Chapter 1

Fixed point theorems

One of the most important instrument to treat (nonlinear) problems with the aid of functional analytic methods is thefixed point approach. This approach is an important part of nonlinear (functional-)analysis and is deeply connected to geometric methods of topology. We consider in this chapter the famous theorems of Banach, Brouwer and Schauder. A more detailed description of thefixed point theory can be found for instance in [1, 2, 6, 8, 9] dar.

1.1 Thefixed point approach – preliminaries

Consider the followingfixed point equation

x=F(x) (1.1)

Here,F is a mapping fromX X. We assume thatX is endowed with the metricd. − A pointz X which satisfiesz=F(z) is called afixed point of F. Fixed point theo- ∈ rems guarantee the existence→ and/or uniqueness whenF andX satisfy certain additional conditions. A simple example of a mappingF which doesn’t posses afixed point is the translation in a vector spaceX:

F:X x x+x 0 X wherex 0 =θ. � �− ∈ �

Many problems of applied mathematics→ may be formulated as afixed point equation or reformulated in this way: Linear equations, differential equations, zeroes of gradients, . . . . The following examples show how the treatment of nonlinear problems may be translated intofixed point problems:

Determination of zeros of nonlinear functionsg:R R: • − g(x) =0 → There are different possibilities to translate this equation into afixed point problem:

F(x) :=x−g(x) simplest method F(x) :=x−ωg(x) linear relaxation with relaxation parameterω>0 −1 F(x) :=x−(g �(x)) g(x) Newton method

1 Consider the determination of a zeroz of a functionf:R n R n . Clearly, such a • − zero is afixed point of the mappingF:R n x x+f(x) R n . The formulation � �− ∈ of equations as afixed point equation is possible in various ways.→ Consider

3 → f:R x x − 13x+ 18 R. � �− ∈ z f z g , i 1, 2, 3, 4, is a zero of iff is afixed point of→ i = where 3 g x x 3 12x 18 g x x + 18 1( ) := − + 2( ) := 13 1 13x− 18 g (x) := (13x− 18) 3 g (x) := 3 4 x2

Ordinary differential equations • 0 y� =f(t, y),y(t 0) =y (1.2)

If the righthandsidef is continuous on an appropriate domainQ R R n then this ⊂ × initial value problem is equivalent to the following integral equation offixed point character: t y(t) =F(y)(t) :=y 0 + f(s, y(s))ds 0 �y (Nonlinear) partial differential equations: • −Δu=f(u), inΩ,u=θ in ∂Ω

f is given and the solutionu has to be found. Under certain conditions one can define the inverse(−Δ) −1 ofΔ on spaces of functionsu which vanish on ∂Ω. Then we obtain the followingfixed point equation:

u=F(u) := (−Δ) −1(f(u))

Suppose we have a mappingF:X X as given above. We want to consider − methods to compute such afixed point. Here are some preliminary remarks. We define

0 →n+1 n F := id,F :=F F , n N 0 . ◦ ∈ n Choosingx X, the sequence(F (x))n N is called the orbit ofx underF. This orbit is ∈ ∈ the result of the iteration

n+1 n 0 x :=F(x ),n N 0;x :=x. (1.3) ∈ This iteration process is in the focus of the following chapters. The main question which we will analyze is under which assumptions (concerningF andX) the orbit converges.

2 1.2 Fixed point theorem of Banach

Consider thefixed point equation (1.1) again. The essential step to prove the existence of a solution of this equation via the iteration

n+1 n 0 x :=F(x ),n N 0;x :=x, (1.4) ∈ n is to prove that the orbit(F (x))n N is a Cauchy sequence (for everyx X). The ∈ ∈ assumption which guarantees this is the assumption thatF is Lipschitz continuous with Lipschitz constantc satisfyingc<1. This means

There existsc [0, 1) such thatd(F(x),F(y)) cd(x, y) for all x, y X. (1.5) ∈ ≤ ∈ If (1.5) is satisfyied,F is called a contraction. Notice thatc n is the Lipschitz constant ofF n . Lemma 1.1. Let(X, d) be a metric space and letF:X X be a contraction. If − c [0, 1) is the Lipschitz constant of F, then we have ∈ 1 → d(x, y) (d(x, F(x)) +d(y, F(y))) , x, y X. (1.6) ≤ 1−c ∈ Proof: Let x, y X. Using the triangle inequality ∈ d(x, y) d(x, F(x)) +d(F(x),F(y)) +d(F(y), y) ≤ and usingd(F(x),F(y)) cd(x, y) the result follows. ≤ � (1.6) is a very strange inequality: it says that we can estimate how far apart any two points x, y are just from knowing how farx is from its imageF(x) and how fary is from its imageF(y).Afirst application is Corollary 1.2. Let(X, d) be a metric space and letF:X X be a contraction. Then − F has at most onefixed point. Proof: → Suppose that x, y arefixed poits ofF. Thend(x, F(x)), d(y, F(y)) are zero, so by the inequality (1.6)d(x, y) is zero too. � Lemma 1.3. Let(X, d) be a metric space and letF:X X be a contraction. Then n − the orbit(F (x))n N0 is a Cauchy sequence. ∈ Proof: → Letc [0, 1) be the Lipschitz constant ofF. From the inequality (1.6) we read off ∈ n m 1 n n 1 m m 1 d(F (x),F (x)) (d(F (x),F + (x)) +d(F (x),F + (x))) , m, n N. ≤ 1−c ∈ Recalling thatc k is the Lipschitz constant ofF k we get n m n m c +c d(F (x),F (x)) d(x, F(x)) , m, n N. ≤ 1−c ∈ k n m Since limk c =0 we obtain lim m,n d(F (x),F (x)) =0. �

→∞ 3 Theorem 1.4 (Contraction Theorem, Banach 1922). Let(X, d) be a complete metric space and letF:X X be a contraction. Ifc [0, 1) is the Lipschitz constant of F, − ∈ then the following assertions hold: → (a) There exists a uniquely determinedfixed pointz ofF.

(b) The iteration n+1 n 0 x :=F(x ),n N 0 , x :=x, (1.7) ∈ n defines a sequence(x )n N0 which converges toz for allx X. ∈ ∈ n n c (c)d(x , z) d(F(x), x),n N 0 . ≤ 1−c ∈ n c n n−1 (d)d(x , z) d(x , x ),n N. ≤ 1−c ∈ (e)d(x n, z) cd(x n−1, z),n N. ≤ ∈ Proof: Ad(a),(b) The uniqueness of afixed point follows from Corollary 1.2. Letx X. From Lemma (1.3) n ∈ we obtain that the orbit(F (x))n N0 converges to a pointz since(X, d) is complete. Since ∈ F is continuous due the fact thatF is a contraction – see (1.5) –z is afixed point. Ad(c) Letx X. We may prove with the help of (1.5) inductively ∈ d(Fn+1(x),Fn(x)) c nd(F(x), x). (1.8) ≤ Then with (1.8)

m m d(Fm(x),Fn(x)) d(Fj(x),Fj−1(x)) cj−1d(F(x), x) (1.9) ≤ ≤ j n 1 j n 1 =�+ =�+ 1−c m−n c n d(F(x), x),m n. (1.10) ≤ 1−c ≥ Sincec [0, 1) we obtain the assertion (c) by going to the limitm in (1.9),(1.10) . ∈ We obtain from (1.8)

m−1 →∞ m 1−c n n 1 d(F (x),F(x)) c d(F (x),F − (x)),m N. (1.11) ≤ 1−c ∈

(d),(e) are simple consequences of these inequalities. � Theorem 1.5 (Banach’sfixed point theorem). Let(X, ) be a Banach space, letV be �·� a closed subset ofX and letF:V V be a contraction, that is − c [0, 1) x, y V(d(F(x),F(y)) cd(x, y)). (1.12) ∃ ∈ ∀ → ∈ ≤ Then the following statements hold:

a) There exists a uniquely determinedfixed pointz ofF inV.

4 b) The successive iteration

n+1 n n+1 0 x :=F(x ) =F (x),n N 0 , x =x, (1.13) ∈ n defines a sequence(x )n N0 which converges toz for allx V. ∈ ∈ n n c c) x −z F(x)−x ,n N. � � ≤ 1−c � � ∈ n c n n−1 (d) x −z x −x ,n N. � � ≤ 1−c � � ∈ (e) x n −z c x n−1 −z ,n N. � � ≤ � � ∈ Proof: Consider the metric space(V, d) withd(x, y) := x−y , x, y V. This space is complete � � ∈ due to the closedness ofV. Now apply Theorem 1.4. � Banach’sfixed point theorem is a perfect theorem“. It formulates an easy to check ” assumption under which we have existence and uniqueness and computatibilty of afixed point is guarateed. Moreover, it presents the order of the convergence of the approximating sequence and gives hints to stop the iteration infinite steps; see below. Remark 1.6. The application of thefixed point theorem of Banach to an initial value problem for ordinary differential equations leads to an existence and uniqueness result, the so called Picard–Lindel¨off theorem; see [3]. The assumption we need to apply Theorem 1.5 is that the righthand sidef in (1.2) is continuous and Lipschitz continuous with respect toy uniformly int. �

In many applications, the mappingF in afixed point equation is dependent on a parameterp P: ∈ x=F (x),x X, p P. (1.14) p ∈ ∈ In this situation we have

Corollary 1.7. Let(X, d) be a complete metric space, let(P, d �) be a metric space and letF :X X be a contraction for eachp P. Letc [0, 1) be the Lipschitz constant p − ∈ p ∈ ofF . We assumec c < 1, p P. Moreover we assume, that there existsp P with p p ≤ ∈ 0 ∈ lim F (x) =F (x) for allx X. Then the equation (1.14) posesses for eachp Pa p p0 p→ p0 ∈ ∈ uniquely determined solutionx p X and we have lim p p0 xp =x p0 . → ∈ Proof: → Letx P be the uniquely determined solution of (1.14). This solution exists due to p ∈ Theorem 1.4. Then

d(xp, xp0 ) =d(F p(xp),Fp0 (xp0 )) d(F (x ),F (x )) +d(F (x ),F (x )) ≤ p p p p0 p p0 p0 p0 cd(x , x ) +d(F (x ),F (x )) ≤ p p0 p p0 p0 p0 and we conclude 1 d(x , x ) d(F (x ),F (x )). p p0 ≤ 1−c p p0 p0 p0

By taking the limitp p 0 we obtain limp p0 xp =x p0 . �

→ → 5 1 Example 1.8.X=R, :=| |,M := [0, ),F:M t t+ M. We �·� · � �− 1+t ∈ have |F(t)−∞F(s)|<|t−s|. → F posesses nofixed point! Notice that

|F(t)−F(s)| c|t−s| , t, s M, ≤ ∈ for noc [0, 1) is possible. ∈ � Corollary 1.9. LetX be a Banach space and letF: B X be a contraction, i. e. r − L [0, 1) x, y B ( F(x)−F(y) L x−y ). (1.15) ∃ ∈ ∀ ∈ r � � ≤ →� � IfF(∂ B ) B thenF has a uniquely determinedfixed point in B . r ⊂ r r Proof: Define 1 G: B x (x+F(x)) X. r � �− 2 ∈ Letx B , x=θ, and consideru := rx x −1 ∂ B B . Then we obtain ∈ r � � �→ ∈ r ⊂ r F(x)−F(u) L x−u =L(r− x ) � � ≤ � � � � and hence

F(x) F(u) + F(x)−F(u) r+L(r− x ) 2r− x � � ≤ � � � � ≤ � � ≤ � � and 1 1 G(x) = x+F(x) ( x + F(x) ) r. � � 2� � ≤ 2 � � � � ≤ Due to the continuity ofG we obtain G(θ) r. � � ≤ This shows thatG: B B andG is a contraction too as we conclude from r − r 1 1 G(x)−G(y) →( x−y +L x−y ) = (1+L) x−y , x, y Br . � � ≤ 2 � � � � 2 � � ∈ Theorem 1.5 implies thatG has afixed point x B . This implies x=F( x). Due to the ∈ r fact thatF is a contraction thisfixed point is uniquely determined.� Remark 1.10. What is a kind of the inverse of the contractionfixed point theorem? The most elegant result in this direction is due to Bessaga [5]. Suppose thatX is an arbitrary nonempty set and suppose thatF:X X has − the property thatF and each of its iteratesT n has a uniquefixed point. Then for each c (0, 1) there exists a metricd on eachX such that(X, d ) is complete and for which ∈ c c → d (F(x),F(y)) cd (x, y) for all x, y X. c ≤ c ∈ �

6 Theorem 1.11 (Edelstein, 1961). LetX be a Banach space and letV X. Suppose that ⊂ F:V V is contractive, i. e. − x, y V, x= y,( F(x)−F(y) < x−y ). (1.16) → ∀ ∈ � � � � � Then we have

a)F has at most afixed point inV.

b) IfV is compact thenF posesses exactly onefixed point.

Proof: Ada) Obvious. Adb) Consider the mappingV v v−F(v) R. SinceV is compact and since � �− � � ∈ the norm is continuous there exists x V with x−F( x) = inf v V v−F(v) . Now we ∈ � � ∈ � � must have x=F( x) because otherwise→ we would have F(x)−(F F)( x) < x−F( x) . � ◦ � � � � Remark 1.12. The completeness that we assume in Theorem 1.5 and in the following results is more or less necessary for the existence offixed points; see for instance [7, 11, 10, 14, 15]. � Remark 1.13. An application of the contraction theorem of Banach is the theorem of the inverse mapping; see for instance [4]. �

1.3 Successive approximation for a contraction

Consider n+1 n 0 x :=F(x ),n N 0 ;x :=x, (1.17) ∈ with a contractionF in a metric space(X, d). From a practical programming point of view, we need a stopping rule for the successive approximation in order to compute the fixed point in afinite number of iteration steps; see above. The goal of a stopping rule is to guarantee for a given accuracy quantityε>0 an approximationx N X with ∈ d(xN, z) ε ≤ n where(F (x))n N ist the orbit generated by the iteration (1.17). To implement such a ∈ stopping rule one may pursue two different concepts.

A priori concept This results from the estimation(c) in Theorem 1.4:

n n c d(x , z) d(F(x), x),n N. ≤ 1−c ∈ A priori stopping rule

7 Ifδ :=d(F(x), x) andN>(log(ε) + log(1−c)− log(δ))/ log(c) thend(x N, z) ε. ≤ Notice that such a numberN exists due to the fact that0 c<1. This number can be n ≤ computed before we compute the orbit(F (x))n N0 . ∈ Suppose we takeε := 10 −m in our stopping rule inequality. In order to obtain one more decimal digit of precision we have to do (roughly) 1/| log(c)| more iterations steps. Stated a little differently, if we needN iterative steps to getm decimal digits of precision, then we need anotherN to double the precision to 2m digits. This kind of error behavior is called linear convergence.

A posteriori stopping concept The basis for this conception is the estimation in(d) of Theorem 1.4.

A posteriori stopping rule

Ifd(x N, xN−1) ε(1−c)/c thend(x N, z) ε. ≤ ≤

Algorithm 1.1 Successive approximation – A posteriori stopping rule Given afixed point equationx=F(x) and an accuracy parameterε>0. This algorithm computes a sequencex 0, x1, , . . . , xN such thatd(x N, z) ε. Herex 0 is a given starting ≤ value. Assumptions: see Theorem 1.4

(1)k :=0

(2)u :=x k (3) Ifu=F(u) setN :=0 and go to line (7)

(4)x k+1 :=F(u) (5) Ifd(x k+1, xk) ε(1−c)/c setN :=k+1 and go to line (7). ≤ (6) Setk :=k+1 and go to line (2).

(7) STOP with an approximationx N for thefixed pointz which satisfiesd(x N, z) ε ≤

For the realization of the stopping rules above we need the Lipschitz constantc. It is obvious that an approximation c˜ with c˜ [c, 1) is sufficient. ∈ Now, we consider the successive approximation method under errors. We follow mainly [12]. We consider in the complete metric space(X, d) thefixed point equation (1.1)

x=F(x) (1.18) and the associated successive approximation

k+1 k 0 x :=F(x ),k N 0 x := x, (1.19) ∈ 8 whereF:X X is a given mapping andx 0 is a given starting value of the iteration. − k This defines an approximating family(x )k N . We assume thatF is a contraction with ∈ k contraction→ constantc [0, 1). Then – by the contraction theorem –(x )k N converges ∈ k ∈ to the uniquefixed pointz of the equation (1.18). Now, instead of(x )k N0 we have at k ∈ hand an approximating sequence( x˜ )k N0 only. This sequence may be constructed by an ∈ iteration of a Lipschitz continuous mapping F˜:X X as follows: − k+1 ˜ k 0 x˜ := F(x˜ ),k N 0 , x˜ :=x. (1.20) ∈ → k ˜ In general, we do not have limn N x˜ =z when F is different fromF. ∈ Theorem 1.14. Let(X, d) be a complete metric space and letF:X X be a − contraction with Lipschitz constantc [0, 1) and afixed pointz. Letδ>0 and suppose k k k ∈ that(x )k N0 ,( x˜ )k N0 = (x˜(δ) )k N0 are sequences such that the following conditions are ∈ ∈ ∈ → satisfied:

n+1 n 0 x :=F(x ),n N 0 , x =x. (1.21) n+1 n ∈ d(x˜ ,F(x˜ )) δ,n N 0 . (1.22) ≤ ∈ d(x˜0, x0) δ. (1.23) ≤ Stopping rule: Suppose there existsN(δ) N with ∈ d(x˜n+1, x˜n) d( x˜n+1, x˜n), n= 0, . . . , N(δ),d( x˜N(δ)+1, x˜N(δ))>d( x˜N(δ), x˜N(δ)−1) (1.24) ≤ Then lim x˜N(δ) =z (1.25) δ 0

Proof: ↓ Letm N(δ). Then ≤ d(x˜m, xm) d( x˜m,F(xm−1) δ+ cd( x˜m−1, xm−1 ≤ . ≤ . δ+cδ+ +c m−1δ+c md(x˜0, x0) ≤ ··· δ +d( x˜0, x0). ≤ 1−c This implies lim d(x˜m, xm) =0,m= 1, . . . , N(δ), lim d(x˜N(δ), xN(δ)) =0. (1.26) δ 0 δ 0

Since ↓ ↓ d(x˜m, z) d( x˜m, xm) +d(x m, z),m= 0, . . . , N(δ), (1.27) ≤ N(δ) we have to show limδ 0 d(x , z) =0. Case 1: limδ 0 N(δ) = . n ↓ Since limn d(x , z) = 0, nothing has to be proved. ↓ Case 2: If limδ 0 N(δ) = does not hold then there exist a sequence(δ l)l N andq N ∞ ∈ ∈ withN l :=N(δ l) q, l N. ↓ ≤ ∈ Letm q. Then ≤ ∞ d(xq, xq−1) cd(x q−1, xq−2) c q−md(xm, xm−1). (1.28) ≤ ≤···≤ 9 Moreover, due toc q−m 1 we have ≤ d(xq, xq−1) d(x Nl , x˜Nl ) +d( x˜Nl , x˜Nl−1) +d( x˜Nl−1, xNl−1). (1.29) ≤ Forl thefirst and the second term in (1.29) converge to zero due to 1.26. From the stopping rule we obtain d(xNl , xNl−1) d(x Nl+1, xNl ) →∞ ≤ and hence

d(x˜Nl , x˜Nl−1) d(x Nl+1,F(x˜Nl )) +d(F( x˜Nl ),F(x˜Nl−1)) +d(F( x˜Nl−1), x˜Nl ). ≤ This implies 2δ d(x˜Nl , x˜Nl−1) . (1.30) ≤ 1−c Therefore, the seond term in (1.29) converges to zero too ifl goes to infinity. From this we conclude that we must have

d(xq, xq−1) =d(F(x q−1), xq−1) =0.

This impliesx q−1 =z and d(xn, z) =0 forn q−1. (1.31) ≥ Forn

d(xn, z) =d(x n, xq−1) d(x n, xn+1) + d(x q, xq−1). ≤ ··· Each term in the sum on the righthand side can be extimated analogous to (1.29) and converges to zero asl goes to infinity. This shows

lim d(xN(δ), z) =0. (1.32) δ 0

From (1.26) and (1.32) we conclude↓ (1.25). � Remark 1.15. The stopping rule may be formulated a little bit more general with a general distance function d˜ which satisfies

|d(x, y)− d˜(x, y)| ε(δ), x, y X, ≤ ∈ with limδ 0 ε(δ) =0. �

↓ 1.4 Brouwer’sfixed point theorem

Probably, the most simplest existence result for afixed point is the following one: Every continuous functionf:[a, b] [a, b] has afixed point. − To prove this, letg(t) :=f(t)− t, t [a, b]. Then, sincef is a mapping from[a, b] ∈ into[a, b], we must haveg(a) a, g(b) b.→ By the mean value theorem there exists ≥ ≤ z [a, b] withg(z) =0. Clearly,z is afixed point off. ∈ This result holds in particular for the interval[−1, 1]. Now, the classical formulation of Brouwer’sfixed point theorem is an analogous“ result in afinite-dimensional normed ” vector space. For the proof we refer to the literature; see for instance [6, 8, 16].

10 Theorem 1.16 (Fixed point theorem of Brouwer, 1912; Poincar´e,1883). Let B1 be the n closed unit ball in the euclidean spaceR and letF: B1 B1 be a continuous mapping. − ThenF has afixed point. → LetC be a closed convex subset ofR n with nonempty interior. We may assume that n θ C. ThenC is homeomorph to the closed unit ball B1 inR . This can easily seen ∈ using the Minkovski functional with respect to the ball which is contained in the interior ofC. IfC is an arbitrary nonempty closed convex subset inR n thenC has a nonempty interior as a subset of the affine hull aff(C) ofC. Therefore,C is homeomorph to the unit k ball B1 inR for somek n. LetT: B1 C be such a homoemorphism. We set ≤ − G :=T −1 F T: B B . ThenG is continuous and has afixed pointz. ThenT(z) ◦ ◦ 1 − 1 is afixed point ofF. This consideration leads to→ a more general formulation of Brouwer’s fixed point theorem. → Theorem 1.17. LetC be a nonempty convex compact subset of afinite-dimensional normed space and letF:C C be a continuous mapping. ThenF has afixed point. − Proof: We present a proof different→ of the considerations above. It is more in the spirit of the main topic in this monograph. Without loss of generality we may assume thatC is a subset of the euclidean spaceR k for somek. SinceC is compact it is bounded. SupposeC B . Consider the mapping ⊂ r k PC :R C defined by the property x−P C(x) = inf x−u . u C − � � ∈ � � Such a map is well defined since the functionf:R d x x−u R is continuous → � �− � � ∈ and the setC is compact; the uniqueness of the minimum off overC is obvious. We will study this in the next chapter much more detailed. Notice→ thatP C is the identity onC. k n The mappingP C is continuous. To prove this, letx R and let(x )n N be a sequence n n ∈ ∈ withx= lim n x . Since the pointsP C(x ) belong to the compact setC for alln N n nl ∈ (PC(x ))n N has a cluster pointy which is contained inC;y := lim l PC(x )). From ∈ nl nl nl x−y = lim x −P C(x ) = lim dist(x ,C) = dist(x, C) � � l � � l

nl we concludeP C(x) =y sinceP C(x) is uniquely determined. For the fact liml dist(x ,C) = dist(x, C) we use that dist( ,C) is Lipschitz-continuous; see Lemma 2.1. · Now considerR :=P C . ThenG :=F R is a continuous mapping from Br into Br and Br ◦ has, due to Brouwer’sfixed point theorem, afixed pointx. SinceG(x) Cx is inC too. � ∈ HenceR(x) =x andF(x�) =x. � Remark 1.18. Deeply connected to the Brouwerfixed point theorem are the following theorems: Hairy ball therorem • This theorem states that there is no nonvanishing continuous tangent vectorfield on evendimensionaln-spheres. Br¨otchensatz“ • ” This ham sandwich theorem states that givenn (measurable) objects“ inn-dimensional ” space can be divided in half (with respect to their measure) with a single(n−1)- dimensional hyperplane.

11 Theorem of Borsuk-Ulam • This theorem states that every continuous function from ann-sphere into an eu- clideann-space maps some pair of antipodal points to the same point. The connection between these results is the following: Theorem of Borsuk-Ulam= Br¨otchensatz“ ” Theorem of Borsuk-Ulam= Brouwersfixed point Theorem The hairy ball Theorem= ⇒ Brouwersfixed point Theorem ⇒ � Example 1.19. The following⇒ example shows that the result does not hold in infinite- dimensional Banach spaces. Let be the space of the square-summable sequences,i.e.: =l . This space is a Hilbert H H 2 space endowed with the inner product

(xn)n N|(yn)n N := xnyn � ∈ ∈ � n N �∈ and norm 1 n 2 2 (x )n N := ( xn) . � ∈ � n N �∈ We define a mappingF onl 2 by n n 2 F((x )n N) := ( 1− (x )n N , x1, x2,...). ∈ � ∈ � n n Clearly, F((x )n N) =1 for all(x )�n N l 2 and henceF: B1 B1 .F is continuous � k ∈ � ∈ ∈ − since limk x =x inl 2 implies lim F(xk)−F(x) 2 = lim(( 1− x k 2 − 1− x 2)2 + x→k −x 2) =0. k � � k � � � � � � � � AssumeF has afixed pointx=(x n)n N B1 then x =1 (see above) and hence ∈ ∈ � � x = 0, x =x = 0, x = 0, . . . , . Thus, x = 0, contradicting the fact that x =1. 1 2 1 3 � � � � � In Example 1.19 we have seen that Brouwer’sfixed point theorem in a infinite-dimen- sional Hilbert space does not hold in general. The followingfixed point theorem of Kaku- tani shows that the generalization of Brouwer’sfixed point theorem without additional assumptions is not possible; we omit the proof. Theorem 1.20. Let be an infinite-dimensional Hilbert space. Then there is a contin- H uous mappingF: which maps B into B and do not have afixed point. H− H 1 1 1.5 Thefixed→ point property

A topological spaceX has thefixed point property, if every continuous mappingf: X X has afixed point. Theorem 1.16 and 1.17 say that the unit ball in the euclidean − spaceR n and every nonempty compact convex subset in afinite-dimensional normed space n has→ thefixed point property. Thus, we see that the unit ball inR , independently of the norm chosen, has thefixed point property. In Chapter 7 we shall prove the followingfixed point theorem. Theorem 1.21 (Browder, G¨ohde, Kirk, 1965). 1 Let be a uniformly convex Banach X 1This theorem has been proved independently by Browder, G¨ohdeand Kirk in 1965

12 space and letC := B be the closed unit ball in . IfF:C C is a nonexpansive 1 X − mapping thenF has afixed point. → Let us consider Theorem 1.21 in a Hilbert space . Such a space is is uniformly H convex; see Section 4.1. Then the result can be proved by considering the contraction F ,F (x) := tu+(1−t)F(x) for someu C and fort [0, 1). t t ∈ ∈ Theorem 1.22. Let be a Hilbert space, letC be a bounded closed convex subset of H H and letF:C C be a nonexpansive mapping. ThenF has afixed point. − Proof: → Consider fort (0, 1] the mapF :C x tu+(1−t)F(x) . SinceC is convex ∈ t � �− ∈H the range ofF t is contained inC. We have → F (x)−F (y) (1−t) x−y , x, y C, � t t � ≤ � � ∈ and we conclude from Banach’s contraction theorem thatF t has a uniquely determined fixed pointx C:x = tu+(1−t)F(x ), t (0, 1]. SinceC is a bounded set the diameter t ∈ t t ∈ diam(C) isfinite. We obtain

x −F(x ) =t u−F(x ) t diam(C). � t t � � t � ≤ This shows lim x −F(x ) =0 and we say thatF has approximatefixed points. t 0 � t t � Consider Fixε(F) :={z C: F(z)−z ε}. We conclude from the fact that lim t 0 xt − ↓ ∈ � � ≤ � F(xt) =0 that Fix ε(F)= for allε>0. Clearly, Fix ε(F) is closed for allε>0 sinceF � � ∅ ↓ is continuous. Moreover, Fixε(F) is convex for allε>0 sinceC is a subset of a Hilbert space. We shall prove this in Chapter 7 in a more general context; see 7.16. Thus we know that Fixε(F) is weakly closed for allε>0 and therefore weakly compact as a subset of the weakly compact subsetC. This implies Fix (F)= . Clearly, everyx Fix (F) ∩ ε>0 ε � ∅ ∈∩ ε>0 ε is afixed point ofF. � The approach which is used in the proof of Theorem 1.22 is a part of a huge theory on the iterative computation offixed points of nonexpansive mappings. For example, Halpern’s iteration n+1 n n x :=α nx + (1−α n)F(x ),n N 0, (1.33) ∈

(see Chapter 7) is used to compute afixed point ofF when we choose the sequence(α n)n N0 ∈ in an appropriate way. Surely, limn αn =0 should be satisfied. We caome back to this iteration method in Chapter 7.

1.6 Fixed point theorem of Schauder

Definition 1.23. Let be a Banach space, letM and letF:M be a X ⊂X − X mapping. → (a)F is called bounded ifF(M B) is bounded for every bounded subsetB of . ∩ X (b)F is called completely continuous ifF maps every weakly convergent sequence (in M) into a convergent sequence.

13 (c)F is called compact if F(M B) is compact for every bounded subsetB of . ∩ X �

Now, we present three versions of Schauder’sfixed point theorem. It is sufficient to proof only one version since they are equivalent. We shall proof the second version in Section 5.8 as a consequence of the considerations concerning the metric projection. Theorem 1.24 (Fixed point Theorem of Schauder, 1930/First Version). LetX be a Banach space and letM be a nonempty bounded closed convex subset of . Then every X completely continuous mappingF:M M posesses afixed point. − Theorem 1.25 (Fixed point Theorem of Schauder, 1930/Second Version). LetX be a Banach space and letM be a nonempty→ compact convex subset of . LetF:M M X − be continuous. ThenF posesses afixed point. Proof: → See the proof of Theorem 5.14. � Theorem 1.26 (Fixed point Theorem of Schauder, 1930/Third Version). LetX be a Banach space and letM be a nonempty closed convex subset of . LetF:M M be X − a continuous mapping and let F(M) be compact. ThenF posesses afixed point. Definition 1.27. A metric space(X, d) is totally bounded if and only if for→ everyr>0 there exists afinite collection of open balls inX of radiusr whose union containsX.� Theorem 1.28. Let , be Banach spaces, letU X open and letF:U . Let X Y ⊂ − Y A U bounded. Then the following conditions are equivalent: ⊂ (a)F:A Y is completely continuous. → − (b) For everyε>0 there exists afinite-dimensional subspaceY of and a continuous ε Y bounded→ mappingF :A such that ε − Y ε sup F(x)−F ε(x) ε. → x A � � ≤ ∈ F can be chosen such thatF (A) co(F(A)) holds. ε ε ⊂ Proof: Ad(a)= (b) SinceF is compact andA is bounded F(A) is compact. Letε>0. Then there exist 1 m y , . . . , y⇒ F(A) with ∈ m i F(A) B (y ). ⊂∪ i=1 ε We defineφ :A [0, 1] by j − j dist(F(x), \B ε(y )) → φj(x) := m Y , x . i ∈X dist(f(x), \B ε(y )) i=1 Y We haveφ (x) + +φ (x) =� 1, x A. Set 1 ··· m ∈ m F := φ yj ,Y := span(y1, . . . , ym) . ε j ε ⊂Y j 1 �= 14 Now we have

1 m sup F(x)−F ε(x) ε,F ε(A) co(y , . . . , y ) co(F(A)), x A � � ≤ ⊂ ⊂ ∈

and since co(F(A)) is boundedF ε(A) is bounded too. Since the function dist( ,M) is Lipschitz continuous in every metric spaceM (see Lemma · 2.1) we conclude due to the continuity ofF that everyφ j is continuous. HenceF ε is continuous. Ad(b)= (a) SinceF is the uniformly limit of a family of continuos mappingsF is continuous. Since in a Banach⇒ space each point possesses a contable basis of neighborhoods it is sufficient to show that F(A) is sequentially compact. Hence it is sufficient to show for a sequence k (F(x ))k N inF(A): there exists a subsequence(F(x kl )l N which is convergent to a point ∈ ∈ kl y := liml F(x ) in F(A). k (Fn(x ))k N is for eachn N a bounded sequence in afinite-dimensional space. Hence ∈ ∈ k n n k there exist subsequences of(x )k N undy withy = limF n(x ), n N. The n ∈ ∈Y ∈ sequence(y )n N is a Cauchy sequence due to ∈ yl −y m yl −F (xk) + F (xk)−F (xk) + F (xk)−y m � � ≤ � l � � l m � � m � n if one choosesk=k(l, m) appropriate. As a consequence the sequence(y )n N; congerges. n ∈ y := limn y . From

F(xk)−y F(x k)−F (xk) + F (xk)−y n + y n −y � � ≤ � n � � n � � � we concludey= lim F(xk). Obviously,y F(A). k ∈ �

1.7 Schauder’sfixed point theorem and successive approximation

Now, we want to consider the successive approximation of afixed point under the as- sumption of thefixed point theorem of Schauder. The idea is to consider the successive approximation in a partial ordered Banach space. This has been done in a rigorous way for thefirst time by J. Schr¨oder[13].

Definition 1.29. Let be Banach space endowed with a partial order . X � We say that a mappingT:D ,D , is syntoneiffv w implies Tv Tw . T − X T ⊂X � � We say that a mappingT:D ,D , is antitoneiffv w implies Tw Tv . T − X T ⊂X � � → � → Consider for a continuous mapT:D ,D convex, an equation T − X T ⊂X x=F(x) :=T(x) +y forx , (1.34) → ∈X and suppose thatT can be decomposed as follows:

T=T 1 +T 2 whereT 1 is syntone andT 2 is antitone. (1.35)

15 Then we may consider the iteration

n+1 n n n+1 n n v :=T 1(v ) +T 2(w ) +y,w :=T 1(w ) +T 2(v ) +y (1.36) with starting valuesv 0, w0 D . If we can verify that ∈ T v0 v 1 w 1 w 0 (1.37) � � � n n then the iteration (1.36) can be continued and we obtain sequences(v )n N0 ,(w )n N0 ∈ ∈ with 0 1 2 n n 2 1 0 v v v v w w w w , n N. (1.38) � � �···� � �···� � � ∈ Under assumptions which imply that the interval“{u :v 0 u w 0} is mapped ” ∈X � � into a relative compact subset of there exists afixed pointx with X ∈X n n v x w for alln N 0 (1.39) � � ∈ by Schauder’sfixed point theorem 1.26. Thus, we may construct approximationsv n, wn for thefixed pointx ofF which allow the inclusion

n n v x w for alln N 0 , (1.40) � � ∈

if we have the additional property thatθ u � u u �� implies u �� −u u �� −u � . � � � � � ≤ � � This approach may be used in various situations. Notice that a matrixA R n,n may ∈ be decomposed intoA=A 1 +A 2 where bothA 1,−A 2 have nonnegative entries. Then the mappingu A u is syntone andu A u is antitone with respect to the �− 1 �− 2 usual partial order inR n . Another area of application→ is the theory of integral→ equationsx=F(x) = Tx+y whenT is an integral operator of (Hammerstein-)type:

Tu(t) := κ(t, s)φ(u(s))ds , t B, ∈ �B We may decompose the kernelκ and/orφ such that the method above may be applied. Example 1.30. Consider the equation 1 1 x(t)−1=T(x)(t) := |t−s|(x(s)− x(s)2)ds , t [0, 1], 2 ∈ �0 in the spaceC[0, 1] endowed with the supremum-norm. The partial order inC[0, 1] is given byf g f(t) g(t) for allt [0, 1].T may be decomposed as follows: � ≤ ∈ 1 1 1 T (u )(⇐⇒t) := |t−s|u(s)ds , T (u)(t) :=− |t−s|u(s) 2ds , t [0, 1]. 1 2 2 ∈ �0 �0 Clearly,T is syntone,T is antitone. Withv 0 0, w0 2 we obtainv 1(t) :=2(t− 1 2 ≡ ≡ t2), w1(t) =2(1−t+t 2),t [0, 1]. We can verfifyv 0 v 1 w 1 w 0 . Then we obtain ∈ � � � a solutionx of the equation above with

v1(t) x(t) w 1(t),t [0, 1]. ≤ ≤ ∈ �

16 1.8 Appendix: Variational principle of Ekeland and applications

Wir beweisen das Variationslemma von Ekeland in einer endlichdimensionalen Version. Der Beweis ist hier elementar im Gegensatz zur Situation in metrischen oder topologischen R¨aumen. Sei nun in diesem und dem n¨achsten Abschnitt| | stets die euklidische Norm · inR n , assoziert zum euklidischen Skalarprodukt | . �· ·� Theorem 1.31 (Variationslemma von Ekeland, 1972). Seif:R n R unterhalbstetig, n − nach unten beschr¨ankt.Seiε > 0, x ∗ R mit ∈ → f(x∗) inf f(x) +ε. (1.41) x Rn ≤ ∈ Dann gibt es zu jedemγ>0 ein x R n mit ∈ (a)f( x) f(x ∗); ≤ ε (b)|x ∗ − x| ; ≤ γ (c)f(x) f( x)−γ|x− x| f¨urallex R n . ≥ ∈ Proof: O. E. inf{f(x)|x R n}=0, alsof(x) 0 f¨urallex R n . Betrachte die Abbildung ∈ ≥ ∈ n F:R x f(x) +γ|x−x ∗| R. � �− ∈ DaF nach unten beschr¨ankt und unterhalbstetig ist, und da die NiveaumengenN (F) := → r x R n|F(x) r} beschr¨ankt und abgeschlossen sind, also kompakt sind, gibt es ein x, � ∈ ≤ das ein Minimum vonF darstellt (siehe etwa [4], Satz 3.12) Nun haben wir

n f(x) +γ| x−x ∗| f(x) +γ|x−x ∗| f¨urallex R ,ε f(x ∗) f( x) +γ| x−x ∗|. ≤ ∈ ≥ ≥ Daraus folgt(b),(c) sofort. � Corollary 1.32. Seif:R n R Fr´echet-differenzierbar und nach unten beschr¨ankt. n − Seiε > 0, x ∗ R mit ∈ →f(x∗) inf f(x) +ε. (1.42) x Rn ≤ ∈ Dann gibt es ein x R n mit ∈ (a)f( x) f(x ∗); ≤ (b)|x ∗ − x| √ε; ≤ (c)| f( x)| √ε. ∇ ≤ Proof: Wende Satz 10.16 an mitγ := √ε. Dann gibt es also x R n mit ∈ n f(x) f(x ∗),|x ∗ − x| √ε, √ε|x− x| (f( x)−f(x)) f¨urallex R . ≤ ≤ ≥ ∈ Aus der dritten Ungleichung folgt

n f(x+ tw)−f( x)) −t √ε|w| f¨uralle t > 0, w R . ≥ ∈ 17 Daraus folgt n f(x), w − √ε|w| f¨urallew R , �∇ � ≥ ∈ was nun(c) impliziert. � F¨urden Beweis des Variationslemmas im Kontext metrische R¨aume/unendlichdimen- ” sionale R¨aume“ stellen wir ein allgemeines Prinzip vor. Erinnerung: SeiX eine Menge. Eine Relation “heisst Halbordnung aufX genau ”≤ dann, wenn gilt:

i)x x (Reflexivit¨at) ≤ ii)x y, y x= y=x (Symmetrie) ≤ ≤ iii)x y, y z= x z x, y, z X (Transitivit¨at) ≤ ≤ ⇒ ≤ ∀ ∈ Wir schreiben dann(X, ), vereinbaren die Schreibweise ⇒ ≤ S(x) :={y X|y x}, ∈ ≥ und k¨onnen f¨ureine Folge(x n)n N inX definieren: ∈

(xn)n N monoton wachsend, fallsx n x n+1 f¨urallen N. ∈ ≤ ∈ Theorem 1.33 (Brezis–Browder, 1976). Sei(X, ) eine halbgeordnete Menge undϕ: ≤ X R monoton wachsend, d.h.ϕ(x) ϕ(y), fallsx y. Es gelte: − ≤ ≤

i) F¨urjede monoton wachsende Folge(x n)n N mit sup{ϕ(xn)|n N}< existiert → ∈ ∈ y X mitx n y f¨urallen N. ∈ ≤ ∈ ∞ ii) F¨urallex X gibt esu X mitx u,ϕ(x)<ϕ(u). ∈ ∈ ≤ Dann gilt: sup{ϕ(y)|y S(x)}= f¨urallex X. ∈ ∈ Proof: ∞ F¨uru X setzeµ(u) := sup{ϕ(y)|y S(u)}. ∈ ∈ Annahme: Es gibtx X mitµ(x)< . ∈ Definiere eine Folge(x n)n N induktiv in folgender Weise:x 1 :=x; sindx 1, . . . , xn bestimmt, ∈ w¨ahlex S(x ) mitµ(x )− 1/n ϕ(x ). n+1 ∈ n n ∞ ≤ n+1 Wegenµ(x ) µ(x)< existiertx und wegenx x giltµ(x ) µ(x ). Nun n ≤ n+1 n+1 ≥ n n+1 ≤ n gilt also ∞ xn x, d. h.x n S(x), n N;(x n)n N ist monoton wachsend;ϕ(x n) µ(x), n N. ≥ ∈ ∈ ∈ ≤ ∈

Wegen i) gibt esy X mitx n y f¨urallen N. W¨ahle nach ii)u X mitu S(y) ∈ ≤ ∈ ∈ ∈ undϕ(y)<ϕ(u). Nun giltx n u,ϕ(u) µ(x n), n N. Also ≤ ≤ ∈ 1 ϕ(u) µ(x n) ϕ(x n+1) + 1/n ϕ(y) + , n N, ≤ ≤ ≤ n ∈ d. h.ϕ(u) ϕ(y)<ϕ(u), was ein Widerspruch ist. ≤ �

18 Corollary 1.34. Sei(X, ) eine halbgeordnete Menge,ψ:X R nach oben ≤ − beschr¨anktund es gelte: → F¨urjede monoton wachsende Folge(x n)n N inX existierty X mitx n y,n N. ∈ ∈ ≤ ∈ (1.43) Dann gilt: (a) Istψ monoton wachsend, so folgt: u X v S(u) w S(v)(ψ(v) =ψ(w)). ∀ ∈ ∃ ∈ ∀ ∈ (b) Istψ streng monoton wachsend, so folgt: u X v S(u) w X(w v= ∀ ∈ ∃ ∈ ∀ ∈ ≥ w=v). Proof: ⇒ Zu(a). Seiu X. Betrachte(S(u), ),ϕ :=ψ| . Dann istϕ monoton wachsend und nach oben ∈ ≤ S(u) beschr¨ankt. Annahme: v S(u) w v(ϕ(v)<ϕ(w)) ∀ ∈ ∃ ≥ Nun sind die Voraussetzungen des Satzes 10.26 f¨ur(S(u), ),ϕ erf¨ullt. Es folgt: ≤ w S(u)(sup{ϕ(y)|y S(w) S(u)}= ). ∀ ∈ ∈ ∩ Dies ist ein Widerspruch zur Beschr¨anktheit vonϕ. ∞ Zu(b). Seiu X. Nach(a) gibt esv S(u) mitϕ(w) =ϕ(v) f¨urallew X, w v. Daϕ ∈ ∈ ∈ ≥ streng monoton wachsend ist, gilt also:

w X, w v= ϕ(w) =ϕ(v)= w=v. ∈ ≥ ⇒ ⇒ � Corollary 1.35. Sei(X, d) ein metrischer Raum und sei(X, ) halbgeordnet. Seiη: ≤ X R streng monoton fallend und nach unten beschr¨ankt.Es gelte: − (a) F¨urallex X istS(x) abgeschlossen. → ∈ (b) F¨urjede monoton wachsende Folge(x n)n N inX ist {xn|n N} kompakt. ∈ ∈ Dann gilt: u X v S(u) w X(w v= w=v). ∀ ∈ ∃ ∈ ∀ ∈ ≥ Proof: ⇒ Seiψ :=−η. Wir weisen (10.19) nach. Sei dazu(x n)n N eine monoton wachsende Folge ∈ inX. Dann existiert wegen(b) einy X und eine Teilfolge(x nk )k N mity= lim k xnk . ∈ ∈ Dax nk S(x nj ) f¨urallek j gilt, folgt wegen(a)y S(x nj ) S(x n), nj n, n N. ∈ ≥ ∈ ⊂ ≥ ∈ Alsoy x n f¨urallen N. Nun k¨onnen wir Folgerung 10.27(b) anwenden. ≤ ∈ �

Theorem 1.36. Sei(X, d) ein vollst¨andigermetrischer Raum und seif:X (− , ] − unterhalbstetig, nach unten beschr¨anktund eigentlich, d. h. es gibtz X mitf(z)< . ∈ Seiε > 0, x ∗ X mit ∈ → ∞ ∞ f(x∗) inf f(v) +ε. (1.44) v X ∞ ≤ ∈ Dann gibt es x X mit ∈ 19 1.f( x) f(x ∗); ≤

2.d( x, x∗) 1; ≤ 3.f(y) +εd(y, x)>f( x) f¨uralley X\{ x}. ∈ Proof: f(x∗) ist endlich wegen (10.20). X :={z X|f(z) f(x ∗)} ist abgeschlossen, daf unter- ∈ ≤ halbstetig ist. Also ist( X, d) ein vollst¨andiger metrischer Raum mitx ∗ X. Haben wir � ∈ das Resultat in( X, d) f¨ur f :=f |X bewiesen, dann ist das Resultat in(X, d) bewiesen, denn die Bedingung 3. folgt� dann mit 1. aus der Zeile � � � � f(y) +εd(y, x)>f(x ∗) +εd(y, y) f(x ∗) f( x),y X\ X. ≥ ≥ ∈ O. E. k¨onnen wir daher nun X=X annehmen. � Wir definieren u v: �f(v)−f(u) −εd(u, v) , u, v X, ≤ ≤ ∈ und zeigen, dass dadurch aufX eine Halbordnung erkl¨artist. Die Reflexivit¨atist klar.Seien ⇐⇒u v, v u, d. h. ≤ ≤ f(v)−f(u) −εd(u, v),f(u)−f(v) −εd(u, v). ≤ ≤ Daraus folgtd(u, v) =0 und daher ist die Symmetrie gezeigt. Seien u, v, w X mit ∈ u v, v w, d. h. ≤ ≤ f(v)−f(u) −εd(u, v),f(w)−f(v) −εd(v, w). ≤ ≤ Mit der Dreiecksungleichung folgt

f(w)−f(u) −εd(u, w) , d. h.u w. ≤ ≤ Damit ist auch die Transitivit¨atklar. Wir zeigen nun, dassf streng monoton fallend ist ist bez¨uglich der Halbordnung . Seien ≤ dazu u, v X, u v, u=v; also ∈ ≤ � f(v)−f(u) −εd(u, v)<0 , d. h.f(v)

n−1 n−1 εd(xn, xm) εd(xi, xi+1) (f(xi+1)−f(x i)) =f(x n)−f(x m) , n, m N, n > m, ≤ ≤ ∈ i m i m �= �=

20 ab, dass(x n)n N eine Cauchyfolge und daher konvergent ist. ∈ Wir k¨onnen nun Folgerung 10.28 anwenden: Es gibt x S(x ∗) mit der Eigenschaft ∈ u x= u= x . (1.45) ≥

Wegen x S(x ∗) folgt ∈ ⇒

0 εd(x ∗, x) f(x ∗)−f( x) inf f(v) +ε−f( x) ε. v X ≤ ≤ ≤ ∈ ≤ Damit ist 1., 2. gezeigt. Seiy X\{ x}. Wegen (10.21) gilt nichty x, d. h.f(y)−f( x)> ∈ ≥ −εd(y, x). � Corollary 1.37. Sei(X, d) ein vollst¨andigermetrischer Raum und seif:X − (− , ] unterhalbstetig, nach unten beschr¨anktund eigentlich. Seiγ > 0, x ∗ X. ∈ Dann gibt es x X mit ∈ → ∞ ∞ i)f( x)

ii)f( x) f(x ∗)−γd(x ∗, x)) ; ≤ Proof: 1 SeiY :={x X|f(x) +γd(x, x ∗) f(x ∗)},g := f| .Y ist abgeschlossen, daf unterhalb- ∈ ≤ γ Y stetig ist. Also ist(Y, d) ein vollst¨andiger metrischer Raum.g ist unterhalbstetig, nach unten beschr¨ankt und eigentlich. Seiα := inf v Y g(v). Es gibtx ∗∗ Y mitg(x ∗∗) α+ 1. ∈ ∈ ≤ Also ist Satz 10.29 anwendbar mitε=1: 1 1 x Y x Y\{ x}( f(x)< f(x) +d(x, x)). ∃ ∈ ∀ ∈ γ γ Also f(x)f(x ∗) und mit ii) folgt ∈

f(x) f(x ∗)−γd(x ∗, x) 0, x ∗ X mit ∈ f(x∗) inf f(v) +ε. (1.46) → ∞ ∞ v X ≤ ∈ Dann gibt es zu jedemγ>0 ein x X mit ∈

(a)f( x) f(x ∗); ≤ (b)d(x ∗, x) γ; ≤ (c)f(x) + ε d(x, x)>f( x) f¨urallex X\{ x}. γ ∈

21 Proof: ˜ 1 Seiγ > 0. Wende Satz 10.29 an unter Verwendung der zud ¨aquivalenten Metrik d := γd . � Beachte: Es liegt folgender Kompromiss f¨urdie Wahl vonγ nahe:γ := √ε; siehe Folgerung 10.17.

Das folgende Resultat stellt eine Existenzaussage im Kontext der Voraussetzungen von Satz 10.29 bereit. Ob es als Existenzsatz Verwendungfinden kann, ist zweifelhaft, wir werden ihn aber verwenden k¨onnen. Theorem 1.39 (Takahashi, 1989). Sei(X, d) ein vollst¨andigermetrischer Raum und sei f:X (− , ] unterhalbstetig, eigentlich und nach unten beschr¨ankt.Es gelte: − F¨uralleu X mitf(u)> inf f(x) gibt esv X mitv= u, f(v) +d(u, v) f(u). → ∞ ∞ x X ∈ ∈ ∈ � ≤

Dann gibt es x X mitf( x) = infx X f(x). ∈ ∈ Proof: Annahme:f(u)>µ := inf x X f(x) f¨uralleu X. ∈ ∈ Nach Satz 10.31 gibt es x X mit ∈ f(y) +d(y, x)>f( x) f¨uralley X\{ x}. ∈ Wegenf( x) > µ, gibt esv X, v= x, mitf(v) +d(v, x) f( x). Dies ist ein Widerspruch. ∈ � ≤ � Die Vollst¨andigkeit der metrischen R¨aume in den obigen Resultaten war wesentlich f¨urdie Beweise. Sie ist auch wesentlich f¨urdie G¨ultigkeit der Resultate, denn man kann zeigen, dass die Vollst¨andigkeit in einem metrischen Raum schon durch die Tatsache charakterisiert wird, dass die Aussage von Satz 10.31 f¨urjedesf mit den dortigen Eigen- schaften gilt. Theorem 1.40. Sei(X, d) ein metrischer Raum. Dann sind ¨aquivalent: (a)(X, d) ist vollst¨andig.

(b) F¨urjede unterhalbstetige Abbildungf:X R mit inf v X f(v) 0 gilt: Ist − ∈ ≥ ε > 0, x∗ X mitf(x ∗) inf v X f(v) +ε, so gibt es x X mit ∈ ≤ ∈ ∈ 1.f( x) f(x ∗); → ≤ 2.d( x, x∗) 1; ≤ 3.f(y) +εd(y, x) f( x) f¨uralley X. ≥ ∈ Proof: (a)= (b). Satz 10.29. (b)= (a).

Sei(x n)n N eine Cauchyfolge inX. Dann existiert offenbar lim n d(xn, y) f¨urjedesy X. ⇒ ∈ ∈ Definieref:X R durch ⇒ −

f(y) := lim d(xn, y), y X. → n ∈ 22 f ist unterhalbstetig, dad(y, ) stetig ist f¨uralley X, und es gilt inf v X f(v) =0. Sei · ∈ ∈ ε (0, 1), x ∗ X mitf(x ∗) ε. Dann gibt es nach Satz 10.29 x X mit ∈ ∈ ≤ ∈ f(x) f(x ∗) ε,f(y) +εd(y, x) f( x) f¨uralley X. ≤ ≤ ≥ ∈ Wir zeigen:f( x) ε l f¨urallel N. ≤ ∈ Der Induktionsanfangl=1 ist klar. l Seiη>0 beliebig. Daf( x) ε ist nach Induktionsvoraussetzung, gibt esx m mit l ≤ d(xm, x)<ε +η, f(x m)<η. Mit 3. folgt

η+ε(ε l +η)>f(x ) +εd(x , x) f( x). m m ≥ Durch Grenz¨ubergangη 0 folgtf( x) ε l+1 . ≤ Daε (0, 1) ist, gilt alsof( x) =0, d. h. x= lim x . ∈ n n � Theorem 1.41. Sei(X,→ ) ein Banachraum und seif:X (− , ] eigentlich, �·� − unterhalbstetig, Gateaux–differenzierbar in dom(f) :={x X|f(x)< } und nach unten ∈ beschr¨anktDann gibt es zu jedemε > 0,γ>0 undx ∗ X mit ∈ → ∞ ∞ ∞ f(x∗) inf f(v) +ε v X ≤ ∈ ein x X mit ∈ 1.f( x) f(x ∗). ≤ 2. x−x ∗ γ. � � ≤ 3. D f(x) ε . � G � ≤ γ Proof: W¨ahle x X gem¨a”sSatz 10.31. Dann gilt ∈ ε f(y) f( x)− y− x , y X. (1.47) ≥ γ� � ∈ Also ε ε f(x+ tw) f( x)− w t f¨uralle t > 0, d. h.D f(x)(w) − w . ≥ γ� � G ≥ γ� � Dies zeigt ε D f(x) . � G � ≤ γ � Remark 1.42. Satz 10.34 zeigt, dass f¨urein Optimierungsproblem

Minimieref(x), x X, ∈ unter den dortigen Voraussetzungen Punkte x X existieren, dief fast minimieren“ und ∈ ” die die notwendige BedingungD f(x) =θ fast erf¨ullen“: G ” ε f(x) inf f(v) +ε, D Gf(x) . v X γ ≤ ∈ � � ≤ �

23 Theorem 1.43 (Kontraktionssatz/Fixpunktsatz von Banach). Sei(X, d) ein vollst¨andi- ger metrischer Raum und seiF:X X eine Kontraktion, d. h. − d(F(x),F(y)) Ld(x, y) f¨uralle x, y X →≤ ∈ mitL [0, 1). Dann besitztF einen eindeutigen Fixpunkt. ∈ Proof: Betrachte f:X x d(x, F(x)) R. � �− ∈ Offenbar istf stetig, nach unten beschr¨ankt und eigentlich. Seix ∗ X,ε :=d(x ∗,F(x∗)),γ := ∈ 1−L. Wende damit Satz 10.29 an: Es→ gibt x X mit ∈ d(x, F(x))

d(x, F(x))−d(z, x) f¨urallez X\{ x}). Also ∈ ∈ f(x)−f(F( x))

24 Theorem 1.46. Sei(X, d) ein vollst¨andigermetrischer Raum und seif:X (− , ] − unterhalbstetig, eigentlich und nach unten beschr¨ankt. SeiT:X POT(X)\{ }. Es − ∅ gelte: → ∞ ∞ x X, x / T(x) y=x(f(y) +d(y, x) f(x)). ∀ ∈ ∈ ∃ � ≤ → Dann gibt esz X mitz T(z). ∈ ∈ Theorem 1.47. Sei(X, d) ein vollst¨andigermetrischer Raum und seif:X (− , ] − unterhalbstetig, eigentlich und nach unten beschr¨ankt. SeiT:X POT(X)\{ }. Es − ∅ gelte f¨urallex X: ∈ → ∞ ∞ → f(y) +d(y, x) f(x) f¨uralley T(x). ≤ ∈ Dann gibt esz X mit{z}=T(z). ∈ Theorem 1.48. Sei(X, d) ein vollst¨andigermetrischer Raum und seif:X (− , ] − unterhalbstetig, eigentlich und nach unten beschr¨ankt. SeiT:X POT(X)\{ }. Es − ∅ gelte f¨urallex X ∈ → ∞ ∞ f(y) +d(y, x) f(x) f¨uralley T(x). ≤ ∈ → Dann gibt esz X mitz T(z). ∈ ∈ Theorem 1.49. Sei(X, d) ein vollst¨andigermetrischer Raumund seif:X (− , ] − unterhalbstetig, eigentlich und nach unten beschr¨ankt. SeiT:X POT(X)\{ } − ∅ abgeschlossen. Es gelte: → ∞ ∞ → d(x, T(x)) f(x)− sup f(y) f¨urallex X. ≤ y T(x) ∈ ∈ Dann gibt esz X mitT(z) ={z}. ∈ Lemma 1.50. Die S¨atze10.32, 10.39, 10.40, 10.41, 10.42 sind ¨aquivalent.

Proof: Aus Satz 10.32 folgt Satz 10.39, denn: Annahme:T hat keinen Fixpunkt, d. h.x/ T(x) f¨urallex X.. ∈ ∈ Seiu X mitf(u)> inf x X f(x). DaT(u)= und dau/ T(u) gibt esv T(u), v= u, ∈ ∈ � ∅ ∈ ∈ � mitf(v) +d(v, u) f(u). Damit liefert nun Satz 10.32 die Existenz von x mitf( x) = ≤ infx X f(x). Sei y T( x). Aus ∈ ∈ 0 < d(x, y) f( x)−f( y) f( y)−f( y) =0 ≤ ≤ lesen wir einen Widerspruch ab. Aus Satz 10.39 folgt Satz 10.40, denn: Annahme: Es gibt keinu X mit{u}=T(u). ∈ Definiereg:X x T(x)\{x} POT(X). Es gilt � �− ∈ f(g(x)) +d(g(x), x) f(x),x X. → ≤ ∈ Aus Satz 10.39 folgt die Existenz vonz X mitz T(z). Dies ist auch ein Fixpunkt von ∈ ∈ g nach Satz 10.37. Aus Satz 10.40 folgt Satz 10.32, denn:

25 DefiniereT:X x {y X|f(y) +d(x, y) f(x)} POT(X). Auf Grund der � �− ∈ ≤ ∈ Voraussetzung in Satz 10.32 giltT(x)= f¨urallex X. � ∅ ∈ Annahme: Es gibt kein x X mitf( x) = infv X f(v). Nun gilt → ∈ ∈ f(y) +d(x, y) f(x),y T(x), x X. ≤ ∈ ∈ Aus Satz 10.40 folgt die Existenz vonz X mit{z}=T(z). Dies zeigt, dass keinv X ∈ ∈ esistiert mitf(y) +d(y, z) f(z), was ein Widerspruch ist. ≤ Aus Satz 10.40 folgt Satz 10.41, denn: Dies ist offensichtlich. Aus Satz 10.41 folgt Satz 10.42, denn: Setzeg(x) := 1 f(x), x X. Istd(x, T(x)) = 0, dann istx T(x), daT(x) abgeschlossen 2 ∈ ∈ ist. Hat alsoT keinen Fixpunkt, dann istd(x, T(x))>0 f¨urallex X. Seiy T(x) so, 1 ∈ ∈ dassd(x, y)< 2 d(x, T(x)). Dann haben wir 1 1 d(x, y) d(x, T(x)) (f(x)− sup f(y)) g(x)−g(y). ≤ 2 ≤ 2 y T(x) ≤ ∈ Dag eigentlich, unterhalbstetig und nach unten beschr¨ankt ist, hatT einen Fixpunkt nach Satz 10.41. Aus Satz 10.42 folgt Satz 10.40, denn Satz 10.40 ist ein Spezialfall von Satz 10.42.�

1.9 Conclusion and comments

Nonlinear functional analysis is the study of operators lacking the property and equations which are governed by such operators. Thefixed point principle is the bunch of methods to translate or reformulate (analytic) problems in the applied sciences to equations of fixed point type, mostly governed by nonlinear mappings. The most earliest observation that afixed point principle is helpful is the method of computing the square root of a nonnegative number known as Babylonian method which is nowadays considered as a Newton-type method. Another method which shows that the fixed point principle is very powerful is the proof that a matrix with nonnegative entries has an eigenvalue which is nonnegative (see exercises). Other examples we have seen in Section 1.1. In the sections of this chapter we have presented the most importantfixed point the- orems in the analytic study of problems in the applied science: Banach’s and Schauder’s fixed point theorem. In our context, Brouwer’sfixed point may be considered as a result which enables us to proof the Schauderfixed point theorem. Thefixed point theorem of Darbo is the central result of a unified approach for Banach’s and Schauder’s theorem. Afirst impression forfixed point theorems for nonexpansive mappings is the proof of a version of thefixed point theorem of Browder, G¨ohde, Kirk. As we will see,fixed point theory for nonexpansive mappings benefit mostly of geometric properties of Banach spaces. There is a lot offixed point theorems mostly located in other topics of mathematics: the fixed point theorems of Borsuk, Krasnoselski, Caristi (see Section 10.6), Knaster-Tarski, Lefschetz, Kakutani; see [2, 8, 9].

26 1.10 Exercises

1.) Leta

(b) If eachF i has afixed point thenF has afixed point. 6.) Consider inC[0, 1], endowed with the supremum norm, the (nonlinear) mapping

1 G(x)(t) := κ(t, s)f(s, x(s))ds , t [0, 1], x C[0, 1]. ∈ ∈ �0 Here,κ C([0, 1] [0, 1]), f C([0, 1] R). Show: If ∈ × ∈ × 1 |f(s, u)−f(s, v)| L|u−v|,s [0, 1], u, v R,L max |κ(t, s)|ds < 1, ≤ ∈ ∈ · t [0,1] ∈ �0 thenG:C[0, 1] C[0, 1] is a contraction. − 7.) Let be a Banach space, letV be a compact subset of and letF:V V be X X − contractive, i. e.→ → x, y V, x= y,( F(x)−F(y) < x−y ). (1.49) ∀ ∈ � � � � � Show: (a)F has a uniquely determinedfixed point x . (b) The iterationx n+1 :=F(x n), x0 = x, converges for eachx to x . ∈X Hint for(b): One can show that the functionx x−F(x) along the iteration �− � � is monotone decreasing. 8.) LetX be a compact metric space. Show:X is separable.→

27 9.) LetC[a, b] be the space of continuous functions on[a, b] endowed with the maximum- norm. Show thatC[a, b] is separable but not compact. 10.) Consider the following system of nonlinear equations:

2 2 2 2x1 −x 2 − 8x1 =0,x 1 +x 1x2 − 4x2 +1=0

Find a reformulation as afixed point equation inR 2 and solve this equation (by Brouwer’sfixed point theorem). 11.) Consider

1 2 2 −x2 2 2 2 F:R (x 1, x2) (x1e +x 1x2 + 3, ln(1+x 1 +x 2)−1) R . � �− 6 ∈ LetR 2 be endowed with the→ maximum norm . �·� (a) Show thatF is Lipschitz-continuous in[0, 1] [∞0, 1] with Lipschitz-constant 5 × c= 6 . (b) Show that there exists a uniquely determinedfixed point ofF. (c) How many stepsk do we need to obtain an accuracy x k −z 10−3 using � � ≤ thefixed point iterationx k+1 :=F(x k) starting withx 0 := (0, 0). ∞ 12.) Show that x+ 1/2 f:[0, ) x [0, ) � �− x+1 ∈ is a contraction. Compute thefixed point off and an approximationx 1 using the ∞ → ∞ starting pointx 0 =1 for thefixed point iteration. 13.) Show that the following system of nonlinear equations

sin(x1 +x 2)−x 2 =0, cos(x 1 +x 2)−x 1 =0

is solvable inR 2 . n,n 14.) LetA=(a ij) R be a matrix with entriesa ij 0. Show thatA has a ∈ ≥ nonnegative eigenvalueλ with associated eigenvectorx=(x ) with entriesx 0. i i ≥ 15.) LetF:R n R n be a continuous mapping with − n F(x)−F(y) c x−y for all x, y R → � � ≥ � � ∈ withc>1. Show: IfF is surjective thenF has a uniquely determinedfixed point. 16.) LetC[0, 1] endowed with the maximum-norm . LetC :={x C[0, 1]:0= �·� ∈ x(0) x(s) x(1) =1} and consider ≤ ≤ ∞ T:C[0, 1] x T(x) C[0, 1] withT(x)(s) := sx(s), s [0, 1]. � �− ∈ ∈ Show: → (1)T is a linear nonexpansive mapping with T =1. � � (2)T(x) C ifx C. ∈ ∈ (3)C is a bounded closed convex set.

28 (4)T posesses a uniquely determinedfixed point belonging toC[0, 1]\C. (5) There existsx C withx = tu+(1−t)T(x ) whereu C, u(s) :=s. t ∈ t t ∈ (6) infx C x−T(x) =0. ∈ � � (7) lim x0 −T n(x0) = diam(C). n � � 17.) Let be a Banach space, letC be compact and letF:X X be X ∞ ⊂X − continuous. Then the following statements are equivalent:

(a) infx X x−F(x) =0. → ∈ � � (b)F posesses afixed point. 18.) Let Xs be a Banach space, letC be compact and convex and letF:X X ⊂X − be nonexpansive. ThenF posesses afixed point. 2 2 2 2 19.) Let be the euclidean spaceR and letC 0 :={(x, y) R :x +→y 1}, H 2 ∈2 ≤ C1 :={(x, y) R :x 0,|y| x},C 2 :={(x, y) R :x 0,|y| |x|} und ∈ ≥ ≤ ∈ ≤ 2 ≤ C := (C0 C 1) (C C 2). Show:C co({P C(x, y):(x, y) R \C}). ∩ ∪ ∩ ⊂ ∈ 20.) Let be a Banach space and letU be a closed subspace of . Consider the X X quotient space

/U :={[x]:x } where[x] :={x+u:u U}. X ∈X ∈ Show that [x] := dist(x, U), x , defines a norm in /U. � � ∈X X 21.) Let be a Banach space nd letU be a closed subspace of , different from . X X X Show that for everyε>= there existsx B with dist(x, U) 1−ε. ∈ 1 ≥

29 Bibliography

[1] R.P. Agarwal, M. Meehan, and D. O’Regan. Fixed point theory and applications. Cambridge University Press, 2004. [2] J. Appell and M. V¨ath. Elemente der Funktionalanalysis. Vieweg, Braunschweig, 2005.

[3] J. Baumeister. Differentialgleichungen. Skriptum 2006, Goethe–Universit¨atFrank- furt/Main.

[4] J. Baumeister. Konvexe Analysis, 2014. Skriptum WiSe 2014/15, Goethe–Universit¨at Frankfurt/Main.

[5] C. Bessaga. On the converse of the banach fixed-point principle“. Colloquio Math., 7:41– ” 43, 1959.

[6] J. Cronin. Fixed points and topological degree in nonlinear analysis. Amer. Math. Soc., Providence, 1964.

[7] M. Edelstein. An extension of Banach’s contraction principle. Proc. Amer. Math Soc., 12:7–10, 1961.

[8] A. Granas and J. Dugundij. Fixed point theory. Springer, Berlin, 2003.

[9] V.I. Istratescu. Fixed point theorems: an introduction. Reidel, Dordrecht, 1988.

[10] O. Kada, T. Suzuki, and W. Takahashi. Non convex minimization theorems andfixed point theorems in complete metric spaces. Math. Japon., 44:381–391, 1996.

[11] E. Karapinar. Edelstein typefixed point theorems. Ann. Funct. Anal., 2:51–58, 2011.

[12] R. Krawczyk. Abbrechkriterium f¨urIterationsverfahren. Z. Angewandte Math. Mech., 52:227–232, 1972.

[13] J. Schr¨oder.Anwendung von Fixpunkts¨atzenbei der numerischen Behandlung nichtlinearer Gleichungen in halbgeordneten R¨aumen. Arch. Rational Mech. Anal., 4:177–192, 1959.

[14] S.L. Singh and S.N. Mishra. Remarks on recentfixed point theorems. Fixed point theory and applications, pages 1–18, 2010.

[15] T. Suzuki and W. Takahashi. Fixed point theorems and characterizations of metric com- pleteness. Topological methods in nonlinear Analysis, 8:371–382, 1996.

[16] P. Wittbold. Funktionalanalysis I. https://fsr6.files.wordpress.com/2014/07/funktional- ana−skript.pdf, 2006. Humboldt Universit¨atBerlin.

30