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1 III. X(U, V), Y(U, V), Z(U, V) Remain Continuous on the Unit Circle

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1 III. X(U, V), Y(U, V), Z(U, V) Remain Continuous on the Unit Circle 242 MA THEMA TICS: T. RAD6PPROC. N. A. S. This completes the proof of Theorem A. 1Proc. Lond. Math. Soc., 30, 1930 (339-346). For definitions of combinatory entities see two papers, Proc. Roy. Soc. Amsterdam, 29, 1926 (610-641), here cited as FI and FlI. 2 Loc. cit., p. 343, par. (d). 3 All simplexes and cells in this paper are closed sets. 4 F II, Theorem 6, Coroll. 1. 'FF, 12. 6 F II, Theorem 8a, Coroll. SOME REMARKS ON THE PROBLEM OF PLA TEA U BY TIBOR RAD6 D&PARTMSNT OF MATHUiATIcs, HARvARD UNIVERSITY Communicated December 19, 1929 1. The following remarks are concerned with a recent paper of R. Garnier on the problem of Plateau.' This problem, as investigated by Gamier according to the ideas of Riemann, Weierstrass and Schwartz, may be stated in the following analytic form: Given, in the xyz-space, a simple closed curve r*. Determine three functions x(u,v), y(u, v), z(u, v) with the following properties. I. x(u, v), y(u, v), z(u, v) are harmonic for u2 + v2 <1 II. Satisfy for u2 + v2 < 1 the relations XU++ y2+ ZU = XD + yV + Z2V, XuX, + YuYD + ZuZv = 0. III. x(u, v), y(u, v), z(u, v) remain continuous on the unit circle u2 + v2= 1 and carry the unit circle in a one-to-one and continuous way into the given curve r*.2 We shall see (§11), that at least for a certain class of curves r* the surface x = x(u, v), y = (u, v), z = z(u, v) is uniquely determined by the above conditions; therefore it is not advisable to restrict the problem by any further conditions. 2. The solution, given by Gamier, of the problem of Plateau, can be greatly simplified by the following Theorem of Approximation.-Let rT be a sequence of simple closed curves in the xyz-space with the following properties. (a) The length l(r*) of r* is uniformly bounded: l(r*) _ L. (,) The problem of Plateau (as stated above) is solvable for rI (n = 1, 2, ...). (-y) The sequence r* converges in the sense of Frechet toward a simple closed curve r*.3 Downloaded by guest on September 23, 2021 Vo.. 16, 1930 MA THEMA TICS: T. RAD6 243' Then the problem of Plkteau is also solvable for the limit curve r*.4 This theorem can be applied to the work of Garnier as follows. Garnier solves the problem of Plateau in the first place for simple closed polygons, by solving the monodromic group problem of Riemann under very difficult supplementary conditions. This being done, he takes up the case of a simple closed curve r1*, approximates it by inscribed polygons H*, and investigates the solution corresponding to 1I for n - co. He does this by means of a discussion of the solution of a monodromic group problem with variable initial conditions.5 This complicated discussion can be avoided by the above theorem of approximation. Indeed, the lengths of the inscribed polygons are bounded by the length of r* and the in- scribed polygons converge toward r * precisely in the sense of Frechet; so the theorem of approximation yields immediately the result that if the problem of Plateau is solvable for any simple closed polygon, then it is solvable for any simple closed rectifiable curve.6 The method of Gamier for the case of a polygon is a method of continuity (Kontinuitatsbeweis). The reader will find several opportunities to simplify this part of the work of Gamier by the use of our theorem of approximation. 3. The method of Gamier is not appropriate for the discussion of the question of the uniqueness; even for the case of a skew quadrilateral he admits the possibility of the existence of several solutions,7 in contra- diction to a statement of Schwartz.8 The theorem of uniqueness, proved in §11 may therefore be of some interest in spite of its very particular character. 4. Lemma.-Let there be given n functions hi(u, v), ..., h,(u, v), continuous for u2 + v2 < 1 and harmonic for u2 + v2 < 1. Let l(p) (O < p < 1) denote the length of the curve xI = hi(p cos 0, p sin 0), . ., xn = h"(p cos 0, p sin0), 0 0 ._ 27r, (1(1) may be infinite). Then l(p) is a monotonically increasing function of p for 0 < v _ 1. In particular, if 1(1) is finite and . L, then l(p) _ L for 0 <p _ 1. If n = 2 and if hi and h2 are conjugate harmonic functions, this statement reduces to a nice theorem concerning conformal mapping, proved inde- pendently by P. Csillag and L. Bieberbach. The proof of Bieberbach9 extends immediately to our general statement; the easy and attractive details may be left to the reader. 5. In order to prove the theorem of approximation, take three distinct points A, B, C on the unit circle u2 + v2 = 1 and three distinct points, A * B*, C* on r*; then, by virtue of condition y (§2), it is possible to take three distinct points A4, B*, CG on r* converging toward A*, B*, C*, respectively. The problem of Plateau being solvable by hypothesis for rF, we have three harmonic functions xt(u, v), y.(u, v), z.(u, v) with Downloaded by guest on September 23, 2021 244 MA THEMA TICS: T. RADP N PROC. N. A. S. the properties listed in the statement of the problem; for the sake of clarity, we denote the boundary values of x^(u, v), yn(u, v), zx(u, v) by tn(O)v tin(O), tn(O), 0 being the polar angle. Applying a one-to-one and conformal mapping of the unit circle upon itself (with preserved or re- versed indicatrix), we can suppose that we have %#(A), -1n(A), t.(A)) =Aln, (tn(B), nn(B), r.(B)) =Bn(1 (en(C), q7n(C), MO(C)) = Cn- The equations x = tn(0), y = %q(O), z = tn(0) carry the unit circle in a one-to-one and continuous way into the curve r*; from a and -y (§2) it follows then that the sequences tn(O), t1n(O), Pn/(O) are uniformly bounded and of uniformly bounded total variation. By virtue of a well-known theorem of Helly, we can pick out from them everywhere convergent subsequences-in order to avoid complicated notations, we denote these subsequences by tn(O), %n(O), rnf(O) again. The limit func- tions t(O), 7(0), t(O) are bounded and of bounded variation. Denote now by x(u, v), y(u, v), z(u, v) the harmonic functions determined by the Poisson integral formula, t(O), t7(0), t(O) being the boundary functions. We shall show that these harmonic functions solve the problem for the limit curve r*. 6. Uniformly bounded sequences being integrable term-by-term, x.(u, v), y.(u, v), z (u, v) converge in the interior of the unit circle toward x(u, v), y(u, v), z(u, v), the convergence being uniform on any closed point- set interior to the unit circle; moreover, the same holds for all of the partial derivatives. From this latter remark, it follows immediately that x(u, v), y(u, v), z(u, v) satisfy the condition II (§1). From the same remark we infer that I(p) = lim l1(p) for 0 < p < 1, (2) where 1(p), ln(p) denote the lengths of the images of the circle u2 + v2 = r2 < 1 by means of x, y, z and xn, Y., z,, respectively. Apply now the lemma of §4. With regard to condition a (§2) it follows that In(p) . L, and therefore we get from (2) 2r l(p) = (Xo + ye + Z42)11'do _L; (3) 0 0 by using the relations XuX+ YUY+ U2U = X2 + Y2 +Z wxox + ytyv + zthv = ( we obtain further from (3) Downloaded by guest on September 23, 2021 Voi. 16, 1930 MA THEMA TICS: T. RAD2 245 2r 2r p f(x2 + y2 + Z2')2dO = p f(x2 + y2 + z!)'12dO . L, (5) 0 0 the integrations being extended over any circle U2 + V2 = p2 < 1. Now we can write x(u, v) = Rf,(w), y(u, v) =Tf2(w), z(u, v) = Rf3(W), (6) where f,(w), f2(w), f3(w) are analytic functions of w = u + iv ( w| <1). Denote by Tk(p) the total variation of fk(p) on the circle w = p < 1 (k = 1, 2, 3). We have 2. 2-r = lflwI T,(p)flp f2 || dw|= S(X2 + XV2)1/2 |Idw |, 0 O etc.; from (5) we readily get that Tk(p) < 2LforO<p< 1, k = 1, 2, 3. (7) The total variation of fk(w) on concentric circles being bounded, fk(w) remains continuous for Iwi = 1, by virtue of a theorem of F. Riesz;'0 with regard to (6) it follows from this that x(u, v), y(u, v), z(u, v) remain continuous for U2 + V2 = 1. Denote their boundary values by x(O), y(O), z(O)- 7. The functions t(O), tq(G), r(O) are of bounded variation; the set G of their common points of continuity is therefore certainly everywhere dense on the unit circle. From a well-known property of the Poisson integral formula it follows that x(@) = t(O), y(O) = i(O), z(O) = t(0) on G. (8) 8. P being any point on the unit circle, put p* = (x(P), y(P), z(P), P* = (U(P), lo(P), t(P), (9) PX = (tn(P) p 7ln (P),s n (P)).- If P describes the unit circle, Pn describes just once the curve rP.
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