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A Model of Following
JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS ARTICLE NO.
222, 547]561 Ž1998.
AY985957
A Model of Following S. ´ Swierczkowski* Department of Mathematics, Uni¨ ersity of Colorado, Boulder, Colorado 80309-0395 Submitted by A. Schumitzky Received October 22, 1997
In the discrete-time models of learning in w5x Žbased on the Kaczmarz projection algorithm w3x. there were obtained estimates ŽTheorems 4.1. and 4.2. of the sum of the errors and the length of the learning path of a learning process. These depended on the size of a time step h, a parameter of the model, and they would yield limits as h ª 0 but the resulting asymptotic estimates were not realizable in any particular model Žwith fixed h.. Here we define a continuous-time analogue of our former discrete-time model of learning and we call it a model of following. It is shown that the asymptotic forms of the main results about learning are true in the models of following. Q 1998 Academic Press Key Words: approximation; analytic movement; convex body; algorithm of following.
1. DESCRIPTION OF THE MODEL We consider here a situation where a convex body Žopen set. S with a regular boundary M is moving in R n, n G 2, occupying at time t the position St . A point moving in R n is following St ; we denote by f Ž t . the position of that point at time t. The speed of the point is never allowed to exceed a fixed constant ¨ , whereas the speed of St may be arbitrary. Thus it may happen that if the point starts from some initial position f Ž t 0 . far from S t 0 , it will remain at a positive distance from St for a while. During that period the point should follow St in the sense that at each moment its velocity vector is directed toward the nearest point of the boundary Mt of St and its speed is ¨ . If the point finds itself on Mt , it should move with the smallest possible speed which will allow it to stay in contact with S t j Mt * E-mail: [email protected]. 547 0022-247Xr98 $25.00 Copyright Q 1998 by Academic Press All rights of reproduction in any form reserved.
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Žthus, for instance, if remaining on Mt , it will move perpendicularly to Mt .. However, if St speeds up sufficiently, the point might be unable to remain in St j Mt Žas its speed cannot exceed ¨ .. At any time, when the point finds itself in Žthe open set. S t , it is supposed to stop and not move for as long as it can stay in St without moving Žall this is equivalent to Definition 3.2.. It is clear from this description that the paths of following, i.e., the curves f Ž t ., will, in general, not even be of class C 1. For example, at the moment when f Ž t . catches up with St , f 9Ž t . will most likely have a discontinuity. However, we shall assume another kind of regularity for all the movements: continuity with right analyticity, the latter meaning that, for any time t, all movements during a sufficiently small time interval beginning at t can be described by real analytic functions of time. This combination of continuity and right analyticity we shall call right regularity ŽDefinition 3.1.. We are motivated to investigate the above model of following because a special case, namely when S is the slice of R n contained between two parallel hyperplanes, is the obvious continuous-time analog of the models of learning studied in w4, 5x. In these, time was discrete and it advanced in equal steps of length h. Estimates, depending on h, were obtained, concerning the quality of the learning process. By letting h approach 0, one could obtain asymptotic estimates. In the present paper the learning model is replaced by a model of following, so that a sequence of consecutive memory states during learning is replaced by a path of following. It turns out then that the asymptotic estimates mentioned above hold exactly in the model of following.
2. SUMMARY OF RESULTS Various assumptions about the movement of St allow us to draw conclusions about the paths of following f : w t 0 , `. ª R n. The least interesting case is when St moves very fast so that a point following St with speed ¨ will never catch up. To exclude this, we shall sometimes assume that St is pinned by a point p* from t 0 onwards, i.e., that p* g St j Mt for all t G t 0 . We show then that for any path of following the distance from f Ž t . to p* is decreasing and the integral of the distance from f Ž t . to St over w t 0 , `. is finite ŽTheorem 4.1.. However, this need not imply that the paths of following have finite length. For example, let S : R 2 be the half-plane x 1 ) 0 and let St be the position of that set after a rotation by the angle t about Ž0, 0.. Then it can be easily verified that every path of following, except the one which starts at Ž0, 0., is, for sufficiently large t, of the form f Ž t . s Ž r cos t, r sin t . for some 0 - r F ¨ . Every such path has
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infinite length in spite of the fact that St is pinned by Ž0, 0.. Under the stronger assumption that St is pinned by a ball from t 0 onwards, i.e., that some ball of center p* and radius r ) 0 is contained in all St , t G t 0 , we show that each path of following has finite length ŽTheorem 4.2.. In fact, `
X r
Ht 5 f Ž t . 5 dt F 5 f Ž t . y p*5
2r
0
2
,
0
where f rX denotes the right derivative of f and 5 w 5 denotes the length of the vector w. In general, a path of finite length need not converge to St , but we show that every such path converges to S t , if, in any bounded part of R n, the transversal velocities of Mt Ži.e., the components of velocities of points of Mt perpendicular to Mt ; Definition 5.1. are bounded for all t ŽTheorem 5.2.. We consider also a u-swelling of St defined for u G 0 by StŽ u . s x q x9: x g St and 5 x9 5 F u 4 .
Ž 2.1.
We show that if some path of following the convex body StŽ u . has finite length and the transversal velocities of Mt are bounded on bounded sets then there is a critical value u G 0 such that all paths of StŽ u .-following for 0 F u - u have infinite length, and all paths of StŽ u .-following for u ) u have finite length ŽTheorem 5.8.. Analogous to models of learning, u corresponds to the threshold of sensiti¨ ity of the learning device. Then the above result says that a device endowed with greater sensitivity Ži.e., with u smaller than a critical value u . will learn with greater difficulty Žhigher precision posing higher demands..
3. MAIN DEFINITIONS Let S be the closure of the nonempty open convex set S : R n, so that M s S _ S is its boundary. Then for each p g M there is a supporting hyperplane P Ž p . : R n passing through p such that S is entirely to one side of P Ž p . w6, Theorem 3.10x. We shall assume that M is an analytic hypersurface in R n w2x. Then it follows that P Ž p . is unique for each p; we call it the tangent hyperplane to M at p. We denote by N Ž p . the unit ¨ ector normal to P Ž p . and directed into S, i.e., satisfying
Ž q y p. ? NŽ p. ) 0
for all q g S
Ž 3.1.
Ž? denotes inner product in R n .. The admissible movements of S and of the point which follows S will be described by right-regular functions. To define these, let us call an interval
S. ´ SWIERCZKOWSKI
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V of R of the form w t, t9. or w t, `. right open or a right neighborhood of t. Let U, V always denote right-open intervals. DEFINITION 3.1. A map f : U ª R n will be called right regular if it is continuous and every t 0 g U has a right neighborhood w t 0 , t 1 . such that f Ž t. s
`
Ý
am Ž t y t 0 .
m
ms0
for some a0 , a1 , . . . g R n and all t g w t 0 , t 1 ., where the series converges in the norm of R n. It is clear that a right-regular map has right derivatives Ži.e., calculated from positive increments of t only. at all points of its domain. We shall denote such derivatives by Ž dfrdt . r or f rX Ž t .. Evidently, f rX Ž t . is continuous from the right. A right-regular map f : U ª R will be called a right-regular real function. To describe the motion of S, we assume that for each time t g R there is given an orientation-preserving isometry It : R n ª R n, i.e., a rotation Ot of R n followed by a translation by a vector wt , such that the coefficients of the rotation matrix and the components of wt are right-regular real functions of t. Now the position of the moving set S at time t is given by St s It S Žthe image of S by It .. We also put Mt s It M and denote by Nt the unit normal vector field on Mt , directed into St Žsee Ž3.1... Given p f St , we denote by p t Ž p . the unique point of Mt which is closest to p and we call p t Ž p . the projection of p on Mt . It is not difficult to see that p t Ž p . y p is orthogonal to Mt at p t Ž p .; thus it is a scalar multiple of Nt Žp t Ž p ... We shall assume that the motion of the point which follows St , i.e., a path of following, is given by a right-regular map, and that there is a fixed upper bound ¨ ) 0 for the speed of this point. DEFINITION 3.2. A right-regular map f : U ª R n will be called a path of following if for all t g U, 5 f rX Ž t .5 F ¨ and f rX
¡¨ N Žp Ž f Ž t . . . Ž t . s~5 f Ž t . 5 N Ž f Ž t . . ¢0 t X r
t
t
if f Ž t . f St , if f Ž t . g Mt , if f Ž t . g S t .
It is by no means obvious that for every t 0 and p 0 there is a unique path of following f : w t 0 , `. ª R n such that f Ž t 0 . s p 0 . It is precisely the need for showing this which has motivated our choice of right-regular functions for describing following. As the existence and unicity proofs for the paths of following turn out to be very tedious, we postpone these to another
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paper. Here we shall explore the length of paths of following and their convergence to St . 4. FOLLOWING A PINNED-DOWN SET The proofs in this section are complete, save for references to the simple facts about right-regular functions which are established in Appendix A. For p g R n, E : R n, let d Ž p, E . be the distance from p to E:
d Ž p, E . s inf 5 p y q 5 : q g E 4 . THEOREM 4.1. If St is pinned by p* for t G t 0 then for every path of following f : w t 0 , `. ª R n : Ži. the function t ¬ 5 f Ž t . y p* 5 is nonincreasing, Žii. 2 ¨ Ht` d Ž f Ž t ., St . dt F 5 f Ž t 0 . y p* 5 2 . 0 Proof. Thus by theorem To show
Let GŽ t . s Ž f Ž t . y p*. ? Ž f Ž t . y p*.. Then GŽ t . is right regular. the proposition in Appendix A, we can establish part Ži. of the by showing that GrX Ž t . F 0. Clearly, GrX Ž t . s 2 f rX Ž t . ? Ž f Ž t . y p*.. that GrX Ž t . F 0, we take any t G t 0 and consider the cases:
Case f Ž t . g St . Then f rX Ž t . s 0 and GrX Ž t . s 0 Žsee Definition 3.2.. Case f Ž t . g Mt . Then f rX Ž t . s 5 f rX Ž t .5 Nt Ž f Ž t .. and hence GrX Ž t . s 2 5 f rX Ž t . 5 Nt Ž f Ž t . . ? Ž f Ž t . y p* . F 0
by Ž 3.1. .
Case f Ž t . f St . Let mŽ t . s p t Ž f Ž t ... Then from ŽB.2. in Appendix B applied to f Ž t . and Mt Žin place of p and M ., f Ž t . s m Ž t . y 5 f Ž t . y m Ž t . 5 Nt Ž m Ž t . . .
Ž ).
It follows now by Definition 3.2 that GrX Ž t . s 2 f rX Ž t . ? Ž f Ž t . y m Ž t . . q 2 f rX Ž t . ? Ž m Ž t . y p* . s y2 ¨ 5 f Ž t . y m Ž t . 5 q 2 ¨ Nt Ž m Ž t . . ? Ž m Ž t . y p* . . Ž )) . In the last sum none of the terms is positive Žfor the second term use Ž3.1... This proves part Ži.. We turn now to the proof of Žii.. Noting that d Ž p, S . is a continuous Ž f Ž t .., S ., we see that t ¬ d Ž f Ž t ., St . function of p and d Ž f Ž t ., St . s d Ž Iy1 t is also continuous. Hence the Riemann integral in Žii. exists. We claim that y2 ¨ d Ž f Ž t . , St . G GrX Ž t . .
Ž ))).
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We have just established that GrX Ž t . F 0. Thus the inequality always holds when the left-hand side vanishes, i.e., when f Ž t . g St . When f Ž t . f St , we have d Ž f Ž t ., St . s 5 f Ž t . y mŽ t .5, by the definition of mŽ t ., whence Ž))). follows from Ž)). and Ž3.1.. From Ž))). we get, for every t G t 0 , 1
t
t
1
X r
Ht d Ž f Ž t . , S . dt F 2 ¨ Ht G Ž t . dt s 2 ¨ Ž G Ž t . y G Ž t . . t
0
0
0
Žsee Appendix A for the integration of GrX .. The above inequality remains valid if the nonnegative GŽ t . on the right-hand side is removed. A passage to the limit as t ª ` gives Žii. of the theorem. THEOREM 4.2. Suppose S t is pinned by a ball of center p* and radius r ) 0 for t G t 0 . Then e¨ ery path of following f : w t 0 , `. ª R n has finite length. In fact, `
X r
Ht 5 f Ž t . 5 dt F 5 f Ž t . y p*5
2r
2
0
.
0
Proof. The assumption is now stronger than in Theorem 4.1. Hence all the parts of the previous proof can be used. We claim that it will be enough to establish y2 r 5 f rX Ž t . 5 G GrX Ž t . .
Ž a.
Indeed, since 5 f rX Ž t .5 is Riemann integrable ŽAppendix A., Ža. implies that, for t G t 0 , t
X r
Ht 5 f Ž t . 5 dt F G Ž t . y G Ž t . F G Ž t . s 5 f Ž t . y p*5
2r
0
0
2
0
.
0
To establish Ža., we consider for each t two possibilities: Case f Ž t . g S t . Then f rX Ž t . s 0 and Ža. follows from GrX Ž t . F 0 Žsee the proof of the previous theorem.. Case f Ž t . f St . In this case f rX Ž t . s 5 f rX Ž t .5 Nt Ž mŽ t ... We claim that Nt Ž m Ž t . . ? Ž m Ž t . y p* . - yr .
Ž aa.
Indeed, since S t is pinned by the ball of center p* and radius r , p* y r Nt Ž m Ž t . . g St , whence Nt Ž m Ž t . . ? Ž p* y r Nt Ž m Ž t . . y m Ž t . . ) 0
by Ž 3.1. .
A MODEL OF FOLLOWING
553
This implies Žaa.. From Ž)). and Žaa. we now get GrX Ž t . - y2 ¨ r F y2 r 5 f rX Ž t . 5 , i.e., Ža. holds.
5. CONVERGENCE OF A PATH OF FOLLOWING, FOLLOWING A SWELLING SET The remaining two theorems involve, for each t, a vector field Ft on Mt which indicates how Mt moves through space: at every m g Mt , Ft is the component of the velocity of m normal to Mt . DEFINITION 5.1. The trans¨ ersal ¨ elocity field Ft Ž m. of Mt at m g Mt is Ft Ž m . s Vt Ž m . Nt Ž m . , where Vt Ž m . s Ž Ž drdt . r It . Ž Iy1 t Ž m . . ? Nt Ž m . . We say that the trans¨ ersal ¨ elocities of Mt are bounded on bounded sets if for each ball B : R n the set of norms
5 Ft Ž m . 5 :
t g R, m g Mt l B 4
is bounded. A path of following f : w t 0 , `. ª R n of finite length always converges as t ª ` Žsee Lemma 5.7. and we shall denote its limit by f` . THEOREM 5.2. If the trans¨ ersal ¨ elocities of Mt are bounded on bounded sets, then e¨ ery path of following of finite length con¨ erges to S t , i.e., lim d Ž f Ž t . , St . s lim d Ž f` , St . s 0.
tª`
tª`
Remark. If in the example of following in Section 2 Žrotating half-plane. we allow the angular velocities of rotation to be arbitrarily large then it is not hard to find paths of following which have finite length but do not converge to St . Proof of Theorem 5.2. This requires a few preparations which come in Lemmas 5.3]5.7. Let us observe first that if a point moves so as to stay on Mt then the component of its velocity perpendicular to Mt equals Ft . LEMMA 5.3. For e¨ ery right-regular map m: V ª R n satisfying mŽ t . g Mt for all t we ha¨ e mXr Ž t . ? Nt Ž m Ž t . . s Vt Ž m Ž t . . .
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Proof. We have here a composite motion: mŽ t . moves relative to Mt and Mt moves in R n. Thus for any time t the velocity, mXr Ž t . is the sum of the velocity of mŽ t . relative to Mt which is orthogonal to Nt Ž mŽ t .. and the velocity of the point of Mt occupying at time t the position mŽ t .. The latter velocity projects on Nt Ž mŽ t .. as Vt Ž mŽ t .., by Definition 5.1. LEMMA 5.4. Let f : U ª R n be a path of following and put mŽ t . s p t Ž f Ž t .. whene¨ er f Ž t . f St . Then, for ¨ ery t 0 g U such that f Ž t 0 . f St 0 ,
Ž drdt . r d Ž f Ž t . , St . s y¨ q Vt Ž m Ž t . . . Proof. By continuity, f Ž t . f St Ži.e., d Ž f Ž t ., St . ) 0. holds on some right neighborhood V of t 0 . Hence mŽ t . is defined for all t g V. Moreover, it is clear that m Ž t . s p t Ž f Ž t . . s It (p ( Iy1 t f Ž t. ,
t g V,
holds for the analytic projection map p : R _ S ª M ŽTheorem B.3.. It follows that mŽ t . is right regular. Thus d Ž f Ž t ., St . s 5 f Ž t . y mŽ t .5 is right regular on V and we have n
Ž drdt . r Ž d Ž f Ž t . , St . . s 2 Ž mXr Ž t . y f rX Ž t . . ? Ž m Ž t . y f Ž t . . . 2
However, by Ž). in Section 4 and Definition 3.2, yf rX Ž t . ? Ž m Ž t . y f Ž t . . s yf rX Ž t . ? 5 m Ž t . y f Ž t . 5 Nt Ž m Ž t . . s y¨ 5 m Ž t . y f Ž t . 5 s y¨ d Ž f Ž t . , S t . . Further, by Ž). and Lemma 5.3, mXr Ž t . ? Ž m Ž t . y f Ž t . . s mXr Ž t . ? 5 m Ž t . y f Ž t . 5 Nt Ž m Ž t . . s Vt Ž m Ž t . . d Ž f Ž t . , St . . From these equations we conclude that 2
Ž drdt . r Ž d Ž f Ž t . , St . . s 2 y¨ q Vt Ž m Ž t . d Ž f Ž t . , St . , which obviously also equals 2 d Ž f Ž t ., S t .Ž drdt . r d Ž f Ž t ., St .. Division by 2Ž d Ž f Ž t ., St . ) 0 gives the required identity. LEMMA 5.5. Let e: w t 0 , `. ª R be a continuous, nonnegati¨ e function such that for each t there is a t ) t satisfying eŽ t . s 0, and suppose further that eŽ t . ¢ 0 as t ª `. Then there is an a ) 0 and an infinite sequence of disjoint inter¨ als w t n , t n x with t n - t n such that, for each n s 1, 2, . . . ,
Ž q. e Ž tn . s a ,
e Ž t n . s 0,
0 - eŽ t . F a
for t n F t - t n .
Proof. An elementary exercise in limits and continuity.
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LEMMA 5.6. If the function e: w t 0 , `. ª R in Lemma 5.5 is right regular on the open set t: eŽ t . ) 04 and for the number a satisfying Žq. there are constants ˜t and C such that
Ž qq.
0 - e Ž t . F a « < eXr Ž t . < - C,
for e¨ ery t ) t :
then each of the inter¨ als w t n , t n x in Lemma 5.5 with t n ) ˜t has length at least arC. Proof. Let t n ) ˜t. Then by Žq., 0 - eŽ t . F a for every t in w t n , t n ., and hence < eXr Ž t .< - C. Consequently, for t g w t n , t n ., e Ž t . y e Ž tn . s
t X
Ht e Ž t . dt G yC Ž t y t . . r
n
n
As t ª t n , we get, by continuity, e Ž t n . y e Ž t n . G yC Ž t n y t n . . Since eŽ t n . s 0 and eŽ t n . s a , it follows that t n y t n G arC. LEMMA 5.7.
E¨ ery path of following of finite length con¨ erges.
Proof. Let f : w t 0 , `. ª R n be a path of following of finite length. Thus `
X r
Ht 5 f Ž t . 5 dt - ` 0
and 5 f Ž t 2 . y f Ž t1 . 5 s
t2 X
Ht
1
f r Ž t . dt F
t2
X r
Ht 5 f Ž t . 5 dt 1
by Appendix A. It follows that 5 f Ž t 2 . y f Ž t 1 .5 ª 0 as t 1 , t 2 ª `. Proof of Theorem 5.2 Ž Concluded.. Suppose f : w t 0 , `. ª R n is a path of following of finite length and d Ž f Ž t ., St . ¢ 0 as t ª `. To deduce from this a contradiction, we put eŽ t . s d Ž f Ž t ., St . and claim then that all the assumptions of Lemmas 5.5 and 5.6 are satisfied. Indeed, we cannot have a t9 such that d Ž f Ž t ., St . ) 0 for all t ) t9 because this would imply 5 f rX Ž t .5 s ¨ for all t ) t9, contradicting the finiteness of the length of this path. Thus Žq. holds for some a ) 0 and w t n , t n x, n s 1, 2, . . . . Now let ˜t be such that 5 f Ž t . y f` 5 - a for all t ) ˜t Žwhere f` s lim t ª` f Ž t .; see Lemma 5.7.. We check that the assumptions of Lemma 5.6 are satisfied. If eŽ t . ) 0, define mŽ t . s p t Ž f Ž t ... Then e Ž t . s d Ž f Ž t . , St . s 5 f Ž t . y m Ž t . 5
S. ´ SWIERCZKOWSKI
556
is right regular Žsee the proof of Lemma 5.4.. Now, to verify the assumption Žqq., let t ) ˜t be such that 0 - eŽ t . F a , i.e., 5 f Ž t . y mŽ t .5 F a . Then, by the choice of ˜t, we have 5 mŽ t . y f` 5 F 2 a . By Lemma 5.4, eXr Ž t . s y¨ q Vt Ž mŽ t .., and since Vt Ž x . is bounded on bounded sets and mŽ t . is in the ball of center f` and radius 2 a , it follows that there is a constant C such that < eXr Ž t .< - C. Thus all the assumptions of Lemma 5.6 are satisfied and we have t n y t n G arC if t n ) ˜t. Moreover, eŽ t . s d Ž f Ž t ., St . is positive on w t n , t n .. Thus 5 f rX Ž t .5 s ¨ by Definition 3.2. It follows that
Ht t 5 f Ž t . 5 dt s ¨ Ž t n
X r
n
y t n . G ¨ arC
for every t n ) ˜t ,
n
contradicting the finite length of the path f Ž t .. For the last theorem we consider for each u ) 0 the u-swelling S Ž u . of S given by S Ž u . s x q x9: x g S and 5 x9 5 F u 4 . Evidently the set S Ž u . is open and convex. We shall show that its boundary S Ž u . is an analytic hypersurface in R n. Thus, for each fixed u G 0, we may consider the model of following with the moving set StŽ u . s It S Ž u .. THEOREM 5.8. Suppose the trans¨ ersal ¨ elocities of Mt are bounded on bounded sets and for some u G 0 there exists a path of S tŽ u .-following of finite length. Then the trans¨ ersal ¨ elocities of e¨ ery MtŽ u ., u G 0, are bounded on bounded sets. Moreo¨ er, there exists a u G 0 such that Ži. For e¨ ery u ) u , e¨ ery path of StŽ u .-following has finite length. Žii. For e¨ ery 0 F u - u , e¨ ery path of StŽ u .-following has infinite length. Proof. We would like to apply the previous results to StŽ u .. Thus it is necessary to show that each boundary StŽ u . is an analytic hypersurface MtŽ u . in R n. Let u ) 0 be fixed. We consider the analytic manifold M = Ry, where Rys s g R: s - 04 and the map K : M = Ryª R n _ S given by K Ž m, s . s m q sN Ž m. for each m g M, s - 0. By Appendix B, K is an analytic bijection. Since M = yu 4 is an analytic submanifold of M = Ry of codimension 1, its image by K is a hypersurface in R n. We denote this image by M Ž u . and show that M Ž u . s S Ž u .. Proof of M Ž u . : S Ž u .. By definition, M Ž u . s m y u N Ž m.: m g M 4 . Let m g M. By Lemma B.1, m q d N Ž m. g S for all sufficiently small
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557
d ) 0; thus m y Ž u y d . N Ž m. g S Ž u .. It follows that m y u N Ž m. is in the closure of S Ž u .. But also, arbitrarily close to m y u N Ž m. there are points not belonging to S Ž u .. Indeed, let us show that m y Ž u q d . N Ž m. f S Ž u . for d ) 0. Assuming the contrary, we get, for some 5 x9 5 F u , m y Ž u q d . N Ž m . q x9 g S. Then, by Ž3.1., ŽyŽ u q d . N Ž m. q x9. ? N Ž m. ) 0, whence 5 x9 5 ) u q d and this is a contradiction. It follows that m y u N Ž m. g d S Ž u ., i.e., the inclusion is proved. Proof of S Ž u . : M Ž u .. Consider any p g S Ž u .. Then p f S Ž u ., hence p f S and m s p Ž p . g M. By ŽB.2., p s m y 5 p y m5 N Ž m. .
Ž (.
Moreover, 5 p y m 5 s d Ž p, S .. We claim now that 5 p y m 5 s u . Indeed, arbitrarily close to p there are points q g S Ž u ., i.e., satisfying d Ž q, S . - u . Thus d Ž p, S . F u , i.e., 5 p y m 5 F u . On the other hand, we cannot have 5 p y m 5 - u because then the triangle inequality would immediately imply that p g S Ž u .. Thus 5 p y m 5 s u and Ž(. implies that p s m y u N Ž m., that is, p g M Ž u .. We now put MtŽ u . s It M Ž u . and show that the transversal velocities of MtŽ u . are bounded on bounded sets. There is a natural embedding of M in M = Ry given by m ¬ Ž m, yu .. Combining this map with K, we get an analytic bijection between hypersurfaces k Žu . : M ª M Žu . ,
where k Ž u . Ž m . s m y u N Ž m . .
Let N Ž u . be the unit normal vector field on M Ž u ., directed into S Ž u .. It is easily checked that, for any smooth curve g Ž t . in M, its image under k Ž u ., i.e., g Ž t . y u N Ž g Ž t .., is orthogonal to N Ž g Ž t ... This shows that the vector fields N and N Ž u . are identical at m and k Ž u . Ž m., respectively. Let NtŽ u . denote the unit normal inward-directed vector field on MtŽ u . and let k tŽ u . Ž p . s p y u Nt Ž p . for all p g Mt . By the above, k tŽ u . : Mt ª MtŽ u . is an analytic bijection between hypersurfaces and NtŽ u . Ž k tŽ u . Ž p .. is Nt Ž p .. Taking any right-regular curve mŽ t . in Mt and its image by k tŽ u ., i.e., mŽ t . y u Nt Ž mŽ t .., we check easily that the derivatives mXr Ž t . and Ž drdt . r k tŽ u . Ž mŽ t .. have identical projections on Nt Ž mŽ t ... By Lemma 5.3, this implies that the transversal velocity of Mt at p is the same as the transversal velocity of MtŽ u . at k tŽ u . Ž p .. Since the distance between these two points is u , it follows that the transversal velocities of MtŽ u . are bounded on bounded sets. Proof of Theorem 5.8 Ž Concluded.. We claim that if, for some u G 0, some path f Ž t . of StŽ u .-following has finite length then, for each u 9 ) u ,
S. ´ SWIERCZKOWSKI
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StŽ u 9. is pinned by a ball from some t onwards. Indeed, by Theorem 5.2, d Ž f` , StŽ u . . ª 0 as t ª `, whence, given r - u 9 y u , the ball of center f` and radius r is contained in StŽ u 9. for sufficiently large t. We conclude now by Theorem 4.2 that if some path of StŽ u .-following has finite length then e¨ ery path of StŽ u 9.-following for each u 9 ) u has finite length. Thus the required u is given by
u s g.l.b. u G 0: some path of StŽ u .-following has finite length 4 .
APPENDIX A: INTEGRABILITY OF RIGHT-REGULAR FUNCTIONS Let f : U ª R n be right regular ŽDefinition 3.1. and consider its right derivative f rX : U ª R n. PROPOSITION.
If f rX is bounded on w t 0 , t 1 x : U then
Ži. the functions f rX and 5 f rX 5 are Riemann integrable on w t 0 , t 1 x and
Ž ii .
f Ž t1 . y f Ž t 0 . s
t1 X
Ht
f r Ž t . dt.
0
Proof of Ži.. Let D denote the set of discontinuities of f rX and let us assign to each t g D a right neighborhood V of t in U such that f rX is continuous on V _ t 4 . ŽBy assumption, on some right neighborhood V of t, f is given by a power series.. Clearly, in any such assignment, right neighborhoods corresponding to different elements of D are disjoint. Thus D is countable, and hence of Lebesgue measure zero. So f rX is Riemann integrable, by the Lebesgue criterion for Riemann integrability of a bounded function Žsee w1x.. Also 5 f rX 5 is Riemann integrable because the set of discontinuities of 5 f rX 5 is a subset of the set of discontinuities of f rX . Proof of Žii.. Let T be the set of t g w t 0 , t 1 x such that f Ž t . y f Ž t0 . s
t X
Ht f Ž t . dt , r
0
whenever t 0 F t - t. Obviously, T s w t 0 , t9. for some t9 g Ž t 0 , t 1 x. Since f rX is bounded, the above integral is a continuous function of t, whence by the continuity of f, f Ž t9 . y f Ž t 0 . s
t9 X
Ht
0
f r Ž t . dt .
Ž A.1.
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Thus, if t9 s t 1 , we are done. Suppose t9 - t 1. Then, since f rX Žt . is given for t ) t9 and t close to t9 by a power series, there is a t0 g Ž t9, t 1 . such that f Ž t0 . y f Ž t9 . s
t0 X
Ht9
f r Ž t . dt .
This, jointly with ŽA.1., contradicts the definition of t9.
APPENDIX B: CONVEXITY IMPLICATIONS Let S and M be as in Section 1. Given m g M and a ) 0, we shall consider the line L m s m q sN Ž m.: s g R4 and its open interval Lam s m q sN Ž m.: 0 - s - a 4 . LEMMA B.1.
For e¨ ery m g M there is an a ) 0 such that Lam : S.
Proof. Consider first the case when n s 2, i.e., M is a convex curve in R 2 and the hyperplane P Ž m. is the line tangent to m at M. Then the whole of S lies to one side of P Ž m. and L m is perpendicular to M at m Ži.e., to P Ž m... Thus, arbitrarily close to m, there are pairs of points of M Žand hence of S . which are on opposite sides of L m . Hence, by convexity, there are arbitrarily close to m points of L m l S. These must be of the form m q sN Ž m., where s ) 0, by Ž3.1.. Using convexity again, we obtain Lam : S for some a ) 0. If n ) 2, it suffices to repeat the above reasoning in a two-dimensional plane passing through L m and some point of S. By Ž3.1., no point m q sN Ž m., where m g M and s - 0, can belong to S. Thus the correspondence Ž m, s . ¬ m q sN Ž m. yields a mapping K : M = Ryª R n _ S Žwhere Rys s g R: s - 04.. Recall that, for p f S, p Ž p . denotes the point of S Žthus of M . which is closest to p. PROPOSITION B.2.
K is a bijection and Ky1 is gi¨ en by
Ky1 Ž p . s Ž p Ž p . , y5p Ž p . y p 5 .
for all p f S.
Ž B.1.
Proof. It is not difficult to see that, for every p f S, p Ž p . y p is perpendicular to M at p Ž p ., and thus p Ž p . y p s "5p Ž p . y p 5 N Žp Ž p ... Applying Lemma B.1, we see that we must take the q sign here, i.e., p s p Ž p . y 5p Ž p . y p 5 N Ž p Ž p . . ,
Ž B.2.
or, equivalently, p s K Žp Ž p ., y5p Ž p . y p 5.. This shows that K is surjective. On the other hand, if p s m q sN Ž m. and s - 0 then Ž3.1. implies
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that p f S. Also, by Ž3.1. and Lemma B.1, P Ž m. separates S from p. Thus m s p Ž p . and s s y5p Ž p . y p 5. This shows the injectivity of K and that the inverse map is given by ŽB.1.. In proving Theorem B.3 we shall use tangent vectors and principal curvatures with the following notation w2x. The coordinate functions x 1 , . . . , x n define at each m g R n the base Ž r x 1 . m , . . . , Ž r x n . m of the tangent space Tm R n. We shall identify Tm R n with R n by identifying the tangent vector X s Ý nis1 a i Ž r x i . m with the n-tuple Ž a1 , . . . , a n .. If J: M ª R n is the inclusion map, we use the Jacobian map J#: Tm M ª Tm R n Žwhich is nonsingular by assumption. to identify the tangent space Tm M with its image J#ŽTm M . in Tm R n. Combining this with the previous identification, we view each Tm M as an Ž n y 1.-dimensional subspace of R n. Recall that a nonzero tangent vector X g Tm M is called principal if the derivative Dx N of the unit normal vector field N in the direction of X is a scalar multiple of X. For a principal X we shall write Dx N s yk X, where k g R is the principal curvature of M at m in the direction of X. The convexity of M implies that all principal curvatures of M are nonnegative. THEOREM B.3. analytic.
Each of the maps K, Ky1 , and p : R n _ S ª M is
Proof. K is analytic because the unit normal vector field N on M is analytic. Once we show that Ky1 is analytic, we conclude from ŽB.1. that p is analytic. We wish to apply now the inverse function theorem to K. That is, we are going to show that the Jacobian map K# is everywhere nonsingular and this will lead to the conclusion that Ky1 is analytic. So our remaining task is to find n tangent vectors to M = Ry at Ž m, s . whose images by K# are linearly independent. We shall choose these vectors of the form
Ž X1 , 0 . , . . . , Ž X ny1 , 0 . , Ž 0, Ž r t . s . , where X 1 , . . . , X ny1 g Tm M and Ž r t .s is the basis vector of the tangent space to R at s. For the X 1 , . . . , X ny1 we take any basis of Tm M composed of principal vectors, i.e., vectors X i such that each covariant derivative D X i N of N in the direction X i is equal to yk i X i , where k i G 0 is a principal curvature at m w2, p. 24x. Thus we have to show that if Yi s K#Ž X i , 0. for i - n and Yn s K#Ž0, Ž r t .s . then Y1 , . . . , Yn are independent. To find Yi , let hi Ž t . be curves in M such that hi Ž0. s m, hiX Ž0. s X i ,
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i - n, and let hnŽ t . s s q t, so that hnŽ0. s s, hnX Ž0. s Ž r t .s . Then, for i - n, Yi s s
d dt d dt
K Ž hi Ž t . , s .
ts0
hi Ž t . q sN Ž hi Ž t . .
ts0
s X i q sD X i N s X i y sk i X i s Ž 1 y sk i . X i and Yn s s
d dt d dt
K Ž m, hn Ž t . .
ts0
m q Ž s q t . N Ž m.
s N Ž m. . ts0
It follows now from s - 0 and k i G 0 that each Yi for i - n is a nonzero multiple of X i ; moreover, Yn is orthogonal to all other Yi . This proves the independence of Y1 , . . . , Yn . ACKNOWLEDGMENT This paper owes its existence to Jan Mycielski; the research originated in conversations with him and some results were in that way jointly obtained.
REFERENCES 1. T. M. Apostol, ‘‘Mathematical Analysis,’’ Addison]Wesley, Reading, MA, 1974. 2. N. Hicks, ‘‘Notes on Differential Geometry,’’ Van Nostrand, Princeton, NJ, 1964. 3. S. Kaczmarz, Angenahrte Auflosung von Systemen linearer Gleichungen, Bull. Acad. ¨ ¨ Polon. Sci. Lett. Ser. A 35 Ž1937., 355]357. 4. J. Mycielski, Linear dynamic approximation theory, J. Approx. Theory 25 Ž1979., 369]383. 5. J. Mycielski and S. Swierczkowski, A model of the neocortex, Ad¨ . in Appl. Math. 9 Ž1988., 465]480. 6. J. van Tiel, ‘‘Convex Analysis,’’ Wiley, Chichester, 1984.
222, 547]561 Ž1998.
AY985957
A Model of Following S. ´ Swierczkowski* Department of Mathematics, Uni¨ ersity of Colorado, Boulder, Colorado 80309-0395 Submitted by A. Schumitzky Received October 22, 1997
In the discrete-time models of learning in w5x Žbased on the Kaczmarz projection algorithm w3x. there were obtained estimates ŽTheorems 4.1. and 4.2. of the sum of the errors and the length of the learning path of a learning process. These depended on the size of a time step h, a parameter of the model, and they would yield limits as h ª 0 but the resulting asymptotic estimates were not realizable in any particular model Žwith fixed h.. Here we define a continuous-time analogue of our former discrete-time model of learning and we call it a model of following. It is shown that the asymptotic forms of the main results about learning are true in the models of following. Q 1998 Academic Press Key Words: approximation; analytic movement; convex body; algorithm of following.
1. DESCRIPTION OF THE MODEL We consider here a situation where a convex body Žopen set. S with a regular boundary M is moving in R n, n G 2, occupying at time t the position St . A point moving in R n is following St ; we denote by f Ž t . the position of that point at time t. The speed of the point is never allowed to exceed a fixed constant ¨ , whereas the speed of St may be arbitrary. Thus it may happen that if the point starts from some initial position f Ž t 0 . far from S t 0 , it will remain at a positive distance from St for a while. During that period the point should follow St in the sense that at each moment its velocity vector is directed toward the nearest point of the boundary Mt of St and its speed is ¨ . If the point finds itself on Mt , it should move with the smallest possible speed which will allow it to stay in contact with S t j Mt * E-mail: [email protected]. 547 0022-247Xr98 $25.00 Copyright Q 1998 by Academic Press All rights of reproduction in any form reserved.
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Žthus, for instance, if remaining on Mt , it will move perpendicularly to Mt .. However, if St speeds up sufficiently, the point might be unable to remain in St j Mt Žas its speed cannot exceed ¨ .. At any time, when the point finds itself in Žthe open set. S t , it is supposed to stop and not move for as long as it can stay in St without moving Žall this is equivalent to Definition 3.2.. It is clear from this description that the paths of following, i.e., the curves f Ž t ., will, in general, not even be of class C 1. For example, at the moment when f Ž t . catches up with St , f 9Ž t . will most likely have a discontinuity. However, we shall assume another kind of regularity for all the movements: continuity with right analyticity, the latter meaning that, for any time t, all movements during a sufficiently small time interval beginning at t can be described by real analytic functions of time. This combination of continuity and right analyticity we shall call right regularity ŽDefinition 3.1.. We are motivated to investigate the above model of following because a special case, namely when S is the slice of R n contained between two parallel hyperplanes, is the obvious continuous-time analog of the models of learning studied in w4, 5x. In these, time was discrete and it advanced in equal steps of length h. Estimates, depending on h, were obtained, concerning the quality of the learning process. By letting h approach 0, one could obtain asymptotic estimates. In the present paper the learning model is replaced by a model of following, so that a sequence of consecutive memory states during learning is replaced by a path of following. It turns out then that the asymptotic estimates mentioned above hold exactly in the model of following.
2. SUMMARY OF RESULTS Various assumptions about the movement of St allow us to draw conclusions about the paths of following f : w t 0 , `. ª R n. The least interesting case is when St moves very fast so that a point following St with speed ¨ will never catch up. To exclude this, we shall sometimes assume that St is pinned by a point p* from t 0 onwards, i.e., that p* g St j Mt for all t G t 0 . We show then that for any path of following the distance from f Ž t . to p* is decreasing and the integral of the distance from f Ž t . to St over w t 0 , `. is finite ŽTheorem 4.1.. However, this need not imply that the paths of following have finite length. For example, let S : R 2 be the half-plane x 1 ) 0 and let St be the position of that set after a rotation by the angle t about Ž0, 0.. Then it can be easily verified that every path of following, except the one which starts at Ž0, 0., is, for sufficiently large t, of the form f Ž t . s Ž r cos t, r sin t . for some 0 - r F ¨ . Every such path has
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A MODEL OF FOLLOWING
infinite length in spite of the fact that St is pinned by Ž0, 0.. Under the stronger assumption that St is pinned by a ball from t 0 onwards, i.e., that some ball of center p* and radius r ) 0 is contained in all St , t G t 0 , we show that each path of following has finite length ŽTheorem 4.2.. In fact, `
X r
Ht 5 f Ž t . 5 dt F 5 f Ž t . y p*5
2r
0
2
,
0
where f rX denotes the right derivative of f and 5 w 5 denotes the length of the vector w. In general, a path of finite length need not converge to St , but we show that every such path converges to S t , if, in any bounded part of R n, the transversal velocities of Mt Ži.e., the components of velocities of points of Mt perpendicular to Mt ; Definition 5.1. are bounded for all t ŽTheorem 5.2.. We consider also a u-swelling of St defined for u G 0 by StŽ u . s x q x9: x g St and 5 x9 5 F u 4 .
Ž 2.1.
We show that if some path of following the convex body StŽ u . has finite length and the transversal velocities of Mt are bounded on bounded sets then there is a critical value u G 0 such that all paths of StŽ u .-following for 0 F u - u have infinite length, and all paths of StŽ u .-following for u ) u have finite length ŽTheorem 5.8.. Analogous to models of learning, u corresponds to the threshold of sensiti¨ ity of the learning device. Then the above result says that a device endowed with greater sensitivity Ži.e., with u smaller than a critical value u . will learn with greater difficulty Žhigher precision posing higher demands..
3. MAIN DEFINITIONS Let S be the closure of the nonempty open convex set S : R n, so that M s S _ S is its boundary. Then for each p g M there is a supporting hyperplane P Ž p . : R n passing through p such that S is entirely to one side of P Ž p . w6, Theorem 3.10x. We shall assume that M is an analytic hypersurface in R n w2x. Then it follows that P Ž p . is unique for each p; we call it the tangent hyperplane to M at p. We denote by N Ž p . the unit ¨ ector normal to P Ž p . and directed into S, i.e., satisfying
Ž q y p. ? NŽ p. ) 0
for all q g S
Ž 3.1.
Ž? denotes inner product in R n .. The admissible movements of S and of the point which follows S will be described by right-regular functions. To define these, let us call an interval
S. ´ SWIERCZKOWSKI
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V of R of the form w t, t9. or w t, `. right open or a right neighborhood of t. Let U, V always denote right-open intervals. DEFINITION 3.1. A map f : U ª R n will be called right regular if it is continuous and every t 0 g U has a right neighborhood w t 0 , t 1 . such that f Ž t. s
`
Ý
am Ž t y t 0 .
m
ms0
for some a0 , a1 , . . . g R n and all t g w t 0 , t 1 ., where the series converges in the norm of R n. It is clear that a right-regular map has right derivatives Ži.e., calculated from positive increments of t only. at all points of its domain. We shall denote such derivatives by Ž dfrdt . r or f rX Ž t .. Evidently, f rX Ž t . is continuous from the right. A right-regular map f : U ª R will be called a right-regular real function. To describe the motion of S, we assume that for each time t g R there is given an orientation-preserving isometry It : R n ª R n, i.e., a rotation Ot of R n followed by a translation by a vector wt , such that the coefficients of the rotation matrix and the components of wt are right-regular real functions of t. Now the position of the moving set S at time t is given by St s It S Žthe image of S by It .. We also put Mt s It M and denote by Nt the unit normal vector field on Mt , directed into St Žsee Ž3.1... Given p f St , we denote by p t Ž p . the unique point of Mt which is closest to p and we call p t Ž p . the projection of p on Mt . It is not difficult to see that p t Ž p . y p is orthogonal to Mt at p t Ž p .; thus it is a scalar multiple of Nt Žp t Ž p ... We shall assume that the motion of the point which follows St , i.e., a path of following, is given by a right-regular map, and that there is a fixed upper bound ¨ ) 0 for the speed of this point. DEFINITION 3.2. A right-regular map f : U ª R n will be called a path of following if for all t g U, 5 f rX Ž t .5 F ¨ and f rX
¡¨ N Žp Ž f Ž t . . . Ž t . s~5 f Ž t . 5 N Ž f Ž t . . ¢0 t X r
t
t
if f Ž t . f St , if f Ž t . g Mt , if f Ž t . g S t .
It is by no means obvious that for every t 0 and p 0 there is a unique path of following f : w t 0 , `. ª R n such that f Ž t 0 . s p 0 . It is precisely the need for showing this which has motivated our choice of right-regular functions for describing following. As the existence and unicity proofs for the paths of following turn out to be very tedious, we postpone these to another
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A MODEL OF FOLLOWING
paper. Here we shall explore the length of paths of following and their convergence to St . 4. FOLLOWING A PINNED-DOWN SET The proofs in this section are complete, save for references to the simple facts about right-regular functions which are established in Appendix A. For p g R n, E : R n, let d Ž p, E . be the distance from p to E:
d Ž p, E . s inf 5 p y q 5 : q g E 4 . THEOREM 4.1. If St is pinned by p* for t G t 0 then for every path of following f : w t 0 , `. ª R n : Ži. the function t ¬ 5 f Ž t . y p* 5 is nonincreasing, Žii. 2 ¨ Ht` d Ž f Ž t ., St . dt F 5 f Ž t 0 . y p* 5 2 . 0 Proof. Thus by theorem To show
Let GŽ t . s Ž f Ž t . y p*. ? Ž f Ž t . y p*.. Then GŽ t . is right regular. the proposition in Appendix A, we can establish part Ži. of the by showing that GrX Ž t . F 0. Clearly, GrX Ž t . s 2 f rX Ž t . ? Ž f Ž t . y p*.. that GrX Ž t . F 0, we take any t G t 0 and consider the cases:
Case f Ž t . g St . Then f rX Ž t . s 0 and GrX Ž t . s 0 Žsee Definition 3.2.. Case f Ž t . g Mt . Then f rX Ž t . s 5 f rX Ž t .5 Nt Ž f Ž t .. and hence GrX Ž t . s 2 5 f rX Ž t . 5 Nt Ž f Ž t . . ? Ž f Ž t . y p* . F 0
by Ž 3.1. .
Case f Ž t . f St . Let mŽ t . s p t Ž f Ž t ... Then from ŽB.2. in Appendix B applied to f Ž t . and Mt Žin place of p and M ., f Ž t . s m Ž t . y 5 f Ž t . y m Ž t . 5 Nt Ž m Ž t . . .
Ž ).
It follows now by Definition 3.2 that GrX Ž t . s 2 f rX Ž t . ? Ž f Ž t . y m Ž t . . q 2 f rX Ž t . ? Ž m Ž t . y p* . s y2 ¨ 5 f Ž t . y m Ž t . 5 q 2 ¨ Nt Ž m Ž t . . ? Ž m Ž t . y p* . . Ž )) . In the last sum none of the terms is positive Žfor the second term use Ž3.1... This proves part Ži.. We turn now to the proof of Žii.. Noting that d Ž p, S . is a continuous Ž f Ž t .., S ., we see that t ¬ d Ž f Ž t ., St . function of p and d Ž f Ž t ., St . s d Ž Iy1 t is also continuous. Hence the Riemann integral in Žii. exists. We claim that y2 ¨ d Ž f Ž t . , St . G GrX Ž t . .
Ž ))).
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We have just established that GrX Ž t . F 0. Thus the inequality always holds when the left-hand side vanishes, i.e., when f Ž t . g St . When f Ž t . f St , we have d Ž f Ž t ., St . s 5 f Ž t . y mŽ t .5, by the definition of mŽ t ., whence Ž))). follows from Ž)). and Ž3.1.. From Ž))). we get, for every t G t 0 , 1
t
t
1
X r
Ht d Ž f Ž t . , S . dt F 2 ¨ Ht G Ž t . dt s 2 ¨ Ž G Ž t . y G Ž t . . t
0
0
0
Žsee Appendix A for the integration of GrX .. The above inequality remains valid if the nonnegative GŽ t . on the right-hand side is removed. A passage to the limit as t ª ` gives Žii. of the theorem. THEOREM 4.2. Suppose S t is pinned by a ball of center p* and radius r ) 0 for t G t 0 . Then e¨ ery path of following f : w t 0 , `. ª R n has finite length. In fact, `
X r
Ht 5 f Ž t . 5 dt F 5 f Ž t . y p*5
2r
2
0
.
0
Proof. The assumption is now stronger than in Theorem 4.1. Hence all the parts of the previous proof can be used. We claim that it will be enough to establish y2 r 5 f rX Ž t . 5 G GrX Ž t . .
Ž a.
Indeed, since 5 f rX Ž t .5 is Riemann integrable ŽAppendix A., Ža. implies that, for t G t 0 , t
X r
Ht 5 f Ž t . 5 dt F G Ž t . y G Ž t . F G Ž t . s 5 f Ž t . y p*5
2r
0
0
2
0
.
0
To establish Ža., we consider for each t two possibilities: Case f Ž t . g S t . Then f rX Ž t . s 0 and Ža. follows from GrX Ž t . F 0 Žsee the proof of the previous theorem.. Case f Ž t . f St . In this case f rX Ž t . s 5 f rX Ž t .5 Nt Ž mŽ t ... We claim that Nt Ž m Ž t . . ? Ž m Ž t . y p* . - yr .
Ž aa.
Indeed, since S t is pinned by the ball of center p* and radius r , p* y r Nt Ž m Ž t . . g St , whence Nt Ž m Ž t . . ? Ž p* y r Nt Ž m Ž t . . y m Ž t . . ) 0
by Ž 3.1. .
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553
This implies Žaa.. From Ž)). and Žaa. we now get GrX Ž t . - y2 ¨ r F y2 r 5 f rX Ž t . 5 , i.e., Ža. holds.
5. CONVERGENCE OF A PATH OF FOLLOWING, FOLLOWING A SWELLING SET The remaining two theorems involve, for each t, a vector field Ft on Mt which indicates how Mt moves through space: at every m g Mt , Ft is the component of the velocity of m normal to Mt . DEFINITION 5.1. The trans¨ ersal ¨ elocity field Ft Ž m. of Mt at m g Mt is Ft Ž m . s Vt Ž m . Nt Ž m . , where Vt Ž m . s Ž Ž drdt . r It . Ž Iy1 t Ž m . . ? Nt Ž m . . We say that the trans¨ ersal ¨ elocities of Mt are bounded on bounded sets if for each ball B : R n the set of norms
5 Ft Ž m . 5 :
t g R, m g Mt l B 4
is bounded. A path of following f : w t 0 , `. ª R n of finite length always converges as t ª ` Žsee Lemma 5.7. and we shall denote its limit by f` . THEOREM 5.2. If the trans¨ ersal ¨ elocities of Mt are bounded on bounded sets, then e¨ ery path of following of finite length con¨ erges to S t , i.e., lim d Ž f Ž t . , St . s lim d Ž f` , St . s 0.
tª`
tª`
Remark. If in the example of following in Section 2 Žrotating half-plane. we allow the angular velocities of rotation to be arbitrarily large then it is not hard to find paths of following which have finite length but do not converge to St . Proof of Theorem 5.2. This requires a few preparations which come in Lemmas 5.3]5.7. Let us observe first that if a point moves so as to stay on Mt then the component of its velocity perpendicular to Mt equals Ft . LEMMA 5.3. For e¨ ery right-regular map m: V ª R n satisfying mŽ t . g Mt for all t we ha¨ e mXr Ž t . ? Nt Ž m Ž t . . s Vt Ž m Ž t . . .
S. ´ SWIERCZKOWSKI
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Proof. We have here a composite motion: mŽ t . moves relative to Mt and Mt moves in R n. Thus for any time t the velocity, mXr Ž t . is the sum of the velocity of mŽ t . relative to Mt which is orthogonal to Nt Ž mŽ t .. and the velocity of the point of Mt occupying at time t the position mŽ t .. The latter velocity projects on Nt Ž mŽ t .. as Vt Ž mŽ t .., by Definition 5.1. LEMMA 5.4. Let f : U ª R n be a path of following and put mŽ t . s p t Ž f Ž t .. whene¨ er f Ž t . f St . Then, for ¨ ery t 0 g U such that f Ž t 0 . f St 0 ,
Ž drdt . r d Ž f Ž t . , St . s y¨ q Vt Ž m Ž t . . . Proof. By continuity, f Ž t . f St Ži.e., d Ž f Ž t ., St . ) 0. holds on some right neighborhood V of t 0 . Hence mŽ t . is defined for all t g V. Moreover, it is clear that m Ž t . s p t Ž f Ž t . . s It (p ( Iy1 t f Ž t. ,
t g V,
holds for the analytic projection map p : R _ S ª M ŽTheorem B.3.. It follows that mŽ t . is right regular. Thus d Ž f Ž t ., St . s 5 f Ž t . y mŽ t .5 is right regular on V and we have n
Ž drdt . r Ž d Ž f Ž t . , St . . s 2 Ž mXr Ž t . y f rX Ž t . . ? Ž m Ž t . y f Ž t . . . 2
However, by Ž). in Section 4 and Definition 3.2, yf rX Ž t . ? Ž m Ž t . y f Ž t . . s yf rX Ž t . ? 5 m Ž t . y f Ž t . 5 Nt Ž m Ž t . . s y¨ 5 m Ž t . y f Ž t . 5 s y¨ d Ž f Ž t . , S t . . Further, by Ž). and Lemma 5.3, mXr Ž t . ? Ž m Ž t . y f Ž t . . s mXr Ž t . ? 5 m Ž t . y f Ž t . 5 Nt Ž m Ž t . . s Vt Ž m Ž t . . d Ž f Ž t . , St . . From these equations we conclude that 2
Ž drdt . r Ž d Ž f Ž t . , St . . s 2 y¨ q Vt Ž m Ž t . d Ž f Ž t . , St . , which obviously also equals 2 d Ž f Ž t ., S t .Ž drdt . r d Ž f Ž t ., St .. Division by 2Ž d Ž f Ž t ., St . ) 0 gives the required identity. LEMMA 5.5. Let e: w t 0 , `. ª R be a continuous, nonnegati¨ e function such that for each t there is a t ) t satisfying eŽ t . s 0, and suppose further that eŽ t . ¢ 0 as t ª `. Then there is an a ) 0 and an infinite sequence of disjoint inter¨ als w t n , t n x with t n - t n such that, for each n s 1, 2, . . . ,
Ž q. e Ž tn . s a ,
e Ž t n . s 0,
0 - eŽ t . F a
for t n F t - t n .
Proof. An elementary exercise in limits and continuity.
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A MODEL OF FOLLOWING
LEMMA 5.6. If the function e: w t 0 , `. ª R in Lemma 5.5 is right regular on the open set t: eŽ t . ) 04 and for the number a satisfying Žq. there are constants ˜t and C such that
Ž qq.
0 - e Ž t . F a « < eXr Ž t . < - C,
for e¨ ery t ) t :
then each of the inter¨ als w t n , t n x in Lemma 5.5 with t n ) ˜t has length at least arC. Proof. Let t n ) ˜t. Then by Žq., 0 - eŽ t . F a for every t in w t n , t n ., and hence < eXr Ž t .< - C. Consequently, for t g w t n , t n ., e Ž t . y e Ž tn . s
t X
Ht e Ž t . dt G yC Ž t y t . . r
n
n
As t ª t n , we get, by continuity, e Ž t n . y e Ž t n . G yC Ž t n y t n . . Since eŽ t n . s 0 and eŽ t n . s a , it follows that t n y t n G arC. LEMMA 5.7.
E¨ ery path of following of finite length con¨ erges.
Proof. Let f : w t 0 , `. ª R n be a path of following of finite length. Thus `
X r
Ht 5 f Ž t . 5 dt - ` 0
and 5 f Ž t 2 . y f Ž t1 . 5 s
t2 X
Ht
1
f r Ž t . dt F
t2
X r
Ht 5 f Ž t . 5 dt 1
by Appendix A. It follows that 5 f Ž t 2 . y f Ž t 1 .5 ª 0 as t 1 , t 2 ª `. Proof of Theorem 5.2 Ž Concluded.. Suppose f : w t 0 , `. ª R n is a path of following of finite length and d Ž f Ž t ., St . ¢ 0 as t ª `. To deduce from this a contradiction, we put eŽ t . s d Ž f Ž t ., St . and claim then that all the assumptions of Lemmas 5.5 and 5.6 are satisfied. Indeed, we cannot have a t9 such that d Ž f Ž t ., St . ) 0 for all t ) t9 because this would imply 5 f rX Ž t .5 s ¨ for all t ) t9, contradicting the finiteness of the length of this path. Thus Žq. holds for some a ) 0 and w t n , t n x, n s 1, 2, . . . . Now let ˜t be such that 5 f Ž t . y f` 5 - a for all t ) ˜t Žwhere f` s lim t ª` f Ž t .; see Lemma 5.7.. We check that the assumptions of Lemma 5.6 are satisfied. If eŽ t . ) 0, define mŽ t . s p t Ž f Ž t ... Then e Ž t . s d Ž f Ž t . , St . s 5 f Ž t . y m Ž t . 5
S. ´ SWIERCZKOWSKI
556
is right regular Žsee the proof of Lemma 5.4.. Now, to verify the assumption Žqq., let t ) ˜t be such that 0 - eŽ t . F a , i.e., 5 f Ž t . y mŽ t .5 F a . Then, by the choice of ˜t, we have 5 mŽ t . y f` 5 F 2 a . By Lemma 5.4, eXr Ž t . s y¨ q Vt Ž mŽ t .., and since Vt Ž x . is bounded on bounded sets and mŽ t . is in the ball of center f` and radius 2 a , it follows that there is a constant C such that < eXr Ž t .< - C. Thus all the assumptions of Lemma 5.6 are satisfied and we have t n y t n G arC if t n ) ˜t. Moreover, eŽ t . s d Ž f Ž t ., St . is positive on w t n , t n .. Thus 5 f rX Ž t .5 s ¨ by Definition 3.2. It follows that
Ht t 5 f Ž t . 5 dt s ¨ Ž t n
X r
n
y t n . G ¨ arC
for every t n ) ˜t ,
n
contradicting the finite length of the path f Ž t .. For the last theorem we consider for each u ) 0 the u-swelling S Ž u . of S given by S Ž u . s x q x9: x g S and 5 x9 5 F u 4 . Evidently the set S Ž u . is open and convex. We shall show that its boundary S Ž u . is an analytic hypersurface in R n. Thus, for each fixed u G 0, we may consider the model of following with the moving set StŽ u . s It S Ž u .. THEOREM 5.8. Suppose the trans¨ ersal ¨ elocities of Mt are bounded on bounded sets and for some u G 0 there exists a path of S tŽ u .-following of finite length. Then the trans¨ ersal ¨ elocities of e¨ ery MtŽ u ., u G 0, are bounded on bounded sets. Moreo¨ er, there exists a u G 0 such that Ži. For e¨ ery u ) u , e¨ ery path of StŽ u .-following has finite length. Žii. For e¨ ery 0 F u - u , e¨ ery path of StŽ u .-following has infinite length. Proof. We would like to apply the previous results to StŽ u .. Thus it is necessary to show that each boundary StŽ u . is an analytic hypersurface MtŽ u . in R n. Let u ) 0 be fixed. We consider the analytic manifold M = Ry, where Rys s g R: s - 04 and the map K : M = Ryª R n _ S given by K Ž m, s . s m q sN Ž m. for each m g M, s - 0. By Appendix B, K is an analytic bijection. Since M = yu 4 is an analytic submanifold of M = Ry of codimension 1, its image by K is a hypersurface in R n. We denote this image by M Ž u . and show that M Ž u . s S Ž u .. Proof of M Ž u . : S Ž u .. By definition, M Ž u . s m y u N Ž m.: m g M 4 . Let m g M. By Lemma B.1, m q d N Ž m. g S for all sufficiently small
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557
d ) 0; thus m y Ž u y d . N Ž m. g S Ž u .. It follows that m y u N Ž m. is in the closure of S Ž u .. But also, arbitrarily close to m y u N Ž m. there are points not belonging to S Ž u .. Indeed, let us show that m y Ž u q d . N Ž m. f S Ž u . for d ) 0. Assuming the contrary, we get, for some 5 x9 5 F u , m y Ž u q d . N Ž m . q x9 g S. Then, by Ž3.1., ŽyŽ u q d . N Ž m. q x9. ? N Ž m. ) 0, whence 5 x9 5 ) u q d and this is a contradiction. It follows that m y u N Ž m. g d S Ž u ., i.e., the inclusion is proved. Proof of S Ž u . : M Ž u .. Consider any p g S Ž u .. Then p f S Ž u ., hence p f S and m s p Ž p . g M. By ŽB.2., p s m y 5 p y m5 N Ž m. .
Ž (.
Moreover, 5 p y m 5 s d Ž p, S .. We claim now that 5 p y m 5 s u . Indeed, arbitrarily close to p there are points q g S Ž u ., i.e., satisfying d Ž q, S . - u . Thus d Ž p, S . F u , i.e., 5 p y m 5 F u . On the other hand, we cannot have 5 p y m 5 - u because then the triangle inequality would immediately imply that p g S Ž u .. Thus 5 p y m 5 s u and Ž(. implies that p s m y u N Ž m., that is, p g M Ž u .. We now put MtŽ u . s It M Ž u . and show that the transversal velocities of MtŽ u . are bounded on bounded sets. There is a natural embedding of M in M = Ry given by m ¬ Ž m, yu .. Combining this map with K, we get an analytic bijection between hypersurfaces k Žu . : M ª M Žu . ,
where k Ž u . Ž m . s m y u N Ž m . .
Let N Ž u . be the unit normal vector field on M Ž u ., directed into S Ž u .. It is easily checked that, for any smooth curve g Ž t . in M, its image under k Ž u ., i.e., g Ž t . y u N Ž g Ž t .., is orthogonal to N Ž g Ž t ... This shows that the vector fields N and N Ž u . are identical at m and k Ž u . Ž m., respectively. Let NtŽ u . denote the unit normal inward-directed vector field on MtŽ u . and let k tŽ u . Ž p . s p y u Nt Ž p . for all p g Mt . By the above, k tŽ u . : Mt ª MtŽ u . is an analytic bijection between hypersurfaces and NtŽ u . Ž k tŽ u . Ž p .. is Nt Ž p .. Taking any right-regular curve mŽ t . in Mt and its image by k tŽ u ., i.e., mŽ t . y u Nt Ž mŽ t .., we check easily that the derivatives mXr Ž t . and Ž drdt . r k tŽ u . Ž mŽ t .. have identical projections on Nt Ž mŽ t ... By Lemma 5.3, this implies that the transversal velocity of Mt at p is the same as the transversal velocity of MtŽ u . at k tŽ u . Ž p .. Since the distance between these two points is u , it follows that the transversal velocities of MtŽ u . are bounded on bounded sets. Proof of Theorem 5.8 Ž Concluded.. We claim that if, for some u G 0, some path f Ž t . of StŽ u .-following has finite length then, for each u 9 ) u ,
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558
StŽ u 9. is pinned by a ball from some t onwards. Indeed, by Theorem 5.2, d Ž f` , StŽ u . . ª 0 as t ª `, whence, given r - u 9 y u , the ball of center f` and radius r is contained in StŽ u 9. for sufficiently large t. We conclude now by Theorem 4.2 that if some path of StŽ u .-following has finite length then e¨ ery path of StŽ u 9.-following for each u 9 ) u has finite length. Thus the required u is given by
u s g.l.b. u G 0: some path of StŽ u .-following has finite length 4 .
APPENDIX A: INTEGRABILITY OF RIGHT-REGULAR FUNCTIONS Let f : U ª R n be right regular ŽDefinition 3.1. and consider its right derivative f rX : U ª R n. PROPOSITION.
If f rX is bounded on w t 0 , t 1 x : U then
Ži. the functions f rX and 5 f rX 5 are Riemann integrable on w t 0 , t 1 x and
Ž ii .
f Ž t1 . y f Ž t 0 . s
t1 X
Ht
f r Ž t . dt.
0
Proof of Ži.. Let D denote the set of discontinuities of f rX and let us assign to each t g D a right neighborhood V of t in U such that f rX is continuous on V _ t 4 . ŽBy assumption, on some right neighborhood V of t, f is given by a power series.. Clearly, in any such assignment, right neighborhoods corresponding to different elements of D are disjoint. Thus D is countable, and hence of Lebesgue measure zero. So f rX is Riemann integrable, by the Lebesgue criterion for Riemann integrability of a bounded function Žsee w1x.. Also 5 f rX 5 is Riemann integrable because the set of discontinuities of 5 f rX 5 is a subset of the set of discontinuities of f rX . Proof of Žii.. Let T be the set of t g w t 0 , t 1 x such that f Ž t . y f Ž t0 . s
t X
Ht f Ž t . dt , r
0
whenever t 0 F t - t. Obviously, T s w t 0 , t9. for some t9 g Ž t 0 , t 1 x. Since f rX is bounded, the above integral is a continuous function of t, whence by the continuity of f, f Ž t9 . y f Ž t 0 . s
t9 X
Ht
0
f r Ž t . dt .
Ž A.1.
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Thus, if t9 s t 1 , we are done. Suppose t9 - t 1. Then, since f rX Žt . is given for t ) t9 and t close to t9 by a power series, there is a t0 g Ž t9, t 1 . such that f Ž t0 . y f Ž t9 . s
t0 X
Ht9
f r Ž t . dt .
This, jointly with ŽA.1., contradicts the definition of t9.
APPENDIX B: CONVEXITY IMPLICATIONS Let S and M be as in Section 1. Given m g M and a ) 0, we shall consider the line L m s m q sN Ž m.: s g R4 and its open interval Lam s m q sN Ž m.: 0 - s - a 4 . LEMMA B.1.
For e¨ ery m g M there is an a ) 0 such that Lam : S.
Proof. Consider first the case when n s 2, i.e., M is a convex curve in R 2 and the hyperplane P Ž m. is the line tangent to m at M. Then the whole of S lies to one side of P Ž m. and L m is perpendicular to M at m Ži.e., to P Ž m... Thus, arbitrarily close to m, there are pairs of points of M Žand hence of S . which are on opposite sides of L m . Hence, by convexity, there are arbitrarily close to m points of L m l S. These must be of the form m q sN Ž m., where s ) 0, by Ž3.1.. Using convexity again, we obtain Lam : S for some a ) 0. If n ) 2, it suffices to repeat the above reasoning in a two-dimensional plane passing through L m and some point of S. By Ž3.1., no point m q sN Ž m., where m g M and s - 0, can belong to S. Thus the correspondence Ž m, s . ¬ m q sN Ž m. yields a mapping K : M = Ryª R n _ S Žwhere Rys s g R: s - 04.. Recall that, for p f S, p Ž p . denotes the point of S Žthus of M . which is closest to p. PROPOSITION B.2.
K is a bijection and Ky1 is gi¨ en by
Ky1 Ž p . s Ž p Ž p . , y5p Ž p . y p 5 .
for all p f S.
Ž B.1.
Proof. It is not difficult to see that, for every p f S, p Ž p . y p is perpendicular to M at p Ž p ., and thus p Ž p . y p s "5p Ž p . y p 5 N Žp Ž p ... Applying Lemma B.1, we see that we must take the q sign here, i.e., p s p Ž p . y 5p Ž p . y p 5 N Ž p Ž p . . ,
Ž B.2.
or, equivalently, p s K Žp Ž p ., y5p Ž p . y p 5.. This shows that K is surjective. On the other hand, if p s m q sN Ž m. and s - 0 then Ž3.1. implies
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S. ´ SWIERCZKOWSKI
that p f S. Also, by Ž3.1. and Lemma B.1, P Ž m. separates S from p. Thus m s p Ž p . and s s y5p Ž p . y p 5. This shows the injectivity of K and that the inverse map is given by ŽB.1.. In proving Theorem B.3 we shall use tangent vectors and principal curvatures with the following notation w2x. The coordinate functions x 1 , . . . , x n define at each m g R n the base Ž r x 1 . m , . . . , Ž r x n . m of the tangent space Tm R n. We shall identify Tm R n with R n by identifying the tangent vector X s Ý nis1 a i Ž r x i . m with the n-tuple Ž a1 , . . . , a n .. If J: M ª R n is the inclusion map, we use the Jacobian map J#: Tm M ª Tm R n Žwhich is nonsingular by assumption. to identify the tangent space Tm M with its image J#ŽTm M . in Tm R n. Combining this with the previous identification, we view each Tm M as an Ž n y 1.-dimensional subspace of R n. Recall that a nonzero tangent vector X g Tm M is called principal if the derivative Dx N of the unit normal vector field N in the direction of X is a scalar multiple of X. For a principal X we shall write Dx N s yk X, where k g R is the principal curvature of M at m in the direction of X. The convexity of M implies that all principal curvatures of M are nonnegative. THEOREM B.3. analytic.
Each of the maps K, Ky1 , and p : R n _ S ª M is
Proof. K is analytic because the unit normal vector field N on M is analytic. Once we show that Ky1 is analytic, we conclude from ŽB.1. that p is analytic. We wish to apply now the inverse function theorem to K. That is, we are going to show that the Jacobian map K# is everywhere nonsingular and this will lead to the conclusion that Ky1 is analytic. So our remaining task is to find n tangent vectors to M = Ry at Ž m, s . whose images by K# are linearly independent. We shall choose these vectors of the form
Ž X1 , 0 . , . . . , Ž X ny1 , 0 . , Ž 0, Ž r t . s . , where X 1 , . . . , X ny1 g Tm M and Ž r t .s is the basis vector of the tangent space to R at s. For the X 1 , . . . , X ny1 we take any basis of Tm M composed of principal vectors, i.e., vectors X i such that each covariant derivative D X i N of N in the direction X i is equal to yk i X i , where k i G 0 is a principal curvature at m w2, p. 24x. Thus we have to show that if Yi s K#Ž X i , 0. for i - n and Yn s K#Ž0, Ž r t .s . then Y1 , . . . , Yn are independent. To find Yi , let hi Ž t . be curves in M such that hi Ž0. s m, hiX Ž0. s X i ,
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561
i - n, and let hnŽ t . s s q t, so that hnŽ0. s s, hnX Ž0. s Ž r t .s . Then, for i - n, Yi s s
d dt d dt
K Ž hi Ž t . , s .
ts0
hi Ž t . q sN Ž hi Ž t . .
ts0
s X i q sD X i N s X i y sk i X i s Ž 1 y sk i . X i and Yn s s
d dt d dt
K Ž m, hn Ž t . .
ts0
m q Ž s q t . N Ž m.
s N Ž m. . ts0
It follows now from s - 0 and k i G 0 that each Yi for i - n is a nonzero multiple of X i ; moreover, Yn is orthogonal to all other Yi . This proves the independence of Y1 , . . . , Yn . ACKNOWLEDGMENT This paper owes its existence to Jan Mycielski; the research originated in conversations with him and some results were in that way jointly obtained.
REFERENCES 1. T. M. Apostol, ‘‘Mathematical Analysis,’’ Addison]Wesley, Reading, MA, 1974. 2. N. Hicks, ‘‘Notes on Differential Geometry,’’ Van Nostrand, Princeton, NJ, 1964. 3. S. Kaczmarz, Angenahrte Auflosung von Systemen linearer Gleichungen, Bull. Acad. ¨ ¨ Polon. Sci. Lett. Ser. A 35 Ž1937., 355]357. 4. J. Mycielski, Linear dynamic approximation theory, J. Approx. Theory 25 Ž1979., 369]383. 5. J. Mycielski and S. Swierczkowski, A model of the neocortex, Ad¨ . in Appl. Math. 9 Ž1988., 465]480. 6. J. van Tiel, ‘‘Convex Analysis,’’ Wiley, Chichester, 1984.