AN INEQUALITY IN THE GEOMETRY OF NUMBERS

AN INEQUALITY IN THE GEOMETRY OF NUMBERS

BY N. O L E R University College, London( x )

I

A t h e o r e m d u e t o L. F e j e s - T d t h [2] s t a t e s t h a t if K1, . . . , K~ a r e n n o n - o v e r l a p p i n g c o n v e x d o m a i n s e a c h of w h i c h arises f r o m a g i v e n c o n v e x d o m a i n K b y a n a r e a - p r e s e r v i n g affine t r a n s f o r m a t i o n a n d H is a c o n v e x p o l y g o n h a v i n g a t m o s t six sides w h i c h c o n t a i n s t h e m t h e n A (H) >1 n h ( K ) w h e r e A (H) d e n o t e s t h e a r e a of H a n d h(K) is t h e a r e a of t h e s m a l l e s t p o l y g o n h a v i n g a t m o s t six sides w h i c h c a n b e c i r c u m s c r i b e d a b o u t K .

R e s t r i c t i n g t h e d o m a i n s K 1 .. . . . K~ t o b e c o n g r u e n t a n d s i m i l a r l y s i t u a t e d , C. A.

R o g e r s [4] o b t a i n s a s i m i l a r r e s u l t in w h i c h H is a n y c o n v e x d o m a i n c o v e r i n g K 1 .. . . . K~

a n d h (K) is r e p l a c e d b y d (K), t h e d e t e r m i n a n t of t h e closest l a t t i c e p a c k i n g of K . R o g e r s ' s r e s u l t s d e p e n d on t h e following t h e o r e m :

THEORV.~ (Rogers). Let G be a plane, strictly convex, Jordan curve containing the origin, O, o] a cartesian coordinate system in its interior. Denote by G (P) the translate o / G which results/tom the translation which takes 0 into P. .Let P0, P1 . . . . , P~ = P 0 , P~+I . . . Pn+m be points which satis[y

(1) the polygon, POP1... P~ is a Jordan polygon, H , bounding a domain, i.e. a closed, bounded, simply-connected set, II * ;

(2) the domains bounded by G(Pr-1) and G(P,) have a common boundary point i[

l <~r<~n;

(3) the points, Pn+l . . . Pn+m lie in l-I*;

(4) the domains bounded by G(Pr) and G(P~) have no interior points in common i~

l <~r < s <~n + m . Then

A (II*) 1

4A(~ ~>m+~n-1

where A (H*) is the area o/ II* and A (G) is the critical determinant o/G.

(1) This work has been carried out in part during t~he tenure of a National Research Council (Canada) Overseas :Fellowship. The author is indebted to the referee for his many helpful suggestions.

2 0 ~ . OLER

Our main object in this paper is the proof of a more general theorem of which the above is a special case.

We consider the set of points {(l(x 1 - x 2 ) , 89 Y2))} where @1, Yl) and @2, Y2) are a n y two points in the domain bounded b y G and denote b y F the boundary of the set ob- tained in this way. F is strictly convex, has 0 as a centre of s y m m e t r y and, defining F (P) in the same way as G(P) above, if G(P) and G(Q) bound domains which touch so do F (P) and F (Q) and conversely (see [4], p. 313).

Let/~ be the Minkowski distance function defined b y F (see, for example, Bonnesen- Fenchel [1], p. 21). Thus, for a n y two points, A and B, in the plane, # (A, B) = I A B I/I 0 R I where I A B I denotes the Euclidean distance between A and B and R is the point of F such t h a t the vector 0 R has the same direction as A B.

The condition t h a t the domains bounded b y F (P) and F (Q) touch is equivalent to

# (P, Q) = 2 while # (P, Q) > 2 is equivalent to their haviIlg no points in common. The same conditions with respect to 89 (P) and 89 (Q) are characterized b y # (P, Q) = 1 and # (P, Q) > 1 respectively where 89 is the set consisting of the mid-points of the segments joining O to the points of F. I t will be simpler to deal with 89 and its translates. Indeed, if we replace G b y 89 G in the above theorem A (H*) and A (G) must each be multiplied b y ~ and the result is the same.

DEFINITION 1. Let II be a Jordan polygon and E a / i n i t e set el points. We shall say that the pair ([[, E) is "weakly admissible" i/the/ollowing conditions are saris/led.

(i) The vertices o / H are contained in E.

(ii) E is contained in II*, the domain whose boundary is H.

(iii) For any two points, P and Q, in E, i/ the segment P Q lies in H* then # ( P , Q) >~ 1.

In particular, if E consists merely of the vertices of II we shall say t h a t II is a "weakly admissible polygon" when (II, E) is a weakly admissible pair.

Our main result is the following:

T H E O ~ E ~ 1. Let # be the Minkowski distance/unction de/ined by F a convex, centrally symmetric, Jordan curve and let (H, E) be a weakly admissible pair then

F ( I I ) A(II*) , M ( I I )

A r - ~ - - + 1 ~>n (In)

where A (1I*) is the area o / H * ,

M(n)

is the #-length o / H , n is the number o/points in E and A is the critical determinant with respect to F.

The existence of inequalities of the type (In) was suggested in a remark b y t{. Zassen-

A N I N E Q U A L I T Y I N T H E G E O M E T R Y O F N U M B E R S ~I

haas [6]. For the "star-shaped domain" ]xy] ~<] it has been shown by N. E. Smith [5]

t h a t

A (II*) N ( I I )

where N ( I I ) is the perimeter of II measured b y the norm-distance

~((Xl,

Yl), (x2, Y2))=

[(x2 - xl)(y~ - y 0 1 t

We shall first prove t h a t Theorem 1 is true for weakly admissible polygons. The general case will then follow b y induction on the number of points of E contained in the interior of II. The former case will occupy us for the greater part and it will be convenient to distinguish it as

T ~ E o ~ E M 2. Let II = P1P~ . . . PnP~+I where P~+I = P1 be a w e a k l y admissible polygon, then

F ( I I ) = A (II*) - - + M ( n ) ~ - + l ~ > n . A

The method we employ in the proof of Theorem 2 is b y induction on n and is based on the following observation. Let P~Ps be a diagonal of II (i.e._P~ and P j are not consecutive and the open segment P ~ P j - P ~ - P j is contained in the interior, II* - I I , of II). Further let #(P~, Pj) = 1. l:'iP j divides II into two polygons, II 1 and II2, which have, say, n 1 and n 2 vertices respectively. The assumption t h a t Theorem 2 holds for polygons with fewer vertices t h a n H yields F([[1)~> n 1 and F(H2)/> n2 since n~ and n 2 are each less than n.

Adding these inequalities and noting t h a t

A (117) + A (II~) = A (II*),

M(111) + M(II2) = M ( I I ) + 2#(P~, Pj) = M ( I I ) + 2

and n 1 + n 2 = n + 2

we obtain -4(II*) ~ M ( I ] ) + 2 + 2 ~ > n + 2 ;

A 2

hence F ( I I ) >~n.

Thus we would wish to show t h a t if, in no weakly admissible polygon with n vertices (n >~ 4), on which F takes the value F ( I I ) , is there a diagonal of/t-length equal to 1, then there exists a w. a. polygon, I I ' , with n vertices such t h a t F ( I I ' ) < F ( I I ) and I I ' does contain such a diagonal. For it would then suffice to prove the Theorem for n = 3 and when n ~> 4 for such polygons as contain a diagonal of unit #-length.

22 N. OLER

The above objective would be realised if we could show that, (i) in the absence of a diagonal of unit/~-length in a n y polygon with n vertices on which 2' takes the value F ( [ [ ) a local variation of the vertices of H exists under which 2"{H) is decreased while I I remains weakly admissible; and (ii) 2" attains a mirSmum on the set of w. a. polygons with n vertices.

The second requirement suggests t h a t we adjoin to the class of w. a. polygons certain limiting ones. Thus, if ( P r l . . . PrnPrll r = 1, 2 . . . . } is a sequence of w. a. polygons and lim Pr~ = P ~ ( i = 1, 2 . . . n) we adjoin the polygon P1 ... P~P~ and prove it to be weakly

r - - ) c ~

admissible.

Such a limiting polygon m a y contain singular vertices, such a vertex being one which is contained in a side other t h a n those of which it is an end point. I n Figures 1 a-1 d we il- lustrate examples of these.

a b c d

Fig. 1.

Certain singular vertices (e.g. Figs. 1 a, 1 c) lead to a decomposition of the polygon and to an application of the inductional assumption in a similar w a y to t h a t described when a diagonal has #-length 1. Others (e.g. Figs. lb, ld), however, present a new difficulty.

The variation of these or of the ends of the sides in which t h e y lie is restricted in so far as a neighbouring polygon may, in a sense, be self-overlapping (Fig. 2). To overcome this difficulty we are led to enlarge the class of weakly admissible polygons still further to include polygons of this type. I t will be seen, however, t h a t we are not obliged to allow the angles, as we shall define them, in these polygons to exceed 2 7e.

I n the next section we shall give a precise definition of a class of polygons which meets the requirements we have just outlined. Furthermore we shall show t h a t these polygons possess such properties as are necessary for the proof of Theorem 2 b y the method de- scribed.

Fig. 2.

A N I N E Q U A L I T Y I N T H E G E O M E T R Y O F N U M B E R S 23 I I

I n w h a t follows we shall regard the angle at a v e r t e x of a triangle, s a y B in A B C, as t h e intersection of the half-plane, p, which is h o u n d e d b y t h e line t h r o u g h A a n d B a n d which contains C a n d t h e half-plane, q, b o u n d e d b y t h e line t h r o u g h C a n d B a n d con- t a i n i n g A. I f A ' B ' C' is t h e image of A B C u n d e r a barycentric, orientation-preserving mapping, t h e angle at B ' is t h e intersection of half-planes which correspond to p a n d q, i.e. t h e half-plane which is on t h e same side of A ' B ' as p is of A B a n d t h a t which is on t h e same side of B'C' as q is of BC. If, in particular, A ' B ' C ' is a n i m p r o p e r triangle, i.e. one in which t w o angles are zero a n d t h e third, s a y / _ B ' , is ~, t h e angle at B ' is well defined b y t h e m a p p i n g as one of t h e half-planes b o u n d e d b y t h e line t h r o u g h A ' , B ' a n d C'.

B y a v e r t e x t r i a n g u l a t i o n of a domain, K*, b o u n d e d b y a convex J o r d a n polygon, K, w i t h n vertices we shall m e a n a set of n - 2 n o n - o v e r l a p p i n g triangles which cover K * a n d whose vertices are t h e vertices of K.

Dv, F I ~ I T I O ~ 2. We shall say that a closed polygon, H, belongs to the class ~, if it is the image o / a convex, Jordan polygon, K, with vertices P1 . . . P , , P~+I = P1, under a single- valued mapping, ~, which is de/ined on K* and has the properties:

1) ~) maps the triangles, T 1 . . . T~_ 2, o/ a vertex trianqulation o / K * barycentrieaUy onto triangles which may be improper, i.e. with angles O, 0 and ~r;

2) ~) is orientation preserving;

3) i / T ~ and Tj (i ~= ~) have a common vertex, say Pr, then the angles at @Pr in 6) T~ and T~ have no common interior points.

I n addition we define

(i) OP~(i = 1 . . . n) to be the vertices o/ l-I;

(ii) / O P ~ , the angle in II at OP~, to be the sum o~ the angles at @P~ in those triangles, 6) Tj, /or which Tj has P~ as a vertex;

(iii) OP~)P~+I (i = 1 . . . n) to be the sides o/ H, and the #-perimeter o / H to be

M (II) = l~ltt (OPt, 0P~+I).

(iv) A (H*), the area associated with I I , as the sum o / t h e areas o / t h e triangles @ T~

(i = 1 . . . n - 2).

W e n o t e in connection with (ii) that, as a consequence of t h e conditions of Defini- tion 1, 0 ~ ~ @ P ~ ~ 27~(i --- 1 . . . n). R e g a r d i n g (iii) we note t h a t M ( I I ) depends o n l y o n

24 N. OLER

OP1 . . . 0 P ~ . W i t h (x, y) as coordinates in a cartesian system we can express A (II*) according to (iv) as

~-21 f ^~ x d y - y d x ,

~=1 OT~

the sense of each p a t h of integration being such t h a t the integrals are each non-negative.

Accordingly, in the sum, common sides of the triangles

0 T 1 ... 0 Tn-2

are traversed back and forth precisely once; thus

A ( H * )

=89 f x d y - y d x

I I

and A (II*) depends only on 0 P 1 . . . 0 P ~ .

D E F I N I T I O N 3.

We de/ine the straight segment O P i O P j ( [ i -

]] =~1)

to be a diagonal o/ II i/there exists a path, ~, in K* whose end points are P~ and Pj and whose inner points are inner points o/ K* such that 0 is a sense preserving mapping o/,~ onto OP~OPj.

We m a y assume t h a t ~ is a simple polygonal p a t h whose vertices are contained in the common sides of those triangles in the vertex triangulation of K * associated with 0 which cover ~. For if A and B are two points of ~ contained in the sides of such a triangle, say Tk, the straight segment, A B, is m a p p e d b y 0 onto the straight segment 0 A 0 B and we can replace the section of ~ between A and B b y A B. Indeed, this section is already A B if 0 Tg is proper.

D E F I N I T I O N

4. A polygon in ~ is de/ined to be weakly admissible i/its sides and dia- gonals are each o/~-length not less than 1. We shall denote the subclass o] polygons in ~ with this property by 9~(#).

T H E O ~ E ~ 3.

Let K =P1 ...P~P~+I,P~+I =P1 be a convex Jordan polygon and | a mapping o / K * such that O K belongs to 9~(/~).

Let OP~@Pj be a diagonal o/ O K and K 1 and K 2 denote the polygons into which P~P~

divides K.

There exist mappings, 01 o] K~ and 02 o/ K~, such that @2 agrees with @ on K 1 - P ~ P j

and O1K1

is in ~(#); similarly/or 02.

Proo].

L e t n = 4 a n d suppose t h a t 0 is barycentrie on the triangles

PIP2P3

and

P1PaPa.

I f the diagonal

@P~OPj

is

OPIOPa

the restrictions of ~) to these triangles arc the mappings 01 a n d 02 of the theorem. I f

OP2@P4

is a diagonal, t h e n

#(OP2, OP4)/>

1 and we m a y take 01 and 02 to be the barycentric mappings of

P1P2P4

onto

@PI@P2@P4

and

P2PaP4

onto

(~P2OPa@Pa

respectively.

A N I N E Q U A L I T Y I N T H E G E O M E T R Y O F N U M B E R S 25 L e t us a s s u m e t h a t t h e t h e o r e m is t r u e for p o l y g o n s in ~ ( / t ) w i t h m vertices, 4 < m < n.

T h e r e is n o loss of g e n e r a l i t y in t a k i n g i < ] a n d l e t t i n g

K, = P, ... -P~PJPs+*... P~P,

a n d K 2 =

P~P~+I ... Pj-IPjP~ 9

W e s h a l l be c o n c e r n e d w i t h t h o s e of t h e d i a g o n a l s of K w h i c h a r e sides of t h e t r i a n g l e s in t h e t r i a n g u l a t i o n of K * a s s o c i a t e d w i t h O a n d s h a l l refer t o t h e s e a s O - d i a g o n a l s of K . L e t 1 be a p o l y g o n a l p a t h in K * such t h a t O t =

OP~OP r To

p r o v e t h a t 02 e x i s t s we e x a m i n e i in r e l a t i o n t o t h o s e 0 - d i a g o n a l s of K w h i c h h a v e a n e n d p o i n t a m o n g s t t h e v e r t i c e s of K s. W e c o n s i d e r s e p a r a t e l y t h e case in w h i c h t h e r e e x i s t s a 0 - d i a g o n a l b o t h of whose e n d p o i n t s a r e v e r t i c e s of K 2 a n d t h a t in w h i c h t h e r e is no such 0 - d i a g o n a l .

(a) L e t

PrPs

be a O - d i a g o n a l , i ~< r < s ~< j; i.e.

Pr

a n d P8 a r e b o t h v e r t i c e s of K 2.

F u r t h e r , l e t

PrP~P t

be t h e t r i a n g l e in t h e v e r t e x t r i a n g u l a t i o n of K * w h i c h h a s

P~.P~

as a side a n d lies in K ' * w h e r e

K ' = Ps... PnP1... PrP~.

W e e x a m i n e t h e p o s s i b i l i t y t h a t c o n t a i n s p o i n t s in

P~P~;

t h i s is c e r t a i n l y t h e case if i t c o n t a i n s p o i n t s in K * - K ' * . T h e r e is a p o i n t , s a y A , in

P~.P~P~

w h i c h is n e a r e s t a l o n g ~ t o P~ a n d a p o i n t , s a y B, in

P~.P~Pt

which is n e a r e s t t o P j . Since 0 m a p s A B o n t o a s t r a i g h t s e g m e n t we c a n r e p l a c e t h e sec- t i o n of )t b e t w e e n A a n d B b y A B. T h e r e s u l t i n g p a t h is one w h i c h coincides w i t h

PTP~

if i = r a n d ?" = s, for t h e n A ~- P~ a n d B - P j , or i t s i n t e r i o r p o i n t s lie i n t h e i n t e r i o r of K ' * . I f 1 c o n t a i n s no p o i n t s of

PrP~

its i n t e r i o r p o i n t s lie in t h e i n t e r i o r of K ' * . T h u s we c a n a s s u m e u n d e r all c i r c u m s t a n c e s t h a t e i t h e r 1 coincides w i t h

P~,Ps

or its i n t e r i o r p o i n t s lie in t h e i n t e r i o r of K ' * .

L e t 0 ' be t h e r e s t r i c t i o n of 0 t o K ' * . C l e a r l y

O'K"

belongs t o a. 0 ' K ' is in f a c t in ~ ( # ) . F o r a d i a g o n a l of

O'.K"

is a d i a g o n a l of 0 K a n d t h e r e f o r e o f / t - l e n g t h n o t less t h a n 1 while t h e sides of

O'K',

b e i n g a m o n g s t t h o s e of O K a n d in a d d i t i o n

OPrOP~

w h i c h is a d i a g o n a l of

OK,

h a v e / t - l e n g t h n o t less t h a n 1. F u r t h e r m o r e

O'P~O'Pj

is a d i a g o n a l of 0 ' K ' or a side of

O'.K'.

T h e l a t t e r is c e r t a i n l y t h e case if

K'

h a s o n l y t h r e e v e r t i c e s for t h e n r = 1 a n d s = n - 1 a n d we m u s t h a v e i = r a n d j = s . I f

O'P~O'Pj

is a side of 0 ' K ' t h e n 0 ' is t h e m a p p i n g 01 w h i c h we seek. I f

O'.PiO'Pj

is ~ d i a g o n a l of

O'K', K"

h a s m o r e t h a n t h r e e v e r t i c e s a n d we a p p l y t h e i n d u c t i o n a l a s s u m p t i o n t o 0 ' K ' t o o b t a i n t h e e x i s t e n c e of 01.

(b) S u p p o s e t h a t t h e r e is no 0 - d i a g o n a l of K b o t h of whose e n d p o i n t s a r e v e r t i c e s of K~.

W e s h a l l show t h a t all of t h e p o i n t s , 0 P , + I , -.-, 0 P j _ ~ , lie in one of t h e o p e n half- p l a n e s b o u n d e d b y L , t h e s t r a i g h t line c o n t a i n i n g 0 P ~ a n d 0 P j or in L itself.

H j - i = 2 t h e a b o v e a s s e r t i o n is t r i v i a l . L e t ] - i > 2 a n d c o n s i d e r

-P~P~,+I

w h e r e i < r < ] - 1. T h e t r i a n g l e in t h e v e r t e x t r i a n g u l a t i o n of K * w i t h side

P~.Pr+I

is s u c h t h a t i t s t h i r d v e r t e x , s a y P , is n o t in K~. I t follows t h a t , since i n n e r p o i n t s of i a r e b y d e f i n i t i o n in t h e i n t e r i o r of K * , t c o n t a i n s ~n i n n e r p o i n t in e a c h of

PrPz

a n d

P~.+IPz.

H e n c e

OP~OP~

26 ~. OLER

contains an inner point of

19Pr19P~

a n d an inner point of

19Pr+119Pl.

I f therefore, 19Pr lies in L t h e n

19Pr19Pl

lies in L and in particular 19Pt lies in 15 a n d conversely. Similarly, if 19Pl lies in L so also does 19Pr+1- Considering each of the sides of K s we see t h a t if one of the vertices 19P~+1 .. . . . 19Pj_1 lies in L so do t h e y all and in a d d i t i o n / 5 contains the images of the vertices of K1, which are end points of the 19-diagonals the other end points of which are 19Pi+1 . . . . ,19PJ-1. If, on the other hand, 19Pr does not lie in L then 19Pl does not lie in L; indeed 19Pr lies in one of the open half-planes bounded b y L and 19Pz in the other.

Similarly, if L does not contain 19Pt then 19P~+1 does not lie in L b u t in the openhalf-plane bounded b y L which does not contain 19Pz namely the same one as contains 19P~. Again considering each of the sides of K~ we see t h a t if one of the vertices 19P~+1 .. . . . 19Pj_l does not lie in L t h e y all lie in one of the open half-planes bounded b y L while the other contains the images of those vertices of K 1 which are end points of the 19-diagonals with end points 19Pi+1

... 19P~_ 1.

W h a t we h a v e lust shown implies t h a t amongst 19P~+1 .. . . ,19Pj_ 1 there is a vertex, s a y 19P~, at which the angle is n o t greater t h a n ~. F o r if 19P~+1 .. . . . 19Pj_l all lie in one of the open half-planes bounded b y L it suffices to choose 19P~ to be one which is furthest from L.

L e t us suppose t h a t 19Pi+1 . . . 19Pj_x all lie i n L , t h e n the angles,/-19P~+1 . . . L19Ps-1 can h a v e the values 0, ~, or 2 ~ and and we m u s t show t h a t t h e y are not all 2 ~. We choose a direction in L according to which 19Pj is to the right of 19P~. L e t 19P~+1, be to the left of OPt. I f /-19P~+1 were equal to 2 rc there would be a 19-diagonal w i t h end point P~+I, say

Pi+IP,,

such t h a t | is to the left of 19P~+1. Thus

19P~+lOP,~

would have no point in common with

19Pi19Pj

which is not possible since

Pf+IP,~

contains points of 2. Hence if 19Pt+1 is to the left of 19Pt t h e n Z-OPt+l is equal to 0 or z. L e t OP~+I be to the right of 19Pt a n d let/-19P~+1 = 2~. Then there is a 19-diagonal, sayP~+~P~, with P~+~ as an end point such t h a t 19Pb is to the right of 19P~+~. Since, moreover,

19P~+I19Pb

and

19P~19P~

h a v e an interior point in common, 19P~+~ is to the left of 19P~ and so 19P~+~ is distinct from 19Pr H ~19P~+z = 2 ~ there would be a 19-diagonal, say

P~+~Pc,

with P~+~ as an end point such t h a t 19Pc is to the left of 19P~+~. Since, however,P~+~Pe contains a point of ~ betweenP~ and those points of 2 in P~+I P~ it follows t h a t 19 P~+~ 19P~ cannot lie wholly to the left of 19 P~+119 P~.

Hence OPr is to the right of 19P~+z and /-19P~+~ cannot be equal to 2 zr. Thus under all circumstances there is a vertex amongst 19P~+~, ..., 19P~_~ say 19Pr, such t h a t

A19Pr <:~.

We fix our attention on

Pr

a n d the union of those triangles in the vertex triangulation of K* which h a v e Pr as a vertex. L e t 8r denote the polygon bounding the domain defined in this way. Amongst the vertices of S~ other t h a n P~_~, P~ and P~+I there is one, say P~, with the following property. The triangles, say

P~P~P,~

and

PrPtP~, in

the vertex triangula-

A N I N E Q U A L I T Y I N T H E G E O M E T R Y O F N U M B E R S 27 tion of K * which have PrPt as a common side are m a p p e d b y 0 onto a quadrilateral no angle of which exceeds ~. Thus, if OP~ does not lie in L, then no vertex of OS~ lies in L and we can choose OPz to be t h a t which is furthest from L. For then OPrOP~OPzOPb lies in the strip hounded b y the two lines parallel to L one of which contains OP~ and the other 0P~. I f OP~ is contained in L t h e n so are all of the vertices of OS~ and from amongst those of S~ other t h a n P~-I, P~ and P~+I we choose P~ to be such t h a t OP~OPI is longest.

We shall show t h a t OPrOP~OPIOPb t h e n has the desired property. L e t P~PtP~ be on the same side of P~Pz as P , and P~PtPo on the same side as Pj. I t suffices to show t h a t at least one of the angles at OPz in OP~OPtOPa and OP~OPtOP~ is 0. L e t O P j be to the right of OP~ as before a n d suppose t h a t 0 P t is to the right of OPt. We m u s t show t h a t O P , is to the left of OPt. Were it otherwise t h e n OP~OPa would be longer t h a n OP~OP~

hence P a - P~-I" However, P~IPI contains a point of i between Pt a n d the points of which P~Pt contains. Hence OP~_IOPI cannot lie to t h e right of OP~OPt. Thus O P a cannot be to the right of O P t and the angle a t O P t in OPrOP~OP,~ is 0. Similarly, if OP~

is to the right of O P t then the angle at OPz in OPrOPtOP~ is 0.

The angles in the quadrilateral OP~OPaOP~OPo being each not greater t h a n ~, there is an orientation preserving m a p p i n g which takes P~PaPt a n d P~P~P~ barycentrically onto OP~OPaOPt and OP~OPbOPz respectively. Thus we can retriangulate the domain bounded b y P~PaPzPo and modify 0 accordingly. As a result the n u m b e r of vertices of S~ is decreased. Repeating the process sufficiently m a n y times P~-~Pr+I becomes a O- diagonal and we again have the situation dealt with in (a).

B y giving similar consideration to the vertices of K~ as we have given to those of K~

we obtain the existence of 0~ also.

COROLLARY. Let K = P 1 ... P~Pn+I, Pn+I=P1 be a convex Jordan polygon and O K a polygon in 9~(~t). Let ,~ be a path in K* one o/ whose end points is Pt and the other an interior point, Q, o/ the side PjPj+I (i~-], ] + 1 ) while the inner points o/

are in the interior o/ K*. Furthermore, let 0 ~ = 0 P~.

There exists a mapping, 01, which agrees with 0 on PtPt+l . . . Pj-IPj such that 0 1 ( P t P i + l . . . Pj-IPjPt) is in 9~(#) provided that ] + i + l; also a mapping, 02, exists which agrees with 0 on Pj+IPj+2 ... P t - l P t such that 02(PiPj+IPJ+2 ... P~-,Pt) is in 9~ (#) provided that ] + 1:4: i - 1.

Proo/. I n the vertex triangulation of K* associated with 0 let the other sides of the triangle with one side PJPJ+I be PjPk and Pj+IPk. The path, i , contains a point, say A, in one of these two sides. Substituting for the section of 1 between A and Q b y A P j we get a p a t h with end points Pt and P j which is m a p p e d b y 0 onto OPtOPj. Thus, if ]~: i + 1,

2 8 N. OLER

O P , O P j is a diagonal and applying the theorem we obtain the existence of Oz. Similarly, substituting A P j + 1 for the section of ~t between A and Q we find t h a t OP~OPj+I is a diagonal provided ] + 1 4= i - 1. Again applying the theorem we obtain the existence of 02, which proves the corollary.

We shall say t h a t O P , is an inner point of O P j O P j + z (i 4= ?', ] + 1) when, as in the above corollary, there exists a path, 2, whose end points are P , and an inner point of PJPJ+I, whose remaining points are in the interior of K* and which is mapped b y 0 onto the single point, 0 P , .

We next prove t h a t certain sequences of polygons in ~ ( # ) have a limit which is again a polygon in ~ (#). Thus

T ~ O R E ~ 4. Let K = P 1 . . . PnP~+I, P~+I =P~ be a convex Jordan polygon and F~ a vertex triangulation o / K * , the domain bounded by K .

Let 01, 02 . . . . be a sequence o/mappings o] K * such that

(i) O~K is in ~(tt), r = 1, 2 . . . . ;

(ii) The vertex triangulation o / K * associated with Or is C, r = 1, 2 . . . . ; (iii) lim O~P~ exists/or i = 1, 2 . . . n.

r....~oo

The mapping, % of K * which takes P~ onto lim O~Pi, i = ] . . . n, and is barycentric on

r,.-~ o o

the triangles o/F~ is such that q)K belongs to 92[(re ).

Proo]. Let PuP~P,o be a triangle i n / : and in a cartesian coordinate system let x~ y~ 1

x~, Yw 1

where (x~, Yu), (xv, Yv) and (xw, Yw) are coordinates of P~, P~ and Pw respectively. T h a t Or(r = I, 2 . . . . ) preserves the orientation of P~PvPw is equivalent to

IP , Po, Pwl" I orP , ORB,, 1>0.

B u t

lira lOrRy, O~P,, O~P~[ = IlimQrP~, limO~P~, limOrP~[ = [~P~, ~P~, ~P~I"

Hence IPu, Pv, Pwl " 15~ vPv, cflPwl >10 and ~v preserves the orientation of PuP~Pw.

We next prove t h a t

#(TP~, ~vP~+l) >/1 (i = 1, ..., n)

AI~ I N E Q U A L I T Y lq~ T H E G E O I ~ E T R Y O F N U M B E R S 29

and ke (~P~, ~P~) ~> 1

whenever cfP~cfP~ is a diagonal.

Since

# (~Pi, ~P~+~) =/~ (lira | lim | = lim # (| |

r - - > r r - - > o| ~ - - ~ o o

and/~ (@rP~, 0rP~+~) >~ 1 (r = l, 2, ...) it follows t h a t # (cfP,, q~P~+I) >~ 1.

Suppose t h a t cfP~cfP~ is a diagonal, the image under ~ of a polygonal path, say

= QoQ~ ... Q~Q~+~ where Q0 = P,, Q~+~ = P~ and Q~ .. . . . ~ are interior points of ~-diagonals, P ~ Pz~ .... P~P,~ respectively. I n order to show t h a t ~t(~P,, ~P~) ~> 1 we consider the sequence 01,t , ~ t . . . Since

it follows t h a t

lim | =cfQ~(h = 0 , 1, ... s, s + 1)

r - - ~ o|

lim # (@rQh, @rQh+l) = # (cfQh, ~Qh+l) (h = O, 1 . . . . s).

T.--> o r

Recalling that, b y definition, ~ preserves the sense of 2 in mapping it onto cfP~cfPj we have t h a t

s

/~ (~vP,, ~vP,) = h=~0/~ (~ Qh, ~0 Q;z+I) = h-0 ~ r->cclim

[~(OrQh, OrQh+l) = r-->qzlim

h~-0/~ (@~ Q ' ~ ' - | Q~+I.

T h a t /~(~0P~, ~Pj)~> 1 will follow therefore if we can show that, for each r, M(@r). ) =

I~ (OrQh,

OrQh+i) ~ 1.

h=0

In order to do so we consider the set, A, of all the polygonal paths with end points P~ and P~ which have vertices which correspond to those of 2 and which lie in the closed segments P~IPh . . . . , P k s P l . Considering the images of these paths under Or, the set, A, as defined, being closed, it contains a path, say i =QoQ1 ... -O~Q~+x, such t h a t M (Or).) is a minimum. I t suffices to show t h a t M(@~t) ~ 1.

L e t t, 0 < t ~< s + 1, be the smallest index for which i~ t =Pkt or ~t =Pit, say Qt =Pkt, where Q~+I =Pk,+l = PJ. I f t = 1 then @r~ c o n t a i n s 0rP~OrP/r 1. Noting t h a t P~PkxPzl is a triangle in the vertex triangulation, s it follows t h a t @rP~@rPkl is either a side or a diagonal of @~K and therefore has/~-length not Iess t h a n 1. Hence M ( @ ~ ) / > 1. Suppose t h a t t > 1. If for some u, 0 < u < t,/@~u-x@~-Q~@~-Q=+l is different from z, then Q= being an interior point of Pk=Pl= we can v a r y it along Pk=Pl~ in such a way t h a t #(@~u-1, O~Q=) + # (O~Qu, @~Q~+I) is decreased. Indeed we obtain thereby a path, say ~' in A such t h a t M (@~t') < M (| which is a contradiction. H e n c e , / @ ~ Q=-I | Q~| = z (0 < u < t)

30 z~. OL~R

implying

that OrP~OrPkt

is a diagonal with

PiQ1

^{. . .}

Qt-lPkt

as antecedent or

OrP~OrP~t

is a side of 0 r K . I n either case #(OrP~,

O~P~ t)

>--1. Hence M(Or~) ~>1.

I t follows that, in particular, the sides of the triangles of s are mapped b y ~0 onto segments of #-length not less than 1. Thus the images under ~ of these triangles have distinct vertices and the angles at these vertices are therefore well defined.

I t remains for us to show t h a t if T q and T~2 are a pair of triangles in ~ which have P~

as a common vertex then the angles at ~0P~ in ~ T q and ~ T~2 have no common interior point. Let us suppose, on the contrary, t h a t A is such a common interior point. There exists r 1 such that, if r > rl, A is an inner point of the angle at OrP~ in Or T q and r 2 such that, if r > r2, A is an inner point of the angle at OrP~ in Or T~ 2. Thus, if r > max (rl, r~), A is an inner point of the angles at OrP~ in both Or T q and Or T~ 2 which contradicts the hypothesis t h a t 0 r K is in 9~(#). Hence the angles at ~vP~ in ~0 T h and ~v Tt~ have no inner points in common. We have therefore shown t h a t ~0K satisfies the conditions necessary for it to belong to 9~(/~).

I n addition to the above we shall have need of

THEO~WM 5.

Let K =P1 ""PnPn+l, -Pn+l =-P1 be a convex Jordan polygon and O K a polygon in ~ (#). I] no vertex o] 0 K is an interior point o / a side there exists a mapping, 0', o] K* which agrees with ~) on K, O' K is in 9~(1~ ) and O' maps the triangles in the vertex triangulation o / K * associated with it onto proper triangles.

Proo/.

If E) maps the triangles in the vertex triangulation of K* associated w i t h it onto proper triangles the theorem is true with @' = | throughout K*.

Let re(O) denote the number of triangles which ~) maps onto improper triangles and suppose

re(O)>0.

I t suffices to show t h a t a mapping, ~0, of K* exists which agrees with 0 on K, such t h a t ~0K is in 9~(#) and m(~) < m(0).

Let

PrPsPt

be a triangle which 0 maps onto an improper triangle the angle, say at OPt, being equal to ~. We can assume without loss of generality t h a t r < s. Since, according to our hypothesis,

PrP s

cannot be a side of K let

PrP~P 8

be the other triangle with

PrPs

as a side. We note t h a t r < u < s if we assume, as we may, t h a t t < r or t > s.

The triangle 0

(PrPuPs)

can be distinguished according to whether (1) 0

(PrPuP~)

is a proper triangle,

(2) 0

(PrP,,P~)

is improper and the angle at 0 P ~ is 0, or (3) O

(PrP~P~)

is improper and the angle at

OPu

is ~.

If (1) is the case we define ~ as a mapping which agrees with 0 on K and on those triangles associated with 0 other than

PrP~P~

and

PrPuPs

while it maps

PtPrP~

and

AN INEQUALITY IN THE GEOMETRY OF NUMBERS 31

P~PuP~

b a r y c e n t r i e a l l y . I n d e e d ~ m a p s t h e s e o n t o p r o p e r t r i a n g l e s a n d we h a v e m ( ~ ) = m(@) - 1, while ~ K is c l e a r l y in 9~(#).

I f @

(PrP~Ps)

is as in (2) t h e a n g l e a t e i t h e r OP~ o r @P~ is 0, s a y a t O P t . L e t us w r i t e t 1 for s, r 1 for r a n d s 1 for u. T h e t r i a n g l e ,

P~IP~IPh,

is m a p p e d b y 6) o n t o a n i m p r o p e r t r i a n g l e in w h i c h t h e angle a t 6 ) P h is H. I n p a r t i c u l a r 0 < s 1 -- r I < s -- r. I f 6)

(PrPuPs)

is as in (3) we r e p l a c e 6) b y 6)1 w h e r e 6)1 is d e f i n e d in t h e s a m e w a y as ~ in t h e p r e c e d i n g p a r a g r a p h , a l t h o u g h in t h i s case m(6)1) = ~n(6)). H o w e v e r , we m a y r e p l a c e 6) b y 6)~ in t h e p r o o f of t h e t h e o r e m . I n one of t h e t r i a n g l e s 6)1

(P~PrP~)

a n d 6)1

(PtPuPs)

t h e a n g l e a t @IP~ is e q u a l t o H a n d in t h e o t h e r 0. W e choose t h e one in w h i c h i t is H, s a y in 6)1

(PtPrP~).

W r i t i n g t 1 f o r t, r 1 for r a n d s 1 for u, t h e t r i a n g l e

PhPsiPtl

is m a p p e d b y @1 o n t o a n i m p r o p e r t r i a n g l e in w h i c h t h e angle a t

6)lPtl

is yr. F u r t h e r m o r e , 0 < 81 - r l < 8 ^-- r.

I n e i t h e r of t h e cases (2) a n d (3) we a p p l y t h e s a m e a r g u m e n t t o

PhP~tPtl

as we h a v e d o n e to

PresPt.

E i t h e r t h e o t h e r t r i a n g l e w i t h side

PhP~I

is as in (1) a n d we o b t a i n t h e m a p p i n g ~0 or we a r e l e d t o a t r i a n g l e

P~2PsPt2

w h i c h is o b t a i n e d in t h e s a m e m a n n e r as

PhP~lPtf

C o n t i n u i n g in t h i s w a y e i t h e r we c o m e t o t h e m a p p i n g ~0 w i t h m (~) = m (6)) - I or t o a sequence of t r i a n g l e s P r l P s l P t l , Pr2Ps2Pt2 . . . . e a c h of w h i c h is o b t a i n e d in t h e s a m e w a y as

PhP~tPtf

B u t since sz - r I > s2 - r e > 9 9 ", for s o m e k, O P t e is t h e i n t e r i o r p o i n t of a side, n a m e l y

6)P~k6)P~k,

w h i c h c o n t r a d i c t s t h e h y p o t h e s i s .

I I I

W e r e t u r n n o w t o t h e p r o o f of T h e o r e m 2 a n d in o r d e r t o e s t a b l i s h (In) we first p r o v e

LEMMA 1. I] T, a triangle in ~(#) with vertices P, Q, R has a pair o/sides o//z-length greater than 1 there exists a triangle T' in 9~(#) such that 2'(T') < 2'(T).

Pro@

I f T is a n i m p r o p e r t r i a n g l e with, say, A P = H,/~ (Q, R) > 2 a n d if, s a y , # (P, R) > 1 a p o i n t R ' in

P R

such t h a t 1 ~<#(P, R ' ) < # ( P , R) satisfies

F(PQR') < F(T).

S u p p o s e t h a t T is a p r o p e r t r i a n g l e a n d l e t # (P, Q) > 1 a n d # (P, R) > 1. Since P lies o u t s i d e F (Q) a n d F (R) (see F i g . 3) t h e r e e x i s t s a n e i g h b o u r h o o d of P w i t h t h e s a m e p r o p e r t y

P

Fig. 3.

3 2 N . O L ] ~ R

m _ _

i)

R= (x,Y2) Fig. 4.

a n d in particular a point, P ' , such t h a t

QP;~=

2 Q P ( 0 < 2 < 1). L e t T ' be the triangle with vertices

P', Q, R.

We have A (T'*) < A (T*) and since # (P', R) ~< # (P', P) + # (P, R)

# ( P ' , R) + # ( P ' , Q) < # ( P ' , P) + / ~ ( P ' , Q) +/~(P, R)

a n d

M ( T') < M ( T).

Hence P ( T ' ) <

F(T).

I t suffices therefore to prove (I3) for a triangle having at m o s t one side of #-length greater t h a n 1.

I n the triangle, T, with vertices P, Q and R let # (P, Q) =/~ (P, R) = 1 and # (Q, R) >1 1.

We choose a coordinate system in the following manner. With P as origin let the y- axis be the line through P parallel to and having the same sense as

RQ.

This meets F (P) a t S and

S',

say. As x-axis we choose the line through P parallel to the tangent (1) to F (P) a t S a n d having t h a t sense b y which the x-coordinate of Q, hence also of R, is non- negative.

Since #(Q, R) >1 1,

QR >~PS

and since Q and R lie on F ( P ) t h e y lie on opposite sides of t h e x-axis.

L e t (x, Yl) a n d (x, y~) be the coordinates of Q and R respectively. We h a v e x >/0, Yl > 0 and y~ < 0. Referring to Fig. 4,

sin 0 + 1

F ( T (x)) = ~ x (Yl - Y~) 2 ~ (Yl - Y2) + 2.

Differentiating twice with respect to x:

sin 0 . ,, , . . . I ,,

F " ( x ) = ~ [x(y~ - y ~ ) + 2 ( y l - - y ~ ) ] + ~ = ( y l --y~).

(1) We shall assume throughout that F is twice differentiable.

A N I N : E Q U A L I T Y I N TH~I G E O M E T R Y O F N U M B E R S 33 Since Yl > 0, y~ ~< 0 and y~' < 0 while y~ < 0 implies t h a t y~ ~> 0 and y~' > 0. I t follows t h a t F"(x) < O. Thus, either #(Q, R) = 2, #(Q, R) = 1 or there exists T ' such t h a t F ( T ' ) < F ( T ) . I t is sufficient therefore to consider the cases: # (Q, R) = 2 and # (Q, R) = 1. I f / t (Q, R) = 2 then F ( T ) = 3 and (I3) is satisfied with equality. I n the case #(Q, R) = 1, (I3) is a conse- quence of the fact (see Mahler [3], p. 693) t h a t if P, Q and R are such t h a t PQ, P R and Q R each have re-length 1 the lattice generated b y P Q and Q R is admissible. For this implies t h a t A (T) in this case is not less t h a n 89 Hence

F(T)~89

~ + ~ + 1 = 3 . ³

This completes the proof of (I3).

We shall complete the proof of Theorem 2 b y means of induction on n. This will be based on

LE~MA 2. Let K be a convex Jordan polygon with vertices P1 . . . Pn and O K = II be a polygon in 9~ (tt). Either

(i) II has a vertex which is an interior point o / a side, (ii) II has a diagonal o/#-length 1

or (iii) there exists a polygon, I I ' , in 9.1(#) with n vertices such that F ( I I ' ) < F ( I I ) or II' has property (i) or (ii) and F ( I I ' ) = F ( I I ) .

I n proving this l e m m a we shall repeatedly a p p l y

LEMMA 3. Let II = 0 P L O P 2 . . - O P n O P 1 be a polygon in ~(tt) whose diagonals each have tt-length greater than 1 and which has no vertex an interior point o] a side. I / i t is possible to vary the vertices o / I I in such a way that II remains in 9~ and its sides remain o/#-length not less than 1 then a su//iciently small such variation exists under which II remains in ~(lt).

Proo[. Let us suppose t h a t the l e m m a is false. There exists a sequence of polygons with II as limit none of which is in a (#). Since, however, each of these, b y hypothesis, is in 9~

and h a s sides of/t-length not less t h a n 1 there m u s t be in each a diagonal of ~-length less t h a n 1. There is therefore a subsequence in which these diagonals are the images of paths in K* all of which have the same end points, say Pr and Ps. L e t these paths be hi, 2~ . . . Recalling t h a t each such p a t h is polygonal with vertices lying in the diagonals of K there exists a subsequence 2~(1), ~(2~, ... with corresponding vertices which occur in the same diagonals of K. The limit of the sequence ~,1), ki~2) .. . . is a path, 4, which is m a p p e d b y O onto OPrOPs. Furthermore since the images of a~{1), 2~ r . . . . each have/~-length less

3 -- 60173047. Acta mathematica, 105. I m p r i m 6 le 20 m a r s 1961

34 N. O L E R

OPi§ OPi+3

OPi_ I OPi+l

Fig. 5.

than 1 it follows t h a t #(OPT, OPt) ~< 1. Either the inner points of 2 lie in the interior of K* or certain of the vertices of 2 coincide with vertices of K.

I n the former case it follows t h a t

OPTOP~

is a diagonal. Since, however, # ( O P t , OPs) ~ 1 this is a contradiction.

I n the latter case either there is a section of 2 which is a p a t h with end points which are distinct vertices of K and inner points which are in the interior of K* this leading to the same contradiction as before or 2 lies entirely in K. I n this last case since PT and P~

are not consecutive 2 consists of at least two consecutive sides of K. This implies t h a t two or more sides of II have/~-lengths the sum of which is less than or equal to 1 and hence there is a side of II whose #-length is less than 1 which is again a contradiction.

Proo/o/Lemma

2. We shall assume t h a t II has neither property (i) nor (ii) and show t h a t this implies (iii).

1. I n virtue of Theorem 5 the negation of (i) allows us to assume t h a t O maps the triangles associated with it onto proper triangles; in particular that / O P i > 0 (i = 1 . . . n).

2. If L O P , < ~ we can assume that L O P , + I < 2 z and / _ O P i _ 1 < 2 ~ . For suppose t h a t L O P , + 1 = 2 ~ (the argument for / O P , - 1 is the same). If

/~ (@P~+~, OP~+~) > # (@P~, OP~+~)

we may, in virtue of Lemma 3, v a r y OPi+ 1 along p (OPi, OP,+I)F (OPi) so as to decrease / - O P t + 1 (see Fig. 5). Since # ( O P i , OPi+I)F(OP,) lies inside /~(OPi+2, OP,+I)F(OPi+~) this variation decreases #(OPi+~,OP,+I) , hence also M ( I I ) . Since, moreover, A ( I I * ) is decreased this variation results in a decrease in F ( I I ) .

Let

tt (OPt+e, OPt+l) = # (OPi, OPi+l).

We m a y v a r y OPf+ 1 along # ( O P i , OP~+I)I'(OP,) so as to increase / O P i . This leaves F (II) unchanged but continuing to v a r y OPi+l in this w a y leads either to a polygon which satisfies (i) or (ii) or to one in which the number of angles in II which are less than ~ is

AN I N E Q U A L I T Y I:N THE GEOMETRY OF NITMB]~RS 3 5

decreased. Thus, if fl-OP~+2 ~< 2 z - A O P ~ we can decrease A O P i § to 0 if necessary while

~ O P ~ remains less t h a n 2 ~ a n d if LOP~+~ > 2 z - / - O P ~ we can increase / - O P ~ t o w i t h o u t L O P , + ~ becoming less t h a n zz.

W e m u s t observe t h a t in t h e course of such a continued v a r i a t i o n if at some stage II is no longer in ~l(#) there is an earlier point at which II is in 9~(#) a n d has p r o p e r t y (i) or (ii). To show this we d e n o t e t h e a m o u n t b y which / - O P , is increased b y t a n d consider I I as a function YI (t) of t. I f there are values of t, in t h e range considered, a t which H is n o t in ~(/~) let t o be t h e g.l.b, of these. Since L e m m a 3 is applicable t o tq ~ [[ (0) it follows t h a t 4 4 0 . W e can therefore find an increasing sequence {t~} such t h a t {II (t~)} converges to I I (to). F u r t h e r m o r e , each p o l y g o n in this sequence being in 9~ (/z), it follows f r o m T h e o r e m 4 t h a t I I (to) is in ~l(#). If, however, H (to) h a d neither p r o p e r t y (i) nor (ii) we could a p p l y L e m m a 3 to contradict t h e fact t h a t t o is t h e g.l.b, of values of t for which II (t) is n o t in 9~(#). H e n c e II (to) has p r o p e r t y (i) or (ii).

L e t # (OPt+2, OP~+l) < # (@P~,

OP~+l).

I f / - O P t + 2 < 2 z we can decrease F ( I I ) b y v a r y i n g OP~+I along # (OPt+2, OP~+I)F (OPt+2) so as t o decrease / O P i . I f ~ 0 P 1 + 2 = 2 ~ we can m o v e

OP~+2OPi+I

along

OP~OPi+a

w i t h o u t changing F ( H ) until OPt+2 coincides with OP~ or (iii) is satisfied. I f at some stage in this variation a p o l y g o n is o b t a i n e d which is n o t in ~ ( # ) there m u s t be one which is o b t a i n e d earlier which satisfies (i) or (ii). This follows b y t h e a r g u m e n t of t h e preceding case in which we n o w t a k e the p a r a m e t e r , t, to be t h e a m o u n t b y which # (OPt, OPt+l) is decreased. Finally, if OPt+2 coincides with OP~ t h e n #(OPi+2,

OPt+l) =#(OPi, OPi+l)

a n d this we h a v e a l r e a d y considered.

Thus, t h e assumption, ~ O P ~ + I = 2 ~r when / - O P ~ < ~r, leads either to a decrease in F ( I I ) , to a polygon satisfying (i) or (ii) or to a decrease in t h e n u m b e r of angles in II which are less t h a n ~. H e n c e we can assume t h a t an angle in I I which is less t h a n ~ is preceded a n d followed b y angles which are each less t h a u 2 ~r.

3. There are a m o n g s t t h e triangles in the v e r t e x t r i a n g u l a t i o n of K at least two h a v i n g t w o sides which are sides of K, since there are n sides of K a n d n - 2 triangles. L e t OP~_~

OP~OP~+~

be t h e image of one of these. According t o w 1 of this proof we m a y assume Z_OP~ > 0.

I f b o t h / t (OP~_i, O P t ) a n d # (OPi, OP~+i) are greater t h a n 1 the m e t h o d of L e m m a 1 enables us t o decrease 2 ' ( I I ) while, according to L e m m a 3, I I remains in 9d(ft). W e m a y assume therefore t h a t at least one of/~ (OPt_x, O P , ) a n d # (OPt, OP~+~) is equal to 1. L e t

tt (OPt, OPi+~) = 1.

W e distinguish two cases according as ~OP~+~ < ;~ or /-OP~+I ~> ~.

36 N . O L E R

OPi. I OPi§

L- . . . 0pi+ 2

0Pi-[ O~_ I

a b

F i g . 6.

4. L e t /@P~+I < ~- According to w 2 we can thus assume t h a t ~@P~+2 < 2 z and, since / O P l < ~r, t h a t L@Pt_I < 2 Jr. This enables us to v a r y OP~ and OPt+l locally while H remains in 9~. If, furthermore, the sides of II remain not less t h a n 1 in #-length the nega- tion of (ii) in addition allows us to apply Lemma 3 to ensure t h a t H remains in 9~(#).

If # ( O P t _ D OPt) > 1 and/~(@Pi+l, @P~+2) > 1 we can decrease F ( H ) as follows: we move | along

@P~-IOP~

towards @P~-I and | varies in such a way as to preserve the vector

@Pt@Pt+ 1.

The change in A (II*) is equal to the change in A (@S~+I) which is equal to the change in the area of the quadrilateral

@Pt_IOP,|

The change in M (H) is equal to t h a t in ju (@P~-I, @Pt) + # (OP~+I, | In Fig.6a ~_@P~ + / O P ~ + I >~r;

in Fig. 6b / @ P ~ + / O P , + 1 <~r. In both cases

# (| @'Pt) + I~ (| @P~+2)

~</~ (OPt-~, O'Pt) + / t (OP,+~, OPt+2) + # (OPt+~,

O'Pt+l)

# (OPt-~, OPt) + # (OPt+~, OPt+2).

T h a t the area of the quadrilateral is decreased is clear in Fig. 6a while in Fig. 6b it is made evident b y placing the new quadrilateral onto the original with O'P~ on OP~ and O'P~+I on OPt+l-

We shall therefore assume that # (OPt-l, OPt), say, is equal to 1 and show next t h a t if/~(@Pi+l, OP~+~) > 1 we can again decrease F ( I I ) .

Let

OP~+~OP,+2

lie outside P (OP,). B y moving

OPt+ 1

towards OP,+2 Mong

OPt+~OPt+2

we increase /~(OP,, OP,+~) which therefore remains not less than 1. However,

# (OPt,

OPi+l) q-

# (OP~+~, OPt+2)

is decreased. For if X is a p o i n t between OP~+~ and OPt+2 we have /t(OP~, X) < # ( O P t , OP~+I) + # ( O P t + D X),

h e n c ( ~

# ( O P t , X) + # ( X , OPt+2) < # ( O P t ,

OPt+l) q-# (OPt+l, OPt+2).

.AN

I N E Q U A L I T Y I N T H E GEOMETI~Y OF :NUMBERS

ePi+l / / OPi+2

~'ig. 7.

37

Also, t h e area of t h e quadrilateral is decreased. Thus we can decrease F ( I I ) . So we shall assume t h a t F(| intersects |174 2 or |174 p r o d u c e d b e y o n d | a n d shall show t h a t F ( H ) can be decreased b y a v a r i a t i o n of b o t h | a n d | u n d e r which

# (| O P t ) and/~ ( | | r e m a i n equal to 1.

W e choose a coordinate s y s t e m as follows. W i t h origin at | we take as x-axis t h e line parallel to |174 a n d as y-axis t h e line which is parallel t o t h e t a n g e n t s t o l~ (| at the points where t h e x-axis cuts F (| L e t A a n d B be points of F (@P~-I) such t h a t O P ~ - I A = | a n d | = | 1 7 4 (see Fig. 7) a n d let the coordinates of | A a n d B be (x, k), @i, Yl) a n d @2, Y2) respectively. W e m a y so direct the y-axis t h a t k > 0. We observe t h a t

x ~ - x l = x } (1)

Y2 - Yl = k.

U n d e r these constraints a n d requiring als0 t h a t A a n d B lie on F (| it follows t h a t xl, Yl, x2 a n d Y2 are d e p e n d e n t on x so t h a t F ( H ) is also:

F (Y[) = ~ -t- fix ~- ~ (xly 2 - x 2 y l ) , where ~ a n d fl are constants a n d ~ i s a positive constant.

Using (1) we find t h a t d F

d x - fl + y {(y~ - x2 y'~) y~ - (y~ - x l Y~)Y'I} (Y'~ - Y~)-I

a n d l d e F 2 y ; y ~ + (y;)2y~' , (y~)~y;" ,

r d x 2 - Y, - Y2 (y; - y~)a [(x~ - x2) yl - - ( Y l - - Y2)] + (y~ _ y~)a [(x2 - x,) y.z - (y~ - Yl)]

where y[ a n d y~' d e n o t e d y d d x ~ a n d d2y~/dx~ respectively (i = 1, 2).

38 ~. OLER

I n o r d e r t o d e t e r m i n e t h e sign of d ~ F / d x ~ we n e e d t o k n o w in p a r t i c u l a r t h e signs of t h e expressions in s q u a r e b r a c k e t s . Since s i m i l a r e x p r e s s i o n s will occur a g a i n i t is con- v e n i e n t for us t o f o r m u l a t e t h e following s i m p l e rule.

L e t P be a p o i n t on a c e n t r a l l y s y m m e t r i c closed c o n v e x c u r v e a n d l e t Q b e a p o i n t d i s t i n c t f r o m P w h i c h is e i t h e r on t h e c u r v e or in its i n t e r i o r . W i t h a c o o r d i n a t e s y s t e m as in F i g . 7 (i.e. w i t h c e n t r e t h e origin a n d y - a x i s p a r a l l e l t o t h e t a n g e n t s t o t h e c u r v e a t its p o i n t s of i n t e r s e c t i o n w i t h t h e x-axis) l e t P ~ (~, ~) a n d Q ~ (a, b). W r i t e 9 ' = d ~ / d ~ . T h e n 9 > 0 i m p l i e s ( ~ - a ) v I ' - ( ~ - b ) < O / (2)

< 0 i m p l i e s ( ~ - a ) ~ ' (~ b ) > 0 .

J

To show t h a t t h i s is so l e t ~ ' ~= 0 a n d (~t, b) be t h e p o i n t a t w h i c h t h e t a n g e n t t o t h e c u r v e a t (~, ~) i n t e r s e c t s t h e line t h r o u g h (a, b) w h i c h is p a r a l l e l t o t h e x-axis. W e h a v e

L e t ~] > 0.

I f ~]' > 0 t h e n ~t < a, hence ~ - a § (b -~])/~]' < 0 a n d (~ - a ) 9 ' - (~ - b) < 0 I f ~' < 0 t h e n ~t > a, h e n c e ~ - a § (b - ~ ) / ~ ' > 0 a n d (~ - a ) ~ ' - (~ - b) < 0.

I f ~' = 0 t h e n ~] is a n a b s o l u t e m a x i m u m a n d - (~ - b) < 0.

T h u s if ~ > 0 t h e n ( ~ - a ) ~ ' - ( ~ - b ) < 0 a n d b y s i m i l a r a r g u m e n t we f i n d t h a t ~ < 0 i m p l i e s (~ - a ) 9 ' - (9 - b) > 0.

R e t u r n i n g t o t h e e x p r e s s i o n for d 2 F / d x ~ we n o t e t h a t y~ > 0 since / - @ P i + l < ~, a n d since F (@Pi) i n t e r s e c t s @Pi+IOP~+2 or @P~+I@Pt+2 p r o d u c e d b e y o n d @Pi+2 i t follows t h a t y ~ > 0 .

L e t Yl > 0. Since YI < Y2 a n d L O P ~ < ~ i t follows t h a t y~ < 0. T h u s

y ~ - y ~ < 0 a n d

A p p l y i n g t h e a b o v e r u l e we f i n d

[(x 1 - x~)y~ - (Yl ^-- Y2)] < 0

- 2 y i y ~ / ( y ; - y~) < O.

a n d [(x 2 - x l ) y ~ - (y~ - Yl)] < 0.

Since f u r t h e r m o r e , y~' a n d y~' a r e e a c h n e g a t i v e i t follows t h a t d 2 F / d x 2 < O.

r p 9 t

L e t Yl < 0 . Since / @ P ~ < T e we h a v e Y l > y 2 > 0 . T h u s Y l - Y 2 > 0 a n d

- 2 y ; y ~ / ( y ~ - y~) < 0. Also, y~' < 0 a n d [(x 1 - x~)y; - (yl - Y2)] > 0 while y ; ' > 0 a n d [(X2 ^-- X l ) y 2 - - (Y~ ^- Yl)] < 0. A g a i n d Z F / d x 2 < O.

F o r Yl in t h e n e i g h b o u r h o o d of zero we see t h a t

d F d F

l i m - - = lira dxx = f l + ? ( x l - x2) y2.

y~--~0 + d x y ~ - - ~ 0 -

A N I N E Q U A L I T Y I N T I t E G E O M E T R Y O F N U M B E R S 3 9

I n t h a t F is defined on an open set of values of x we can choose

dx

with either sign a n d in v i r t u e of t h e a b o v e with such sign t h a t F (II) c a n always be decreased.

L e t us assume, therefore, t h a t / t ( 0 P i + l ,

OPi+e)

- 1 also a n d consider the change in F ( I ] ) resulting f r o m a v a r i a t i o n of 0 P ~ a n d

OPt+ 1

u n d e r which ~u

(OPi-1, OPt), # (OPt,

O P ~ + i ) a n d # (0P~+I,

OPi+2)

each remain equal t o 1.

W e choose a coordinate s y s t e m as follows. W i t h origin at 0 P i - 1 we t a k e t h e x-axis t o h a v e its positive half contain 0P~+e a n d t h e y-axis to be parallel to the t a n g e n t s to P ( 0 P , - 1 ) at the points where the x-axis cuts F

(OPi-1).

L e t A a n d B be points on F

(OPt-l)

such t h a t

@P~_IA = ~ P ~ O P ~

a n d

OP~_lJ~=@P~+2OP~

a n d let t h e coordinates of A, B, @P~ a n d OPt+2 be (xl,

Yl), (x2, Y2), (x, y)

a n d (r, 0) respectively (see Fig. 8).

W e note t h a t ~ P ~ a n d @P~+I c a n n o t lie on opposite sides of t h e x-axis. Otherwise

@P~@P~+~

would intersect t h e x-axis to t h e left of @P~_~ or to the right of @P~+~. I n t h e first case we would find t h a t OP~_~ lies inside t h e triangle

@P~@P~+~@P~+~

hence also inside F (~P~+~) implying t h a t ~u (OP~+~, @P~-I) < 1. Recalling, however, t h a t we chose P~

for which

P~_~P~P~+~

is a triangle in t h e v e r t e x triangulation associated with O it follows t h a t

@P~+~@P~_~

is a diagonal a n d therefore f~

(~P~+I,

@P~_I) >~ 1. T h u s t h e first possibility is ruled out. As to t h e second, we would find in t h a t case t h a t @P~+~ lies inside the triangle

@P~_~@P~@P~+~.

This is ruled o u t b y condition 3 of Definition 2 when we again recall t h a t

P~_IP~P~+I

is a triangle in t h e v e r t e x triangulation associated with @. U n d e r t h e circum- stances we can choose the direction of the y-axis to be such t h a t @P~ a n d @P~+I b o t h lie in t h e u p p e r half-plane. Thus, in particular, y > 0.

W e observe t h a t x~ - x~ = x - r, ]

(3)

Y~ - Yl = Y. J

(

W e can express F ( I ] ) as

F ( I I ) = u §

fl[(r + xI)y + (r - x)yl] ,

where ~ a n d fl are constants a n d

fi,

in particular, is positive. Subject t o (3) a n d t h e restric- t i o n of A, B a n d OP~ to lie on F ( 0 P i - 1 ) we can r e g a r d F ( I I ) as a function of x only. Using y[ a n d y [ ' as before a n d y' a n d y" t o denote

d y / d x

a n d

d2y/dx ~

respectively we

r 9

find ~1 d-~d2' =

( r + x l ) y ' - y ~ + { ( r - x ) y ~ + y } y _ y ~ " Y~ -

a n d ^{1 d~F} [3 dx2

~ 2 t 9 p t 9 t 9 9

- - - (y - yl) (y

_ y ~ ) ( y ~ _ y ~ ) - l + { y + ( r § 2 _ y~),,-1 y,,

- [(x~ - xl) y~ - (y~ - y~)]

(y' - ys (ys _ y~)-Sy~9

- [(% - x2) Y~ - (Yl - Y2)]

(Y' - Yl) 2 (Y~ - y;)-3y~,.

40 ~. OLER

B l 8Pi§

] 0Pi-t 8Pi+2

I

Fig. s.

I t suffices t o a s s u m e t h a t Yl >~ 0, t h a t is t h a t / O P ~ § >1 ~ for o t h e r - wise / O P ~ + I § >~ z a n d we can i n t e r c h a n g e t h e roles of O P t _ 1 a n d OPi+2 t a k i n g t h e l a t t e r as origin a n d so on.

W e o b s e r v e t h a t # ( O P t _ l , OPt+e) > 1. O t h e r w i s e OP~-IOP~+2 w o u l d n o t b e a d i a g o n a l a n d t h e r e w o u l d n e c e s s a r i l y be a O - d i a g o n a l w i t h P~+I as a n e n d p o i n t t h e i m a g e of whose o t h e r e n d p o i n t , s a y O P s , lies inside t h e q u a d r i l a t e r a l OPi-IOP~OPi+IOP~+2. R e c a l l i n g c o n d i t i o n 3 of D e f i n i t i o n 2 a n d t h e f a c t t h a t P~_IP~P~+ 1 is a t r i a n g l e in t h e v e r t e x t r i a n g u l a - t i o n a s s o c i a t e d w i t h 0 we o b s e r v e t h a t OP~ m u s t in f a c t lie in t h e t r i a n g l e OP~_IOP~+IOP~+2.

A m o n g s t such v e r t i c e s as OP~ t h e r e is one, s a y O P t , for w h i c h /_OPtOP~+IOP~+ 2 is a m i n i m u m . I t follows t h a t OPtOP~+2 is a d i a g o n a l b u t , since OP~-IOPi+2OPi+I lies inside F (OPt+z), # (OPt, OPi+~) < 1 w h i c h is a c o n t r a d i c t i o n . T h u s # ( O P t - l , OPt+2) > 1, i m p l y i n g

Since, therefore, xe - x = x 1 - r < 0 we h a v e x > x2 f r o m w h i c h i t follows t h a t y~ > y ' . F u r t h e r m o r e , since • < ~, we h a v e y ' > y~. H e n c e (y' - y~) (y' - y~) (y~ - y~)-i < 0.

F o r Yl > 0 a c c o r d i n g t o t h e r u l e s t a t e d e a r l i e r [(x I - x e ) y ~ - ( Y l - Y2)] < 0 a n d since y~ < 0 i t f o l l o w s t h a t - [ ( x l - x 2 ) y ~ - ( y l - y ~ ) ] ( y ' - y l ) 2 ( y ~ - ~ ) Y2 o.

Since y~ > 0 we h a v e [ ( x ~ - x l ) y ~ - ( y ~ - y ~ ) ] < 0 a n d since y l ' < 0 i t follows t h a t

- [ ( x 2 - x ~ ) y ~ - ( y ~ - y ~ ) ] ( y ' - y ~ ) ~ ( y ~ - y'~)-~y'~" < O.

Since (x~ - x l ) y ~ - (y~ - y ~ ) < O, y + (r - x)y~ > O; t h e r e f o r e y + (r + x l ) y ~ - (x + x~)y~ > (x + x~)(y~ - y~) = (r + x~)(y~ - y~) > 0 a n d since y" < 0 i t follows t h a t {y + (r + Xl) Y2 - - (X + Xl) y~} (y~ - y i ) - l y " < 0.

Thus, for y~ > 0 we h a v e s h o w n t h a t d ~ F / d x ~ < 0. F o r y~ in t h e n e i g h b o u r h o o d of 0 we o b s e r v e t h a t

l i m d Yl l i m d-ff -yl = y~ - y ' > 0.

y~.-).O+ ~ ~ y~--~0- ~ X Also d F / d x exists:

l i m d F l i m ~ d F = (r - x) y2 + (r + x~) y ' . , y~o+ d x = yl.-.o- d x

A N I N E Q U A L I T Y I N T H E G E O M E T R Y O F NUI~[BERS 41

:Fig. 9.

We can therefore adjoin to the values of x for which Yl >~ 0 those corresponding to YI in the neighbourhood of 0. The result is an open set and from w h a t we h a v e shown it follows t h a t F ( I I ) can always be decreased b y choosing d x with suitable sign.

5. There remains for us to consider the case in which / O P ~ < ~ , # ( O P t , OPi+l) = 1 and J_OP~+ 1 ~> ~.

We consider the variation in F ( I ] ) which results from varying @Pi and @Pi+l in such a w a y t h a t # (OPt_l, OPi), # (OPt, OPt+l) and # (OPt+l, OPt+2) remain constant.

We choose the following coordinate system. With origin at OP~-I we take the x-axis to contain OPt+2 in its positive half and the y-axis to be parallel to the tangents to F (OPt-l) at the points where the x-axis cuts F(OP~_~). L e t A and B be points such t h a t OP~_~A = OP~+IOPi+2 and OP~_I B = O P t + l O P ~ (see Fig. 9) and let the coordinates of OPi, A , B and OPt+2 be (x, y), (xl, Yl), (x2, Y2) and (r, 0) respectively. Then

x 2 - v x l = x - r } (3)

Y~ - Yl = Y.

We can write

F ( I I ) = ~ + fl {ry - (x 1 Y2 - - x 2 y ~ ) } = o~ + fi {(x - x 2 ) y - (x~ - x 2 ) Y l

},

where ~ and fl are constants and fi, in particular, is positive.

Choosing x 1 as independent variable we find t h a t

t

1 d F Y~ - Y~ [(x~ - x2) y; - ( y l - y~)].

d x---1 = [ ( x - - ( y - _--

Since # (OPi-1, O P t ) / > 1 and # (OP~_I, A) ~> 1 while # (OPi_l, B) = 1 we can determine the sign of each of the square brackets b y applying the rule established ear]ier to/~ (OPi-1, OP~)F (OP~_~) and/~ (OP~_~, A ) F (OPt-l).