University of Alberta
EXTENSION OF THE PERRON-FROBENIUS THEOREM TO SEMIGROUPS OF POSITIVE OPERATORS ON ORDER-CONTINUOUS BANACH LATTICES
by Aaron Levin
A thesis submitted to the Faculty of Graduate Studies and Research in partial fulfillment of the requirements for the degree of Master of Science
in
Mathematics
Department of Mathematical and Statistical Sciences Edmonton, Alberta Spring 2009 Library and Archives Bibliotheque et 1*1 Canada Archives Canada Published Heritage Direction du Branch Patrimoine de I'edition 395 Wellington Street 395, rue Wellington Ottawa ON K1A 0N4 Ottawa ON K1A 0N4 Canada Canada
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•+• Canada To my mother for being a constant source of inspiration. ABSTRACT
In 2000, Heydar Radjavi provided an extension of the Perron-Probenius theorem to semigroups of positive compact operators on Lp spaces. We care fully go through Radjavi's proof, rigourously expanding many arguments in an effort to provide a clear and concise treatment. We also extended Rad javi's proof to semigroups of positive, compact operators on order-continuous Banach lattices. This extension uses many techniques not used in Radjavi's proof. ACKNOWLEDGEMENT
First and foremost I would like to thank my co-supervisor Vladimir Troit- sky for his exceptional abilities as a teacher, supervisor, and friend. I am also indebted to the functional analysis group at the University of Alberta for their generous feedback during the initial stages of this project. In addition, I would like to thank Matthew Mazowita for his scholarly insight and Peter Pivovarov for his tex-nical assistance. Finally, the completion of this thesis would have been impossible without the editorial assistance and emotional support of Marie Leblanc Flanagan. Table of Contents
1 Introduction 1
2 Preliminaries 4 2.1 Banach Spaces 4 2.2 Spectral Theory for Banach Spaces 5 2.3 Banach Lattice Theory 11
3 Decomposable Semigroups 17
4 Positive Projections 22
5 Perron-Frobenius Extension 39
6 Examples and Conclusions 43
Bibliography 47 Chapter 1
Introduction
The Invariant Subspace Problem remains one of the most prominent open conjectures in analysis. As of late [11], order structures have been used as an attempt to approach the problem from a new direction. Though activity in the area of positivity has burgeoned only recently, the roots of this approach have been present since the turn of the century. The Perron-Frobenius theo rem provides much insight into the possible techniques the invariant subspace problem can be tackled using ordering structures. To introduce the traditional Perron-Frobenius theorem we need two con cepts: indecomposability and a partial order on any finite-dimensional space. An n x n matrix is said to be indecomposable if there exist no nontrivial in variant norm-closed order-ideals. For an n-dimensional space X there is a naive partial-order accomplished by first picking a basis and ordering coordinate- wise (i.e. x = (xi, • • • , xr) < y = (y\, • • • ,yr) if and only if Xj < yi for all i (after an appropriate basis is chosen)). With this partial order, all norm- closed order-ideals are standard subspaces. The traditional finite-dimensional Perron-Frobenius Theorem for a positive matrix is:
Theorem 1.1. [10] Let A be an n x n indecomposable matrix. Then there is a vector x with positive entries, unique up to scalar multiples, such that Ax = r(A)x (where r(A) is the spectral radius).
In sum, an indecomposable positive matrix has a largest positive eigenvalue
1 with positive eigenvector. More recently, Heydar Radjavi and Peter Rosenthal extended the finite-dimensional Perron-Probenius theorem to semigroups of positive matrices:
Theorem 1.2. [8, Theorem 5.2.6] Let 6 be an indecomposable semigroup of positive matrices on X and denote the minimal positive rank in K+(5 by r. If R+(5 has a unique minimal right ideal, then the following hold:
(i) there is a vector x with positive entries, unique up to scalar multiples, such that Sx — r(S)x for all SG6;
(ii) every S £ & has at least r eigenvalues of modulus r(S), counting multi plicities; these are all of the form r(S)Q with 0r! = 1;
(Hi) after a permutation of the basis, & has an r x r block partition such that
the block matrix {Sij)lj=1 of each nonzero member S of & has exactly one nonzero block in each block-row and each block-column;
(iv) if, for any S 6 6, the block matrix Sij has a cyclic pattern (i.e., there ex
ists a permutation {ii, • • • ,ir} of {1, • • • ,r} such that the nonzero blocks
of & are precisely (Siui2, Si2ti3, • • • , S^^), then o~(S) is invariant under the rotation about the original by the angle ^-;
(v) r = 1 if and only if some member of & has at least one positive column.
This theorem has been extended from the original Perron-Probenius the orem only by the usage of semigroups of positive matrices rather than single matrices. The block partition of the matrix allows for the existence of invariant norm-closed ideals. In 1999 Heydar Radjavi [7] successfully extended the finite-dimensional
Perron-Probenius theorem to semigroups of positive compact operators on Lp spaces. The presence of compact operators in Radjavi's extension is somewhat intuitive since they are a very close infinite-dimensional analog to matrices.
2 We know that Lp spaces are a large subclass of order-continuous Banach lattices and therefore the goal of this thesis is an extension of the Perron- Frobenius theorem to order-continuous Banach lattices using Radjavi's ap proach. At first glance, Radjavi's theorem seemed to translate with ease to the language of order-continuous Banach lattices. However, along the way many novel deviations from this path were found and a much more clear, con cise, and self-contained proof is presented. As well, these results now apply to a much broader class of objects like Orlicz-Lorentz spaces and other interesting order-continuous Banach lattices.
3 Chapter 2
Preliminaries
2.1 Banach Spaces
Let us briefly visit some essential concepts in functional analysis that are pertinent to this thesis. For a more thorough dissection of the material please see [4]. A normed space is a pair (X, ||-||) where X is a vector space and ||-|| : X —> K+ is a norm, i.e. for all x, y G X and A G K (where K is E or C):
(i) ||x|| = 0 4=>- x = 0 (ii) ||Ax|| = |A|||x|| ("i) ll* + i/ll The presence of a norm allows us to induce a metric on the vector space X by defining a distance between elements x,y G X as d(x,y) = \\x — y\\. If the induced metric is complete, i.e. all Cauchy sequences have a limit in the vector space, then we call the pair (X, ||-||) a Banach space. We denote the norm-closure of a set A C X by A. If Z, Y are normed spaces then we call an operator T : Z —» Y linear if for all a,p G IK and x,y G Z -we have T (ax + /3y) = aT(x) + (3T(y). A linear operator is bounded if there exists K 3 M > 0 such that for all 4 x G Z \\Tx\\ < M||x||. The space of all linear operators from Z into Y is denoted by C (Z,Y). The space of all linear bounded operators from Z into Y is denoted B(Z,Y). If Y — Z we denote this space by B(Z) for brevity. All operators in this thesis will be assumed linear and bounded. We can turn B (Z, Y) into a normed space by defining the norm of a linear operator ||T|| = sup{||Tx|| | x G X, \\x\\ = 1}. We also say an operator T G B(X) is a projection if T2 = T, which also means that X can be decomposed into X = Range T © ker T. A subspace Y of a vector space X is T-invariant if TY C Y. We denote the restriction of T to any subspace ZClas T\z. Example 2.1. It can be shown that if Y is a Banach space then B(Z, Y) is a Banach space. For a normed space X we call a function / G B(X, K) a linear functional. The space of all linear functionals on X is called the dual of X and is denoted X* := B(X,K). One of the important results in functional analysis is the Hahn-Banach theorem, which allows us to extend functionals from subspaces of our normed space to functionals on the whole space. Theorem 2.2 (Hahn-Banach Theorem). Let X be a normed space and let Y C X be a subspace and f G Y*. Then there exists F e X* such that F]Y = f and\\F\\ = \\f\\. The following corollary to the Hann-Banach theorem will be useful for our main result. Corollary 2.3. Let X be a normed space, Y a closed subspace of X, and XQ G X \ Y. Then there exists f £ X* such that f\Y = 0, f(xo) = 1, and = where H/ll dist{X0,Y) dist{x0,Y) = inf{\\x0 - y\\ \ y G Y}. 2.2 Spectral Theory for Banach Spaces The Perron-Frobenius theorem determines certain properties of the spectrum of operators. It is therefore relevant to discuss some of the elementary aspects of spectral theory. 5 For an operator T on a complex Banach Space, the spectrum of T is the set 0-(T) := {A G C | (XI — T) is not invertible}, where / is the identity operator. For an operator T on a real Banach space, we define cr(T) := o~(Tc) where Tc is the complexification of T (see [1] for more details). In either case, cr(T) is a non-empty compact subset of C. When X is a finite dimensional space the spectrum is the set of all eigenvalues. The spectral radius is defined and denoted as r(T) := sup{|A| | A 6 a(T)} and it is known that r(AB) = r(BA) for all operators A, B e B(X). Theorem 2.4. [5j[Gelfand] Let T be a linear operator on a Banach space X. n Then r(T) = limn||T ||n = infn ||T"||". In addition, for each A G K. satisfying 1 n+1 n A > r(T) we have (A - T)" = £n A"( )T . For certain types of operators the spectrum has a very distinct form. A linear operator is called strictly singular if its restriction to any closed, infinite dimensional subspace of X is not isomorphism. A linear operator T is called compact if the closure of T {x € X | ||a;|| < 1} is compact. Lemma 2.5. If a linear operator on a Banach space is compact then it is strictly singular. Lemma 2.6. [1, Theorem 7.11] Let T be a bounded, strictly singular operator on a Banach space X. Then the spectrum ofT is either a finite set or can be written as a sequence converging to 0. Moreover, if X is infinite dimensional thenOe a(T). Quasinilpotent operators also have a very distinct spectrum. A linear op erator T is called nilpotent if Tk = 0 for some positive integer k. We broaden the notion of nilpotency in the following way: a linear operator T is called quasinilpotent if r(T) = 0. Finding invariant spaces to decompose strictly singular operators will be imperative for our main result. It is therefore necessary to state some of the conclusions from functional calculus. For a more thorough review see Chapter 6.4 of [1]. 6 Let T be a linear operator on a normed space X. A subset of a(T) is called a spectral set if it is both open and closed in the relative topology (the topology formed by the intersecting of all open sets with cr(T)). Lemma 2.7. Let T be a bounded operator on a Banach space X. Let o~\ be a nontrivial spectral set of a(T). Then T admits a unique pair of nontrivial T- invariant pair of subspaces (Yai, Zai) such that a (T\Y„ ) = &i and a {T\za ) = o~(T) \<7i. We call Y0l the spectral subspace corresponding to o\. The projection of X onto Yai along Zai is called the spectral projection, which we use to decompose X into X = Yai © Zai. With Lemma 2.7 we have a nice result on the decomposition of strictly singular operators. Lemma 2.8. Suppose X is a Banach space and T G B(X) is strictly singular with r{T) = 1. Define o\ :— o~(T) f) {z G C | \z\ = 1} and let m be the dimen sion of the corresponding spectral subspace, denoted X\. Then m < oo and there exists subsequences (nj) C N and (rj) G M+ such that either (i) rjTni —> N in the norm topology and N ^ 0, N2 = 0, and rank(N) < m; (ii) orTni —> E in the norm topology, where E ^ 0 is a projection, rank(E) = m, and Range E = X\. Proof. Since T is strictly singular X\ and X2 are T-invariant and a (7]x3) = &±, a (T\x2) = °~2- Note that the spectral radius of T\x2 is strictly less than one. Claim: lin^^oo 77^ = 0 in the norm topology. Choose b G R such that r(T\x2) < b < 1. Then by Theorem 2.4 for all sufficiently large n G N we have ||Tj^J| < bn -* 0. 7 Since 0 G" o\ and T is strictly singular then X\ must be finite dimensional otherwise T would be invertible on an infinite dimensional subspace. Let l m = dim(Xi). By choosing a basis for X\ we can write T\x1 = P~ JP where J is a Jordan form, that is J = U + N where U is unitary, N is nilpotent and UN = NU. Since (j\ is on the unit circle then elements along the diagonal of i27Tai 27raini 2 mni U, have the form e . Let an = (e , • • • , e ™ ). The following claim is similar to Kronecker's Approximation Theorem [3]. 1 Claim: There exists a subsequence {rn} C Z such that aTn —> (1, • • • , l) For every n G N and j = 1, • • • , m let bj>n = (nctj) mod 1. It suffices to show that VA; G N 3n G N such that Vj = 1, • • • , m we have bj>n < | (as we then set r^ = n). Fix fceN. We know that 0 < &,-,„ < 1 for all n G N and j = 1, • • • , m. By the pigeon-hole principle, 3ii G {1, • • • , m} such m that (fei,n) has an infinite subsequence, say, (&i,nr) [|S ^^j • Applying we the pigeon-hole principle to (&2,nr) find an infinite subsequence (62,nrs) contained in [^, !2jjp] for some 22 € {1, • • • , m}. Proceeding inductively A; times we get a subsequence (ne) such that V j < m 3ij < m such that In (bj,ne) Q fc".^ • particular, V j < m we have \bj,ni - bjjTl2\ < |. Hence, if n = \n\ — 77,2) then b^n < £. Case 1: N = 0 In this case T is diagonalizable since T = P~lUP and £/ is diagonal. Thus, by the above claim we can find a subsequence such that the matrix nj representation of T,^ converges to IXl • Then this subsequence forces T to converge to E, the projection onto Xi, and clearly rank(-E) = m. Case 2: N^O Let fc G N be such that Nk ^ 0 but Nk+1 = 0. Then for any n > k we have n n n k k ([/ + TV)" = U + (^U^N + • • • + ( k)U - N . From the above claim mj we can find a subsequence m^ such that U —> IXl (this convergence is JThe idea of this proof was provided by Adi Tcaciuc. 8 in the norm-topology since X\ is finite dimensional). Let rij := rrij + k n k ni fc then U i~ -> IXl as j -* oo. Let r,- := -^-. Then r, (E/ + N) -• iV . V fc / Claim: rjTnj converges to Nk in the norm-topology. Recall that with respect to X = X\ © X2 we can decompose T into T = 7i 0T2. Then r^' - rjT? er^. On Xx, we have r^^ -»• k k 1 2 PN P'\ where {PN P~ ) = 0. Since X2 is finite dimensional the convergence is in norm. On X2, \\rjT?x || < HTj^ || —> 0. Therefore, r^T^ -> Nk © 0. Since A^fc is nilpotent and therefore noninvertible it cannot have full rank, hence rank(iVfc) < m. a Remark 2.9. Suppose X is a Banach space and A, B € B(X). If AB is a strictly singular operator with r(AB) = 1, AB and BA are both in case (i) or both in case (ii) of Lemma 2.8, and both AB and £M are strictly singular then the sequence of integers (rij) found in Lemma 2.8 for the operator AB will be the same sequence of integers for the strictly singular, nonquasinilpotent operator BA. However, the projection or nilpotent operator that AB converges to may differ from the projection or nilpotent operator that BA converges to. Proof. Suppose A, B & B(X) such that both AB and BA are strictly singular operators with r(AB) = 1 = r(BA). Let (rij) denote the sequence of integers used in either case (i) or (ii) of Lemma 2.8 applied to the operator AB. Since a(AB) \ {0} = a(BA) \ {0}, we can construct the same spectral sets (a\ and cr2) for BA as we would have constructed for AB (we do not need to worry about the sets not being equal at 0 since we concern ourselves only with the part of the spectrum on the unit circle). However, the corresponding spectral subspaces X[ and X'2 may differ. By picking a basis for X[ we can write l (BA)\X[ = P'~ (U' + N')P' where U' is unitary and N' is nilpotent. Since a(AB) = a(BA) then the entries for U' will be the same as the diagonalized 9 representation for AB (up to repetition and ordering of entries which will ultimately not affect the produced sequence). Hence, the sequence of integers used for the convergence of BA in case (i) or (ii) will be (rij). Note that the projection or nilpotent operator BA converges to may not be the same since X[ and X'2 may be different than X\ and X2. • Remark 2.10. We proved Lemma 2.8 for the case K = C. In the case that K = R the Lemma is still correct, though we must consider the complexification of X and then take the real parts of the operators. The spectral radius behaves in a very nice way on the space of compact operators. Lemma 2.11. Let X be a Banach space. Then spectral radius is an upper semi-continuous function on B(X). Proof. Let A G C and T G B(X). Let e > 0 and define the following set: S{a(T),e) := \J\ea(T) B(\,e). Then, for |A| > r(T) [4, Corollary 3 in VII.3.4] 1 1 1 1 tells us ||(A - T)" !!" < dist(A*g(T)) < f Pick 8 such that ||(A - T)" !!" < 8 < \. Let Tx G B(X) such that \\TX - T\\ < 8. We know the set of invertible operators Q C B(X) is open and for any A G Q the open ball {B | \\{X - A)-1 - B\\ < IKA-^)"1!!-1} eg. By our choice of 5, H^-TH = l l ||(A - T) - (A - Ti)|| < ||(A - T)- \\~ . So, (A - Tx) is invertible, implying c A G piT,). Therefore, (S(a(T), e)) C p{Tx) implying a(Tx) C S(a(T),e). Hence, for all e > 0 there exists 6 > 0 such that for all T\ G B(X) such that ||Ti — T|| < 8 then r(7\) < r(T) + e. Consequently, the spectral radius is upper semi-continuous. • Corollary 2.12. Let X be a Banach space. Then spectral radius is continuous on the space of strictly singular operators on X. Proof. Suppose 7] is a sequence of strictly singular operators converging to T. Then T is strictly singular since the space of strictly singular operators on X is closed[l]. Since the only possible accumulation point in the spectrum of a strictly singular operator is 0 let A0 G cr{T) be such that |AQ| = r(T). 10 Since T is strictly singular, then Ao is an isolated point of a(T). Without loss of generality we assume Ao 7^ 0 otherwise upper semi-continuity will provide continuity. Claim: 3(\) C C such that A, (E o-(Ti) and A, —> X0. We prove by contradiction and assume the negation of the claim. Then up to a subsequence 3 e > 0 such that Vz, cr(Tj) n B(\0, e) = 0. Define T := dB(X0, |) (we've chosen this ball to be small enough so that the only element contained in both T and the spectrum of T is A0. Then, for all i, (A — Tj)"1 exists and is analytic on I\ By Cauchy's theorem 1 l /r(A - Ti)- = 0 for all i. Also by Cauchy's theorem /r(A - T)~ ^ 0 owing to the pole at A0. Since the inverse function is continuous via the l l we e a identity \\S~ — T" \\ < I_\\T_S\\\YT-\ § ^ contradiction. Hence, by the above claim r(Ti) > |Aj| —> |Ao| = r(T). By upper semi- continuity of spectral radius we get limr(Tj) < r(T) and thus limr(Tj) = r(T). n Although we will not use it in this thesis, the continuity of spectrum has been extended to operators whose spectrum is totally disconnected. Theorem 2.13. [6] Let A be a Banach Algebra and (Ti) C A and T £ A where Ti —•> T. Suppose the spectrum of T is totally disconnected. Then o~(Ti) —> cr(T) in the induced Hausdorff topology. 2.3 Banach Lattice Theory Here we will define the basic concepts in Banach lattice theory which are useful for our main result. When possible, proofs will be given in an effort to make the material self-contained. For a more thorough discussion of this material see [1]. A set S is said to be partially ordered if it is equipped with a partial order < such that: 11 (i) Vi6 5 x < x (reflexive) (ii) Vx, y & S x < y, y < x => x = y (antisymmetric) (Hi) Vx, y, z G S x Example 2.14. Partially ordered sets are essential for defining nets. We say a partially ordered set Y is directed upwards if for all a, (3 G T there exists 7 G T such that a < 7 and (3 < 7. We define a net in a partially ordered set S to be a map from a directed upwards set to S. If 5 is a partially ordered set and T is a directed upwards set, then we denote nets as (xa)aer C S. Suppose S is a partially ordered set and A C S. We say that A has an upper bound if there exists u G S such that a < u for all a G A. Similarly, vl has a lower bound if there exists v G S such that v < a for all aeA We say that A is order bounded if A has both an upper and lower bound. If A has a least upper bound we denote it sup A and if A has a greatest lower bound we denote it inf A. For two element sets we denote x V y := sup {x, y} and x A y := inf {x, y}. We say 5 is a lattice if for all x,y Example 2.15. Suppose X is a vector lattice and x,y,z,E X. Then it is easy to show that x + y A z = (y + x) A (z + x). Claim: x + (y A z) < (x + y) A (x + z) We know that y A z < y and y A z < z. Therefore, x + (y A z) < (x + y) 12 and x + (y A z) < (x + z). Consequently, x + (y A z) < (x + y) A (x + z). Claim: x + (y A z) > (x + y) A (x + z) First note that (x + y) > (x + y) A (x + z) and (x + z) > (x + y) A (x + z). Hence, y > (x + y) A (x + z) — x and z > (x + y) A (x + z) — x. This implies y A z > (x + y) A (x + z) — x. Therefore, y A z + x > (x + y) A (x + z). Two essential notions in lattice theory are order-completeness and order- convergence. We say a partially ordered set S is order-complete if every bounded non-empty subset of S has a supremum and infimum. If (xa) is a net in S, then we say it's increasing and write xa f if a < (3 => xa < xp. We broaden the notation to xa f x which means xa | and x = supxa. We use similar notation for decreasing sets (i.e. xa [ x). Suppose (xn) C S is a sequence in our partially ordered set. We say (xn) is order-convergent to an element x and write xn —> x if there exists two sequences (an), (bn) C 5 such that for all n G N we have an < xn < bn and an] x, bn [ x. The definition of order-convergence can be extended to nets as follows. A net (xa)Q6r is order convergent to x if there are two nets (aa)a€A and (bp)p&B such that aa T x, bp i x and for all a G A, (5 G B there exists 70 G T such that for all 7 > 70, 1 aa < x1 < bp. A subset A C S is called order-closed if it is closed with respect to this order-convergence. We would like to understand the notion of ideals in this lattice setting (which we will soon define as order-ideals). To do this we need an operation similar to what serves as multiplication in rings. Suppose X is a vector lattice. Then we say x,y G X are disjoint if \x\ A \u\ = 0. Riesz [2] was able to prove a very useful theorem allowing us to decompose elements of a vetor lattice into the sum of positive, disjoint elements. Theorem 2.16. Suppose X is a vector lattice and x G X. Then there exists + + + + x ,x~ G X+ such that x A x~ = 0 and x = x — x~. Furthermore, x and x~ are unique in that if you can decompose x into the sum of two positive, disjoint elements x = 11 — v then u = x+ and v = x~. 13 Notice that while the inf operation is not multiplicative in an algebraic sense, it does allow us to draw similarities between lattice theory and ring theory. We say that e G X+ is a weak order-unit if for all x G X, \x\ A e = 0 implies x = 0 (i.e. the only element disjoint with e is 0). Similarly, we say + e G X+ is a strong order-unit if for all x G X there exists A G M such that |x| < Ae (i.e. a scaling of e dominates all elements in X). Suppose that A is a subset of our vector lattice X, and let x G A and j/eX. Then >1 is called solid if \y\ < \x\ implies y G A. With all of this in mind, we can now define order-ideals. We say V is an order-ideal if V is a solid subspace of a vector lattice. Example 2.17. Suppose we can decompose X into pair-wise disjoint closed subspaces X = X1 © X2 © X3 and x G (Xi)+, y G (A"2)+ and z G (X3)+. Claim: x A (y + z) — 0 (i.e. X; is disjoint from Xj © -X*., where {i,j, k} = {1,2,3}) We know that y < y V z and z < y V z. Hence, x A (y + z) < 2(y V 2). Therefore, x A (y + z) < x A 2(y V z) < 2[(x A(yV z)] = 0. Example 2.18. For a subset A of a vector lattice X, the smallest ideal con taining A is called the ideal generated by A and denoted I A- Then, I A = a + {x G X I |x| < XT=i ^l il> where n G N, al5 • • • , an G A, Ai, • • • , An G IR }. For the purpose of defining decomposability for operators on a Banach lattice, it is important to define bands. A band is an order-closed ideal (i.e. an ideal that is closed with respect to order-convergence). Example 2.19. For a subset A C. X the disjoint complement of A, denoted Ad, is the set of all elements in X disjoint from all elements in A. That is, Ad = {x G X I \x\ A \y\ = 0 for all y G A}. Then Ad is a band. Example 2.20. A vector lattice X is said to be Archimedean if for all x G X the set {nx \ n G N} being bounded above implies x < 0. For a subset B of a vector lattice X, the smallest band containing B is called the 14 band generated by B. If X is also Archimedean then Bdd = {Bd} is the band generated by B. Most vector lattices are Archimedean and counter examples are highly pedantic, and thus we will assume all our vector lattices are Archimedean. A band B in a vector lattice X is said to be a projection band if X — B © Bd. That is, for all x G X there exists a unique y G B and z G Bd such that x = y + z. By defining an operator P : X —>• X such that Px = y, then P is a linear operator and a projection. Furthermore, P is called the band projection for the band B. We can now produce results which will motivate us to work in an order-continuous Banach lattice setting defined below. Lemma 2.21. Let X be a vector lattice and B a band in X. Then B is a projection band if and only if for all x G X+, sup {y G B | 0 < y < x) exists. In this case sup {y£B\0 Corollary 2.22. Suppose that X is an order-complete vector lattice. Then every band in X is a projection band. Suppose that (X, ||-||) is a normed lattice. Then we say that X is order- continuous if the norm is order-continuous. Proposition 2.23. In an order-continuous normed lattice every norm-closed ideal is a band. Proposition 2.24. Let X be an order-continuous normed lattice. Then X is order-complete. We summarize Corollary 2.22 and Proposition's 2.23 and 2.24 in the fol lowing useful theorem. Theorem 2.25. Let X be an order-continuous Banach lattice. Then every norm-closed ideal is a projection band. In particular, for A C X, I A = Add (the norm-closure of the ideal generated by A is equal to the band generated by A) and is a projection band. 15 We will now turn our attention to defining the basic properties of bounded operators on vector lattices. We say a linear operator T is positive if x > 0 implies Tx > 0. We say T is strictly positive if x > 0 implies Tx > 0. If T is a positive operator of rank one then T is strongly positive if T = x Proof. Let x £ X and suppose ip € X* vanishes on X+. By Theorem 2.16 we + + + know x = x — x~ where x , x~ G X+. Then ij)(x) = ip(x ) — ip{x~) = 0. • Theorem 2.27 (Kantorovich Lemma). Suppose that X and Y are vector lat tices and Y is Archimedean and T e B(X,Y). If swpT[0,x] exists for every x > 0 then T+, T~, and \T\ exist are are defined as: T+x = supT[0,x] T~x = -infT[0,x] \T\x = supT[—x,x] A very basic but useful result stemming from a clever usage of functionals on a lattice is the following lemma: Lemma 2.28. Suppose X is a Banach lattice. If x,y G X+ are linearly independent then there exists z G Span{x,y} such that z is neither positive nor negative. Proof. We will prove by contradiction. Suppose every z G Span {x, y} is either positive or negative. Then Span {x, y} C X+fl(—X+). By Theorem 12.4 in [1], there exists 0 G X+ such that (f> is strictly positive on [0,x + y\. Since ker^ has co-dimension one while Span {x, y} has dimension two we have ker 0 n Span{x,y} ^ {0}. Let z G ker (finSpan {x,y}, 2^0. By assumption, either z or —z is positive. Without loss of generality, z > 0. Hence, 0 < z = \x + ny < max {|A|, |/i|} (x + y). Therefore, a multiple of z is in [0, x + y]. Consequently, 4>(z) 7^ 0, a contradiction since z G ker cfi. • 16 Chapter 3 Decomposable Semigroups We take a naive approach to extending the Perron-Frobenius theorem. We will look for a finite dimensional invariant subspace of our semigroup which corresponds to the range of a projection and then apply the finite-dimensional Perron-Frobenius theorem. This will require a number of results pertaining to decomposable (or indecomposable) semigroups. Throughout this section we will assume all operators are linear and bounded on an order-continuous Banach lattice X. Definition 3.1. A subset & C B(X) is a semigroup if it is closed under multiplication (i.e. for all A,Be& we have AB e &). We say J C & is an algebraic-ideal if J& C J and &J C J. Definition 3.2. For a semigroup of operators & we denote and define the Radjavi closure as M+& = {rS \ r E R+, S € &}. Definition 3.3. A semigroup & is decomposable if there exists a nontrivial band invariant under all members of &. Lemma 3.4. For an arbitrary semigroup & of bounded, positive operators on X, the following statements are equivalent; (i) & is decomposable; (ii) A&B = {0} for some nonzero positive operators A, B e B(X); 17 (Hi) There exist positive x £ X, 0 £ X* such that (iv) Some nonzero algebraic-ideal of & is decomposable. Proof, (i) =>• (ii): Let Y" be an invariant projection band. Then B = Py and A = PYd will satisfy the required conditions. They are nonzero positive operators by the definition of projection bands. (ii) =>• (in): Suppose there exist positive A, B such that A&B = {0}. Pick any y > 0 such that x = By ^ 0. Case 1: &x = {0} Then any nonzero Case 2; 6x ^ {0} Since ^4 is positive then A* is positive. Hence, there exist ip £ X+ such that := A*ip ^ 0. Therefore, 0(6z) = (A*il>)(&By) = ^(A&By) = 0. (iii) =4> (%): Let x £ X, 0 £ X* be positive and nonzero such that 0 (5x) = 0 for all SeG. Case 1: &x = {0}. Obviously ©7X = {0}, and since S is continuous for all S £ 6, ©4 = {0}. Let y := 4. Therefore, by Theorem 2.25 Y = {x}dd C flses kerS*, giving us an invariant projection band. Case 2: &x ^ {0} 18 Let I6x = {y € X | \y\ < £"=1 \z%, for Ai, • • • , A„ G R+, Zi, • • • , 2„ G be the ideal generated by Gx. Obviously IQX is invariant under 6, and since © is a semigroup of positive operators, hence continuous, I&x is invariant under &. Therefore, by Theorem 2.25 IQX = {Gx} is an ©-invariant projection band. Furthermore, {Gx} is nontriv- ial since 6a; C {Gx} and {Gx} ^ X owing to positive, and vanishing on IQX (we know IQX is a band by Theorem 2.25). (iv) <*= (i): Trivial. (iv) => (i): Let J be a nonzero, decomposable algebraic-ideal of G. By (ii) there exists A, B, nonzero, positive such that AJB = {0}. Case 1: JB = {0} Pick any nonzero J £ J. Then J(5I? = {0} and by (ii) & is decom posable. Case 2: JB ^ {0} If JB is a nonzero member of JB, then A&JB = {0} yields the desired result. • With this result we can already characterize the decomposability of some semigroups. Lemma 3.5. Suppose A is a nonzero, positive operator on X such that either A or A* is not strictly positive. Then D = {5e B{X) \ S>Q,AS = SA} is decomposable. Proof. Case 1: A\s not strictly positive. 19 Let x > 0 be such that Ax = 0. Then ASx = SAx = 0 for all S G T). Choose (j)^X*+ such that Case 2: Suppose A*4> = 0 for some 0 > 0. Since A* This next result is a combination of an early finding by Drnovsek and a recent result of Turovskii [12]. Theorem 3.6. [1, Corollary 10.47] If & is a semigroup of quasinilpotent com pact positive operators on a Banach lattice, then 6 has a nontrivial closed invariant ideal. Theorem 3.6 in combination with Proposition 2.23 give us an extremely useful corollary. Corollary 3.7. If © is a semigroup of quasinilpotent compact positive opera tors on X, then & is decomposable. Definition 3.8. Let 21 C B(X) and r(A) be the spectral radius for A G 21. We say r is submultiplicative on 21 if r{AB) < r(A)r(B) for all A, B G 21. We say r is permutable on 21 if r(ABC) = r(CBA) for all A,B,C G 21. The following result by H. Radjavi in chapter 8.6 of [8] relates submulti- plicativity of the spectral radius to its permutibility. We will use this to relate submultiplicativity to decomposability. Theorem 3.9. [8, Theorem 8.6.3] Let & be a semigroup of compact operators on a Banach space. Then spectral radius is submultiplicative on & if and only if it is permutable on S. 20 Theorem 3.10. If the spectral radius is submultiplicative but not multiplicative on a semigroup & of positive compact operators on X, then & is decomposable. Proof. We assume that & is Radjavi closed, otherwise replace 6 with ~R+& since, by Corollary 2.12, spectral radius is continuous. Note that the set of all quasinilpotent elements in & is an algebraic-ideal since the spectral radius is submultiplicative. If the ideal consisting of quasinilpotent operators is non- trivial then we are done by Lemma 3.4 and Corollary 3.7. Therefore, without loss of generality, assume that & has no nonzero quasinilpotent operators. By the hypothesis there exists A,Be& such that r(AB) < r(A)r(B) and by scaling r(A) = 1 = r(B). Therefore A, B are not quasinilpotent. Since 6 contains no nilpotent operators, by Lemma 2.8 there exist subsequences {nj} and {rrij} such that linx^^-, Ani = E and lim^oo Bmi = F where E and F are positive projections. Let kj = mm(nj,rrij). By Theorem 3.9 spectral radius is permutable on 6. Therefore, nj mj kj n k mj kj r(A B ) = r {{AB) A i~ iB ~ ) < r (AB)ki r{A)ni-kir{B)mi-ki = r(ABp -> 0 As a result lim, r (An'Bm') = 0 which implies r(EF) = 0 by continuity of the spectral radius. Since no other quasinilpotent operators in & are 0 we have EF = 0. It follows that (F&E)2 = {0} and hence F&E = {0} (since no nonzero elements are nilpotent). An application of Lemma 3.4 shows that & is decomposable. • 21 Chapter 4 Positive Projections Positive projections play an important role in the proof of the Perron-Frobenius theorem. Let us state some very nice results about positive projections and their relation to indecomposability. Throughout this chapter X will denote an order-continuous Banach lattice. Recall that if T is a positive operator of rank one, then T is strongly positive if T — x ® 4> where x 6 X is a weak order-unit and 0 is a strictly positive functional. Lemma 4.1. Suppose E is a positive projection of rank one on X. Then E is indecomposable if and only if E is strongly positive. Proof. E is indecomposable =>• E is strongly positive: Suppose E is a positive projection of rank one. Pick any nonzero, positive x G Range E. Then there exists 0 G X* such that Ey = strictly positive we consider y G X+ such that cp(y) = 0. Thus the band generated by y is contained in the kernel of E and hence invariant, which is a contradiction. 22 E is indecomposable <= E is strongly positive: Suppose that E is strongly positive so that E = x tg> Lemma 4.2. Suppose E is a strictly positive projection on X such that rank(E) > 1. Then Range E contains two positive disjoint vectors. Proof. Claim: There exist at least two positive, linearly independent ele ments in Range E We will prove by contradiction and assume the negation of the claim. We know E > 0 so there exists at least one positive element, x, in Range E. If every positive vector in Range E is a multiple of x then Range E P) X+ C Span (x). For any y € Range E we have y = Ey = Ey+ — Ey~ € Span {x}, contradicting the rank of E (which is strictly greater than one). Let x, y > 0 be linearly independent vectors in Range E. The existence of a z G Range E where z is neither positive nor negative follows from Lemma 4.2. Then z+ and z~ are positive and disjoint. By [9, Proposition 3.11.5], Ranged is a sublattice and therefore z+,z~ € Ranged. Thus, Ranged contains two disjoint nonzero positive vectors. • Lemma 4.3. Suppose E is a positive projection of finite rank r on X. Then: (i) Suppose both E and E* are strictly positive. Then there exists r pairwise disjoint projection bands Xi and indecomposable positive rank-one pro jections Ei such that X = X\ © • • • © Xr and E = E\ © • • • © Er where Ei : Xi —> Xi; 23 (ii) In general, there are three disjoint projection bands Yi,Y2, and Y"3; such that X = Yi © Y2 © Y3 and E has an operator matrix ' 0 AC ACD ' 0 C CD 0 0 0 relative to X — Y\®Y2®Yz, where C is a strictly positive rank r pro jection, with C* also strictly positive, and A and D are positive. In this case C can be decomposed into C = C\ ffi • • • © Cr as in (i). We also have rank(A) = rank(C). Proof. We begin by proving (i). Case 1: r = 1. E is a rank one projection so let £ X* and x £ X be such that E = x that (p is strictly positive. Suppose y £ X+ is such that 4>(y) = 0. Then Ey = 0, which contradicts the assumption that E is strictly positive. Case 2: r > 1 By induction we need only to show that X = X\ © X2 and E = F © G where F and G are positive projections of rank < r, each satisfying the hypothesis. Since rank(E') > 1 then by Lemma 4.2 there exist two pos itive disjoint vectors Zi,z2 £ Range E. Since z\ E RangeE then Ez\ = z\. + Consider the ideal generated by zx, IZl = {y £ X \ \y\ < \z\, for some A £ R }. 24 + Let yeIZl. Then for some A G R , \y\ < Az2. Thus Ey < E\y\ < \Ezx = Xzi. Furthermore, IZl is invariant under E. Prom the continuity of E, EIZl C 72l. Since X is order-continuous then IZl = {zi} =: X\ and X\ is a projection band, which implies we can decompose X as X = X\ (&X2, where X2 = Xf. In addition, we know that X2 is nonempty since z2 is disjoint from Z\. We write E = [% Q] relative to X = X\ @X2 (the entry 0 comes from the fact that Xj is invariant under E). Because E — E2 it is evident that FRG = 0. This also implies that RG = 0 otherwise there is an x G X2 such that x > 0 and RGx 7^ 0. This would imply the existence of a z G X\ such that z > 0 and Fz = 0, which is a contradiction as then Ez = 0 implying 2 G ker i? or 2 = 0, neither of which is true. To show that R — 0 is a little more work, but is very similar to Lemma 3.4. Claim: G ^ 0. Let 0O G (^2)+- Extend 0O to 0 G (-X"*)+ via 0(x) = ^0(^2) where 1 x = xi + x2 relative to X = X!©X2. Since -E * is strictly positive then E*(j) ^ 0, so there exists x e X±,y £ X2 such that 0 ^ (E* Claim: R = 0. Suppose R y£ 0. Let x G X2 such that x > 0 and y = Gx ^ 0. Recall that kexR = (R*X*)± = {z £ X \ (R*(j)) {z) = 0 for all 0 G X*}. Let 0 be any positive functional not in ker R*. Then R*(j> is a positive functional such that {R*4>) (z) = 0 for all z G keri?. Furthermore, = the range of G is contained in ker it!, thus (-R*0)iRangeG 0- Extend (i?*0) to if? := O©(i?*0) defined on X = XX@X2. Then V>|Ranged = 0, contradicting the assumption that E* is strictly positive. Now we need to show that F and G have the properties of the hypothesis. Suppose x G (Xi)+ such that Fx = 0. Then Ex = 0, so x = 0. If 0 G (^i)+ such that 0 vanishes on Range F; extend it to 0 on X by 25 setting it to zero on X2, implying that Now to prove (ii). First we will show the existence of a maximum projection band generated by positive elements annihilated by E, denoted Y\. Define the following: dd = (kerEnX+) dd Y2 = Yfn {Range E) c Y3 = Yfn (Range £) Then clearly X = Yi © Y2 © Y3, and since Y\ C ker£", then relative to this o A B decomposition we can write E as o c D Lo G H\ Claim: G = 0 = H. Ax Let x e {Y2)+. Then Ex = Cx But Ex 6 Range E and Range £Tl Y" = Gx 3 {0}, so Gx = 0. Then G vanishes on (^2)+, implying G = 0. The same procedure will yield H = 0. Since E is a projection then E2 = E and thus A = AC, B = AD = ACD, and D = CD. With these relations we can write E as: 0 AC ACD E 0 C CD 0 0 0 Now we show that C has the properties of (i) in the Lemma. Claim: ranki? = rankC. If x G Range E then with respect to X Y1 © Y2 © y3, x = and 0 AC ACD ~ V~ ACz" Ex = x. Thus 0 C CD z ^ Cz which implies Cz = z and 0 0 0 0 0 finally that z G Range C and y — Az. This means we can write Range E = 26 A { v \ve Range CJ. Now, let T : Y2 -> X denned as Tv := If] for 1 any v G Y2. This means Range E can also be described as Range .E = T (Range C). It is clear that T is linear and one-to-one since if Tv = Tw then and therefore v — w. Since T is one-to-one . o J L o J then rank(i?) < rank(C). Finally, because T\RangeC is a bijection, then rank(£) = rank(C). Finally, we show that C and C* are strictly positive. Since E is positive, so are ACx A, C and D. Suppose x G (Y2)+ such that Cx = 0. Then Ex — Cx 0, o implying x E ker E. Hence x G ker E n X+ C. Yi so x £ Yi. Therefore, x e Yi fl y2 implying a: = 0. Suppose now that (j> e (1^)+ such that |RangeC = 0. Extend 0 to a positive functional tp on X by setting it to zero on Yi and Y3. Claim: ip vanishes on Range E. t r Av 1 \ Let v e Range C. Then V ( « ) = Hv) = °- Therefore, ip\ RsmgeE = 0. Claim: •0 = 0. It is clear that -0 vanishes on Yi and Y3. Suppose y € (/Range£;)+- Then for some z G RangeE we have 0 < y < \Ez\. By the positivity of tp we get 0 < if>(y) < ip(\Ez\) < ip (\E\\z\) = ^ (E\z\) = 0. Hence, -0 vanishes on all the positive elements of /Ranged and by Remark 2.26 tp vanishes on ^Ranges = (Range E) dd . Therefore ip = 0. D Corollary 4.4. Suppose E is a positive projection of rank r on X. Then the range of E contains r positive vectors x\, • • • ,xr such that {x\, • • • , xr} form a basis of Range E. Proof If E is of the form in Lemma 4.3 (i), then there exists E\,--- ,Er positive rank one projections such that E = E\@- • -®Er on X = Xi@- • -®Xr, where Xi,- • • ,Xr are disjoint projection bands. For each i choose xi to be the positive vector generating the band Xi as constructed in the proof of Lemma 27 4.3. If E is not of the form in Lemma 4.3 (i), then by Lemma 4.3 (ii) we can write E in block matrix form. If x G X+ such that Ex ^ 0, then x G X2, ACx which implies Ex = Cx Hence, C can be decomposed into C\ © • • • © Cr 0 where Cj is a rank-one projection on the decomposition of X2 = Y\ © • • • © Yr. As above, we pick xt to be the positive vector generating the band Yi and therefore RangeE will be spanned by x* . In both cases we end up with r linearly independent, positive vectors that span a r-dimensional vectors space and are thus a basis. • Remark 4.5. For a positive projection of rank r > 0, Corollary 4.4 allows us to construct an order on Range E with respect to the basis {xi, • • • , xr}. Suppose x e Range i?. Then x = SixH \-srxr where the st are scalars. We say x >z 0 if and only if s; > 0 for all i. This way Range E is linearly isomorphic to W and thus we can induce the same lattice structure on Range E. Subspaces with compatible orders but different lattice structures are called lattice subspaces. Hence, Range E is a lattice subspace of X with this newly induced order. Lemma 4.6. Let & be a semigroup of positive compact operators on X, sup pose that E is a positive projection of finite rank r > 0, and {xi, • • • , xr} is a basis of Range E as in Lemma 4-4- (i) Then EL'-E| Range E *S positive in the order y for every positive operator Tee; (ii) If G is indecomposable then E&E\ RangeE is indecomposable with respect to the new lattice structure on Range .E1. Proof, (i): Case 1: Both E and E* are strictly positive. If both E and E* are strictly positive then Corollary 4.4 provides a basis for RangeE denoted {xi,--- ,xr}. By a simple induction, example 2.17 shows that any vector in this basis is disjoint from any sum of vectors disjoint from it. 28 Claim: The original order on X and the new order >z are the same. Suppose that O^iG X+. Then x — t\X\ + • • • + trxr for some scalars £j, • • • , tr. Suppose that for some j we have tj < 0. Then x = t\Xi + • • • + tj-iXj-i + tj+iXj+i + • • • + trxr — \tj\xj. For brevity, denote z := tiXi H h tj-iXj-i + tj+iXj+\ H— • + trxr. Thus x A Xj = Xj f\[z — \tj\xj\ < \tj\xj + Xj A [z — \tj\xj]. By example 2.15 and a simple induction on example 2.17 we get x A Xj < (1 + \tj\)xj A [z] = 0, indicating that x A Xj < 0, which provides the contradictory result that x A Xj = 0. Hence by an easy induction, x is positive in the original order only if all scalars U are positive. The contrapositive is easily seen (any positive linear combination of positive vectors is positive). , , For any positive S G & the above claim shows that -E5 JE|RangeJB is positive in the order y. Case 2: Either E or E* is not strictly positive. Corollary 4.4 provides a basis for Range E which we will denote {zi, • • • , zr}. For clarity we explain the differences between this basis and the basis in case 1. Lemma 4.3 allows us to decompose the space X into X = X1®X2®X3 where X2 = Yx © • • • © Yr (disjoint projec tion bands) and with respect to this decomposition E — o c CD . We let {yi, • • • , yr} be a basis for X2 consisting of positive, disjoint, linearly independent vectors. Then each Z{ is a vector of the following Ayi form Zi = Vi , which we can also write as zi = Ay + yi. 0 t Claim: The original order on X and the new order y are the same. Suppose that 0/i£ X+. Then x = txxi + • • • + trxr for some scalars tx, • • • ,tr. Suppose that for some j we have tj < 0. Then x — t\Z\ -\ h tj_iZj-i + tj+iZj+i H h trzr — \tj\zj. By using the expansion of z^ we can write x = p — \tj\yj — \tj\Ayj where p = ti(Ayi+yi)-\ htj-i(>tyj-i +yj-i) + tj+i(Ayj+i +yj+i) + 29 • • • + tr(Ayr + yr). Since A is a positive operator then x < p. By example 2.15 we have xAi/j = 0, which is a contradiction. Hence by an easy induction, x is positive in the original order only if all scalars £j are positive. The contrapositive is easily seen (any positive linear combinations of positive vectors is positive). (ix): Let us prove the contrapositive. Suppose E&E\Ranged is decomposable. By Lemma 3.4 there exist positive nonzero members x and (f> of Range E and (Range E)*, respectively such that 4>{ESEx) = 0 for all S G 6. Let ip G X* be defined as 4>(y) = (f>{Ey) for all y G X*. Note that ip(SEx) = 4>{ESEx) — 0 for all S G &. Therefore & is decomposable by Lemma 3.4. • We now have the following results about indecomposable semigroups of positive, compact operators. First, we define a partial order on projections by: E < F if Range E C Range F and ker F C ker E. Definition 4.7. Let F € B{X) be a projection. With respect to the above partial order we say that F is a minimal projection if E •< F implies E = F for any projection E € B(X). Proposition 4.8. Let & = R+6 be an indecomposable semigroup of positive compact operators on X with minimal rank r < oo. Suppose A G (3 such that rank A = r and r{A) = 1. Then only case (ii) is possible in Lemma 2.8, i.e. the subsequence (nj) found in Lemma 2.8 causes powers of A to converge to a projection of rank r. Proof. By Lemma 2.8 there exist subsequences (rij) and (rj) such that rjAnj —>• Q where Q is either a nilpotent operator of rank less than r or a projection of rank equal to r. We will show that Q cannot be nilpotent. Note that for all j, r,A^) e ©. Also, since & = R+6 then Q E &. Thus Q cannot be nilpotent 30 otherwise its rank would contradict the minimality of r. Therefore by Lemma 2.8 Q must be a projection of rank r. • Lemma 4.9. Let & = M+6 be an indecomposable semigroup of positive com pact operators on X. Let r < oo be the minimal positive rank of operators in & and let A £ & have rank r. Then there exists 5 6 & and (nj) C N such that F := \m\j{AS)nj is a projection and FA = A. Proof. By Lemma 3.4 A& ^ {0} and we can choose S0 e & such that ASo ^ 0. Similarly pick 50o E & such that 50oA50 ^ ^. Therefore the ideal 6A6 is nonzero and indecomposable. It follows from Corollary 3.7 that there exists 5*1,52 G & such that SiAS^ is not quasinilpotent. Furthermore, recall that 1 a(S1AS2) = a(AS2Si) up to 0. Let 5 = S2S2, B = AS and note that rankjB = rankA Without loss of generality we can assume r(B) = 1 by scaling A. By Lemma 2.8 and Proposition 4.8 there exists a subseqence of integers (rij) such that we produce the projection F = lira, BUi. For all integers n, Range Bn C Range B. Consequently Range F C Range I?. However, B has finite-rank and thus Range B is finite dimensional, implying Range B = Range.B and with the minimality of the rank of B, RangeF = Ranged. Therefore, FB = B. Claim: FA = A Since B ^ 0 then Range B C Ranged, and with the minimality of rank A, we have Range B = Range A Given that F is a projection and FB = B, then Ranged = Range!? C RangeF, and therefore FA = A. a Lemma 4.10. Let & — R+(3 be an indecomposable semigroup of positive compact operators on X. Then & contains nonzero operators of finite rank. Let r > 0 be the minimal positive rank of elements in &. We have the following properties of rank r operators in G: 1Note that it is not a problem that the spectrums may differ at 0 since their usage in reference to Lemma 2.8 relies only on their similarity on the unit circle, which we have. 31 (i) A "projection E G 6 is of rank r if and only if E is minimal; (ii) For each A of rank r in & there exist minimal projections F,E G &, such that FA = A and EA = A; (Hi) For each x > 0 in X there exists a minimal projection E £ & with Ex 7^ 0. Also, for each X* 9 SUCh that (j)\ Range F ¥" 0/ (iv) If E is a minimal projection in &, then {ESE \ S G (5, r(ESE) = 1} is a group with identity E. Proof. By Theorem 3.7, & contains operators that are not quasinilpotent. Futhermore, by Lemma 2.8, & contains an operator that is of finite rank. (i): Let E 6 & be a projection. Since E is a compact projection then it has finite rank. Claim: E is of rank r => E is minimal. Suppose E is of rank r and let F be a projection such that F ^ E, then Range F C Range E. However, by the minimality of the rank of E, Range F = Range E. To show their kernels are equal, suppose they are not. Since F •< E then ker E C ker F. Pick x € ker F\ker E such that x ^ 0. We know E is a projection so X — Range E® ker E. Thus x = Ex + y where 0 ^ y £ ker E C ker F. Hence Fx = FEx + Fy and since Range E = Range F and x, y G ker F then Fx = 0, Fy = 0 and FEx = Ex. Therefore Ex = 0, which gives us a contradiction. Claim: E is of rank r <= E is minimal. Suppose E is a minimal projection. Let the rank(F) = s > r, and &r denote the ideal of & consisting of members with rank < 32 r. By Lemma 3.4 &r is indecomposable since &r is an algebraic- ideal. Apply Lemma 4.6 to get a basis of Range i? consisting of vectors {xi, • • • , xs} such that -ES-Ej Range £ is a positive operator for all S £ 6. Denote 60 = E&rE^KangeE. Similarly, by Lemma 4.6 ©o is indecomposable. Furthermore, by Theorem 3.7 there exists As6o such that A is not quasinilpotent and by scaling we have r(A) = 1. By Lemma 2.8 and Proposition 4.8 there is a subsequence {rij} C N such that AnJ —> F, where F is a projection of rank r and F e 60. Since F e 60 we know F = EGE\ Range £ for SOme G 6 6. , , 1 Since F = -E G£ |Range£; is a rank r projection on Range-E then its extension to X, simply EGE, is also a projection of rank r. We note that EGE •< E, hence by minimmality E — EGE. However, rank E = rank EGE = rank F = r. Therefore, s = r. (n): By Lemma 4.9 we produce the subsequence of integers (rij), operator S (Hi): Let x e X+ and 0 G X*+. Consider 6r = {S e & \ rank(S) = r}U{0}. &r is an ideal and indecomposable by Lemma 3.4. Also by Lemma 3.4 there exists A £ &r such that (iv): 33 Note that for every 5e6 with r(ESE) — 1, it follows from the minimal ity of r that ESE\ Range £ is a full-rank operator on a finite-dimensional space and thus invertible. Hence, for all S G 6, (ESE\Ranges)-1 exists. By Lemma 2.8 in combination with the fact that r(ESE) = 1 we get the existence of (rrij) C N such that (ESE)mJ —• G. Owing to the closure of 1+6, G G R+S. By Proposition 4.8 G is not nilpotent. Thus by Lemma 2.8 G is a positive projection of rank r. Let S E & such that r(ESE) = 1. Claim: ^(ESE)™* = E. Suppose x G Range G. Since G is a projection, Gx = lim.j(ESE)mJ = x. It follows that EGx = Ex = EYim^ESE^x = lim^ E(ESE)m^x = m: lirn?(i?,S'.E) >£ — x. Hence, Ex = x and thus RangeG C Ranged. Now, suppose y G ker i?. Then clearly y G ker G. Hence G < E and since E is a minimal projection G = E. Therefore lim, (ESE)mi = E. mj 1 Claim: \imj(ESEijiangeE) ~ exists. By the above claim we know that \im.j(ESE\Ranges)"^ exists. We also l , know that -E5jE|Range£; is a full-rank operator on a finite-dimensional space and thus invertible. Since ESE\Yia,ngeE is a continuous opera 1 rn _1 tor then so is (ESE\RangeE)~ and therefore lim:,(SS'E|Range£;) J exists. Define an operator A : Range E —> Range E as A := lim^(ESE\ Rangee)™3-1 • For all x G X, Ex G Ranged. Thus, {ESE)mi~lx = (ESE)mi~lEx = m l 1 1 {ESE\K&ngeE) ^ Ex -> AEx = EAEx. Therefore, (ESE)" ^ -> EAE. However, 6 is norm-closed, implying EAE G &. Therefore, there exists T G 6 such that ETE = £M£. Then clearly, #r£ • £S£ = ESE • ETE = E. Furthermore, by Corollary 2.12 the spectral radius is continu ous, implying (ESE)-1 = ETE is in the group {ESE \ S G 6, r(ESE) = 1} with identity E. O 34 Lemma 4.11. Let & = R+© be an indecomposable semigroup of positive compact operators on X. Then: (i) Every nonzero right ideal of & contains a minimal right ideal; (ii) Every minimal right ideal is of the form E& for some minimal projection E€S. Proof. Let J be a right ideal and let J be some nonzero member of J. By Lemma 4.10, 6 contains operators of finite-rank. Let r be the minimal nonzero rank in 6 and define (50 := {S 6 & | rank(5) < r}. By Lemma 3.4 &0 is inde composable and J&o 7^ {0}. Therefore, without loss of generality, rank J = r. By Lemma 4.9 there exists S € & and (nf) C N such that E := \imj(JS)nj is a projection and EJ = J. Since JS € J C\ &o, without loss of generality JS = J. Claim: EJE ^ 0. Because E = linij JnJ and J is continuous, EJ = J implies lim^ J"J+1 = J + 0. Therefore, EJE = JE = lim,- J^+1 = J ^ 0. We can scale EJE so that r(EJE) = 1. By (iv) of Lemma 4.10 there exists T £ E&E such that J 3 JET = (EJE)T = E. Therefore E G J and E&CJ. Claim: E6 is a minimal right ideal. Since E& is a right ideal, suppose there exists T C E& such that X is a right ideal. By performing the same operations to T as we did to J in this lemma, there exists a minimal projection F G & such that F6CIC£6. It follows that F = FF = EA for some A 6 6. Hence RangeF = Ranged, implying FE = E. Therefore, E& = FE& C F6CIC £©. The above claim also proves (ii). • Lemma 4.12. Let & = R+6 6e an indecomposable semigroup of positive compact operators on X, and let £ denote the set of all its minimal projections. The following conditions are mutually equivalent: 35 (i) EF = F for every E and F in 8 (i.e., all minimal projections have the same range); (ii) all nonzero minimal-rank members of & have the same range; (in) GE = E&E for some E e £; (iv) &E = E&E for all E e £; (v) GE C E& for some E e £; (vi) GE C E& for all E e £; (vii) & leaves the range of some E € £ invariant; (viii) & leaves the range of every E € £ invariant; (ix) some minimal right ideal is an ideal; (x) every minimal right ideal is an ideal; (xi) & has a unique minimal right ideal. Proof, (i) =3- (xi): By Lemma 4.11, & has a minimal right ideal of the form E& for some E G £. Let J\ and Ji be two right ideals. Then by the same Lemma Jx = E& and J2 = F&. It is evident by (i) that Jx = J2. (i) <= (xi): Let E,F e£. By (xi) and Lemma 4.11 it follows that EF = F. (i) <* (ii): This equivalence follows directly from an application of Lemma 4.11. (ii) => (iv): Let E E £. Then every nonzero member of &E has the same rank and range as E. Thus, &E = E&E. 36 The implications (iv) =>• (vi) and (vi) => (viii) are obvious. (viii) =>• (x): Consider any minimal right ideal EG and observe that the hypothesis of invariance gives &E = E6E, which means that &E& = E&E& C EG. (x) => (xi): Let E,F e E. Then E&F ^ {0} by Lemma 3.4. Pick A G & with EAF ^ 0. Subsequently EAFG is a right ideal contained in the minimal ideal EG and thus EAF& — E&. Furthermore, (x) implies that F& is an ideal and EG = (EA)F& C FG. Therefore E<5 = F&. (ix) =>• (v): If some minimal right idea is an ideal, then by Lemma 4.11 there exists E 6 & such that J = E& where J is a minimal right ideal. Therefore, GE&QE&, and in particular &E Q EG. (v) => (vii): By (v) there exists a minimal projection E G £ such that &E C EG. Suppose x G Range E and let S, S' G G such that S1^ = ES'. Thus, S:r = S'.&r = ES'x G Range £. The implication (vii) => (iii) is clear. (Hi) => fir,).' Let E G £ be the minimal projection such that &E = E&E. Define J := EG. By construction J is a right-ideal. However, ^7 is also a minimal right-ideal because E is a minimal projection. Let S G (3. Then SJ = S£S = (ESE)S = E(SES) e J". Hence, J is also a left-ideal and thus an ideal. Since (x) =>• (ix), the proof will be complete by showing (ix) => (i). (ix) =*• ^;.- 37 Consider a minimal idea in 6. By Lemma 4.11, it is of the form E& for some E G 6. Let F G £. By Lemma 3.4 (ii) there is a T G & with FTE ^ 0, so that FTE has the same range as F. By the relation FTE £6£C BE& = E& we deduce that the ranges of E and F coincide, and thus EF = F. Therefore, RangeF — Range!? for all Fe£. • Lemma 4.13. Let & = R+6 be an indecomposable semigroup of positive compact operators on X, and let £ denote the set of all its minimal projections. Suppose all minimal projections in & have the same range. Then for all F G £, F* is strictly positive. Proof. Define r to be the smallest rank of operators in &. Let F G £ and define &r := {S G 6 | rank(S') = r or 0}. Then 6r is an ideal in & and hence indecomposable by Lemma 3.4. Let x > 0 and Since ©r is indecomposable then by Lemma 3.4 we know there exists A G ©r such that cf)(Ax) ^ 0. Since rank(A) = r then Ranged = RangeF by Lemma 4.11 (ii), which implies A = FA. Hence 0 ^ <£(AE) = (j){AFx) = (F* Lemma 4.14. Let & be an indecomposable semigroup of positive compact operators on X such that M+(5 has a unique minimal right ideal. Then & contains no quasinilpotent elements other than 0. Proof. Denote r to be the smallest rank of operators in & and let TZ be the common range of the minimal projections. Let S G & be quasinilpotent. Since 71 has finite dimension r and is invariant under all S G &, then S\TZ is nilpotent and has rank less than r, implying S\n = 0. We write S\n = SE = ESE and conclude that SE = 0 for any minimal projection E. Claim: 5 = 0. Suppose 5^0. Then there exists x G X+ such that Sx ^ 0. Note that Sx > 0 and therefore by Lemma 4.10 there exists a minimal projection G such that SGx ^ 0, giving us a contradiction. • 38 Chapter 5 Perron-Frobenius Extension Theorem 5.1. Let & be an indecomposable semigroup of positive compact operators on X such that R+(5 has a unique minimal right ideal. Denote the minimal positive rank of elements in R+© by r and the common invariant range of the minimal projections 1Z. Then the following hold: (i) There is an x G X such that x > 0 and Sx = r(S)x for every S G &. The vector x is unique up to scalar multiples; (ii) Every S G & has at least r eigenvalues of modulus r(S), counting mul tiplicities; these are all of the form r(S)a where ar] = 1; (Hi) If the block matrix of any S G & has a cyclic pattern, then a(S) is invariant under rotation about the origin by the angle —. Proof. Without loss of generality we assume & = K+(E5 as any conclusions found for R+(5 will surely hold for 6. By Lemma 4.12 TL is invariant for all members S E & and thus we will restrict our entire semigroup to TZ; in doing so, all our operators are r xr matrices. Let E G & be a positive projection of rank r. Claim: &\n consists of positive matrices. By Lemma 4.6 there exist r positive vectors X\, • • • ,xr that form a basis for 1Z. With respect to this basis, introduce the order used in Lemma 39 4.6. This order turns 1Z into a lattice subspace and makes all operators &\iz = ESE positive matrices. By remark 4.5 we note that vectors are positive in the new order if and only if they are positive in the old order. However, while both orders are the same, the lattice structure may be different. With the previous claim established all notions of positivity are carried over to the finite dimensional space 1Z, which we restrict our operators to. Claim: For all S e © we have r(S|TC) = r(S). Let See. We know that r(S\n) < r(S). By Corollary 4.14 there are no quasinilpotent elements in © other than 0; we assume without loss of generality that r(S) = 1. From the finite-dimensional Perron-Frobenius theorem we know there exists x G 1Z+ such that Sx = SEx — r(S\n)x. Suppose r(Syji) < 1. Since © contains no quasinilpotent elements, S is not quasinilpotent and by Lemma 2.8 there exists (m,j) C N such that Smj —> F where F is a projection of finite rank r. Therefore SmjEx = r(S\-n)mix. By taking the limit we find FEx = 0, which implies Range E % Range F, giving us a contradiction since all minimal projections have the same range. Claim: &\-JI is indecomposable. Suppose ©IT?, is decomposable. By Lemma 3.4 (ii) there exist nonzero, positive operators A,B e 3(11) such that A&\nB = {0}. Let £ £ 6 be any minimal positive projection of rank r and note that S\-n = ESE. Thus, EAE&EBE = {0} and by Lemma 3.4 (ii) © is decomposable. The above three claims demonstrate that ©j^ is an irreducible semigroup of positive matrices and (i), (ii) and (iv) in the theorem follow from the finite dimensional analog of the Perron-Frobenius theorem (Theorem 1.2). • Remark 5.2. If, in addition to the conditions of the lemma, there exists a positive projection P of rank r in the closed convex hull of 6 such that 40 Range P = 1Z and both P and P* are strictly positive, then there are r disjoint projection bands Xi such that X = X\ © • • • © Xr. As well, the r x r operator matrix of every nonzero S E & relative to the partition X = Xi © • • • © Xr has exactly one nonzero block in each block-row and each block-column. Proof. Note that the existence of P and Theorem 4.3 allow us to decompose our space X into r disjoint projection bands, i.e., X = X\ ffi • • • © Xr. With respect to this decomposition we can write P = P\ © • • • © Pr where Pi is a one-dimensional projection on X^ By construction Range P = 1Z since all the minimal projections have the same range. Let 0^5G6. Claim: S\-R ^ 0. By Theorem 4.4 we know there exists a basis of 1Z consisting of positive vectors Xi, • • • ,xr G X. Suppose S\n = 0. Let xt be an element in the basis. Then Sxt = 0. Let IXi be the order-ideal generated by Xj. Since S is positive then SIXi = 0. Furthermore, since S is continuous then SIXi — 0 and by Theorem 2.25, S {x^ = 0. Since 6 is indecomposable dd dd dd then {xi} = {0} or {xi) = X. We know that xt ^ 0 so {xi} ^ {0}. Therefore, {x^} = X. However, this would imply that weak-order unit and since Xi € X^ then Xi = X, contradicting the decomposition of X in Lemma 4.3. As there are no quasinilpotent elements in 6 and Range P = TZ for all S 6 6, S\TI is a full rank operator and S\-n = PSP. • Remark 5.3. We say that a set of operators A on a Banach lattice X is sep arating if for all x e X+ there exists /I G .4 such that Ax > 0. Therefore, instead of assuming the existence of a strictly positive projection of rank r in the closed convex hull of &, we can assume the existence of a dense, countable separating subset of £ (the set of all minimal projections). If {E{\ C £ is such a set, then we can take P = ^2n anEn where for every n, an > 0, ^2n an = 1 and X^n0™!!-^"!! < °°- Since X is a Banach space then P is well-defined and is a projection with range 1Z. By construction P is strictly positive. Claim: P* is strictly positive 41 Suppose P* is not strictly positive. Then there exists (f> € X* such that P*4> = 0. As a result E*4> = 0 for all i e N (i.e. ^RangeE, = 0 for all i G N). By Lemma 4.10 (iii) there exists F G £ such that 0|RangeF 7^ 0. Note that {Ei} is dense in S and therefore F = lira,- £'„r By continuity, 4>\ Range F = 0, a contradiction. 42 Chapter 6 Examples and Conclusions (i) Semigroup generated by a single indecomposable operator Suppose X is an order-continuous Banach lattice and A G B(X) is a compact indecomposable operator. We can apply the Perron-Frobenius theorem directly to this operator by generating the semigroup of powers n of A, followed by taking the Radjavi-closure. Let ©0 = {A | n G N}. We then define & = &0. This semigroup is indecomposable since A is indecomposable. It is a standard fact that there is only one minimal projection in this semigroup. Claim: There is only one minimal projection in & Claim: There is a positive eigenvector with eigenvalue r(A) By [1, Corollary 9.20] we know that r(A) > 0. By [1, Theorem 7.10] there exists a positive eigenvector with eigenvalue r(A). Let x be the eigenvector of A with eigenvalue r(A) and let Ix denote the order-ideal generated by x. Claim: Ix is dense Suppose Ix is not dense. Let y € Ix, then \y\ < \xx for some \x > 0. Then \Ay\ < A\y\ < \xAx = jir(A)x, implying that Ty G Ix. Hence Ix is T-invariant and therefore Ix is also T- invariant, contradicting the indecomposability of T. 43 Let E and F be nonzerzo minimal projections in (3. Claim: Either E = F or FF = FF = 0 Let P = .EF. Then P is a projection. Note that & is a com mutative semigroup and therefore EF = FE. Hence, it fol lows from P = EF = FE that kerECkerP and RangeP C Range E, so that P < E. Similarly, P Since E e & then a^™' —• F for some increasing sequence (n*) Cfj and a sequence of positive scalars (ai). Note that Ex ^ 0, otherwise E would vanish on Ix (which is dense by the above claim) and we would have E = 0. Hence, ttjX = a.iAnix —> Ex, so that Fx = Ax for some A > 0. This also implies that a; —> A. By squaring we find afx = (ctiAni) x —> £'2x = Fx = Ax and hence af —* A. It follows that A = 1. Therefore, Ex — x so that x G Range E. We follow the same steps for F and find that Fx = x, implying that EFx = x and therefore EF ^ 0. It follows from the above claim that E = F. As a result, Theorem 5.1 holds for & and therefore A. (ii) Finite Dimensional Let X = R4. Then X is an order-continuous Banach lattice. Let A = [}}] bea2x2 matrix on M2. Using A we form the semigroup e a = {[a«2aa$] I n> «i2, «2i, «22 £ K}- This semigroup has only compact operators but does not have a unique minimal ideal as the hA0 o o do not have the same range. minimal projections o o and However, the theorems proved in Chapter 4 still apply to this semigroup. A simple example of an indecomposable semigroup of matrices that is not generated by a single element is the semigroup generated by the Radjavi-closure of {[} §] > [o i]}• This semigroup is generated by minimal projections that have the same range. Therefore, Theorem 5.1 holds. (iii) The order-continuous Banach Lattice CQ 44 The Banach Lattice c$ := {(an) | «•« G 1, limn an = 0} is order-continuous and not isomorphic to any Lp space. It therefore provides an exten sion of the Perron-Frobenius theorem not covered in Radjavi and Rosen thal's book Simultaneous Triangularization. First, decompose the natu ral numbers into an infinite union of disjoint, infinite subsets (we can do this by separating the natural numbers into even numbers and those di visible by primes), N = \J^=1 An. Note that for all i G N, c0^A. '—* c0. Let P be any indecomposable projection on CQ. Let F be a block operator + acting on CQ composed of blocks a^P : CQ\A. —» c0\Ai, where a^ G R for alH, j G N and chose such that a^- —> 0 as j —> oo. Then the operator F is indecomposable and compact and therefore Theorem 5.1 holds. The extension of the Perron-Frobenius theorem to the realm of order- continuous Banach lattices will help to establish positivity as an appropriate means to approach the invariant subspace problem. (i) Relevance of Order-Continuity Of worthy investigation is the relevance of order-continuity in this the sis. Can a proof be constructed that does not use properties of order- continuity? Throughout this thesis norm-closed ideals are usually con structed to show decomposability. Thus, one may ask why we used the band approach at all? Why not define decomposability in terms of norm-closed ideals? Unfortunately, a proof for Theorem 3.4, one of the most useful theorems in this thesis, could not be found without using projection bands, and therefore order-continuity. (ii) Semigroups With Rank-one Projections Only one case of Radjavi's extension was not extended to this new set ting. In [8][Theorem 8.7.20] case (v) states: r = 1 if and only if given a Borel subset XQ of X with /i(Xo) < oo, a positive e, and any nonzero g > 0 in CP(X), there exists S E & such that the function Sg is positive on XQ except for a subsest of measure less than e. 45 (iii) Invariant Sublattices Vladimir Troitsky and Heydar Radjavi provided a number of nice results in the area of invariant sublattices in LP spaces [11]. Radjavi's extension of the Perron-Probenius theorem was used extensively in their paper, and thus, with this new extension of said theorem, it is possible that their paper may also be extended to the setting of order-continuous Banach lattices. (iv) Generalizing Beyond Banach Lattices How important is lattice structure? For an ordered Banach space we have similar ideas of ideals and bands. Let Y be an ordered Banach space and A C Y. Let us define I A '•= \JaeA ^ where Ia := (JneN [—na,na]. When X is an order-continuous Banach lattice then the set I A is the ideal generated by A and IA = {A} . Is it possible to generalize the Perron-Frobenius theorem to simply ordered Banach spaces? 46 Bibliography [1] Y. A. Abramovich and C. D. Aliprantis, An invitation to operator the ory, first ed., American Mathematical Society, Providence, Rhode Island, 2002. [2] Charalambos D. Aliprantis and Owen Burkinshaw, Positive operators, first ed., Springer, Netherlands, 2006. [3] Tom M. Apostol, Modular functions and Dirichlet series in number theory, second ed., Graduate Texts in Mathematics, vol. 41, Springer-Verlag, New York, 1990. [4] J.B. Conway, A course in functional analysis, Springer-Verlag, New York, 1990. [5] I. M. Gelfand, Normierte ringe, Matematicheskii Sbornik 9 (1941), 3-24. [6] J. D. Newburgh, The variation of spectra, Duke Mathematical Journal 44 (1951), 165-176. [7] Heydar Radjavi, The Perron-Frobenius theorem revisited, Positivity 3 (1999), 317-332. [8] Heydar Radjavi and Peter Rosenthal, Simultaneous triangularization, Springer-Verlag, New York, 2000. [9] Helmut H. Schaefer, Banach lattices and positive operators, Springer- Verlag, New York, 1974. 47 [10] E. Seneta, Non-negative matrices and Markov chains, second ed., Springer, New York, 2006. [11] V. Troitsky and H. Radjavi, Invariant sublattices, Illinois Journal of Math ematics, to appear. [12] Y. V. Turovskii, Volterra semigroups have invariant subspaces, Journal of Functional Analysis 162 (1999), 313-322. 48 (fi where x G X is a weak order-unit and 0 is a strictly positive functional. Remark 2.26. Let X be a vector lattice and ip a linear functional on X. If ip vanishes on X+ then ip = 0.