ON LINEAR INEQUALITY SYSTEMS WITH SMOOTH

ON LINEAR INEQUALITY SYSTEMS WITH SMOOTH COEFFICIENTS M.A. GOBERNA

, L. HERNÁNDEZ

y , AND

M.I. TODOROV

z

Abstract. A linear inequality system with in…nitely many constraints is polynomial (analytical) if its index set is a compact interval of the real line and all its coe¢ cients are polynomial (analytical, respectively) functions of the index on this interval. This paper provides su¢ cient conditions for a given closed convex set to be the solution set of a certain polynomial (or at least analytical) system. Key words. Linear systems, closed convex sets, linear semi-in…nite programming

1. Introduction. A linear system = fa0t x bt ; t 2 T g, where bt = b(t) 2 R and at = a(t) = (a1 (t); :::; an (t))0 2 Rn for all t 2 T , is called analytical (polynomial) if T is a compact interval in R and all the coe¢ cients are analytical (polynomial) functions of the index t. In many linear semi-in…nite programming problems, the constraints system are analytical or even polynomial (see Chapter 1 in Ref. [1]). Di¤erent simplex-like (or puri…cation) and feasible directions methods have been proposed for these problems (see, e.g., Refs. [2]-[5], and Refs. [6]- [7], respectively). All these methods require, at each iteration, the calculation of the set of active indices T (x) := ft 2 T j a0t x = bt g, which also allows to check the optimality and the strong uniqueness of x (see Ref. [1], Theorems 7.1 and 10.6, respectively). If is polynomial, then deg := max fdeg ai ; i = 1; :::; n; deg bg (upper bound for jT (x)j if T (x) 6= T ) can be seen as an indicator of the computational di¢ culty of computing T (x). This paper gives su¢ cient conditions for the existence of polynomial (or at least analytical) representation of closed convex sets, providing a polynomial representation of minimal degree when it is possible. The …rst results of this type were given in Ref. [5], where it was proved that any polyhedral convex set admits polynomial representation whereas the closed balls in Rn do not admit analytical representation if n 3, and the same is true for a large class of closed convex sets in Rn , n 3 (see Refs. [8]-[9]). For convex analysis concepts we have adopted the notation of Ref. [10]. Moreover, we denote by 0n , Bn , Sn and span X the null element, the unit closed ball, the unit sphere and the linear span of X Rn . If 0 2 X R, we denote by X (T ) the set of functions : T ! R such that t = 0 for all t 2 T except on a …nite set of indices, supp = ft 2 T j t 6= 0g (called supporting set of ). Given a function h of a single variable t, we denote by h(j) t the jth derivative of h at t. Finally, given a nonzero polynomial p 2 R [t], we shall denote by deg p the degree of p, whereas it is also convenient to de…ne deg 0 = 1. We …nish this introduction recalling some results that will be used in the sequel. The extended Farkas Lemma (Ref. [11]) establishes that an inequality a0 x b is consequence of a consistent system = fa0t x bt ; t 2 T g if and only if Professor, Departamento de Estadística e Investigación Operativa, Universidad de Alicante, Alicante, Spain. His research was supported by DGES of Spain and FEDER of CE, Grant BFM200204114-C02-01. y PhD Student, Departamento de Matemáticas, Benemérita Universidad Autónoma de Puebla, Puebla, Mexico. z Professor, Departmento de Física y Matemáticas, Universidad de las Américas, Cholula, Puebla, Mexico. His research was supported by Generalitat Valenciana, Grant CTESIN/2002/23. 1

a b

2 cl cone

at bt

;t 2 T;

0n 1

;

which is the so-called reference cone of . Hence, two linear systems are equivalent (i.e., they have the same solution set) if and only if their reference cones coincide. If F is the solution set of , 0+ F is the solution set of the homogeneous system fa0t x 0; t 2 T g. The next two results are a consequence of Corollary 5.9.1 in Ref. [1] and a variant (with similar proof) of Theorem 2.1 in Ref. [5], respectively. Lemma 1.1 Let = fa0t x bt ; t 2 T g be an analytical representation of F Rn at such that 6= 0n+1 for every t 2 T: Then x 2 bd F if and only if there exists bt t 2 T such that a0t x = bt : Lemma 1.2 Let = fa0t x bt ; t 2 T g be a polynomial representation of F at such that dim F = n. If there exists t 2 R such that = 0n+1 , then there bt a0

t x exists p 2 R [t], with deg p > 0, such that f p(t) representation of F .

bt p(t) ;

t 2 T g is another polynomial

2. Polynomial Representations of Minimal Degree. Let F Rn be a given nonempty closed convex set. If F admits polynomial representation, we shall denote by deg F the minimal degree of all its polynomial representations. Obviously, deg F = 1 (0) if F is Rn (a halfspace, respectively). Similarly, deg f02 g = 2 (consider tx1 + t2 1 x2 0; t 2 [ 2; 2] ). It is easy to show that, given a natural number d deg F , there exists a polynomial representation of F of degree d. Proposition 2.1 If F admits polynomial representation and it is not the result of translating a convex cone, then deg F

n

min fdim C j C 6= ; is a face of F g :

In particular, deg F n if F does not contain lines. Proof. Let = fa0t x bt ; t 2 T g be a polynomial representation of minimal degree of F , let C 6= ; be a face of F of minimal dimension, and let x 2 rint C. The geometric assumption on F entails jT (x)j < 1. According to Corollary 2.1 in Ref. [5], n o n dim C = dim span a(j) (t) ; j = 0; :::; d(t); t 2 T (x) ; where d(t) + 1 is the order of multiplicity of t 2 T (x) as zero of the slack function at x, a0t x bt . Hence, n

dim C

deg fa0t x

bt g

deg

= deg F:

If F does not contain lines there exists an extreme point of F , say x. Taking C := fxg, we get n deg F . Since deg R2+ = 1, the assumption in Proposition 2.1 is not super‡uous. In order to characterize deg F for polyhedral convex sets we need a lemma about parametric curves contained in a polyhedral convex cone. A parametric curve fz (t) j t 2 T g, where z 2 C (T ), T being a proper interval in R, is polynomial (rational, analytical, C 1 ) if all the components of z are polynomial (rational, analytical, C 1 , respectively). 2

Lemma 2.1 Assume that y de…nes an extreme direction of the polyhedral convex cone K Rn . Let fz (t) j t 2 T g K be a C 1 parametric curve, and let t 2 int T such that z t belongs to the ray generated by y. Then there exists 2 R such that the tangent vector at z t satis…es z (1) t = y. Proof. Let K = fx 2 Rn j wi0 x 0; i 2 Ig, jIj < 1. By assumption, y 2 K and dim span fwi j wi0 y = 0; i 2 Ig = n 1: Let 0 such that z t = y (contact point of the curve with cone fyg) and (1) 0 denote v = z t . We shall prove that wi0 v = 0 for all i such ! that wi y = 0. In fact, if z (t) z t wi0 v < 0, then there exists " > 0 such that wi0 < 0 for all t 2 t; t + " t t T . Then wi0 (z (t) y) = wi0 z (t) < 0 so that z (t) 2 = K, and we get a similar contra? 0 diction if we assume wi v > 0. Consequently, v 2 [span fwi j wi0 y = 0; i 2 Ig] and so we have v 2 span fyg. Proposition 2.2 If p is the minimal number of inequalities in the family of all the linear representations of a polyhedral convex set F , then deg F max f0; 2p 3g. The equality holds if F is full dimensional (so that deg F is odd if p > 1). Proof. We can assume F 6= Rn and p > 2. Let F = fx 2 Rn j c0i x di ; i = 1; ::; pg and let 0 = fc0i x di ; i = 1; ::; pg. The minimality of 0 entails ci 6= 0n , i = 1; ::; p. First, we shall construct a polynomial representation of F , 1 , such that deg 1 = 2p 3, and then we shall prove that deg F = 2p 3 if dim F = n. To do that we associate with 0 and any p arbitrary scalars t1 ; :::; tp such that t1 < t2 < ::: < tp , the family, , of polynomial systems depending on p positive parameters, 1 ; :::; p , and p 2 scalars, 2 ; :::; p 1 , of the form ( p !0 ) p X X = 'i (t) ci x 'i (t) di ; t 2 [t1 ; tp ] ; (2.1) i=1

i=1

where rip

'i (t) = ( 1)

[

i

+

i

(t

ti )]

Y

rij

(t

tj )

; i = 1; ::; p;

1 j p

with

1

=

p

= 0, and 8 < 0; if j = i; 1; if (i; j) 2 f(i; 1) ; i 6= 1; (i; p) ; i 6= pg ; rij = : 2; otherwise.

It can be realized that deg 'i = 2p

2p 2p

deg 'i = for i = 2; :::; p 1. Accordingly, deg for all j 6= i and 'j (tj ) =

3 for i = 1; p and

j

3; if 4; if

= 2p Y

i=1;:::;p;i6=j

i i

6= 0; = 0;

3 for all jti

rji

tj j

2

. Moreover, 'i (tj ) = 0

>0

for all j = 1; :::; p. Taking t1 < t2 < ::: < tp arbitrarily in R, and i = 1 and for all i = 2; ::; p 1 we obtain a polynomial representation of F , say 1 2 Theorem 4.1 in Ref. [5]). Hence, deg F 2p 3. 3

i

=0 (see

Now we shall prove that, if 2 = fa0t x bt ; t 2 T g is a polynomial representation of F of minimal degree and dim F = n, then deg 2 = 2p 3. In order to do this, we shall determine scalars t1 < t2 < ::: < tp and suitable parameters, , say 3 , has the same 1 ; :::; p ; 2 ; :::; p 1 , so that the corresponding element of (polynomial) coe¢ cients as 2 , in which case deg 2 = deg 3 = 2p 3. The assumption on 2 entails, by Lemma 1.2, at bt

6= 0n+1 for every t 2 T;

(2.2)

so that, according to Theorem 5.3 and Corollary 5.9.1 in Ref. [1], we have at bt

K := cone

;t 2 T;

0n 1

cj dj

= cone

; j = 1; ::; p;

0n 1

:

Then for each j 2 f1; ::; pg we can write cj dj

=

X

at bt

j t

t2T

+

0n 1

j

;

j

(T )

2 R+ ;

j

2 R+ ;

(2.3)

with supp j 6= ; (recall that cj 6= 0n ). Since 0 is a minimal representation of the full dimensional set F , each face x 2 F j c0j x = dj , j = 1; ::; p, is actually a facet of F . Reordering 0 if it is necessary, we can assume that t1 < t2 < ::: < tp and we consider the corresponding family of polynomial systems . In order to select a suitable 3 2 , we must obtain the relevant information from 2 . xj Let xj be a point of the corresponding facet. Multiplying by both 1 P j 0 j members of 2.3, we get 0 = bt + j , so that j = 0 and 2.3 becomes t at x t2T

cj dj

X

=

t2supp

j t j

at bt

:

(2.4)

cj de…nes an extreme direction of K (due to the minimality dj of 0 ) for each j, there exist tj 2 supp j T , with ti 6= tj if i 6= j, and j > 0, j = 1; ::; p, such that From 2.4, since

atj btj

=

j

cj dj

; j = 1; ::; p:

(2.5)

On the other hand, given j 2 f2; :::; p 1g we have tj 2 ]t1 ; tp [ and the curve at ; t 2 [t1 ; tp ] satis…es the conditions of Lemma 2.1 at tj , so that there exist bt scalars j , j = 2; :::; p 1, such that a(1) (tj ) b(1) (tj )

=

j

cj dj

; j = 2; :::; p

1:

(2.6)

Now we are ready to select a system 3 2 (de…ned by means of 2.1) satisfying the same conditions as 2 , 2.5 and 2.6 (without any requirement on its solution set). 4

Let us write etj ftj

= fe0t x

ft ; t 2 [t1 ; tp ]g.Since 2 p X Y ci = 'i (tj ) =4 j di 3

i=1

there exist

j

i=1;:::;p;i6=j

> 0, ; j = 1; :::; p, such that etj ftj

=

j

3

cj dj

3

cj dj

rji 5

jti

tj j

; j = 1; :::; p;

veri…es 2.5, i.e., ; j = 1; :::; p:

(2.7)

On the other hand, for j 2 f2; :::; p 1g, we have rij = 0 if i = j and rij = 2 (1) otherwise. In order to calculate 'i (tj ) we shall distinguish two cases: (1) 2 If i 6= j, (t tj ) divides 'i , so that 'i (tj ) = 0. Q r If i = j, we can write 'j (t) = j + j (t tj ) pj (t), where pj (t) = ( 1) jp

(t

rkj

tk )

.

1 k p;k6=j

(1)

Then, 'j (tj ) = Therefore e(1) (tj ) f (1) (tj )

j pj

(1)

(tj ) +

= 'i (tj )

(1) j pj

ci di

(tj ).

=

h

j pj

with pj (tj ) 6= 0 for all j, and so there exist e(1) (tj ) f (1) (tj )

=

j

(1) j pj

(tj ) +

cj dj

2 ; :::;

i (tj )

p 1

such that

; j = 2; :::; p

Finally, we shall prove that each component of

cj dj

; j = 2; :::; p 3

1:

1;

veri…es 2.6, i.e., (2.8)

at et is the null polynom, bt ft 3 and the proof will be complete.

so that we can conclude that deg 2 = deg 3 = 2p Let q (t) 2 R [t] be any of these components. Each of the points t1 < t2 < ::: < tp is a zero of q (compare 2.5 with 2.7), so that (by Rolle’s Theorem) there exist 1 ; :::; p 1 such that t1 < 1 < t2 < ::: < tp 1 < (1) ( j ) = 0, j = 1; :::; p 1. Moreover, q (1) (tj ) = 0, j = 2; :::; p 1 p 1 < tp and q (compare now 2.6 with 2.8). Since 1 < t2 < ::: < tp 1 < p 1 are 2p 3 di¤erent zeros of q (1) , and deg q (1) 2p 4, q (1) must be identically zero and q will be constant. Since q vanishes at any tj , q will be the null polynom. Observe that p = 3 for f02 g, but deg f02 g = 2 < max f0; 2p 3g. So the assumption dim F = n in Proposition 2.2 is not super‡uous.

3. Plane Sets with Smooth Representation. Proposition 3.1 Let f : Rn ! ] 1; +1] be a polynomial convex function such that F := fx 2 Rn j f (x) 0g int dom f and let fz (t) ; t 2 T g be an analytical (rational) curve which is dense in bd F . Assume that T is a proper compact interval in R and there exists x b 2 Rn such that f (b x) < 0. Then F admits an analytical (polynomial, respectively) representation. Proof. Convexity arguments yield bd F = fx 2 Rn j f (x) = 0g and F = fx 2 Rn j u0 (x z) 0 8u 2 @f (z) 8z 2 bd F g 0 = x 2 Rn j rf (z) (x z) 0 8z 2 bd F : By the continuity of rf on z 2 bd F and the density of fz (t) j t 2 T g in bd F , rf (z (t)) rf (z) 0 ; t 2 T is dense in ; z 2 bd F , so that rf (z (t)) (x 0 0 rf (z (t)) z (t) rf (z) z 5

z (t))

0; t 2 T

is another representation of F by the extended Farkas Lemma. The conclusion is straightforward. 2 Example 3.1 Applying Proposition 3.1 to f (x) = kxk 1, we get B 2 = x 2 R2 j z 0 x

1; z 2 S2 :

(3.1)

Choosing the analytical curve cos t sin t

; t 2 [0; 2 ]

= S2 ;

(3.2)

we get the well-known analytical representation of B2 : f(cos t) x1 + (sin t) x2

1; t 2 [0; 2 ]g :

We can obtain a polynomial representation of the disk from any rational curve dense in S2 . For instance, the substitution w = tan 2t in 3.2 yields ! ) ( 1 w2 1 1+w2 ; w 2 R = S2 n : 2w 0 1+w2 A similar curve can be obtained from a particular case of the Fermat´s Theorem for polynomials: given three polynomials in one variable without common factors, p; q and r, p2 + q 2 = r2 if and only if there exist two polynomials without common factors, u and v, such that p = uv; q = (v 2 u2 )=2 and r = (v 2 + u2 )=2. Consequently, if these polynomials exist, then r(t) 6= 0 for every t 2 R and deg r is even. Taking u(t) = t + 1 and v(t) = 1 t we get p(t) = 1 t2 , q(t) = 2t, and r(t) = t2 + 1. Then, it can be easily realized that ! ) ( 1 t2 1 1+t2 ; t 2 R = S2 n : (3.3) 2t 0 1+t2 We shall use 3.3 in order to get the aimed polynomial representation of the disk. The s is bijective, so that the substitution function ' : ] 1; 1[ ! R de…ned as ' (s) = 1 s2 t = ' (s) will allows us to replace R by a bounded interval. In fact, from 3.3 we get ! ) ( s4 3s2 +1 1 4 2 s s +1 ; s 2 ] 1; 1[ = S2 n : 2s(s2 1) 0 4 2 s

s +1

Then, according to 3.1, we get a linear representation of B2 : s4 s4

3s2 + 1 s2 + 1

x1 +

2s(s2 1) s4 s2 + 1

x2

1; s 2 ] 1; 1[ :

(3.4)

x2

1; s 2 [ 1; 1]

(3.5)

By continuity of the coe¢ cients in 3.4, s4 s4

3s2 + 1 s2 + 1

x1 +

2s(s2 1) s4 s2 + 1

is an alternative representation of B2 . Finally, multiplying both members of the inequality of index s in 3.5 by s4 s2 + 1 > 0 for all s 2 [ 1; 1] we get the equivalent polynomial system s4

3s2 + 1 x1 + 2s(s2

1)x2 6

s4

s2 + 1; s 2 [ 1; 1] :

(3.6)

Proposition 3.2 B2 admits polynomial representation and deg B2 = 4. Proof. Since 3.6 is a polynomial representation of B2 , deg B2 4. Now we assume that = fp(t)x1 + q(t)x2 r(t); t 2 T g is a polynomial representation of B2 such that deg < 4 and we shall get a contradiction . According to Lemma 1.2, satis…es the conditions of Lemma 1.1. Hence, for all ' 2 [0; 2 [ there exists t' 2 T such that p(t' )x1 +q(t' )x2 = r(t' ) is the tangent line at 0 (cos '; sin ') . Since B2 is a smooth convex set, there is a one-to-one correspondence between S2 and the set of tangent lines, i.e., for each ' 2 [0; 2 [ there exists a unique p(t ) t' 2 T such that r(t' ) 6= 0 (the tangent lines do not contain 02 ), r(t'' ) = cos ', and q(t' ) r(t' )

= sin '. Since p2 (t) + q 2 (t) = r2 (t) for every t 2 ft' ; ' 2 [0; 2 [g, this set being in…nite, the equation holds for all t 2 R, i.e., p2 + q 2 = r2 . Since p, q and r cannot have common linear factors (by Lemma 1.2), r(t) 6= 0 for all t 2 R and deg r is even by the Fermat´s Theorem. Assuming ! deg r = 2, we shall obtain a contradiction. The vector z (t) =

p(t) r(t) q(t) r(t)

is well de…ned for all t 2 R. Obviously, fz (t) ; t 2 T g

S2 . On the other hand, we have shown that z (t' ) =

cos ' sin '

for all ' 2 [0; 2 [,

so that S2 = fz (t' ) ; ' 2 [0; 2 [g

fz (t) ; t 2 T g ;

and we have fz (t) ; t 2 T g = S2 . Since the degree of p and q is at most 2, we can write p (t) = p2 t2 + p1 t + p0 , q (t) = q2 t2 + q1 t + q0 , and r (t) = r2 t2 + r1 t + r0 , with r2 6= 0. It is obvious, that p22 +q22 = r22 : Without loss of generality, we may assume that jp2 j jq2 j ; which implies that pr22 < 1 and q2 6= 0: Now consider the following points in S2 : z 1 = ( pr22 ;

q2 0 r2 )

and z 2 = ( pr22 ; rq22 )0 .

p2 Since z 1 6= z 2 , there exist t1 6= t2 2 T which satisfy the equation p(t) r(t) = r2 , i.e., the linear algebraic equation (p1 r1 pr22 )t = r0 pr22 p0 : But this equation cannot have two di¤erent solutions. The proof is complete. Proposition 3.3 If f : R !R is an analytical (rational) proper convex function such that dom f is open, then epi f admits analytical (polynomial) representation. Proof. Since f is continuous on the open interval dom f , f is closed, i.e., epi f is a closed convex set and so it is the intersection of all the closed halfplanes tangent to it by Theorem 18.8 in Ref. [10]. Since f is di¤erentiable at every x 2 dom f , the tangent halfplanes to epi f are of the form x2 f (t) f (1) (t) (x1 t), with t 2 dom f , i.e., 1

:=

n

f (1) (t) x1 + x2

f (t)

tf (1) (t) ; t 2 dom f

o

is a representation of epi f (not analytical since dom f is not even closed). Now assume that f is rational (the proof is simpler if f is analytical). Let f (t) = g (t) 2 , where g and h are polynomial functions. Then, taking into account that h (t) > h (t) 0 for all t 2 dom f , we have 2

epi f = fx 2 R2 j g (t) h(1) (t) h (t) g (1) (t) x1 + h (t) x2 g (t) h (t) + tg (t) h(1) (t) th (t) g (1) (t) ; t 2 dom f g; 7

so that

2

:= fa0t x

bt ; t 2 dom f g, where at =

g (t) h(1) (t) h (t) g (1) (t) 2 h (t)

and bt = g (t) h (t) + tg (t) h(1) (t)

th (t) g (1) (t) ;

is another representation of epi f with polynomial coe¢ cients whose degrees are non greater than d := deg h + max fdeg g; deg hg. In order to replace the index set of 2 by a compact interval, we shall discuss three possible cases. If dom f is bounded, then T := cl dom f is a compact interval and the polynomial system fa0t x bt ; t 2 T g is another representation of epi f by the continuity of the coe¢ cients. If dom f is a hal‡ine in R bounded from below we can write dom f = ] ; +1[, , which de…nes a bijection between 2 R. Consider the function (s) = (1 1 )s+ s ]0; 1[ and ] ; +1[. The substitution t = (s) in 2 yields the equivalent system 0 ds ; s 2 ]0; 1[g, where 3 = fcs x c1 (s) = g ( (s)) h(1) ( (s))

2

h ( (s)) g (1) ( (s)) ; c2 (s) = h ( (s))

and d (s) = g ( (s)) h ( (s)) +

(s) g ( (s)) h(1) ( (s))

(s) h ( (s)) g (1) ( (s)) :

d

Multiplying by (1 s) > 0 both members of the inequality of index s in 3 , we obtain a representation of epi f with polynomial coe¢ cients and index set ]0; 1[. Aggregating the inequalities corresponding to s = 0; 1, we obtain the aimed polynomial representation of epi f . The argument is similar if dom f is a hal‡ine in R bounded from above. Finally, assume that dom f = R and consider the function (s) = 1 ss2 , which establishes a bijection between ] 1; 1[ and R. Reasoning as above, we can obtain a polynomial representation of epi f just multiplying both members of the inequality of d index s, in 3 , by 1 s2 > 0. The next example illustrates the …rst case in the last proof. The other two cases will arise in the proof of Proposition 3.4. o n 1 Example 3.2 Let F = x 2 R2 j x2 1 x21 ; jx1 j < 1 . Taking f (x) =

at =

2t 1 t2

2

, bt = 1

n F = x 2 R2 j

1 x2 +1; jxj

1

; jxj < 1; 1;

3t2 and T = [ 1; 1], we get 2tx1 + 1

t2

2

x2

1

o 3t2 ; t 2 [ 1; 1] :

Proposition 3.4 If F is the intersection of a solid circular right cone in R3 with x3 = 0 (identi…ed with R2 ), then F admits a polynomial representation in R2 and deg F is given in Table 1. 8

F singleton hal‡ine plane angle convex hull of an ellipse convex hull of a parabola convex hull of a branch of hyperbola

deg F 2 2 1 4 4 2

Table 1 Proof. The key of the proof is the following observation (also valid in Rn ): if F admits a polynomial representation and g is a nonsingular a¢ ne transformation in R2 , then g (F ) belongs to the same class of sets and deg g (F ) = deg F . In fact, if fa0t x bt ; t 2 T g is a polynomial representation of F of minimal degree and g (x) = M x + c, M (2 2) regular and c 2 R2 , then g (F ) = x 2 Rn j a0t M

1

x

bt + a0t M

1

c; t 2 T ;

(3.7)

and this is a polynomial representation of g (F ) of degree at most deg F . The argument can be repeated with g 1 (g (F )) = F , so that deg F deg g (F ) and the equality holds. We shall discuss the seven possible cases re‡ected in the table. Case 1: Any singleton set is the result of translating f02 g and we have seen that deg f02 g = 2: Case 2: If F is a hal‡ine, then it is the image of G := x 2 R2 j x2 0; x1 = 0 through a suitable a¢ ne transformation. Since tx1 + t2 x2 0; t 2 [ 1; 1] is a polynomial representation of G, deg F = deg G = 2: Case 3: If F is a plane angle, then it is the intersection of two halfplanes, so that deg F = 1. Case 4: If F is the convex hull of an ellipse whose center of symmetry is c, then there exists an a¢ ne transformation in R2 , say g (x) = M x + c, such that bd F = fg (x) j x 2 S2 g. Given y 2 int F , any line containing y intersects bd F at two points. Let y 1 ; y 2 2 bd F such that y 2 y 1 ; y 2 . Let 2 ]0; 1[ such that y = (1 ) y1 + y2 . Let xi := M 1 y i c 2 S2 , i = 1; 2, and let x := (1 ) x1 + x2 2 B2 nS2 . Then y = (1

) g x1 + g x2 = g (x) ;

so that int F fg (x) j x 2 B2 nS2 g and the reverse inclusion also holds. Hence F = fg (x) j x 2 B2 g and deg F = deg B2 = 4 by Proposition 3.2. Case 5: Let F be the convex hull of a parabola. Given y 2 int F , there exists a line through y intersecting bd F at two points, and it is easy to prove that F is the image of the convex hull of the parabola x2 = x21 by means of a suitable a¢ ne transformation of R2 . So we have to prove that G := x 2 R2 j x2 x21 9

admits polynomial representation of minimal degree 4. Applying Proposition 3.3 2t with f (x) = x2 , we get at = and bt = t2 , and we obtain 1 G = x 2 R2 j 2tx1 + x2 n = x 2 R2 j 2s 1 s2 x1 + 1 s2

2

t2 ; t 2 R

o s2 ; s 2 [ 1; 1] ;

x2

so that deg G 4. Denote by 1 the above polynomial representation of G. Now we assume that 2 = fp (t) x1 + q (t) x2 r (t) ; t 2 T g is a representation of G of minimal degree, 2 deg 2 3, and we shall obtain a contradiction. For each s 2 [ 1; 1] the corresponding constraint in 1 is binding at a unique point of bd G, and these points are di¤erent for di¤erent values of s. Since G is smooth, applying Lemmas 1.1 and 1.2 to 2 , for each s 2 [ 1; 1] there exist ts 2 T and s > 0 such that 1 0 0 1 2s 1 s2 p (ts ) 2 A; @ q (ts ) A = s @ 1 s2 2 r (ts ) s 2

with p (ts ) + 4q (ts ) r (ts ) = 0. Since fts ; s 2 [ 1; 1]g is in…nite, we have p2 =

4qr;

(3.8)

so that q and r are nonzero polynomials without common linear factors (by Lemma 1.2), and the order of any zero (real or complex) in both polynomials will be even. 0n On the other hand, since 02 2 F and 0+ F = , we have 1 r (t)

0 and q (t)

0 for all t 2 T: (3.9) n o Let S := ft 2 T j q (t) 6= 0g. We shall see that p(t) q(t) ; t 2 S is unbounded from above p(t) q(t)

and from below. In fact, if k1 if

p(t) q(t)

k2 for all t 2 S, then

1

1 k1

for all t 2 S, then

1 k2

2 0+ F . Analogously,

2 0+ F . Then, taking into account the com-

pactness of T , there exist skj k=1 S and tj 2 T , j = 1; 2, such that limk!1 skj = tj p skj j and limk!1 = ( 1) 1, j = 1; 2. If t1 6= t2 , since limk!1 q skj = q (tj ) = 0, k q sj 2 2 j = 1; 2, (t t1 ) (t t2 ) divides q (t) contradicting deg 2 3. Then the sign of p(t) q(t) changes in any neighborhood of t1 = t2 , so that t1 2 int T and we can write m

p (t) = (t

t1 ) p1 (t) ; p1 (t1 ) 6= 0; 0

m

q (t) = (t

t1 ) q1 (t) ; q1 (t1 ) 6= 0; m < k

3

(3.10)

3:

(3.11)

and k

Since p (t) p (t1 ) = (t q (t) q (t1 ) 10

m k

t1 )

;

k m must be a positive odd number, and this entails k = 2 and m = 1. From 3.11, the even number deg q = 2 + deg q1 3, so that deg q1 = 0. Then, recalling 3.9, we can write q (t) =

2

(t

t1 ) ;

> 0:

Concerning the even number deg r 3, if deg r = 0, then r is a negative number (again by 3.9) and 02 is a Slater point for 2 , contradicting 02 2 bd G. Hence deg r = 2. Since the order of all the zeros of r is even, and recalling 3.8, there exist t2 2 R and < 0 such that 2

r (t) = (t

t2 ) and p (t) =

(t

t1 ) (t

t2 ) ;

where 2 = 4 > 0. We …nish this part of the proof showing that 2 cannot be a representation of G. t t2 Denoting c := t t1 , the boundary of the halfplanes de…ned in 2 can be c2 . This family of lines can be 4 interpreted as the general solution of the Claireaut’s equation y = xy (1) + f y (1) written as x2 = cx1 + f (c), with f (c) =

v 2 v2 4

whose singular solution is the curve t t

V =

2

c2 =

; v 2 V , where t2 t1

; t 2 T n ft1 g

can be expressed as V = ] 1; ] and [ ; +1[. Hence the solution set of 2 is the intersection of the supporting halfplanes for G at the points of abscissa x1 2 V , and this set is the union of G with the triangle formed by a chord of the parabola and the tangent lines at the end points. So 2 cannot be a representation of G and deg G = 4. Case 6: Let F be the convex hull of a branch of hyperbola, i.e., the image by a suitable a¢ ne transformation in R2 of G = x 2 R2 j x2 x1 1 ; x1 > 0 . Taking f (x) =

in Proposition 3.3, we get at =

1 t2

x 1 ; x > 0; +1; x 0; and bt = 2t, so that

2 2 n G = x 2 R j x1 + t x2 2 = x 2 R2 j (1 s) x1 + s2 x2

2t; t 2 ]0; +1[

2s (1

o s) ; s 2 [0; 1] :

By Proposition 2.1, deg F = deg G = 2. This completes the proof. It is known that given an algebraic curve R2 , there exists a rational curve covering except …nitely many points if and only if the number of multiple points of 2) (counting d(d2 1) times those points of multiplicity d) is exactly (p 1)(p , where 2 p denotes the degree of (see Ref. [12] and references therein). In the case of the conic curves both numbers are zero, so that there exists a rational parameterization of bd F which could be used in order to obtain a polynomial representation of F . The advantage of our proof of Proposition 3.4 is that it provides deg F and, even more, a polynomial representation of minimal degree of F , as the following example shows. 11

Example 3.3 Let F = x 2 R2 j 2x21 + x22 + 2x1 x2 + 2x1

0 . Since g (S2 ) = 1 0 x 2 R2 j 2x21 + x22 + 2x1 x2 + 2x1 = 0 if g (x) = M x + c, with M = and 1 1 1 c= , we get a polynomial representation of F by means of 3.6 and 3.7: 1

t4 + 2t3

3t2

2t + 1 x1 + 2t(t2

1)x2

2t2 ; t 2 [ 1; 1] :

References

REFERENCES

[1] G OBERNA, M. A. and L ÓPEZ, M. A., Linear Semi-In…nite Optimization, Wiley, Chichester, England, 1998. [2] A NDERSON, E. J. and L EWIS, A. S., An Extension of the Simplex Algorithm for SemiIn…nite Linear Programming, Mathematical Programming, Vol. 44, pp. 247-269, 1989. [3] A NDERSON, E. J. and N ASH, P., Linear Programming in In…nite Dimensional Spaces, Wiley, Chichester, England, 1987. [4] L EÓN, T. and V ERCHER, T., A Puri…cation Algorithm for Semi-In…nite Programming, European Journal of Operations Research, Vol. 57, pp. 412-420, 1992. [5] G OBERNA, M. A., J ORNET, V., P UENTE, R., and T ODOROV, I. M., Analytical Linear Inequality Systems and Optimization, Journal of Optimization Theory and Applications, Vol. 103, pp. 95-119, 1999. [6] L EÓN, T. and V ERCHER, T., New Descent Rules for Solving the Linear Semi-In…nite Optimization Problem, Operations Research Letters, Vol. 15, pp. 105-114, 1994. [7] G OBERNA, M. A. and L ÓPEZ, M. A., Optimality Theory for Semi-In…nite Linear Programming, Numerical Functional Analysis and Optimization, Vol. 16, pp. 669-700, 1995. [8] J AUME, D. and P UENTE, R., Conjugacy for Closed Convex Sets, Contributions to Algebra and Geometry, to appear. [9] J AUME, D. and P UENTE, R., Representability of Convex Sets by Analytical Linear Inequality Systems, Linear Algebra and Its Applications, Vol. 380, pp 135-150, 2004. [10] R OCKAFELLAR, R. T., Convex Analysis, Princeton University Press, Princeton, New Jersey, 1970. [11] Z HU, Y. J., Generalizations of Some Fundamental Theorems on Linear Inequalities, Acta Mathematica Sinica, Vol. 16, pp. 25-40, 1966. [12] A BHYANKAR, S. S., Algebraic Geometry for Scientists and Engineers, Mathematical Surveys and Monographs, American Mathematical Society, Providence, Rhode Island, Vol. 35, 1990.

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