Harmonic Mappings onto Convex Domains

Lemma 2.1. Let $g_n$ and g belong to $L^1[0, 2\pi]$ , $$\lim_{n\to\infty} g_n = g \text{ a.e.}, \quad and$$ $$\int_{0}^{2\pi} |g_{n}(x)| dx \le M \quad \text{for all } n = 1, 2, 3, \dots$$ Then there is a subsequence $\{g_{n_k}\}$ of $\{g_n\}$ such that $g_{n_k}dx$ converges in the weak \* topology as $k \to \infty$ to gdx + ds, where ds is a singular measure on $[0, 2\pi]$ with respect to the Lebesgue measure dx.

Lemma 2.2. Assume that $$f_n(z) = \int_0^{2\pi} P(z, t) \hat{f}_n(e^{it}) dt$$ and that $$\lim_{n\to\infty} \hat{f}_n = g \quad \text{a.e.}$$ where $\hat{f}_n$ , $g \in L^1$ and $$\int_{0}^{2\pi} |\hat{f}_{n}(e^{it})| dt \leq M \text{ for all } n = 1, 2, 3, \dots$$ If $f_n$ converges locally uniformly on U to a function $f \in h^1$ , then $\hat{f} = g$

Theorem 2.3. Let D be a convex domain. Fix $w_0 \in D$ , and let $a \in H(U)$ satisfy $a(U) \subset U$ . Then there exists a univalent, harmonic, orientationpreserving mapping f with the following properties. - (a) $f(U) \subset D$ , $f(0) = w_0$ , and $f_z(0) > 0$ ; - (b) f is a solution of $\overline{f_z} = af_z$ ; (c) the limits $\lim_{r \to 1} f(re^{it})$ exist and belong to $\partial D$ for a.e. t.

Theorem 2.4. Let f be a univalent, harmonic, orientation-preserving mapping from U onto an unbounded convex domain D which is neither a strip nor a half-plane. Then - (a) $f \in h^1$ ; - (b) there is only one point $e^{i\lambda}$ that corresponds to $\infty$ : (c) $$f(z) = \int_0^{2\pi} P(z, t) \hat{f}(e^{it}) dt + AP(z, \lambda)$$ for some constant $A \in \mathbb{C}$ ; - (d) there is a countable set $E \subset \partial U \setminus \{e^{i\lambda}\}$ such that - (i) the unrestricted limit $\lim_{z \to 0} f(z)$ exists as $z \to e^{i\theta}$ , $z \in U$ , and is continuous for all points $e^{i\theta} \in \partial U \setminus [E \cup \{e^{i\lambda}\}]$ , (ii) $$\lim_{t \uparrow \theta} \hat{f}(e^{it})$$ and $\lim_{t \downarrow \theta} \hat{f}(e^{it})$ exist and are different for $e^{i\theta} \in E$ , (iii) and the cluster set of f at $e^{i\theta} \in E$ is the line segment joining $$\lim_{t \uparrow \theta} \hat{f}(e^{it}) \quad to \quad \lim_{t \mid \theta} \hat{f}(e^{it});$$ (e) the cluster set at $e^{i\lambda}$ is either the point at $\infty$ , a half-line, or two parallel half-lines. By correspondence in (b), we mean that there is a sequence $\{z_k\} \subset U$ so that $$\lim_{k \to \infty} z_k = e^{i\lambda} \quad \text{and} \quad \lim_{k \to \infty} f(z_k) = \infty.$$

Theorem 2.6. Let f be a univalent, harmonic, orientation-preserving mapping from U onto an unbounded convex domain D. Choose X and a so that there are points zk ΠU for which $$z_k \to e^{i\lambda}, |f(z_k)| \to \infty, \text{ and}$$ $f(z_k)/|f(z_k)| \to e^{i\alpha} \text{ as } k \to \infty.$ Then for each r > 0 $$f + re^{i\alpha}P(\cdot,\lambda)$$ is a univalent, harmonic, orientation-preserving mapping of U onto an unbounded convex domain contained in D. Proof Write / = h + g where h and g are analytic, and define $$H(z) = h(z) + \frac{r}{4\pi}e^{i\alpha}\frac{e^{i\lambda} + z}{e^{i\lambda} - z}$$ and $$G(z) = g(z) + \frac{r}{4\pi}e^{-i\alpha}\frac{e^{i\lambda} + z}{e^{i\lambda} - z}.$$ Then $F = H + \overline{G}$ satisfies $$F(z) = f(z) + re^{i\alpha}P(z, \lambda).$$ By [2, Theorem 5.7], the harmonic function F is univalent, orientation-preserving, and maps onto a convex domain if and only if for each $\theta \in [0, \pi)$ the function $$\Phi = ie^{-i\theta}[H - e^{2i\theta}G]$$ is univalent and maps onto a domain that is convex in the vertical direction. This is the case [13, Theorem 1] if and only if $\Phi$ is nonconstant and there are parameters $\mu \in [0, 2\pi)$ and $\nu \in [0, \pi]$ such that (1) $$\operatorname{Re}\left\{-ie^{i\mu}[1-(2\cos\nu)e^{-i\mu}z+e^{-2i\mu}z^2]\Phi'(z)\right\} \ge 0.$$ Choose $\mu$ and $\nu$ so that (1) is satisfied for the function $$\varphi = ie^{-i\theta}[h - e^{2i\theta}g],$$ which does map onto a domain convex in the vertical direction because f is convex. Since $$\Phi = \varphi - \frac{r}{2\pi} \sin(\alpha - \theta) \frac{e^{i\lambda} + z}{e^{i\lambda} - z},$$ it is sufficient to verify that (2) $$\operatorname{Re}\left\{ie^{i\mu}[1-(2\cos\nu)e^{-i\mu}z+e^{-2i\mu}z^2]\sin(\alpha-\theta)\frac{e^{i\lambda}}{(e^{i\lambda}-z)^2}\right\}$$ is nonnegative. It is known [13] that the point $e^{i(\mu+\nu)}$ corresponds to $\inf_U \operatorname{Re} \varphi$ and that $e^{i(\mu-\nu)}$ corresponds to $\sup_U \operatorname{Re} \varphi$ . We consider two cases. If $$\operatorname{Re}\{ie^{-i\theta}e^{i\alpha}\}\geq 0,$$ then the point $e^{i\lambda}$ corresponds to $\sup_U \operatorname{Re} \varphi$ because Re $$\varphi = \text{Re}\{ie^{-i\theta}f\}.$$ In this case $$e^{i\lambda} = e^{i(\mu-\nu)}$$ and (2) reduces to $$\sin(\alpha - \theta) \operatorname{Re} \left\{ i \frac{e^{i\mu} - e^{-i\nu}z}{e^{i(\mu-\nu)} - z} \right\}$$ $$= -\sin(\alpha - \theta) \frac{(1 - |z|^2)\sin\nu}{|e^{i(\mu-\nu)} - z|^2},$$ which is nonnegative. Similarly, if $$\operatorname{Re}\{ie^{-i\theta}e^{i\alpha}\}<0,$$ then the point $e^{i\lambda}$ corresponds to $\inf_U \operatorname{Re} \varphi$ , and so $$e^{i\lambda} = e^{i(\mu+\nu)}$$ In this case (2) reduces to $$\sin(\alpha - \theta) \operatorname{Re} \left\{ i \frac{e^{i\mu} - e^{i\nu}z}{e^{i(\mu+\nu)} - z} \right\}$$ $$= \sin(\alpha - \theta) \frac{(1 - |z|^2)\sin \nu}{|e^{i(\mu+\nu)} - z|^2},$$ which is nonnegative. To see that $\Phi$ is nonconstant, choose $z_k$ as in the hypothesis. Then $$F(z_k) = f(z_k) + re^{i\alpha}P(z_k, \lambda)$$ approaches infinity in the direction $e^{i\alpha}$ as $k \to \infty$ . For this reason $$\operatorname{Re} \Phi = \operatorname{Re} \{ i e^{-i\theta} F \}$$ could be constant only if $$\operatorname{Re}\{ie^{-i\theta}e^{i\alpha}\}=0.$$ However, in this case $$\operatorname{Re} \Phi = \operatorname{Re} \{ i e^{-i\alpha} f \}$$ is not constant because f is open. Finally, the domain F(U) is contained in D since the addition of $re^{i\alpha}P(z,\lambda)$ to f(z) is a translation along a ray in the convex domain D. In addition, F(U) is unbounded since the points $F(z_k)$ mentioned earlier approach infinity. Except at the point $e^{i\lambda}$ the Poisson kernel $P(z, \lambda)$ has boundary values zero. Therefore, in Theorem 2.6, the only place where the boundaries of D and F(U) can differ is at the cluster sets of f and F at $e^{i\lambda}$ . 3. Mappings onto a convex wedge. In this section, we consider the set $S_W$ of all univalent, harmonic, orientation-preserving mappings f of U onto the wedge $$W = \{w : |\arg w| < \pi/4\},$$ with normalization f(0) = 1. Let $\overline{S_W}$ denote the closure of $S_W$ in the topology of locally uniform convergences. If $\mathscr W$ is any convex wedge, then there is an affine transformation that maps $\mathscr W$ onto W. So the properties of $S_W$ will be representative. By Theorem 2.4, every $f \in S_W$ is in $h^1$ and can be expressed in the form (3) $$f(z) = \int_0^{2\pi} P(z, t) \hat{f}(e^{it}) dt + AP(z, \lambda)$$ where the radial limit values $\hat{f}(e^{it})$ are finite and belong to $\partial W$ for every point $e^{it} \neq e^{i\lambda}$ . The following theorem shows that this representation and other properties persist for $\overline{S_W}$ , and it identifies some degeneracies.

Theorem 3.1. If $f \in \overline{S_W}$ , then - (a) $f \in h^1$ , $f(U) \subset W$ , f is not constant, and $\hat{f}(e^{it}) \in \partial W$ for almost all t; - (b) $\overline{f}_{\overline{z}} = af_z$ for some analytic function a with $a(U) \subset \overline{U}$ ; - (c) if $|a| \not\equiv 1$ , then f is univalent and orientation preserving, f(U) is convex, and f has the representation (3); - (d) if $|a| \equiv 1$ , then f(U) is either a line segment or half-line through the point 1 with endpoint(s) on $\partial W$ .

Theorem 3.2. For $\alpha \leq \beta \leq \gamma \leq \alpha + 2\pi$ the functions $T_{(\alpha,\beta,\gamma)}$ belong to $\overline{S_W}$ .

Theorem 3.3. $HS_W = \mathcal{T}$ .

Corollary 3.4. $HS_W(\kappa, \lambda) = \mathcal{T}(\kappa, \lambda)$ . We shall add the prefix E to denote the set of extreme points. THEOREM 3.5. $$EHS_W = \{T_{(\alpha,\beta,\gamma)} : 0 \le \alpha < 2\pi, \alpha \le \beta \le \gamma \le \alpha + 2\pi\}$$ and $$EHS_{W}(\kappa, \lambda) = \{T_{(\lambda, \beta, \gamma)} : \lambda \leq \beta \leq \kappa \leq \gamma \leq \lambda + 2\pi\}.$$

Lemma 3.6. The function $$G(x, y) = \frac{1}{2}s(x)^2 + \frac{1}{2}s(y)^2 - s(x)s(y)\sin(x + y)$$ satisfies $0 \le G(x, y) \le 1$ for all $x, y \ge 0$ , and G(x, y) = 1 only when x = y = 0.

Lemma 3.7. For $x, y \ge 0$ , the function $$H(x, y) = \frac{1}{2}s(x)^2 + \frac{1}{2}s(y)^2 + s(x)s(y)\sin(x + y)$$ is nonnegative and assumes its maximum at a unique point $(x_0, x_0)$ . In particular, $$\sup_{(\pi/8) \le x \le (\pi/4)} H(x, y) = \max_{(\pi/8) \le x \le (\pi/4)} H(x, x) = H(x_0, x_0) \approx 1.7114$$ where $x_0 \approx 0.5875$ . Proof. As before, $$H(x, y) \ge \frac{1}{2} [|s(x)| - |s(y)|]^2 \ge 0.$$ If $x \ge \pi$ , then $$H(x, y) \le \frac{1}{2\pi^2} + \frac{1}{2} + \frac{1}{\pi}$$ which is less than the given $H(x_0, x_0)$ , and so we may restrict attention to $0 \le x \le \pi$ . In fact, since s(x) is decreasing it follows that $$H(x, y) \le \frac{1}{2} \sin^2(1) + \frac{1}{2} + \sin(1) = 1.6955...$$ for 1 ^ x ^ 77. This bound is also less than H(x0, x0), and so by symmetry, we may further restrict attention to 0 ^ x, y < 1. Now restrict (x, y) also to the line segment x + y = c, 0 ^i x, y < 1, and parametrize x = / and 7 = c — t. Then C = sin c = 0 and $$\frac{d^2H}{dt^2}(x,y) = \frac{1}{2}[s(x)^2]'' + \frac{1}{2}[s(y)^2]''$$ $$+ C[s''(x)s(y) - 2s'(x)s'(y) + s(x)s''(y)].$$ By considering power series, one observes that s(x) > 0, s'(x) < 0, s"(x) < 0, and [s(x)² ]" < 0, at least for 0 < x < 1. Therefore H is strictly concave on the line segment x + y = c, O^x , y ta 1. Since i/ is symmetric, the maximum of H on this segment occurs only when x = y. This is true for each line segment x + y = c, 0 ^i x, y ^ 1. Therefore $$\max_{x,y \ge 0} H(x, y) = \max_{0 \le x \le 1} H(x, x) = \max_{0 \le x \le 1} s(x)^{2} [1 + \sin(2x)].$$ Since sin(2x) and s(x) are decreasing for 77/4 < x ^ 1, the maximum cannot occur there. Similarly, if 0 ^ x ^ TT/8, we have $$H(x, x) \leq 1 + \frac{1}{\sqrt{2}},$$ which is less than H(x0, x0). Therefore the maximum of //(x, x) occurs in the interval 77/8 ^ x ^ 77/4. The derivative of H(x, x) is $$2s(x)\{s'(x)[1 + \sin(2x)] + s(x)\cos(2x)\}.$$ This can be zero only if $$w(x) = s'(x)[1 + \sin(2x)] + s(x)\cos(2x)$$ is zero. Since W(TT/8) > 0, w(77/4) < 0, and the first two terms of $$w'(x) = s''(x)[1 + \sin(2x)] - 2s(x)\sin(2x) + 3s'(x)\cos(2x)$$ are negative and the third is nonpositive, the equation w(x) = 0 has a unique solution x0 in [77/8, 77/4]. We conclude that $$\max H(x, y) = H(x_0, x_0) = s(x_0)^2 [1 + \sin(2x_0)].$$ Numerical computations give x0 « 0.5875 and max H « 1.7114.

Theorem 3.8. Let f belong to Sw, and suppose that $$f(z) = 1 + \sum_{n=1}^{\infty} a_n z^n + \sum_{n=1}^{\infty} b_n z^n.$$ Then $$|a_n| \le \sqrt{H(x_0, x_0)} \approx 1.3082$$ for all $n \ge 1$ where $H(x_0, x_0)$ is defined in Lemma 3.7. Equality for any n is possible only for the functions $T_{(\alpha,\beta,\gamma)}$ where $$\beta = \alpha + \frac{2x_0}{n}$$ and $\gamma = \alpha + 2\pi - \frac{2x_0}{n}$ . In addition, $|b_n| \le 1$ for all $n \ge 1$ . Equality in this case for any n is possible only for the functions $2\pi P(\cdot, \alpha)$ .

Theorem 3.9. If $f = h + \overline{g}$ belongs to $\overline{S_W}$ , then $$|h'(0)| \ge 2/\pi$$ . Equality occurs only when $$f = T_{(\alpha,\alpha+\pi,\alpha+\pi)}, \quad 0 \le \alpha < 2\pi.$$

Theorem 3.10. If $f = h + \overline{g}$ belongs to $\overline{S_W^0}$ , then $$\frac{4}{\pi + 2} \le |h'(0)| \le \frac{4}{\pi}.$$ The lower bound is sharp only for the functions $$f = \frac{\pi}{\pi + 2} T_{(\alpha, \alpha + \pi, \alpha + \pi)} + \frac{4\pi}{\pi + 2} P(\cdot, \alpha), \quad 0 \le \alpha < 2\pi,$$ and the upper bound is sharp only for the functions $$f = T_{(\alpha,\alpha+\pi/2,\alpha+\pi/2)}, \quad 0 \le \alpha < 2\pi.$$

Lemma 4.1. If $f \in S_R$ , then Re $f = 2\pi P(\cdot, \lambda)$ for some $\lambda, 0 \le \lambda < 2\pi$ .

Theorem 4.2. If $f \in S_R$ , then there are a probability measure $\mu$ and real numbers $\lambda$ , b > 0, and c so that (13) $$f(z) = 2\pi P(z, \lambda)$$ + $$i \left[ \int_0^{2\pi} bK(z, t, \lambda) d\mu(t) + 2c \operatorname{Re} \left\{ \frac{z}{e^{i\lambda} - z} \right\} \right].$$ Let $\mathscr{K}$ denote the set of all functions of the form (13) for real numbers $\lambda$ , $b \ge 0$ , and c and probability measures $\mu$ . Then $\mathscr{K}$ is closed, but not compact since b and c are not bounded. However, the following is true.

Theorem 4.3. If $f \in \mathcal{K}$ with b > 0, then f is a univalent, harmonic, orientation-preserving mapping of U onto a convex domain.

Theorem 4.5. $$\mathscr{K} = \overline{S_R}$$ . In the remaining part of this section, we shall study the subclasses $S_R^0$ and $\mathcal{K}^0$ of $S_R$ and $\mathcal{K}$ , respectively, of functions $f = h + \overline{g}$ that satisfy the additional condition $$f_{\overline{z}}(0) = g'(0) = 0$$ or, equivalently, p(0) = 1. It is clear from the definitions that $$S_R^0 \subset \mathscr{K}^0$$ and $\overline{S_R^0} = \mathscr{K}^0$ . Since b=1 and c=0 in (13), it is also clear that $\mathscr{K}^0$ is compact. Next, we characterize the extreme points of $\mathscr{K}^0$ .

Theorem 4.6. $$E\mathcal{K}^0 = \{2\pi P(\cdot, \lambda) + iK(\cdot, t, \lambda) : 0 \le t, \lambda < 2\pi\}.$$

Theorem 4.9. If $$f(z) = 1 + \sum_{n=1}^{\infty} a_n z^n + \sum_{n=2}^{\infty} b_n z^n$$ belongs to $\mathcal{K}^0$ , then $|a_1| = 2$ , $|a_n| \le n + 1$ , and $|b_n| \le n - 1$ for $n \ge 2$ . Equality in either of the inequalities occurs only for the functions from Example 4.7.

Proposition 5.1. S is a nonparametric minimal surface over a simply-connected domain $D \neq \mathbb{C}$ if and only if there are analytic functions $\Phi_1, \Phi_2, \Phi_3$ in U such that (a) $$\Phi_1^2 + \Phi_2^2 + \Phi_3^2 \equiv 0$$ in U, (b) $$|\Phi_1^2(z)| + |\Phi_2^2(z)| + |\Phi_3^2(z)| \neq 0$$ for all $z \in U$ , (c) the mapping $$z \rightarrow \text{Re} \int \Phi_1 dz + i \text{ Re} \int \Phi_2 dz$$ is harmonic and univalent from U onto D, and (d) $$S = \left\{ \left( \operatorname{Re} \int \Phi_1 dz, \operatorname{Re} \int \Phi_2 dz, \operatorname{Re} \int \Phi_3 dz \right) : z \in U \right\}.$$ Of course, in (c) and (d) appropriate constants of integration must be chosen for the antiderivatives $\int \Phi_i dz$ . If S is given as in (d), then the mapping $$z \rightarrow \left( \text{Re } \int \Phi_1 dz, \text{ Re } \int \Phi_2 dz, \text{ Re } \int \Phi_3 dz \right)$$ of U onto S is conformal, that is, angles are preserved, and z is an isothermal parameter. In other words, are length on S can be obtained by $ds = \rho |dz|$ where $\rho > 0$ . In fact, one has $$\rho^2 = \frac{1}{2} ( |\Phi_1|^2 + |\Phi_2|^2 + |\Phi_3|^2 ).$$ We are interested in the univalent harmonic mapping of U onto D that is given in (c). Denote $$f = \text{Re} \int \Phi_1 dz + i \text{ Re} \int \Phi_2 dz.$$ It is no loss of generality to assume that f is orientation-preserving. Hence, if we write $f = h + \overline{g}$ , then a = g'/h' satisfies |a(z)| < 1 for $z \in U$ and we also have $$\Phi_1 = h' + g', \quad \Phi_2 = -i(h' - g'), \text{ and}$$ $\Phi_3^2 = -4h'g' = -4a(h')^2.$ The latter shows that a has a single-valued square root in U. For our purposes the following is a useful restatement of Proposition 5.1.

Proposition 5.2. S is a nonparametric minimal surface over a simply-connected domain $D \neq \mathbb{C}$ if and only if there is a univalent, harmonic, orientation-preserving mapping $f = h + \overline{g}$ of U onto D such that a = g'/h' has a single-valued square root in U and $$S = \left\{ \left( \operatorname{Re} f(z), \operatorname{Im} f(z), 2 \operatorname{Im} \int \sqrt{a}h' dz \right) : z \in U \right\}.$$ In terms of $f = h + \overline{g}$ , the conformal factor $\rho$ becomes simply $\rho = |h'| + |g'|$ . Therefore the Gaussian curvature of S is (14) $$k = \frac{-\Delta \log \rho}{\rho^2} = \frac{-|h'g'' - h''g'|^2}{|h'g'| (|h'| + |g'|)^4} = \frac{-|a'|^2}{|a| (1 + |a|)^4 |h'|^2}$$ where a = g'/h'. We summarize as follows.

Lemma 5.3. Let S be a nonparametric minimal surface over a simply-connected domain $D \neq \mathbb{C}$ . Let $f = h + \overline{g}$ and a be as in Proposition 5.2. Then for $z \in U$ the Gaussian curvature of S at the point (Re $$f(z)$$ , Im $f(z)$ , 2 Im $\int \sqrt{a}h'dz$ ) is $$k(z) = \frac{-|a'(z)|^2}{|a(z)|(1+|a(z)|)^4|h'(z)|^2}.$$ In addition, one has $$(15) |k(z)| \leq \frac{4(1 - |a(z)|)^2}{(1 - |z|^2)^2 (1 + |a(z)|)^4 |h'(z)|^2}$$ $$\leq \frac{4}{(1 - |z|^2)^2 (|g'(z)| + |h'(z)|)^2}$$ and $$(16) |k(0)| \le \frac{4}{(|g'(0)| + |h'(0)|)^2} \le \frac{4}{|g'(0)|^2 + |h'(0)|^2} \le \frac{4}{|h'(0)|^2}.$$ Proof. Since the function a has a single-valued analytic square root, which is bounded by one, the invariant form of Schwarz's lemma implies $$|a'| \le 2\sqrt{|a|} (1 - |a|)/(1 - |z|^2).$$ Use this estimate in (14) to obtain the first inequality in (15). The second inequality in (15) follows from $$1 - |a| \le 1 + |a|$$ . The inequalities in (16) are simple consequences. Remark 5.4. If D is a convex domain, then the analytic part h of $f = h + \overline{g}$ maps U homeomorphically onto a domain $\Omega$ [2, Theorem 5.7]. Hence, the function $F = f \circ h^{-1}$ is a univalent harmonic mapping of $\Omega$ onto D, and in terms of F, the surface S has the attractive conformal parametrization $$S = \left\{ \left( \operatorname{Re} F(w), \operatorname{Im} F(w), 2 \operatorname{Im} \int \sqrt{A} dw \right) : w \in \Omega \right\}$$ where $A = \overline{F_w}/F_w$ has a single-valued analytic square root and |A| < 1. Furthermore, the Gaussian curvature becomes $$k = \frac{-|A'|^2}{|A| (1 + |A|)^4}.$$ For the rest of this section, we shall estimate the Gaussian curvature k for nonparametric minimal surfaces over certain domains D. E. Heinz [6] proved that if $D = \{w: |w| < R\}$ , then the curvature of S at the point P above the origin satisfies $$|k| \leq 4\pi^2/(3R^2).$$ Improvements were considered by E. Hopf [9] and many others. This estimate can be improved to $$|k| \le 16\pi^2/(27R^2)$$ by combining R. Hall's [5] determination of the "Heinz constant", which implies $$|g'(0)|^2 + |h'(0)|^2 \ge 27R^2/(4\pi^2),$$ and inequality (16). Unfortunately, this curvature estimate is still not sharp. However, if in addition, S has a horizontal tangent plane at P, then at this point $$|k| \le \pi^2/(2R^2)$$ and this estimate is sharp [12, p. 108]. Furthermore, for a general domain D it is known [12, p. 108] that if S has a horizontal tangent plane at P, then $$|k| \le 64/(9d^2)$$ where d is the distance along the surface from P to $\partial S$ . This is also known to be sharp for a certain domain D. Suppose, in the framework of Proposition 5.2, that the surface S has a horizontal tangent plane at the point P above f(0). Then we have $$\frac{\partial}{\partial x} \left[ 2 \operatorname{Im} \int \sqrt{a} h' dz \right] = \frac{\partial}{\partial y} \left[ 2 \operatorname{Im} \int \sqrt{a} h' dz \right] = 0$$ at z = 0. This implies that $\sqrt{a(0)}h'(0) = 0$ and, consequently, that g'(0) = 0. The following theorem is the result of combining the estimates for |h'(0)| from below in Theorems 3.9, 3.10, 4.9 and [8, Theorem 2.3] with inequality (16) in Lemma 5.3.

Theorem 5.5. Let S be a nonparametric minimal surface in $\mathbb{R}^3$ that lies above a domain D in C. Let k denote the Gaussian curvature of S at a point P that lies above $w_0 \in D$ . - (a) If $D = W = \{w : |\arg w| < \pi/4\}$ and $w_0 = 1$ , then $|k| \le \pi^2$ . In addition, assume that S has a horizontal tangent plane at P. - (b) If D = W and $w_0 = 1$ , then $|k| \le (\pi + 2)^2/4$ . - (c) If $D = R = \{w : \text{Re } w > 0\}$ and $w_0 = 1$ , then $|k| \le 1$ . - (d) If $D = \Omega = \{w : |\text{Im } w| < \pi/4\} \text{ and } w_0 = 0, \text{ then } |k| \le 4.$ The estimate (16) that we have used in proving Theorem 5.5 is sharp only when $$a(z) = e^{i\alpha}z^2$$ for some $\alpha$ . Therefore the curvature estimates in this theorem are sharp only if the lower bounds for |h'(0)| arise from harmonic mappings $f = h + \overline{g}$ with $$g'(z)/h'(z) = e^{i\alpha}z^2$$ . This is not the case in (a) or (b). For (c) equality occurs for the surface $$S = \left\{ \left( \operatorname{Re} \left[ \frac{1+z}{1-z} \right], \operatorname{Im} \left[ \frac{z}{(1-z)^2} \right] + \frac{1}{2} \operatorname{arg} \left[ \frac{1+z}{1-z} \right], \right.$$ $$\left. \operatorname{Re} \left[ \frac{z}{(1-z)^2} \right] - \frac{1}{2} \operatorname{log} \left| \frac{1+z}{1-z} \right| \right) : z \in U \right\}$$ and for (d) equality occurs for the surface $$S = \left\{ \left( \operatorname{Re} \left[ \frac{z}{1 - z^2} \right], \frac{1}{2} \operatorname{arg} \left[ \frac{1 + z}{1 - z} \right], \operatorname{Im} \left[ \frac{z^2}{1 - z^2} \right] \right) : z \in U \right\}.$$