On the Number of Doublepoints of Analytic Curves

MATHEMA TICS

ON THE NUMBER OF DOUBLEPOINTS OF ANALYTIC CURVES BY

R. J. WILLE (Communicated by Prof. H. D.

KLOOSTEHMAN

at the meeting of February 24,1951)

Let w = f(z) be a uniform analytic function (holomorphic or meromorphic); then we shall prove the following theorem: The number of doublepoints of the image 1= f(O) of the circumference 0 of a circle is finite.

We define doublepoints in the following way: A point of I will be doublepoint if 1° it has more than one corresponding z-value on 0, 2° its corresponding z-values have no disjoint neighbourhoods on 0 with equal images. Before proving the theorem we give in fig. 1 an example

B

A Fig. 1

of an analytic curve with parts falling along each other. Point A has three corresponding z-values but will be no doublepoint according to our second condition. Evidently the points Bare doublepoints. In fig. 2 point A will be doublepoint. It has four corresponding

Fig. 2

z-values and if we choose four disjoint neighbourhoods of the corresponding z-values at least two of them will have different images.

179 We could imagine the curve I to have an infinity of doublepoints. In fig. 3 we give a suggestion.

A

Fig. 3

If we consider the riemannien surface covering the w-plane; then this surface will have a branchpoint above point A. Point B will be the projection of two different points of the to I corresponding curve on this riemannien surface.

Now we give the proof: By means of a homography we transform C onto the real axis (x-axis) plus the infinite point. From now on I is the image of the closed x-axis under w = I(z), which may be defined on some neighbourhood of the closed x-axis. Assumption: Suppose the number of doublepoints of I to be infinite. The corresponding z-values of the doublepoints lie on the closed x-axis and they will have at least one point of accumulation xO. Let Xi be a monotone sequence of corresponding values which converges to XO' then Wi = I(xi ) converges to Wo = I(xo). Without restriction we may assume that Xo = 0; Xi 0 and Wo = 1(0) = o. If not then we apply a homography on the z-plane and one on the w-plane.

>

1. Suppose moreover: 1'(0) 7'=- o. Then there is a 1 - 1 correspondence between some neighourhood N of z = 0 and a neighbourhood M of w = o. Hence the image I(a) of a sufficiently small right segment a of x = 0 will have no doublepoints (if we consider only I(a) instead of the whole curve I). But the points Wi are doublepoints of the whole curve I and therefore have, beyond the corresponding values Xi in a, also corresponding values x~ outside a. The points x~ will have at least one point of accumulation, say x~. Now we shall show that the image I(a') of either a suitable right or left segment a' of x~ must fall along the image I(a). Of the function 1 defined on N the inverse has a branch 10\ uniform and analytic on M; as I(xi ) = Wi and Wi converges to w = 0 the function 101(f(z)) is uniform and analytic also on a sufficiently small neighbourhood of z = x~. We now have to prove 101(f(x)) to be real for x near x~. As 10 1(f (x~)) = 10 1(Wi) = 10 1(f (Xi)) = Xi the function 101(f(x)) is real and

>

0 for x

=

x~.

180 Suppose (an 0:/= 0)

10 1(f (Z) = an (z = an(z -

x~)" x~)n

+ ... = (an + i (3n) (z - x~)n + ... = + .. + i {(3n(z - x~)n + ... },

then

10 1(f (x))

=

an (x -

x~)n

+ ... + i {(3n (x-

x~)n

+ ... }.

But 101(/(x;)) is real, hence

(3n (Xi' -

Xo, )n

I

T

-

••. -

o· ,

x~ accumulate in x~, so (3n = (3n+1 = ... = 0 and consequently lol(f(x)) is real for real x near x~. We have for such x's:

and

If n is odd we can choose a suitable segment a' of x~ (if an > 0 a right segment, if an < 0 a left segment) and a suitable right segment a of x = 0 in such a way that hence

I(a)

=

t(/OI (/(a')))

=

I(a').

If n is even we can choose a suitable segment a' of x~ (a left as well as a right segment will suite; an will be > 0) and a suitable right segment a of x = 0 in such a way that 101(/(a')) = a and hence

I(a)

=

1(/0 1(/(a'))) = I(a').

Now to construct a contradiction we need our second condition for doublepoints. Suppose there are more points of accumulation like x~. As Wi converges to W = 0 we have I(x~) = 0 for each x~. But I(z) takes the value 0 only a finite number of times; therefore the number of such points x~, and consequently the number of such pairs of suitable segments a, a', is finite. Hence the common part of the images I(a) (= I(a')) for each pair a, a', is a continuum which has W = 0 as endpoint and therefore contains an infinite number of Wi' say w{. Only a finite number of wi may have one or more corresponding x-values outside the interior of the segments a, a'; only these wi may be doublepoints. All other wi will have all their corresponding x-values (for each wi finite in number) in the interior of the segments a, a'. These x-values will have disjoint intervals with equal images. Hence there are wi which are no doublepoints. This contradicts our assumption.

181

II. We gave the proof under the supposition that 1'(0) =1= o. Now let 1'(0) = 0 and also I'(x~) = 0 for each x~ (if I'(x~) =1= 0 for one of the points we could choose this We suppose

x~,

x~

for the point x o).

Then there is a 1 - 1 correspondence between some neighbourhood N of z = 0 and an m-sheeted coveringsurface M of a neighbourhood M of W = 0, where W = 0 is the only branchpoint. The image t(a) in M of a sufficiently small right segment a of x = 0 has no doublepoints (if we consider only t(a) instead of I). For suppose it had an infinity of doublepoints to which correspond Xi > 0, then there would be an infinite number of doublepoints in a certain sector 15 of M with opening less than 2n. Consider the m sectors 6 of M, which cover c5; then at least two different sectors ~ and ~ would contain an infinite number of points covering doublepoints of t(a) in M. Consider one of the m

Vw.

This function transforms the sectors 61 and ~2 onto two sectors c5i and c5; in the t-plane which meet each other only in t = O.

m functions t

=

Now we choose two sequences xt

>

0 and xt*

>

m

0 with Vt(xt) in c5i and

m m m Vt(xt*) in c5;; then lim Vt(x)/x does not exist, which is absurd, for Vt(z) •.

a:->O

is uniform and analytic. Hence t(a) has no doublepoints (if we consider only t(a) instead of the whole curve I). The doublepoints Wi of I will have, beyond the corresponding values Xi in a, also corresponding values x~ outside a. The points x~ will have at least one point of accumulation x~ (x~ may be equal to 0 as will be the case for the curve in fig. 3 with Wi converging to A). To simplify the notations we put t1(Z) = t(z + x~) when there is an infinite number of x~ > x~ and put xX = x~ - x~ for these x; > x~. Otherwise we have an infinite number of x; < x~; then we put x; = - x; + x~ and t1(Z) = t( -z + x~). (If x~ = 0 we have t1(Z) = t(-z)). In any case > o. Now we shall show that we can choose an other right segment a" of x = 0 in such a way that Ma") = t(a). As t~(O) = 0 we have

x;

(a~

We may suppose n accumulation). We may write:

~

=1= 0).

m (if not, we interchange the two points of

t1(Z) = z,,-m. zm(a~

+

a~z

+ ... ) =

z,,-m. t2(Z).

Now W = t2(Z) gives a 1- 1 correspondence between some neighbourhood N' of z = 0 and an m-sheeted coveringsurface M' of a neighbourhood .llI' of W = 0, where W = 0 is the only branchpoint. The arguments of 13 Series A

182

I(z) and Mz), now considered on M resp. M', run from 0 to 2 m n. As known there are m possibilities to fix the arguments on the coveringsurfaces. In any case, since is real and

x;

(1)

the arguments of I(xi ) on M and of Mxn on lW only differ 2k n (k = = 0, 1, ... , m- 1). So we are able to fix the arguments on M and M' in such a way that for an infinite number of Xi and x~ we have: (2)

The arguments of I(z) and Mz) being fixed in this way, the functions ·m

m

VI(z) and V/2(Z) are well defined. From (1), (2) and the fact that x~ follows: m m m VI(xi ) = vTx~)'·-m. VI2(X~), mi-:--;;-:-::--:::where V(x~)n m is defined positive as usual.

>

0 it

m

The function gJ(z) = VI(z) being uniform and analytic on Nand gJ'(O) #- 0, the inverse gJ-l is uniform and analytic on the image gJ(N). Consider the function (3)

for x As

~

0, x sufficiently small.

m

m

m

gJ-l(V(x~)n=m. VI2(X~)) =gJ-l(Vl(xi )) =Xi> 0

the function (3) is real and positive for m

Since VMz) jtnd

gJ-l

are analytic the function (3) can be expanded in 1

a series of powers of

xm :

m m gJ-IU1xn=m. VI2 (x))

(ak

k

k

=

x = x;.

+ i{3/c) xm + ... =

k k akxm + ... + i ({3kxm + ... ).

1

Hence {3k(X;y:m + ... = o. But (x~)m accumulates in x = 0, hence PI = P2 = ... = 0 and the function (3) is real for x ~ 0, x sufficiently small. We have (ak #- 0):

The function is positive for x;; hence ak > o. Now we can chose two right segments (Jff and (J of x = 0 in such a way that:

183 hence
m

=

'In

V(a,,)ii=m. Vf2 (a")

and If a' is the right or left segment of segment a" of 0, then .

x~

which corresponds to the right

We now can derive as in I a contradiction with the assumption that there would be an infinite number of doublepoints; this proves the theorem. Remark: Our theorem holds also for quasi analytic functions (fonctions presque analytiques). This is a direct consequence of the following lemma: Lemma: Let f(x) be quasi analytic on the closed x-axis then we can transform the x-axis bicontinuous on itself, say by x =
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