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
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DIPLOMARBEIT - ASC: Aktuelles · DIPLOMARBEIT Spectral Theory for Nonlinear Operators ausgef uhrt am Institut f ur Analysis und Scienti c Computing der Technischen Universit at Wien
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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‖Tx‖ : ‖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 ∈ Kwith |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
r(T ) = limn→∞
n√‖Tn‖.
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
polynomial p(λ) = anλn + · · ·+ a1λ+ a0 we get the identity
σ(p(T )) = p(σ(T ))
Here p(T ) denotes the operator anTn + · · ·+ a1T + a0I 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 := infλ∈F
1
‖R(λ, T )‖> 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.
2
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 − Tis 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
ρA(T ) := λ ∈ K : λI − T has property A
and the A-spectrum
σA(T ) = K\ρA.
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.
3
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‖x‖→∞
‖T (x)‖‖x‖
and
[T ]q := lim inf‖x‖→∞
‖T (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
In particular, the fact that [T ]Q = λ or [T ]q = λ implies that there exists an unbounded sequence
(xn)n∈N in X such that limn→∞ ‖T (xn)‖/‖xn‖ = λ. Furthermore, the inequality [T ]q ≤ [T ]Q
is obviously true. Consequently, [T ]Q = 0 actually implies that limn→∞ ‖T (xn)‖/‖xn‖ = 0 for
every sequence (xn)n∈N with ‖xn‖ → ∞.
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‖x‖→∞
‖T (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. In particular,
[T − S]Q = 0 implies [T ]q = [S]q.
(iv) [T−1]Q = [T ]−1q 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 large ‖x‖ there exists a k > 0 such that ‖T (x)‖ ≥ k ‖x‖.This verifies (i).
For (ii) consider
[T + S]q = lim inf‖x‖→∞
‖T (x) + S(x)‖‖x‖
≤ lim inf‖x‖→∞
‖T (x)‖+ ‖S(x)‖‖x‖
≤ lim inf‖x‖→∞
‖T (x)‖‖x‖
+ lim sup‖x‖→∞
‖S(x)‖‖x‖
= [T ]q + [S]Q.
For the second inequality replace T by T + S and S by −S.
Similarly, consider
[T − S]Q = lim sup‖x‖→∞
‖T (x)− S(x)‖‖x‖
≥ lim sup‖x‖→∞
∣∣∣∣‖T (x)‖ − ‖S(x)‖‖x‖
∣∣∣∣ ≥ lim sup‖x‖→∞
(‖T (x)‖‖x‖
− ‖S(x)‖‖x‖
)≥ lim sup‖x‖→∞
‖T (x)‖‖x‖
− lim inf‖x‖→∞
‖S(x)‖‖x‖
≥ lim inf‖x‖→∞
‖T (x)‖‖x‖
− lim inf‖x‖→∞
‖S(x)‖‖x‖
= [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 ‖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)‖‖x‖
)−1=
1
[T ]q.
5
Finally, to see (v), consider
[R T ]q = lim inf‖x‖→∞
‖R(T (x))‖‖x‖
= lim inf‖x‖→∞
‖R(T (x))‖‖T (x)‖
‖T (x)‖‖x‖
≥ lim inf‖x‖→∞
‖R(T (x))‖‖T (x)‖
lim inf‖x‖→∞
‖T (x)‖‖x‖
≥ lim inf‖y‖→∞
‖R(y)‖‖y‖
lim inf‖x‖→∞
‖T (x)‖‖x‖
= [R]q[T ]q.
The last inequality holds true, because if ‖T (x)‖9∞, then [T ]q = 0 and the inequality holds,
and if ‖T (x)‖ → ∞ 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, Bε(z) will always denote the open ball with radius ε and center z.
Definition 2.3 Let X be a Banach space and M ⊆ X. The Hausdorff measure of noncompact-
ness of M is defined as
α(M) := infε > 0| ∃m ∈ N, z1, . . . , zm ⊆ X : M ⊆ Bε(z1) ∪ · · · ∪Bε(zm) ∈ [0,∞].
A finite set z1, . . . , zm ⊆ X with M ⊆ Bε(z1) ∪ · · · ∪Bε(zm) is called a finite ε-net for M .
So if M has a measure of noncompactness ε, it can be covered by finitely many open balls of
any radius greater than ε.
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) = maxα(M), α(N).
(ix) α(co(M)) = α(M), where co(M) denotes the convex hull of M .
(x) α(Br(z)) = 0 if dimX <∞, and α(Br(z)) = r if dimX =∞.
6
Proof: The first four assertion are straightforward to verify. To see (v), observe, that if
z1, . . . , zm is a finite ε-net for M , and w1, . . . , wn is a finite η-net for N , then zi +wj |i =
1, . . . ,m; j = 1, . . . , n is a finite (ε+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 z1, . . . , zm is a
finite ε-net for M then λz1, . . . , λzm is a finite |λ|ε-net for λM , so (vi) follows. Similarly,
(vii) follows from the observation that z1, . . . , zm is a finite ε-net for M if and only if z1 +
z, . . . , zm + z is a finite ε-net for M + z. For (viii) we use the fact that if z1, . . . , zm is a
finite ε-net for M and w1, . . . , wn is a finite η-net for N , then z1, . . . , zm ∪ w1, . . . , wnis a finite δ-net for M ∪N , where δ = maxε, 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 z1, . . . zm and define N = co(z1 . . . zm). Any x ∈ co(M) can be
written as a convex combination x =∑aixi with xi ∈ M and
∑ai = 1. For every xi there is
a zj(i) with |xi − zj(i)| < η. Setting z =∑aizj(i) we get
‖x− z‖ =∥∥∥∑ aixi −
∑aizj(i)
∥∥∥ =∥∥∥∑ ai(xi − zj(i))
∥∥∥ ≤∑ |ai|∥∥xi − zj(i)∥∥ ≤∑ |ai|η = η.
Because z ∈ N , we have dist(x,N) ≤ η. Since N is compact, we can find a finite ε-net
w1 . . . wn for N and arbitrary ε > 0. This is then a finite (η + ε)-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 have α(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 radius η2, which gives us a
finite cover of B1(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 ε for every ε > 0. This shows, that B1(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 de-
fined for an arbitrary subset of a Banach space.
Definition 2.5 Let Z ⊆ X. For T ∈ C(Z, Y ) we define
[T ]A := infk : k > 0, α(T (M)) ≤ kα(M) for all bounded M ∈ Z
and
[T ]a := supk : 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
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 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. In particular,
[T − S]A = 0 implies [T ]a = [S]a.
(vi) [T−1]A = [T ]−1a 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.
8
Thus T−1(N) is precompact. Since T is continuous, T−1(N) is also closed and therefore com-
pact. 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]A = sup∞>α(N)>0
α(T−1(N))
α(N)= sup∞>α(M)>0
α(M)
α(T (M))=
(inf
∞>α(M)>0
α(T (M))
α(M)
)−1=
1
[T ]a.
Finally, let kT > 0 such that α(T (M)) ≥ kTα(M) for all bounded M ∈ X. Further, let kR > 0
such that α(R(M)) ≥ kRα(M) for all bounded M ∈ Y . Then α(R T (M)) ≥ kRα(T (M)) ≥kRkTα(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 bi-
jectivity 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 z1, . . . , zmis a finite ε-net for a bounded set M , then Tz1, . . . , T zm is obviously a finite ‖T‖ ε-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 6= y ∈ X with Tx = Ty. Let (xn)n∈N
and (yn)n∈N be sequences with xn → x and yn → y. Then ‖n(xn − yn)‖ → ∞ for n → ∞.
Using the linearity of T we get the contradiction
[T ]q = lim inf‖z‖→∞
‖Tz‖‖z‖
≤ limn→∞
‖T (n(xn − yn))‖‖n(xn − yn)‖
= limn→∞
n ‖T (xn − yn)‖n ‖xn − yn‖
=‖Tx− Ty‖‖x− y‖
= 0.
9
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
[T ]q = [T−1]−1Q =∥∥T−1∥∥−1 =
1
‖T−1‖> 0,
and using Proposition 2.7,(vi) we get
[T ]a =1
[T−1]A≥ 1
‖T−1‖> 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 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 = (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 with Bj ⊆ Ai.
(ii) Let X be a Hausdorff space and U = (Ui)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 = (Ui)i∈I be an open cover of X. A system of real valued continuous
functions (fi)i∈I is called a partition of unity subordinate to U if
(a) fi(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 all i ∈ I,
10
(d)∑
i∈I fi(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]. 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 that T (x) = T (x) for x ∈ C.
Proof: For each x in X\C set rx = 13dist(x,C). Then diamBrx(x) ≤dist(Brx , 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 (Oi)i∈I which is an open cover of X\C.
Define q : X\C → (0,∞) by
q(x) =∑i∈I
dist(x,X\Oi).
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 (Oi)i∈I form an open cover of X\C, q(x) > 0.
Next we define for i ∈ I and x ∈ X\C
ρi(x) =dist(x,X\Oi)
q(x).
We then have 0 ≤ ρi(x) ≤ 1 and∑
i∈I ρi(x) = 1.
Note that since (Oi)i∈I is a refinement of (Brx(x))x∈X\C , for any Oi there exists x0 ∈ X\C with
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 ⊆ Xis 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. 2
12
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) :=⋃x∈M
T (x)
and the preimage of a set A as
T−1(A) := x ∈ X : T (x) ∩A 6= ∅.
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 6= ∅. 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 conti-
nuity. 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 neighbour-
hood V (y) of y there exists a neighbourhood U(x)of x such that
T (u) ∩ V (y) 6= ∅, 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) ⊆ AC . Since AC is open, there exists a neighbourhood U(x) of x, such that
13
T (U(x)) ⊆ AC . 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) 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 of x. It follows that T (u)∩V (y) 6= ∅ for all u ∈ 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, let X be linear space and p be a seminorm on X. For x ∈ X we denote by
Uδ(x) the p-neighbourhood Uδ(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. yn ∈ T (xn), yn → y and p(xn − 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 Vy ⊆ K such that y ∈ Vy and T (Uδ(x)) ∩ Vy = ∅.
Proof: Assume the assertion does not hold true for x ∈ X and y ∈ K\T (x). Let (δn)n∈N be
a null sequence. Since T (Uδn(x)) ∩ Bδn(y) 6= ∅, we can find xn ∈ Uδn(x) and yn ∈ Bδn(y) such
that yn ∈ T (xn). Further, we get that yn → y and p(xn − x)→ 0. Since T is closed, this shows
that y ∈ T (x), which contradicts our choice of y. 2
Lemma 2.19 Let T : X →P(K) be a closed map. If
supy∈T (x)
|y| ≤ p(x) for all x ∈ X,
then T is upper semi-continuous.
14
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 (Uη(x))\V 6= ∅ (if this is not possible, there is nothing to prove). For
z ∈ Uη(x) we have
supλ∈T (z)
|λ| ≤ p(z) ≤ p(x) + η.
Consequently, T (Uη(x)) is bounded, and so the set C := T (Uη(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 Vy ⊆ K with y ∈ Vλ and T (Uδ(y)(x)) ∩ Vy = ∅. Obviously, Vy : y ∈ C is an open
cover of C. Since C is compact, we get C ⊆ Vy1 ∪ · · · ∪Vym for suitable y1, . . . , ym ∈ C. Putting
δ = minη, δ(y1), . . . , δ(ym) we see that T (Uδ(x))∩C = ∅. Since T (Uδ(x)) ⊆ T (Uη(x)), we get
T (Uδ(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) = ∅. 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) 6= ∅. 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) = ∅. 2
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) 6= ∅ 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 ε > 0 such that f(x) + Bε(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) +Bε/2(o)) 6= ∅.Since f is continuous, there exists a neighbourhood U(x) of x, such that ‖f(x)− f(u)‖ < ε/2
for all u ∈ U(x). Let a ∈ f(x) + Bε/2(o). Then for all u ∈ U(x) we get ‖a− (f(u) + o)‖ ≤‖a− f(x)− o‖+ ‖f(x) + o− f(u)− o‖ < ε, i.e. a ∈ f(u) +Bε(o) ⊆ (f(u) +O).
Set U(x) = U(x) ∩ U(x). Then for any u ∈ U(x) we have T (u) ∩ (f(x) + Bε/2(o)) 6= ∅, since
15
u ∈ U(x). But for any a ∈ T (u)∩(f(x)+Bε/2(o)) we have a ∈ (f(x)+Bε/2(o)), hence a ∈ V (y).
Further, u ∈ U(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 and f(x) ≥ 0 for all x ∈ X. Define S : X →P(Y ) by
S(x) =
T (x) ∩Bf(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: 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 ε > 0 with f(x) > ‖y‖ + ε. Since f is continuous, there
exists an open neighbourhood U(x) of x such that f(u) > ‖y‖+ ε for all u ∈ U(x). Moreover,
there exists an open neighbourhood U(x) of x such that T (u) ∩ B‖y‖+ε(0) ∩ V (y) 6= ∅ for all
u ∈ U(x), because T is lower semi-continuous and B‖y‖+ε(0) ∩ V (y) is an open neighbourhood
of y. So for u in the open set U(x) = U(x) ∩ U(x) we get
S(u) ∩ V (y) = T (u) ∩Bf(u) ∩ V (y) ⊇ T (u) ∩B‖y‖+ε ∩ V (y) 6= ∅.
Now suppose f(x) = 0. Then y = 0 and V (y) is a neighbourhood of 0, i.e. there is an ε > 0 such
that Bε(0) ⊆ V (y). Since f is continuous, there is a open neighbourhood U(x) of x such that
f(u) < ε for u ∈ U(x). For f(u) > 0 we have S(u) = T (u)∩Bf(u)(0) ⊆ Bf(u)(0) ⊆ Bε(0) ⊆ V (y).
For f(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 all x ∈ 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, 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
Proof: For y ∈ Y and A ⊆ Y set d(y,A) = infa∈A ‖y − a‖. As a first step, we show that for
each ε > 0 there exists a continuous map f : X → Y such that
d(f(x), T (x)) < ε for all x ∈ X. (2.1)
Fix ε > 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) ∩Bε(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 an xi ∈ X such that supp(fi) ⊆ U(xi) if supp(fi) 6= ∅ and set
f(x) =∑i∈I
fi(x)m(xi).
Then f is a continuous function of X into Y . If fi(x) > 0 for some i, then x ∈ U(xi) and by
(2.2)
m(xi) ∈ T (x) +Bε(0).
Since T (x) + Bε(0) is convex, we get f(x) ∈ T (x) + Bε(0). Hence, d(f(x), T (x)) < ε, i.e. f
satisfies (2.1).
In the second step we construct the requested selection. Set εn = 2−n. We will inductively
define a sequence (fn)n∈N of continuous mappings fn : X → Y with
d(fn(x), T (x)) < εn, x ∈ X, n = 1, 2, . . . (2.3)
d(fn(x), fn−1(x)) < εn−1, x ∈ X, n = 2, 3, . . . (2.4)
As we showed in the first step, there exists f1 with d(f1(x), T (x)) < 1/2, x ∈ X.
Assume that n ≥ 2 and we already have constructed f1, . . . , fn−1. For each x ∈ X we define
G(x) = (fn−1(x) +Bεn−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 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 fn : X → Y such that (2.3)
holds. By construction, fn also satisfies (2.4).
Since∑∞
n=1 εn 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
17
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
S(x) =
T (x) ∩Bf(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
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). 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),
• ‖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 B1(0) ⊆ X under T is an open neighbourhood of 0 ∈ Y . In particular,
there exists a δ > 0 such that Sδ = y ∈ Y : ‖y‖ = δ ⊆ T (B1(0)). For an arbitrary 0 6= y ∈ Ywe have y0 = δy ‖y‖−1 ∈ Sδ. Hence, there exists a x0 ∈ B1(0) with T (x0) = y0 and, further,
T (x0 ‖y‖ δ−1) = y. Therefore, m(y) ≤ ‖x0 ‖y‖ /δ‖ ≤ δ−1 ‖y‖. Now set M = λδ−1 with a λ > 1.
Define S : S1 →P(X) by S(x) = T−1(x). Then for any open U ⊆ X the set S−1(U) = y ∈S1 : T−1(y) ∩ U 6= ∅ = T (U) ∩ S1 is open in S1 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) = M ‖x‖ and get a continuous function s : S1 → X
with s(y) ∈ T−1(y) and ‖s(y)‖ ≤M ‖y‖. Setting
s(y) =
‖y‖ s( y
‖y‖) if y 6= 0,
0 if y = 0,
we get a map from the whole space Y into X with the mentioned properties. 2
18
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 contin-
uous 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) 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∈∂G ‖x− T (x)‖ > 0.
(iii) If supx∈∂G
‖T (x)− S(x)‖ < infx∈∂G
‖x− T (x)‖, 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, let T ∼= S by
H1 and S ∼= R by H2, then T ∼= R by
H(x, t) =
H1(x, 2t) for 0 ≤ t ≤ 1
2 ,
H2(x, 2t− 1) for 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 T on X. Since ∂G is closed and bounded,
and [I− T ]a ≥ [I]a− [T ]A = 1 > 0 by Lemma 2.7,(iii), point (ii) of the same lemma tells us that
(I − T ) is proper on ∂G, so I − T is also proper there. Because (yn)n∈N ∪ y is compact, also
(I − T )−1((yn)n∈N ∪ y) is compact. If we choose xn ∈ (I − T )−1(yn) ∩ ∂G, then (xn)n∈N has
a convergent subsequence xnk→ x ∈ ∂G. Because I − T is continuous, we get (I − T )(x) = y.
Thus, y ∈ (I − T )(∂G).
19
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
The next lemma is especially interesting in connection with point (iii) of the previous proposi-
tion.
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 ε > 0, there exists a compact operator
P : M → Y such that supx∈M ‖T (x)− P (x)‖ ≤ ε and P (M) is contained in finite dimensional
subspace of Y .
Proof: Fix ε > 0. Since M is bounded, T (M) is precompact. Hence, there exist yi ∈ T (M),
i = 1, . . . , N , such that mini=1,...,N ‖T (x)− yi‖ < ε for all x ∈ M . Define continuous functions
pi : M → R by pi(x) = max(ε− ‖T (x)− yi‖ , 0). Then for each x ∈M there exists at least one
i such that pi(x) 6= 0. Therefore, we can define P : M → Y by
P (x) =
∑Ni=1 pi(x)yi∑Ni=1 pi(x)
.
For all x ∈M we then have
‖P (x)− T (x)‖ =
∥∥∥∥∥∑N
i=1 pi(x)(yi − T (x))∑Ni=1 pi(x)
∥∥∥∥∥ ≤∑N
i=1 pi(x)ε∑Ni=1 pi(x)
≤ ε.
By construction, P (M) lies in the finite dimensional subspace spanned by yi : 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. 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 open Gi ⊆ X, i = 1, . . . , n, such that G =⋃ni=1Gi and Gi ∩Gj = ∅ for i 6= j,
then
i(T,G) =n∑i=1
i(T,Gi)
20
whenever T ∈ V (G,X) and T |Gi∈ V (Gi, 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) 6= x for all (x, t) ∈ A× [0, 1].
If H(., 1) has a fixed point free compact extension H1 : B → X, then H has a compact extension
H : B × [0, 1]→ X with H(x, t) 6= x for all (x, t) ∈ B × [0, 1]
Proof: Set
H0(x, t) =
H1(x) for x ∈ B, t = 0
H(x, t) for 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] with H0(x, t) = x. Then B0 is closed and A∩B0 = ∅. Therefore, by Corollary 2.13,
we can extend the map
a(x) =
0 for x ∈ B0
1 for x ∈ A
to a continuous map a : 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 of H. In fact, if H(x, t) = x, then H0(x, τ) = x for some τ , 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
21
Lemma 2.35 Let A and B be compact subsets of a proper linear subspace of RN with N ≥ 2
and A ⊆ B. Then every continuous fixed point free map f : A → RN 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 (ζ1, . . . , ζM , 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 → RN by Corollary 2.13. By the Weierstrass approximation theorem, there exists a map
G : B → RN with supx∈B |F −G| < α/3, and such that the components of G are polynomials.
Since ∂Gi(x)/∂ζj = 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 x0 such that x0 /∈ G(B) and |x0| < α/3.
We show that there exists a continuous function S : BR(0)→ RN with
S(x) = T (x) for all x ∈ ∂BR(0), (2.5)
S(x) = 0 for all x ∈ Br(0), (2.6)
1This is the multidimensional equivalent of the following easy construction in R2: Divide the plane into its fourquadrants. Choose Q1 and Q2 as the first and second quadrant, respectively. Then −Q1 is the third quadrantand −Q2 the fourth. So R2 = Q1 ∪Q2 ∪ −Q1 ∪ −Q2.
22
S(x) 6= x for all x ∈ ∂(±Bj), j = 1, . . . , 2N−1, (2.7)
S(−x) = −S(x) for all x ∈ BR(0). (2.8)
Because of (2.6) we only have to define S on ±Bj . There S is already defined on ∂Bj ∩ ∂BR(0)
and ∂Bj ∩ ∂Br(0) by (2.5) and (2.6), and S is fixed point free there. The remaining border of
Bj can be divided into N parts, each of which lies in a N − 1 dimensional hyperplane of RN .
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 define S on −B1 by S(x) = −S(−x) for x ∈ −B1, such that (2.8) is
fulfilled. We repeat this procedure consecutively with B2 to B2N−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 of Bj .
Now (2.6) implies that i(S,Br(0)) = 1 by axiom (A1) of the fixed point index. Using the home-
omorphism h : x 7→ −x we can use Theorem 2.32 (with Ω = BR(0)) to see that i(S(x), Bj) =
i(−S(−x),−Bj) = i(S(x),−Bj). Axiom (A3) of the fixed point index then gives
i(S,BR(0)) = i(S,Br(0)) +2N−1∑j=1
(i(S,Bj) + i(S,−Bj)) = i(S,Br(0)) + 22N−1∑j=1
i(S,Bj),
i.e. i(S,BR(0)) is odd. Since S(x) = T (x) for x ∈ ∂BR(0), Proposition 2.28 together with
axiom (A4) of the fixed point index show that i(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 ∈ ∂BR(0). By Proposition 2.28 and (A4), this yields i(T (x), BR(0)) = i((T (x) −T (−x))/2, BR(0)). By Lemma 2.29, there is a compact operator S : BR(0)→ Y , where Y is a
finite dimensional subspace of X, such that supx∈BR(0) ‖T (x)− S(x)‖ < infx∈∂BR(0) ‖T (x)− x‖.As above, for P (x) = (S(x) − S(−x))/2 we have i(P,BR(0)) = i((T (x) − T (−x))/2, BR(0)).
But since P maps BR(0) into a finite dimensional subspace Y , Theorem 2.33 shows that
i(P,BR(0)) = i(P,BR(0) ∩ Y ), the latter of which is odd by the first part of the proof, since
P (−x) = −P (x). 2
23
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 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. 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 k′-stably solvable for 0 ≤ k′ < k. This motivates
the following definition.
24
Definition 3.4 For T ∈ C(X,Y ) we call
µ(T ) := infk : 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) 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. For n ∈ N define the operator
Sn(x) = dn(‖x‖)S(x), where
dn(t) =
1 if 0 ≤ t ≤ n,2− 1
n t if n ≤ t ≤ 2n,
0 if t ≥ 2n.
Then x ∈ X : Sn(x) 6= 0 is bounded and Sn compact. Hence, by assumption, there exists an
xn ∈ X with Sn(xn) = T (xn). If ‖xn‖ ≤ n for some n ∈ N, then T (xn) = Sn(xn) = S(xn) and
we are finished.
So assume ‖xn‖ > n for all n ∈ N. Then ‖xn‖ → ∞ as n→∞, and we get the contradiction
[T ]q = lim inf‖x‖→∞
‖T (x)‖‖x‖
≤ limn→∞
‖T (xn)‖‖xn‖
= limn→∞
dn(‖xn‖)‖S(xn)‖‖xn‖
≤ limn→∞
‖S(xn)‖‖xn‖
≤ lim sup‖x‖→∞
‖S(x)‖‖x‖
= 0.
2
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) 6= 0is 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
25
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 x ∈ X.
Proof: We construct a sequence (xn)n∈N as follows. Let x0 ∈ B be arbitrary. Due to
Lemma 3.2 we can choose inductively xn ∈ T−1(H(xn−1)) for n ≥ 1, so T (xn) = H(xn−1).
By construction, the set A = x0, x1, . . . satisfies T (A) = T (xo) ∪H(A). Therefore, we get
α(T (A)) = α(T (xo) ∪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′ 6= ∅ since A is precompact. So for x ∈ A′ we find a subsequence (xnk)k∈N of
(xn)n∈N such that xnk→ x, hence, also H(xnk−1) = T (xnk
) → T (x) as k → ∞. Therefore, if
we choose y ∈ A′ to be a limit point of (xnk−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 M0 the intersection of all M ∈M. Note that
M0 is nonempty and closed, since M ⊇ A′ 6= ∅ for each M ∈M and each M is closed.
Set M1 = T−1(co(H(M0))). Then for all M ∈M we get
M1 = T−1(co(H(M0))) ⊆ T−1(co(H(M))) ⊆M,
hence M1 ⊆M0.
Further, M1 is closed and satisfies both
M1 = T−1(co(H(M0))) ⊇ T−1(co(H(A′))) ⊇ A′
and
T−1(co(H(M1))) ⊆ T−1(co(H(M0))) = M1.
Therefore, M1 ∈ M and M0 ⊆ M1. Consequently, T−1(co(H(M0))) = M1 = M0. Since T is
stably solvable and, therefore, surjective by Lemma 3.2, this in turn implies T (M0) = co(H(M0))
26
and, hence,
α(T (M0)) = α(co(H(M0))) = α(H(M0)).
Thus, by assumption, M0 is compact, and by the continuity of H also H(M0) is compact.
By Corollary 2.13, we find a continuous operator G : X → X with G(x) = H(x) for x ∈ M0
and G(X) ⊆ co(H(M0)). This shows that G(X) is precompact, and in turn that G is a
compact operator. Furthermore, [G]Q = 0 since G(X) is bounded. Since T is stably solv-
able, there exists an x ∈ X with T (x) = G(x). But from T (x) ∈ co(H(M0)) it follows that
x ∈ T−1(co(H(M0))) = M0. Thus, T (x) = G(x) = H(x). 2
Corollary 3.8 (Darbo fixed point theorem) Let B ⊆ X be nonempty, bounded, closed,
and convex. Let S ∈ C(B,B) and suppose [S]A < 1. Then S has a fixed point.
Proof: Choose H = S and T = I in Theorem 3.7, the latter of which is possible because of
Corollary 3.6 2
Corollary 3.9 Let T ∈ C(X) with [T ]A < 1 and [T ]Q < 1. Then T has a fixed point.
Proof: Let b ∈ ([T ]Q, 1). Then ‖T (x)‖ ≤ b‖x‖ for all x ∈ X with ‖x‖ ≥ r for sufficiently large
r > 0. Because of [T ]A <∞, T (Br(0)) is bounded. Hence, there exists a d > 0 with ‖T (x)‖ < d
for x ∈ Br(0). Altogether we have ‖T (x)‖ ≤ b‖x‖+ d for all x ∈ X.
Setting R = d/(1− b), for x ∈ BR(0) we get
‖T (x)‖ ≤ b‖x‖+ d ≤ bR+ d =bd+ d(1− b)
1− b=
d
1− b= R.
This shows, that T maps BR(0) into itself and the assertion follows from Corollary 3.8. 2
Theorem 3.10 Let T ∈ B(X,Y ). Then T is stably solvable if and only if T is surjective.
Proof: By Lemma 3.2, we only have to prove that a surjective operator is stably solvable. So
let T ∈ B(X,Y ) be surjective. By Theorem 2.26 we may find a continuous function s : Y → X
such that s(y) ∈ T−1(y) for all y ∈ Y and ‖s(y)‖ ≤M‖y‖ for some M > 0.
Now let G : X → Y be compact with [G]Q = 0. Since s maps bounded sets onto bounded sets,
the map G s : Y → Y is compact.
Since [G]A = [G]Q = 0, we can use the same technique as at the beginning of the proof of
Corollary 3.9 to show that for every ε > 0 we can find a d > 0 such that ‖G(x)‖ ≤ ε‖x‖+ d for
27
all x ∈ X. For any sequence (yn)n∈N with ‖yn‖ → ∞ we then get
limn→∞
‖G s(yn)‖‖yn‖
≤ limn→∞
ε‖s(yn)‖+ d
‖yn‖≤ lim
n→∞
εM‖yn‖+ d
‖yn‖= lim
n→∞
(εM +
d
‖yn‖
)= εM.
Since ε > 0 was arbitrary, we get limn→∞‖Gs(yn)‖‖yn‖ = 0. Hence, [G s]Q = 0.
Thus, we can use Corollary 3.9 to see that Gs has a fixed point y ∈ Y . Consequently, x = s(y)
satisfies
Tx = Ts(y) = y = G(s(y)) = G(x),
i.e. T is stably solvable. 2
3.1.2 FMV-regularity
Definition 3.11 An operator T ∈ C(X,Y ) is called FMV-regular, if T is stably solvable and
fulfils [T ]q > 0 and [T ]a > 0.
By what we have shown in chapter 2, amongst the properties of FMV-regular operators are,
that they are proper and coercive. Both of these properties are due to the fact, that [T ]q > 0,
and [T ]a > 0. But the positivity of these two characteristics also has an important influence on
the stable solvability of the operator.
Lemma 3.12 Every FMV-regular operator is strictly stably solvable. More precisely, the esti-
mate
µ(T ) ≥ min [T ]q, [T ]a
holds true.
Proof: Fix k with k < [T ]q and k < [T ]a. We have to show that the equation T (x) = S(x)
has a solution x ∈ X for any S ∈ C(X,Y ) with [S]Q ≤ k and [S]A ≤ k.
Choose two numbers b and c such that [S]Q < b < c < [T ]q. Then ‖S(x)‖ ≤ b‖x‖ and
‖T (x)‖ ≥ c‖x‖ for ‖x‖ ≥ r with sufficiently large r > 0. Since [S]A <∞, S(Br(0)) is bounded
by some constant R > 0. Hence, ‖S(x)‖ ≤ R+ b‖x‖ for all x ∈ X. Put
ρ =R
c− b.
By choosing R sufficiently large, we may assume without loss of generality that ρ ≥ r. We
now want to apply Theorem 3.7 with B = Bρ(0). It is clear that S(Bρ(0)) is bounded, because
S(Bρ(0)) ⊆ BR+bρ(0).
Fix x ∈ X with T (x) ∈ BR+bρ(0). If ‖x‖ > ρ were true (which would also imply ‖x‖ > r), we
Furthermore, for all M ⊆ Bρ(0) with α(M) > 0, by assumption we have
α(T (M)) ≥ [T ]aα(M) > [S]Aα(M) ≥ α(S(M)).
Thus, all hypotheses of Theorem 3.7 are satisfied. Hence the equation T (x) = S(x) has a solu-
tion x ∈ X. 2
Lemma 3.13 Let T, S ∈ C(X,Y ), let T be k-stably solvable. Further, assume that k ≥ [S]A
and k ≥ [S]Q. Then T +S is k′-stably solvable for 0 ≤ k′ ≤ k−max[S]A, [S]Q. In particular,
µ(T + S) ≥ µ(T )−max[S]A, [S]Q.
Proof: Let H ∈ C(X,Y ) with [H]A ≤ k −max[S]A, [S]Q and [H]Q ≤ k −max[S]A, [S]Q.We have to show that the equation T (x) +S(x) = H(x) has a solution x ∈ X. But [H −S]A ≤[H]A + [S]A ≤ k and [H − S]Q ≤ [H]Q + [S]Q ≤ k. Since T is k-stably solvable, the equation
T (x) = (H − S)(x) has a solution x ∈ X. 2
Theorem 3.14 Let T, S ∈ C(X,Y ). If T is FMV-regular with min[T ]q, [T ]a > [S]A and
min[T ]q, [T ]a > [S]Q, then T + S is also FMV-regular.
Proof: By Lemma 3.12, T is k-stably solvable for k < min[T ]q, [T ]a. By Lemma 3.13, T +S
is stably solvable. Moreover, [T + S]q ≥ [T ]q − [S]Q > 0 and [T + S]a ≥ [T ]a − [S]A > 0, by
Lemma 2.2,(ii) and Lemma 2.7,(iv) respectively. 2
Heuristically, this theorem shows that FMV-regularity is stable under perturbations that are
sufficiently compact and quasibounded.
FMV-regularity is also invariant under linear isomorphisms.
29
Theorem 3.15 Let T ∈ C(X,Y ) and let S ∈ B(Y,Z) be bijective. Then the operator S T ∈C(X,Z) is FMV-regular if and only if T is FMV-regular.
Proof: First, let T be FMV-regular. By definition [T ]a > 0 and [T ]q > 0. Moreover [S]a > 0
and [S]q > 0 by Lemma 2.8,(iv). So Proposition 2.2,(v) and Proposition 2.7,(vii) show that
[S T ]a ≥ [S]a[T ]a > 0
and
[S T ]q ≥ [S]q[T ]q > 0,
respectively. It remains to show that S T is stably solvable. So let G ∈ C(X,Z) satisfy
[G]A = [G]Q = 0. Then obviously S−1 G is still compact and [S−1 G]Q ≤ ‖S−1‖[G]Q = 0.
Since T is stably solvable, there exists an x ∈ X with T (x) = S−1G(x). Hence, ST (x) = G(x)
and S T is stably solvable.
Conversely, suppose S T is FMV-regular. Since S−1 is a linear bijection, we can use the first
part of the proof to see that T = S−1 (S T ) is FMV-regular. 2
Finally, we want to show, that for linear operators FMV-regularity reduces to a very simple
property.
Theorem 3.16 Let T ∈ B(X,Y ). Then T is FMV-regular if and only if T is bijective.
Proof: If T is FMV-regular, then [T ]q > 0. By Lemma 2.8,(iii), T is injective. Since T is
also stably solvable, T is surjective by Theorem 3.10.
Conversely, let T be a bijection. Then [T ]a > 0 and [T ]q > 0 by Lemma 2.8,(iv). Again by
Theorem 3.10, T is also stably solvable. 2
3.2 Topological properties
Definition 3.17 For T ∈ C(X) we call the set
ρFMV (T ) := λ ∈ K : λI − T is FMV-regular
the FMV-resolvent set and its complement
σFMV (T ) := K\ρFMV (T )
the FMV-spectrum of T .
Thanks to our extensive preparations, the following proofs concerning topological properties of
σFMV (T ) are very simple. One of the most important properties that we wanted a nonlinear
30
spectrum to have, has already been proven for the FMV-spectrum.
Proposition 3.18 σ(T ) = σFMV (T ) for T ∈ B(X).
Proof: This is a simple consequence of Theorem 3.16. 2
Proposition 3.19 The FMV-spectrum σFMV (T ) is closed for all T ∈ C(X).
Proof: Fix λ ∈ ρFMV (T ) and let 0 < δ < min[λI − T ]a, [λI − T ]q. Now let µ ∈ K with
|µ− λ| < δ. Because of
[(µ− λ)I]A = |µ− λ| < [λI − T ]a
and
[(µ− λ)I]Q = |µ− λ| < [λI − T ]q,
it follows from Theorem 3.14 that µI − T = (λI − T ) + (µ − λ)I is FMV-regular. This shows
that λ is an interior point of ρFMV (T ). Thus, ρFMV (T ) is open in K. 2
Proposition 3.20 Let T ∈ C(X) and suppose [T ]A < ∞ and [T ]Q < ∞. Then σFMV (T ) is
bounded. In fact, it is contained in Bmax[T ]A,[T ]Q(0).
Proof: Let λ ∈ K with |λ| > [T ]Q and |λ| > [T ]A. We will show, that λ ∈ ρFMV (T ).
First of all, we have [λI − T ]q ≥ |λ| − [T ]Q > 0 and [λI − T ]a ≥ |λ| − [T ]a > 0 by Lemma
2.2,(ii) and Lemma 2.7,(iv), respectively. It remains to show that λI − T is stably solvable. So
let S ∈ C(X) be compact with [S]Q = 0. Then the operator H := (T + S)/λ satisfies both
[H]A =
[T + S
λ
]A
=1
|λ|[T + S]A ≤
1
|λ|([T ]A + [S]A) =
[T ]A|λ|
< 1
and
[H]Q =
[T + S
λ
]Q
=1
|λ|[T + S]Q ≤
1
|λ|([T ]Q + [S]Q) =
[T ]Q|λ|
< 1.
By Corollary 3.9, H has a fixed point x ∈ X, from which the assertion follows. 2
The previous proposition motivates the following definitions.
Definition 3.21 We define the set
AQ(X) := A(X,X) ∩Q(X,X) = T ∈ C(X) : [T ]A <∞ and [T ]Q <∞.
31
For T ∈ AQ(X) define
pAQ(T ) := max[T ]A, [T ]Q.
Further, for T ∈ C(X) we call
rFMV (T ) := sup|λ| : λ ∈ σFMV (T )
the FMV spectral radius of T . If σFMV (T ) = ∅, we set rFMV (T ) = 0.
The following assertion has been proven in Proposition 3.20 and 3.19
Corollary 3.22 Let T ∈ AQ(X). Then σFMV (T ) is compact. Furthermore,
rFMV (T ) ≤ pAQ(T ).
It is easy to see, that pAQ is a seminorm on AQ. Also, note that B(X) ⊆ AQ(X) and pAQ(T ) =
‖T‖ for T ∈ B(X). This follows from Lemma 2.8,(i) and (ii). Hence, Corollary 3.22 includes
the well known result that for a linear operator the spectral radius of the classical spectrum is
bounded by its operator norm.
Further, AQ(X) can be equipped with a topology defined by the seminorm pAQ. The discussion
above then also shows that the following proposition includes the result of Proposition 1.1.
Proposition 3.23 The multivalued map σFMV : AQ(X) → P(K), which assignes to each
T ∈ AQ(X) its FMV-spectrum, is upper semi-continuous.
Proof: First, we want to show, that the map σFMV is closed. To do so, choose a sequence
(Tn)n∈N in AQ(X) and a sequence (λn)n∈N in ⊆ K such that
the sequence (λnI − Tn)n∈N converges to λI − T in AQ(X). If λI − T were FMV-regular, then
for sufficiently large n, λnI−Tn = (λI−T )+((λnI−Tn)−(λI−T )) would also be FMV-regular
by Theorem 3.14, contradicting our choice of λn. Therefore, λ ∈ σFMV (T ) and σFMV is closed.
The assertion now follows from Lemma 2.19, since by Corollary 3.22 for all T ∈ AQ(X) we have
supλ∈σFMV (T )
|λ| = rFMV (T ) ≤ pAQ(T ).
2
32
3.3 Subdivision of the FMV-spectrum
Definition 3.24 For T ∈ C(X) we set
σδ(T ) := λ ∈ K : λI − T is not stably solvable,
σa(T ) := λ ∈ K : [λI − T ]a = 0,
σq(T ) := λ ∈ K : [λI − T ]q = 0,
and
σπ(T ) := σa(T ) ∪ σq(T ).
By definition, we then have
σFMV = σδ ∪ σπ = σδ ∪ σa ∪ σq.
However, this decomposition need not be disjoint.
The seminorm pAQ can be usefull to compare the spectrum und subspectra of two operators.
Lemma 3.25 Let T, S ∈ C(X) such that pAQ(T−S) = 0. Then σq(T ) = σq(S), σa(T ) = σa(S),
and σδ(T ) = σδ(S), and hence σFMV (T ) = σFMV (S).
Proof: By Proposition 2.2,(iii), [T − S]Q = 0 implies σq(T ) = σq(S), while [T − S]A implies
σa(T ) = σa(S) by Proposition 2.7,(v).
Assume λI−S is stably solvable and let H ∈ C(x) with pAQ(H) = 0. Since pAQ(H+(T−S)) = 0,
there exists an x ∈ X with λx−S(x) = H(x) +T (x)−S(x), i.e. λx−T (x) = H(x). So λI −Tis also stably solvable. Hence, σδ(T ) = σδ(S). 2
We now want to show that these subdivisons have their counterparts in the linear theory.
Definition 3.26 For T ∈ B(X) we define the approximate point spectrum
σapp(T ) := λ ∈ K : ∃xn, n ∈ N, ‖xn‖ = 1, ‖(λI − T )(xn)‖ → 0
and the defect spectrum
σd(T ) := λ ∈ K : λI − T is not surjective.
Just as the FMV-spectrum is the union of σδ(T ) and σπ(T ), for linear operators we have
σ(T ) = σapp(T ) ∪ σd(T ). We show that σapp(T ) and σd(T ) correspond with σδ(T ) and σπ(T ),
respectively. In particular, they are the same in the case of linear operators.
Lemma 3.27 If T ∈ B(X), then σa(T ) ⊆ σq(T ).
Proof: If dim(X) < ∞, then σa(T ) = ∅. So we can restrict ourselves to infinite dimensional
spaces. Let λ ∈ σa(T ). Then for every n ∈ N there exists a bounded set Mn with α(Mn) > 0
33
and α((λI − T )(Mn)) < α(Mn)/(2n).
If (λI − T ) is not injective on Mn, choose x1, 6= x2 ∈Mn with (λI − T )(x1) = (λI − T )(x2) and
set yn = x1 − x2.If (λI − T ) is injective on Mn, define
K = infx 6=y
x,y∈Mn
‖(λI − T )(x)− (λI − T )(y)‖‖x− y‖
.
Assume K > 0 and fix k ∈ (0,K). Then for x, y ∈Mn we have
‖(λI − T )(x)− (λI − T )(y)‖ ≥ k‖x− y‖.
If z1, . . . , zn is a finite ε-net for (λI − T )(Mn), then (λI − T )−1(z1), . . . , (λI − T )−1(zn) is
a finite ε/k−net for Mn, i.e. α((λI − T )(Mn)) ≥ kα(Mn). This in turn means 1/(2n) > k.
Hence, 1/n > K.
We therefore can choose x1, x2 ∈ Mn with ‖(λI − T )(x1) − (λI − T )(x2)‖ < ‖x1 − x2‖/n. Set
yn = x1 − x2. This is of course also possible in the case of K = 0.
We need to show that
lim inf‖y‖→∞
‖(λI − T )(y)‖‖y‖
= 0.
For now we have
limn→∞
‖(λI − T )(yn)‖‖yn‖
= 0.
If (yn)n∈N has a subsequence (ynk)k∈N with ‖ynk
‖ → ∞, then λ ∈ σq(T ) and we are done. If it
has subsequence (ynk)n∈N that is both bounded and bounded away from zero, set xk = kynk
to
see that λ ∈ σq(T ). If ‖ynk‖ → 0, set xk = ynk
/‖ynk‖2. We then get ‖xk‖ → ∞ and
limk→∞
‖(λI − T )(xk)‖‖xk‖
= limk→∞
‖ynk‖−2‖(λI − T )(ynk
)‖‖ynk‖−1
= limk→∞
‖(λI − T )(ynk)‖
‖ynk‖
= 0,
hence, λ ∈ σq(T ). 2
Theorem 3.28 If T ∈ B(X), then σδ(T ) = σd(T ) and σπ(T ) = σapp(T ).
Proof: That σδ(T ) = σd(T ) is a simple consequence of Theorem 3.10. To see that σq(T ) =
σapp(T ) consider the equation
lim inf‖x‖→∞
‖λx− T (x)‖‖x‖
= lim inf‖x‖→∞
∥∥∥∥λ x
‖x‖− T
(x
‖x‖
)∥∥∥∥ = lim inf‖y‖=1
‖λy − T (y)‖.
34
Since σq(T ) = σπ(T ) by Lemma 3.27, we are done. 2
One of the important properties of the approximate point spectrum is that it is closed and
contains the boundary of the classical spectrum. The same is true for its nonlinear counterpart.
Lemma 3.29 Let T ∈ C(X), then σa(T ), σq(T ), and σπ(T ) are closed.
Proof: Let (λn)n∈N ⊆ σq(T ) converge to a λ. By Proposition 2.2,(iii), we get
[λI − T ]q = |[λI − T ]q − [λnI − T ]q| ≤ [λI − T − λnI + T ]Q = |λn − λ| → 0.
Hence, σq(T ) is closed. Similarly, σa(T ) is closed and, therefore, also σπ(T ) = σa(T )∪σq(T ). 2
Theorem 3.30 Let T ∈ C(X). Then
∂σFMV (T ) ⊆ σπ(T ).
Proof: First, we want to show, that σFMV (T )\σπ(T ) is an open subset of K. So fix a
λ ∈ σFMV (T )\σπ(T ). It suffices to show that there exists a δ > 0, such that µI − T is not
stably solvable for |µ− λ| < δ.
If this is not true, we find a sequence (λn)n∈N with λn → λ such that λnI −T is stably solvable
for all n. Since λI − T is not stably solvable, there exists a compact operator S with [S]Q = 0
such that λx − T (x) 6= S(x) for all x ∈ X. On the other hand, since all λnI − T are stably
solvable, we get a sequence (xn)n∈N with λnxn − T (xn) = S(xn).
In case ‖xn‖ → ∞ we get
‖λxn − T (xn)‖‖xn‖
≤ ‖λxn − λnxn‖‖xn‖
+‖λnxn − T (xn)‖
‖xn‖= |λ− λn|+
‖S(xn)‖‖xn‖
→ 0.
But this means [λI−T ]q = 0, contradicting λ /∈ σπ. Hence, there exists a bounded subsequence
(xnk)k∈N. This implies
‖(λ− T − S)(xnk)‖ = ‖λxnk
− (T − S)(xnk)‖ = ‖λxnk
− λnkxnk‖ ≤ |λ− λnk
| ‖xnk‖ → 0.
Furthermore, by Proposition 2.4,(v), [λ−T −S]a = [λ−T ]a > 0. Thus, by Proposition 2.7,(ii),
the preimage of the compact set (λ−T −S)(xnk)∪0, which in particular includes (xnk
)k∈N,
is compact. We can therefore assume without loss of generality, that xnk→ x for some x ∈ X.
By continuity, λx− T (x)− S(x) = 0. This contradiction shows that our assertion was true.
Now let λ ∈ ∂σFMV (T ), and assume λ /∈ σπ(T ). Then λ ∈ σFMV (T )\σπ(T ) which is an in Kopen subset of σFMV (T ), contradicting λ ∈ ∂σFMV (T ). 2
35
Corollary 3.31 Let T ∈ C(X) and let C ⊆ K\σπ(T ) be a connected set. Then either C ⊆σFMV (T ) or C ⊆ ρFMV (T ).
Proof: Let K be the connected component of K\σπ(T ) which contains C. Set
K0 = ν ∈ K : νI − T is FMV-regular = K ∩ ρFMV (T )
and assume K0 is not empty. By Theorem 3.30, the boundary of K0 relative to K is empty.
Therefore, K0 is open and closed in K. Since K is connected, it follows, that K0 = K. 2
3.3.1 Special classes of operators
We now look at two special classes of operators for which we can give more detailed information
about their FMV-spectrum. First, we look at compact operators in infinite dimensional Banach
spaces. The restriction to infinite dimensional spaces is to be expected, since in finite dimen-
sional spaces every continuous operator is compact. Hence, compactness only yields additional
properties in infinite dimensions. In the following proposition, for a closed set Σ ⊆ K with
0 /∈ Σ, we denote by c0[Σ] the connected component of K\Σ containing 0.
Proposition 3.32 Let X be an infinite dimensional Banach space and suppose T ∈ C(X) is
compact. Then the following is true:
(i) σa(T ) = 0, hence σπ = 0 ∪ σq(T ).
(ii) T is not surjective. In particular, 0 ∈ σδ(T )
(iii) Either 0 ∈ σq(T ), or c0[σq(T )] ⊆ σδ(T ).
(iv) If 0 /∈ σq(T ) and σFMV (T ) is bounded, then σq(T ) contains a positive and a negative value.
(v) If σFMV (T ) 6= K, then σq(T ) 6= ∅.
Proof: Assertion (i) simply follows from the fact, that [λI−T ]a = |λ|, which is a consequence
of Proposition 2.4,(v). For (ii), assume that T is surjective. For each y ∈ X we get y ∈ T (Bn(0))
for some n ∈ N. We therefore have the representation
X =∞⋃n=1
T (Bn(0)) =∞⋃n=1
T (Bn(0)).
By Baire’s catgegory theorem, the countable union of closed sets without interior points has no
interior points. Since X does have an interior point, at least one of the compact sets T (Bn(0))
contains an interior point. In particular, T (Bn(0)) contains a closed ball, which in turn is
compact. Since closed balls in a space X are compact if and only if X is finite dimensional,
this contradicts our choice of X as an infinite dimensional Banach space. 0 ∈ δδ(T ) now follows
36
from Lemma 3.2.
To see that (iii) holds true, first observe that the condition 0 /∈ σq(T ) together with (i) implies
that 0 is an isolated point of σπ(T ) since σq(T ) is closed. Below we will show that λI−T is not
surjective for |λ| small enough. By Lemma 3.2, this implies that 0 is an interior point of σδ(T )
and thus also of the FMV-spectrum. Theorem 3.30 and (i) then tell us that ∂σFMV (T ) ⊆ σq(T ).
Hence, σδ(T ) has no boundary in c0[σq(T )]. σδ(T )∩ c0[σq(T )] is therefore both open and closed
in the connected set c0[σq(T )], i.e. σδ(T )∩ c0[σq(T )] is either empty or coincides with c0[σq(T )].
Since 0 ∈ σδ(T ) by (ii), the latter must be true, which proves (iii).
To show that λI − T is not surjective for |λ| small enough if 0 /∈ σq(T ), assume that this is
not the case. Then there exists a sequence (λn)n∈N in K converging to zero, such that λnI − Tis surjective for all n. Fix a ∈ (0, 12 [T ]q), and choose R > 0 such that ‖T (x)‖ ≥ 2a‖x‖ for
‖x‖ > R. Taking b = 2aR we then have ‖T (x)‖ ≥ 2a‖x‖ − b for all x ∈ X. Consequently, for
|λ| ≤ a we have ‖λx− T (x)‖ ≥ a‖x‖ − b.Now fix y ∈ B1(0). By assumption, we find a sequence (xn)n∈N in X such that λnxn−T (xn) = y
for all n. Without loss of generality we may assume that |λn| ≤ a for all n, and thus
1 > ‖y‖ = ‖λnxn − T (xn)‖ ≥ a‖xn‖ − b.
Hence, xn ∈ Br(0) with r = (1 + b)/a. Since λnxn → 0, we conclude that T (xn) → −y. But
since y ∈ B1(0) was arbitrary, we see that B1(0) ⊆ T (Br(0)), which is impossible as T (Br(0))
is precompact.
(iv) is an immediate consequence of (iii), since for instance R+∩σq(T ) = ∅ yields the unbound-
edness of σδ(T ) ⊆ σFMV (T ). Finally, if σq(T ) = ∅, then σδ(T ) = K by (iii). Thus, (v) holds
true. 2
Definition 3.33 An operator T ∈ C(X) is called asymptotically linear if there exists an operator
T ′ ∈ B(X) with [T − T ′]Q = 0. T ′ is called the asymptotic derivative of T .
If an operator is asymptotically linear, its asymptotic derivate is uniquely defined, since for two
asymptotic derivates T ′1 and T ′2 we get
‖T ′1 − T ′2‖ = [T ′1 − T ′2]Q ≤ [T ′1 − T ]Q + [T − T ′2]Q = 0.
For asymptotically linear operators the spectrum of its asymptotic derivate can be used to gain
information about its FMV-spectrum.
Proposition 3.34 Let T ∈ C(X) be an asymptotically linear operator with asymptotic derivate
T ′. Then the following is true:
(i) σq(T ) = σq(T′).
(ii) ∂σ(T ′) ⊆ σq(T ), in particular, σq(T ) is not empty for K = C.
37
(iii) If T − T ′ is compact, then σFMV (T ) = σ(T ′), σa(T ) = σa(T′), σq(T ) = σq(T
′), and
σδ(T ) = σδ(T′).
Proof: First, from the definition of the asymptotic derivative and Proposition 2.2,(iii) we get
∣∣[λI − T ]q − [λI − T ′]q∣∣ ≤ [λI − T − λI + T ′]Q = [T ′ − T ]Q = 0,
from which (i) follows immediately . To see that (ii) holds, use Proposition 3.18, Theorem 3.30,
Proof: Fix λ ∈ σq(T ) and let (xn)n∈N be a sequence satisfying the conditions ‖xn‖ → ∞ and
‖λxn − T (xn)‖/‖xn‖ → 0. Without loss of generality we may assume, that ‖xm − xn‖ > 2 for
m 6= n. Put
ηn(x) = max1− ‖x− xn‖, 0,
so that ηn(xm) = δmn. Now define S : X → X by
S(x) = T (x) +
∞∑n=1
ηn(x)(λxn − T (xn)).
By definition, ηn(x) 6= 0 for only one n, because if ηn(x) 6= 0, then ‖x − xn‖ < 1. So ‖x −xm‖ ≥ |‖x− xn‖ − ‖xn − xm‖| > 1 and, hence, ηm(x) = 0 for m 6= n, i.e. S is well defined.
Furthermore, [S]Q <∞. In fact, because of [T ]Q <∞ and our choice of (xn)n∈N, we get
[S]Q = lim sup‖y‖→∞
‖S(y)‖‖y‖
≤ lim sup‖y‖→∞
‖T (y)‖‖y‖
+ lim sup‖y‖→∞
‖∑∞
n=1 ηn(y)(λxn − T (xn))‖‖y‖
= [T ]Q + lim sup‖y‖→∞
ηn(y)(y)‖λxn(y) − T (xn(y))‖‖y‖
≤ [T ]Q + limn→∞
‖λxn − T (xn)‖‖xn‖ − 1
= [T ]Q,
where n(y) is the one n where ηn(y) 6= 0, if such an n exists. Next, let (yk)k∈N be a bounded
sequence. Then ‖xn − yk‖ < 1 can only be true for finitely many n. Hence, (S − T )(yk) ∈spanλxn1 − T (xn1), · · · , λxnm − T (xnm) for some m ∈ N. It therefore has a convergent
subsequence, i.e. S − T is compact. Thus, we get
[S]A = sup∞>α(M)>0
α(S(M))
α(M)≤ sup∞>α(M)>0
(α(T (M))
α(M)+α((S − T )(M))
α(M)
)= [T ]A.
In summary, S ∈ AQ(X).
By definition, S(xn) = λxn for all n, i.e. λ ∈ σ0p(S). Further, pAQ(S − T ) = 0. In fact,
[T − S]Q ≤ lim sup‖y‖→∞
‖∑∞
n=1 ηn(y)(λxn − T (xn))‖‖y‖
= 0,
as has been shown above, and [T − S]A = 0 since S − T is compact. Thus, we have proven the
if ‖w(t)‖ > N . This implies that w(t) is bounded. 2
Note that the condition [(A T A−1)⊥]Q < ∞ is in particular met if [T ]Q < 0. Because
in this case we can set y = A−1x and get
[A T A−1]Q‖A−1‖
=1
‖A−1‖lim sup‖x‖→∞
‖(A T A−1)(x)‖‖x‖
= lim sup‖y‖→∞
‖(A T )(y)‖‖A−1‖‖Ay‖
≤ lim sup‖y‖→∞
‖A‖ ‖T (y)‖‖y‖
= ‖A‖[T ]Q <∞,
since ‖x‖ → ∞ if and only if ‖A−1x‖ → ∞. Lemma 3.45,(iv) then gives [(A T A−1)⊥]Q ≤[A T A−1]Q ≤ ‖A−1‖‖A‖[T ]Q <∞.
If T is linear, it is a well known fact that the condition σ(T ) ⊆ K− implies that every solution
to the differential equation z = Tz is bounded. But σ(T ) ⊆ K− does not necessarily imply
that W (T ) ⊆ K−. However, it can be shown that there exists a linear ismorphism A, such that
WFMV (A T A−1) = W (A T A−1) ⊆ K−. Since also [T ]Q <∞, this shows that Theorem
4.12 is indeed an extension of this assertion about linear differential equations.
54
4.3 The nonlinear Fredholm alternative
The Fredholm alternative is an important and very useful theorem in the theory of linear
operators on a Banach space X. Its proof is in essence contained in Corollary 3.42.
Theorem 4.13 (Fredholm alternative) Let T ∈ B(X) be compact and 0 6= λ ∈ K. Then
either
• λ ∈ σ(T ), i.e. λ is an eigenvalue of T ,
or
• λI − T is bijective.
Proof: If λ is not an eigenvalue of T , then λ ∈ ρ(T ) by Corollary 3.42. Hence λI − T is
bijective. 2
The interesting implication here is, of course, that if λ is not an eigenvalue, then the equa-
tion λx− Tx = y has a unique solution for each y ∈ X.
In our theory the role of bijectivity has been taken over by FMV-regularity. The property of
FMV-regular operators, which can be used to guarantee solutions to certain equations, is stable
solvability. This motivates the following definition.
Definition 4.14 An operator T ∈ C(X) is called alternative if
σFMV (T ) = σπ(T ).
Again, the interesting implication here is that if λ /∈ σπ(T ), then λI − T is FMV-regular and,
in particular, stably solvable. In order to assert that λ /∈ σπ(T ), we have to check that both
λ /∈ σq(T ) and λ /∈ σa(T ).
The first of these conditions is the one that is easier to handle. By definition, λ ∈ σq(T ) if and
only if λ is an asymptotic eigenvalue. For λ ∈ σq(T ), we have to check whether there exists a
sequence (xn)n∈N in X with ‖xn‖ → ∞ such that
limn→∞
‖λxn − T (xn)‖‖xn‖
= 0.
Furthermore, according to Proposition 3.47,(vi), σq(T ) ⊆WFMV (T ). Thus, the implication
λ /∈WFMV (T )⇒ λ /∈ σq(T )
holds true and gives us a further tool to see whether λ ∈ σq(T ) or not.
The set σa(T ) is more complicated to handle. It is difficult to calculate [T ]a, as even the
calculation of the quotient α(T (M))/α(M) for fixed M can be very challenging. We give some
conditions which indicate that λ /∈ σa(T ).
Lemma 4.15 Let T ∈ C(X) fulfil [T ]a > 0. Then the following holds true
55
(i) There is a k > 0 such that α(T (M)) ≥ kα(M) for all bounded sets M ⊆ X.
(ii) T is proper on closed bounded sets.
(iii) If (xn)n∈N is bounded and (T (xn))n∈N converges, then (xn)n∈N has a convergent subse-
quence.
In particular, if λI − T does not meet one of these three conditions, then λ /∈ σa(T ).
Proof: (i) follows directly from the definition of [T ]a. (ii) has been proven in Proposition 2.7.
We show that (ii) ⇒ (iii). So let (xn)n∈N be bounded by R > 0 and assume that (T (xn))n∈N
converges to some y ∈ X. Then the set T (xn) : n ∈ N ∪ y is compact. Since T is proper
on BR(0) by assumption, the preimage of T (xn) : n ∈ N ∪ y is also compact. In particular,
(xn)n∈N lies in a compact set and, therefore, has a convergent subsequence. 2
Furthermore, we want to identify some classes of alternative operators for which there is no
need to be concerned with the subspectrum σa(T ) at all.
Proposition 4.16 Let T ∈ C(X) be alternative. Assume that one of the conditions
(i) X is finite dimensional,
(ii) T is compact,
(iii) T is asymptotically linear with asymptotic derivate T ′ such that T − T ′ is compact,
holds true. Then σFMV (T ) = σq(T ). In particular, for any λ ∈ K either
• λ ∈ σq(T ), i.e. λ is an asymptotic eigenvalue of T ,
or
• λI − T is FMV-regular.
Proof: Since T is alternative we have σFMV (T ) = σπ(T ) = σq(T ) ∪ σa(T ). Thus, we have to
show that σa(T ) ⊆ σq(T ).
If X is finite dimensional then σa(T ) = ∅ and we are done. If T is compact, then σa(T ) = 0by Proposition 3.32,(i), so we have to show that 0 ∈ σq(T ). But if this were not the case, then
Proposition 3.32,(iii) would give
c0[σq(T )] ⊆ σδ(T ) ⊆ σFMV (T ) ⊆ σq(T ) ∪ 0,
i.e. c0[σq(T )] ⊆ 0. But this is impossible since c0[σq(T )] is both open and nonempty.
Finally, if T fulfills (iii), then Proposition 3.34,(iii) and Lemma 3.27 show that
σa(T ) = σa(T′) ⊆ σq(T ′) = σq(T ).
2
56
Proposition 4.17 Let T ∈ B(X) be alternative. Then σ(T ) = σq(T ). In particular, either
• λ ∈ σq(T ), i.e. λ is an asymptotic eigenvalue of T ,
or
• λI − T is bijective.
Proof: By Lemma 3.27 we have σa(T ) ⊆ σq(T ). Since T is linear and alternative, this yields
σ(T ) = σFMV (T ) = σq(T ). 2
Now that we have explored the usefulness of the notion of alternative operators, we want to
identify some classes of operators as alternative. One such class has essentially already been
identified.
Theorem 4.18 Let T ∈ C(X) be compact and asymptotically odd. Then T is alternative.
The last inequality holds true since λ /∈ σq(T + S) and [L−1]q > 0 by Lemma 2.8,(iv). Now,
L−1 (S + K) is not only compact, but also asymptotically odd. Thus, it is alternative by
Theorem 4.18. Hence, 1− L−1 (S +K) is FMV-regular. 2
We are now interested in identifying classes of balanced operators
Every linear operator T on a finite dimensional Hilbert space is balanced, since it can be easily
seen that σa(T ) = σes(T ) = ∅. In infinite dimensional Hilbert spaces the operators λI for λ ∈ Kare simple examples of balanced operators, since σa(λI) = σes(λI) = λ.To identify more interesting examples of balanced operators the following result will be very
useful.
58
Lemma 4.24 Let T ∈ B(H). Then [T ]a > 0 if and only if ran(T ) is closed and fulfils
dim ker(T ) <∞.
Proof: Suppose [T ]a > 0 and let (xn)n∈N be an arbitrary sequence in B1(0) ∩ ker(T ). Then
T (xn) = 0 for all n. In particular (T (xn))n∈N converges and by Lemma 4.15,(iii), the sequence
(xn)n∈N has a convergent subsequence. Consequently, the closed unit ball in the subspace ker(T )
is compact. Thus, ker(T ) is finite dimensional.
This in turn also implies that there exists a closed subspace H1 of H such that H = H1⊕ker(T ).
As T (H) = T (H1) we only need to show that T (H1) is closed. Note that the restriction T |H1 :
H1 → T (H1) is bijective. It is obviously surjective, but also injective, since T |H1(x) = T |H1(y)
for x 6= y would imply T |H1(x− y) = 0, which leads to the contradiction x− y ∈ ker(T ). Thus,
T |H1 has an inverse.
If we can show that this inverse is bounded, i.e. T |H1 fulfils ‖T |H1x‖ ≥ m‖x‖ for all x ∈ H1
with ‖x‖ = 1 and some m > 0, then T (H1) = T (H) is closed as the preimage of a closed set
under a continuous map. Assume the contrary. Then there exists a sequence (xn)n∈N in H1
with ‖xn‖ = 1 such that T |H1xn → 0. Using Lemma 4.15,(iii), we can assume without loss
of generality that (xn)n∈N converges to some x ∈ H1 with ‖x‖ = 1. Obviously Tx = 0, which
contradicts H1 ∩ ker(T ) = 0.Conversely, we have to show that if ran(T ) is closed and dim ker(T ) <∞, then [T ]a > 0. Since
dim ker(T ) < ∞, there exists a closed subspace H1 of H such that H = H1 ⊕ ker(T ). Let
P1 : H → H1 be the projection along ker(T ) onto H1 and set P2 = I − P1. Let T |H1 be as
above.
Since ran(T ) is closed, T |H1 is an isomorphism of H1 onto ran(T ) = ran(T |H1) by the open map-
ping theorem. Since T = T |H1P1 we have [T ]a ≥ [T |H1 ]a[P1]a by Proposition 2.7,(vii). Since P2
maps H into the finite dimensional space ker(T ), it is compact. Thus, [P1]a = [I−P2]a = [I]a =
(ii): Let n be the smallest integer such that Tn is compact. As above, we get [T ]na ≤ [Tn]a ≤[Tn]A = 0 and thus 0 ∈ σa(T ). Since σa(T ) ⊆ σes(T ) by Corollary 4.25, we are finished once
we show that λ 6= 0 implies λ /∈ σes(T ).
If n > 1, we can write λnI − Tn = A (λI − T ), where
A = λn−1I + λn−2T + · · ·+ Tn−1.
So if λ 6= 0, we get 0 < |λ|n = [λnI − Tn]a ≤ [A]A[λ− T ]a by Proposition 2.7,(vii). This yields
[λI − T ]a > 0, which is also true if n = 1. Since T ∗n is also compact, we get [λI − T ∗]a > 0.
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By Lemma 4.24 and Theorem 4.27,(ii), ran(λI − T ) is closed and both dim ker(λI − T ) and
dim(ran(λI − T ))⊥ = dim ker(λI − T ∗) are finite. Thus, λI − T is Fredholm.
Since (cT )n is compact for any c ∈ R, we get that λI− cT is Fredholm for any c ∈ R and λ 6= 0.
Theorem 4.21 tells us that the Fredholm index is continuous. Hence, ind(λI−T ) = ind(λI) = 0
and λ /∈ σes(T ).
(iii): Let λ /∈ σa(T ). By Lemma 4.24, dim ker(λI − T ) < ∞ and ran(λI − T ) is closed. As
λI − T is normal by assumption, Theorem 4.27,(iii) gives H = ker(λI − T )⊕ ran(λI − T ). For
the Fredholm index we get
ind(λI − T ) = dim ker(λI − T )− dim(ran(λI − T ))⊥ = dim ker(λI − T )− dim ker(λI − T ) = 0.
So we have σes(T ) ⊆ σa(T ) and by Corollary 4.25, σes(T ) = σa(T ). 2
Next, we want to show a third class of alternative operators. Since every linear selfadjoint
operator is obviously normal, the previous theorem shows that they are alternative. We can, in
a certain manner, extend the notion of selfadjoint operators to nonlinear operators.
Definition 4.29 An operator T ∈ C(H) is called selfadjoint, if 〈T (x), x〉 ∈ R for all x ∈ H.
By SA(H) we denote the set of all selfadjoint operators on H.
By Theorem 4.27,(iv), this coincides with the usual definition of selfadjoint linear operators.
Note that the equation 〈T (x), y〉 = 〈x, T (y)〉 can not be used for nonlinear operators, as its
validity for all x, y ∈ H already implies the linearity of T .
We will now give a condition under which selfadjoint operators are alternative.
Lemma 4.30 If T ∈ SA(H), then WFMV (T ) ⊆ R
Proof: For λ ∈WFMV (T ) = σq(T⊥) we have
0 = lim inf‖x‖→∞
∥∥∥λx− 〈T (x),x〉‖x‖2 x∥∥∥
‖x‖= lim inf‖x‖→∞
|λ− 〈T (x), x〉‖x‖2
|.
Since 〈T (x),x〉‖x‖2 ⊆ R, λ must also lie in R. 2
Theorem 4.31 Let T ∈ SA(H). Suppose T fulfils the following two conditions.
(i) σa(T ) ⊆ R.
(ii) There exists a λ1 with Imλ1 < 0 and a λ2 with Imλ2 > 0 such that λ1I − T and λ2I − Tare stably solvable.
Then T is alternative.
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Proof: Because T is selfadjoint, we have WFMV (T ) ⊆ R by Lemma 4.30. Therefore, σπ(T ) =
σq(T )∪σa(T ) ⊆WFMV (T )∪σa(T ) ⊆ R by Proposition 3.47,(vi). Since ∂σFMV (T ) ⊆ σπ(T ) ⊆ Rby Theorem 3.30, this leaves only four possibilites: σFMV (T ) = C, σFMV (T ) = λ ∈ C : Imλ ≥0, σFMV (T ) = λ ∈ C : Imλ ≤ 0, and σFMV (T ) ⊆ R. The first three cases are impossible by
assumption (ii). Thus, σFMV (T ) ⊆ R. Since R has no interior points in C, every point of the
FMV-spectrum must lie in its boundary. Hence, σFMV (T ) ⊆ ∂σFMV (T ) ⊆ σπ(T ). 2
The final two results give special cases in which the conditions of the previous theorem are
met.
Proposition 4.32 Let T ∈ SA(H). Suppose T fulfils one of the following two conditions.
(i) T is assymptotically linear with asymptotic derivate T ′ and T − T ′ is compact.
(ii) T = R+ S where R ∈ B(H) is selfadjoint and S ∈ C(H) is compact.
Then T fulfils condition (i) in Theorem 4.31.
Proof: In the first case, we can use Proposition 3.34,(iii), Lemma 3.27, Proposition 3.47,(vi),