Spectral Theory for Nonlinear Operators

(1)

DIPLOMARBEIT

Spectral Theory for Nonlinear Operators

ausgef¨uhrt am Institut f¨ur

Analysis und Scientific Computing

der Technischen Universit¨at Wien

unter der Anleitung von

Ao.Univ.Prof. Dipl.-Ing. Dr.techn. Michael

Kaltenb¨ ack

durch

Andreas Widder, B.Sc.

Matr. Nr. 0625002 Otto Gl¨ockel-Gasse 7

7210 Mattersburg

Wien, 6.05.2012

(2)

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 kTk= sup{kT xk:kxk= 1}.

We start with defining the resolvent set

ρ(T) :={λ∈K:λI−T is bijective},

whereI is the identity operator on X. It can easily be shown thatρ(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|<kR(λ, T)k⁻¹ implies µ∈ρ(T), ρ(T) is an open subset of K.

The spectrum of the operatorT 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

r(T) = lim

n→∞

pn

kTⁿk.

The estimater(T)≤ kTkis true, even in the caseK=R. Consequentlyσ(T) is always bounded and, therefore, compact.

A very important property of the spectrum is the spectral mapping theorem. That is, for any

(4)

polynomial p(λ) =a_nλⁿ+· · ·+a₁λ+a₀ we get the identity σ(p(T)) =p(σ(T))

Herep(T) denotes the operatora_nTⁿ+· · ·+a₁T +a₀I andp(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 ForT ∈ B(X) let G⊆K be an open set withσ(T)⊆G. Then there exists a δ >0 such thatσ(S)⊆G for everyS ∈ B(X) withkS−Tk< δ.

Proof: We define the closed set F =K\G. Since every λin F belongs to ρ(T),R(λ, T) lies inB(X). We are going to show that alsoλ∈ρ(S) for all S∈ B(X) satisfying

kS−Tk< 1 kR(λ, T)k.

To do so, we use the well known fact from the theory of Neumann series thatI−Ris invertible for any R withkRk<1. Therefore, the formula

I−(λI−T)⁻¹(λI−S)

=kR(λ, T) ((λI−T)−(λI−S))k ≤

≤ kR(λ, T)k k(λI−T)−(λI−S)k=kR(λ, T)k kS−Tk<1

implies that (λI−T)⁻¹(λI−S) is invertible. Consequently,λI−S is invertible, i.e. λ∈ρ(S).

The fact thatkR(λ, T)k →0 as |λ| → ∞ implies that actually δ:= inf

λ∈F

1

kR(λ, T)k >0.

2 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.

(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 pointsλsuch thatλI−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 λsuch thatλI−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

ρA(T) :={λ∈K:λI−T has propertyA}

and theA-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 FMV- spectrum 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.

(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 [T]Q:= lim sup

kxk→∞

kT(x)k kxk and

[T]q := lim inf

kxk→∞

kT(x)k kxk ,

as elements of [0,∞].If [T]Q <∞, we call T quasibounded. By Q(X, Y) we denote the set of all quasibounded continuous maps ofX into Y.

(7)

In particular, the fact that [T]_Q =λor [T]_q=λimplies that there exists an unbounded sequence (xn)n∈N in X such that limn→∞kT(xn)k/kx_nk = λ. Furthermore, the inequality [T]q ≤[T]Q

is obviously true. Consequently, [T]_Q = 0 actually implies that limn→∞kT(x_n)k/kx_nk= 0 for every sequence (x_n)n∈N withkx_nk → ∞.

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 thatT is coercive, i.e. lim

kxk→∞kT(x)k=∞.

(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. In particular,

[T−S]_Q= 0 implies [T]_q= [S]_q.

(iv) [T⁻¹]_Q = [T]⁻¹_q if T is a homeomorphism and either T is linear or X and Y are finite dimensional.

(v) [R◦T]_q≥[R]_q[T]_q.

Proof: If [T]_q >0, then for sufficiently largekxkthere exists ak >0 such thatkT(x)k ≥kkxk.

This verifies (i).

For (ii) consider

[T+S]_q= lim inf

kxk→∞

kT(x) +S(x)k

kxk ≤lim inf

kxk→∞

kT(x)k+kS(x)k kxk

≤lim inf

kxk→∞

kT(x)k

kxk + lim sup

kxk→∞

kS(x)k

kxk = [T]q+ [S]_Q. For the second inequality replace T by T+S and S by −S.

Similarly, consider [T−S]_Q= lim sup

kxk→∞

kT(x)−S(x)k

kxk ≥lim sup

kxk→∞

kT(x)k − kS(x)k kxk

≥lim sup

kxk→∞

kT(x)k

kxk −kS(x)k kxk

≥lim sup

kxk→∞

kT(x)k

kxk −lim inf

kxk→∞

kS(x)k

kxk ≥lim inf

kxk→∞

kT(x)k

kxk −lim inf

kxk→∞

kS(x)k

kxk = [T]_q−[S]_q, which for symmetry reasons proves (iii).

In (iv), the two assumptions that T is linear and that X and Y are finite dimensional both assure that kxk → ∞ ⇔ kT(x)k → ∞. We therefore can consider the chain of equalities

[T⁻¹]Q = lim sup

kyk→∞

T⁻¹(y)

kyk = lim sup

kxk→∞

kxk kT(x)k =

lim inf

kxk→∞

kT(x)k kxk

−1

= 1

[T]_q.

(8)

Finally, to see (v), consider [R◦T]q = lim inf

kxk→∞

kR(T(x))k

kxk = lim inf

kxk→∞

kR(T(x))k kT(x)k

kT(x)k kxk

≥ lim inf

kxk→∞

kR(T(x))k kT(x)k lim inf

kxk→∞

kT(x)k

kxk ≥lim inf

kyk→∞

kR(y)k kyk lim inf

kxk→∞

kT(x)k

kxk = [R]q[T]q. The last inequality holds true, because ifkT(x)k9∞, then [T]q = 0 and the inequality holds, and if kT(x)k → ∞we can set T(x) =y to see that the inequality holds. 2

Next, we are going to introduce a tool that is often used in nonlinear functional analysis. In the following, B(z) will always denote the open ball with radius and centerz.

Definition 2.3 Let X be a Banach space andM ⊆X. The Hausdorff measure of noncompactness of M is defined as

α(M) := inf{ >0| ∃m∈N,{z₁, . . . , z_m} ⊆X :M ⊆B(z₁)∪ · · · ∪B(z_m)} ∈[0,∞].

A finite set {z₁, . . . , zm} ⊆X withM ⊆B(z1)∪ · · · ∪B(zm) is called a finite -net for M. So if M has a measure of noncompactness , it can be covered by finitely many open balls of any radius greater than .

Proposition 2.4 Let X be a Banach space,M, N ⊆X, λ∈K, andz∈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 ifdimX <∞, and α(B_r(z)) =r if dimX=∞.

(9)

Proof: The first four assertion are straightforward to verify. To see (v), observe, that if {z₁, . . . , zm} is a finite-net forM, and{w₁, . . . , wn}is a finiteη-net forN, then{z_i+wj|i= 1, . . . , m;j= 1, . . . , n}is a finite (+η)-net forM+N, which showsα(M+N)≤α(M)+α(N).

The first inequality in (v) immediately follows from this one. Moreover, if {z₁, . . . , z_m} is a finite -net for M then {λz₁, . . . , λzm} is a finite |λ|-net for λM, so (vi) follows. Similarly, (vii) follows from the observation that{z₁, . . . , z_m}is a finite-net for M if and only if {z₁+ z, . . . , z_m+z} is a finite-net for M+z. For (viii) we use the fact that if{z₁, . . . , z_m} is a finite -net for M and {w₁, . . . , wn} is a finiteη-net for N, then {z₁, . . . , zm} ∪ {w₁, . . . , wn} is a finite δ-net forM ∪N, where δ= max{, η}.

To see that (ix) holds true, by (iv) it suffices to show thatα(co(M))≤α(M). So forη > α(M) choose a finite η-net {z₁, . . . zm} and define N = co({z₁ . . . zm}). Any x ∈ co(M) can be written as a convex combination x =P

a_ix_i with x_i ∈M and P

a_i = 1. For every x_i there is a z_j(i) with |x_i−z_j(i)|< η. Settingz=P

aiz_j(i) we get kx−zk=

Xa_ix_i−X a_iz_j(i)

=

Xa_i(x_i−z_j(i)) ≤X

|a_i|

x_i−z_j(i) ≤X

|a_i|η =η.

Because z ∈ N, we have dist(x, N) ≤ η. Since N is compact, we can find a finite -net {w₁ . . . wn} forN and arbitrary >0. This is then a finite (η+)-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(B₁(0)) = 1. Since B1(0) can be covered by itself, we haveα(B1(0)) ≤1. Assume α(B1(0)) <1. Then B1(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 radiusη², which gives us a finite cover ofB1(0) by such sets. Since ηⁿ →0, by iterating this process we get a finite cover of B₁(0) with open balls with a radius smaller than for every >0. This shows, thatB₁(0) is precompact, i.e. B1(0) is compact. Thus, we get a contradiction to the well known fact that

this is only the case in finite dimensional spaces. 2

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]Athe measure of noncompactness ofT and denote byA(Z, Y)the set of all continuous maps T from Z into Y with [T]_A<∞.

(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

[T]A= sup

∞>α(M)>0

α(T(M)) α(M) and

[T]a= inf

∞>α(M)>0

α(T(M)) α(M) .

Sets withα(M) = 0 can be left out here, since the continuity ofT assures that alsoα(T(M)) = 0.

This can be seen by consideringα(T(M))≤α(T(M)) = 0. From this representation it is also clear that an operatorT 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 operatorT 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. ForT, S ∈ C(X, Y)andR ∈ 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. In particular,

[T−S]A= 0 implies [T]a= [S]a.

(vi) [T⁻¹]A = [T]⁻¹_a if T is a homeomorphism and either T is linear or X and Y are finite dimensional.

(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

kα(T(T⁻¹(N)))≤ 1

kα(N) = 0.

(11)

Thus T⁻¹(N) is precompact. SinceT 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 ifα(M) = 0. We therefore get the chain of equalities [T⁻¹]A= sup

∞>α(N)>0

α(T⁻¹(N))

α(N) = sup

∞>α(M)>0

α(M) α(T(M)) =

∞>α(M)>0inf

α(T(M)) α(M)

−1

= 1

[T]a

.

Finally, let kT >0 such thatα(T(M))≥kTα(M) for all bounded M ∈X. Further, let kR>0 such that α(R(M))≥ k_Rα(M) for all bounded M ∈ Y. Then α(R◦T(M))≥k_Rα(T(M))≥ k_Rk_Tα(M) for all bounded M ⊆ X. This shows the first inequality in (vii). For the second inequality, consider

[R◦T]a = inf

∞>α(M)>0

α(R◦T(M))

α(M) = inf

∞>α(M)>0

∞>α(T(M))6=0

α(R◦T(M)) α(T(M))

α(T(M)) α(M)

≤ sup

∞>α(N)>0

α(R(N)) α(N) inf

∞>α(M)>0

α(T(M))

α(M) = [R]A[T]a.

2

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=kTk.

(ii) [T]A≤ kTk.

(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 ofT. Also, if{z₁, . . . , z_m} is a finite -net for a bounded set M, then {T z₁, . . . , T zm} is obviously a finite kTk-net for T(M), i.e. α(T(M))≤ kTkα(M). Hence, [T]A≤ kTk.

For (iii), assume T is not injective. Then there exist x 6=y ∈X with T x =T y. Let (x_n)n∈N

and (yn)n∈N be sequences with xn → x and yn → y. Then kn(x_n−yn)k → ∞ for n → ∞.

Using the linearity ofT we get the contradiction [T]q= lim inf

kzk→∞

kT zk

kzk ≤ lim

n→∞

kT(n(x_n−y_n))k

kn(x_n−yn)k = lim

n→∞

nkT(x_n−y_n)k

nkx_n−ynk = kT x−T yk kx−yk = 0.

(12)

To see (iv), note that becauseT 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

[T]q= [T⁻¹]⁻¹_Q = T⁻¹

−1 = 1

kT⁻¹k >0, and using Proposition 2.7,(vi) we get

[T]_a= 1

[T⁻¹]_A ≥ 1

kT⁻¹k >0.

2

2.2 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 versionXis allowed to be an arbitrary metric space andY 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= (Ai)i∈I and B= (Bj)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 withB_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 spaceX is called paracompact if for every open cover U of X there exists an open cover V of X, such thatV is a locally finite refinement ofU.

The following definition will be needed in the next section, but is closely related to the definitions above.

Definition 2.10 Let U = (Ui)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 toU if

(a) f_i(x)≥0 for all x∈X and all i∈I,

(b) (supp(fi))i∈I is a locally finite system, where supp(f) :={x∈X:f(x)6= 0}, (c) supp(fi)⊆Ui for alli∈I,

(13)

(d) P

i∈If_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 coverU of a paracompact space there exists a a partition of unity subordinate to U.

Proof: For example [Q]. 2

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 thatT˜(x) =T(x) for x∈C.

Proof: For eachx inX\C setrx= ¹₃dist(x, C). Then diamBrx(x)≤dist(B_r_x, C).

The collection (Brx(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) =X

i∈I

dist(x, X\O_i).

Since (Oi)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 fori∈I and x∈X\C

ρi(x) = dist(x, X\O_i) q(x) . We then have 0≤ρi(x)≤1 andP

i∈Iρi(x) = 1.

Note that since (O_i)i∈I is a refinement of (B_r_x(x))_x∈X\C, for anyO_i there existsx₀∈X\Cwith dist(C, Oi)≥dist(C, Brx0(x0))≥diamBrx0(x0)>0.

For each i ∈ I, we can therefore choose an xi ∈ C such that dist(xi, Oi) ≤ 2dist(C, Oi), and define

T˜(x) =

( T(x) f or x∈C, P

i∈Iρi(x)T(xi) f or 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.

(14)

Letx∈∂C. SinceT is continuous, for given >0 we find aδ >0 such thatkT(x)−T(y)k ≤ fory∈C withkx−yk ≤δ. Fory ∈X\C we have

T˜(x)−T(y)˜ =

T(x)−X

i∈I

ρi(y)T(xi)

≤X

i∈I

ρi(y)kT(x)−T(xi)k.

If ρi(y) 6= 0 then dist(y, X\O_i) > 0, i.e. y ∈ Oi. Taking the infimum over all w ∈ Oi in ky−xik ≤ ky−wk+kw−xik we obtain

ky−x_ik ≤diamO_i+ dist(x_i, O_i).

Now, Oi ⊆Brx0(x0) for some x0 ∈X\C. Since diamO_i≤diamB_r_x

0(x₀)≤dist(B_r_x

0, C)≤dist(C, O_i), we get

ky−xik ≤3dist(C, Oi)≤3ky−xk.

Thus, for i such that ρ_i(y) 6= 0 we get kx−x_ik ≤ ky−xk +ky−x_ik ≤ 4kx−yk. Hence, kx−yk ≤δ/4 implieskx−xik ≤δ, and, therefore, kT(x)−T(xi)k ≤. Finally,

T˜(x)−T(y)˜ ≤X

i∈I

ρ_i(y)kT(x)−T(x_i)k ≤X

i∈I

ρ_i(y) =.

2

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 toX 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 ofT as in Theorem 2.12 is also compact. 2

(15)

2.3 Set-valued maps

For a setX we will denote the power set ofX byP(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 T⁰ :X → P(Y) by settingT⁰(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) := [

x∈M

T(x)

and the preimage of a set A as

T⁻¹(A) :={x∈X:T(x)∩A6=∅}.

Note that unlike for single-valued maps the inclusionT(T⁻¹(A))⊆A need not hold. However, unlessT⁻¹(A) =∅, we haveT(T⁻¹(A))∩A6=∅. 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 ofT.

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 ifT⁻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 thatT(U(x))⊆V.

(ii) T is lower semi-continuous if and only if for everyx∈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)6=∅, for allu∈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

(16)

T(U(x))⊆A^C. ThereforeU(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 satisfiesT(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. SinceA is open, there exists a neighbourhood ofy withV(y)⊆A.

Because of our assumption, there exists a neighbourhoodU(x) ofxwithT(u)∩V(y)6=∅ 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 ofx. It follows that T(u)∩ V(y)6=∅ for allu∈U(x).

2 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, letX be linear space and p be a seminorm onX. For x∈X we denote by U_δ(x) thep-neighbourhoodU_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 ⊆Ksuch 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 (δn)n∈N be a null sequence. SinceT(Uδn(x))∩Bδn(y)6=∅, we can find xn ∈Uδn(x) and yn∈Bδn(y) such thaty_n∈T(x_n). Further, we get thaty_n→yand p(x_n−x)→0. Since T is closed, this shows

thaty ∈T(x), which contradicts our choice ofy. 2

Lemma 2.19 Let T :X→P(K) be a closed map. If sup

y∈T(x)

|y| ≤p(x) for all x∈X,

thenT is upper semi-continuous.

(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(Uδ(x))⊆V.

Choose η > 0 with T(U_η(x))\V 6= ∅ (if this is not possible, there is nothing to prove). For z∈U_η(x) we have

sup

λ∈T(z)

|λ| ≤p(z)≤p(x) +η.

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 findδ(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 ofC. SinceCis compact, we getC ⊆V_y₁∪ · · · ∪V_y_m for suitabley₁, . . . , y_m∈C. Putting δ= min{η, δ(y₁), . . . , δ(ym)}we see thatT(Uδ(x))∩C=∅. SinceT(Uδ(x))⊆T(Uη(x)), we get

T(U_δ(x))⊆V. 2

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 semi- continuous.

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) =∅. SinceV(y) is open, this also means T(u)∩V(y) = S(u)∩V(y) =∅, which consequently impliesT(u)∩V(y) =∅. Ify∈T(x), this contradicts the lower semicontinuity of T.

Ify /∈T(x), we havey∈T(x). Hence,V(y)∩T(x)6=∅. Further,V(y) is an open neighbourhood V(z) of every z∈V(y)∩T(x). As above we getT(u)∩V(z) =T(u)∩V(y) =∅. 2

Lemma 2.21 LetT :X→P(Y), whereY is a Banach space. LetT be lower semi-continuous, O ⊆ Y open, f : X → Y a continuous map, and suppose that T(x)∩(f(x) +O) 6=∅ for all x∈X. Then S :X→P(Y), defined byS(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 > 0 such that f(x) +B(o) ⊆ (f(x) +O) ∩V(y). Since T is lower semi-continuous, there exists a neighbourhood ˜U(x) of x such that for all u ∈ U˜(x) we have T(u)∩(f(x) +B_/2(o))6=∅.

Since f is continuous, there exists a neighbourhood ˆU(x) of x, such thatkf(x)−f(u)k < /2 for all u ∈ Uˆ(x). Let a ∈ f(x) +B_/2(o). Then for all u ∈ Uˆ(x) we get ka−(f(u) +o)k ≤ ka−f(x)−ok+kf(x) +o−f(u)−ok< , i.e. a∈f(u) +B(o)⊆(f(u) +O).

Set U(x) = ˜U(x)∩Uˆ(x). Then for any u ∈ U(x) we have T(u)∩(f(x) +B_/2(o)) 6=∅, since

(18)

u∈U˜(x). But for anya∈T(u)∩(f(x) +B_/2(o)) we havea∈(f(x) +B_/2(o)), hencea∈V(y).

Further,u∈Uˆ(x) yields a∈(f(u) +O). Hence, a∈T(u)∩(f(u) +O)∩V(y) =S(u)∩V(y).

Therefore, S(u)∩V(y)6=∅ for all u∈U(x). 2

Lemma 2.22 Let T = X → P(Y) be lower semi-comtinuous and Y be a Banach space.

Suppose f :X→R is continuous andf(x)≥0 for all x∈X. Define S :X→P(Y) by S(x) =

( T(x)∩B_f(x)(0) if f(x)>0,

{0} if f(x) = 0.

If S(x)6=∅ for all x∈X, then S is lower semi-continuous.

Proof: Letx∈X,y ∈S(x), andV(y) be open withy∈V(y). First, supposef(x)>0. Then f(x) > kyk. Hence, there exists an > 0 with f(x) > kyk+. Since f is continuous, there exists an open neighbourhood ˆU(x) of x such that f(u) >kyk+for allu∈Uˆ(x). Moreover, there exists an open neighbourhood ˜U(x) of x such that T(u)∩B_kyk+(0)∩V(y) 6= ∅ for all u∈U˜(x), becauseT is lower semi-continuous and Bkyk+(0)∩V(y) is an open neighbourhood of y. So for u in the open setU(x) = ˆU(x)∩U˜(x) we get

S(u)∩V(y) =T(u)∩B_f_(u)∩V(y)⊇T(u)∩B_kyk+∩V(y)6=∅.

Now supposef(x) = 0. Theny= 0 andV(y) is a neighbourhood of 0, i.e. there is an >0 such that B(0)⊆V(y). Sincef is continuous, there is a open neighbourhoodU(x) of x such that f(u)< foru∈U(x). Forf(u)>0 we haveS(u) =T(u)∩B_f_(u)(0)⊆B_f(u)(0)⊆B(0)⊆V(y).

Forf(u) = 0 we have S(u) ={0} ⊆V(y). In any case, S(u)∩V(y)6=∅. 2 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 allx∈X.

The existence of a selection is obviously equivalent to the fact, that T(x) 6= ∅ 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 Ba- nach spaces, andT :X→P(Y) a lower semi-continuous set-valued map. IfT(x) is nonempty, closed, and convex for all x∈X, then there exists a continuous selection t:X →Y of T.

(19)

Proof: For y∈Y and A⊆Y setd(y, A) = infa∈Aky−ak. As a first step, we show that for each >0 there exists a continuous map f :X →Y such that

d(f(x), T(x))< for all x∈X. (2.1) Fix >0 and choose any selection m :X →Y. Since T is lower semi-continuous, for each x∈X there exists an open neighbourhoodU(x) ofx such that

T(u)∩B(m(x))6=∅ for allu∈U(x). (2.2) Let (fi)i∈I be a partition of unity subordinate to the open cover (U(x))x∈X of X. For every i∈I choose anx_i∈X such that supp(f_i)⊆U(x_i) if supp(f_i)6=∅and set

f(x) =X

i∈I

f_i(x)m(x_i).

Then f is a continuous function of X intoY. If f_i(x) >0 for some i, then x ∈U(x_i) and by (2.2)

m(x_i)∈T(x) +B(0).

Since T(x) +B(0) is convex, we get f(x) ∈ T(x) +B(0). Hence, d(f(x), T(x)) < , i.e. f satisfies (2.1).

In the second step we construct the requested selection. Set _n = 2⁻ⁿ. We will inductively define a sequence (fn)n∈N of continuous mappingsfn:X →Y with

d(f_n(x), T(x))< _n, x∈X, n= 1,2, . . . (2.3) d(fn(x), fn−1(x))< n−1, x∈X, n= 2,3, . . . (2.4) As we showed in the first step, there existsf₁ withd(f₁(x), T(x))<1/2, x∈X.

Assume thatn≥2 and we already have constructedf₁, . . . , fn−1. For eachx∈X we define G(x) = (fn−1(x) +Bn−1(0))∩T(x)

By the induction hypothesis, G(x) is not empty. SinceT(x) convex, so is G(x). Furthermore, G:X→P(Y) is lower semi-continuous by Lemma 2.21. 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,fn also satisfies (2.4).

Since P∞

n=1n converges, (fn)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. 2

(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{kyk :y ∈ T(x)} and suppose f : X → R is continuous weith f(x) ≥ 0 for all x, andf(x)> m(x) whenever m(x)>0. Then there exists a continuous selectiont ofT such that kt(x)k ≤f(x) for allx∈X.

Proof: The mapS :X→P(Y) defined by S(x) =

( T(x)∩B_f(x)(0) if f(x)>0,

{0} if f(x) = 0,

is lower semi-continuous by Lemma 2.22. Hence, R :X → P(Y), defined by R(x) =S(x), is lower semi-continuous by Lemma 2.20. Furthermore,R(x) is nonempty, closed, and convex for allx∈X. By Theorem 2.24, there exists a selectiontofR. By constructiontis also a selection

of T and fulfilskt(x)k ≤f(x). 2

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),

• ks(y)k ≤Mkyk.

Proof: First, we show that there exists anM >0, such thatm(y) = inf{kxk:x∈T⁻¹(y)} ≤ Mkyk and m(y) < Mkyk whenever m(y) > 0. Since T is surjective, by the open mapping theorem the image of B₁(0) ⊆X under T is an open neighbourhood of 0 ∈Y. In particular, there exists a δ >0 such thatSδ ={y ∈Y :kyk=δ} ⊆T(B1(0)). For an arbitrary 06=y∈Y we have y₀ = δykyk⁻¹ ∈ S_δ. Hence, there exists a x₀ ∈ B₁(0) with T(x₀) = y₀ and, further, T(x₀kykδ⁻¹) =y. Therefore,m(y)≤ kx₀kyk/δk ≤δ⁻¹kyk. Now set M =λδ⁻¹ with a λ >1.

DefineS :S1 →P(X) byS(x) =T⁻¹({x}). Then for any openU ⊆X the setS⁻¹(U) ={y∈ S₁ :T⁻¹({y})∩U 6=∅} =T(U)∩S₁ is open inS₁ by the open mapping theorem. Hence,S is lower semi-continuous. Moreover, S(y) is nonempty, closed, and convex for all y ∈S₁. So we can apply Corollary 2.25 to S with f(x) = Mkxk and get a continuous function ˜s: S1 → X with ˜s(y)∈T⁻¹(y) and k˜s(y)k ≤Mkyk. Setting

s(y) =

( kyks(˜ _kyk^y ) if y6= 0,

0 if y= 0,

we get a map from the whole spaceY intoX with the mentioned properties. 2

(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 Gbe a nonempty and open bounded set in a Banach space X. ThenV(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 mapH 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)6=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) infx∈∂Gkx−T(x)k>0.

(iii) If sup

x∈∂G

kT(x)−S(x)k< inf

x∈∂Gkx−T(x)k, then T ∼=S.

Proof: The relation ∼= is reflexive, since T ∼= T by H(x, t) = T(x). If T ∼= S by H, then S ∼=T by H1 =H(x,1−t), so ∼= is symmetric. To see that ∼= is also transitive, letT ∼=S by H₁ and S∼=R by H₂, thenT ∼=R by

H(x, t) =

( H1(x,2t) f or 0≤t≤ ¹₂, H2(x,2t−1) f or ¹₂ < 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 (yn)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−Tˆ]_a≥[I]_a−[ ˆT]_A= 1>0 by Lemma 2.7,(iii), point (ii) of the same lemma tells us that (I−Tˆ) is proper on∂G, soI−T is also proper there. Because (yn)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 subsequencex_n_k →x∈∂G. Because I−T is continuous, we get (I−T)(x) =y.

Thus,y∈(I−T)(∂G).

(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) andH(x,1) =S(x) for x∈∂G, and for all (x, t)∈∂G×[0,1] we have

kH(x, t)−xk=kT(x)−x−t(T(x)−S(x))k ≥ kT(x)−xk+tkT(x)−S(x)k>0.

2 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 > 0, there exists a compact operator P :M →Y such that sup_x∈MkT(x)−P(x)k ≤and P(M) is contained in finite dimensional subspace of Y.

Proof: Fix >0. Since M is bounded,T(M) is precompact. Hence, there exist y_i ∈T(M), i= 1, . . . , N, such that min_i=1,...,NkT(x)−y_ik< for allx∈M. Define continuous functions pi :M →Rby pi(x) = max(− kT(x)−yik,0). Then for each x∈M there exists at least one isuch thatp_i(x)6= 0. Therefore, we can define P :M →Y by

P(x) = PN

i=1pi(x)yi

PN

i=1pi(x) . For allx∈M we then have

kP(x)−T(x)k=

PN

i=1pi(x)(yi−T(x)) P_N

i=1p_i(x)

≤ PN

i=1pi(x) P_N

i=1p_i(x) ≤.

By construction, P(M) lies in the finite dimensional subspace spanned by {y_i :i= 1, . . . , N}.

Finally, since T(M) is bounded, so isP(M). Since a bounded set in a finite dimensional space

is precompact,P is compact. 2

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) =x₀ for all x∈G and some fixed x₀∈G, then i(T, G) = 1.

(A2) If i(T, G)6= 0, then there exists an x∈G such that T(x) =x.

(A3) If there are openG_i ⊆X, i= 1, . . . , n, such that G=Sn

i=1G_i and G_i∩G_j =∅ for i6=j, then

i(T, G) =

n

X

i=1

i(T, Gi)

(23)

whenever T ∈V(G, X) and T|_G

i ∈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 spaceX 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 withG⊆Ωandh: Ω→X be a linear map such thatI−h is compact. Suppose h: Ω→h(Ω)is a homeomorphism, T(G)⊆Ω, and0∈/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, theni(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)6=x for all (x, t)∈A×[0,1].

IfH(.,1)has a fixed point free compact extensionH₁ :B →X, thenH has a compact extension H˜ :B×[0,1]→X withH(x, t)˜ 6=x for all (x, t)∈B×[0,1]

Proof: Set

H˜0(x, t) =

( H₁(x) f or x∈B, t= 0 H(x, t) f or x∈A, t∈[0,1],

and extend ˜H₀ by Corollary 2.14 to a compact map H₀ :B ×[0,1] → X. Let B₀ ={x ∈ B :

∃t∈[0,1] withH0(x, t) =x}. ThenB0 is closed andA∩B0 =∅. Therefore, by Corollary 2.13, we can extend the map

˜ a(x) =

( 0 f or x∈B0

1 f or x∈A

to a continuous mapa:B →[0,1]. Finally, we set ˜H(x, t) =H0(x, a(x)t) for (x, t)∈B×[0,1].

Then ˜H is the desired extension ofH. In fact, if ˜H(x, t) =x, then H0(x, τ) =xfor someτ, i.e.

x ∈B₀. Therefore, a(x) = 0 and further H₀(x,0) =x, i.e. H₁(x) =x, which is impossible by

hypothesis. 2

(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˜toB.

Proof: We may assume without loss of generality that A and B lie in the subspace with coordinates (ζ₁, . . . , ζ_M,0, . . . ,0). We also set F(x) =f(x)−x.

Since F has no zeroes on A, we get α = infx∈A|F(x)| > 0. We extend F continuously to F :B →R^N by Corollary 2.13. 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 ofGare polynomials.

Since∂Gi(x)/∂ζj = 0 forj=M+ 1, . . . , N, we have det(G⁰(x)) = 0 onB. By Sard’s Theorem, G(B) has no interior points. Hence, there exists an x₀ such that x₀ ∈/G(B) and|x₀|< α/3.

For (x, t)∈(A×[0,1])∪(B×1) defineh(x, t) = (1−t)F(x) +t(G(x)−x₀). Thenh(x, t)6= 0 since|h(x, t)| ≥ |F(x)| − |F(x)−G(x)| − |x₀|>0. This also impliesh(x,1)6= 0 forx∈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. 2

Theorem 2.36 (Antipodal Theorem) Let X be a Banach space and let T : BR(0) → X be compact. If T has no fixed points on ∂BR(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 someN ≥1.

Qis called an orthant of R^N if

Q={x∈R^N : 0≤ejxj forj= 1, . . . , N},

where (e₁, . . . e_N) ∈ {−1,1}^N is fixed. Per definition, there are 2^N orthants in R^N. Let Q1, . . . , Q₂^N−1 be the 2^N−1 orthants of R^N with x_N ≥ 0 for all x ∈ Qj. Then R^N = Q1∪ · · · ∪Q₂^N−1 ∪ −Q₁∪ · · · ∪ −Q₂N−1.¹

Now fixr ∈(0, R) and define B_j as the interior of Q_j∩B_R(0)\B_r(0). Then BR(0) =Br(0)∪B1∪ · · · ∪B₂^N−1 ∪ −B₁∪ · · · ∪ −B₂N−1. We show that there exists a continuous function S:B_R(0)→R^N with

S(x) =T(x) for allx∈∂B_R(0), (2.5) S(x) = 0 for all x∈Br(0), (2.6)

1This is the multidimensional equivalent of the following easy construction inR²: Divide the plane into its four quadrants. ChooseQ1 andQ2 as the first and second quadrant, respectively. Then−Q1 is the third quadrant and−Q2the fourth. SoR²=Q1∪Q2∪ −Q1∪ −Q2.

(25)

S(x)6=x for allx∈∂(±B_j), j= 1, . . . ,2^N−1, (2.7) S(−x) =−S(x) for allx∈BR(0). (2.8) Because of (2.6) we only have to defineS on±B_j. ThereS is already defined on∂Bj∩∂B_R(0) and ∂Bj∩∂Br(0) by (2.5) and (2.6), andS 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 ofR^N. We start by applying Lemma 2.35 to each of these parts in order to extend S continuously to ∂B1 such that (2.7) is fulfilled. We now can extend S continuously from ∂B1 to B1 using Theorem 2.12. Further, we defineS on−B₁ byS(x) =−S(−x) forx∈ −B₁, such that (2.8) is fulfilled. We repeat this procedure consecutively withB2 toB₂^N−1. Note that when considering Bj, S is already defined on Bj ∩Bi for i < j, so we can skip defining S on that part of the border ofB_j.

Now (2.6) implies thati(S, Br(0)) = 1 by axiom (A1) of the fixed point index. Using the homeomorphism h :x7→ −x we can use Theorem 2.32 (with Ω = B_R(0)) to see thati(S(x), B_j) = i(−S(−x),−B_j) =i(S(x),−B_j). Axiom (A3) of the fixed point index then gives

i(S, BR(0)) =i(S, Br(0)) +

2^N−1

X

j=1

(i(S, Bj) +i(S,−B_j)) =i(S, Br(0)) + 2

2^N−1

X

j=1

i(S, Bj),

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 thati(T, BR(0)) =i(S, BR(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 :BR(0)→Y, whereY is a finite dimensional subspace ofX, such that sup_x∈B_R₍₀₎kT(x)−S(x)k<inf_x∈∂B_R₍₀₎kT(x)−xk.

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 BR(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). 2

(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.

3.1.1 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 andY 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 solutionx∈X.

Lemma 3.2 Let T ∈ C(X, Y) be stably solvable. ThenT 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 anx∈X withT(x) =S(x) =y. 2

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 k⁰-stably solvable for 0 ≤k⁰ < k. This motivates the following definition.

(27)

Definition 3.4 For T ∈ C(X, Y) we call

µ(T) := inf{k:k≥0, T is notk-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 isk-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)6= 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. Forn ∈N define the operator S_n(x) =d_n(kxk)S(x), where

d_n(t) =







1 if 0≤t≤n, 2−_n¹t if n≤t≤2n,

0 if t≥2n.

Then {x∈X:S_n(x)6= 0} is bounded andS_n compact. Hence, by assumption, there exists an x_n ∈X withS_n(x_n) =T(x_n). Ifkx_nk ≤nfor somen∈N, thenT(x_n) =S_n(x_n) =S(x_n) and we are finished.

So assumekx_nk> n for alln∈N. Thenkx_nk → ∞ asn→ ∞, and we get the contradiction [T]_q = lim inf

kxk→∞

kT(x)k

kxk ≤ lim

n→∞

kT(x_n)k

kx_nk = lim

n→∞d_n(kx_nk)kS(x_n)k kx_nk

≤ lim

n→∞

kS(x_n)k

kx_nk ≤lim sup

kxk→∞

kS(x)k kxk = 0.

2

Corollary 3.6 The identity operator is stably solvable.

Proof: Since [I]_q = 1 we can restrict ourselves to compact mapsS with{x ∈X:S(x)6= 0}

is bounded to prove stable solvability. For such S, there obviously exists an r >0 such that S(Br(0)) ⊆Br(0). But then Schauders fixed point theorem tells us that S has a fixed point,

i.e. S(x) =I(x). 2

Spectral Theory for Nonlinear Operators

DIPLOMARBEIT