C OMPOSITIO M ATHEMATICA
E. R. V AN K AMPEN
On the argument functions of simple closed curves and simple arcs
Compositio Mathematica, tome 4 (1937), p. 271-275
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curves
and simple
arcsby
E. R. van
Kampen
Baltimore, Maryland
I. In his papers on oriented line elements
1),
Ostrowskiproved
certain theorems about directions oftangents
toplane
curves. In an additional paper
2 ), Hopf
showed that a betterinsight
into the matter treated can be obtainedby emphasizing
the directions of secants. The purpose of this note is to
point
out the
great advantage
that can be obtainedby considering
secants
only, discarding, together
with thetangents,
the restric-tive condition that the curves considered are differentiable.
After
all,
any theorem on the direction oftangents
can be ob-tained
by
a directlimiting
process from a theorem on the direction of secants. Agood example
of thesimplification
that is thussometimes obtained is Ostrowski’s theorem 1
(p. 178),
sincethis theorem does not state more than the existence of the function
99(s, t)
of III below(i.e.,
of the functiont(s1, s2), Hopf, l.c.,
p.55).
In II we prove the
,,Umlaufsatz"
forarbitrary simple
closedcurves in
essentially
the same way asHopf (p. 51). However, slightly
more care is necessary because nodifferentiability
con-dition is used. It should be noted that
by fixing
the orientation of the curve in advance we eliminate theambiguity
insign usually occurring
in the statement of this theorem.In III we prove a theorem on
argument
functions ofsimple
arcs. The Rolle-Ostrowski theorem in its
generalized
form fornon-differentiable curves is now derived very
easily
in IV. In V we state a modified version of Ostrowski’s theorem 2(l.c.,
p.178)
and show how this may be used togive
a secondproof
ofthe theorem in III.
l) Three notes. Compositio Mathematica 2 (1935), 26-49, and 177-200.
2) Compositio Mathematica 2 (1935), 50-62. See these papers for further references.
272
II. In order to determine a
positive
direction on asimple
closed curve
C,
we construct three half linesli’ i
= 1, 2, 3,each li having
in common with Conly
itsendpoint Pi,
and suchthat no
two li
have apoint
in common. The half lines are num-bered in such a way that the orientation of the
plane 3)
deter-mined
by
the orderl1, l2’ 13
ispositive.
We then determine thepositive
orientation of Cby
the orderPl, P2, P 3.
The curve C can be determined
by
a continuous funtionP(s),
- oo s + oo, whose value
corresponds
to apoint
in theplane.
We suppose thatP(s) = P(t),
if andonly
if s - t(mod 1),
and that the orientation of C determined
by inereasing s
is thepositive
orientation. Then we can findnumbers ti, i
= 1, 2, 3, such thatPi = P(ti ) and t1 C t2 t3 t1
+ 1.It is
easily seen 4)
that vce can find a continuous functionqJ(s, l)
defined for s t s + 1 andrepresenting
one of thepossible
values for theargument
of the orientedsegment
P(s) P(t).
We prove now.The
right
hand side of(1)
isobviously
an oddmultiple
ouf nand is
independent
of sand t,
so that weonly
need to prove(1)
for fixed values of s and t.
Replacing s and t
in(1) by t
and8 + 1 and
adding
we findcp(s + l, t + 1) := cp(s, t) + 2n.
Thisformula
gives
the"Umlaufsatz"
without anydifferentiability
condition for C.
Another conséquence of
(1)
is that the orientation of C isindependent
of theparticular
halflines 1,
used in its definition.In order to prove
(1),
we remark that the differencecp(t1, t3)- 99(t,, t2 )
can be determinedby letting
theendpoint
P of asegment PtP
move fromP2
toP3 along
anypath
in theplane
that doesnot meet the halfline
l1.
Similar remarks hold for’T(t2l t3)
-
9(tl, t3)
andcp(i2, tl
-f --1) - cp(t2, t3).
The threepaths
may be chosen aspolygons,
with a total of not more than four sides.Thus it is necessary to prove
(1 ) only
in case C is aquadrilateral 5 ).
The passage from a
quadrilateral
to atriangle
istrivial,
so thatthe rest of the
proof
of(1)
is evident.3) It is clear that the ordered triple of half lines determines the same orien- tation on every circle w-hich has the three endpoints in its interior.
1) A detailed proof is given bij Hopf in ,,Vorbemerkungen", I.C. p. 51.
5) It is also possible to reduce (1) to the case were C is one of the circles men-
tioned in 3).
A continuation of this
argument
leads to aproof
of the Jordancurve
theorem, closely
related to the onegiven by
E.Schmidt 6),
the main difference
being
that the curve is oriented apriori,
sothat,
forinstance,
the index of an interiorpoint
withrespect
to the curve is + 1 instead
of +
1. Now if the orientation of thecurve is known and the curve contains a line
segment,
it ispossible
to decide which side of the line
segment
is interior and which side is exterior to the curve. This will be used inIII,
c.III. A
simple
arc D can be determinedby
a continuousfunction
P., 0 s
1, of which the valuescorrespond
topoints
in the
plane
and such thatP s * Pl whenever s t.
We can find
4)
a continuous function99(s, t), 0 s
t 1,representing
forgiven s and t
a value of theargument
of theoriented
segment PsPt.
Thefollowing
theorem(2)
remains trueif the word
"maximum"
isreplaced by "minimum".
It imme-diately implies
the Rolle-Ostrowskitheorem,
as shown in IV.(2 ) 1 f
D is asimples
arcdetermined by the
continuousfunetion Ps, 0 8
1, andif cp(s, t),
0 st -:S
1, is ilsargument function, finally if q;( 0, 1) is a
maximumf or
bothq;( 0, s)
andp(s, 1),
then D is the linesegment PüP1.
a. We assume that D is not the
segment P, )Pl,
thatp(O, 1)
= 0and that
99(0, s )
andq;(s,l)
never take apositive
value. Wemust then prove a contradiction.
The line
PoP1
and the opensegment POP1
we call m and 1respectively.
The open half line determined on mby Po
and notcontaining P,
wecall lo;
the halfline li
issimilarly
defined. The halfplane containing
allpoints
S’ such that theargument
of theoriented
segment RS, R
on m, is between 0 and n we call oc, the other halfplane
determinedby
m we callfl.
b.. If a
point P,
of D is noton m and the arc t s 1
of Ddoes not meet
lo,
then - n ç(0, t )
0, so thatPl
is inf3
and
cp(t, 1)
iscongruent
mod 2n to a number between 0 and n.Since
99(t, 1)
0, it follows thatp(t, 1)
- ncp(O, t).
Simi-larly
we may prove thatcp(t, 1)
> - Je> p(0, t )
for certain values of tsufficiently
near to 0. Since both99(s, 1)
and99(o, s)
are
continuous,
it follows thatlp(O, s )
-p(s, 1)
for certain values of s.Now if
9(0, s)
=99(s, 1)
thenP s
is a commonpoint
of 1 and D.On the other hand
99(0, s)
=ç(s, 1)
= 0 isimpossible,
since if6) Sitz. Ber. Preuss. Akad. 1923, 318-329.
274
this
equality
is true forP 8
and s is variedcontinuously
in sucha way that
p(O, s)
becomesnegative,
thenp(s, 1 )
becomespositive.
Thus it follows that there exists a common
point Pg
of 1 andD,
such thatq;(0, s )
isnegative.
Thegreatest
value of s with thisproperty
we call a,then q;(0, a)
=- - 2nn 0.c. The least
parameter
value s > a for whichPg
is on thesegment PaP,
of m will be denotedby
b and the arcPaPb
ofD will be denoted
by
D’. Thesegment
k =PaPb
of m,together
with
D’,
forms asimple
closed curve C. Sinceq;(0, a)
0 andq;(0, b )
= 0, the index ofPo
withrespect
to C is notequal
to zero,and hence is
equal
to 1, so that9’(0, a) = -
2n. Thus thepositive
orientation on C is determined on D’
by
the orderPaPb,
henceon k
by
the orderPbPa.
It follows that apoint Q
in the halfplane
a near interiorpoints
of k has index 0 withrespect
to C.Since
q;(0, b )
= 0, thepoints
on D nearPb
are in the halfplane fl,
so that thepoint Q
maybe joined
toPi by
apolygon
in xnot
meeting
C. ThusPl
is either on C or exterior to C.d. The last statement
implies,
in view of the Jordan curvetheorem,
that there may be determined a continuous functionf(R) giving
anargument
of thesegment RP1,
where R is either in the interior of C or on C(but
not inP1
if b =1)
andf(PO) =
0.Since
q;(0, s )
= 0 for any commonpoint Ps =F Pa
of D’ and thesegment POPb,
whilealways ç(o, s )
0, it is clear thatf ( R )
isdefined,
henceequal
to 0, for any R on thesegment PoPb.
It is clear that
f(Pi)
is the value ofq;(t, 1)
if nopoint Ps,
o 8 t,
is exterior to C. Hence99(c, 1)
= 0, ifPc
is the commonpoint
of D and k which has the lowestparameter
value. Unlessc = a, it follows
that,
forslightly
lower values of s, we haveq;(8,I)
> 0.But,
if c = a,then f(PI) = 99(t, 1)
for0 s C b,
while
obviously f ( P t )
> 0 for somepoints D of
D’. So in bothcases we find values of s for which
q;(8, 1)
> 0. This contradiction proves(2).
IV. The Rolle-Ostrowski theorem can be derived from
(2) by
a verysimple argument.
In ourterminology
it says:(3)
For any e > 0 the rangeof q;(s, t)
over s t s + 8is
equal
to the total rangeo f p(s, t).
If this is not true, then two numbers and
b, 0 a b 1,
can be
determined,
suchthat 99(a, b)
is eithergreater
than or less than any value ofq;(s, t),
for whichs t 8 + b -
a.This
immediately
contradicts(2).
V. Let D be a
simple
arc determinedby
the functionPs,
o s
1, letq:;(s, t), 0 s t
1, be itsargument function,
and let m be the line
POP,.
The second theorem in the second note of Ostrowski(l. c.
p.178)
is to the effect that D consists of twospirals,
one around eachendpoint,
if thefollowing
two con-ditions are satisfied:
(1)
At any intersection of D and m thecurve D crosses the line m
(except
of course inPo
and inPl);
(II )
IfP,.
andPu
are two such intersections which are successiveon the arc
D,
then eitherq:;(u, 1) #- q:;(v, 1)
or99(0, u) =A 99(0, v).
It is
easily
seen that the statement remains true if(II)
is modifiedso as to concern intersections successive on the line m. In the
new form this theorem may be used to
give
a secondproof
ofthe theorem in III. In
fact,
inproving (2)
for anarbitrary
arc Dwe may
replace
thepart
of D forwhich a 8
bby
the linesegment PaPb
whenever the new curve isagain
asimple
arc andthe
change
in valueof 99(0, s )
andq:;(s, 1)
is the samealong
theline
segment
asalong
the arc.(If Pa
orPb
is anendpoint,
we maydrop
the conditioncorresponding
to thatendpoint. )
Forby
sucha
change
the rangeof p(O, s )
and ofq:;(s, 1)
is never increased.By
a finite number of thesechanges
we can reduce(2)
to thecase of a
polygon 1Ii
which satisfies condition(I), except
ifthe
original
arc is a linesegment. By
a further finite number ofchanges
we canreplace III by
apolygon 1-12
that satisfies the modified condition(II),
so thatII2
has the doublespiral
formdescribed
by
Ostrowski.Finally, by
means of two morechanges
of the
type described, II2
may bereplaced by
apolygon 77g
which has
exactly
two sides. For77g
the statement of(2)
istrivial,
so that(2)
follows forarbitrary
arcs.The Johns Hopkins University.
(Received January 18th, 1936; received with modifications October 29th, 1936.)