Spectral Theory for Nonlinear Operators

Transcription

1 DIPLOMARBEIT Spectral Theory for Nonlinear Operators ausgeführt am Institut für Analysis und Scientific Computing der Technischen Universität Wien unter der Anleitung von Ao.Univ.Prof. Dipl.-Ing. Dr.techn. Michael Kaltenbäck durch Andreas Widder, B.Sc. Matr. Nr Otto Glöckel-Gasse Mattersburg Wien,

2 Contents 1 Introduction Spectral theory for linear operators Spectral theory for nonlinear operators Preliminary considerations Characteristics of nonlinear operators The Dugundji Extension Theorem Set-valued maps The Antipodal Theorem The FMV-spectrum FMV-regular operators Stable solvability FMV-regularity Topological properties Subdivision of the FMV-spectrum Special classes of operators Eigenvalues and approximate eigenvalues Numerical range of operators Applications Maps on spheres Differential equations The nonlinear Fredholm alternative

3 Chapter 1 Introduction 1.1 Spectral theory for linear operators We first want to recapitulate some basic facts about spectral theory for a bounded linear operator T that operates on a Banach space X over the field K = R or K = C. The set of all such operators will be denoted as B(X) and will be equipped with the operator norm T = sup{ T x : x = 1}. We start with defining the resolvent set ρ(t ) := {λ K : λi T is bijective}, where I is the identity operator on X. It can easily be shown that ρ(t ) coincides with the set of points such that λi T is invertible and its inverse is also a bounded operator. Therefore, it also coincides with the set of points where λi T is a homeomorphism. The inverse operator R(λ, T ) := (λi T ) 1 is called the resolvent operator of T at λ. Since λ ρ(t ) and µ K with µ λ < R(λ, T ) 1 implies µ ρ(t ), ρ(t ) is an open subset of K. The spectrum of the operator T is definded as σ(t ) := K\ρ(T ), which is in turn closed. In the case of K = C the set ρ(t ) is also always nonempty and the spectral radius r(t ) := sup{ λ : λ ρ(t )} can be calculated by the Gelfand formula n r(t ) = lim T n. n The estimate r(t ) T is true, even in the case K = R. Consequently σ(t ) is always bounded and, therefore, compact. A very important property of the spectrum is the spectral mapping theorem. That is, for any 1

4 polynomial p(λ) = a n λ n + + a 1 λ + a 0 we get the identity σ(p(t )) = p(σ(t )) Here p(t ) denotes the operator a n T n + + a 1 T + a 0 I and p(σ(t )) = {p(λ) : λ σ(t )}. This theorem is one of the starting points for the spectral theorem, which allows the representation of certain linear operators as integrals over the spectrum. Another important property of the spectrum is its upper semi-continuity. Since this fact often remains unmentioned, we will give a short proof. Proposition 1.1 For T B(X) let G K be an open set with σ(t ) G. Then there exists a δ > 0 such that σ(s) G for every S B(X) with S T < δ. Proof: We define the closed set F = K\G. Since every λ in F belongs to ρ(t ), R(λ, T ) lies in B(X). We are going to show that also λ ρ(s) for all S B(X) satisfying S T < 1 R(λ, T ). To do so, we use the well known fact from the theory of Neumann series that I R is invertible for any R with R < 1. Therefore, the formula I (λi T ) 1 (λi S) = R(λ, T ) ((λi T ) (λi S)) R(λ, T ) (λi T ) (λi S) = R(λ, T ) S T < 1 implies that (λi T ) 1 (λi S) is invertible. Consequently, λi S is invertible, i.e. λ ρ(s). The fact that R(λ, T ) 0 as λ implies that actually δ := inf λ F 1 R(λ, T ) > 0. Heuristically, upper semi-continuity assures that the spectrum does not expand suddenly. However it is not lower semi-continuous, which means that even under small perturbations it can collapse. 1.2 Spectral theory for nonlinear operators Since spectral theory is so fruitful in the case of linear operators, it is natural to try to extend its principles to nonlinear operators. In order to justify the label spectral theory for nonlinear operators, we would like the spectrum of nonlinear operators to have similar properties as in the linear case. It should even be identical to the classical spectrum when applied to a linear operator. Since this task is complicated enough, we will restrict ourselves to continuous operators. 2

5 The first question when talking about a spectral theory for nonlinear operators is then, of course, how to define the spectrum. A first straightforward attempt could simply be, as in the linear case, to define the spectrum of an operator T as the set of all points λ such that λi T is not bijective. But simple examples show that in that case the spectrum would fail to have even any basic properties like being closed, bounded or nonempty. For linear operators, the linearity guarantees the linearity of its inverse and the open mapping theorem its continuity. Thus, bijectivity is equivalent to being a homeomorphism. Since this tool is not available for nonlinear operators, these two properties are not equivalent anymore. Defining the spectrum as all λ such that λi T is not a homeomorphism will consequently lead to a different kind of spectrum. Although this approach turns out to be just as dissapointing as the previous one, it points out an important difference between the spectral theory for linear and nonlinear operators. The linearity of an operator is responsible for the fact that many different properties are equivalent to bijectivity. So in the linear theory the spectrum contains information about all these properties, while at the same time one has to check only bijectivity, which is comparatively easy to handle. If we want to deal with nonlinear operators, we have to look at all these different properties seperately. So for any property A that makes sense for a nonlinear operator, we can define the A-resolvent ρ A (T ) := {λ K : λi T has property A} and the A-spectrum σ A (T ) = K\ρ A. The only restriction for this property A is that it should be equivalent to bijectivity in case of linear operators, so that the spectrum coincides with the classical one in the linear case. Therefore, there are many different spectra to be considered. None of them has yet lead to results of the same extent as the linear spectrum. By the above considerations this is not completely unexpected. It is possible, and not very unlikely, that there is no particular nonlinear spectrum that is as all-encompassing. However, the study of these many different spectra is not in vain, and leads to interesting results. In this thesis we will present one of these spectra, the Furi-Martelli-Vignoli-spectrum or FMVspectrum for short. This example will show how a spectral theory for nonlinear operators can be developed and we will provide some results from this new theory. 3

6 Chapter 2 Preliminary considerations In this chapter we discuss some results which will be used throughout this paper. We will restrict ourselves to results pertaining to the nonlinearity of operators. Well known results that strictly deal with linear operators (e.g. the open mapping theorem) will be cited without proofs throughout this thesis. We will also assume knowledge of basic theorems about topological properties of Banach spaces, like the Baire category theorem or the fact that closed balls are compact only in finite dimensional spaces. 2.1 Characteristics of nonlinear operators Let X and Y be two Banach spaces and T : X Y a continuous operator, which in general will be nonlinear. By C(X, Y ) we denote the set of all continuous operators from X into Y. Of course, this set forms a linear space. C(X) := C(X, X) is an algebra with respect to the composition. However, in the case of linear operators the space B(X, Y ) is normed by the operator norm. As indicated in the introduction, the significance of the operator norm here is due to the linearity of the operator. In the case of nonlinear operators one has to consider multiple seminorms or other characteristics. We only present the four characteristics that we will use in the following. Definition 2.1 For T C(X, Y ) we define and [T ] Q := lim sup x T (x) x T (x) [T ] q := lim inf x x, as elements of [0, ]. If [T ] Q <, we call T quasibounded. By Q(X, Y ) we denote the set of all quasibounded continuous maps of X into Y. 4

7 In particular, the fact that [T ] Q = λ or [T ] q = λ implies that there exists an unbounded sequence (x n ) n N in X such that lim n T (x n ) / x n = λ. Furthermore, the inequality [T ] q [T ] Q is obviously true. Consequently, [T ] Q = 0 actually implies that lim n T (x n ) / x n = 0 for every sequence (x n ) n N with x n. We now gather some more properties of these characteristics. Proposition 2.2 Let T, S C(X, Y ) and R C(Y, Z). Then the following holds: (i) [T ] q > 0 implies that T is coercive, i.e. lim T (x) =. x (ii) One of the quantities on the left being finite, [T ] q [S] Q [T + S] q [T ] q + [S] Q. (iii) One of the quantities on the left being finite, [T ] q [S] q [T S] Q. [T S] Q = 0 implies [T ] q = [S] q. In particular, (iv) [T 1 ] Q = [T ] 1 q dimensional. if T is a homeomorphism and either T is linear or X and Y are finite (v) [R T ] q [R] q [T ] q. Proof: This verifies (i). For (ii) consider If [T ] q > 0, then for sufficiently large x there exists a k > 0 such that T (x) k x. T (x) + S(x) T (x) + S(x) [T + S] q = lim inf lim inf x x x x T (x) S(x) lim inf + lim sup = [T ] q + [S] Q. x x x x For the second inequality replace T by T + S and S by S. Similarly, consider [T S] Q = lim sup x lim sup x T (x) S(x) x T (x) x lim sup x which for symmetry reasons proves (iii). T (x) S(x) x ( T (x) lim sup S(x) ) x x x S(x) T (x) S(x) lim inf lim inf lim inf = [T ] q [S] q, x x x x x x In (iv), the two assumptions that T is linear and that X and Y are finite dimensional both assure that x T (x). We therefore can consider the chain of equalities [T 1 ] Q = lim sup y T 1 (y) y = lim sup x ( x T (x) = lim inf x ) T (x) 1 = 1. x [T ] q 5

8 Finally, to see (v), consider R(T (x)) [R T ] q = lim inf = lim inf x x x R(T (x)) T (x) T (x) x R(T (x)) T (x) R(y) T (x) lim inf lim inf lim inf lim inf = [R] q [T ] q. x T (x) x x y y x x The last inequality holds true, because if T (x), then [T ] q = 0 and the inequality holds, and if T (x) we can set T (x) = y to see that the inequality holds. Next, we are going to introduce a tool that is often used in nonlinear functional analysis. In the following, B ɛ (z) will always denote the open ball with radius ɛ and center z. Definition 2.3 Let X be a Banach space and M X. The Hausdorff measure of noncompactness of M is defined as α(m) := inf{ɛ > 0 m N, {z 1,..., z m } X : M B ɛ (z 1 ) B ɛ (z m )} [0, ]. A finite set {z 1,..., z m } X with M B ɛ (z 1 ) B ɛ (z m ) is called a finite ɛ-net for M. So if M has a measure of noncompactness ɛ, it can be covered by finitely many open balls of any radius greater than ɛ. Proposition 2.4 Let X be a Banach space, M, N X, λ K, and z X. Then the measure of noncompactness has the following properties: (i) α(m) = α(m). (ii) α(m) = 0 if and only if M is precompact, i.e. has a compact closure. (iii) α(m) < if and only if M is bounded. (iv) M N α(m) α(n). (v) α(m) α(n) α(m + N) α(m) + α(n), where for the first inequality to hold either α(m) or α(n) needs to be finite. (vi) α(λm) = λ α(m). (vii) α(m + z) = α(m). (viii) α(m N) = max{α(m), α(n)}. (ix) α(co(m)) = α(m), where co(m) denotes the convex hull of M. (x) α(b r (z)) = 0 if dim X <, and α(b r (z)) = r if dim X =. 6

9 Proof: The first four assertion are straightforward to verify. To see (v), observe, that if {z 1,..., z m } is a finite ɛ-net for M, and {w 1,..., w n } is a finite η-net for N, then {z i + w j i = 1,..., m; j = 1,..., n} is a finite (ɛ+η)-net for M +N, which shows α(m +N) α(m)+α(n). The first inequality in (v) immediately follows from this one. Moreover, if {z 1,..., z m } is a finite ɛ-net for M then {λz 1,..., λz m } is a finite λ ɛ-net for λm, so (vi) follows. Similarly, (vii) follows from the observation that {z 1,..., z m } is a finite ɛ-net for M if and only if {z 1 + z,..., z m + z} is a finite ɛ-net for M + z. For (viii) we use the fact that if {z 1,..., z m } is a finite ɛ-net for M and {w 1,..., w n } is a finite η-net for N, then {z 1,..., z m } {w 1,..., w n } is a finite δ-net for M N, where δ = max{ɛ, η}. To see that (ix) holds true, by (iv) it suffices to show that α(co(m)) α(m). So for η > α(m) choose a finite η-net {z 1,... z m } and define N = co({z 1... z m }). Any x co(m) can be written as a convex combination x = a i x i with x i M and a i = 1. For every x i there is a z j(i) with x i z j(i) < η. Setting z = a i z j(i) we get x z = ai x i a i z j(i) = ai (x i z j(i) ) a i x i z j(i) a i η = η. Because z N, we have dist(x, N) η. Since N is compact, we can find a finite ɛ-net {w 1... w n } for N and arbitrary ɛ > 0. This is then a finite (η + ɛ)-net for co(m). Since every bounded set in a finite dimensional space is precompact, the first assertion of (x) is trivial. To see the second one, by (vi) and (vii) it suffices to show that α(b 1 (0)) = 1. Since B 1 (0) can be covered by itself, we have α(b 1 (0)) 1. Assume α(b 1 (0)) < 1. Then B 1 (0) can be covered by finitely many open balls with radius η (0, 1). Again by (vi) and (vii) we can in turn cover each of those balls with finitely many open balls with radius η 2, which gives us a finite cover of B 1 (0) by such sets. Since η n 0, by iterating this process we get a finite cover of B 1 (0) with open balls with a radius smaller than ɛ for every ɛ > 0. This shows, that B 1 (0) is precompact, i.e. B 1 (0) is compact. Thus, we get a contradiction to the well known fact that this is only the case in finite dimensional spaces. Unlike the characteristics we introduced before, the following characteristics can also be defined for an arbitrary subset of a Banach space. Definition 2.5 Let Z X. For T C(Z, Y ) we define [T ] A := inf{k : k > 0, α(t (M)) kα(m) for all bounded M Z} and [T ] a := sup{k : k > 0, α(t (M)) kα(m) for all bounded M Z} as elements of [0, ]. We call [T ] A the measure of noncompactness of T and denote by A(Z, Y ) the set of all continuous maps T from Z into Y with [T ] A <. 7

10 Note that in finite dimensional spaces we always have [T ] A = 0 and [T ] a =. In infinite dimensional spaces, where this characteristic is of more use, we get the equivalent representations and α(t (M)) [T ] A = sup >α(m)>0 α(m) [T ] a = inf >α(m)>0 α(t (M)). α(m) Sets with α(m) = 0 can be left out here, since the continuity of T assures that also α(t (M)) = 0. This can be seen by considering α(t (M)) α(t (M)) = 0. From this representation it is also clear that an operator T with [T ] A < maps bounded sets into bounded sets. These two characteristics are also closely related to two important properties of operators. Definition 2.6 Let T C(X, Y ). The operator T is called compact, if T (M) is precompact for every bounded set M X. The operator T is called proper, if the preimage T 1 (N) is compact for every compact set N Y. Proposition 2.7 Let X, Y, and Z be Banach spaces. For T, S C(X, Y ) and R C(Y, Z) the following assertions hold true: (i) T is compact if and only if [T ] A = 0. (ii) If [T ] a > 0 and [T ] q > 0, then T is proper. (iii) [T ] a > 0 implies that T is proper on closed bounded sets. (iv) One of the quantities on the left being finite, [T ] a [S] A [T + S] a [T ] a + [S] A. (v) One of the quantities on the left being finite, [T ] a [S] a [T S] A. [T S] A = 0 implies [T ] a = [S] a. In particular, (vi) [T 1 ] A = [T ] 1 a dimensional. if T is a homeomorphism and either T is linear or X and Y are finite (vii) [R] a [T ] a [R T ] a [R] A [T ] a, where the second inequality holds if [R] A <. Proof: The first assertion follows immediately from the definition of a compact operator and the definition of the measure of noncompactness of an operator. In (ii), because [T ] a > 0, we may find a k > 0 such that α(t (M)) kα(m) for each bounded M X. As [T ] q > 0, Lemma 2.2,(i) shows that T is coercive. Therefore, for any compact set N Y, T 1 (N) is bounded and α(t 1 (N)) 1 k α(t (T 1 (N))) 1 α(n) = 0. k 8

11 Thus T 1 (N) is precompact. Since T is continuous, T 1 (N) is also closed and therefore compact. The same reasoning shows that T is proper on closed bounded sets if only [T ] a > 0. (iv) and (v) are proven similarly as in Lemma 2.2. (vi) is trivial if X and Y are finite dimensional. If one is infinite dimensional, the assumption that T is linear assures that T maps bounded sets into bounded sets. Also, due to T being a homeomorphism, α(t (M)) = 0 if and only if α(m) = 0. We therefore get the chain of equalities [T 1 α(t 1 (N)) ] A = sup = sup >α(n)>0 α(n) >α(m)>0 ( α(m) α(t (M)) = inf >α(m)>0 ) α(t (M)) 1 = 1. α(m) [T ] a Finally, let k T > 0 such that α(t (M)) k T α(m) for all bounded M X. Further, let k R > 0 such that α(r(m)) k R α(m) for all bounded M Y. Then α(r T (M)) k R α(t (M)) k R k T α(m) for all bounded M X. This shows the first inequality in (vii). For the second inequality, consider α(r T (M)) [R T ] a = inf = inf >α(m)>0 α(m) >α(m)>0 >α(t (M)) 0 α(r T (M)) α(t (M)) α(t (M)) α(m) α(r(n)) α(t (M)) sup inf = [R] A [T ] a. >α(n)>0 α(n) >α(m)>0 α(m) For linear operators the characteristic [T ] q and [T ] a are linked to the injectivity and bijectivity of the operator T. [T ] A and [T ] Q on the other hand can be linked to the operator norm. Lemma 2.8 Let T : X Y be linear and bounded. Then the following holds (i) [T ] Q = T. (ii) [T ] A T. (iii) If [T ] q > 0, then T is injective. (iv) If T is bijective, then [T ] q > 0 and [T ] a > 0. Proof: The first assertion follows immediately from the linearity of T. Also, if {z 1,..., z m } is a finite ɛ-net for a bounded set M, then {T z 1,..., T z m } is obviously a finite T ɛ-net for T (M), i.e. α(t (M)) T α(m). Hence, [T ] A T. For (iii), assume T is not injective. Then there exist x y X with T x = T y. Let (x n ) n N and (y n ) n N be sequences with x n x and y n y. Then n(x n y n ) for n. Using the linearity of T we get the contradiction T z [T ] q = lim inf z z lim T (n(x n y n )) n T (x n y n ) T x T y = lim = = 0. n n(x n y n ) n n x n y n x y 9

12 To see (iv), note that because T is linear, bijective, and bounded, it is invertible and its inverse is also a linear and bounded operator. Using Proposition 2.2,(iv) we get and using Proposition 2.7,(vi) we get [T ] q = [T 1 ] 1 Q = T 1 1 = 1 T 1 > 0, [T ] a = 1 [T 1 1 ] A T 1 > The Dugundji Extension Theorem Dugundjis extension theorem assures that for any continuous map T : A Y that is defined on a closed set A X, there exists a continuous extension T of T to X, i.e. T : X Y and T (x) = T (x) for x A, with the additional restriction that the range of T is in some sense constrained by the original range of T. In its original version X is allowed to be an arbitrary metric space and Y a locally convex linear space. We will give a simpler version of this theorem for mappings between two Banach spaces X and Y. First we need some definitions. Definition 2.9 (i) Let A = (A i ) i I and B = (B j ) j J be systems of subsets of a set X. B is called a refinement of A, if for every j J there is an i I with B j A i. (ii) Let X be a Hausdorff space and U = (U i ) i I be a cover of X. U is called locally finite, if every point x X has a neighborhood O(x) such that O(x) intersects at most finitely many elements of U. (iii) A Hausdorff space X is called paracompact if for every open cover U of X there exists an open cover V of X, such that V is a locally finite refinement of U. The following definition will be needed in the next section, but is closely related to the definitions above. Definition 2.10 Let U = (U i ) i I be an open cover of X. A system of real valued continuous functions (f i ) i I is called a partition of unity subordinate to U if (a) f i (x) 0 for all x X and all i I, (b) (supp(f i )) i I is a locally finite system, where supp(f) := {x X : f(x) 0}, (c) supp(f i ) U i for all i I, 10

13 (d) i I f i(x) = 1 for all x X. The following theorem contains two well known topological results. Theorem 2.11 (i) Every metric space is paracompact. (ii) To every open cover U of a paracompact space there exists a a partition of unity subordinate to U. Proof: For example [Q]. We are now able to formulate our version of Dugundjis theorem. Theorem 2.12 (Dugundji Extension Theorem) Let X and Y be Banach spaces and let T : C K be a continuous map, where C X is closed and K Y is convex. Then there exists a continuous mapping T : X K such that T (x) = T (x) for x C. Proof: For each x in X\C set r x = 1 3 dist(x, C). Then diamb r x (x) dist(b rx, C). The collection (B rx (x)) x X\C is an open cover of X\C. By Theorem 2.11,(i) it has a locally finite refinement (O i ) i I which is an open cover of X\C. Define q : X\C (0, ) by q(x) = dist(x, X\O i ). i I Since (O i ) i I is locally finite, the sum contains only finitely many not vanishing terms and q is a continuous function. Further, since the (O i ) i I form an open cover of X\C, q(x) > 0. Next we define for i I and x X\C ρ i (x) = dist(x, X\O i). q(x) We then have 0 ρ i (x) 1 and i I ρ i(x) = 1. Note that since (O i ) i I is a refinement of (B rx (x)) x X\C, for any O i there exists x 0 X\C with dist(c, O i ) dist(c, B rx0 (x 0 )) diamb rx0 (x 0 ) > 0. For each i I, we can therefore choose an x i C such that dist(x i, O i ) 2dist(C, O i ), and define T (x) = { T (x) for x C, i I ρ i(x)t (x i ) for x / C. Obviously T : X K and T is an extension of T. Further, T is continuous on the interior of C as well as on X\C. In order to show that T is continuous, it suffices to prove that T is continuous on C. 11

14 Let x C. Since T is continuous, for given ɛ > 0 we find a δ > 0 such that T (x) T (y) ɛ for y C with x y δ. For y X\C we have T (x) T (y) = T (x) ρ i (y)t (x i ) ρ i (y) T (x) T (x i ). i I i I If ρ i (y) 0 then dist(y, X\O i ) > 0, i.e. y O i. Taking the infimum over all w O i in y x i y w + w x i we obtain y x i diamo i + dist(x i, O i ). Now, O i B rx0 (x 0 ) for some x 0 X\C. Since we get diamo i diamb rx0 (x 0 ) dist(b rx0, C) dist(c, O i ), y x i 3dist(C, O i ) 3 y x. Thus, for i such that ρ i (y) 0 we get x x i y x + y x i 4 x y. Hence, x y δ/4 implies x x i δ, and, therefore, T (x) T (x i ) ɛ. Finally, T (x) T (y) i I ρ i (y) T (x) T (x i ) ɛ i I ρ i (y) = ɛ. This theorem has two very useful corollaries. Corollary 2.13 Let X and Y be Banach spaces and let T : C Y be continuous, where C X is closed. Then there exists a continuous extension T of T to X with T (X) co(t (C)). Corollary 2.14 Let X and Y be Banach spaces and let T : C Y be compact, where C X is closed and bounded. Then there exists a compact extension T of T to X with T (X) co(t (C)). Proof: Since T is compact and C bounded, T (C) is precompact. By Lemma 2.4(ix), so is co(t (C)). Therefore, an extension T of T as in Theorem 2.12 is also compact. 12

15 2.3 Set-valued maps For a set X we will denote the power set of X by P(X). By a multivalued (or set-valued) map between two sets X and Y we mean a map T : X P(Y ). It assigns to each point of X a subset of Y. Every map T : X Y can be identified with a set-valued map T : X P(Y ) by setting T (x) = {T (x)}. Such maps are then referred to as single-valued maps. For a set-valued map we define the image of a set M as T (M) := T (x) and the preimage of a set A as x M T 1 (A) := {x X : T (x) A }. Note that unlike for single-valued maps the inclusion T (T 1 (A)) A need not hold. However, unless T 1 (A) =, we have T (T 1 (A)) A. For single-valued maps this definition coincides with the classical preimage of a set. Furthermore, we call the set G(T ) := {(x, y) : x X, y T (x)} the graph of T. For set-valued maps we have the following notions of continuity: Definition 2.15 Let X and Y be topological spaces and T : X P(Y ) a set-valued map. (i) T is called upper semi-continuous if T 1 (A) is closed for all closed sets A Y. (ii) T is called lower semi-continuous if T 1 (A) is open for all open sets A Y. For single-valued maps both upper and lower semi-continuity are obviously equivalent to continuity. In the following we will also need a different characterization of semi-continuity. Proposition 2.16 Let X and Y be topological spaces and T : X P(Y ) a set-valued map. (i) T is upper semi-continuous if and only if for every x X and every open set V in Y with T (x) V there exists a neighbourhood U(x) such that T (U(x)) V. (ii) T is lower semi-continuous if and only if for every x X, y T (x) and every neighbourhood V (y) of y there exists a neighbourhood U(x)of x such that T (u) V (y), for all u U(x). Proof: (i) Suppose T fulfills the assumptions and A Y is closed. Choose x (T 1 (A)) C, then T (x) A C. Since A C is open, there exists a neighbourhood U(x) of x, such that 13

16 T (U(x)) A C. Therefore U(x) (T 1 (A)) C and it follows that T 1 (A) is closed. Conversely, suppose T 1 (A) is closed for all closed A Y. Let x X and V Y be open with T (x) V. Then V C is closed and by assumption so is T 1 (V C ). Moreover, x / T 1 (V C ). Hence, there exists a neighbourhood of x with U(x) (T 1 (V C )) C. This neighbourhood apparently satisfies T (U(x)) V. (ii) Suppose T fulfills the assumptions and A Y is open. Assume that x T 1 (A) and choose y T (x) A. Since A is open, there exists a neighbourhood of y with V (y) A. Because of our assumption, there exists a neighbourhood U(x) of x with T (u) V (y) for all u U(x). This means U(x) T 1 (V (y)) T 1 (A). It follows that T 1 (A) is open. Conversely, suppose T 1 (A) is open for every open set A Y. Assume that x X, y T (x), and V (y) is an open neighbourhood of y. Then T 1 (V (y)) is open and x T 1 (V (y)). Therefore, U(x) = T 1 (V (y)) is a neighbourhood of x. It follows that T (u) V (y) for all u U(x). With this characterization of upper semi-continuity it is easy to see that Proposition 1.1 indeed shows the upper semi-continuity of the classical spectrum. It also clarifies what we meant with the heuristical explanation that the values of a lower semi-continuous map cannot collapse and the values of an upper semi-continuous map cannot suddenly expand. We now give a condition for a set-valued map to be upper semi-continuous, which will be needed later on. For this, let X be linear space and p be a seminorm on X. For x X we denote by U δ (x) the p-neighbourhood U δ (x) := {y X : p(x y) < δ}. Definition 2.17 A set-valued map T : X P(K) is called closed if the graph of T is closed in X K, i.e. y n T (x n ), y n y and p(x n x) 0 imply that y T (x). Lemma 2.18 Let T : X P(K) be closed. Then for every x X and y K\T (x) there exists δ > 0 and an open set V y K such that y V y and T (U δ (x)) V y =. Proof: Assume the assertion does not hold true for x X and y K\T (x). Let (δ n ) n N be a null sequence. Since T (U δn (x)) B δn (y), we can find x n U δn (x) and y n B δn (y) such that y n T (x n ). Further, we get that y n y and p(x n x) 0. Since T is closed, this shows that y T (x), which contradicts our choice of y. Lemma 2.19 Let T : X P(K) be a closed map. If then T is upper semi-continuous. sup y p(x) for all x X, y T (x) 14

17 Proof: Let x X and let V K be open with T (x) V. By Proposition 2.16,(i), we have to show that there exists a δ > 0 with T (U δ (x)) V. Choose η > 0 with T (U η (x))\v (if this is not possible, there is nothing to prove). For z U η (x) we have sup λ p(z) p(x) + η. λ T (z) Consequently, T (U η (x)) is bounded, and so the set C := T (U η (x))\v is compact. Let y C be arbitrary. Since y / T (x) and T is closed, by Lemma 2.18 we find δ(y) > 0 and an open V y K with y V λ and T (U δ(y) (x)) V y =. Obviously, {V y : y C} is an open cover of C. Since C is compact, we get C V y1 V ym for suitable y 1,..., y m C. Putting δ = min{η, δ(y 1 ),..., δ(y m )} we see that T (U δ (x)) C =. Since T (U δ (x)) T (U η (x)), we get T (U δ (x)) V. For the rest of this section we will only deal with lower semi-continuity. Lemma 2.20 Let T : X P(Y ) be lower semi-continuous, where X and Y are topological spaces. If S : X P(Y ) is such that T (x) = S(x) for all x X, then S is lower semicontinuous. Proof: Assume S is not lower semi-continuous. Then there exists an x X, y S(x), and an open set V (y) Y with y V (y) such that for all neighbourhoods U(x) of x there exists an u U(x) with S(u) V (y) =. Since V (y) is open, this also means T (u) V (y) = S(u) V (y) =, which consequently implies T (u) V (y) =. If y T (x), this contradicts the lower semicontinuity of T. If y / T (x), we have y T (x). Hence, V (y) T (x). Further, V (y) is an open neighbourhood V (z) of every z V (y) T (x). As above we get T (u) V (z) = T (u) V (y) =. Lemma 2.21 Let T : X P(Y ), where Y is a Banach space. Let T be lower semi-continuous, O Y open, f : X Y a continuous map, and suppose that T (x) (f(x) + O) for all x X. Then S : X P(Y ), defined by S(x) = T (x) (f(x) + O), is lower semi-continuous. Proof: Let x X, y S(x), and V (y) be an open set with y V (y). Then y T (x) and (f(x) + O) V (y) is an open neighbourhood of y. Therefore, there exists an o O with y = f(x) + o and an ɛ > 0 such that f(x) + B ɛ (o) (f(x) + O) V (y). Since T is lower semi-continuous, there exists a neighbourhood Ũ(x) of x such that for all u Ũ(x) we have T (u) (f(x) + B ɛ/2 (o)). Since f is continuous, there exists a neighbourhood Û(x) of x, such that f(x) f(u) < ɛ/2 for all u Û(x). Let a f(x) + B ɛ/2(o). Then for all u Û(x) we get a (f(u) + o) a f(x) o + f(x) + o f(u) o < ɛ, i.e. a f(u) + B ɛ (o) (f(u) + O). Set U(x) = Ũ(x) Û(x). Then for any u U(x) we have T (u) (f(x) + B ɛ/2(o)), since 15

18 u Ũ(x). But for any a T (u) (f(x)+b ɛ/2(o)) we have a (f(x)+b ɛ/2 (o)), hence a V (y). Further, u Û(x) yields a (f(u) + O). Hence, a T (u) (f(u) + O) V (y) = S(u) V (y). Therefore, S(u) V (y) for all u U(x). Lemma 2.22 Let T = X P(Y ) be lower semi-comtinuous and Y be a Banach space. Suppose f : X R is continuous and f(x) 0 for all x X. Define S : X P(Y ) by { T (x) B S(x) = f(x) (0) if f(x) > 0, {0} if f(x) = 0. If S(x) for all x X, then S is lower semi-continuous. Proof: Let x X, y S(x), and V (y) be open with y V (y). First, suppose f(x) > 0. Then f(x) > y. Hence, there exists an ɛ > 0 with f(x) > y + ɛ. Since f is continuous, there exists an open neighbourhood Û(x) of x such that f(u) > y + ɛ for all u Û(x). Moreover, there exists an open neighbourhood Ũ(x) of x such that T (u) B y +ɛ(0) V (y) for all u Ũ(x), because T is lower semi-continuous and B y +ɛ(0) V (y) is an open neighbourhood of y. So for u in the open set U(x) = Û(x) Ũ(x) we get S(u) V (y) = T (u) B f(u) V (y) T (u) B y +ɛ V (y). Now suppose f(x) = 0. Then y = 0 and V (y) is a neighbourhood of 0, i.e. there is an ɛ > 0 such that B ɛ (0) V (y). Since f is continuous, there is a open neighbourhood U(x) of x such that f(u) < ɛ for u U(x). For f(u) > 0 we have S(u) = T (u) B f(u) (0) B f(u) (0) B ɛ (0) V (y). For f(u) = 0 we have S(u) = {0} V (y). In any case, S(u) V (y). Finally, we define the objects that will be studied for the rest of this section. Definition 2.23 Let T : X P(Y ) be a set-valued map. A single-valued map t : X Y is called a selection of T, if t(x) T (x) for all x X. The existence of a selection is obviously equivalent to the fact, that T (x) for all x X. Selections are an important tool when dealing with set-valued maps. Therefore, it is of interest to show the existence of selections with additional properties. The following important theorem by Michael provides conditions for the existence of a continuous selection. Theorem 2.24 (Michael s Selection Theorem) Let X be a paracompact space, Y a Banach spaces, and T : X P(Y ) a lower semi-continuous set-valued map. If T (x) is nonempty, closed, and convex for all x X, then there exists a continuous selection t : X Y of T. 16

19 Proof: For y Y and A Y set d(y, A) = inf a A y a. As a first step, we show that for each ɛ > 0 there exists a continuous map f : X Y such that d(f(x), T (x)) < ɛ for all x X. (2.1) Fix ɛ > 0 and choose any selection m : X Y. Since T is lower semi-continuous, for each x X there exists an open neighbourhood U(x) of x such that T (u) B ɛ (m(x)) for all u U(x). (2.2) Let (f i ) i I be a partition of unity subordinate to the open cover (U(x)) x X of X. For every i I choose an x i X such that supp(f i ) U(x i ) if supp(f i ) and set f(x) = f i (x)m(x i ). i I Then f is a continuous function of X into Y. If f i (x) > 0 for some i, then x U(x i ) and by (2.2) m(x i ) T (x) + B ɛ (0). Since T (x) + B ɛ (0) is convex, we get f(x) T (x) + B ɛ (0). Hence, d(f(x), T (x)) < ɛ, i.e. f satisfies (2.1). In the second step we construct the requested selection. Set ɛ n = 2 n. We will inductively define a sequence (f n ) n N of continuous mappings f n : X Y with d(f n (x), T (x)) < ɛ n, x X, n = 1, 2,... (2.3) d(f n (x), f n 1 (x)) < ɛ n 1, x X, n = 2, 3,... (2.4) As we showed in the first step, there exists f 1 with d(f 1 (x), T (x)) < 1/2, x X. Assume that n 2 and we already have constructed f 1,..., f n 1. For each x X we define G(x) = (f n 1 (x) + B ɛn 1 (0)) T (x) By the induction hypothesis, G(x) is not empty. Since T (x) convex, so is G(x). Furthermore, G : X P(Y ) is lower semi-continuous by Lemma So we can apply the first part of our proof also to G, since the only additional property of T is that T (x) is closed, which was not used in the first part. Therefore, there exists a continuous map f n : X Y such that (2.3) holds. By construction, f n also satisfies (2.4). Since n=1 ɛ n converges, (f n ) n N is a uniform Cauchy sequence and, therefore, converges to a continuous map t : X Y. Since T (x) is closed, d(t(x), T (x)) = 0 implies that t is a selection of T. 17

20 Corollary 2.25 Let X be paracompact, Y a Banach space, and T : X P(Y ) a lower semi-continuous map such that T (x) is nonempty, closed, and convex for every x X. Set m(x) = inf{ y : y T (x)} and suppose f : X R is continuous weith f(x) 0 for all x, and f(x) > m(x) whenever m(x) > 0. Then there exists a continuous selection t of T such that t(x) f(x) for all x X. Proof: The map S : X P(Y ) defined by { T (x) B S(x) = f(x) (0) if f(x) > 0, {0} if f(x) = 0, is lower semi-continuous by Lemma Hence, R : X P(Y ), defined by R(x) = S(x), is lower semi-continuous by Lemma Furthermore, R(x) is nonempty, closed, and convex for all x X. By Theorem 2.24, there exists a selection t of R. By construction t is also a selection of T and fulfils t(x) f(x). Theorem 2.26 Let T : X Y be a continuous linear surjection from a Banach space X onto a Banach space Y. Then there exists a continous function s : Y X and a constant M > 0 such that for every y Y s(y) T 1 (y), s(y) M y. Proof: First, we show that there exists an M > 0, such that m(y) = inf{ x : x T 1 (y)} M y and m(y) < M y whenever m(y) > 0. Since T is surjective, by the open mapping theorem the image of B 1 (0) X under T is an open neighbourhood of 0 Y. In particular, there exists a δ > 0 such that S δ = {y Y : y = δ} T (B 1 (0)). For an arbitrary 0 y Y we have y 0 = δy y 1 S δ. Hence, there exists a x 0 B 1 (0) with T (x 0 ) = y 0 and, further, T (x 0 y δ 1 ) = y. Therefore, m(y) x 0 y /δ δ 1 y. Now set M = λδ 1 with a λ > 1. Define S : S 1 P(X) by S(x) = T 1 ({x}). Then for any open U X the set S 1 (U) = {y S 1 : T 1 ({y}) U } = T (U) S 1 is open in S 1 by the open mapping theorem. Hence, S is lower semi-continuous. Moreover, S(y) is nonempty, closed, and convex for all y S 1. So we can apply Corollary 2.25 to S with f(x) = M x and get a continuous function s : S 1 X with s(y) T 1 (y) and s(y) M y. Setting { y s( y y ) if y 0, s(y) = 0 if y = 0, we get a map from the whole space Y into X with the mentioned properties. 18

21 2.4 The Antipodal Theorem To formulate and prove the Antipodal Theorem we first need to establish the fixed point index and other preliminary results. Definition 2.27 Let G be a nonempty and open bounded set in a Banach space X. Then V (G, X) denotes the set of all compact maps T : G X such that T has no fixed points on the boundary G of G, i.e. x G : T (x) = x. Two maps T, S V (G, X) are called compactly homotopic on G if there exists a continuous map H with the following properties (i) H : G [0, 1] X is compact, (ii) H(x, 0) = T (x) and H(x, 1) = S(x) for x G, (iii) H(x, t) x for all (x, t) G [0, 1]. We write T = S. The map H is called a compact homotopy. Proposition 2.28 Let G be a nonempty open bounded subset of a Banach space X and let T, S V (G, X). Then the following holds true: (i) The relation = is an equivalence relation. (ii) inf x G x T (x) > 0. (iii) If sup T (x) S(x) < inf x T (x), then T = S. x G x G Proof: The relation = is reflexive, since T = T by H(x, t) = T (x). If T = S by H, then S = T by H 1 = H(x, 1 t), so = is symmetric. To see that = is also transitive, let T = S by H 1 and S = R by H 2, then T = R by { H 1 (x, 2t) for 0 t 1 H(x, t) = 2, 1 H 2 (x, 2t 1) for 2 < t 1. To prove (ii), we show that (I T )( G) is closed. Becaus of 0 / (I T )( G), this gives the assertion. So let (y n ) n N be a sequence in (I T )( G) that converges to a y X. By Corollary 2.14 we can extend T to a compact map ˆT on X. Since G is closed and bounded, and [I ˆT ] a [I] a [ ˆT ] A = 1 > 0 by Lemma 2.7,(iii), point (ii) of the same lemma tells us that (I ˆT ) is proper on G, so I T is also proper there. Because (y n ) n N {y} is compact, also (I T ) 1 ((y n ) n N {y}) is compact. If we choose x n (I T ) 1 (y n ) G, then (x n ) n N has a convergent subsequence x nk x G. Because I T is continuous, we get (I T )(x) = y. Thus, y (I T )( G). 19

22 For (iii) let H(x, t) = (1 t)t (x) + ts(x). Then H is continuous and compact on G [0, 1], H(x, 0) = T (x) and H(x, 1) = S(x) for x G, and for all (x, t) G [0, 1] we have H(x, t) x = T (x) x t(t (x) S(x)) T (x) x + t T (x) S(x) > 0. The next lemma is especially interesting in connection with point (iii) of the previous proposition. Lemma 2.29 Let X and Y be Banach spaces and M X be nonempty and bounded. Let T : M Y be a compact operator. Then to each ɛ > 0, there exists a compact operator P : M Y such that sup x M T (x) P (x) ɛ and P (M) is contained in finite dimensional subspace of Y. Proof: Fix ɛ > 0. Since M is bounded, T (M) is precompact. Hence, there exist y i T (M), i = 1,..., N, such that min i=1,...,n T (x) y i < ɛ for all x M. Define continuous functions p i : M R by p i (x) = max(ɛ T (x) y i, 0). Then for each x M there exists at least one i such that p i (x) 0. Therefore, we can define P : M Y by P (x) = N i=1 p i(x)y i N i=1 p i(x). For all x M we then have N i=1 P (x) T (x) = p i(x)(y i T (x)) N i=1 p i(x) N i=1 p i(x)ɛ N i=1 p i(x) ɛ. By construction, P (M) lies in the finite dimensional subspace spanned by {y i : i = 1,..., N}. Finally, since T (M) is bounded, so is P (M). Since a bounded set in a finite dimensional space is precompact, P is compact. Definition 2.30 Let X be a Banach space. An integer valued function i(t, G), where G is a nonempty open bounded subset of X and T V (G, X), is called the fixed point index of T on G if the following axioms are satisfied. (A1) If T (x) = x 0 for all x G and some fixed x 0 G, then i(t, G) = 1. (A2) If i(t, G) 0, then there exists an x G such that T (x) = x. (A3) If there are open G i X, i = 1,..., n, such that G = n i=1 G i and G i G j = for i j, then i(t, G) = n i(t, G i ) i=1 20

23 whenever T V (G, X) and T Gi V (G i, X) for all i. (A4) If T = S, then i(t, G) = i(s, G). Despite their importance, we will not give a proof of the following three results, since they are beyond the scope of the present thesis. The proof of the first of these theorems includes a lengthy construction of the fixed point index, while the other two proofs necessitate a detailed knowledge of this construction. For a proof of these theorems see for example Zeidler [Z]. Theorem 2.31 For every Banach space X there exists a unique fixed point index. Theorem 2.32 Let G be a nonempty open bounded region and T V (G, X). Let Ω X be a bounded region with G Ω and h : Ω X be a linear map such that I h is compact. Suppose h : Ω h(ω) is a homeomorphism, T (G) Ω, and 0 / h (I T )( G). Then i(h T h 1, h(g)) = i(t, G). Theorem 2.33 If T V (G, X) and if T (G) lies entirely in a closed linear subspace Y of X, then i(t, G) = i(t, G Y ), where the right hand side is the fixed point index in Y. The last two lemmas that we need before we are able to proof the Antipodal Theorem deal with fixed point free extensions of fixed point free maps. Lemma 2.34 Let A and B be closed and bounded subsets of a Banach space X with A B. Let H : A [0, 1] X be compact and H(x, t) x for all (x, t) A [0, 1]. If H(., 1) has a fixed point free compact extension H 1 : B X, then H has a compact extension H : B [0, 1] X with H(x, t) x for all (x, t) B [0, 1] Proof: Set H 0 (x, t) = { H 1 (x) for x B, t = 0 H(x, t) for x A, t [0, 1], and extend H 0 by Corollary 2.14 to a compact map H 0 : B [0, 1] X. Let B 0 = {x B : t [0, 1] with H 0 (x, t) = x}. Then B 0 is closed and A B 0 =. Therefore, by Corollary 2.13, we can extend the map ã(x) = { 0 for x B 0 1 for x A to a continuous map a : B [0, 1]. Finally, we set H(x, t) = H 0 (x, a(x)t) for (x, t) B [0, 1]. Then H is the desired extension of H. In fact, if H(x, t) = x, then H 0 (x, τ) = x for some τ, i.e. x B 0. Therefore, a(x) = 0 and further H 0 (x, 0) = x, i.e. H 1 (x) = x, which is impossible by hypothesis. 21

24 Lemma 2.35 Let A and B be compact subsets of a proper linear subspace of R N with N 2 and A B. Then every continuous fixed point free map f : A R N has a continuous fixed point free extension f to B. Proof: We may assume without loss of generality that A and B lie in the subspace with coordinates (ζ 1,..., ζ M, 0,..., 0). We also set F (x) = f(x) x. Since F has no zeroes on A, we get α = inf x A F (x) > 0. We extend F continuously to F : B R N by Corollary By the Weierstrass approximation theorem, there exists a map G : B R N with sup x B F G < α/3, and such that the components of G are polynomials. Since G i (x)/ ζ j = 0 for j = M + 1,..., N, we have det(g (x)) = 0 on B. By Sard s Theorem, G(B) has no interior points. Hence, there exists an x 0 such that x 0 / G(B) and x 0 < α/3. For (x, t) (A [0, 1]) (B 1) define h(x, t) = (1 t)f (x) + t(g(x) x 0 ). Then h(x, t) 0 since h(x, t) F (x) F (x) G(x) x 0 > 0. This also implies h(x, 1) 0 for x B. Thus we can apply Lemma 2.34 to H(x, t) = x + h(x, t). In particular, we get that the extention H(x, 0) : B R N is fixed point free and H(x, 0) = f(x) for x A. Theorem 2.36 (Antipodal Theorem) Let X be a Banach space and let T : B R (0) X be compact. If T has no fixed points on B R (0) and if T satisfies the antipodal condition T ( x) = T (x) for x B R (0), then the fixed point index i(t, B R (0)) is odd. Proof: First, we proof this theorem for dim(x) <. For this we can assume that X = R N for some N 1. Q is called an orthant of R N if Q = {x R N : 0 e j x j for j = 1,..., N}, where (e 1,... e N ) { 1, 1} N is fixed. Per definition, there are 2 N orthants in R N. Let Q 1,..., Q 2 N 1 be the 2 N 1 orthants of R N with x N 0 for all x Q j. Then R N = Q 1 Q 2 N 1 Q 1 Q 2 N 1. 1 Now fix r (0, R) and define B j as the interior of Q j B R (0)\B r (0). Then B R (0) = B r (0) B 1 B 2 N 1 B 1 B 2 N 1. We show that there exists a continuous function S : B R (0) R N with S(x) = T (x) for all x B R (0), (2.5) S(x) = 0 for all x B r (0), (2.6) 1 This is the multidimensional equivalent of the following easy construction in R 2 : Divide the plane into its four quadrants. Choose Q 1 and Q 2 as the first and second quadrant, respectively. Then Q 1 is the third quadrant and Q 2 the fourth. So R 2 = Q 1 Q 2 Q 1 Q 2. 22

25 S(x) x for all x (±B j ), j = 1,..., 2 N 1, (2.7) S( x) = S(x) for all x B R (0). (2.8) Because of (2.6) we only have to define S on ±B j. There S is already defined on B j B R (0) and B j B r (0) by (2.5) and (2.6), and S is fixed point free there. The remaining border of B j can be divided into N parts, each of which lies in a N 1 dimensional hyperplane of R N. We start by applying Lemma 2.35 to each of these parts in order to extend S continuously to B 1 such that (2.7) is fulfilled. We now can extend S continuously from B 1 to B 1 using Theorem Further, we define S on B 1 by S(x) = S( x) for x B 1, such that (2.8) is fulfilled. We repeat this procedure consecutively with B 2 to B 2 N 1. Note that when considering B j, S is already defined on B j B i for i < j, so we can skip defining S on that part of the border of B j. Now (2.6) implies that i(s, B r (0)) = 1 by axiom (A1) of the fixed point index. Using the homeomorphism h : x x we can use Theorem 2.32 (with Ω = B R (0)) to see that i(s(x), B j ) = i( S( x), B j ) = i(s(x), B j ). Axiom (A3) of the fixed point index then gives 2 N 1 2 N 1 i(s, B R (0)) = i(s, B r (0)) + (i(s, B j ) + i(s, B j )) = i(s, B r (0)) + 2 i(s, B j ), j=1 i.e. i(s, B R (0)) is odd. Since S(x) = T (x) for x B R (0), Proposition 2.28 together with axiom (A4) of the fixed point index show that i(t, B R (0)) = i(s, B R (0)). Now let dim(x) =. First, we observe that (T (x) T ( x))/2 is compact and coincides with T (x) for x B R (0). By Proposition 2.28 and (A4), this yields i(t (x), B R (0)) = i((t (x) T ( x))/2, B R (0)). By Lemma 2.29, there is a compact operator S : B R (0) Y, where Y is a finite dimensional subspace of X, such that sup x BR (0) T (x) S(x) < inf x BR (0) T (x) x. As above, for P (x) = (S(x) S( x))/2 we have i(p, B R (0)) = i((t (x) T ( x))/2, B R (0)). But since P maps B R (0) into a finite dimensional subspace Y, Theorem 2.33 shows that i(p, B R (0)) = i(p, B R (0) Y ), the latter of which is odd by the first part of the proof, since P ( x) = P (x). j=1 23

26 Chapter 3 The FMV-spectrum 3.1 FMV-regular operators FMV-regularity, named after its inventors M. Furi, M. Martelli, and A. Vignoli, will play the essential role in the definition of the FMV-spectrum as explained in the introduction Stable solvability Stable solvability is a property of operators that is important in the definition of FMV-regularity and of the FMV-spectrum. It assures solvability of certain types of equations. In the following, X and Y will always denote Banach-spaces. Definition 3.1 A continuous operator T : X Y is called stably solvable, if, given any compact operator S : X Y with [S] Q = 0, the equation T (x) = S(x) has a solution x X. Lemma 3.2 Let T C(X, Y ) be stably solvable. Then T is surjective. Proof: For y Y the operator S(x) y fulfils [S] A = [S] Q = 0. By the stable solvability of T, there is an x X with T (x) = S(x) = y. In general, the converse of this lemma does not hold true. We will however show later that in the case of linear operators stable solvability reduces to surjectivity. The notion of stable solvability can be extended further. This more general definition will be usefull in the study of FMV-regular operators. Definition 3.3 For k 0, we call an operator T C(X, Y ) k-stably solvable, if, given any operator S C(X, Y ) with [S] A k and [S] Q k, the equation T (x) = S(x) has a solution x X. Obviously, 0-stably solvable operaters are exactly the stably solvable operators. Moreover, every k-stably solvable operator is certainly k -stably solvable for 0 k < k. This motivates the following definition. 24

27 Definition 3.4 For T C(X, Y ) we call µ(t ) := inf{k : k 0, T is not k-stably solvable} the measure of stable solvability of T. We call an operator T C(X, Y ) strictly stably solvable if µ(t ) > 0, i.e. T is k-stably solvably for some k > 0. For the rest of this section we return to stably solvable operators and show some of their properties. Lemma 3.5 Let T C(X, Y ) with [T ] q > 0. Then T is stably solvable if and only if the equation T (x) = S(x) has a solution x X for every compact operator S : X Y for which the set {x X : S(x) 0} is bounded. Proof: Since every operator S with S(x) = 0 outside a bounded set fulfils [S] Q = 0, this direction of the proof is trivial. So let S be a compact operator from X into Y with [S] Q = 0. For n N define the operator S n (x) = d n ( x )S(x), where 1 if 0 t n, d n (t) = 2 1 nt if n t 2n, 0 if t 2n. Then {x X : S n (x) 0} is bounded and S n compact. Hence, by assumption, there exists an x n X with S n (x n ) = T (x n ). If x n n for some n N, then T (x n ) = S n (x n ) = S(x n ) and we are finished. So assume x n > n for all n N. Then x n as n, and we get the contradiction T (x) T (x n ) [T ] q = lim inf lim = lim x x n x n d n( x n ) S(x n) n x n S(x n ) S(x) lim lim sup = 0. n x n x x Corollary 3.6 The identity operator is stably solvable. Proof: Since [I] q = 1 we can restrict ourselves to compact maps S with {x X : S(x) 0} is bounded to prove stable solvability. For such S, there obviously exists an r > 0 such that S(B r (0)) B r (0). But then Schauders fixed point theorem tells us that S has a fixed point, i.e. S(x) = I(x). 25

28 Theorem 3.7 Let T C(X) be stably solvable, B X a nonempty closed subset, and let H : B X be a continuous operator. Assume that H(B) is bounded and T 1 (co(h(b))) B, i.e. T (y) co(h(b)) y B. Moreover, suppose that the equality α(t (M)) = α(h(m)) implies the precompactness of M for any M B. Then the equation T (x) = H(x) has a solution ˆx X. Proof: We construct a sequence (x n ) n N as follows. Let x 0 B be arbitrary. Due to Lemma 3.2 we can choose inductively x n T 1 (H(x n 1 )) for n 1, so T (x n ) = H(x n 1 ). By construction, the set A = {x 0, x 1,... } satisfies T (A) = {T (x o )} H(A). Therefore, we get α(t (A)) = α({t (x o )} H(A)) = α(h(a)). Hence, A is precompact by assumption. Moreover, by construction A B and consequently H(A) H(B) is bounded. Next, we show that A T 1 (H(A )), where A denotes the set of all accumulation points of A. Note that A since A is precompact. So for x A we find a subsequence (x nk ) k N of (x n ) n N such that x nk x, hence, also H(x nk 1) = T (x nk ) T (x) as k. Therefore, if we choose y A to be a limit point of (x nk 1) k N, we get H(y) = T (x). Next, we consider the family M = {M : M A, M = M, T 1 (co(h(m))) M}. Since B M, this family is nonempty. Denote by M 0 the intersection of all M M. Note that M 0 is nonempty and closed, since M A for each M M and each M is closed. Set M 1 = T 1 (co(h(m 0 ))). Then for all M M we get M 1 = T 1 (co(h(m 0 ))) T 1 (co(h(m))) M, hence M 1 M 0. Further, M 1 is closed and satisfies both M 1 = T 1 (co(h(m 0 ))) T 1 (co(h(a ))) A and T 1 (co(h(m 1 ))) T 1 (co(h(m 0 ))) = M 1. Therefore, M 1 M and M 0 M 1. Consequently, T 1 (co(h(m 0 ))) = M 1 = M 0. Since T is stably solvable and, therefore, surjective by Lemma 3.2, this in turn implies T (M 0 ) = co(h(m 0 )) 26