An application of Ahlfors's theory of covering surfaces

Zuiman YÛJÔBÔ

doi:10.2969/jmsj/00410059

Zuiman YÛJÔBÔ

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1952 年 4 巻 1 号 p. 59-61

DOI https://doi.org/10.2969/jmsj/00410059

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Published: 1952 Received: 1948/05/02 Available on J-STAGE: 2006/08/29 Accepted: - Advance online publication: - Revised: -
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訂正日: 2006/08/29 訂正理由: - 訂正箇所: 引用文献情報訂正内容: Right : 1) L. Ahlfors, Sur les domaines dans lesquels une fonction méromorphe prend des valeurs appartenant à une région donnée. (Acta Soc. Sci. Fenn. N. s. 2 Nr. 2 (1933)).
2) L. Ahlfors, Zur Theorie der Überlagerungsflächen (Acta Math. 65 (1935)); or R. Nevanlinna, Eindeutige analytische Funktionen.
3) By this expression we mean that the Riemann image of |z|<R by f (z) contains no connected island above D_i whose number of sheets is<μ_i.
4), 5) L. Ahlfors or R. Nevanlinna, loc. cit.
6) This can be proved as follows: Let λ be the image of |z|=γ_f by f on the Riemann sphere. Then the complement of λ consists of a finite number of connected domains. If the southern pole lie on λ, then λ is obviously contained in the lower hemisphere on account of (3); then no point of the upper hemisphere is assumed by f in |z|<γ_f, for otherwise the whole upper hemisphere would be covered by the image of |z|<γ_f by f, which contradicts (4). So we may consider only the case where the southern pole does not lie on λ. Now let G be the one containing the southern pole among the above-mentioned domains. Then G is obviously simply connected and all the points of G are assumed by f on account of f(0)=0. Now let us suppose that λ is not contained in the lower hemisphere. Then there must exist intersection points of λ with the equator (|w|=1), for otherwise G must cover the whole lower hemisphere which is impossible on account of (4). Let us denote one of them by P and by C_P the locus of the points with spherical distance π/4 from P. Then C_P meets with the boundary of G, since C_P is a great circle which passes through the southern pole and G cannot cover a whole hemisphere. But this is clearly impossible on account of (3). Now since λ is contained in the lower hemisphere, either all the points in the upper hemisphere or none of them are assumed by f. But the former case is impossible on account of (4).

訂正日: 2006/08/29 訂正理由: - 訂正箇所: PDFファイル訂正内容: -

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