Chapter 1Analysis in Bana h spa es1.1 Di�erentiation and integration in Bana h sp-a esWe �rst review some basi fa ts from al ulus in Bana h spa es.LetX and Y be two Bana h spa es and denote by C(X; Y ) the set of ontinuousfun tions from X to Y and by L(X; Y ) � C(X; Y ) the set of (bounded) linearfun tions. Let U be an open subset of X. Then a fun tion F : U ! Y is alleddi�erentiable at x 2 U if there exists a linear fun tion dF (x) 2 L(X; Y ) su h thatF (x + u) = F (x) + dF (x) u+ o(u); (1.1)where o, O are the Landau symbols. The linear map dF (x) is alled derivative ofF at x. If F is di�erentiable for all x 2 U we all F di�erentiable. In this ase weget a map dF : U ! L(X; Y )x 7! dF (x) : (1.2)If dF is ontinuous, we all F ontinuously di�erentiable and write F 2 C1(U; Y ).Let Y = Qmj=1 Yj and let F : X ! Y be given by F = (F1; : : : ; Fm) withFj : X ! Yi. Then F 2 C1(X; Y ) if and only if Fj 2 C1(X; Yj), 1 � j � m, andin this ase dF = (dF1; : : : ; dFm). Similarly, if X = Qmi=1Xi, then one an de�nethe partial derivative �iF 2 L(Xi; Y ), whi h is the derivative of F onsidered asa fun tion of the i-th variable alone (the other variables being �xed). We havedF v =Pni=1 �iF vi, v = (v1; : : : ; vn) 2 X, and F 2 C1(X; Y ) if and only all partialderivatives exist and are ontinuous. 1

2 Chapter 1. Analysis in Bana h spa esIn the ase of X = Rm and Y = Rn ,the matrix representation of dF withrespe t to the anoni al basis in Rm and Rn is given by the partial derivatives�iFj(x) and is alled Ja obi matrix of F at x.We an iterate the pro edure of di�erentiation and write F 2 Cr(U; Y ), r � 1,if the r-th derivative of F , drF (i.e., the derivative of the (r � 1)-th derivative ofF ), exists and is ontinuous. Finally, we set C1(U; Y ) = Tr2N Cr(U; Y ) and, fornotational onvenien e, C0(U; Y ) = C(U; Y ) and d0F = F .It is often ne essary to equip Cr(U; Y ) with a norm. A suitable hoi e isjF j = max0�j�r supx2U jdjF (x)j: (1.3)The set of all r times di�erentiable fun tions for whi h this norm is �nite forms aBana h spa e whi h is denoted by Crb (U; Y ).If F is bije tive and F , F�1 are both of lass Cr, r � 1, then F is alled adi�eomorphism of lass Cr.Note that if F 2 L(X; Y ), then dF (x) = F (independent of x) and drF (x) = 0,r > 1.For the omposition of mappings we note the following result (whi h is easy toprove).Lemma 1.1 (Chain rule) Let F 2 Cr(X; Y ) and G 2 Cr(Y; Z), r � 1. ThenG Æ F 2 Cr(X;Z) andd(G Æ F )(x) = dG(F (x)) Æ dF (x); x 2 X: (1.4)In parti ular, if � 2 Y � is a linear fun tional, then d(�ÆF ) = d�ÆdF = �ÆdF .In addition, we have the following mean value theorem.Theorem 1.2 (Mean value) Suppose U � X and F 2 C1(U; Y ). If U is onvex,then jF (x)� F (y)j �M jx� yj; M = max0�t�1 jdF ((1� t)x + ty)j: (1.5)Conversely, (for any open U) ifjF (x)� F (y)j �M jx� yj; x; y 2 U; (1.6)then supx2U jdF (x)j �M: (1.7)

1.1. Di�erentiation and integration in Bana h spa es 3Proof. Abbreviate f(t) = F ((1 � t)x + ty), 0 � t � 1, and hen e df(t) =dF ((1 � t)x + ty)(y � x) implying jdf(t)j � ~M = M jx � yj. For the �rst part itsuÆ es to show �(t) = jf(t)� f(0)j � ( ~M + Æ)t � 0 (1.8)for any Æ > 0. Let t0 = maxft 2 [0; 1℄j�(t) � 0g. If t0 < 1 then�(t0 + ") = jf(t0 + ")� f(t0) + f(t0)� f(0)j � ( ~M + Æ)(t0 + ")� jf(t0 + ")� f(t0)j � ( ~M + Æ)"+ �(t0)� jdf(t0)"+ o(")j � ( ~M + Æ)"� ( ~M + o(1)� ~M � Æ)" = (�Æ + o(1))" � 0; (1.9)for " � 0, small enough. Thus t0 = 1.To prove the se ond laim suppose there is an x0 2 U su h that jdF (x0)j =M + Æ, Æ > 0. Then we an �nd an e 2 X, jej = 1 su h that jdF (x0)ej = M + Æand hen e M" � jF (x0 + "e)� F (x0)j = jdF (x0)("e) + o(")j� (M + Æ)"� jo(")j > M" (1.10)sin e we an assume jo(")j < "Æ for " > 0 small enough, a ontradi tion. 2As an immediate onsequen e we obtainCorollary 1.3 Suppose U is a onne ted subset of a Bana h spa e X. A mappingF 2 C1(U; Y ) is onstant if and only if dF = 0. In addition, if F1;2 2 C1(U; Y )and dF1 = dF2, then F1 and F2 di�er only by a onstant.Next we want to look at higher derivatives more losely. Let X = Qmi=1Xi,then F : X ! Y is alled multilinear if it is linear with respe t to ea h argument.It is not hard to see that F is ontinuous if and only ifjF j = supx:Qmi=1 jxij=1 jF (x1; : : : ; xm)j <1: (1.11)If we take n opies of the same spa e, the set of multilinear fun tions F : Xn ! Ywill be denoted by Ln(X; Y ). A multilinear fun tion is alled symmetri providedits value remains un hanged if any two arguments are swit hed. With the normfrom above it is a Bana h spa e and in fa t there is a anoni al isometri iso-morphism between Ln(X; Y ) and L(X;Ln�1(X; Y )) given by F : (x1; : : : ; xn) 7!

4 Chapter 1. Analysis in Bana h spa esF (x1; : : : ; xn) maps to x1 7! F (x1; :). In addition, note that to ea h F 2 Ln(X; Y )we an assign its polar form F 2 C(X; Y ) using F (x) = F (x; : : : ; x), x 2 X. If Fis symmetri it an be re onstru ted from its polar form usingF (x1; : : : ; xn) = 1n!�t1 � � ��tnF ( nXi=1 tixi)jt1=���=tn=0: (1.12)Moreover, the r-th derivative of F 2 Cr(X; Y ) is symmetri sin e,drFx(v1; : : : ; vr) = �t1 � � ��trF (x+ rXi=1 tivi)jt1=���=tr=0; (1.13)where the order of the partial derivatives an be shown to be irrelevant.Now we turn to integration. We will only onsider the ase of mappings f :I ! X where I = [a; b℄ � R is a ompa t interval and X is a Bana h spa e. Afun tion f : I ! X is alled simple if the image of f is �nite, f(I) = fxigni=1,and if ea h inverse image f�1(xi), 1 � i � n is a Borel set. The set of simplefun tions S(I;X) forms a linear spa e and an be equipped with the sup norm.The orresponding Bana h spa e obtained after ompletion is alled the set ofregulated fun tions R(I;X).Observe that C(I;X) � R(I;X). In fa t, onsider fn = Pn�1i=0 f(ti)�[ti;ti+1) 2S(I;X), where ti = a+i b�an and � is the hara teristi fun tion. Sin e f 2 C(I;X)is uniformly ontinuous, we infer that fn onverges uniformly to f .For f 2 S(I;X) we an de�ne a linear map R : S(I;X)! X byZ ba f(t)dt = nXi=1 xi�(f�1(xi)); (1.14)where � denotes the Lebesgue measure on I. This map satis�esZ ba f(t)dt � jf j(b� a): (1.15)and hen e it an be extended uniquely to a linear map R : R(I;X)! X with thesame norm (b� a). We even haveZ ba f(t)dt � Z ba jf(t)jdt: (1.16)

1.2. Contra tion prin iples 5In addition, if � 2 X� is a ontinuous linear fun tional, then�(Z ba f(t)dt) = Z ba �(f(t))dt; f 2 R(I;X): (1.17)We use the usual onventions R t2t1 f(s)ds = R ba �(t1;t2)(s)f(s)ds and R t1t2 f(s)ds =� R t2t1 f(s)ds.If I � R, we have an isomorphism L(I;X) � X and if F : I ! X we willwrite _F (t) in stead of dF (t) if we regard dF (t) as an element of X. In parti ular,if f 2 C(I;X), then F (t) = R ta f(s)ds 2 C1(I;X) and _F (t) = f(t) as an be seenfromjZ t+"a f(s)ds� Z ta f(s)ds� f(t)"j = jZ t+"t (f(s)� f(t))dsj � j"j sups2[t;t+"℄jf(s)� f(t)j:(1.18)This even shows that F (t) = F (a) + R ta ( _F (s))ds for any F 2 C1(I;X).1.2 Contra tion prin iplesA �xed point of a mapping F : C � X ! C is an element x 2 C su h thatF (x) = x. Moreover, F is alled a ontra tion if there is a ontra tion onstant� 2 [0; 1) su h that jF (x)� F (~x)j � �jx� ~xj; x; ~x 2 C: (1.19)Note that a ontra tion is ontinuous. We also re all the notation F n(x) =F (F n�1(x)), F 0(x) = x.Theorem 1.4 (Contra tion prin iple) Let C be a losed subset of a Bana hspa e X and let F : C ! C be a ontra tion, then F has a unique �xed pointx 2 C su h that jF n(x)� xj � �n1� � jF (x)� xj; x 2 C: (1.20)Proof. If x = F (x) and ~x = F (~x), then jx � ~xj = jF (x) � F (~x)j � �jx � ~xjshows that there an be at most one �xed point.Con erning existen e, �x x0 2 C and onsider the sequen e xn = F n(x0). Wehave jxn+1 � xnj � �jxn � xn�1j � � � � � �njx1 � x0j (1.21)

6 Chapter 1. Analysis in Bana h spa esand hen e by the triangle inequality (for n > m)jxn � xmj � nXj=m+1 jxj � xj�1j � �m n�m�1Xj=0 �jjx1 � x0j� �m1� � jx1 � x0j: (1.22)Thus xn is Cau hy and tends to a limit x. Moreover,jF (x)� xj = limn!1 jxn+1 � xnj = 0 (1.23)shows that x is a �xed point and the estimate (1.20) follows after taking the limitn!1 in (1.22). 2Next, we want to investigate how �xed points of ontra tions vary with respe tto a parameter. Let U � X, V � Y be open and onsider F : U � V ! U . Themapping F is alled a uniform ontra tion if there is a � 2 [0; 1) su h thatjF (x; y)� F (~x; y)j � �jx� ~xj; x; ~x 2 U; y 2 V: (1.24)Theorem 1.5 (Uniform ontra tion prin iple) Let U , V be open subsets ofBana h spa es X, Y , respe tively. Let F : U � V ! U be a uniform ontra tionand denote by x(y) 2 U the unique �xed point of F (:; y). If F 2 Cr(U � V; U),r � 0, then x(:) 2 Cr(V; U).Proof. Let us �rst show that x(y) is ontinuous. Fromjx(y + v)� x(y)j = jF (x(y + v); y + v)� F (x(y); y + v)+ F (x(y); y + v)� F (x(y); y)j� �jx(y + v)� x(y)j+ jF (x(y); y + v)� F (x(y); y)j (1.25)we infer jx(y + v)� x(y)j � 11� � jF (x(y); y + v)� F (x(y); y)j (1.26)and hen e x(y) 2 C(V; U). Now let r = 1 and let us formally di�erentiate x(y) =F (x(y); y) with respe t to y,d x(y) = �xF (x(y); y)d x(y) + �yF (x(y); y): (1.27)Considering this as a �xed point equation T (x0; y) = x0, where T (:; y) : L(Y;X)!L(Y;X), x0 7! �xF (x(y); y)x0+�yF (x(y); y) is a uniform ontra tion sin e we have

1.2. Contra tion prin iples 7j�xF (x(y); y)j � � by Theorem 1.2. Hen e we get a unique ontinuous solutionx0(y). It remains to showx(y + v)� x(y)� x0(y)v = o(v): (1.28)Let us abbreviate u = x(y + v)� x(y), then(1� �xF (x(y); y))(u� x0(y)v) == F (x(y) + u; y + v)� F (x(y); y)� �xF (x(y); y)u� �yF (x(y); y)v= o(u) + o(v) (1.29)sin e F 2 C1(U�V; U) by assumption. Moreover, j(1��xF (x(y); y))�1j � (1��)�1and u = O(v) (by (1.26)) implying u� x0(y)v = o(v) as desired.Finally, suppose that the result holds for some r � 1 � 1. Thus, if F isCr, then x(y) is at least Cr�1 and the fa t that d x(y) satis�es (1.27) impliesx(y) 2 Cr(V; U). 2As an important onsequen e we obtain the impli it fun tion theorem.Theorem 1.6 (Impli it fun tion) Let X, Y , and Z be Bana h spa es and letU , V be open subsets of X, Y , respe tively. Let F 2 Cr(U � V; Z), r � 1, and �x(x0; y0) 2 U � V . Suppose �xF (x0; y0) 2 L(X; Y ) is an isomorphism. Then thereexists an open neighborhood U1 � V1 � U � V of (x0; y0) su h that for ea h y 2 V1there exists a unique point (�(y); y) 2 U1 � V1 satisfying F (�(y); y) = F (x0; y0).Moreover, the map � is in Cr(V1; Z) and ful�llsd�(y) = �(�xF (�(y); y))�1 Æ �yF (�(y); y): (1.30)Proof. Using the shift F ! F � F (x0; y0) we an assume F (x0; y0) = 0.Next, the �xed points of G(x; y) = x � (�xF (x0; y0))�1F (x; y) are the solutionsof F (x; y) = 0. The fun tion G has the same smoothness properties as F andsin e j�xG(x0; y0)j = 0, we an �nd balls U1 and V1 around x0 and y0 su h thatj�xG(x; y)j � � < 1. Thus G(:; y) is a uniform ontra tion and in parti ular,G(U1; y) � U1, that is, G : U1 � V1 ! U1. The rest follows from the uniform ontra tion prin iple. Formula (1.30) follows from di�erentiating F (�(y); y) = 0unsing the hain rule. 2Note that our proof is onstru tive, sin e it shows that the solution �(y) anbe obtained by iterating x� (�xF (x0; y0))�1F (x; y).Moreover, as a orollary of the impli it fun tion theorem we also obtain theinverse fun tion theorem.

8 Chapter 1. Analysis in Bana h spa esTheorem 1.7 (Inverse fun tion) Suppose F 2 Cr(U; Y ), U � X, and let dF (x0)be an isomorphism for some x0 2 U . Then there are neighborhoods U1, V1 of x0,F (x0), respe tively, su h that F 2 Cr(U1; V1) is a di�eomorphism.Proof. Apply the impli it fun tion theorem to G(x; y) = y � F (x). 21.3 Ordinary di�erential equationsAs a �rst appli ation of the impli it fun tion theorem, we prove (lo al) existen eand uniqueness for solutions of ordinary di�erential equations in Bana h spa es.The following lemma will be needed in the proof.Lemma 1.8 Suppose I � R is a ompa t interval and f 2 Cr(U; Y ). Thenf� 2 Cr(Cb(I; U); Cb(I; Y )), where(f�x)(t) = f(x(t)): (1.31)Proof. Fix x0 2 Cb(I; U) and " > 0. For ea h t 2 I we have a Æ(t) > 0 su hthat jf(x) � f(x0(t))j � "=2 for all x 2 U with jx � x0(t)j � 2Æ(t). The ballsBÆ(t)(x0(t)), t 2 I, over the set fx0(t)gt2I and sin e I is ompa t, there is a �nitesub over BÆ(tj )(x0(tj)), 1 � j � n. Let jx � x0j � Æ = min1�j�n Æ(tj). Thenfor ea h t 2 I there is ti su h that jx0(t) � x0(tj)j � Æ(tj) and hen e jf(x(t)) �f(x0(t))j � jf(x(t))� f(x0(tj))j+ jf(x0(tj))� f(x0(t))j � " sin e jx(t)� x0(tj)j �jx(t)� x0(t)j+ jx0(t)� x0(tj)j � 2Æ(tj). This settles the ase r = 0.Next let us turn to r = 1. We laim that df� is given by (df�(x0)x)(t) =df(x0(t))x(t). Hen e we need to show that for ea h " > 0 we an �nd a Æ > 0 su hthat supt2I jf�(x0(t) + x(t))� f�(x0(t))� df(x0(t))x(t)j � "Æ (1.32)whenever jx� x0j � Æ. By assumption we havejf�(x0(t) + x(t))� f�(x0(t))� df(x0(t))x(t)j � "Æ(t) (1.33)whenever jx(t) � x0(t)j � Æ(t). Now argue as before. It remains to show that df�is ontinuous. To see this we use the linear map� : Cb(I;L(X; Y )) ! L(Cb(I;X); Cb(I; Y ))T 7! T�x ; (1.34)

1.3. Ordinary di�erential equations 9where (T�x)(t) = T (t)x(t). Sin e we havejT�xj = supt2I jT (t)x(t)j � supt2I jT (t)jjx(t)j � jT jjxj; (1.35)we infer j�j � 1 and hen e � is ontinuous. Now observe df� = � Æ (df)�.The general ase r > 1 follows from indu tion. 2Now we ome to our existen e and uniqueness result for the initial value prob-lem in Bana h spa es.Theorem 1.9 Let I be an open interval, U an open subset of a Bana h spa e Xand � an open subset of another Bana h spa e. Suppose F 2 Cr(I � U � �; X),then the initial value problem_x(t) = F (t; x; �); x(t0) = x0; (t0; x0; �) 2 I � U � �; (1.36)has a unique solution x(t; t0; x0; �) 2 Cr(I1� I2�U1��1; X), where I1;2, U1, and�1 are open subsets of I, U , and �, respe tively. The sets I2, U1, and �1 an be hosen to ontain any point t0 2 I, x0 2 U , and �0 2 �, respe tively.Proof. If we shift t ! t � t0, x ! x � x0, and hen e F ! F (: + t0; : + x0; �),we see that it is no restri tion to assume x0 = 0, t0 = 0 and to onsider (t0; x0)as part of the parameter � (i.e., � ! (t0; x0; �)). Moreover, using the standardtransformation _x = F (�; x; �), _� = 1, we an even assume that F is independent oft. We will also repla e U by a smaller (bounded) subset su h that F is uniformly ontinuous with respe t to x on this subset.Our goal is to invoke the impli it fun tion theorem. In order to do this weintrodu e an additional parameter " 2 R and onsider_x = "F (x; �); x 2 Dr+1 = fx 2 Cr+1b ((�1; 1); U)jx(0) = 0g; (1.37)su h that we know the solution for " = 0. The impli it fun tion theorem will showthat solutions still exists as long as " remains small. At �rst sight this doesn'tseem to be good enough for us sin e our original problem orresponds to " = 1.But sin e " orresponds to a s aling t ! "t, the solution for one " > 0 suÆ es.Now let us turn to the details.Our problem (1.37) is equivalent to looking for zeros of the fun tionG : Dr+1 � �� (�"0; "0) ! Crb ((�1; 1); X)(x; �; ") 7! _x� "F (x; �) : (1.38)

10 Chapter 1. Analysis in Bana h spa esLemma 1.8 ensures that this fun tion is Cr. Now �x �0, then G(0; �0; 0) = 0and �xG(0; �0; 0) = T , where Tx = _x. Sin e (T�1x)(t) = R t0 x(s)ds we anapply the impli it fun tion theorem to on lude that there is a unique solutionx(�; ") 2 Cr(�1 � (�"0; "0); Dr+1). In parti ular, the map (�; t) 7! x(�; ")(t=") isin Cr(�1; Cr+1((�"; "); X)) ,! Cr(�� (�"; "); X). Hen e it is the desired solutionof our original problem. 2