On the existence of a harmonic mapping

Lemma 1. The family $\Omega$ is normal in each singular simplex of $\Sigma$ . Proof. Let |z| < 1 be a local coordinate disk, which is a topological image of the interior of a simplex $S_j \in \Sigma$ . Let $\{\omega_n\}_{n=1}^{\infty}$ be an arbitrary sequence chosen out of $\Omega$ . Set $$T_n(z) = \int\limits_0^z \tau_n(\zeta) d\zeta$$ , $\tau_n(z) dz = \omega_n$ for |z| < 1. Then $\{T_n(z)\}_{n=1}^{\infty}$ is equicontinuous in |z| < 1. For, otherwise, there exist a constant c > 0, a pair of sequences $\{z'_{\nu}\}_{\nu=1}^{\infty}$ , $\{z'_{\nu}\}_{\nu=1}^{\infty}$ in |z| < 1 and a subsequence $\{T_{n_{\nu}}(z)\}_{\nu=1}^{\infty}$ , such that $|z''_{\nu} - z'_{\nu}| \to 0$ as $\nu \to \infty$ , while $$|T_{n_{\nu}}(z_{\nu}^{\prime\prime})-T_{n_{\nu}}(z_{\nu}^{\prime\prime})| \geq c, \quad \nu=1,2,\cdots.$$ Let us denote by $z_0$ one of the accumulation points of $\{z'_i\}_{\nu=1}^{\infty}$ . Our proof dispenses with the case when $|z_0| < 1$ , since it is trivial. Assume $|z_0| = 1$ . Draw a circular arc $C_r$ with centre $z_0$ and with radius r < 1 on |z| < 1. However small $\varepsilon > 0$ may be, there can be found a pair of points $z'_{\nu}$ and $z''_{\nu}$ separated from z=0 by $C_r$ , such that $$|T_{n_{\nu}}(z_{\nu}')-T_{n_{\nu}}(z_{\nu}'')| \geq c$$ . Hence the length L(r) of the image curve $T_{n_v}(C_r)$ ( $\varepsilon \leq r < 1$ ) is not smaller than c. Thus the well-known length-area principle for holomorphic functions will lead to a contradiction: $$c^2 \leq [L(r)]^2 \leq 2\pi r \int\limits_0^{2\pi} |\tau_{n_{\mathcal{V}}}(re^{i\theta})|^2 rd\theta$$ , whence $$\int_{\frac{s}{r}}^{1} \frac{dr}{r} \leq \frac{2\pi}{c^2} \int_{|z| \leq 1} |\tau_{n_{\nu}}(re^{i\theta})|^2 r d\theta dr.$$ On the other hand, the right-hand member is uniformly bounded on account of the normalization. Therefore $\{T_n(z)\}_{n=1}^{\infty}$ contains a subsequence, say again $\{T_n(z)\}_{n=1}^{\infty}$ , which converges on every compact set in |z| < 1. Accordingly $\{\tau_n(z)\}_{n=1}^{\infty}$ is uniformly convergent there. Repeating this process on every simplex $S_j$ $(j=1,2,\dots,\kappa)$ we finally get a sequence of 1-forms belonging to $\Omega$ , whose coefficients constitute a uniformly convergent sequence on every compact comprised in arbitrarily fixed local coordinate disk corresponding to $S_j$ $(j=1,2,\dots,\kappa)$ , determining a differential holomorphic in $R-\bigcup_{j=1}^{\infty} S_j$ . REMARK 1. It is easily seen that $\{T_n(z)\}_{n=1}^{\infty}$ is normal on $|z| \leq 1$ too.

Lemma 2. It is possible to introduce the distance between two points on R by means of any $\omega$ belonging to $\Omega$ . Proof. We integrate an arbitrary holomorphic differential $\tau(z)dz = \omega$ along a path in a simplex $S_j$ : we set $T(z) = \int_0^z \tau(\zeta) d\zeta$ for |z| < 1. Then by Lemma 1, T(z) is uniformly continuous in |z| < 1. Hence we see that T(z) is continuously prolongable up to $|z| \le 1$ . In fact, let $\zeta$ be an arbitrary point on |z| = 1 and let $\{z_n\}_{n=1}^{\infty}$ an arbitrary sequence in |z| < 1 tending to $\zeta$ . Given any $\varepsilon > 0$ , there is $\delta_1 = \delta_1(\varepsilon)$ depending only on $\varepsilon$ , such that (1) $$|T(z)-T(z')| < \varepsilon$$ so far as $|z-z'| < \delta_1$ by uniform continuity of T(z). Taking $n_0 = n_0(\delta_1)$ sufficiently large, we shall have $|\zeta - z_n| < \delta_1/2$ $(n \ge n_0)$ and so $$(2) |z_n-z_m| \leq |z_n-\zeta|+|\zeta-z_m| < \delta_1, n, m \geq n_0.$$ From (1), (2) it follows that $|T(z_n)-T(z_m)| < \varepsilon$ $(n, m \ge n_0)$ , which implies the existence of $Z = \lim_{n \to \infty} T(z_n)$ . Z depends only on $\zeta$ : let another sequence $\{z_n'\}_{n=1}^{\infty}$ tend to $\zeta$ . Then $Z' = \lim_{n \to \infty} T(z_n')$ exists. Since the mixed sequence $z_1, z_1', z_2, z_2', \dots, z_n, z_n', z_{n+1}, \dots$ also tends to $\zeta$ , $T(z_1)$ , $T(z_1')$ , $T(z_2)$ , $T(z_2')$ , $\dots$ , $T(z_n)$ , $T(z_n')$ , $T(z_{n+1})$ , $\dots$ has a limit, say Z''. $\{T(z_n)\}_{n=1}^{\infty}$ , as a subsequence of the last convergent sequence, must have the same limit; Z = Z''. In analogy Z' = Z''. Thus we can determine the local distance on each singular simplex by means of this integral of $\omega$ , although the line elements may fail to be defined on a certain sebset of $\bigcup_{i=1}^{\kappa} \partial S_i$ .

Proposition 1. $\mathfrak{F}_{\omega}$ is non-void. Proof. Let g be a quasi-conformal mapping with maximal dilatation K homotopic to $\mathcal{X}$ above mentioned. g clearly fulfills the conditions (I), (II), (III) and (V). Denoting by w(z) its local realization, we have $$\iint_{\mathbb{R}} \rho(w(z)) \left( \left| \frac{\partial w}{\partial z} \right|^{2} + \left| \frac{\partial w}{\partial \bar{z}} \right|^{2} \right) |dz \wedge d\bar{z}|$$ $$= \iint_{\mathbb{R}} \rho(w(z)) \left( \left| \frac{\partial w}{\partial z} \right|^{2} + \left| \frac{\partial w}{\partial \bar{z}} \right|^{2} \right) \left( \left| \frac{\partial w}{\partial z} \right|^{2} - \left| \frac{\partial w}{\partial \bar{z}} \right|^{2} \right)^{-1} \left( \left| \frac{\partial w}{\partial z} \right|^{2} - \left| \frac{\partial w}{\partial \bar{z}} \right|^{2} \right) |dz \wedge d\bar{z}|$$ $$= \iint_{\mathbb{R}'} \left( \left| \frac{\partial w}{\partial z} \right|^{2} + \left| \frac{\partial w}{\partial \bar{z}} \right|^{2} \right) \left( \left| \frac{\partial w}{\partial z} \right|^{2} - \left| \frac{\partial w}{\partial \bar{z}} \right|^{2} \right)^{-1} \rho(w) |dw \wedge d\bar{w}| \leq \frac{1}{2} \left( K + \frac{1}{K} \right).$$ Hence $g \in \mathfrak{F}_{\omega}$ , Q.E.D. Now we can formulate our first extremum problem: Minimize the Dirichlet integral $$I[f] = \frac{1}{4} \iint_{\mathbb{R}} \rho(w(z)) \left( \left| \frac{\partial w}{\partial z} \right|^2 + \left| \frac{\partial w}{\partial \bar{z}} \right|^2 \right) |dz \wedge d\bar{z}|$$ within the family $\mathfrak{F}_{\omega}$ .

Lemma 3. & is equicontinuous. Proof. We show, for any $\varepsilon > 0$ , there exists a suitable positive number $\delta_2 = \delta_2(\varepsilon)$ independent of an individual $f \in \mathfrak{F}_{\omega}$ , such that, dist $(f(\mathfrak{p}_1), f(\mathfrak{p}_2)) < \varepsilon$ so far as dist $(\mathfrak{p}_1, \mathfrak{p}_2) < \delta_2$ . Here the distances on R' and R are to be measured with regard to the fixed conformal metric $ds^2 = \rho(w) |dw|^2$ and the metric induced by $\omega = \tau(z) dz$ respectively. Suppose, on the contrary, there exist a positive number $c_1$ , two sequences of points $\{\mathfrak{p}_n^{(1)}\}_{n=1}^{\infty}$ , $\{\mathfrak{p}_n^{(2)}\}_{n=1}^{\infty}$ on R and a sequence $\{f_n\}_{n=1}^{\infty}$ belonging to $\mathfrak{F}_{\omega}$ , such that dist $(\mathfrak{p}_n^{(1)}, \mathfrak{p}_n^{(2)}) \to 0$ as $n \to \infty$ , while dist $(f_n(\mathfrak{p}_n^{(1)}), f_n(\mathfrak{p}_n^{(2)})) \ge c_1$ . We may assume that $\mathfrak{p}_n^{(j)}$ (j=1,2) are contained in a local coordinate neighbourhood $\mathfrak{U}$ for all $n \ge n_0$ . Let |z| < 1 be a local coordinate disk corresponding to $\mathfrak{U}$ and set $z(\mathfrak{p}_n^{(j)}) = z_j$ (j=1,2) simply for some fixed n. We draw a circular disk $|z_1-z| < r_0$ which contains $z_2$ in its interior and is contained in |z| < 1. A domain on R corresponding to the circular annulus $\mathfrak{E} = |z_2-z_1| < |z-z_1| < r_0$ is mapped by $f_n(\mathfrak{p}(z))$ onto an annular domain on R'. Let L(r) be the length of the image curve of $|z-z_1| = r$ by $f_n(\mathfrak{p}(z))$ , which is defined for almost all values of r in the interval $[\varepsilon, r_0]$ . Set $$I(r) = \int_{0}^{r} \int_{0}^{2\pi} \rho(w(z)) \left( \left| \frac{\partial w}{\partial z} \right|^{2} + \left| \frac{\partial w}{\partial \bar{z}} \right|^{2} \right) t d\theta dt \qquad z = te^{i\theta},$$ where w(z) is a local realization of $f_n(\mathfrak{p})$ . Then we would have by the condition (III) $$c_{1}^{2} \leq (L(r))^{2} = \left( \oint \sqrt{\rho(w)} |dw| \right)^{2} = \left( \int_{\theta=0}^{2\pi} \sqrt{\rho(w(z))} \left| \frac{\partial w}{\partial \theta} \right| d\theta \right)^{2}$$ $$\leq \left( \oint_{|z|=r} \sqrt{\rho(w(z))} \left( \left| \frac{\partial w}{\partial z} \right| + \left| \frac{\partial w}{\partial \overline{z}} \right| \right) |dz| \right)^{2}$$ $$\leq \left( \oint_{|z|=r} \rho(w(z)) \left( \left| \frac{\partial w}{\partial z} \right| + \left| \frac{\partial w}{\partial \overline{z}} \right| \right)^{2} |dz| \right) \cdot \oint_{|z|=r} |dz|$$ $$\leq 2 \left( \oint_{|z|=r} \rho(w(z)) \left( \left| \frac{\partial w}{\partial z} \right|^{2} + \left| \frac{\partial w}{\partial \overline{z}} \right|^{2} \right) |dz| \right) \cdot \oint_{|z|=r} |dz| = 4\pi r \frac{dI(r)}{dr}.$$ Therefore $$\frac{dr}{r} \leq \frac{4\pi}{c_1^2} dI(r) .$$ Integrating from $r=\delta$ to $r=r_0$ , we obtain $$\log \frac{r_{\scriptscriptstyle 0}}{\varepsilon} \leq \frac{4\pi}{c_{\scriptscriptstyle 1}^{\, 2}} [I(r_{\scriptscriptstyle 0}) - I(\varepsilon)] \leq \frac{M}{c_{\scriptscriptstyle 1}^{\, 2}} (K + \frac{1}{K}).$$ This is a contradiction, since $\varepsilon$ can be chosen arbitrarily small.

Theorem 1. The family $\mathfrak{F}_{\omega}$ is normal on R. Proof. Let $\{f_n(\mathfrak{p})\}_{n=1}^{\mathfrak{F}}$ be an arbitrary sequence, such that $f_n \in \mathfrak{F}_{\omega}$ $(n=1,2,\cdots)$ . Since R is separable, we can choose a countable subset $E=\{\mathfrak{p}_n\}_{n=1}^{\mathfrak{F}}$ which is everywhere dense on R. On account of compactness of R', its infinite subset $\{f_n(\mathfrak{p}_1)\}_{n=1}^{\mathfrak{F}}$ accumulates at one point on R'. Hence its suitable subsequence $\{f_{1n}(\mathfrak{p}_1)\}_{n=1}^{\mathfrak{F}}$ converges. In analogous way we can choose a suitable subsequence out of $\{f_{1n}(\mathfrak{p}_2)\}_{n=1}^{\mathfrak{F}}$ , which we denote by $\{f_{2n}(\mathfrak{p}_2)\}_{n=1}^{\mathfrak{F}}$ . Repeating this procedure, we get a sequence $\{f_{ln}(\mathfrak{p}_l)\}_{n=1}^{\mathfrak{F}}$ convergent at $\mathfrak{p}_l$ . Since $\{f_{ln}(\mathfrak{p})\}_{n=1}^{\mathfrak{F}}$ is a subsequence of $\{f_{mn}(\mathfrak{p})\}_{n=1}^{\mathfrak{F}}$ provided l>m, $\{f_{ln}(\mathfrak{p})\}_{n=1}^{\infty}$ converges at $\mathfrak{p}=\mathfrak{p}_{j}$ $(j=1,2,\cdots,l)$ . If we consider the diagonal sequence $\{f_{nn}(\mathfrak{p})\}_{n=1}^{\infty}$ it is obviously a subsequence of $\{f_{ln}(\mathfrak{p})\}_{n=1}^{\infty}$ . Therefore it converges at $\mathfrak{p}=\mathfrak{p}_{j}$ $(j=1,2,\cdots l)$ . Here, since l may be arbitrarily large, we see that $\{f_{nn}(\mathfrak{p})\}_{n=1}^{\infty}$ converges at every point of the set E. With this in mind, we shall show that the subsequence $\{f_{nn}(\mathfrak{p})\}_{n=1}^{\infty}$ of $\{f_{n}(\mathfrak{p})\}_{n=1}^{\infty}$ is uniformly convergent in the whole surface R. In virtue of equicontinuity of $\mathfrak{F}_{\omega}$ , there exists a positive number $\delta_3$ , such that for any $\varepsilon > 0$ the inequalities $$\mathrm{dist}\left(f_{nn}(\mathfrak{p}),\ f_{nn}(\mathfrak{p}')\right) < \varepsilon/3\ , \quad \mathrm{dist}\left(f_{mm}(\mathfrak{p}),\ f_{mm}(\mathfrak{p}')\right) < \varepsilon/3$$ hold, whenever dist $(\mathfrak{p}, \mathfrak{p}') < \delta_3$ . We can choose a finite subset $\{\mathfrak{p}_j\}_{j=1}^l$ of E, such that for any point $\mathfrak{p}$ on R $$\inf_{1 \leq j \leq l} \operatorname{dist}(\mathfrak{p}, \mathfrak{p}_j) = \operatorname{dist}(\mathfrak{p}, \mathfrak{p}_{j_0}) < \delta_3$$ . Hence we have dist $$(f_{nn}(\mathfrak{p}), f_{nn}(\mathfrak{p}_{j_0})) < \varepsilon/3$$ , dist $(f_{mm}(\mathfrak{p}), f_{mm}(\mathfrak{p}_{j_0})) < \varepsilon/3$ . Since $\{f_{nn}(\mathfrak{p})\}_{n=1}^{\infty}$ converges at $\mathfrak{p}_{j}$ $(j=1, 2, \dots, l)$ , we can choose $n_0 = n_0(\varepsilon)$ large enough, so that dist $$(f_{nn}(\mathfrak{p}_i), f_{mm}(\mathfrak{p}_i)) < \varepsilon/3$$ $j = 1, 2, \dots, l$ as far as $n, m \ge n_0(\mathcal{E})$ . Therefore, if $n, m \ge n_0(\mathcal{E})$ , we have dist $$(f_{nn}(\mathfrak{p}), f_{mm}(\mathfrak{p}))$$ $$\leq \operatorname{dist}(f_{nn}(\mathfrak{p}), \ f_{nn}(\mathfrak{p}_{j_0})) + \operatorname{dist}(f_{nn}(\mathfrak{p}_{j_0}), \ f_{mm}(\mathfrak{p}_{j_0})) + \operatorname{dist}(f_{mm}(\mathfrak{p}_{j_0}), \ f_{mm}(\mathfrak{p})) \leq \varepsilon.$$ Setting $q_n = f_{nn}(\mathfrak{p})$ , we see that $\{q_n\}_{n=1}^{\infty}$ form a Cauchy sequence on R'. In virtue of compactness of R', a point $\mathfrak{q}$ is uniquely determined, so that $$\lim_{n\to\infty} \operatorname{dist}(q, q_n) = 0.$$ Let the correspondence $\mathfrak{p} \rightarrow \mathfrak{q}$ be denoted by $f_{\mathfrak{o}}$ . Then, for any $\mathfrak{E} > 0$ dist $$(f_0(\mathfrak{p}), f_{nn}(\mathfrak{p})) < \varepsilon$$ so far as $n \ge n_0(\varepsilon)$ and $n_0(\varepsilon)$ is clearly independent of choice of $\mathfrak{p} \in R$ . It implies that the subsequence $\{f_{nn}(\mathfrak{p})\}_{n=1}^{\infty}$ of $\{f_n(\mathfrak{p})\}_{n=1}^{\infty}$ converges uniformly on R towards $f_0(\mathfrak{p})$ , Q.E.D. Now it is asked whether the functional I[f] can be minimized within the family $\mathfrak{F}_{\omega}$ . First of all, we can form a minimizing sequence $\{f_n\}_{n=1}^{\infty}$ for this extremum problem: $$\lim_{n\to\infty} I[f_n] = \inf_{\mathfrak{F}_{\omega}} I[f] = I_{\omega} \ge 0, \qquad f_n \in \mathfrak{F}_{\omega}.$$ By Theorem 1, the minimizing sequence contains a subsequence, say $\{f_n\}_{n=1}^{\infty}$ again for brevity, uniformly convergent on R. Let the limiting map be denoted by $f_{\infty}(\mathfrak{p})$ . Then $f_{\infty}(\mathfrak{p})$ is, by definition, one-valued as well as continuous.

Proposition 2. $f_{\infty}(\mathfrak{p})$ is schlicht. Proof. Suppose, on the contrary, there exist a pair of points $\mathfrak{p}_1$ and $\mathfrak{p}_2$ on R, such that $f_\infty(\mathfrak{p}_1)=f_\infty(\mathfrak{p}_2)=\mathfrak{q}$ . Then there could be found, for each n, a continuum $\mathfrak{c}_n \in R$ , such that $\mathfrak{c}_n \ni \mathfrak{p}_1$ , $\mathfrak{p}_2$ and the diameter of the closed set $f_n(\mathfrak{c}_n)$ tends to zero as $n\to\infty$ . Let $\Gamma$ be a simple closed Jordan curve enclosing $\mathfrak{p}_1$ in its interior, which is comprised in a local coordinate neighbourhood of $\mathfrak{p}_1$ and separates $\mathfrak{p}_2$ from $\mathfrak{p}_1$ , Then $\mathfrak{c}_n \cap \Gamma$ is always nonempty, on which we take an arbitrary point $\mathfrak{p}^{(n)}$ . Let z be some local coordinate about $\mathfrak{p}_1$ , w some local coordinate about $f_n(\mathfrak{p}_1)$ and $z=z_n(w)$ the local realization of $f_n^{-1}$ in the neighbourhood of $f_n(\mathfrak{p}_1)$ . Setting $w(f_n(\mathfrak{p}^{(n)}))=w'$ , we may assume without loss of generality that $w(f_n(\mathfrak{p}_1))=0$ . When n is sufficiently large, we see clearly $|w'| < \varepsilon$ , while dist $(z_n(w'), z_n(0)) \ge c_2 > 0$ . Draw a circle |w| = r $(\varepsilon \le r \le r_0)$ in the local coordinate disk |w| < 1, where $\log(r_0/\varepsilon) \ge \pi M(K^2+1)/c^2K$ . Putting $$J_n(r) = \int_0^r \int_0^{2\pi} |\tau(z_n(w))|^2 \left( \left| \frac{\partial z_n}{\partial w} \right|^2 + \left| \frac{\partial z_n}{\partial \overline{w}} \right|^2 \right) |dw \wedge d\overline{w}|,$$ we have just as in the proof of Lemma 3 $$\begin{aligned} c_{\mathbf{z}}^{2} & \leq \left( \int |\tau(z)| |dz| \right)^{2} \leq \left( \int |\tau(z_{\mathbf{n}}(w))| \left( \left| \frac{\partial z_{\mathbf{n}}}{\partial w} \right| + \left| \frac{\partial z_{\mathbf{n}}}{\partial \overline{w}} \right| \right) |dw| \right)^{2} \\ & \leq 2 \left[ \int |\tau(z_{\mathbf{n}}(w))|^{2} \left( \left| \frac{\partial z_{\mathbf{n}}}{\partial w} \right|^{2} + \left| \frac{\partial z_{\mathbf{n}}}{\partial \overline{w}} \right|^{2} \right) |dw| \right] \int |dw| \\ & \leq 4\pi r \frac{dJ_{\mathbf{n}}(r)}{dr} \end{aligned}$$ for every r in $[\mathcal{E}, r_0]$ . Hence we get successively $$\frac{dr}{r} \leq \frac{4\pi}{c_2^2} dJ_n(r) ,$$ $$\frac{\pi M(K^2+1)}{c_2^2 K} \leq \log \frac{r_0}{\varepsilon} \leq \frac{4\pi}{c_2^2} [J_n(r_0) - J_n(\varepsilon)] < \frac{4\pi J_n(r_0)}{c_2^2} .$$ The last inequality contradicts the condition (V): indeed, let $w_n(z)$ be the inverse function of $z_n(w)$ . Then, by the condition (III) $$|dw \wedge d\overline{w}| = \left( \left| \frac{\partial w_n}{\partial z} \right|^2 - \left| \frac{\partial w_n}{\partial \overline{z}} \right|^2 \right) |dz \wedge d\overline{z}| = \frac{|dz \wedge d\overline{z}|}{\left| \frac{\partial z_n}{\partial w} \right|^2 - \left| \frac{\partial z_n}{\partial \overline{w}} \right|^2},$$ whence $$\begin{split} J_{\mathbf{n}}(r) &= \iint\limits_{\mathbb{R}} \left( \left| \frac{\partial z_{\mathbf{n}}}{\partial w} \right|^2 + \left| \frac{\partial z_{\mathbf{n}}}{\partial \overline{w}} \right|^2 \right) \left( \left| \frac{\partial z_{\mathbf{n}}}{\partial w} \right|^2 - \left| \frac{\partial z_{\mathbf{n}}}{\partial \overline{w}} \right|^2 \right)^{-1} |\tau(z)|^2 |dz \wedge d\overline{z}| \\ &= \iint\limits_{\mathbb{R}} \left( \left| \frac{\partial w_{\mathbf{n}}}{\partial z} \right|^2 + \left| \frac{\partial w_{\mathbf{n}}}{\partial \overline{z}} \right|^2 \right) \left( \left| \frac{\partial w_{\mathbf{n}}}{\partial z} \right|^2 - \left| \frac{\partial w_{\mathbf{n}}}{\partial \overline{z}} \right|^2 \right)^{-1} |\tau(z)|^2 |dz \wedge d\overline{z}| \leq \frac{M}{2} \left( K + \frac{1}{K} \right). \end{split}$$

Lemma 4. The sequence $\{f_n^{-1}\}_{n=1}^{\infty}$ converges uniformly on R'. Proof. Let $\{q_n\}_{n=1}^{\infty}$ be a convergent sequence of points on R', such that $\lim_{n\to\infty}q_n=q_{\infty}$ . Then the sequence of points $\{p_n\}_{n=1}^{\infty}$ satisfying the relation $f_n(p_n)=q_n$ $(n=1,2,\cdots)$ accumulates at some points, one of which we denote by $p_{\infty}$ . A suitable subsequence $\{p_n\}_{\nu=1}^{\infty}$ will converge to $p_{\infty}$ . If $\nu$ is sufficiently large for any given $\varepsilon>0$ , $$\begin{aligned} &\operatorname{dist}\left(f_{n_{\mathsf{V}}}(\mathfrak{p}_{n_{\mathsf{V}}}),\ f_{\scriptscriptstyle{\infty}}(\mathfrak{p}_{\scriptscriptstyle{\infty}})\right) \\ &\leq &\operatorname{dist}\left(f_{n_{\mathsf{V}}}(\mathfrak{p}_{n_{\mathsf{V}}}),\ f_{n_{\mathsf{V}}}(\mathfrak{p}_{\scriptscriptstyle{\infty}})\right) + &\operatorname{dist}\left(f_{n_{\mathsf{V}}}(\mathfrak{p}_{\scriptscriptstyle{\infty}}),\ f_{\scriptscriptstyle{\infty}}(\mathfrak{p}_{\scriptscriptstyle{\infty}})\right) \leqslant \varepsilon, \end{aligned}$$ that is, $$f_{\infty}(\mathfrak{p}_{\infty}) = \lim_{{\scriptscriptstyle \mathcal{V}} \to \infty} f_{n_{\mathcal{V}}}(\mathfrak{p}_{n_{\mathcal{V}}}) = \mathfrak{q}_{\infty}$$ . This implies that any accumulation point of the sequence $\{\mathfrak{p}_n\}_{n=1}^{\infty}$ coincides with $f_{\infty}^{-1}(\mathfrak{q}_{\infty})$ ; the original sequence $\{\mathfrak{p}_n\}_{n=1}^{\infty}$ accumulates only at a single point $f_{\infty}^{-1}(\mathfrak{q}_{\infty})$ , since $f_{\infty}$ is schlicht. Hence $$\lim_{n} f_n^{-1}(\mathfrak{q}_n) = f_\infty^{-1}(\mathfrak{q}_\infty)$$ , in particular $$\lim_{n\to\infty}f_n^{-1}(\mathfrak{q}_\infty)=f_\infty^{-1}(\mathfrak{q}_\infty)$$ . Since $q_{\infty}$ was arbitrary, it follows that for any $q \in R'$ $$\lim_{n\to\infty}f_n^{-1}(\mathfrak{q})=f_\infty^{-1}(\mathfrak{q})$$ . If the last convergence were not uniform, there would exist a positive number $\varepsilon$ , a subsequence $\{f_{nm}^{-1}(\mathfrak{q})\}_{m=1}^{\infty}$ of $\{f_{n}^{-1}(\mathfrak{q})\}_{m=1}^{\infty}$ and a sequence $\{q'_{m}\}_{m=1}^{\infty}$ on R' converging to q', such that dist $$(f_{n_m}^{-1}(\mathfrak{q}'), f_{\infty}^{-1}(\mathfrak{q}')) \geq \varepsilon > 0$$ , which is contrary to the above relation $$\lim_{n\to\infty}f_{nm}^{-1}(\mathfrak{q}'_m)=f^{-1}(\mathfrak{q}'_\infty).$$

Proposition 3. $f_{\infty}^{-1}$ is continuous. Proof. $f_{\infty}^{-1}(\mathfrak{q})$ is the uniform limit of the sequence of continuous maps $f_{n}^{-1}(\mathfrak{q})$ . Proposition 4A. $f_{\infty}$ is $L^2$ -derivable and measurable (i.e., absolutely continuous in 2-dimensional sense) on R. Proof. Let $w=w_{\nu}(z)$ be a local realization of $f_{\nu}(\mathfrak{p})$ near a fixed point $\mathfrak{p}_{0}$ ( $\nu=n,\infty$ ). There exists a sufficiently large number $n_{0}$ and a suitable neighbourhood $\mathfrak{U}(\mathfrak{p}_{0})$ , such that for any $n\geq n_{0}$ $f_{n}(\mathfrak{U}(\mathfrak{p}_{0}))$ is contained in a fixed local coordinate neighbourhood of $f_{\infty}(\mathfrak{p}_{0})$ . Let us denote by |z|<1 the local coordinate disk corresponding to $\mathfrak{U}(\mathfrak{p}_{0})$ , by S the square |x|<1/2, |y|<1/2 and by $\varphi_{0}(z)$ a continuously differentiable function vanishing outside S. Then we have by Green's formula $$0 = \int_{\partial S} [w_n(z) - w_{n+m}(z)] \varphi_0(z) dx = \int_{S} \frac{\partial}{\partial y} \{ [w_n(z) - w_{n+m}(z)] \varphi_0(z) \} dx dy,$$ where m is an arbitrary positive integer. Hence $$\iint_{S} \left( \frac{\partial w_{n}}{\partial y} - \frac{\partial w_{n+m}}{\partial y} \right) \varphi_{0}(z) dxdy = -\iint_{S} \left[ w_{n}(z) - w_{n+m}(z) \right] \frac{\partial \varphi_{0}}{\partial y} dxdy.$$ Since the right hand integral tends to zero as $n \to \infty$ , the definite integrals $$\iint_{S} \frac{\partial w_{n}}{\partial y} \varphi_{0}(z) dx dy \qquad n = 1, 2, \dots$$ constitute a Cauchy sequence. Thus the existence of $\lim_{n\to\infty} \iint_S (\partial w/\partial y) \varphi_0(z) dxdy$ is concluded. This limiting value is obviously a linear functional in $\varphi_0$ . So we shall denote if by $\Lambda(\varphi_0)$ . If $n_0$ is large enough comparatively with $1/\varepsilon$ , it follows from Schwarz's inequality that for $n \ge n_0(\varepsilon)$ $$\big[ \, | \, \Lambda(\varphi_{\scriptscriptstyle 0}) - \mathcal{E} \big]^2 \! \leq \! \int_{\mathcal{S}} \left| \frac{\partial w_{\scriptscriptstyle \boldsymbol{n}}}{\partial y} \right|^2 \! dx dy \int_{\mathcal{S}} \, | \, \varphi_{\scriptscriptstyle 0}(z) \, |^2 dx dy \leq A \int_{\mathcal{S}} \, | \, \varphi_{\scriptscriptstyle 0}(z) \, |^2 dx dy \, ,$$ where A is a constant independent of n. Let $\varepsilon \to 0$ . Then the classical theorem due to F. Riesz allows us to extend the domain of arguments to the class of square-summable functions $\varphi(z)$ and to write $$\Lambda(\varphi) = \iint_{\mathbb{S}} w_{\infty(y)} \varphi(z) dx dy$$ with a function $w_{\omega(y)}(z)$ , which is of summable square and uniquely determined almost everywhere in S (cf. Riesz [7]). It follows that $$\iint_{S} w_{\infty}(z) \frac{\partial \varphi_{0}}{\partial y} dx dy = \lim_{n \to \infty} \iint_{S} w_{n}(z) \frac{\partial \varphi_{0}}{\partial y} dx dy = -\lim_{n \to \infty} \iint_{S} \frac{\partial w_{n}}{\partial y} \varphi_{0}(z) dx dy$$ $$= -\Lambda(\varphi_{0}) = -\iint_{S} w_{\infty(y)} \varphi_{0}(z) dx dy.$$ Since S was arbitrarily taken in |z| < 1, we may write according to the definition by Sobolev-Friedrichs $$\frac{\partial w_{\infty}}{\partial y} = w_{\infty(y)}(z)$$ in the $L^2$ -sense. Analogously we have $$\frac{\partial w_{\infty}}{\partial x} = w_{\infty(x)}(z)$$ in the $L^2$ -sense, where $w_{\infty(x)}$ is the weak limit of $\{\partial w_n/\partial x\}_{n=1}^{\infty}$ with respect to $\varphi_0(z)$ : $$\iint\limits_{S} w_{\infty(x)}(z) \, \varphi_0(z) \, dx dy = \lim_{n \to \infty} \iint\limits_{S} \frac{\partial w_n}{\partial x} \varphi_0(z) \, dx dy.$$ As for the demonstration of measurability of the mapping $w_{\infty}(z)$ , we refer to the skillful device due to Ahlfors [1] and give here an expository version of it. Since $w_n(z)$ is topological in |z| < 1, the image of any Borel set X under each mapping $w_n(z)$ is again a Borel set and $\max w_n(X)$ is a nonnegative additive set-function defined for all Borel sets in |z| < 1. Therefore, by Lebesgue's theorem, the ratio $\max w_n(X)/\max X$ tends to a finite value, when X shrinks to a point z except possibly for a set of measure zero. We can see by making use of the total differentiability of $w_n(z)$ that this limiting value equals $|\partial w_n/\partial z|^2 - |\partial w_n/\partial \bar{z}|^2$ . From the absolute continuity of $\max w_n(X)$ follows the relation $$\operatorname{mes} w_n(X) = \iint\limits_X (|\partial w_n/\partial z|^2 - |\partial w_n/\partial \bar{z}|^2) dx dy.$$ If we take the disk $|z| \le r < 1$ for X in particular and denote its image under every $w_{\nu}$ by $E_{\nu}$ ( $\nu = \infty, 1, 2, \cdots$ ), we see readily $$\limsup_{n\to\infty} E_n \subseteq E_{\infty} \subseteq \liminf_{n\to\infty} E_n,$$ which, together with the well known inequalities mes ( $\liminf_{n\to\infty} E_n$ ) $\leq \liminf_{n\to\infty} \max E_n \leq \limsup_{n\to\infty} \max E_n \leq \max (\limsup_{n\to\infty} E_n)$ yields $$\lim_{n \to \infty} \operatorname{mes} E_n = \operatorname{mes} E_{\infty}.$$ Consider the contour integral $$A_{mn}(r) = \frac{1}{2i} \oint_{|z|=r} [w_m(z) - w_n(z)] d\overline{[w_m(z) - w_n(z)]}.$$ On account of the condition (III), this integral can be defined for almost all r in the interval (0, 1). By Green's formula $$A_{mn}(r) = \iint\limits_{|z| \le r} \left[ \left| \frac{\partial w_m}{\partial z} - \frac{\partial w_n}{\partial z} \right|^2 - \left| \frac{\partial w_m}{\partial \bar{z}} - \frac{\partial w_n}{\partial \bar{z}} \right|^2 \right] dx dy.$$ Given $\varepsilon > 0$ , we can choose $n_0(\varepsilon)$ so large that $|w_m(z) - w_n(z)| < \varepsilon$ uniformly on 0 < |z| = r < 1, if $n, m \ge n_0(\varepsilon)$ . Then $$A_{mn}(r) \leq \varepsilon \oint \left( \left| \frac{\partial w_m}{\partial z} \right| + \left| \frac{\partial w_n}{\partial z} \right| + \left| \frac{\partial w_m}{\partial \bar{z}} \right| + \left| \frac{\partial w_n}{\partial \bar{z}} \right| \right) d \arg z$$ is valid uniformly for a.a. r < 1. From the condition (IV) and the Schwarz's inequality follows $$\lim_{m,n\to\infty}\int_0^1 \left[A_{mn}(r)\right]^2 dr = 0.$$ On the other hand $$A_{mn}(r) = \iint_{|z| \le r} \left( \left| \frac{\partial w_m}{\partial z} \right|^2 - \left| \frac{\partial w_m}{\partial z} \right|^2 \right) dx dy + \iint_{|z| \le r} \left( \left| \frac{\partial w_n}{\partial \bar{z}} \right|^2 - \left| \frac{\partial w_n}{\partial \bar{z}} \right|^2 \right) dx dy - 2\Re \left\{ \iint_{|z| \le r} \left[ \frac{\partial w_m}{\partial z} \left( \frac{\overline{\partial w_n}}{\partial z} \right) - \frac{\partial w_m}{\partial \bar{z}} \left( \frac{\overline{\partial w_n}}{\partial \bar{z}} \right) \right] dx dy \right\}.$$ Letting first $m \to \infty$ then $n \to \infty$ , we obtain by (3) and the weak convergence of the complex derivatives $$(5) \qquad \lim_{n\to\infty} \lim_{m\to\infty} A_{mn}(r) = 2 \operatorname{mes} E_{\infty} - 2 \iint_{|z| < r} \left( \left| \frac{\partial w_{\infty}}{\partial z} \right|^2 - \left| \frac{\partial w_{\infty}}{\partial \bar{z}} \right|^2 \right) dx dy.$$ By making use of the uniform boundedness of $A_{mn}(r)$ on (0, 1), we substitute (5) into (4) after application of Lebesgue's theorem. Then we see that the right-hand side in (5) equals zero for almost every r in (0, 1). Thus it has been concluded that (6) $$\operatorname{mes} E_{\infty} = \iint_{|z| \leq r} \left( \left| \frac{\partial w_{\infty}}{\partial z} \right|^2 - \left| \frac{\partial w_{\infty}}{\partial \bar{z}} \right|^2 \right) dx dy$$ for a.a. r and by continuity (6) holds good for all r on the interval [0, 1]. This reasoning is valid not only for the disk $|z| \le r$ but also for arbitrary disks contained in |z| < 1. The density of the set-function is defined almost everywhere and equal to $|\partial w_{\infty}/\partial z|^2 - |\partial w_{\infty}/\partial \bar{z}|^2$ . Integral expression analogous to (6) on an arbitrary Borel set implies the measurability of the mapping $w_{\infty}(z)$ . Since the point $\mathfrak{p}_0$ was arbitrarily taken on R, we complete the proof. Proposition 4B. $f_{\infty}^{-1}$ is $L^2$ -derivable and measurable (i.e., absolutely continuous in 2-dimensional sense) on R'. Proof. By Lemma 4, the sequence $\{f_n^{-1}\}_{n=1}^{\infty}$ converges to $f_{\infty}^{-1}$ uniformly on R'. Hence the proof for $L^2$ -derivability follows the same line as that in Proposition 4A. Let w(z) be a local realization of $f_{\infty}$ around a point $\mathfrak{p} \in R$ . Let $E_0$ be the subset of R, where any local realization of $f_{\infty}$ ceases to be totally differentiable. Since $\operatorname{mes} E_0 = 0$ (cf. Gehring-Lehto [5]), $f_{\infty}(E_0)$ is also of measure zero by the preceding proposition, whence it is seen that both w(z) and z(w) are totally differentiable in respective variables at almost every point on the surfaces. Concerning the local realization w(z) of $f_{\infty}$ near such a point $\mathfrak{p}$ , we set $$w(z) = u(x+iy)+iv(x+iy), \quad \Delta z = \Delta x+i\Delta y.$$ $w(z+\Delta z)-w(z) = \Delta w = \Delta u+i\Delta v.$ Then $$\Delta x = x_u(u, v) \Delta u + x_v(u, v) \Delta v + o(|\Delta u| + |\Delta v|),$$ $$\Delta y = y_u(u, v) \Delta u + y_v(u, v) \Delta v + o(|\Delta u| + |\Delta v|).$$ In particular, putting $\Delta v = 0$ , we have $$\Delta x = x_u(u, v) \Delta u + o(\Delta u), \quad \Delta y = y_u(u, v) \Delta u + o(\Delta u),$$ whence $$\Delta u = u_x(x, y) \Delta x + u_y(x, y) \Delta y + o(|\Delta x| + |\Delta y|)$$ $$= u_x(x, y) [x_u(u, v) \Delta u + o(\Delta u)] + u_y(x, y) [y_u(u, v) \Delta u + o(\Delta u)]$$ $$+ o(|\Delta x| + |\Delta y|).$$ Letting $\Delta u \rightarrow 0$ , we see $$1 = u_{x}(x, y) \cdot x_{u}(u, v) + u_{v}(x, y) \cdot y_{u}(u, v)$$ and analogously $$0 = v_{n}(x, y) \cdot x_{n}(u, y) + v_{n}(x, y) \cdot v_{n}(u, y).$$ Therefore (7) $$x_{u}(u, v) = \frac{v_{y}(x, y)}{u_{x}(x, y)v_{y}(x, y) - u_{y}(x, y)v_{x}(x, y)},$$ $$y_{u}(u, v) = -\frac{v_{x}(x, y)}{u_{x}(x, y)v_{y}(x, y) - u_{y}(x, y)v_{x}(x, y)},$$ and similarly (8) $$x_{v}(u, v) = -\frac{u_{y}(x, y)}{u_{x}(x, y)v_{y}(x, y) - u_{y}(x, y)v_{x}(x, y)},$$ $$y_{v}(u, v) = \frac{u_{x}(x, y)}{u_{x}(x, y)v_{y}(x, y) - u_{y}(x, y)v_{x}(x, y)}.$$ The equivalents to (7), (8) are $$\frac{\partial z}{\partial w} = \frac{\overline{\left(\frac{\partial w}{\partial z}\right)}}{\left|\frac{\partial w}{\partial z}\right|^2 - \left|\frac{\partial w}{\partial \bar{z}}\right|^2}, \qquad \frac{\partial z}{\partial \overline{w}} = \frac{-\frac{\partial w}{\partial \bar{z}}}{\left|\frac{\partial w}{\partial z}\right|^2 - \left|\frac{\partial w}{\partial \bar{z}}\right|^2},$$ which shows that, $|\partial w/\partial z|^2 - |\partial w/\partial \bar{z}|^2 \neq 0$ at all points where w(z) as well as its inverse function z(w) are totally differentiable. In other words, $f_{\infty}$ never transforms any set of positive measure on R to a set of measure zero on R', which implies the absolute continuity of $f_{\infty}$ .

Proposition 5. $f_{\infty}$ is homotopic to X. Proof. Let $J_n = [1-2^{1-n}, 1-2^{-n}]$ denote the closed interval $(n=1, 2, \dots)$ . Since $f_n \approx f_{n+1}$ , there exists such a function $f(\mathfrak{p}, t)$ continuous on the product set $R \times J_n$ , that $f(\mathfrak{p}, 1-2^{1-n}) = f_n(\mathfrak{p})$ , $f(\mathfrak{p}, 1-2^{-n}) = f_{n+1}(\mathfrak{p})$ . Moreover it is possible to take $f(\mathfrak{p}, t)$ , so that (9) $$\sup_{R} \underset{J_{n}}{\operatorname{osc}} f(\mathfrak{p}, t) = \sup_{R} \sup_{J_{n}} \operatorname{dist} (f(\mathfrak{p}, t_{1}), f(\mathfrak{p}, t_{2})) \leq \sup_{R} \operatorname{dist} (f_{n}(\mathfrak{p}), f_{n+1}(\mathfrak{p})).$$ To this end we shall have only to define as follows: $f(\mathfrak{p}, t)$ is the point on a uniquely determined geodesic arc (e.g., with respect to the Poincaré metric) connecting $f_j(\mathfrak{p})$ (j=n, n+1), such that dist $(f_n(\mathfrak{p}), f(\mathfrak{p}, t))/\text{dist}$ $(f_n(\mathfrak{p}), f_{n+1}(\mathfrak{p})) = t$ . We first show that $f(\mathfrak{p}, t)$ defined in $R \times \bigcup_{\nu=1}^{\infty} J_{\nu}$ is continuous on $R \times \bigcup_{\nu=1}^{m} J_{\nu}$ for every m. It is evidently true for m=1. Suppose the assertion holds for m=n. Then the analogous conclusion is also valid for m=n+1. Because, in general (10) $$\begin{aligned} \operatorname{dist} \left( f(\mathfrak{p}_{1}, \ 1 - 2^{-n} - \delta), \ f(\mathfrak{p}_{2}, \ 1 - 2^{-n} + \delta) \right) \\ &\leq \operatorname{dist} \left( f(\mathfrak{p}_{1}, \ 1 - 2^{-n} - \delta), \ f(\mathfrak{p}_{1}, \ 1 - 2^{-n}) \right) \\ &+ \operatorname{dist} \left( f(\mathfrak{p}_{1}, \ 1 - 2^{-n}), \ f(\mathfrak{p}_{1}, \ 1 - 2^{-n} + \delta) \right) \\ &+ \operatorname{dist} \left( f(\mathfrak{p}_{1}, \ 1 - 2^{-n} + \delta), \ f(\mathfrak{p}_{2}, \ 1 - 2^{-n} + \delta) \right) . \end{aligned}$$ Since $f(\mathfrak{p}, t)$ is uniformly continuous on $R \times \bigcup_{\nu=1}^{m} J_{\nu}$ as well as on $R \times J_{m+1}$ , we can find suitable positive numbers $\delta_{j} = \delta_{j}(\mathcal{E})$ for any $\mathcal{E} > 0$ (j=4, 5), such that the inequality (11) $$\operatorname{dist}(f(\mathfrak{p},t), f(\mathfrak{p}',t')) < \varepsilon$$ holds in the two cases: (1°) when $$0 \le t$$ , $t' \le 1-2^{-n}$ , dist $(\mathfrak{p}, \mathfrak{p}') + |t-t'| < \delta_4$ , (2°) when $$1-2^{-n} \le t$$ , $t' \le 1-2^{1-n}$ , dist $(\mathfrak{p}, \mathfrak{p}') + |t-t'| < \delta_{\mathfrak{s}}$ . Take $\mathfrak{p}_1$ , $\mathfrak{p}_2$ and $\delta_6$ , so that dist $(\mathfrak{p}_1, \mathfrak{p}_2) + \delta_6 < \min(\delta_4, \delta_5)$ . Then by (10), (11) dist $(f(\mathfrak{p}_1, 1-2^{-n}-\delta_6), f(\mathfrak{p}_2, 1-2^{-n}+\delta_6) < 3\varepsilon$ . Hence $f(\mathfrak{p}, t)$ is continuous on $R \times \bigcup_{\nu=1}^{m+1} J_{\nu}$ . We can conclude inductively that $f(\mathfrak{p}, t)$ is defined and continuous on R [0, 1). Next, if we set $f(\mathfrak{p}, 1) = f_{\infty}(\mathfrak{p})$ , $f(\mathfrak{p}, t)$ is continuously prolonged up to $R \times [0, 1]$ . Actually, we have $$\begin{aligned} &\operatorname{dist}\left(f(\mathfrak{p},1),\ f(\mathfrak{p}',t)\right) \\ &\leq \operatorname{dist}\left(f(\mathfrak{p},1),\ f(\mathfrak{p}',1)\right) + \operatorname{dist}\left(f(\mathfrak{p}',1),\ f(\mathfrak{p}',t)\right) \\ &\leq \operatorname{dist}\left(f(\mathfrak{p},1),\ f(\mathfrak{p}',1)\right) + \operatorname{dist}\left(f_{\omega}(\mathfrak{p}'),\ f_{n}(\mathfrak{p}')\right) + \operatorname{dist}\left(f_{n}(\mathfrak{p}'),\ f(\mathfrak{p}',t)\right). \end{aligned}$$ Since $\{f_n\}_{n=1}^{\infty}$ converges to $f_{\infty}$ uniformly on R, we can choose $n_0(\varepsilon)$ , so that dist $(f_{\infty}(\mathfrak{p}'), f_n(\mathfrak{p}')) < \varepsilon/3$ so far as $n \ge n_0(\varepsilon)$ . If t belongs to $f_n$ , we have in virtue of (9) dist $$(f_n(\mathfrak{p}'), f(\mathfrak{p}', t)) < \varepsilon/3$$ for any $\mathfrak{p}' \in R$ . Taking $\mathfrak{p}, \mathfrak{p}'$ so near, that the relation dist $(f(\mathfrak{p}, 1), f(\mathfrak{p}', 1)) < \varepsilon/3$ holds, we obtain finally dist $$(f(\mathfrak{p}, 1), f(\mathfrak{p}', t)) < \varepsilon$$ . Thus we see $f_{\infty}(\mathfrak{p}) \approx f_{\mathfrak{i}}(\mathfrak{p})$ , which was to be proved.

Proposition 6. $f_{\infty}$ is a sense-preserving map. Proof. Let $w_n(z)$ be a local realization near $\mathfrak{p}_0$ (|z| < 1; $n = \infty, 1, 2, \cdots$ ), such that z = 0 corresponds to $\mathfrak{p}_0$ and $w_{\infty}(0) = 0$ . If n is large enough, the image of a fixed annulus $0 < a \le |z| \le b < 1$ under the mapping $w_n(z)$ is bounded away from zero. Suitable choice of a branch enables us to define a one-valued function $\log w_{\infty}(z) - \log w_n(z)$ on $a \le |z| \le b$ , which tends to zero uniformly there. We have by Green's formula and Riesz's theorem $$\lim_{n\to\infty} \iint_{a\leq |z|\leq b} \frac{z}{w_n} \frac{\partial w_n}{\partial z} |dz \wedge d\bar{z}| = \iint_{a\leq |z|\leq b} \frac{z}{w_\infty} \frac{\partial w_\infty}{\partial z} |dz \wedge d\bar{z}|,$$ $$\lim_{n\to\infty} \iint_{a\leq |z|\leq b} \frac{z}{w_n} \frac{\partial w_n}{\partial \bar{z}} |dz \wedge d\bar{z}| = \iint_{a\leq |z|\leq b} \frac{z}{w_\infty} \frac{\partial w_\infty}{\partial \bar{z}} |dz \wedge d\bar{z}|,$$ whence follows $$\lim_{n\to\infty} \iint_{a<|z|< b} \frac{\partial \arg w_n}{\partial \arg z} |dz \wedge d\bar{z}| = \iint_{a<|z|< b} \frac{\partial \arg w_\infty}{\partial \arg z} |dz \wedge d\bar{z}|.$$ Since $w_n(z)$ is sense-preserving, it is valid that $$\oint_{|z|=r} \frac{\partial \arg w_n}{\partial \arg z} d \arg z = 2\pi$$ on almost all circumferences |z|=r where arg w is absolutely continuous function in arg z. Thus we get $$\iint_{a \le |z| \le b} \frac{\partial \arg w_n}{\partial \arg z} |dz \wedge d\bar{z}| = 2\pi (b^2 - a^2),$$ accordingly $$\iint\limits_{z\leq |a|\leq b}\frac{\partial \operatorname{arg} w_{\infty}}{\partial \operatorname{arg} z}|dz\wedge d\bar{z}|=2\pi(b^2-a^2),$$ and $$\lim_{a^c\to 0}\frac{1}{\Delta c}\iint\limits_{0\leq |z|\leq c+dc}\frac{\partial \arg w_\infty}{\partial \arg z}|dz\wedge d\bar{z}|=4\pi c\qquad \text{for every }c\in [a,b]\,.$$ It follows that $$2\pi = \oint_{|z|=c} \frac{\partial \arg w_{\infty}}{\partial \arg z} d \arg z = \oint_{|z|=c} \frac{\partial \arg w_{n}}{\partial \arg z} d \arg z \quad \text{for every } c \in [a, b].$$ This shows that the assigned orientation on |z|=c is preserved by $w_{\infty}(z)$ as well as by $w_{n}(z)$ . $f_{\infty}$ is a sense-preserving map, since it is topological. Q.E.D. Owing to the Propositions 4A and 4B, the dilatation $D_{f_{\infty}}(\mathfrak{p})$ of the mapping $f_{\infty}(\mathfrak{p})$ is finitely defined at almost every point of R.

Lemma 5. The family $\Omega$ is compact on R. Proof. Take any infinite sequence $\{\omega_n\}_{n=1}^{\infty}$ out of $\Omega$ . Let $|z| \leq 1$ (resp. $|z'| \leq 1$ ) be the homeomorphic image of $S_j$ (resp. $S_{j+1}$ ) under a coordinate map $z = z(\mathfrak{p})$ (resp. $z' = z'(\mathfrak{p})$ ). Set $$\omega_n = \left\{ \begin{array}{ll} \tau_n(z)dz & \text{in} \quad |z| < 1, \\ \tau'_n(z')dz' & \text{in} \quad |z'| < 1, \end{array} \right. \quad T_n(z) = \int_0^z \tau_n(\zeta)d\zeta \quad |z| \le 1.$$ Then by the Remark to Lemma 1, $\{T_n(z)\}_{n=1}^{\infty}$ contains a suitable subsequence $\{T_n(z)\}_{n=1}^{\infty}$ converging uniformly on $|z| \le 1$ . Write $$T_{\scriptscriptstyle{\infty}}(z) = \lim_{\scriptscriptstyle{\rm N} \subset {\rm I}} T_{\scriptscriptstyle{\rm N}_{\scriptscriptstyle{\rm N}}}\!(z) \,\,\,{\rm on}\,\,\,\,|z| \,{\leq}\, 1, \,\,\,\,\,\,\tau_{\scriptscriptstyle{\rm N}}\!(z) = d\,T_{\scriptscriptstyle{\rm N}}\!(z)/dz \,\,\,{\rm in}\,\,\,\,|z| \,{<} 1\,.$$ Then $\{\tau_{n_{\nu}}(z)\}$ converges uniformly to $\tau_{\infty}(z)$ on $|z| \leq r < 1$ , whence $$\int\limits_0^{2\pi} |\tau_{\scriptscriptstyle{\infty}}(re^{i\theta})| rd\theta = \lim_{\scriptscriptstyle{\nu\rightarrow \infty}} \int\limits_0^{2\pi} |\tau_{n_{\scriptscriptstyle{\nu}}}(re^{i\theta})| rd\theta \leq M' < M$$ by the condition imposed on $\Omega$ . Since r < 1 is arbitrary, $\tau_{\infty}(z)$ has the Fatou property in |z| < 1. The boundary values $\tau_{\infty}(e^{i\theta})$ can be defined for almost all $\theta \in (0, 2\pi)$ . Since the total variation $V[T_{n_{\nu}}(e^{i\theta}); 0, 2\pi]$ of the function $T_{n_{\nu}}(e^{i\theta})$ over the interval $(0, 2\pi)$ ( $\nu = 1, 2, \cdots$ ) is uniformly bounded, we get by Helly's theorem that $$(14) \lim_{\nu \to \infty} \int_{0}^{2\pi} |\tau_{n_{\nu}}(e^{i\theta})| d\theta = \lim_{\nu \to \infty} V[T_{n_{\nu}}(e^{i\theta}); 0, 2\pi] = V[T_{\infty}(e^{i\theta}); 0, 2\pi].$$ On the other hand, on account of the monotonicity of concentrically circumferential integrals and the uniform convergence, we have $$\int\limits_{0}^{2\pi}|\tau_{\infty}(e^{i\theta})|d\theta\geq\int\limits_{0}^{2\pi}|\tau_{\infty}(re^{i\theta})|rd\theta=\lim_{\nu\rightarrow\infty}\int\limits_{0}^{2\pi}|\tau_{n\nu}(re^{i\theta})|rd\theta$$ for every r < 1, whence $$\lim_{\nu\to\infty}\int\limits_0^{2\pi}|\tau_{n_\nu}\!(e^{i\theta})|\,d\theta\!\le\!\int\limits_0^{2\pi}|\tau_\infty(e^{i\theta})|\,d\theta\,.$$ This inequality together with (14) afford (15) $$\int_{0}^{2\pi} |\tau_{\infty}(e^{i\theta})| d\theta = \lim_{\nu \to \infty} \int_{0}^{2\pi} |\tau_{n_{\nu}}(e^{i\theta})| d\theta = V[T_{\infty}(e^{i\theta}); 0, 2\pi],$$ which implies the absolute continuity of the absolute variation $V[T_{\infty}(e^{i\theta}); 0, \Theta]$ for every $\Theta \in (0, 2\pi)$ . Taking into consideration the uniform convergence of $\{T_{n_{\nu}}(e^{i\theta})\}_{\nu=1}^{\infty}$ towards $T_{\infty}(e^{i\theta})$ on every subinterval $(\theta_1, \theta_2)$ of $(0, 2\pi)$ , we obtain $$T_{\scriptscriptstyle \infty}(e^{i\theta_2})\!-T_{\scriptscriptstyle \infty}(e^{i\theta_1})=\int\limits_{\theta_{\scriptscriptstyle +}}^{\theta_2}\tau_{\scriptscriptstyle \infty}(e^{i\theta})ie^{i\theta}d\theta$$ and from (15) (16) $$\int_{\theta_1}^{\theta_2} |\tau_{\infty}(e^{i\theta})| d\theta = \lim_{\nu \to \infty} \int_{\theta_1}^{\theta_2} |\tau_{n_{\nu}}(e^{i\theta})| d\theta.$$ Consider the case in which $S_j \cap S_{j+1}$ contains a 1-simplex. Then setting $\widehat{\zeta_1}$ , $\widehat{\zeta_2} = z(\gamma)$ and $\widehat{\zeta_1'}$ , $\widehat{\zeta_2'} = z'(\gamma)$ for any arc $\gamma$ contained in $S_j \cap S_{j+1}$ , we can see by (16) that concerning the integration along $\gamma$ $$\int\limits_{\zeta_1}^{\zeta_2} |\tau_{\infty}(\zeta)| \, |d\zeta| = \lim_{\nu \to \infty} \int\limits_{\zeta_1}^{\zeta_2} |\tau_{n_{\nu}}(\zeta)| \, |d\zeta| = \lim_{\nu \to \infty} \int\limits_{\zeta_1'}^{\zeta_2'} |\tau'_{n_{\nu}}(\zeta')| \, |d\zeta'| = \int\limits_{\zeta_1'}^{\zeta_2'} |\tau'_{\infty}(\zeta')| \, |d\zeta'| \; .$$ The assertion $|\tau_{\omega}(\zeta)| |d\zeta| = |\tau'_{\omega}(\zeta')| |d\zeta'|$ for a.a. $\mathfrak{p} \in S_j \cap S_{j+1}$ $(j=1,2,\cdots,\kappa-1)$ follows at once, where $\zeta = z(\mathfrak{p})$ and $\zeta' = z'(\mathfrak{p})$ . Thus we have had the linear differential $\omega_{\infty} = \tau_{\infty}(z)dz$ holomorphic in $R - \bigcup\limits_{j=1}^{\kappa} \partial S_j$ , and $|\omega_{\infty}| = |\tau_{\infty}(z)| |dz|$ is well-defined at almost all points on $\bigcup\limits_{j=1}^{\kappa} \partial S_j$ and clearly $\sum_{j=1}^{\kappa} \int\limits_{\partial S_j} |\omega_{\infty}| \leq M$ . Next, let |z|=r be the circle on which $\tau_{\infty}(z) \neq 0$ . If $|\zeta| < r$ , we have $$\log |\tau_{n_{\mathcal{V}}}(\zeta)| = \frac{1}{2\pi} \int\limits_0^{2\pi} (\log |\tau_{n_{\mathcal{V}}}(z)|) \frac{r^2 - |\zeta|^2}{|z - \zeta|^2} d \operatorname{arg} z \ .$$ The right-hand integral tends to $(1/2\pi)\int_{0}^{2\pi}(\log|\tau_{\infty}(z)|)[(r^{2}-|\zeta|^{2})/|z-\zeta|^{2}]$ $d \arg z$ as $\nu \to \infty$ , while $\tau_{n_{\nu}}(\zeta)$ converges to $\tau_{\infty}(\zeta)$ uniformly in $|\zeta| < r$ . Hence $$|\tau_{\omega}(\zeta)| = \exp\left(\frac{1}{2\pi}\int\limits_{0}^{2\pi}(\log|\tau_{\omega}(z)|)\frac{r^2-|\zeta|^2}{|z-\zeta|^2}d\arg z\right) + 0, \qquad |\zeta| < r.$$ Since r is taken arbitrarily near 1, it follows that $\tau_{\infty}(\zeta) \neq 0$ in $|\zeta| < 1$ . One concludes that $\omega_{\infty}$ belongs to $\Omega$ .

Lemma 6. Let $\mathfrak{p}_j$ (j=1,2) be arbitrary points on a singular simplex $S_l$ and let $\Gamma$ the class of all rectifiable simple arcs which connects the two points $\mathfrak{p}_j$ on $S_l$ . Then for any $\gamma \in \Gamma$ and for any $\omega = \tau(z) dz \in \Omega$ , the quantity $$\int_{\gamma} |\omega| = \int_{\gamma} |\tau(z)| |dz|$$ is bounded away from zero. Proof. Suppose on the contrary $$\inf_{\substack{\gamma \in \Gamma \\ \omega \in \Omega}} \int_{\gamma} |\omega| = 0.$$ Then there would exist $\omega_n = \tau_n(z) dz \in \Omega$ and $\gamma_n \in \Gamma$ $(n=1, 2, \dots)$ , such that (17) $$\lim_{n\to\infty}\int_{\gamma_n}|\omega_n|=0.$$ Let z be a local coordinate defined in |z| < 1 corresponding to the interior to $S_l$ , such that $z(\mathfrak{p}_0) = 0$ , $\mathfrak{p}_0$ being a fixed interior point of $S_l$ . Set $$T_n(z) = \int_0^z \tau_n(\zeta) d\zeta \quad \text{for} \quad |z| < 1.$$ Then, according to the Remark after Lemma 1, $T_n(z)$ can be prolonged continuously onto $|z| \le 1$ and the sequence $\{T_n(z)\}_{n=1}^\infty$ contains at least one subsequence, say $\{T_{n_\nu}(z)\}_{\nu=1}^\infty$ converging uniformly on $|z| \le 1$ . Put $T_\infty(z) = \lim_{\nu \to \infty} T_{n_\nu}(z)$ . Then $$|T_{\infty}(z_2) - T_{\infty}(z_1)| = \lim_{\nu \to \infty} |T_{n_{\nu}}(z_2) - T_{n_{\nu}}(z_1)| \le \lim_{\nu \to \infty} \int_{\gamma_{n_{\nu}}} |\tau_{n_{\nu}}(z)| |dz|,$$ $z_j = z(\mathfrak{p}_j), \qquad j = 1, 2.$ The right-hand side is equal to zero by (17). Hence $T_{\infty}(z_1) = T_{\infty}(z_2)$ . $\tau_{\infty}(z) = dT_{\infty}/dz$ is defined and possesses the Fatou property in |z| < 1. $\gamma_n$ are analytic arcs which are uniformly bounded. So by Vitali's theorem there exists a suitable path $\gamma_{\infty} \in \Gamma$ from $\mathfrak{p}_1$ to $\mathfrak{p}_2$ , such that 198 K. SHIBATA (18) $$\int_{\gamma_{\infty}} |\tau_{\infty}(z)| |dz| = 0.$$ If we take on γ^ any point p, we have by (18) Teo (z(p))=Tββ (z1 )=Tco (z2 ). Therefore Too (z) = const, on |^|<1 , whence ωco = roo (z)dz=0 on |z|=l . Since ωTO belongs to ί2, it follows from the requirement concerning Ω that ω^ΞΞO on every simplex adjacent to Sly accordingly on the whole i?, which is absurd.

Proposition 9. g^p) is schlicht. Proof. Suppose, on the contrary, gjp^=gjp^ = (\ for some p19 p2€ R (Pi Φ P² )> q£ R'- Let z=zωn (w) (IM;|<1) be a local realization of / ^ about fω^M (Pi)f suc h tha ^t fωJPi) corresponds to w = 0. Set $$A_{n}(r) = \int\limits_{0}^{r} \int\limits_{0}^{2\pi} |\tau_{n}(z_{\omega_{n}}(w))|^{2} \left( \left| \frac{\partial z_{\omega_{n}}}{\partial w} \right|^{2} + \left| \frac{\partial z_{\omega_{n}}}{\partial \overline{w}} \right|^{2} \right) t d\theta dt \qquad w = te^{i\theta} \,.$$ Then clearly $$A_{n}(r) \leq \frac{M}{4} \left(K + \frac{1}{K}\right).$$ We can find, by Lemma 6, a positive c independent of t and n, such that $$c \leq \int\limits_0^{2\pi} |\tau_{\mathbf{n}}(z_{\omega_{\mathbf{n}}}(te^{i\theta}))| |dz_{\omega_{\mathbf{n}}}(te^{i\theta})| \ .$$ Hence follows the inequality $$c^2 \leq 4\pi t \ dA_n(t)/dt$$ , leading to a contradiction.

Proposition 10. goo(P) is a homeomorphism between R and R'. Proof. This is an immediate consequence from the continuity of goo(p) and Proposition 9. Proposition 11A. g^P) is U-deriυable and absolutely continuous in 2-dirnensional sense on R. Proposition 11B. gZ¹ ^) is U-derivable and absolutely continuous in 2-dimensional sense on R'. The proofs are similar to that of Proposition 4A (resp. Proposition 4B), since g^ (resp. gz 1 ) is the uniform limit of {f } (resp. {/']} ).

Proposition 12. $g_{\infty}$ is homotopic to $\chi$ .

Proposition 13. $g_{\infty}$ is a sense-preserving map. The proofs are similar to that of Propositions 5 and 6, since $g_{\infty}$ is the uniform limit of $\{f_{\omega_n}\}$ . The class $\Omega$ of the locally holomorphic differentials $\omega$ is compact by Lemma 5, so the sequence $\{\omega_n\}_{n=1}^{\infty}$ contains at least one subsequence uniformly convergent in $R-\bigcup\limits_{j=1}^{\kappa} \partial S_j$ , the limit of which we denote by $\omega_{\infty}=\tau_{\infty}(z)\,dz$ ; it belongs again to $\Omega$ . Let $z_{\omega_n}(w),\,Z_{\infty}(w)$ be local realizations of $f_{\omega_n}^{-1}(\mathfrak{q}),\,g_{\infty}^{-1}(\mathfrak{q})$ respectively. Then, using the similar reasoning to that in the proof of Proposition 7, we see $$\begin{split} & \iint\limits_{\mathbb{R}'} \left( \left| \frac{\partial Z_{\infty}}{\partial w} \right|^2 + \left| \frac{\partial Z_{\infty}}{\partial \overline{w}} \right|^2 \right) |\tau_{\infty}(Z_{\infty}(w))|^2 |dw \wedge d\overline{w}| \\ \leq & \liminf\limits_{n \to \infty} \iint\limits_{\mathbb{R}'} \left( \left| \frac{\partial z_{\omega_n}}{\partial w} \right|^2 + \left| \frac{\partial z_{\omega_n}}{\partial \overline{w}} \right|^2 \right) |\tau_n(z_{\omega_n}(w))|^2 |dw \wedge d\overline{w}| \leq \frac{M}{2} \left( K + \frac{1}{K} \right). \end{split}$$ Thus we have obtained, concerning the dilatation $D_{g_{\infty}}$ of the mapping $g_{\infty}$ the following

Proposition 15. If M is sufficiently large, there exists at least one locally holomorphic differential $\tilde{\omega}_M \in \Omega_M$ , such that $$(20) \qquad \qquad \iint\limits_{\mathbb{R}} \left[ D_{f_{M}}(\mathfrak{p}) + \frac{1}{D_{f_{M}}(\mathfrak{p})} \right] \tilde{\omega}_{M} \star \tilde{\omega}_{M} < M \left( K + \frac{1}{K} \right).$$ Proof. We see trivially that the left-hand side of (20) cannot be greater than $M\left(K+\frac{1}{K}\right)$ . If the inequality sign enters in this circumstance for some large value of M, the conclusion is immediate by posing $\tilde{\omega}_M = \omega_M$ . So we may assume (21) $$\iint_{R} \left[ D_{f_{M}}(\mathfrak{p}) + \frac{1}{D_{f_{M}}(\mathfrak{p})} \right] \omega_{M} * \omega_{M} = M \left( K + \frac{1}{K} \right)$$ for every $M$ . Under these hypotheses our purpose is to derive $\tilde{\omega}_M$ from $\omega_M$ through some approximation process for sufficiently large M. In spite of an extreme naïvety of the underlying plan, the author has regrettedly failed to avoid a certain cumbersomeness in the details of proof. Let us set $$\begin{split} E^+ &= \left\{ \mathfrak{p} \,|\, D_{f_M}(\mathfrak{p}) + \frac{1}{D_{f_M}(\mathfrak{p})} \geq M \Big( K + \frac{1}{K} \Big) \right\}\,, \\ E^- &= \left\{ \mathfrak{p} \,|\, D_{f_M}(\mathfrak{p}) + \frac{1}{D_{f_M}(\mathfrak{p})} \leq K + \frac{1}{K} \right\}\,. \end{split}$$ In view of the conditions (IV), (V) and (21), both $E^+$ and $E^-$ are sure to be of positive measure. In order to dispense with unnecessary lengthiness of the context, we shall deal with the case when there exists at least one simplex $S_l \in \Sigma$ , such that $\text{mes}(E^+ \cap S_l) \text{mes}(E^- \cap S_l) > 0$ . The proof for the other case follows almost the similar line. Given an arbitrarily small $\varepsilon > 0$ , let F be a closed subset of $E^- \cap S_t$ , such that (22) $$\iint_{(\mathcal{B}^{-} \cap S_{\ell})^{-F}} \left[ D_{f_{M}}(\mathfrak{p}) + \frac{1}{D_{f_{M}}(\mathfrak{p})} \right] \omega_{M} * \omega_{M} < \varepsilon/M.$$ Let $B_z$ be a simply connected subdomain of a local coordinate disk $U_z = \{z \mid |z| < 1\}$ corresponding to int $S_t$ with piecewise analytic boundary which encloses z(F), such that (23) $$\iint_{B_{\sigma^{-\bar{z}}(F)}} \left[ D_{f_{M}}(z) + \frac{1}{D_{f_{M}}(z)} \right] |\tau_{M}(z)|^{2} |dz \wedge d\bar{z}| < \varepsilon/M.$$ Suppose, for the time being, that $\bar{B}_z$ contains a subarc $\gamma_z \in \partial U_z$ of length $2\delta > 0$ , whose middle point we denote by $z_0$ . Let a < a' be two positive numbers smaller than 1 and let $\tilde{B}_z$ a Jordan subregion confined in $B_z$ . How to fit them shall be found soon below. Employing them, we are going to define two domains $\Delta_z$ , $\Delta_z'$ and the positive quantity $\bar{c}$ : $$\Delta_{z} = \{z \mid |\arg z - \arg z_{0}| < \delta\} \land \{z \mid |z| > a'\},$$ $$\Delta'_{z} = \Delta_{z} \cup \{z \mid a < |z| < a'\},$$ $$\overline{c} = \iint_{|z| < 1} |\tau_{M}(z)|^{2} |dz \wedge d\overline{z}| / \iint_{\widetilde{B}z} |\tau_{M}(z)|^{2} |dz \wedge d\overline{z}|.$$ There exists a polynomial $\varphi(\zeta)$ which satisfies the inequalities (24) $$\begin{vmatrix} \frac{1}{2}\log c - \varphi(\zeta) | \leq (\varepsilon/M)^2 & \text{ on } |\zeta| \leq a, \\ \left| 3\log \frac{1}{M} - \varphi(\zeta) | \leq (\varepsilon/M)^2 & \text{ for } \zeta \in U_{\zeta} - \Delta_{\zeta}' \text{ belonging to } |\zeta| \geq a' \end{aligned}$$ (cf. Walsh [11], p. 15). Set $A_0 = \max |\exp \varphi(z)|$ . Through an adequate choice of a, a' and $\tilde{B}_z$ it is possible to find such a linear transformation $\zeta(z)$ of $U_z$ onto itself, that $\zeta(\tilde{B}_z)$ is contained in $|\zeta| < a$ , and that (25) $$a' < \min_{\overline{U_z - B_z}} |\zeta(z)| < \max_{\overline{U_z - B_z}} |\zeta(z)| < 1, \ \zeta(U_z - B_z) \cap \Delta'_z = \phi,$$ $$\iint_{d_{\zeta'}} \left[ D_{f_M}(\zeta) + \frac{1}{D_{f_M}(\zeta)} \right] |\sigma_M(\zeta)|^2 |d\zeta \wedge d\overline{\zeta}| < \varepsilon / A_0^2 M^2,$$ $$\int_{\partial \overline{U_\zeta} \cap A_\zeta} |\sigma_M(\zeta)| |d\zeta| < \frac{1}{A_0} \oint_{|z|=1} |\tau_M(z)| |dz|,$$ where $\sigma_M(\zeta)d\zeta = \tau_M(z)dz$ . Put $$P^{(l)}(\zeta) = \sigma_M(\zeta) \exp \varphi(\zeta), \qquad |\zeta| < 1.$$ Then by (24), (25) (26) $$\int \int \limits_{|\mathcal{E}| < 1} |P^{(I)}(\zeta)|^2 |d\zeta \wedge d\bar{\zeta}| \ge (1 - 2(\mathcal{E}/M)^2) \int \int \limits_{|z| < 1} |\tau_M(z)|^2 |dz \wedge d\bar{z}| ,$$ (26') $$\oint_{|\zeta|=1} |P^{(I)}(\zeta)| |d\zeta| < (1 + e^{(\epsilon/M)^2}/M^3) \oint_{|z|=1} |\tau_M(z)| |dz|.$$ In possession of (22), (24), (25) we assert that there exists a positive constant $c_0$ , such that (26") $$\iint_{|\zeta| < 1} \left[ D_{f_{M}}(\zeta) + \frac{1}{D_{f_{M}}(\zeta)} \right] |P^{(I)}(\zeta)|^{2} |d\zeta \wedge d\overline{\zeta}|$$ $$\leq \iint_{|z| < 1} \left[ D_{f_{M}}(z) + \frac{1}{D_{f_{M}}(z)} \right] |\tau_{M}(z)| |^{2} dz \wedge d\overline{z}| - c_{o}.$$ Even in case in which $B_z$ lies in $U_z$ entirely apart from $\partial U_z$ , one is sure to succeed in deriving the same results (26) etc. by means of linear transformations and an analytic prolongation across the boundary. Next set $$P(z)=P^{(I)}(z)\bigg[\iint\limits_{|z|<1}|\tau_{M}(z)|^{2}|dz\wedge d\bar{z}| \bigg/\iint\limits_{|z|<1}|P^{(I)}(z)|^{2}|dz\wedge d\bar{z}|\,\bigg].$$ Then (27) $$\iint_{|z|<1} |P(z)|^2 |dz \wedge d\bar{z}| = \iint_{|z|<1} |\tau_M(z)|^2 |dz \wedge d\bar{z}| ,$$ $$(27') \ \limsup_{r \to 1} \oint\limits_{|z| = r} |P(z)| \, |dz| \le ((1 + e^{(\varepsilon/M)^2}/M^3)/(1 - 2(\varepsilon/M)^2)) \oint\limits_{|z| = 1} |\tau_M(z)| \, |dz|,$$ $$(27'') \qquad \iint_{|z| < 1} \left[ Df_{M}(z) + \frac{1}{Df_{M}(z)} \right] |P(z)|^{2} |dz \wedge d\bar{z}|$$ $$\leq \iint_{|z| < 1} \left[ Df_{M}(z) + \frac{1}{Df_{M}(z)} \right] |\tau_{M}(z)|^{2} |dz \wedge d\bar{z}| - c'_{0}.$$ with the constant $c_0 \ge c_0/(1-2(\varepsilon/M)^2)$ . Let $S_{lj}$ be the simplex adjacent to $S_l$ and let $s_j$ a small subarc of $\partial S_{li}$ adjacent to $S_{li} \cap S_l$ , such that $\int_{s_j} |\omega_M| = a_0 > 0$ (j=1, 2, 3). We take suitably a simply connected subregion $S_{lj}^*$ of $S_{li}$ , partly bounded by $s_j \cup (S_{lj} \cap S_l)$ , so that $$(28) \qquad \int_{S_{I_{j}}^{}} \omega_{M} \omega_{M} < \delta' \iint_{S_{I_{j}}} \omega_{M} * \omega_{M},$$ $$\iint_{S_{I_{j}}^{}} \left[ D_{f_{M}}(\mathfrak{p}) + \frac{1}{D_{f_{M}}(\mathfrak{p})} \right] \omega_{M} \omega_{M} < \frac{\varepsilon}{\max(A_{0}^{2}, 1)}.$$ Let $|z'| \leq 1$ be a local coordinate disk corresponding to $S_{ij}$ . The above composed linear differential |P(z)| |dz| induces on the circular arc $z'(S_{ij} \cap S_i)$ the summable linear differential |P'(z')| |dz'| which is identical with |P(z)| |dz|. In view of the concavity of the logarithm $$\frac{1}{b-a}\int_{a}^{b}\log u(x)dx \leq \log\left\{\frac{1}{b-a}\int_{a}^{b}u(x)dx\right\} \quad \text{for summable } u(x)>0 \text{ on } (a,b),$$ we assign the boundary value $$u'(\zeta') = \begin{cases} \log |P'(\zeta')| - \log |\tau_M'(\zeta')| & \text{on } \zeta'(S_{lj} \cap S_l) \\ -\varepsilon & \text{on } \zeta'(s_j) \\ 0 & \text{elsewhere on } |\zeta'| = 1 \end{cases}$$ to define the Poisson integral $$u'(z') = \frac{1}{2\pi} \int_{0}^{2\pi} u'(\zeta') \frac{1 - |z'|^2}{|\zeta' - z'|^2} d \arg \zeta'$$ in |z'| < 1. After an adequate choice of $S_{ij}^*$ and z', we may have (Euclidean) dist. $$(z'(\partial S_{li}^*), \ z'(s_j) \cup z'(S_{lj} \cap S_l)) \ge \delta > 0$$ , $$\int_{z'(S_{I_j} \cap S_{I_j})} (|\log|P'(z')|| + |\log|\tau(z'_M)||) |dz'| \leq \delta^3, \int_{z'(S_{I_j})} |dz'| \leq \delta^3,$$ where $|\tau'_{M}(z')| |dz'| = |\tau_{M}(z)| |dz|$ . Then (29) $$u'(z') \leq \frac{\delta}{2\pi} (1+\varepsilon) \quad \text{in } z'(S_{lj} - S_{lj}^*).$$ With the harmonic conjugate v'(z') to u'(z') we put $$P'(z') = \exp (\log |\tau'_{M}(z')| + u'(z') + iv'(z')).$$ Then $$|P'(z')| = \left\{ \begin{array}{ll} |P'(z')| & \text{on } z'(S_{lj} \cap S_l) \ e^{-e} \, |\tau_M'(z')| & \text{on } z'(s_j) \ |\tau_M'(z')| & \text{elsewhere on } |z'| = 1 \,. \end{array} \right.$$ It follows from (28), (29) that $$\begin{split} &(1-\delta') \exp\left(-\frac{\delta}{\pi}(1+\varepsilon)\right) \iint\limits_{S_{ij}} |\tau_{M}'(z')|^2 |dz' \wedge d\bar{z}'| \leq \iint\limits_{S_{ij}} |P'(z')|^2 |dz' \wedge d\bar{z}'| \;, \\ &\iint\limits_{S_{ij}} \left[ D_{f_M}(z') + \frac{1}{D_{f_M}(z')} \right] |P'(z')|^2 |dz' \wedge d\bar{z}'| \\ &\leq \max\left(A_0^2, \; 1\right) \iint\limits_{S_^*} \left[ D_{f_M}(z') + \frac{1}{D_{f_M}(z')} \right] |\tau_{M}'(z')|^2 |dz' \wedge d\bar{z}'| < \varepsilon \;. \end{split}$$ If we select $\delta$ , $\delta'$ , in particular, so that $$(30) \qquad (1-\delta')\exp\left(-\frac{\delta}{\pi}(1+\varepsilon)\right) \geq 1-(\varepsilon/M)^2, \ \delta(1+\varepsilon)/\pi < \varepsilon/M \,,$$ we get (31) $$\iint_{S_{I_{i}}} |P'(z')|^{2} |dz' \wedge d\bar{z}'| \ge (1 - (\varepsilon/M)^{2}) \iint_{S_{I_{i}}} |\tau'_{M}(z')|^{2} |dz \wedge d\bar{z}'| ,$$ while $$(31') \quad \int\limits_{s_{j}} |P'(z')| \, |dz'| - e^{-z} \int\limits_{s_{j}} |\tau'_{M}(z')| \, |dz'| \leq \int\limits_{s_{j}} |\tau'_{M}(z')| \, |dz'| - (\varepsilon a_{0}/2)$$ $$(a_{0} > 0),$$ $$(31'') \qquad \iint_{S_{I_{j}}} \left[ D_{f_{M}}(z') + \frac{1}{D_{f_{M}}(z')} \right] |P'(z')|^{2} |dz' \wedge d\bar{z}'|$$ $$\leq \exp\left(\frac{\delta}{\pi} (1+\varepsilon)\right) \iint_{S_{I_{j}}} \left[ D_{f_{M}}(z') + \frac{1}{D_{f_{M}}(z')} \right] |\tau'_{M}(z')|^{2} |dz' \wedge d\bar{z}'| + \varepsilon$$ $$\leq \iint_{S_{I_{j}}} \left[ D_{f_{M}}(z') + \frac{1}{D_{f_{M}}(z')} \right] |\tau'_{M}(z')|^{2} |dz' \wedge d\bar{z}'| + 3\varepsilon.$$ Let $S''_{lj}$ ( $\pm S_l$ ) $\in \Sigma$ be the possible simplex adjacent to $S_{lj}$ and let $S^{}_{lj}$ its simply connected subregion, such that $\partial S^{}_{lj} \supset s_j$ and that $$(32) \qquad \iint\limits_{S_{I_{i}}^{}} \omega_{M} \omega_{M} \leq \delta' \iint\limits_{S_{I_{i}}^{}} \omega_{M} \omega_{M}, \ \iint\limits_{S_{I_{i}}^{}} \left[ D_{f_{M}}(\mathfrak{p}) + \frac{1}{D_{f_{M}}(\mathfrak{p})} \right] \omega_{M} \omega_{M} \leq \varepsilon \ .$$ If we take a local coordinate disk $|z''| \le 1$ corresponding to $S_{ij}''$ , we have the summable linear differential |P''(z'')| |dz''| = |P'(z')| |dz'|, $|\tau_M''(z'')| |dz''| = |\tau_M'(z')| |dz'|$ for almost all points on $S_{ij}'' \cap S_{ij}$ . It is possible to adopt $S_{ii}^{**}$ and z'' so as to be subject to the conditions: $$\int\limits_{z''(z_j)} |\log |P''(z'')|||dz''| \leq \delta^3,$$ (Euclidean) dist. $(z''(s_i), z''(\partial S_{li}^{**} \cap \operatorname{int} S_{li}')) \geq \delta$ . If $\delta$ , $\delta'$ are chosen so as to fulfill the requirement (30), the straightforward computation yields (33) $$\iint\limits_{s_{l_{i}}^{\prime\prime}} |P''(z'')|^{2} |dz'' \wedge d\bar{z}''| \ge (1 - (\varepsilon/M)^{2}) \iint\limits_{s_{l_{i}}^{\prime\prime}} |\tau_{M}^{\prime\prime}(z'')|^{2} |dz'' \wedge d\bar{z}''|$$ $$(33'') \qquad \iint\limits_{s_{l_{j}}''} \left[ D_{f_{M}}(z'') + \frac{1}{D_{f_{M}}(z'')} \right] |P''(z'')|^{2} |dz'' \wedge d\bar{z}''| \\ \leq \iint\limits_{s_{l_{j}}''} \left[ D_{f_{M}}(z'') + \frac{1}{D_{f_{M}}(z'')} \right] |\tau_{M}''(z'')|^{2} |dz'' \wedge d\bar{z}''| + 3\varepsilon.$$ The differential $$\tilde{\omega} = \begin{cases} P(z)dz & \text{for } \mathfrak{p}(z) \text{ in int } S_I, \ P'(z')dz' & \text{for } \mathfrak{p}(z') \text{ in } \bigcup_{j=1}^3 \text{int } S_{I_j}, \ P''(z'')dz'' & \text{for } \mathfrak{p}(z'') \text{ in } \bigcup_{j=1}^3 \text{ int } S_{I_j}', \ \omega_M & \text{in } R - S_I - \bigcup_{j=1}^3 S_{I_j} - \bigcup_{j=1}^3 S_{I_j}' \end{cases}$$ satisfies not only the local holomorphy conditions (i), (ii), (iii), (iv), but also $$\iint\limits_{\mathbb{R}} \left[ D_{f_{M}}(\mathfrak{p}) + \frac{1}{D_{f_{M}}(\mathfrak{p})} \right] \tilde{\omega} * \tilde{\omega} \leq M \left( K + \frac{1}{K} \right) - c_{0}^{\prime\prime}$$ and $$|1-||\tilde{\omega}||^2| \leq b(\varepsilon/M)^2$$ for some positive constants $c_0''$ and b (independent of $\varepsilon$ and M) on account of (27), (31), (33), (27"), (31"), (33"). Let it be normalized into $\tilde{\omega}_M = \tilde{\omega}/||\tilde{\omega}||$ : we have $||\tilde{\omega}_M|| = 1$ and the inequality $$\sum_{j=1}^{\kappa} \int_{\partial S_j} |\widetilde{\omega}_M| \leq \frac{1}{\sqrt{1 - b(\varepsilon/M)^2}} \sum_{j=1}^{\kappa} \int_{\partial S_j} |\widetilde{\omega}| \leq \frac{M - \frac{a_0 \varepsilon}{4}}{1 - b(\varepsilon/M)^2} = \frac{M \left(1 - \frac{a_0}{4} (\varepsilon/M)\right)}{1 - b(\varepsilon/M)^2} \leq M$$ is valid in virtue of (27'), (31') for sufficiently large M, which enables us to conclude that $$\tilde{\omega}_M \in \Omega_M$$ for $M \gg 1$ . Finally it can be verified that $$\begin{split} &\iint_{R} \left[ D_{f_{M}}(\mathfrak{p}) + \frac{1}{D_{f_{M}}(\mathfrak{p})} \right] \omega_{M} \omega_{M} : \iint_{R} \left[ D_{f_{M}}(\mathfrak{p}) + \frac{1}{D_{f_{M}}(\mathfrak{p})} \right] \widetilde{\omega}_{M} \widetilde{\omega}_{M} \\ & \geq (1 - b(\mathcal{E}/M)^{2}) \iint_{R} \left[ D_{f_{M}}(\mathfrak{p}) + \frac{1}{D_{f_{M}}(\mathfrak{p})} \right] \omega_{M} \omega_{M} : \iint_{R} \left[ D_{f_{M}}(\mathfrak{p}) + \frac{1}{D_{f_{M}}(\mathfrak{p})} \right] \widetilde{\omega} \widetilde{\omega} \\ & \geq (1 - b(\mathcal{E}/M)^{2}) M \left( K + \frac{1}{K} \right) : \left( M \left( K + \frac{1}{K} \right) - c_{0}^{\prime \prime} \right). \end{split}$$ When M is large enough, the last term will exceed 1, which proves the assertion. Q.E.D.

Theorem 2. There exists at least one topological mapping from R to R' which belongs to the given homotopy class and is harmonic relative to the given conformal metric on R'. EXAMPLE. Consider the simplest case in which R and R' are both tori $(\eta, \mathbf{A})$ being naturally fixed). Then the harmonic map of R onto R' is unique except the conformal mappings of R onto itself. Suppose there exist two such mappings, say, f and $\tilde{f}$ . Let $\mathfrak{q}$ be an arbitrary point on R' and $\alpha_f$ , $\alpha_{\tilde{f}}$ the holomorphic quadratic differentials on R associated with f, $\tilde{f}$ respectively. Since the dimension of the vector space of analytic quadratic differentials on R is one and $\tilde{f}$ is homotopic to f, the nets on R woven by their trajectories and orthogonal trajectories $\Re\{\alpha_f\}=0$ and $\Re\{\alpha_{\tilde{f}}\}=0$ coincide with each other and it is also the case with $\Re\{\alpha_f\}=0$ and $\Re\{\alpha_{\tilde{f}}\}=0$ . Further we can see without so much difficulty that these families of analytic curves correspond to each other by the mapping $f^{-1}(\mathfrak{q}) \to \tilde{f}^{-1}(\mathfrak{q})$ and that composite map $f^{-1}\tilde{f}$ reduces to a conformal transformation of R onto itself. REMARK 2. To the limited knowledge of the present author, the question whether or not the uniqueness in Theorem 2 holds good seems to be still unanswered for the general case. So at the end of the paper we should like to pose the PROBLEM. We restrict ourselves to the pair (R, R') of (topologically equivalent compact Riemann) surfaces with genera superior to 1. Then, is the harmonic mapping relative to the given conformal metric on R' unique within the preassigned homotopy class? University of Osaka Prefecture (Received February 4, 1963)