Baker, Irvine Noel

English

Baker Irvine Noel

University of Szeged

The value distribution of composite entire functions By I. N. BAKER in London (England) 1. If the entire function F(z) is expressible in the form f°g(z), where / and g are transcendental entire functions, it is called composite; otherwise F(z) is said to be pseudo-prime. O Z A W A [3] proved various results about the value distribution of composite entire functions, including the following: If F(z) is entire and offinite order and if there exists a constant A such that F(z) = A has only real roots, then F(z) is not composite. Thus a composite entire function F{z) of finite order has none of its ^4-values distributed entirely on a line and, a fortiori none is distributed on a ray. One can strengthen this last statement and assert that there is no direction which is the sole limiting direction of the ,4-points: T h e o r e m 1. If F(z) is an entire function offinite order and there exist complex A and real a such that for any <5 > 0 all but a finite number of roots of F(z) = A He in the angle | a rgz — a| < S, then F(z) is pseudo-prime. In Section 3 similar arguments to those used in the proof of Theorem 1 are applied to a question of iteration theory. 2. P r o o f of T h e o r e m 1. (i) Without loss of generality, we may suppose a = 7r. Suppose F(z) satisfies the conditions of the theorem and that, nevertheless, F=f(g), f and g are transcendental. Then by a result of PÓLYA [4] , / has zero order and g has finite order (less than that of F). Now f(w) = A has an infinity of solutions w = w1, w2, ..., wn, ... and |w„| — For any ¿>-0, the roots of g(z) = w„ ( « > « 0 ) all lie in the angle A(5): | a rgz — n\ < 5 and so g(z) omits the values w„ in B{n — S): ¡arg z| S n — ö. BIEBERBACH [2] has shown that if the entire function h(z) takes two different finite values at most finitely often in an angle of aperture an, then in every smaller angle l/(2)l = 0 { e x p (K\z\lla)} • for a suitable constant K. 88 I. N. Baker We deduce that g{z) is of order S %n/(n — 5) in B(jz — 25). Since g(z) is of some finite order, say q, in the whole plane, in particular in A(25) of aperture 48, 5 arbitrary, it follows from the Phragmen—Lindelof principle that (ii) Choose wkT±g(Q) and 0 < 5 < 7 r / 1 6 . Then g(z) may be expressed g(z)-wk =.X JF 1 - - \ = P(z) JJ 1 - - , O ^ A constant, P(z) polynomial. Since f(g(z„)) — A we may assume that for given ¿ > 0 , zn (E A(S) when n S n0. F o r z£B\--5 7Z 71 : | a r g z | < — — 5 and for «j=w0 w e then have |arg(—z/z„)| < —, and so 1 1. Thus as z —°° in B\ — — 5 , |g(z)| — °° faster than any power of |z|. (iii) Next we show that for all large enough z, ( | z | > / Q , i.e. in |argz[<:c>, we have for (1) that (2) . D = zg' (z)/{g(z) - Wk} = 2 z/(z - z„) n = l \D\>4n5~l, |arg D\<25. First note that the bilinear function t = z/(z — /?) maps the line joining p to ~P on to the real axis and the angle |arg z— .arg ( — P)\ < 25 into the region E bounded by the two circular arcs joining 0 to 1 and making angles ± 2 5 with the positive real axis at 0. Hence for / iS/7 0 , when zn £ A((5), t = z/(z — z„) maps B(5), which belongs to |arg z —arg ( —z„)| < 25, into E, and so for each ( 3 ) |Im z/(z - z , ) | == tan (25) • Re z/(z - zk) in B(8). Since, for each fixed n, z/(z—z„) — 1 as z — o n e has for all z£B(<5) with suffici-ently large \z\, that (3) holds for all n. Hence, f rom (1), and |Im D\ \D\ sRe D >4n5~1, (tan 25)-Re D, |arg£>| -25 for z£B(5), say. ' ! V (iv) Choose wn, w > « 0 , such that f{w„) = A and so large that \g(z) — wk\"< \w„ — for \z\-^K, where K i s the constant which occurs in (iii). The component C of the set {z: \g(z) — wk vn — w^l}, which contains the origin, has a bounded intersection with B(5) and this intersection contains B(5) Pi {|z| s K}. Then the Composite entire functions 89> boundary of Cf):fi(<5) contains an arc of a level curve y of g{z) — wk which joins a point of arg z = — 5 to a point of a r g z = <5 and lies in \z\>K. On y one has (1) and (2) of (iii). Hence the arc y contains no zerps of g'(z). If, moreover, an. increment. 5z on y corresponds to an increment 5w on — w t | = \w„ — wk\ under w =g (z ) „ then 5w 5 z zg'(z) (w-wky z g(z)-wk (4) so that '5z arg I — % 2 5z arg I — w—wk dw :{l+o(<5z)}, zg'(z) - a r g .{l+o(<5z)}, and by (2) arg zg'(z) j g(z)-wkf 2d < * (g(z)-wk) so the arc y ean be expressed as z = r(0)e10, -Ssd^S. Putting w — wk• = \wn — Wk¡e"p, we have in (4): whence d(p > - zg'W do g(z)-wk 4 n s - 1 , by (2). As z traverses y in the direction of increasing 0, w traverses the circle T : \w — wk\ = \wn — wk\ in the .positive direction and (p increases by at least 4nd~1-2S = 87c. Thus w traverses the whole of r and in particular :g(z) = w = wn for some point z£ycB(S). But this contradicts the fact, established in (i), that g(z) = w>n,n>/70,has no roots outside A(5).Thus the assumption that F(z) is compos-ite must be false. 3. A related question in iteration theory. Let / ( z ) be an entire function and / t ( z ) = / ( z ) , / 2 ( z ) = / ( / ( z ) ) , . . . , / „ ( z ) , . . . be its sequence of iterates. Regarding the Fatou set SCO of those points of the complex plane where {/„(z)} does not form a normal family, it was shown in [1] that if f ( z ) is entire and transcendental, then 5 ( / ) cannot be contained in any finite set of lines but on the other hand, for any constant A > 0 there exists an entire transcendental function for which # ( / ) >s contained in the region {|Im z| < A , R e z > 0 } . The function used to show this last result was of infinite order. In fact, using the arguments of Section 2 we can show: T h e o r e m 2. If f is entire transcendental and for every ¿ > 0 the set %{f)~ - {z, (arg z) < <5} is bounded, then f is of infinite order. •90 I. N. Baker: Composite entire functions P r o o f . Suppose/sat isf ies the hypotheses of.the theorem, but is of finite order. $ ( / ) has the properties (cf. [1]): (i) 5 ( / ) ' s non-empty and perfect, (ii) i f / ( z ) = « € 5 , then We take two different values a, ß in g ( / ) which are not Picard exceptional for f ( z ) . The solutions of / ( z ) = a, ß lie in g and so, with finitely many exceptions in |argz|<<5. Noting that <5>0 is arbitrary and proceeding as in § 2 (i), we see that f ( z ) has order at most The method of Section 2 (ii)—(iv) then shows that in the angle B: |(arg z)~n\ < 5 obtained f rom |argz|<<5 by reflection in the origin, / ( z ) takes all arbitrarily large values, in particular large values z„ for which /(z„) = a, i.e. values for which z„£5-I f f ( t n ) = z n > w e have since Taking a sequence z „ £ 5 for which 'knl~*°°> w e have |i„|— °° and hence is unbounded or g —{z, | a rgz | < <5} is unbounded, against the assumptions of the theorem. Hence / must be of finite order. In Theorem 2 the transcendence of / i s essential. Polynomials have bounded 5-References [1] I. N. BAKER, Sets of non-normality in iteration theory, J. London Math. Soc., 40 (1965), 499—502. 12] L. BIEBERBACH, Über eine Vertiefung des Picardschen Satzes bei ganzen Funktionen endlicher Ordnung, Math. Z„ 3 (1919), 175—190. [3] M. OZAWA, On the solution of the functional equation fog (z) = F(z). III, Ködai Math. Sem. Rep., 20 (1968), 257—263. >{4] G. PÓLYA, On an integral function of an integral function, J. London Math. Soc., 1 (1926), 12—15. (Received March 13, 1970) 

The value distribution of composite entire functions

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The value distribution of composite entire functions

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