The Eigenvalue Problem for Linear and Affine

8/10/2019 The Eigenvalue Problem for Linear and Affine

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The Eigenvalue Problem for Linear and Affine

Iterated Function SystemsMichael Barnsleya

a Mathematical Sciences Institute, Australian National UniversityCanberra, ACT 0200, [email protected]

Andrew Vinceb (corresponding author)b University of Florida, Department of Mathematics

358 Little Hall, PO Box 118105, Gainesville, FL 32611-8105, [email protected]

phone: 352-392-0281 ext 246, fax: 352-392-4958

Abstract

The eigenvalue problem for a linear function L centers on solving the eigen-equation Lx= x. This paper generalizes the eigenvalue problem from a singlelinear function to an iterated function system Fconsisting of possibly an infinitenumber of linear or affine functions. The eigen-equation becomes F(X) = X,

where >0 is real, Xis a compact set, andF(X) =fFf(X). The main result

is that an irreducible, linear iterated function system Fhas a unique eigenvalue equal to the joint spectral radius of the functions in F and a correspondingeigenset Sthat is centrally symmetric, star-shaped, and full dimensional. Resultsof Barabanov and of Dranishnikov-Konyagin-Protasov on the joint spectral radiusfollow as corollaries.

Keywords: eigenvalue problem, iterated function system, joint spectral radiusMathematical subject codes: 15A18; 28A80

1 Introduction

Let L : R2 R2 be a linear map with no nontrivial invariant subspace, equivalentlyno real eigenvalue. We use the notationL(X) := {Lx : x X}. Although L has noreal eigenvalue, L does have an eigen-ellipse. By eigen-ellipsewe mean an ellipse E,centered at the origin, such that L(E) = E, for some real > 0. An example of aneigen-ellipse appears in Example1of Section2and in Figure1. Although easy to prove,the existence of an eigen-ellipse appears not to be well known.

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arXiv:1004.50

40v1[math.MG]

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WhenFconsists of a single linear map on R2, the eigen-ellipse is an example of aneigenset. Section2contains other examples of eigenvalues and eigensets of linear IFSs.Section3contains background results on the joint spectral radius of a set of linear mapsand on contractive IFSs. Both of these topics are germane to the investigation of theIFS eigenvalue problem. Section4contains the main result on the eigenvalue problem

for a linear IFS.

Theorem 2 A compact, irreducible, linear IFSF has exactly one eigenvalue which isequal to the joint spectral radius (F) of F. There is a corresponding eigenset that iscentrally symmetric, star-shaped, and full dimensional.

IfF = {Rn;fi, i I} is an IFS, let F := 1F = {R

n; 1fi, i I}. Another way toview the above theorem is to consider the family {F : > 0} of IFSs. If > (F),then the attractor of F (defined formally in the next section) is the trivial set {0}.If < (F), then F has no attractor. So = (F) can be considered as a phasetransition, at which point a somewhat surprising phenomenon occurs - the emergence

of the centrally symmetric, star-shaped eigenset.Theorems of Dranisnikov-Konyagin-Protasov and of Barabanov follow as corollaries

of Theorem2. These results are discussed in Section5.

No such transition phenomenon occurs in the case of an affine, but not linear, IFS.A result for the affine case is the following, whose proof appears in Section6.

Theorem 3 For a compact, irreducible, affine, but not linear, IFS F, a real number > 0 is an eigenvalue if > (F) and is not an eigenvalue if < (F). There areexamples where(F) is an eigenvalue and examples where it is not.

The transition phenomenon resurfaces in the context of projective IFSs, which willbe the subject of a subsequent paper.

2 Examples

Example 1 Figure1shows the eigen-ellipse for the the IFSF = (R2; L), where

L=

65.264 86.116156.98 62.224

.

The eigenvalue is approximately97.23.

Example 2 Figure2shows an eigenset for the IFSF = (R2; L1, L2), where

L1 =

10 108 0

, L2 =

8 0

10 10

.

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Figure 1: The eigen-ellipse for Example1

Figure 2: The eigenset of Example 2

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Figure 3: The eigenset of Example 3

The eigenvalue is approximately14.9. The picture on the right is the image of the pictureof the left after applying both transformations, then shrinking the result about its centerby a factor 14.9. The green and brown colors help to show how the image is acted onby the two transformations. The dots are an artifact of rounding errors, and serve toemphasize that the pictures are approximate.

Example 3 Figure3shows the eigenset for the the IFSF = (R2; L1, L2), where

L1 =

0.02 00 1

, L2 =

0.0594 1.980.495 0.01547

.

The eigenvalue is approximately1.

3 Background

This section concerns the following three basic notions: (1) the joint spectral radius ofan IFS, (2) contractive properties of an IFS, and (3) the attractor of an IFS. Theorems 4and5 provides the relationship between these three notions for a linear and an affineIFS, respectively.

3.1 Norms and Metrics

Any vector norm on Rn induces a matrix norm on the space of linear maps takingRn to Rn:

L = max

Lx

x : x Rn

.

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Since it is usually clear from the context, we use the same notation for the vector norm asfor the matrix norm. This induced norm is sub-multiplicative, i.e., L L L Lfor any linear mapsL, L.

Two norms 1and 2 are equivalentif there are positive constantsa, bsuch thatax1 x2 bx1 for all x R

n. Two metrics d1(, ) and d2(, ) are equivalent

if there exist positive constants a, b such that a d1(x, y) d2(x, y) b d1(x, y) for allx, y Rn. It is well known that any two norms on Rn are equivalent [1]. This impliesthat any twon nmatrix norms are equivalent. Any norm on Rn induces a metricd(x, y) = x y. Therefore any two metrics induced from two norms are equivalent.

A set B Rn is called centrally symmetric ifx B whenever x B. A convexbodyin Rn is a convex set with nonempty interior. IfCis a centrally symmetric convexbody, define the Minkowski functionalwith respect to C by

xC= inf{ 0 : x C}.

The following result is well known.

Lemma 1 The Minkowski functional is a norm onRn. Conversely, any norm onRn is the Minkowski functional with respect to the closed unit ball{x : x 1}.

Given a metricd(, ), there is a corresponding metric dH, called theHausdorff metric,on the collection H(Rn) of all non-empty compact subsets ofRn:

dH(B, C) = max

supbB

infcC

d(b, c), supcC

infbB

d(b, c)

.

3.2 Joint Spectral Radius

The joint spectral radius of a set L = {Li, i I} of linear maps was introduced byRota and Strang [2] and the generalized spectral radius by Daubechies and Lagarias[3]. Berger and Wang [4] proved that the two concepts coincide for bounded sets oflinear maps. The concept has received much attention in the recent research literature;see for example the bibliographies of [5] and [6]. What follows is the definition of the

joint spectral radius ofL. Let k be the set of all words i1i2 ik, of length k , whereij I, 1 j k. For = i1i2 ik k, define

L :=Li1 Li2 Lik .

A set of linear maps is bounded if there is an upper bound on their norms. Note that

ifL is compact, then L is bounded. For a linear mapL, let (L) denote the ordinaryspectral radius, i.e., the maximum of the moduli of the eigenvalues ofL.

Definition 3 For any setLof linear maps and any sub-multiplicative norm, the jointspectral radius ofL is

= (L) := lim supk

1/kk where k := sup

k

L.

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Thegeneralized spectral radiusofL is

= (L) := lim supk

1/kk where k := sup

k

(L).

The following are well known properties of the joint and generalized spectral radius [6].

1. The joint spectral radius is independent of the particular sub-multiplicative norm.

2. For an IFS consisting of a single linear map L, the generalized spectral radius isthe ordinary spectral radius ofL.

3. For any real >0 we have (L) = (L) and (L) = (L).

4. For allk 1 we have1/kk

1/kk ,

independent of the norm used to define .

5. IfL is bounded, then the joint and generalized spectral radius are equal.

From here on we always assume that L is bounded. So, in view of property 5, wedenote by(L) the common value of the joint and generalized spectral radius.

IfF is an affine IFS, then each f Fis of the form f(x) =Lx+a, where L is thelinear partand a is the translational part. Let LFdenote the set of linear parts ofF.

Definition 4 The joint spectral radius of an affine IFS F is the joint spectralradius of the setLFof linear parts ofFand is denoted(F).

Definition 5 A set{Li, i I} of linear maps is calledreducible if these linear mapshave a common nontrivial invariant subspace. The set isirreducible if it is not re-ducible. An IFS isreducible (irreducible) if the set of linear parts is reducible (irre-ducible).

As shown in [6], a set of linear maps is reducible if and only if there exists an invertiblematrixTsuch that each Li can be put simultaneously in a block upper-triangular form:

T1LiT =

Ai

0 Bi

,

withAiand B

isquare, and

is any matrix with suitable dimensions. The joint spectral

radius(F) is equal to max (({Ai}), ({Bi})).

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3.3 A Contractive IFS

Definition 6 (contractive IFS) A functionf : X Xis acontractionwith respectto a metric d if there is an s,0 s < 1, such that d(f(x), f(y)) s d(x, y) for allx, y Rn. An IFS F = (X; fi, i I) is said to be contractive if there is a metricd : Rn Rn [0, ), equivalent to the standard metric onRn, such that eachf Fis a contraction with respect to d.

Definition 7 (attractor) A nonempty compact setA Rn is said to be an attractorof the affine IFSF if

1. F(A) =A and

2. limk Fk(B) = A, for all compact setsB Rn, where the limit is with respect

to the Hausdorff metric.

Basic to the IFS concept is the relationship between the existence of an attractor and

the contractive properties of the functions of the IFS. The following result makes thisrelationship explicit in the case of a linear IFS. A proof of this result for an affine, butfinite, IFS appears in [8]. For completeness we provide the proof for the infinite linearcase. The notation int(X) will be used to denote the interior of a subset X ofRn. Thenotationconv(X) is used for the convex hull of the set X. In Rn the Minkowski sum andscalar product are defined byY+ Z= {y + z : y Y, z Z} and Y = { y : y Y},respectively.

Theorem 4 For a compact, linear IFSF = (Rn; Li, i I)the following statements areequivalent.

1. [contractive] There exists a norm on Rn and an 0 s < 1 such thatLx s x for allL F and allx Rn.

2. [F-contraction] The map F : H(Rn) H(Rn) defined byF(B) =

LFL(B)is a contraction with respect to a Hausdorff metric.

3. [topological contraction]There is a compact, centrally symmetric, convex bodyCsuch thatF(C) int(C).

4. [attractor] The origin is the unique attractor ofF.

5. [JSR] (F)< 1.

Proof: (attractor topological contraction) LetAbe the attractor ofF. LetA= {x Rm :dH({x} , A) } denote the dilation ofA by radius >0. By the definition of the

attractor, limk dH(Fk(A), A) = 0,so there is an integer m so thatdH(F

m(A1), A)0, the metric dC is equivalent to the standard metric.

(contractive F-contraction) In the case of an IFS F = (Rn; fi, i I), where I isfinite (and the fi are assumed only to be continuous), this is a basic result whose proofcan be found is most texts on fractal geometry, for example [7]. Since F is assumedcontractive,

supd(fi(x), fi(y))d(x, y)

: x =y =si< 1,for each i I. The only sticking point in extending the proof for the finite IFS case tothe infinite IFS case is to show that sup{si : i I}


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contradicting the assumption that each function in Fis a contraction.

(F-contraction attractor) The existence of a unique attractor follows directly fromthe Banach contraction mapping theorem. WhenF is linear, uniqueness immediatelyimplies that the attractor is {0}.

(contractive JSR) First assume thatF is contractive. Hence there is an 0 s


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Theorem 5 If F = (Rn; fi, i I) is a compact, affine IFS, then the following state-ments are equivalent.

1. [contractive] The IFSF is contractive onRn.

2. [topological contraction] There exists a compact set C such that F(C)

int(C).3. [attractor] F has a unique attractor, the basin of attraction beingRn.

4. [JSR] (F)< 1.

Proof: As explained above, we prove only the equivalence of statement (4) to the otherstatements. Assuming(F) 0. Moreover,

ifX and X

are eigensets corresponding to the same eigenvalue, then X X

is also acorresponding eigenset. For an eigenvalue of a linear IFS, call a corresponding eigensetXdecomposable ifX = X1 X2, where X1 = X and X2 = Xare also correspondingeigensets. Call eigenset XindecomposableifSis not decomposable.

Example. It is possible for a linear IFS to have infinitely many indecomposableeigensets corresponding to the same eigenvalue. Consider F = {R2; L1, L2}where

L1 =

0 11 0

, L2=

1 00 0.5

.

Let

S(r1, r2) = { (r1, r2/2k

), (r1, r2/2k

), (r2/2k

, r1, ), (r2/2k

, r1, ) : k 0}.It is easily verified that, for anyr1 r2> 0, the setS(r1, r2) is an eigenset correspondingto eigenvalue 1. In addition, the unit square with vertices (1, 1), (1, 1), (1, 1), (1, 1)is also an eigenset corresponding to eigenvalue 1.

The proof of the following lemma is straightforward. A setB Rn is called starshaped if x B for all for all x B and all 0 1.

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Lemma 2 1. If{Ak} is a sequence of centrally symmetric, convex, compact sets andA is a compact set such that limk Ak =A, thenA is also centrally symmetricand convex.

2. IfF is a compact, linear IFS, B a centrally symmetric, convex, compact set andA= limk F

k(B), thenA is a centrally symmetric, star-shaped, compact set.

Lemma 3 IfF is an compact, irreducible, linear IFS with(F) = 1, then there existsa compact, centrally symmetric, convex bodyA such thatF(A) A.

Proof: Since, for each k 2, we have ((1 1k

)F) = 1 1k

0. So, without loss of generality, it can be assumed that max{x :x Ak} = 1for all k 2. Since the sequence of sets{Ak} is bounded in H(R

n), this sequence hasan accumulation point, a compact set A. Therefore, there is a subsequence {Aki} suchthat limi Aki =A with respect to the Hausdorff metric. Since

1 1

ki

F(Aki) int(Aki),

it is the case that

1 1ki

f(Aki) int(Aki) for all f F. From this is is straightfor-

ward to show that f(A) A for all f F and hence that F(A) A. Moreover, by

Lemma2, since the Aki are centrally symmetric and convex, so is A. Notice also thatAis a convex body, i.e., has nonempty interior; otherwise A spans a subspace E Rn

with dimE < nandF(A) Aimplies F(E) E, contradicting that F is irreducible.

The affine spanaff(B) of a set B is the smallest affine subspace of Rn containingB. Call a set B Rn full dimensional if dim(aff(B)) = n. Given an affine IFSF = (Rn; fi, i I) let

F =

Rn;

1

fi, i I

.

Lemma 4 If an irreducible, affine IFS F has an eigenset X, then X must be fulldimensional.

Proof: Suppose that F(X) = X, i.e. F(X) = X. For x X, let g be a translationbyx. For the IFSF, let Fg = {R

n; gf g1, f F}. IfY = g(X), then 0 Y andFg(Y) =Y. In particular, Yis full dimensional if and only ifXis full dimensional, andthe affine span ofYequals the ordinary (linear) span E=span(Y) ofY . Moreover, the

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linear parts of the affine maps in Fg are just scalar multiples of the linear parts of theaffine maps in F. ThereforeFg is irreducible if and only ifF is irreducible.

Letf(x) =Lx + abe an arbitrary affine map inFg. FromFg(Y) Y Eit followsthatL(Y) + a= f(Y) E. Since 0 Y, also a = L(0)+ a= f(0) Y E. ThereforeL(Y) a+E = E. Since E = span(Y), also L(E) E. Because this is so for all

f Fg, the subspace Eis invariant under all linear parts of maps in Fg. Because Fg isirreducible,dim(E) =n. Therefore Y, and hence X, must be full dimensional.

Lemma 5 IfF = {Rn; Li, i I} is a bounded linear IFS, then there is an >0 suchthat F = {Rn; Li, i I} is contractive.

Proof: By the boundedness ofFthere is anRsuch that, for anyL F, Lxx

L Rfor all x Rn. Therefore, ifDr denotes a disk of radius r centered at the origin, thenF(D1) DR. Hence

12R

F(D1) int(D1). By Theorem4the IFS 12R

F is contractive.

Proof of Theorem2: GivenF = (Rn; Li, i I), consider the family {F} of IFSs for > 0. Recall that F =

Rn; 1

fi, i I

.

It is first proved that Fhas no eigenvalue > (F). By way of contradiction assumethat > (F), which implies that (F) < 1. According to Theorem4the IFS F iscontractive. By Corollary1the only invariant set ofF is {0}, which means that theonly solution to the eigen-equation F(X) = X is X ={0}. But by definition, {0} isnot an eigenset.

The proof thatFhas no eigenvalue < (F) is postponed because the more generalaffine version is provided in the proof of Theorem3in Section6.

We now show that(F) is an eigenvalue ofF. Again letF = 1

F, so that(F) = 1.

WithA as in the statement of Lemma 3, consider the nested intersection

S=k0

Fk (A) = limk

Fk (A).

ThatSis compact, centrally symmetric, and star-shaped follows from Lemma2. Also

F(S) =F

k0

Fk (A)

=k1

Fk (A) =S,

the last equality because A F(A) F

(2)

(A) . FromF(S) =Sit follows thatF(S) = S.

It remains to show that Scontains a non-zero vector. Since A is a convex body anddetermined only up to scalar multiple, there is no loss of generality in assuming that Acontains a ballB of radius 1 centered at the origin. Then

sup { L(x) : k, x B} = k(F) ((F))k = 1.

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So there is a point ak Fk (A) such that ak 1. Ifa is an accumulation point of

{ak}, then a 1, and there is a subsequence {aki}of{ak}such that

limi

aki =a.

Since the sets F(ki) (A) are closed and nested, it must be the case that a F

(ki) (A) for

alli. Thereforea S.

ThatSis full dimensional follows from Lemma 4.

5 Theorems of Dranisnikov-Konyagin-Protasov and

of Barabanov

Important results of Dranisnikov-Konyagin-Protasov and of Barabanov on the jointspectral radius turn out to be almost immediate corollaries of Theorem 2. The first

result is attributed to Dranisnikov and Konyagin by Protasov, who provided a proof in[10]. Barabanovs theorem appeared originally in [11].

Corollary 2 (Dranisnikov-Konyagin-Protasov) If F = (Rn; Li, i I) is a com-pact, irreducible, linear IFS with joint spectral radius := (F), then there exists acentrally symmetric convex bodyKsuch that

conv F(K) =K.

Proof: According to Theorem2there is a centrally symmetric, full dimensional eigensetSsuch that F(S) = S. IfK=conv(S), then Kis also centrally symmetric and

conv F(K) =conv F(conv S) =conv F(S) =conv ( S) = conv S= K.

The second equality is routine to check. Since Sis full dimensional, Kis a convex body,i.e., has nonempty interior.

The original form of the Barabanov theorem is as follows:

Theorem 6 (Barabanov) If a setFof linear maps onRn is compact and irreducible,then there exists a vector norm B such that

for allx and allL F LxB

(F) xB

,

for anyx Rn there exists anL F such that LxB =(F) xB.

Such a norm is called a Barabanov norm. The first property says thatF isextremal,meaning that

LB (F) (2)

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for allL F. It is extremal in the following sense. By property (4) of the joint spectralradius in Section3,

supLF

L (F)

for any matrix norm. Therefore, the joint spectral radius (F) can be characterized

as the infimum over all possible matrix norms of the largest norm of linear maps inF. Since F is assumed compact, the inequality (2 ) cannot be strict for all L F.Hence there exists an L Fwhose Barabanov norm achieves the upper bound (F).Furthermore, the second property in the statement of Barabanovs Theorem says that,for any x Rn, there is such an L achieving a value equal to the joint spectral radiusat the point x. See[12] for more on extremal norms.

In view of Lemma1, Barabanovs theorem can be restated in the following equivalentgeometric form. Here denotes the boundary.

Corollary 3 IfF is a compact, irreducible, linear IFS with joint spectral radius :=(F), then there exists a centrally symmetric convex bodyK such that

F(K) K,

and, for anyx K, there is anL F such thatLx ( K).

Proof: Let Ft = (Rn;Lti, i I), where Lt denotes the adjoint (transpose matrix) ofL.

For a compact set Y, the dualofY(sometimes called the polar) is the set

Y = {z Rn : y, z 1 for all y Y}.

The first two of the following properties are easily proved for any compact set B .

1. B is convex.

2. IfB is centrally symmetric, then so isB.

3. IfL is linear and Lt(S) S, then L(S) S.

To prove the third property above, assume that Lt(S) S. and let x S. Then

x S x, y 1 for ally S

x, Lty 1 for ally S

Lx,y 1 for ally S

Lx S

SinceFis a compact, irreducible, linear IFS, so isFt. LetSbe a centrally symmetriceigenset for Ft as guaranteed by Theorem 2. By properties 1 and 2 above, S is acentrally symmetric convex body. From the eigen-equation Ft(S) = S, it follows that

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1L

t(S) S for all L F. From property 3 above it follows that 1F(S) S or

F(S) S. SettingK=S yields

F(K) K.

Concerning the second statement of the corollary, assume that x K=S. Thenx, y 1 for all y S and x, y = 1 for some y S. Since F(S) = S, the lastequality implies that there is an L F such that 1Lx, z = x,

1L

tz = 1 for some

z S. Now we have 1Lx, y 1 for all y S and 1Lx, z = 1 for some z S.

Therefore, 1Lx S =Kor Lx (K) =( K).

6 The Eigenvalue Problem for an Affine IFS

For an affine IFS F, there is no theorem analogous to Theorem 2. More specifically,there are examples where (F) is an eigenvalue ofFand examples where (F) is not

an eigenvalue ofF. For an example where (F) is an eigenvalue, let

F1= {R2; f}, f(x) =Lx+ (1, 0), L=

0 11 0

.

Note that L, a 90o degree rotation about the origin, is irreducible and (F1) = 1. If S is the unit square with vertices (0, 0), (1, 0), (0, 1), (1, 1), then F1(S) = S. Therefore(F1) = 1 is an eigenvalue ofF1. On the other hand let

F2= {R;f}, f(x) =x+ 1.

In this case (F2) = 1, but it is clear that there exists no compact set X such that

F(X) = X. For the affine case, Theorem3, as stated in the introduction, does holds.The proof is as follows.

Proof of Theorem3: If > (F), then(F)< 1. According to Theorem5, the IFSF has an attractorA so thatF(A) =A. Since at least one function in F is not linear,A = {0}. Since F(A) =A, also F(A) = A. Therefore is an eigenvalue ofF.

Concerning the second statement in the theorem assume, by way of contradiction,that such an eigenvalue < (F) exists, with corresponding eigensetS. ThenF(S) =Sand(F)>1. According to Lemma4, sinceF is assumed irreducible, the eigenset S isfull dimensional. Exactly as in the proof of Lemma4, using conjugation by a translation,there is an affine IFSF and a nonempty compact set S such that

1. F(S) =S,

2. 0 int(conv(S)),

3. The set LF of linear parts of the functions in F is equal to the set LF of linear

parts of the functions in F,

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4. (F) =(F)> 1,

5. F is irreducible

In item 2 above, int(conv(S)) denotes the interior of the convex hull of S. If K =conv(S) and f(x) =Lx+a is an arbitrary affine function such that f(S) S, then

f(K) K.

This follows from the fact thatf(S) S as follows. Ifz K, thenz= x + (1 ) ywhere 0 1 and x, y S. Therefore

f(z) = Lx+ (1 ) Ly+a= (Lx+a) + (1 )(Ly+a)

= f(x) + (1 ) f(y) conv(f(S)) conv(S) =K.

Let r >0 be the largest radius of a ball centered at the origin and contained in KandR the smallest radius of a ball centered at the origin and containing K. Let x K

such that 0< x r. Iff(x) =Lx +ais any affine function such that f(S) S, thenwe claim that Lx R +r. To prove this, first note thatx K. Fromf(K) Kit follows that

Lx+a = f(x) R

Lx+a = L(x) +a = f(x) R

2a = (Lx+a) + (Lx+a) Lx+a + L(x) +a 2R

Lx = f(x) a f(x) + a R+r.

From the definition of the joint spectral radius, (F) > 1 implies that there is an > 0 such that (k(F))

1/k > 1 + for infinitely many values of k. This, in turn,implies that, for each such k, there is an affine map fk {f : k} and its linearpart Lk {L : k} such that Lk (1 +)

k. Choose k = k0 sufficientlylarge that Lk (1 +)

k0 > R+rr

. Then there is a y K with y = r such thatLk0y > r

R+rr = R+r. Since Lk0 is the linear part of an affine function f with the

propertyf(S) S (property 1 above), this is a contradiction to what was proved inthe previous paragraph.

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