$ , $(\alpha+\beta)/2=\varphi_1$ we have (3.1) $$w = f(z) = 2e^{i\varphi_1} \frac{\cos k(\varphi - \varphi_0/k)}{r^k} + \text{lower order terms.}$$ By a rotation we may ignore the term $e^{i\varphi_1}$ in (3.1). We require some notation. Let $\varepsilon > 0$ , and $0 < \delta < 1$ . Let $\Delta = \Delta(\varepsilon, \delta)$ be the portion of U between $|z - 1| = \varepsilon$ and $|z - 1| = \varepsilon^{\delta}$ . The boundary of $\Delta$ is a simple closed curve and, for small $\varepsilon$ , $\varphi$ ranges on an interval only slightly smaller than $(\pi/2, 3\pi/2)$ . Therefore, since $k \geq 4$ , on the side where $|z - 1| = \varepsilon$ there will be at least 3 consecutive intervals of the form $\alpha < k(\varphi - \varphi_0/k) < \alpha + \pi$ . Let $I_0$ be the middle one of a set of 3 consecutive intervals, and $$W = \{w : |\Re e \, w| < \varepsilon^{-(1+\delta)/2}.$$ For small $\varepsilon$ , a portion of $f(I_0)$ extends outside of W and the rest inside. We assume that it is on the right side; the proof for the left side would be similar. As $\varphi$ increases, the portion of $f(I_0)$ outside W has an initial value $x+iy_1$ and terminal value $x+iy_2$ . Since $f(\partial \Delta)$ is sense preserving, it must be that $y_1 > y_2$ . Regarding the images of the two intervals adjacent to $I_0$ , again because of the sense preserving nature of $f(\partial \Delta)$ , the portions of their images as they exit and reenter W on the left side must turn towards each other. This means that there can be no accommodation for another portion of the image of $f(\partial \Delta)$ to exit again on the right side without crossing. Thus it cannot be that $k \geq 4$ ..