Spectral Theory

cel-00376389, version 1 - 17 Apr 2009 Spectral Theory M. Jazar Lebanese University Faculty of Sciences Department of Mathematics May 13, 2004 cel-00376389, version 1 - 17 Apr 2009 2 cel-00376389, version 1 - 17 Apr 2009 Contents 1 Hilbert spaces 1.1 Scalar product . . . . . . . . 1.2 Projection Theorem . . . . . 1.3 Adjoint of a linear continuous 1.4 Hilbert basis . . . . . . . . . . . . . . . . . . . . . mapping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 5 9 11 14 2 Spectrum of a bounded operator 19 2.1 Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.2 Hilbert case . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 3 Symbolic Calculus 27 3.1 Case of bounded operator . . . . . . . . . . . . . . . . . . . . 27 3.2 Case of a bounded self-adjoint operator . . . . . . . . . . . . 29 4 Compact operators 4.1 General properties . . . . . . . . . . . . 4.2 Spectral properties of compact operators 4.3 Hilbert-Schmidt operators . . . . . . . . 4.4 Compact self-adjoint operators . . . . . 4.5 Fredholm equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Unbounded self-adjoint operators 5.1 Closed operators . . . . . . . . . . . . . . . . . . 5.2 Adjoint of an operator . . . . . . . . . . . . . . . 5.3 The L∞ -spectral theorem . . . . . . . . . . . . . 5.4 The L2 -spectral theorem . . . . . . . . . . . . . . 5.5 Stone’s theorem . . . . . . . . . . . . . . . . . . . 5.6 Laplace operator on bounded open domain of RN 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 33 35 37 40 43 . . . . . . 45 45 48 52 55 57 58 cel-00376389, version 1 - 17 Apr 2009 4 CONTENTS Chapter 1 cel-00376389, version 1 - 17 Apr 2009 Hilbert spaces 1.1 Scalar product Let E and F be two C-vector spaces. A mapping f : E −→ F is said to be antilinear if, for all x, y ∈ E and all λ ∈ C we have f (x + y) = f (x) + f (y) and f (λx) = λf (x). Definition 1.1.1 Let E be a complex vector space. We call sesqui-linear form on E a mapping B: E × E −→ C such that, for all y ∈ E the mapping x 7−→ B(x, y) is linear and the mapping x 7−→ B(y, x) is anti-linear. Proposition 1.1.1 (polarization identity) 1. Let E be a complex vector space and B a sesqui-linear form on E. For all x, y ∈ E we have 4B(x, y) = B(x+y, x+y)−B(x−y, x−y)+iB(x+iy, x+iy)−iB(x−iy, x−iy). 2. Let E be a real vector space and B a bilinear form on E. For all x, y ∈ E we have 4B(x, y) = B(x + y, x + y) − B(x − y, x − y). Proof. We have B(x + y, x + y) − B(x − y, x − y) = 2B(x, y) + 2B(y, x). Replacing y by iy, we find B(x + iy, x + iy) − B(x − iy, x − iy) = 2B(x, iy) + 2B(iy, x) = −2iB(x, y) + 2iB(y, x). ¤ In particular, to determine a symmetric sesqui-linear form B, it suffices to determine B(x, x) for all x ∈ E. Corollary 1.1.1 Let E be a complex vector space and B a sesqui-linear form on E. The following are equivalent: 5 6 CHAPTER 1. HILBERT SPACES (i) For all x, y ∈ E we have B(y, x) = B(x, y). (ii) For all x ∈ E, B(x, x) ∈ R. Proof. Set S(x, y) = B(x, y)−B(y, x). This define a sesqui-linear form. By the polarization identity, S is zero if and only if, for all x ∈ E, S(x, x) = 0. ¤ cel-00376389, version 1 - 17 Apr 2009 Definition 1.1.2 Let E be a complex vector space. We call hermitian form on E a sesqui-linear form verifying any of the equivalent conditions of corollary 1.1.1. A hermitian form B on E is said to be positive if, for all x ∈ E, B(x, x) ≥ 0. A symmetric bilinear form B on a real vector space E is said to be positive if, for all x ∈ E, B(x, x) ≥ 0. We call semi-scalar product, often denoted by (x, y) 7−→< x, y >, any symmetric positive form on a real vector space or any positive hermitian form on a complex vector space. It is called scalar product if, it verify in addition the following property: for all x ∈ E, < x, x >= 0 if and only if x = 0. On appelle espace prhilbertien (rel ou complexe) un espace vectoriel (rel ou complexe) muni d’un produit scalaire. Exemples. 1. Let E = RN . If a1 , · · · , aN are positive real numbers, the relation X < x, y >:= aj xj yj 1≤i≤N define on E a semi-scalar product, which is a scalar product if and only if all aj are strictly positive. 2. Let X be metric space locally compact and separable, µ a positive Radon measure on X and E := D0 (X, K). The relation Z < f, g >:= f (x)g(x) dµ(x) define a semi-scalar product, which is a scalar product if and only if Suppµ = X. 3. The space E := C2π = {f : R 7−→ K continuous and 2π − periodical} with the relation Z 2π 1 f (x)g(x) dx < f, g >:= 2π 0 is a prehilbert space. 1.1. SCALAR PRODUCT 7 4. Let I be a set. Denote, for p ≥ 1, by `p (I) ⊂ KI the set of sequences (xi )i∈I such that |xi |p is summable. Put on `p (I) the discrete measure m, Z X X x dm = xi := sup xi < ∞, i∈I J∈Pf (I) i∈J where Pf (I) is the set of finite parts of I. The case p = 2 is very particular, `2 (I) with the scalar product defined by X < x, y >:= xi yi i∈I cel-00376389, version 1 - 17 Apr 2009 is a prehilbert space. The classical proof is applicable to the prehilbert case for: Proposition 1.1.2 (Cauchy-Schwarz inequality) Let E a prehilbert space. For all x, y ∈ E we have | < x, y > |2 ≤< x, x > < y, y > . Corollary 1.1.2 Let E be a prehilbert space. the mapping x 7−→ define a semi-norm on E. √ < x, x > Proof. For all x, y ∈ E, we have < x + y, x + y >=< x, x > + < y < y > + < x, y > +< x, y > ≤< x, x > + < y, y > +2| < x, y > | ≤ √ √ [ < x, x > + < y, y >]2 , by Cauchy-Schwarz inequality. ¤ Proposition 1.1.3 Let E be a prehilbert space. For all x ∈ E, the linear form fx : y 7−→< y, x > is continuous. Moreover the mapping x 7−→ fx is anti-linear and isometric from E into E ∗ . Proof. Let p be the semi-norm of corollary 1.1.2. For y ∈ E we have |fx (y)| ≤ p(x)p(y) (Cauchy-Scwarz). So fx ∈ E ∗ and kfx k ≤ p(x). Now since p(x)2 = fx (x) ≤ kfx kp(x), we get that kfx k = p(x). ¤ In the following we give a case where equality in the Cauchy-Schwarz inequality occur. Proposition 1.1.4 Let x, y ∈ E a prehilbert space. Then | < x, y > | = kxk · kyk if and only if x and y are linearly dependent. 8 CHAPTER 1. HILBERT SPACES Proof. The condition is clearly sufficient. Assume that | < x, y > | = kxk kyk and let ε ∈ C, |ε| = 1 such that Re[ε < x, y >] = | < x, y > |. Then kykxk − εkykxk2 = 0. ¤ A direct consequence of the definition of the norm is Proposition 1.1.5 (parallelogram identity) For all x, y ∈ E we have cel-00376389, version 1 - 17 Apr 2009 ° ° ° ° ° x + y °2 ° x − y °2 1 ¡ ¢ 2 2 ° ° +° ° ° 2 ° ° 2 ° = 2 kxk + kyk . Definition 1.1.3 Let E be a prehilbert space. We say that two elements x and y of E are orthogonal if < x, y >= 0. We say that the subsets A and B are orthogonal if every element of A is orthogonal to every element of B. We call orthogonal of a part A of E the set A⊥ of elements of E orthogonal to A. \ It is clear that A⊥ = kerfx . Hence its is a closed sub-vector space of E. x∈A A direct consequence is Proposition 1.1.6 (Pythagore’s theorem) If x, y ∈ E are orthogonal in a prehilbert space, then kx + yk2 = kxk2 + kyk2 . Definition 1.1.4 A Hilbert space is a complete prehilbert space for the norm defined by its scalar product. Fundamental examples. 1. Every finite dimensional prehilbert space is a Hilbert space. 2. If µ is a measure on a measured space, the space L2 (µ) define a Hilbert space with the following scalar product: Z < f, g >:= f g dµ. 1.2. PROJECTION THEOREM 1.2 9 Projection Theorem One of the fundamental tools of the Hilbert structure is projection theorem. In the following H is a Hilbert space endowed with the scalar product <, > and the associated norm k · k. Theorem 1.2.1 Let C be a nonempty closed and convex set of H. Then for all x ∈ H, there exists a unique y ∈ C such that cel-00376389, version 1 - 17 Apr 2009 kx − yk = d(x, C). This point y, called projection of x on C and denoted by PC (x), is characterized by y∈C and for all z ∈ C Re < x − y, z − y >≤ 0. (1.1) Proof. Denote by d := inf{kx − yk; y ∈ C} the distance to C. Let y, z ∈ C y−z y+z and set b := x − y+z 2 and c := 2 . Then d ≤ kbk since 2 ∈ C. Since x − y = b − c and x − z = b + c, we have £ ¤ ky − zk2 . kx − yk2 + kx − zk2 = 2 kbk2 + kck2 ≥ 2d2 + 2 Thus ky − zk2 ≤ 2[kx − yk2 − d2 ] + 2[kx − zk2 − d2 ]. For n ∈ N, set Cn := {y ∈ C; kx − yk2 ≤ d2 + n1 }. Cn is nonempty closed √ set of H and the diameter of Cn , δ(Cn ) ≤ 2/ n hence tends to zero. Since H is complete, the intersection of Cn , that is equal to {y ∈ C; kx − yk = d}, contains a unique point y0 . Let y ∈ C. For t ∈ [0, 1], we have y0 + t(y − y0 ) ∈ C, hence ky0 + t(y − y0 ) − xk ≥ ky − y0 k. Set f (t) := ky0 + t(y − y0 ) − xk2 = ky0 − xk2 + 2tRe < y0 − x, y − y0 > +t2 ky − y0 k2 . Since f (0) ≤ f (t) for all t ∈ [0, 1], f 0 (0) ≥ 0, i.e. Re < y0 − x, y − y0 >≥ 0. ¤ Denote by PC the projection on C. Condition (1.1) permits to show that PC is a contraction: Proposition 1.2.1 Under the same hypothesis, for all x, y ∈ H, we have kPC (x) − PC (y)k ≤ kx − yk. Proof. Set u := PC x and v := PC y. We have Re < x − y, u − v > = Re < x − v, u − v > +Re < v − x, u − v > = Re < x − u, u − v > +ku − vk2 + Re < v − x, u − v > ≥ ku − vk2 . 10 CHAPTER 1. HILBERT SPACES Hence by Cauchy-Schwarz inequality, ku − vk2 ≤ kx − yk ku − vk. ¤ In the case of sub-vector space: cel-00376389, version 1 - 17 Apr 2009 Proposition 1.2.2 Let E be a closed sub-vector space of H. Then PE is a linear operator from H to E. If x ∈ H, then PE (x) is the unique element y ∈ H such that y∈E and x − y ∈ E ⊥ . Proof. Condition (1.1) could be written as y ∈ E, and for all z ∈ E, Re < x − y, z − y >≤ 0. But if y ∈ E and λ ∈ C∗ , the mapping z 0 7−→ z = y + λ̄z 0 is a bijection from E onto itself. Condition (1.1) is then equivalent to y ∈ E, and for all z 0 ∈ E, and all λ ∈ C, Re[λ < x − y, z 0 >] ≤ 0 which is equivalent to and x − y ∈ E ⊥ . y∈E ¤ Corollary 1.2.1 For all closed sub-vector space E of H, we have H = E ⊕ E⊥ and the projector on E associated to this direct sum is PE . PE is called orthogonal projector on E. Proof. If x ∈ F , x = PE x + (x − PE x) and by proposition 1.2.2, PE x ∈ E and x − PE x ∈ E ⊥ . From the other hand, if x ∈ E ∩ E ⊥ , then < x, x >= 0, so x = 0. ¤ Corollary 1.2.2 For all sub-vector space E of H, we have H = E ⊕ E⊥. In particular, E is dense in H if and only if E ⊥ = {0}. Proof. Remember that E ⊥ = Ē ⊥ . Corollary 1.2.3 For all sub-vector space E of H, we have E = E ⊥⊥ . ¤ 1.3. ADJOINT OF A LINEAR CONTINUOUS MAPPING 11 Proof. Clearly E ⊂ E ⊥⊥ and hence, since E ⊥⊥ is closed, Ē ⊂ E ⊥⊥ . From the other hand we have H = Ē ⊕ E ⊥ and H = E ⊥⊥ ⊕ E ⊥ . ¤ Proposition 1.2.3 The anti-linear isometric mapping x 7−→ fx of proposition 1.1.3 is a bijection from H onto H ∗ . cel-00376389, version 1 - 17 Apr 2009 Proof. Let x∗ ∈ H ∗ and denote by E its kernel. If x∗ 6= 0 then E 6= H and E ⊥ 6= {0} (corollary 1.2.2). Let then x ∈ E ⊥ , x 6= 0. So fx is zero on E. Since fx (x) 6= 0, there exists λ ∈ K such that x∗ (x) = λfx (x). Since E is a hyperplane and x 6∈ E, we have H = E ⊕ Kx. Thus x∗ and λfx that coincide on E and on x are equal. Therefore x∗ = fλ̄x . ¤ Corollary 1.2.4 Every Hilbert space is reflexive. Proof. Let H be a Hilbert space and ` ∈ H ∗∗ . The mapping x 7−→ `(fx ) belongs to H ∗ . By the last proposition, there exists y ∈ H such that for all x ∈ H we have `(fx ) = fy (x) =< y, x >= fx (y). Thus for every x∗ ∈ H ∗ we have `(x∗ ) = x∗ (y), i.e ` is the image of y by the canonical injection from H to H ∗∗ . ¤ 1.3 Adjoint of a linear continuous mapping Recall that L(E, F ) denote the space of linear continuous (operator) from E into F and that L(E) = L(E, E). In what follows E and F are Hilbert spaces. Theorem 1.3.1 (Riesz) The mapping ½ E −→ E ∗ y 7−→ φy : y ∗ ∈ E ∗ 7−→ φy (y ∗ ) :=< y ∗ , y > is surjective isometry. In other words, for all linear continuous form φ on E, there exists a unique y ∈ E such that φ = φy and kyk = kφy k. In the following we study some important applications of Riesz’s Theorem. Proposition 1.3.1 Let T ∈ L(E, F ). There exists a unique T ∗ ∈ L(F, E) such that for all x ∈ E and all y ∈ F we have < T x, y >=< x, T ∗ y > . T ∗ is called the adjoint of T . 12 CHAPTER 1. HILBERT SPACES Proof. For all y ∈ F , the mapping x 7−→< T x, y > is linear and continuous. There exists then a unique element y ∗ denoted by T ∗ y ∈ E such that for all x ∈ E we have < T x, y >=< x, T ∗ y >. Clearly T ∗ is linear. Now, for all x ∈ E and all y ∈ F we have | < x, T ∗ y > | = | < T x, y > | ≤ kT xk kyk ≤ kT k kxk kyk. Thus kT ∗ yk ≤ kT k kyk. Therefore T ∗ is continuous and kT ∗ k ≤ kT k. ¤ cel-00376389, version 1 - 17 Apr 2009 Hereafter some properties of adjoint operator: Proposition 1.3.2 The mapping T 7−→ T ∗ is anti-linear and isometric from L(E, F ) into L(F, E): for all T ∈ L(E, F ) we have T ∗∗ = T and kT ∗ ◦ T k = kT k2 . For all Hilbert space G, all S ∈ L(E, F ) and all T ∈ L(F, G) we have (T ◦ S)∗ = S ∗ ◦ T ∗ Proof. kT ∗ T k ≤ kT ∗ k kT k ≤ kT k2 . Now, for x ∈ E, with kxk ≤ 1 we have kT xk2 =< T x, T x >=< x, T ∗ T x >≤ kT ∗ T k (Cauchy-Schwarz). Hence kT k2 ≤ kT ∗ T k. ¤ Proposition 1.3.3 Let T ∈ L(E, F ). Then ker T ∗ = Im T ⊥ and T ∗ (F ) = (ker T )⊥ . Proof. Let y ∈ F . y ∈ kerT ∗ if and only if for all x ∈ E, < T x, y >=< x, T ∗ y >= 0 if and only if y ∈ T⊥ . From corollary 1.2.3, ImT = kerT ∗ ⊥ , replace then T by T ∗ . ¤ Definition 1.3.1 An element U ∈ L(E, F ) is said to be unitary if U ∗ ◦U = IdE and U ◦ U ∗ = IdF . T ∈ L(E) is said to be normal if T ∗ ◦ T = T ◦ T ∗ , self-adjoint if T = T ∗ and positive if it is self-adjoint and, for all x ∈ E we have < T x, x >≥ 0. Examples. 1. Let H be a Hilbert space and P ∈ L(H) an orthogonal projector and E := Im(P ) its image. For all x, x0 ∈ E and y, y 0 ∈ E ⊥ we have < P (x + y), x0 + y 0 >=< x, x0 >=< x + y, P (x + y) >; hence P is self-adjoint. Moreover, < P (x + y), x + y >=< x, x >≥ 0 hence P is positive. 2. For all T ∈ L(H), T T ∗ and T ∗ T are self-adjoint. 1.3. ADJOINT OF A LINEAR CONTINUOUS MAPPING 13 3. Consider the Hilbert space H := L2 (Ω, µ) where Ω is a measurable space and µ a σ-finite measure (i.e. Ω is countable union of subset of finite measure for µ). Let K ∈ L2 (µ × µ). For f ∈ H define Z TK (f ) := K(x, y)f (y) dµ(y) for µ-a.e. x. By Cauchy-Schwarz inequality, TK f ∈ H and TK is a linear continuous operator on H whose norm verify kTK k ≤ kKkL2 (µ×µ) . cel-00376389, version 1 - 17 Apr 2009 By Fubini’s theorem, one can verify that < TK f, g >= f, TK ∗ g >, ∗ = T ∗ . It is easy to verify that where K ∗ (x, y) := K(y, x). Thus TK K TK is self-adjoint if and only if, K(x, y) = K(y, x) for µ-a.e x and y. Proposition 1.3.4 Let T be a self-adjoint operator on H, then kT k = sup{< T x, x > with x ∈ E, kxk = 1}. Proof. Let γ := sup{< T x, x > with x ∈ E, kxk = 1}. We have γ ≤ kT k and for all x ∈ H, | < T x, x > | ≤ γkxk2 . Let y, z ∈ H nad λ ∈ R, then | < T (y±λz), y±z > | = | < T y, y > ±2λRe < T y, z > +λ2 < T z, z > | ≤ γky±zk2 . Hence £ ¤ £ ¤ 4|λ|Re < T y, z >≤ γ ky + λzk2 + ky − λzk2 = 2γ kyk2 + λ2 kzk2 , this is true for all real λ, hence |Re < T y, z > | ≤ γkyk kzk. Choose now z = T y. ¤ Proposition 1.3.5 Let T ∈ L(E, F ). The following conditions are equivalent: (i) T is unitary. (ii) T is surjective and T ∗ ◦ T = IdE . (iii) T is an isometry from E to F . 14 CHAPTER 1. HILBERT SPACES Proof. (i)⇒(ii): Since T ∗ T = IdF , T is surjective. (ii)⇒(iii): If T ∗ T = IdF , then for all x ∈ E, we have kT xk2 =< T x, T x >=< x, T ∗ T x >=< x, x >= kxk2 . (iii)⇒(i): Since (x, y) 7−→< x, T ∗ T y >=< T x, T y > is a scalar product on E, by polarization identity we get that, for all x, y ∈ E, < x, T ∗ T y >=< x, y >. Hence T ∗ T y − y ∈ E ⊥ = {0}. Thus T ∗ T = IdF and since T is bijective, T ∗ = T −1 . ¤ Definition 1.3.2 (Weak convergence) We say that a sequence (xn ) ⊂ E converges weakly in E if for all y ∈ E we have cel-00376389, version 1 - 17 Apr 2009 lim < xn , y >=< x, y > . n→∞ x is called weak limit of the sequence (xn ). It is clear that a weak limit of a sequence is unique, and by CauchySchwarz inequality, strong convergence implies weak convergence. As a direct application of Riesz’s Theorem one can deduce the following version of Banach-Alaoglu’s Theorem in Hilbert space. Theorem 1.3.2 From every bounded sequence of E one can extract a weakly convergent subsequence. The existence of the adjoint of an arbitrary linear continuous operator gives the following property. Proposition 1.3.6 Let (xn ) be a sequence of E that converges weakly to x ∈ E. Then for all T ∈ L(E), the sequence T xn converges weakly to T x. Proof. For all y ∈ E we have lim < T xn , y >=< xn , T ∗ y >=< x, T ∗ y >=< T x, y > . n ¤ 1.4 Hilbert basis In this section E will denote a prehilbert space. A system (xi )i∈I of E is said to be orthogonal system if for all i 6= j, xi ⊥ xj . Recall that, by Pythagore’s theorem, we have, for all finite subset J of I ° ° °X °2 X ° ° kxi k2 . xi ° = ° ° ° i∈J i∈J We get then directly the following proposition. 1.4. HILBERT BASIS 15 Proposition 1.4.1 An orthogonal system in which all elements are non zero is a free system. Proof. Let J ⊂ I a finite part and (λj )j∈J ⊂ K such that Then °2 ° °X ° X ° ° |λj |2 kxj k2 , 0=° λj xj ° ° ° = ° j∈J ° j∈J P j∈J λj xj = 0. cel-00376389, version 1 - 17 Apr 2009 and so λj = 0 for all j ∈ J. ¤ Definition 1.4.1 An orthogonal system whose elements are of norm 1 is called orthonormal basis (or orthonormed). A total orthonormal basis of E is called Hilbert basis of E. Examples. 1. Let T > 0 and CT the space of T -periodic continuous functions from R into K which is a prehilbert space. For n ∈ Z set en (x) := e 2iπnx T . It is easy to see that the class (en )n∈Z is an orthonormal system of CT . Moreover this system is total in CT endowed with supremum norm. Since the norm associated to the scalar product is less than or equal to the supremum norm, this system is a Hilbert basis. 2. Consider the space E = `2 (I). Define for j ∈ I, the element ej ∈ E by ej (j) = 1 et ej (i) = 0 for i 6= j. The system (ej )j∈I is orthonormal (evident). Let’s show P that it is total. For this, let x ∈ E and ε > 0. By definition, and since i∈I |xi |2 < ∞, there exists a finite part J ⊂ I such that X X X |xi |2 = |xi |2 − |xi |2 ≤ ε2 . i∈I,i6=J This implies that i∈I i∈J °2 ° ° ° X ° ° xi ei ° ≤ ε2 . °x − ° ° i∈J Thus `2 (I) is a Hilbert space and (ei )i∈I is a Hilbert basis of `2 (I). 16 CHAPTER 1. HILBERT SPACES Proposition 1.4.2 Let (ei )i∈I a finite orthonormal system of E and let F be the vector space generated by this system. For all x ∈ E, the orthogonal projection PF (x) is given by PF (x) = X < x, ei > ei . i∈I Consequently, °2 ° ° ° X X ° ° kxk = °x − < x, ei > ei ° + | < x, ei > |2 . ° ° 2 cel-00376389, version 1 - 17 Apr 2009 i∈I i∈I P Proof. For the first point, it suffices to show that y := j∈J < x, ej > ej verify the properties of proposition 1.2.2. It is clear that y ∈ F and for all j ∈ J, < x − y, ej >= 0, so x − y ∈ F ⊥ . For the rest apply Pythagore’s theorem. ¤ A first consequence: Proposition 1.4.3 Bessel’s inequality Let (ei )i∈I be an orthonormal system of E. Then for all x ∈ E we have X | < x, ei > |2 ≤ kxk2 . i∈I In particular, (< x, ei >)i∈I is an element of `2 (I). The equality in the previous inequality is characterized by Theorem 1.4.1 Bessel-Parseval Let (ei )i∈I an orthonormal system of E. The following properties are equivalent: (i) The system (ei )i∈I is a Hilbert Basis. (ii) For all x ∈ E, kxk2 = P i∈I (iii) For all x, y ∈ E, < x, y >= | < x, ei > |2 (Bessel’s equality). P i∈I < x, ei > < ei , y >. Thus, if (ei )i∈I is a Hilbert basis of E, the mapping from E into `2 (I) defined by x 7−→ (< x, ei >)i∈I is a linear isometry. This isometry is surjective if and only if E is Hilbert space. 1.4. HILBERT BASIS 17 Proof. (i)⇒(ii): Let x ∈ E. For all ε > 0, there exists a finite subset J ⊂ I s.t. the distance between x and span(ej , j ∈ J) is less than ε or equal. By proposition 1.4.2, X X | < x, ej > |2 ≤ | < x, ej > |2 . kxk2 − ε2 ≤ j∈i cel-00376389, version 1 - 17 Apr 2009 j∈J Making ε → 0 and using Bessel’s inequality we get the result. (ii)⇒(i): Conversely, for all x ∈PE, and all ε > 0, there exists a finite subset J ⊂ I such that kxk2 − ε2 ≤ j∈J | < x, ej > |2 and then by proposition 1.4.2 ° ° ° ° X ° ° ° ≤ ε. °x − < x, e > e j j ° ° ° ° j∈J Thus (ei ) is total. The equivalence between (ii) and (iii) is direct from the definition of the scalar product in terms of the norm: < x, y >:= 1 2 £ kx + yk2 − kxk2 − kyk2 ¤ + ¤ i£ kx + iyk2 − kxk2 − kyk2 . 2 If the isometry is surjective then E is isometric to `2 (I) and hence complete. P Now assume that E is a Hilbert space and let (xi )i∈I ∈ `( I). Set a := |xi |2 . There exists then sequence (Jn ) P of finite subsets of I such that P an increasing 2 −n for all n ∈ N, Jn |xi | ≥ a − 2 . Set un := Jn xi ei . Then, if p < n, X |xi |2 ≤ 2−n . kup − un k2 = j∈Jp ,j6∈Jn Thus (un ) converges to some x ∈ E. Since a = X |xi |2 , for all i 6∈ ∩Jn , i∈∩n Jn xi = 0 and < x, ei >= limn→∞ < un , ei >= 0. If i ∈ ∩Jn , then < x, ei >= limn→∞ < un , ei >= xi . Thus < x, ei >= xi for all i, which proves the surjectivity. ¤ As a consequence we get Theorem 1.4.2 Let (ei )i∈I a Hilbert system of E. Then for all all x ∈ E we have X x= < x, ei > ei . i∈I Proof. By proposition 1.4.2 we know that for every finite subset J ⊂ I we have °2 ° ° ° X X ° ° 2 ° °x − | < x, ej > |2 . < x, e > e j j ° = kxk − ° ° ° j∈J j∈J 18 CHAPTER 1. HILBERT SPACES It suffices then to apply the definitions and the second property of the last theorem. ¤ Proposition 1.4.4 Schmidt orthonormalization procedure Let N ∈ {1, 2, · · ·} ∪ {+∞} and (fn )0≤n≤N a free system of E. There exists an orthonormal system (fn )0≤n≤N of E, such that, for all p < N , the systems (fn )0≤n≤p and (en )0≤n≤p generate the same sub-vector spaces of E. Proof. Left as an exercise to the reader. cel-00376389, version 1 - 17 Apr 2009 Using this procedure, one can directly show the following Corollary 1.4.1 A prehilbert space is separable if and only if it admits a countable Hilbert basis. Two prehilbert spaces are said to be isometric if there exists a surjective isometry from one of them to the other. Another consequence of theorem 1.4.1: Corollary 1.4.2 An infinite dimensional Hilbert space is separable if and only if it is iometric to the Hilbert space `2 . cel-00376389, version 1 - 17 Apr 2009 Chapter 2 Spectrum of a bounded operator In this chapter we give elementary definitions and properties concerning the spectrum of a linear operator on a Banach or Hilbert space. 2.1 Spectrum If E is a Banach space on K = C, denote by L(E) the Banach algebra (non commutative) of linear continuous mappings from E into itself. The product of two elements T, S is the composition: T S := T ◦ S. An element T ∈ L(E) is said to be invertible, if it admits an inverse in L(E). In other terms, if T is invertible, it is bijective and its inverse in L(E) is unique and equal to T −1 . Indeed, a direct application of the open mapping theorem is that the inverse of a linear bijective continuous operator is always continuous. We start by a simple but useful lemma. Lemma 2.1.1 Let T ∈ L(E) with kT k < 1. Then Id + T is invertible. The series of general term (−T )n converges and its sum is (Id + T )−1 . Proof. Set Sn := P k 0≤k≤n (−T ) . Since kT n k ≤ kT kn , we have, for p ≤ q, ° ° ° ° X ° ° kT kp k ° ° (−T ) )° ≤ kSq − Sp k = ° , ° (1 − kT k) °p+1≤k≤q thus Sn is a Cauchy sequence in the complete space L(E). Let S be its limit. For all n, Sn+1 = Id − T Sn = Id − Sn T . Making n → ∞ we get the equality: S = I − ST = Id − T S, thus Id + T is invertible and (Id + T )−1 = S. ¤ 19 20 CHAPTER 2. SPECTRUM OF A BOUNDED OPERATOR cel-00376389, version 1 - 17 Apr 2009 Proposition 2.1.1 Let E, F be two Banach spaces. The set U ⊂ L(E, F ) of linear continuous and invertible mapping is an open of L(E, F ). The mapping Φ: T 7−→ T −1 is continuous is differentiable from U into L(F, E) and its differential is (dΦ)T : S 7−→ −T −1 ST −1 . Proof. If F = E and by the last lemma, V the set of linear continuous invertible mapping in L(E), is a neighborhood of Id and the mapping ψT 7−→ T −1 is continuous and differentiable and dψId h = −h. In general. Let T ∈ U . Observe that S is invertible if and only if T −1 S is invertible. In this case, S −1 = (T −1 S)−1 T −1 . In other terms, denoting by f : L(E, F ) → L(E), f (S) := T −1 S and g: L(E) → L(E, F ), g(S) := ST −1 , we have U = f −1 (V ) and for all S ∈ U , φ(S) = g(ψ(f (S))). Therefore, U is a neighborhood of T and since f and g are linear and continuous, φ is differentiable at T and dφT = g ◦ dψId ◦ f , i.e. dφT (h) = −T −1 hT −1 . ¤ Definition 2.1.1 Let T ∈ L(E). We call resolvant of T , denoted by ρ(T ) the set of λ ∈ C such that λId − T is invertible. We call spectrum of T , denoted by σ(T ), the complementary of the resolvant: σ(T ) := C\ρ(T ). Finally, we call resolvent of T the mapping that to λ ∈ ρ(T ) associate (λId − T )−1 , denoted by R(λ) or R(λ, T ). Proposition 2.1.2 (Resolvent equation) Let T ∈ L(E). Then, for all λ, µ ∈ ρ(T ), we have R(λ) − R(µ) = (µ − λ)R(λ)R(µ) = (µ − λ)R(µ)R(λ). Proof. Direct calculation. ¤ Theorem 2.1.1 Let E be a non trivial Banach space and T ∈ L(E). The spectrum of T is a nonempty compact of C, the resolvent is analytic from ρ(T ) into L(E) and for all λ ∈ ρ(T ), we have R0 (λ) = R(λ)2 . Proof. Let U ⊂ L(E) be the set of invertible operators. The mapping fλ : λ 7−→ T − λ is continuous, hence the inverse image of U is open. Thus σ(T ) is closed. Let φ: U → L(E) defined by g(S) := S −1 . We have R(λ) = φ ◦ fλ hence by proposition 2.1.1, Rλ is continuous and differentiable and R0 (λ) = dφfλ (λ) fλ0 (λ), so since fλ0 (λ) = −Id then R0 (λ) = −R(λ)(−Id)R(λ) = R(λ)2 . Now let |λ| > kT k. By lemma 2.1.1, Id − λ−1 T is invertible, hence λ − T is invertible and λR(λ) = −R(λ−1 ). Therefore σ(T ) is bounded and so a compact of C. Moreover limλ→∞ λR(λ) = −Id. It is clear that λ 7−→ R(λ) is analytic on ρ(T ). If σ(T ) is empty then R would be entire, and since 2.1. SPECTRUM 21 limλ→∞ R(λ) = 0, by Liouville’s theorem, R ≡ 0, so kIdk ≤ kλ − T kk(λ − T )−1 k = 0, which is impossible unless E is trivial. ¤ Remark. If dim E < ∞, the spectrum of T could be empty in the case where K = R. For this, it suffices that the characteristic polynomial does not admit real solutions, but this is false in the case where K = C. Example. Let E := C([0, 1]) and T the operator defined for all f ∈ E by Z x T f (x) := f (t) dt. cel-00376389, version 1 - 17 Apr 2009 0 It is easy to see that kerT = {0} and ImT = {g ∈ C 1 ([0, 1]); g(0) = 0}. T is injective but not surjective, in other terms 0 ∈ σ(T ) and 0 6∈ σp (T ). Let’s show that 0 is the unique spectral value of T : For this take λ 6= 0 and g ∈ E. If f verify the equation λf − T f = g, (2.1) then the function h := T f ∈ C 1 ([0, 1]) and verify h(0) = 0 and λh0 − h = g. (2.2) Conversely, if h ∈ C 1 ([0, 1]) verify (2.2), then the function f := h0 is solution of (2.1). One can see directly that the unique solution of the differential equation (2.2) is given by Z ex/λ x h(x) = g(t)e−t/λ dt. λ 0 Therefore, " # Z ex/λ x 1 g(x) + g(t)e−t/λ , λf − T f = g ⇐⇒ f (x) = λ λ 0 hence λ ∈ ρ(T ) and " # x/λ Z x e 1 g(x) + g(t)e−t/λ . (λ − T )−1 g (x) = λ λ 0 Proposition 2.1.3 Let T ∈ L(E). The limit limn→∞ kT n k1/n exists and lim kT n k1/n = inf∗ kT n k1/n . n→∞ n∈N This value will be denoted by r(T ) and called spectral radius of T . Moreover, for all λ ∈ σ(T ), we have |λ| ≤ r(T ) and r(T ) = max{|λ|; λ ∈ σ(T )}. In particular r(T ) ≤ kT k and for all λ ∈ σ(T ), |λ| ≤ kT k. 22 CHAPTER 2. SPECTRUM OF A BOUNDED OPERATOR Proof. Set a := inf n∈N∗ kT n k1/n . We have a ≤ lim inf kT n k1/n . n→∞ Let ε > 0 and n0 > 0 such that kT n0 kn0 ≤ a + ε. Let n > 0 and p, q integers with 0 ≤ q ≤ n0 and n = n0 p + q. Hence kT n k ≤ kT n0 kp kT kq . Since limn→∞ q n = 0 and lim np = 1 n0 , we deduce that lim sup kT n k1/n ≤ kT n0 k1/n0 ≤ a + ε. cel-00376389, version 1 - 17 Apr 2009 n→∞ Since this is valid for all ε > 0, we get limn→∞ kT n k1/n = a. Now let λ with |λ| > r(T ) and r ∈]r(T ), |λ|[. Since r P > r(T ), there is n0 > 0 such that for all n ≥ n0 , kT n k ≤ rn . The series n≥0 λ−n−1 T n is then normally convergent in L(E) and it is easy to see that     X X (λ − T )  λ−n−1 T n  =  λ−n−1 T n  (λ − T ) = Id n≥0 n≥0 hence λ ∈ ρ(T ). Let ρ := max{|λ|; λ ∈ ρ(T )}. We know that ρ ≤ r(T ). Set for n > 0 and t>ρ Z 2π 1 Jn (t) := (t exp(iθ))n+1 R(teiθ ) dθ. 2π 0 Since i i ∂ h ∂ h (t exp(iθ))n+1 R(teiθ ) = it (t exp(iθ))n+1 R(teiθ ) , ∂θ ∂θ we see that Z i ∂ h (t exp(iθ))n+1 R(teiθ ) dθ = 0 ∂θ 0 P on ]ρ, ∞[. Hence (expanding R(λ) = λ−n−1 T n ) for all t > ρ, Jn (t) = T n , thus kT n k = kJn (t)k ≤ tn+1 Mt , where Mt is the maximum of kR(teiθ )k, for θ ∈ [0, 2π]. Therefore, for all t > ρ, r(T ) ≤ t, since r(T ) = limn→∞ kT n k1/n , and so r(T ) ≤ ρ(T ). ¤ dJn 1 = dt 2itπ 2π We will often use the following (simple) proposition Proposition 2.1.4 Let E, F two Banach spaces and T ∈ L(E, F ). The following are equivalent: 2.1. SPECTRUM 23 (i) T is injective and its image is closed. (ii) There exists K > 0 such that for all x ∈ X we have kT xk ≥ Kkxk. (iii) There is no sequence (xn ) ⊂ E such that kxn k = 1 and limn→∞ kT xn k = 0. cel-00376389, version 1 - 17 Apr 2009 Proposition 2.1.5 Let E, F be two Banach spaces and T ∈ L(E, F ). Then t T ∈ L(F ∗ , E ∗ ) is invertible if and only if T is invertible. Proof. If T is invertible then T −1 T = IdE and T T −1 = IdF . This gives that t T t (T −1 ) = IdE ∗ and t (T −1 )t T = IdF ∗ . Hence t T is invertible and (t T )−1 = t (T −1 ). Conversely, if t T is invertible. Let x ∈ E and x∗ ∈ E ∗ (by Hahn-Banach) with kx∗ k ≤ 1 and x∗ (x) = kxk. Set y ∗ := (t T )−1 x∗ . Then x∗ = t T y ∗ = y ∗ ◦ T and ky ∗ k ≤ Kkx∗ k ≤ K, where K := k(t T )−1 k. Thus kxk = x∗ (x) = y ∗ (T x) ≤ KkT xk. Hence T is injective and its image is closed in F . By Hahn-Banach theorem there exists A ⊂ F ∗ such that ImT = ∩y∗ ∈A kery ∗ . So, for all y ∗ ∈ A, y ∗ is zero on ImT . Hence t T y ∗ (= y ∗ ◦ T ) is zero, and since t T is bijective, y ∗ = 0. Therefore A ⊂ {0}, i.e. ImT = F . ¤ We get directly: Corollary 2.1.1 σ(T ) = σ(t T ). Definition 2.1.2 Let T ∈ L(E) and λ ∈ σ(T ). We distinguish three possibilities: 1. λ is an eigenvalue, i.e. λ − T is not injective. We say that λ is in the point spectrum σp (T ) of T . 2. λ − T is injective but Im(λ − T ) is not dense in E. We say that λ is in the residue spectrum σr (T ) of T . 3. λ − T is injective but its image is not closed. We say that λ is in the continuous spectrum σc (T ) of T . Remarks 1. λ ∈ σr (T ) means that λ is an eigenvalue of t T , but not of T , i.e λ − T is injective but λ − t T is not: there exists then x∗ ∈ E ∗ such that (λ − t T )x∗ = 0 hence x∗ ◦ (λ − T ) = 0 which implies that Im(λ − T ) ⊂ kerx∗ . Then Im(λ − T ) is not dense in E. 2. λ ∈ σc (T ) means that λ ∈ σ(T ) but λ is not eigenvalue of T or of t T . 3. We have σ(T ) = σp (T ) ∪ σr (T ) ∪ σc (T ). 24 CHAPTER 2. SPECTRUM OF A BOUNDED OPERATOR 2.2 Hilbert case In this section we consider the particular case where H is a non trivial Hilbert space. Some properties of bounded self-adjoint operators are given. From proposition 1.3.3 we deduce directly: Corollary 2.2.1 Let T ∈ L(H), then σ(T ∗ ) = σ(T ) = {λ, λ ∈ σ(T )}. cel-00376389, version 1 - 17 Apr 2009 If λ ∈ ρ(T ), then λ ∈ ρ(T ∗ ) and R(λ, T ) = [R(λ, T )]∗ . Moreover σr (T ) = {λ ∈ C\σp (T ); λ̄ ∈ σp (T ∗ )} Proposition 2.2.1 The residue spectrum of a normal operator is empty. Proof. Let T ∈ L(H) a normal operator. For all x ∈ H, we have kT ∗ xk2 =< T ∗ x, T ∗ x >=< x, T T ∗ x >=< x, T ∗ T x >=< T x, T x >= kT xk2 . So kerT ∗ = kerT . Since for all λ, λ − T is normal, we have ker(λ̄ − T ∗ ) = ker(λ − T ). Thus λ̄σp (T ∗ ) if and only if λ ∈ σp (T ). We get the result applying the last corollary. ¤ There is no relation between eigenvalues of T and those of T ∗ : Example. Let E = `2 (N) and T the operator right shift, defined by (T u) = v where v is the sequence defined by v0 = 0 and for all i ≥ 1, vi = ui−1 . T does not admit eigenvalues: σp (T ) = ∅. It is easy to verify that the adjoint of T is the conjugate of the operator left shift and that σp (T ∗ ) = D(0, 1) the open unit disc. Proposition 2.2.2 For all T ∈ L(H) we have kT ∗ T k = kT T ∗ k = kT k2 . Proof. Since kT ∗ k = kT k we have kT ∗ T k ≤ kT k2 . From the other hand, kT xk2 =< T x, T x >=< x, T ∗ T, x >≤ kxk2 kT ∗ T k. Hence kT k2 ≤ kT ∗ T k. Thus kT k2 = kT ∗ T k. ¤ Proposition 2.2.3 The spectral radius of a normal operator is equal to its norm. 2.2. HILBERT CASE 25 Proof. If T is self-adjoint, then < T x, T x >=< x, T 2 x >, thus kT 2 k = kT k2 n n n −n and kT 2 k = kT k2 . Hence r(T ) = limn→∞ kT 2 k2 = kT k. Now if T is normal then we have kT xk2 =< T x, T x >=< x, T ∗ T x >, hence kT ∗ T k = kT k2 . By induction, k(T ∗ T )n k = kT n k2 and then kT ∗ T k = ρ(T ∗ T ) = ρ(T )2 = kT k2 . ¤ This gives directly cel-00376389, version 1 - 17 Apr 2009 Corollary 2.2.2 Let T ∈ L(H), then p p kT k = r(T T ∗ ) = r(T ∗ T ). Proposition 2.2.4 Let T be a self-adjoint operator on H. Then 1. σp (T ) ⊂ R. 2. For all λ ∈ C, Im(λ − T ) = [ker(λ − T )]⊥ . 3. Eigen-spaces associated to distinct eigenvalues are orthogonal. Proof. 1. Let λ ∈ σp (T ) and x ∈ Hλ , i.e. x 6= 0, T x = λx. Then λkxk2 =< T x, x >∈ R since T is self-adjoint, hence λ ∈ R. 2. direct from proposition 1.3.3. 3. If λ 6= µ are two eigenvalues of T and x ∈ Hλ and y ∈ Hµ , then λ < x, y >=< T x, y >=< x, T y >= µ < x, y >. Thus < x, y >= 0. ¤ The following theorem states that, in fact, the whole spectrum is real. Theorem 2.2.1 Let T be a bounded self-adjoint operator on H. Then σ(T ) ⊂ [m, M ], m ∈ σ(T ) and M ∈ σ(T ), where m = inf{< T x, x >, with x ∈ E, kxk = 1} and M = sup{< T x, x >, with x ∈ E, kxk = 1} Proof. Set, for λ ∈ C, d(λ) the distance from λ to the interval [m, M ]. For all x ∈ H, x 6= 0, we have < λx − T x, x >= kxk2 [λ− < T y, y >] , where y := x/kxk. Then by Cauchy-Schwarz inequality we have d(λ)kxk2 ≤ | < λx − T x, x > | ≤ kxk kλx − T xk. (2.3) 26 CHAPTER 2. SPECTRUM OF A BOUNDED OPERATOR Now if λ 6∈ [m, M ], then d(λ) > 0 and then λ − T is injective. Let’s show that ImT is closed. If (λxn − T xn ) is a sequence that converges to y ∈ H, then by equation (2.3), (xn ) is a Cauchy sequence hence convergent to some x ∈ H. Clearly λx − T x = y hence y ∈ Im (λ − T ). By proposition 2.2.4, we have Im(λ − T ) = ker(λ̄ − T )⊥ . Since λ̄ 6∈ [m, M ], Im(λ − T ) = H and hence λ − T is bijective. Therefore λ ∈ ρ(T ). Remainder to show that m, M ∈ σ(T ). Let’s show, for example, that m ∈ σ(T ) (for M consider −T ). Set S := T −m, then S is positive. The mapping (x, y) 7−→< Sx, y > is a scalar product on H. Cauchy-Schwarz inequality for this scalar product gives, for all x, y ∈ H cel-00376389, version 1 - 17 Apr 2009 | < Sx, y > |2 ≤ | < Sx, x > | | < Sy, y > |. (2.4) Now by definition of m, there is a sequence (xn ), kxn k = 1, with limn→∞ | < Sxn , xn > | = 0. Hence by (2.4) 1 1 1 1 kSxn k2 ≤ | < Sxn , xn > | 2 | < Sxn , Sxn > | 2 ≤ | < Sxn , xn > | 2 kSk 2 kSxn k. 1 1 Therefore kSxn k ≤ | < Sxn , xn > | 2 kSk 2 hence tends to zero. If m 6∈ σ(T ), S is invertible and hence xn → 0 which is impossible. ¤ Using proposition 1.3.4 we get Corollary 2.2.3 Let T be a self-adjoint operator on H. Then T is positive if and only if σ(T ) ⊂ R+ . In this case kT k ∈ σ(T ). Proof. Since T is self-adjoint, kT k = supkxk=1 | < T x, x > | hence kT k = max{|m|, |M |}. T positive implies that 0 ≤ m ≤ M and so kT k = M ∈ σ(T ). ¤ Chapter 3 cel-00376389, version 1 - 17 Apr 2009 Symbolic Calculus One of the most important aims of spectral theory is the symbolic calculus: Given a linear operator A, find the functional space A (the best possible) on which one can define f (A), f ∈ A. A good functional space is for example H(O) the space of analytic functions on the open O of the complex plane that contains the spectrum of A. But also, in the case where the spectrum is real, the space C(R). With a functional space we can “translate properties of functions to the operators”. In this chapter we will define such symbolic calculus in the case where A is bounded, and then in the Hilbert case where A is self-adjoint. Later we will deal with the case of unbounded self-adjoint operator... 3.1 Case of bounded operator In all this section X is a Banach space and A a bounded operator, A ∈ L(X). Denote by RA (X) the set of rational fractions without poles in σ(A), i.e. the set of fractions pq , where p, q ∈ C(X) with Zero(q) ∩ σ(A) = ∅. This space will play an important role, since we can define p(A) in a naturel way and hence pq (A). Note that RA is a ring with identity (1) and for all p, q ∈ RA , p(A)q(A) = q(A)p(A). Proposition 3.1.1 There exists a unique linear mapping Φ: RA → L(X) homomorphism of rings verifying Φ(1) = Id and Φ(X) = A. P Proof. existence. For all polynomial p(x) = ak X k ∈ C[X], set Φ(p) := P k ak A ∈ L(X). It is obvious that Φ is linear and Φ(pq) = Φ(p)Φ(q). Now if p is a polynomial (6= 0) with Zero(p) ∩ σ(A) = ∅, then Φ(A) is invertible: Indeed, it suffices to write p(X) = aΠ(X − rk ), where the rk ’s are the roots of p (counted with their multiplicity). Since the rk ’s are not in σ(A), 27 28 CHAPTER 3. SYMBOLIC CALCULUS each A − rk is invertible and so Φ(A). Now if f = pq ∈ RA then Φ(f ) = p(A)[q(A)]−1 . Of course Φ(f ) is independent of the choice of p and q. The linearity of Φ as well as the homomorphism is direct. uniqueness. If Ψ: RA → L(X) is another mapping verifying the same properties, we can show by induction that Ψ(xn ) = An , so by linearity Ψ and Φ coincide on C[X]. ¤ The uniqueness of the mapping Φ justify the following notation: cel-00376389, version 1 - 17 Apr 2009 Notation. The operator Φ(f ) will be denoted: f (A). Remark. This justify the appellation symbolic calculus. In fact, if for all nonnegative integer n, xn is the function xP 7−→ xn , then xn (A) = An . From this we get that for all polynomial p(x) = ai xi , p(A) defined by use of Φ is the same as the “classical” p(A). Theorem 3.1.1 Spectral mapping theorem For all f ∈ RA we have σ(f (A)) = f (σ(A)), and for all g ∈ Rf (A)) , we have g(f (A)) = [g ◦ f ](A). Proof. Let λ ∈ C. If f − λ does not vanish on the spectrum of A then h : (f − λ)−1 ∈ RA and since (f − λ)h = 1 then (f − λ)(A)h(A) = Id. Thus (f − λ)(A) = f (A) − λId is invertible and so σ(f (A)) ⊂ f (σ(A)). Now let λ ∈ C, that is not a pole of f , there exists then h ∈ RA such that f − f (λ) = (x − λ)h. Then f (A) − f (λ) = (A − λ)h(A) = h(A)(A − λ). If f (λ)−f (A) is invertible of inverse R then (T −λ)h(A)S = Id = Sh(A)(A−λ) and so A − λ is invertible, i.e. λ 6∈ σ(A). Thus f (σ(A)) ⊂ σ(f (A)). To terminate, notice that the two mappings Rf (A) → L(X) defined by g 7−→ g(f (A)) and g 7−→ [g ◦ f ](A) verify the conditions of proposition 3.1.1, thus they coincide. ¤ Other type of symbolic calculus could be defined in this framework: Since the spectrum of A is compact hence bounded, let γ be an arbitrary path, that is bounded and turns around σ(T ). γ oriented positively. Briefly, note that the theory of integrals on paths could be generalized for analytic functions defined on a neighborhood O of Imγ into L(X). Notice also that the residue formula (Cauchy) still valid. Therefore, if f ∈ H(O), an analytic function on O valued in L(X) then the formula Z f (z) dz, γ 3.2. CASE OF A BOUNDED SELF-ADJOINT OPERATOR 29 define a bounded operator in L(X). In this framework we can show that the Dunford integral: Z 1 Φ(f ) := f (z)R(z, A) dz 2iπ γ define symbolic calculus on H(O) (that extend the one defined above). 3.2 Case of a bounded self-adjoint operator cel-00376389, version 1 - 17 Apr 2009 In this section, H will be a Hilbert space. Proposition 3.2.1 Let A ∈ L(X) and f ∈ RA . Then f (A)∗ = f˜(A∗ ), where f˜ is defined by f˜(λ) = f (λ) for all λ ∈ C that is not a pole of f . Proof. The mapping RA → L(H) defined by f 7−→ f˜(A∗ )∗ verify the conditions of proposition 3.1.1. ¤ Proposition 3.2.2 If A ∈ L(H) is normal then for all f ∈ RA , f (A) is normal. Proof. For all Y ⊂ L(H) denote by Y 0 := {S ∈ L(H); ST = T S ∀T ∈ Y }. Y 0 is a closed subspace and a sub-ring of L(H). Moreover, if S ∈ Y 0 and S invertible then S −1 T = S −1 T SS −1 = S −1 ST S −1 = T S −1 , for all T ∈ Y . In other words, S −1 ∈ Y 0 . Therefore, if S ∈ Y 0 and f ∈ RS , then f (S) ∈ Y 0 . Let Y = {A, A∗ } and Z = Y 0 . Since all elements of Y commutes with all elements of Z, we see that Y ⊂ Z 0 . So f (A), f˜(A∗ ) ∈ Z. Since A is normal, Y ⊂ Z, then Z 0 ⊂ Y 0 = Z. Thus f (A) ∈ Z 0 and f˜(A∗ ) = f (A)∗ ∈ Z 0 ⊂ Z so they commute, i.e. f (A) is normal. ¤ Proposition 3.2.3 1. The spectrum of any unitary operator of L(H) is included into the unit circle C(0, 1) of the complex plane. 2. The spectrum of any self-adjoint operator of L(H) is included into the real line R. Proof. 1. Let U ∈ L(H) and λ ∈ σ(U ). Since kU k ≤ 1 the spectral radius of U is less than 1 or equal, hence |λ| ≤ 1, and since U is bijective λ 6= 0 and by theorem 3.1.1, λ−1 ∈ σ(U −1 ). But U −1 = U ∗ so kU −1 k ≤ 1 thus |λ−1 | ≤ 1, i.e. |λ| = 1. 2. Let A ∈ L(H) a self-adjoint operator. For all real t, with t > kAk, A ± tId are invertible. Denote by f the mapping X 7−→ (X + ti)/(X − ti). Since f˜ = f −1 , by proposition 3.2.1, f (A)∗ = f˜(A) = f (A)−1 thus f (A) is 30 CHAPTER 3. SYMBOLIC CALCULUS unitary and by the first point σ(f (A)) ⊂ C(0, 1) = f (R). Using theorem 3.1.1 we get the result. ¤ cel-00376389, version 1 - 17 Apr 2009 Notations. If K is a compact space, denote by C(K) the Banach space of continuous functions from K into C with the supremum norm: kf k∞ := sup{|f (x)|; x ∈ K}. If K is a compact of C, denote by z n , for all integer n, the mapping λ 7−→ λn . Theorem 3.2.1 Let H be a Hilbert space and A ∈ L(H) a self-adjoint or unitary operator. There exists a unique linear continuous mapping Φ: C(σ(A)) → L(H) verifying Φ(1) = Id, Φ(z) = A and for all f, g ∈ C(σ(A)), we have Φ(f g) = Φ(f )Φ(g). For all f ∈ RA , we have Φ(f ) = f (A). Moreover, Φ is an isometry, and for all f ∈ C(σ(A)) we have Φ(f )∗ = Φ(f ). Proof. Let φ: RA → C(σ(A)) defined by φ(f ) = f|σ(A) and Ψ: RA → L(H) defined by Ψ(f ) = f (A). For all f ∈ RA , f (A) is normal by proposition 3.2.2 hence kf (A)k is equal to its spectral radius. Thus by theorem 3.1.1, kf (A)k = sup{|λ|, λ ∈ σ(f (A))} = sup{|f (λ)|, λ ∈ σ(A)}. Therefore kΨ(f )k = kφ(f )k∞ . Now if A is unitary then z̄ = z −1 = φ(X −1 ) ∈ φ(RA ), and if A is self-adjoint then z̄ = z = φ(X) ∈ φ(RA ). In both cases z̄ ∈ φ(RA ). Now for all f ∈ RA , f (z) = f˜(z̄) ∈ φ(RA ). Therefore φ(RA ) is sub-vector space and sub-ring of C(σ(A)) that contains constants (since 1 = φ(1)) stable under conjugate and separate points of σ(A) (since z ∈ φ(RA )), so by Stone-Weierstrass theorem, φ(RA ) is dense in C(σ(A)). Therefore there exists a unique linear continuous mapping Φ: C(σ(A)) → L(H) such that Ψ = Φ ◦ φ. We have Φ(1) = Φ(φ(1)) = Ψ(1) = Id. Φ(z) = Φ(φ(X)) = Ψ(X) = A. The mappings that to (f, g) ∈ C(σ(A)) × C(σ(A)) associates respectively Φ(f g) and Φ(f )Φ(g) coincide on Φ(RA ) × Φ(RA ) so they are equals. Moreover the set of functions f ∈ C(σ(A)) with kf (A)k = kf k∞ is closed and contains Φ(A). Hence Φ is an isometry. Finally, for f ∈ RA , we have f (A)∗ = f˜(A∗ )Φ(f˜(z̄)) = Φ(f (z)). The set of functions f ∈ C(σ(A)) such that Φ(f )∗ = Φ(f¯) is closed and contains φ(A), hence this true for all f ∈ C(σ(A)). Remains to show uniqueness. If Φ1 is another one, then Φ ◦ φ and Φ1 ◦ φ coincide on φ(RA ). By density we get Φ = Φ1 . ¤ Notation. For all f ∈ C(σ(A)), denote by Φ(f ) = f (A). Theorem 3.2.2 Spectral mapping theorem Let A be a self-adjoint, or unitary operator and f ∈ C(σ(A). Then 3.2. CASE OF A BOUNDED SELF-ADJOINT OPERATOR 31 1. f (A) is normal and σ(f (A)) = f (σ(A)). cel-00376389, version 1 - 17 Apr 2009 2. If f (σ(A)) ⊂ R then f (A) is self-adjoint. If f (σ(A)) ⊂ C(0, 1) then f (A) is unitary. Moreover, in these cases, for all g ∈ C(σ(f (A))) we have g ◦ f (A) = g(f (A)). Proof. 1. We have f (A)∗ = f¯(A), so f (A)f (A)∗ = [f f¯](A) = f (A)∗ f (A), so f (A) is normal. Now, if λ 6∈ f (σ(A)), let h ∈ C(σ(A)) the function s 7−→ 1/[f (s) − λ]. Since h(f − λ) = (f − λ)h = 1, then h(A)[f (A) − λ] = [f (A) − λ]h(A) = Id and so λ 6∈ σ(f (A)). Conversely, if λ ∈ f (σ(A)), for ε > 0, set f1 := f − ε and g := ε/(|f1 | + ε). Notice that kgk∞ = 1, and since |f1 g|(t) = ε|f1 (t)|/[|f1 (t)| + ε] so kf1 gk < ε. Since Φ is isometry we have kg(A)k = 1 and kf1 (A)g(A)k < ε. Since kg(A)k = 1 > kf1 gk∞ /ε, there exists x ∈ H such that kg(A)xk > kf1 gk∞ kxk/ε and so kf1 (A)g(A)xk ≤ kf1 (A)g(A)kkxk < εkg(A)xk. Thus there is y = g(A)x such that kf1 (A)yk < εkyk. Thus f1 (A) = f (A) − λ is not injective hence λ ∈ σ(f (A)). 2. If f = f¯ then f (A) = f¯(A) = f (A)∗ . If f (σ(A)) ⊂ C(0, 1) then f f¯ = 1, hence f (A)f (A)∗ = f (A)∗ f (A) = [f f¯](A) = Id; f (A) is unitary. Finally the mapping g 7−→ [g ◦ f ](A) verifies the conditions of the last theorem, hence coincides with g 7−→ g(f (A)). ¤ Theorem 3.2.3 Let A ∈ L(H). The following conditions are equivalent: (i) For all x ∈ H, < Ax, x >∈ R+ . (ii) There exists S ∈ L(H), A = S ∗ S. (iii) There exists S ∈ L(H) self-adjoint, A = S 2 . (iv) A is self-adjoint and σ(A) ⊂ R+ . Proof. (ii)⇒(i): < T x, x >=< S ∗ Sx, x >=< Sx, Sx >≥ 0. (iii)⇒(ii) is direct. √ (iv)⇒(iii): Assume that A∗ = A and σ(T ) ⊂ R+ . Denote by f : t 7−→ t. Then by the last theorem, we have f (A) = f (A)∗ , moreover f (A)2 = A. (i)⇒(iv): The mapping (x, y) 7−→< Ax, y > is sesqui-linear and < Ay, x >= < Ax, y > hence A is self-adjoint. Thus σ(A) ⊂ R. Let t < 0 and let’s show that A − t is bijective. For all x ∈ H, we have −tkxk2 ≤ −tkxk2 + < Ax, x >=< (A − t)x, x >≤ k(A − t)xkkxk so −tkxk ≤ k(A − t)xk and so A − t is injective with closed graph. Now since the residual spectrum of every normal operator is empty, we get the result. ¤ 32 CHAPTER 3. SYMBOLIC CALCULUS Definition 3.2.1 (Fractional powers) If A ∈ L(H) is self-adjoint and positive and α ∈]0, +∞[, set Aα = fα (A) where fα is the mapping t 7−→ tα . Remark. By theorem 3.2.2, we have, for all α, β > 0 • A1 = A, • (Aα )β = Aαβ , cel-00376389, version 1 - 17 Apr 2009 • Aα Aβ = Aα+β . Corollary 3.2.1 (Square root) For all positive self-adjoint operator A ∈ 1 L(H), square root of A, A 2 is a positive self-adjoint operator. Chapter 4 cel-00376389, version 1 - 17 Apr 2009 Compact operators In this chapter we will study spectral properties of some particular type of operators: compact operators and Hilbert-Schmidt operators. We will see also Fredholm alternative. 4.1 General properties In all this section, E and F are two Banach spaces. Definition 4.1.1 A ∈ L(E, F ) is called compact if the image of the closed unit ball of E, A(BE (0, 1)) is relatively compact in F . Denote by K(E, F ) the set of compact operators from E into F and K(E) = K(E, E). Remarks. 1. A ∈ K(E, F ) if and only if the image by A of any bounded subset of E is relatively compact in F . 2. A ∈ K(E, F ) if and only if the image by A of any bounded sequence of E is a sequence of F with convergent subsequences. 3. Riesz theorem becomes: Id ∈ K(E, E) if and only if the dimension of E is finite. Examples. 1. Every operator T of finite rank, i.e. dim ImT < ∞ is compact. In fact, the image T (B̄) is bounded in a finite dimensional space hence relatively compact in ImT hence relatively compact in F . 33 34 CHAPTER 4. COMPACT OPERATORS 2. Let X, Y be two compact metric spaces, K ∈ C(X × Y ) and µ any Radon measure on Y . Define the kernel operator TK , for all f ∈ C(Y ) by Z (TK f )(x) := K(x, y)f (y) dµ(y). TK is compact operator. 3. Let a < b, K ∈ C([a, b]2 ) and α, β two continuous functions from [a, b] into itself. For f ∈ C([a, b]) and x ∈ [a, b] set Z β(x) T f (x) := K(x, y)f (y) dy. cel-00376389, version 1 - 17 Apr 2009 α(x) The operator T is compact: T ∈ K(C([a, b]). In fact, for all f ∈ E where E := C([a, b]), we have kT f k ≤ M kKk kf k, where M := supx∈[a,b] |β(x) − α(x)|. Hence T (B̄) is a bounded in E. From the other hand, for all x, y ∈ [a, b] and all f ∈ E we have |T f (x) − T f (y)| ≤ Mx,y kf k, where Mx,y := kKk (|β(x) − β(y)| + |α(x) − α(y)|) +(kαk∞ + kβk∞ ) sup |K(x, z) − K(y, z)|. z∈[a,b] Uniform continuity of K on [a, b]2 implies that T (B̄) is equicontinuous in E. We conclude using Ascoli’s theorem. 4. Integration operator Z T f (x) := x f (t) dt a is a compact operator on C([a, b]). Proposition 4.1.1 Let R ∈ K(E, F ), T ∈ L(E1 , E), S ∈ L(F, F1 ) where E1 and F1 are normed spaces. Then SRT is a compact operator. Proof. Indeed, ³ ´ SRT (BE ) ⊂ kT kS R(BE ) . Continuous image of a compact being a compact, we get the result. ¤ 4.2. SPECTRAL PROPERTIES OF COMPACT OPERATORS 35 Proposition 4.1.2 K(E, F ) is a closed sub-vector space of L(E, F ). Proof. Let T, S be two compact operators from E to F and λ, µ ∈ K. Then cel-00376389, version 1 - 17 Apr 2009 (λT + µS)(BE ) ⊂ λT (BE ) + µS(BE ), and this last set is compact since if K and H are two compacts then λK +µH is compact as continuous image of K ×H. To show that K(E, F ) is closed, let (Tn ) be a sequence of compact operators that converges to T (in L(E, F )). It suffices to show that T B̄E . Let ε > 0, and n ∈ N such that kT − Tn k ≤ ε/3. Let f1 , · · · , fk ∈ B̄E such that the balls B(Tn fi , ε/3) is a cover of Tn B̄E . Let then f ∈ B̄E and let j ≤ k such that kTn f − Tn fj k ≤ ε/3. By triangle inequality we get kT f − T fj k < ε. Hence [ B(T fj , ε), T B̄E ⊂ 1≤j≤k thus T B̄E is precompact. ¤ Since every finite rank operator is compact, we get Corollary 4.1.1 Every limit of operators of finite rank is a compact operator. We terminate this section by the Schauder Theorem: Theorem 4.1.1 Let T ∈ L(E, F ). T is compact if and only if t T is compact. Proof. 4.2 Spectral properties of compact operators In all this section E is a Banach space and T a compact operator. Lemma 4.2.1 Let F be a closed sub-vector space of a normed vector space E, F 6= E, then there exists u ∈ E, kuk = 1 with d(u, F ) ≥ 12 . Proof. Let v ∈ E\F and δ := d(v, F ). Let w ∈ F with kv − wk < 2δ. Take v−w u := kv−wk . ¤ Proposition 4.2.1 Let T ∈ K(E). Then 36 CHAPTER 4. COMPACT OPERATORS 1. the sub-vector-space ker(I − T ) is of finite dimension. 2. the sub-vector-space Im(I − T ) is closed cel-00376389, version 1 - 17 Apr 2009 3. the operator I − T is invertible in L(E) if and only if it is injective. Proof. 1. Denote by F := ker(I − T ). F is a closed sub-vector-space of E and B̄F = T B̄F = T B̄E ∩F hence compact, so by Riesz theorem dimF < ∞. 2. Let y ∈ Im(I − T ) and (xn ) a sequence of E with lim xn − T xn = y. First case: The sequence (xn ) is bounded. Since T is compact, by choosing a subsequence we can assume that (T xn ) converges to z ∈ E. Then lim xn = y+z and by continuity of T , z = T (y+z) hence y = (y+z)−T (y+z). Second case: The sequence is not bounded. Set, for n ≥ 0, dn := d(xn , ker(I− T )). Since, by the first point, dimker(I−T ) < ∞, there exists zn ∈ ker(I−T ) with dn = kxn − zn k (since the continuous function distance will attain its minimum on the nonempty compact B(xn , kxn k) ∩ ker(I − T )). If the sequence (dn ) is bounded we can replace (xn ) by (xn − zn ) (since T zn = zn ) and apply the first case. If not, using a subsequence, we can suppose that lim dn = ∞. Since the sequence ((xn − zn )/dn is bounded, we can assume, by use of subsequence, that T [(xn − zn )/dn ] is convergent to some u ∈ E. We deduce that y xn − zn = u + lim = u, lim n→∞ dn dn which implies that T u = u and for n large, kxn − zn − dn uk < dn which is impossible and so the sequence (dn ) is bounded and y ∈ Im(I − T ). 3. Assume that I − T is injective, set E1 := Im(I − T ) and suppose that E1 6= E. Set for all n, En := Im(I − T )n with E0 := E. Let’s show by induction that for all n, En is closed and En+1 ( En . This is true for n = 0. Assume it true for n. Clearly T En ⊂ En and hence T induces Tn ∈ L(En ). Since En is closed Tn BEn ⊂ T B̄E ∩ En which is compact. Hence Tn is compact on En . Since En+1 = (IdEn − Tn )En , then by the second point, En+1 is closed in En and hence in E. It is obvious that En+1 ⊂ En+2 . Now since I − E is injective we get, En 6= En+1 implies that En+1 6= En+2 since En+1 = (I − T )(En ) and En+2 = (I − T )(En+1 ). To find a contradiction, by the last lemma, there is a sequence (un ) such that for all n, un ∈ E, kun k = 1 and d(un , En+1 ) ≥ 21 . Then for n < m, T un − T um = un − vn,m with vn,m = T um +(I −T )un ∈ En+1 . Thus for all n 6= m, kT un −T um k ≥ 12 . This is in contradiction with the compacity of T B̄. Thus I − T is surjective. Remainder to show continuity of (I − T )−1 . By contradiction, suppose that there is a sequence (xn ) 6→ 0 with lim xn − T xn = 0. By use of subsequence, we can assume that for all n, kxn k ≥ ε, for some ε > 0. Set un := xn /kxn k. Again, since T is compact, we can assume that (T un ) converges to some v ∈ E. But this will imply that lim un = v and so kvk = 1 and then by continuity T v = v, so (I −T )v = 0 which is impossible since I −T is injective. ¤ 4.3. HILBERT-SCHMIDT OPERATORS 37 Theorem 4.2.1 Let T ∈ K(E). 1. If dimE = ∞ then 0 ∈ σp (T ). 2. σ(T )\{0} = σp (T ) and for all λ ∈ σp (T ) dim Eλ < ∞. cel-00376389, version 1 - 17 Apr 2009 3. σ(T ) is countable. Proof. 1. If 0 is not an eigenvalue then by proposition 4.1.1 I = T T −1 is compact and so dim E < ∞. 2. Let λ ∈ K, λ 6= 0. λ ∈ σ(T ) if and only if I − T /λ is not injective and ker(λ − T ) = ker(I − T /λ). On the other hand, λ ∈ σ(T ) if and only if I − T /λ is not invertible in L(E). Apply then the last proposition. 3. For this it suffices to show that for all ε > 0 there is a finite number of λ ∈ σ(T ) with |λ| ≥ ε. If not, assume that there is a sequence (λn ) ⊂ σ(T ) of distinct elements with |λn | ≥ ε. By the last point λn are eigenvalues. Let then (en ) corresponding eigenvectors with ken k = 1. Thus the (en ) is a free system. For all n, set En := span{e0 , · · · , en }. The (En is a sequence of strictly increasing of finite dimension spaces. From lemma 4.2.1 there exists a sequence of (un ), kun k = 1 and un ∈ En+1 with d(un , En ) ≥ 12 . n Set vn := λun+1 . This sequence is bounded by 1ε and for n > m we have 1 T vn − T vm = un − vn,m with vn,m = T vm + λn+1 (λn+1 − T )un . Since T vm ∈ Em+1 ⊂ En and (λn+1 − T )En+1 ⊂ En , we get vn,m ∈ En and kT vn − T vm k ≥ 21 which is impossible since T is compact. ¤ 4.3 Hilbert-Schmidt operators In this section E and F are two separable Hilbert spaces (of infinite dimensions). Lemma 4.3.1 Let B and B 0 be two Hilbert bases of E and F respectively. For all T ∈ L(E, F ) we have: X X X kT bk2 = kT ∗ b0 k2 ≤ +∞, | < b0 , T b > |2 = b∈B, b0 ∈B 0 b0 ∈B 0 b∈B and this value does not depends on the choice of B or B 0 . P 2 Proof. PFor x ∈ E and y ∈ F we have kxk2 P = b∈B | < x, b > | and 2 0 2 2 kyk = bP 0 ∈B 0 | < y, b > | . Now it is clear that b∈B kT bk is independent of B 0 and b0 ∈B 0 kT ∗ b0 k2 is independent of B. ¤ Notation. For all T ∈ L(E, F ), set " kT k2 := X b∈B #1 2 kT bk2 38 CHAPTER 4. COMPACT OPERATORS where B is any base of E. Set L2 (E, F ) the set L2 (E, F ) := {T ∈ L(E, F ); kT k2 < ∞}. Examples. 1. Finite dimensional case. If E = F has finite dimension n, and (ej ) a basis formed of eigenvectors of T ∗ T , then cel-00376389, version 1 - 17 Apr 2009 kT k22 = n X < T ∗ T ek , ek >= k=1 n X λk , k=1 where (λk ) are the eigenvalues of T ∗ T . If T ∗ = T then n X kT k22 = βk2 , k=1 where (βk ) are the eigenvalues of T . 2. Let H := L2 (0, 2π) and define the Volterra operator, for all f ∈ H by Z x f (t) dt. V f (x) := 0 By the example 4, this operator is compact. Consider the basis en (t) := √1 eint , n ∈ Z. It is easy to verify that kV en k2 ≤ 2 2 and so V is a πn 2π Hilbert-Schmidt operator. Theorem 4.3.1 Let E, F be two separable Hilbert spaces. 1. L2 (E, F ) is sub-vector space of L(E, F ). P 2. For al S, T ∈ L2 (E, F ) and all Hilbert basis P B of E, b∈B < T b, Sb > is finite and the mapping (S, T ) 7−→ b∈B < T b, Sb > is a scalar 2 product on L (E, F ) (independent of the choice of B). 3. With this scalar product L2 (E, F ) is a Hilbert space. 4. L2 (E, F ) ⊂ K(E, F ). Proof. 1 Let S, T ∈ L(E, F ) and B a Hilbert basis of E. For all b ∈ B we have | < Sb, T b > | ≤ kSbkkT bk ≤ 12 [kSbk2 + kT bk2 ]. We deduce that P < Sb, T b > is finite. Since kSb+T bk2 = kSbk2 +kT bk2 +2Re < Sb, T b >, S + T ∈ L2 and so the first point is proved. cel-00376389, version 1 - 17 Apr 2009 4.3. HILBERT-SCHMIDT OPERATORS 39 P 2. It is clear that (S, T ) 7−→ < Sb, T b > is a scalarPproduct. Now by the polarization identity (proposition 1.1.1) we have 4 < Sb, T b >= kS + T k22 − kS − T k22 + ikS + iT k22 − ikS − iT k22 we get the independence of the basis. 3. From the second point L2 is a pre-Hilbert space. For all T ∈ L2 and all x ∈ E, kxk = 1, by taking a Hilbert basis containing x we get that kT k2 ≥ kT xk, so kT k2 ≥ kT k. Thus L2 is separate. By this inequality, if (Tn ) is a Cauchy sequence in L2 , it is also a Cauchy sequence in L which is complete, so (Tn ) converges to an operator T ∈ L(E, F ). Now for ε > 0 notice that the set Cε := {S ∈ LP 2 ; kSk2 ≤ ε} is the intersection on all finite subset I ⊂ B, of {S ∈ L2 ; b∈I kSbk2 ≤ ε2 }, hence Cε is a closed 2 set in L(E, F ) (Indeed, this last set is the inverse P image of2 [0, ε ] by the 2 continuous mapping which to S ∈ L associates b∈I kSbk ). Fix ε > 0. There is N > 0 such that for all m, n ≥ N we have kTn − Tm k2 ≤ ε. Fix n ≥ N and since Tm − Tn → T − Tn , we get that T − Tn ∈ Cε . Thus T ∈ L2 and that lim kT − Tn k2 = 0. 4. Let T ∈ L2 and (en ) a basis of E. For all P k, consider the operator Tk : E −→ F defined, for all x ∈ E, by Tk x := n≤k < x, en > T en . Since Tk is of finite rank, using corollary 4.1.1, it suffices to show that Tk → T . For this, we write 1   k(T − Tk )xk = k X < x, en > T en k ≤  n≥k n≥k 2 2 | < x, en > |  1 X 2 2 kT en k n≥k 1  ≤ kxk X X 2 2 kT en k . n≥k ¤ Definition 4.3.1 Operators in L2 (E, F ) are called Hilbert-Schmidt operators. Proposition 4.3.1 Let E, F, H Hilbert spaces. For all S ∈ L(E, F ) and T ∈ L(F, H) we have: 1. kSk2 = kS ∗ k2 . 2. If T or S is a Hilbert-Schmidt operator then T S is a Hilbert-Schmidt operator also and kT Sk2 ≤ kT kkSk2 or kT Sk2 ≤ kT k2 kSk. 40 CHAPTER 4. COMPACT OPERATORS Proof. 1. Direct from the definition of kSk2 . 2. Let B be a Hilbert basis of E. For all Pb ∈ B we have kT Sbk ≤ kT k kSbk P hence kT Sk22 = b∈B kT Sbk2 ≤ kT k2 kSbk2 = kT k2 kSk22 . The second point could be obtained substituting S and T by their adjoints. 3. Similarly as 2. ¤ cel-00376389, version 1 - 17 Apr 2009 4.4 Compact self-adjoint operators A classic theorem of linear algebra shows that every normal matrix, i.e. a matrix that commutes with its adjoint, in a finite dimensional complex Hilbert space, is diagonalizable in an othonormal base. We will generalize this result to infinite dimensional case, but for compact self adjoint operators. Generalization to normal compact operators could be done. To omit compacity of the operator we need a very powerfull theory as spectral measures or distributions. Assume that T is an operator of finite rank. Since kerT = (ImT )⊥ and since dimImT < ∞ we have H = ImT ⊕ kerT . Thus T induce on the finite dimensional space ImT an invertible self-adjoint operator, whose eigenvalues are those (6= 0) of T . Since we can diagonalize in finite dimension, we get that ImT is direct sum of (orthogonal) eigen-sub-spaces of T , associated to nonzero eigenvalues of T and then M ker(λ − T ). H= λ∈σp (T ) We have proved the diagonalization of a finite rank operator. In the following we will generalize this result to the case of a compact self-adjoint operator. In the following H is a Hilbert space, and T a compact self-adjoint operator on H (not of finite rank). Lemma 4.4.1 T admits at least one eigenvalue and kT k = max{|λ|; λ ∈ σp (T )}. Proof. Clearly, if λ ∈ σp (T ), then |λ| ≤ kT k. Now, by theorem 2.2.1 there is λ ∈ σ(T ) such that |λ| supkxk=1 | < T x, x > |, which is equal to kT k. ¤ Theorem 4.4.1 Let H be a Hilbert space and T a compact self-adjoint operator. For all λ ∈ σp (T ) denote by Hλ the eigen-space associated to λ. Then 4.4. COMPACT SELF-ADJOINT OPERATORS 41 1. σp (T ) is bounded, countable and infinite subset of R, whose unique accumulation point is 0. 2. For all λ ∈ σp (T )\{0}, dimHλ < ∞. 3. For all λ 6= µ ∈ σp (T ), Hλ and Hµ are orthogonal. 4. Spectral decomposition of the identity. Denote for all λ ∈ σp (T )\{0}, Pλ the orthogonal projection on Hλ . Then X T = λPλ . cel-00376389, version 1 - 17 Apr 2009 λ∈σp (T )\{0} Proof. a. Assume that T is not of finite rank. The fact that eigenvalues of T are real and the orthogonality of eigen-spaces was shown in proposition 2.2.4 b. Let’s show that Λ∗ := σp (T )\{0} is infinite. By lemma 4.4.1, there exists λ ∈ σp (T ), |λ| = kT k. Since T is not trivial, then λ 6= 0 and so Λ∗ is not empty. Assume that Λ∗ is finite, Λ∗ = λ1 , · · · λk . Set then G := ⊕kj=1 Hλj and F := G⊥ . Since G is of finite dimension, H = F ⊕ G. It is clear that T G ⊂ G and since T is self-adjoint, T F ⊂ F . T induces then an operator TF from F into itself, and since F is closed, TF is compact also. If TF = 0 then ImT ⊂ G and so T is of finite rank. Thus TF is a self-adjoint non trivial operator on F . By lemma 4.4.1, TF has a non zero eigenvalue µ. But this means that µ ∈ σp (T )\Λ∗ , since for example there is x ∈ F , x 6= 0, TF x = T x = µx (so x 6∈ G). This gives a contradiction and Λ∗ is infinite and by theorem 4.2.1, σp (T ) is countable, and so 0 is the unique accumulation point. c. Let J be a finite subset of Λ∗ and GJ := ⊕λ∈J Hλ , FJ := G⊥ J . T induces on FJ a compact self-adjoint operator, whose norm kTFJ k = max{|λ|, λ ∈ σp (TFJ )}. But every eigenvalue λ of TFJ is an eigenvalue of T does not belongs to J, since by construction Fj ∩ Hµ = {0} for all µ ∈ J. Therefore, σp (TFJ ) ⊂ σp (T )\J. Conversely, if λ ∈ σp (T )\J, then (by orthogonality), Hλ ⊂ G ⊥ λ = FJ and hence λ is an eigenvalue of TFJ . Thus σp (TFJ ) = σp (T )\J and kTFJ k = max{|λ|, λ ∈ σp (T )\J}. P Moreover, P the orthogonal projection on GJ is λ∈J Pλ . Hence, for all x ∈ E, xJ := x − λ∈J Pλ x ∈ FJ and kT xJ k = kTFJ xJ k ≤ kTFJ k kxk ≤ kTFJ k kxk. We deduce that, ° ° ° ° X ° ° T Pλ ° ≤ max |λ|, °T − ° λ∈σp (T )\J ° λ∈J 42 CHAPTER 4. COMPACT OPERATORS and so ° ° ° ° X ° ° λPλ ° ≤ max |λ|. °T − ° ° λ∈σp (T )\J λ∈J Let ε > 0. Since 0 is an accumulation point of σp (T ), the set K := {λ ∈ σp (T ), |λ| ≥ ε} is finite. Thus, for all finite part J ⊂ σp (T )\{0} that contains K, we have ° ° ° ° X ° ° λPλ ° ≤ max |λ| ≤ max |λ|, °T − ° ° λ∈σp (T )\J λ∈σp (T )\K λ∈J cel-00376389, version 1 - 17 Apr 2009 which terminate the proof. ¤ Corollary 4.4.1 With the same notations we have M ImT = Hλ . λ∈σp (T )\{0} P Proof. We know that, for all x ∈ H, T x = λ∈σ(T ) λPλ x. Thus ImT ⊂ ⊕λ∈σp (T )\{0} Hλ . Conversely, if λ ∈ σp (T )\{0}, then Hλ ⊂ ImT . ¤ We can express the last theorem and corollary in the following Corollary 4.4.2 The space ImT admits a countable Hilbert basis (fn )n∈N formed of eigenvectors of T associated to nonzero eigenvalues (µn )n∈N . The sequence (µn )n∈N tends to zero and, for all x ∈ H, we have X Tx = µn < x, fn > fn . n∈N Corollary 4.4.3 For all x ∈ ImT X x= Pλ x. λ∈σp (T )\{0} Corollary 4.4.4 Let P0 be the orthogonal projection on H0 := kerT . Then for all x ∈ E X x= Pλ x, λ∈σp (T ) and H= M λ∈σp (T ) Hλ . 4.5. FREDHOLM EQUATION 43 ⊥ Proof. Since T is self-adjoint, H0 = kerT = ImT . Hence, H = H0 ⊕ImT .¤ Corollary 4.4.5 If H is a separable Hilbert space, then it admits a Hilbert basis formed of eigenvectors of T . Proof. By corollary 4.4.2, ImT admits a countable Hilbert basis. Complete it by a basis of H0 (formed of eigenvectors associated to 0) to get a Hilbert basis of H formed of eigenvectors. ¤ cel-00376389, version 1 - 17 Apr 2009 4.5 Fredholm equation In this case, in the case of a compact self-adjoint operator, for all bounded function f on the set σp (T ), we can define the operator f (T ) on H as X f (T )x := f (λ)Pλ x λ∈σp (T ) for all x ∈ H. By the orthogonality of the spaces Eλ we get the following (Bessel) equalities X kf (T )xk2 = |f (λ)|2 kPλ xk2 , λ∈σp (T ) X kxk2 = kPλ xk2 . λ∈σp (T ) We deduce then that kf (T )k = sup |f (λ)|. λ∈σp (T ) This shows that this symbolic calculus is an extension of the previous one. In particular, if µ ∈ K∗ 6∈ σp (T ), then for all x ∈ H X (µ − T )−1 x = (µ − λ)−1 Pλ x. (4.1) λ∈σp (T ) Now if µ ∈ σp (T ), µ 6= 0, then Im(µ − T ) = Eµ⊥ . Hence the operator T induces on Eµ⊥ a compact self-adjoint operator Tµ with σp (Tµ ) = σp (T )\{µ}, to Tµ we can again apply the formula (4.1) and deduce that, if x ∈ Eµ⊥ , then for all u ∈ Eµ⊥ we have the equivalence µu − T u = x ⇐⇒ u = X (µ − λ)−1 Pλ x. λ∈σp (T )\{µ} 44 CHAPTER 4. COMPACT OPERATORS Now if x ∈ Eµ⊥ and y ∈ E, then y = u + v with u ∈ Eµ⊥ and v ∈ Eµ . Thus X µy − T y = x ⇐⇒ ∃v ∈ Eµ s.t. y = u + (µ − λ)−1 Pλ x. λ∈σp (T )\{µ} In short, if we consider the Fredholm equation µy − T y = x, (4.2) cel-00376389, version 1 - 17 Apr 2009 with µ ∈ K∗ and x ∈ E, then we can distinguish two cases (Fredholm alternative): • µ is not an eigenvalue of T . Then the equation (4.2) admits a unique solution y, given by X (µ − λ)−1 Pλ x. y = (µ − T )−1 x = λ∈σp (T ) • µ is an eigenvalue of T . Then the equation (4.2) – admits an infinite number of solutions if x ∈ ker(µ − T )⊥ , in this case those solutions are given by X (µ − λ)−1 Pλ x, y =u+ λ∈σp (T )\{µ} with u ∈ ker(µ − T ). – does not admit any solution if not, i.e. if x 6∈ ker(µ − T )⊥ . cel-00376389, version 1 - 17 Apr 2009 Chapter 5 Unbounded self-adjoint operators In this chapter we start by giving some properties of closed operators, then general properties of symmetric and self-adjoint operators on a Hilbert space. We terminate by defining symbolic calculus of unbounded self-adjoint operators. In all this chapter H will be a Hilbert space and X a Banach space. 5.1 Closed operators Definition 5.1.1 Let D ⊂ X be a sub-vector space. A linear unbounded operator is a linear mapping from D to X. Remarks. 1. An operator is always a couple (A, D). D, denoted sometimes by D(A) or DA , is called domain of A. 2. Changing the domain could change considerably the operator. See examples below. 3. In all this chapter we will always use densely defined operators, i.e. such that D(A) = X. 4. Two operators (A, DA ) and (B, DB ) are equals if and only if DA = DB and for all x ∈ DA we have Ax = Bx. And we say that B is an extension of A, A ⊂ B, if DA ⊂ DB and for all x ∈ DA we have Ax = Bx. 45 46 CHAPTER 5. UNBOUNDED SELF-ADJOINT OPERATORS 5. The appellations ”bounded”, ”unbounded” are due to the fact that for linear operators, continuity is equivalent to the inequality: kAxk ≤ Ckxk for some C > 0 and all x ∈ X, i.e. to boundedness on the closed unit ball. Examples. cel-00376389, version 1 - 17 Apr 2009 1. Let X := BC 1 (R) the space of continuously bounded differentiable d functions and A := dx . Clearly A is a linear bounded operator. Observe that for all n > 1, the operator (An , Dn ) defined by: Dn := C n (R) and An f := f (n) is an unbounded, densely defined operator. 2. Let X := BC ∞ (R). For all n ≥ 1, An := (d/dx)n is a linear bounded operator. 3. Let X := L2 (]0, 1[) and define the operator (A, DA ) with DA := {f ∈ C 1 ([0, 1]); f (0) = f (1) = 0} and A := d/dx is linear unbounded densely defined operator (since D(0, 1) ⊂ DA ). 4. On the same space X := L2 (]0, 1[) define the operator (B, DB ) with DB := {f ∈ C 1 ([0, 1]); f (0) = 0, f (1) = 1} and B := d/dx is linear unbounded (but not densely defined) operator. The notion of operators whose graph is closed will play an important role: Definition 5.1.2 The operator (A, DA ) is called a closed operator if and only if for any (xn ) ⊂ DA such that xn → x ∈ X and Axn → y ∈ X it follows that x ∈ DA and y = Ax. Remarks. 1. (A, DA ) closed is equivalent to G(A) := {(x, Ax); x ∈ DA } (the graph of) is closed in X × X. 2. By linearity this definition is equivalent to the following: for any (xn ) ⊂ DA such that xn → 0 then Axn → 0. 3. The closure of an operator (if it exists) (A, DA ) is the least closed extension of A. We say in this case that A is closable. It is denoted by Ā. It is the operator whose graph is G(A). 4. If D ⊂ DA is a sub-vector space denote by A|D, called the part of A on D, the operator such that A|D ⊂ T with domain D(A|D) = {x ∈ D; T x ∈ D}. 5.1. CLOSED OPERATORS 47 If an operator (A, DA ) is injective, the operator A−1 : ImA 7→ X is defined. Definition 5.1.3 Let (A, DA ) be a closed linear operator on X and λ ∈ C. We say that λ ∈ ρ(A), the resolvant of A, if λ − A admits a bounded inverse on Im(λ − A). We call spectrum of A, σ(A) the complementary in C of ρ(A): ρ(A) = C\σ(A). cel-00376389, version 1 - 17 Apr 2009 Proposition 5.1.1 The inverse of a closed injective operator is closed. Proof. Let A: DA ⊂ X → Y be a closed injective operator, where X and Y are Banach spaces. The graph G(A−1 ) = Φ(G(A)) hence closed, where Φ : E × F → F × E is the homeomorphism Φ(x, y) = (y, x). ¤ Remarks. 1. Let A be a closed operator on X. If λ − A is bijective from DA to X for some λ then (λ − A)−1 is continuous from Imλ − A = X to X since closed (by the last proposition and the closed graph theorem). Hence λ ∈ ρ(A). 2. The spectrum of A is union of the three disjoint following sets: (a) σp (A) the point spectrum: the set of all eigenvalues. (b) σr (A) the residue spectrum: the set of all λ that are not eigenvalues and such that the image of λ − T is not dense in X. (c) σc (A) the continuous spectrum: the complementary of σp (A) and σr (A) it is also the set of λ such that λ − A is injective with dense image, but (λ − A)−1 is not continuous. Lemma 5.1.1 Let A be an injective closed operator and λ ∈ ρ(A), λ 6= 0. Then 1/λ ∈ ρ(A−1 ) and −1 (λ−1 − A−1 ) = λA(λ − A)−1 = −λ − λ2 (λ − A)−1 . Proof. λ−1 − A−1 = −λ−1 (λ − A)A−1 (they have the same domain ImA). Thus λ−1 −A−1 is bijective from D(A−1 ) onto X and its inverse is −λAR(λ, A). But AR(λ, A) − λR(λ, A) = Id, so we get the result. ¤ Proposition 5.1.2 Let A be a closed operator. 48 CHAPTER 5. UNBOUNDED SELF-ADJOINT OPERATORS 1. The spectrum σ(A) is a closed set of C. 2. The mapping λ ∈ ρ(A) 7−→ R(λ, A) ∈ L(X) is analytic. cel-00376389, version 1 - 17 Apr 2009 Proof. If σ(A) = C there is nothing to show. Otherwise, rescalling A by some λ ∈ ρ(A) we can assume that 0 ∈ ρ(A). Set B := A−1 . 1. By lemma 5.1.1, σ(A) = {λ 6= 0; λ−1 ∈ σ(B)} and since σ(B) is compact, σ(A) is closed. 2. By lemma 5.1.1, R(λ, A) = −λ−1 BR(λ−1 , B), so the mapping λ 7−→ R(λ, A) is analytic on ρ(A)\{0}. Since σ(A) is closed, there is λ0 ∈ ρ(A), λ0 6= 0. Rescalling we get that the mapping λ 7−→ R(λ, A) is analytic on ρ(A)\{λ0 }. ¤ Remark. For all nonempty closed set S of C we can construct a closed operator whose spectrum is S: Since S is not empty, let (λn ) be a dense sequence in S. Consider the operator A on H := `2 (N), with domain the set of sequences (xn ) ∈ H such that (λn xn ) ∈ H, and A(xn ) = (λn xn ) (A is called the multiplication operator see the section forthcoming). It is not difficult to verify that A is closed, densely defined, and σ(A) = σp (A) = S. 5.2 Adjoint of an operator In this section H will denote a Hilbert space and < ·, · > its scalar product. Lemma 5.2.1 Let (A, DA ) be a linear densely defined operator on H. Let y ∈ H, and assume that there exists y ∗ ∈ H such that for every x ∈ DA < Ax, y >=< x, y ∗ > . (5.1) Then y ∗ is unique. Proof. If there is z ∈ H s.t. < Ax, y >=< x, y ∗ >=< x, z > for all x ∈ DA , ⊥ ¤ we get that z − y ∗ ∈ DA which is trivial since D(A) = H. Definition 5.2.1 Let (A, DA ) be a linear densely defined operator on H. Define the (unbounded) operator A∗ , adjoint of A by D(A∗ ) := {y ∈ H, so that ∃y ∗ ∈ H s.t. (5.1) is verified} and A∗ y = y ∗ . 5.2. ADJOINT OF AN OPERATOR 49 The adjoint could be characterized by for all x ∈ DA and all y ∈ DA∗ < Ax, y >=< x, A∗ y > . (5.2) Definition 5.2.2 We say that A is a symmetric operator if DA = H and for every x, y ∈ DA we have < Ax, y >=< x, Ay > . We say that A is self-adjoint if A = A∗ . cel-00376389, version 1 - 17 Apr 2009 In the following we give direct properties: Proposition 5.2.1 Let (A, DA ) be a linear densely defined operator on H. Then 1. The adjoint of A is always closed. 2. If B is an extension of A, A ⊂ B then B ∗ ⊂ A∗ . ∗ 3. If A is closable, then (Ā) = A∗ . 4. If D(A∗ ) = H then A is closable and Ā ⊂ A∗∗ . 5. If A is a symmetric operator then every symmetric extension B of A verify: A ⊂ B ⊂ A∗ . 6. ImA⊥ = kerA∗ . Proof. 1. If yn → y, yn∗ → y ∗ and < Ax, yn >=< x, yn∗ > for every x ∈ DA then < Ax, y >=< x, y ∗ > so y ∈ DA∗ and A∗ y = y ∗ . Points 2, 3, 4 and 5 are obvious. 6. y ⊥ ImA means that < Ax, y >=< x, A∗ y >= 0 for all x ∈ DA , and so A∗ y = 0. ¤ Theorem 5.2.1 let (A, DA ) be a symmetric operator on a Hilbert space H. If DA = H then A is bounded. Proof. For all x, y ∈ H we have | < Ax, y > | = | < x, Ay > | ≤ kxk kAyk. So by The Banach- Steinhauss theorem A is bounded. ¤ Let’s see some examples: Examples. 50 CHAPTER 5. UNBOUNDED SELF-ADJOINT OPERATORS 1. Let H := L2 (0, 1) and define the operator (A, D) by f ∈ C 1 and f (0) = f (1) = 0}, D(A) := {f ∈ H; Af (t) = if 0 (t), d in other terms A = i dx . This operator is symmetric and A ⊂ A∗ : Rt Indeed, integrating by parts (G(t) := 0 g(s) ds) we get Z ∗ < Af, g > = < g, g >= 1 f g¯∗ dt cel-00376389, version 1 - 17 Apr 2009 0 Z = f (1)Ḡ(1) − f (0)Ḡ(0) − Z 1 = (if 0 )(t)(−iG)(t) dt. 1 f 0 (t)Ḡ(t) dt 0 0 Since ImA is dense in E, we get g = −iG. Therefore there is y 0 ∈ L2 with −y 0 = iy ∗ . Hence D(A∗ ) = {y ∈ H; y 0 ∈ L2 } and y ∗ = A∗ y = iy 0 . Thus A is symmetric and A ⊂ A∗ . A is not closed (because of the boundary conditions) but is closable, it is closure A1 := Ā is defined by D(A1 ) := {f ∈ H; f 0 ∈ L2 and f (0) = f (1) = 0}, A1 f (t) = if 0 (t). 2. Define A2 on the same space H by D(A2 ) := {f ∈ H; f 0 ∈ L2 and f (0) = f (1)}, A2 f (t) = if 0 (t). Thus A1 ⊂ A2 and then A∗2 ⊂ A∗1 . Let’s show that A2 is self adjoint: For this let’s calculate Z 1 < A2 f, g > = (if 0 )ḡ dt 0 Z 1 = i[f (1)ḡ(1) − f (0)ḡ(0)] + f (t)ig 0 (t) dt 0 Z 1 = if (1)[ḡ(1) − ḡ(0)] + f (t)ig 0 (t) dt. 0 Since A∗2 ⊂ A∗1 then < A2 f, g >=< f, A∗1 g >=< f, iy 0 > so if (1)[ḡ(1)− ḡ(0)] = 0. If g(1) 6= g(0) then choosing a sequence fn → 0 with fn (0) = fn (1) = 1 we get that < A2 fn , g >6→ 0 but < fn , A∗2 g >→ 0. So g(1) = g(0) and then A∗2 = A2 . 5.2. ADJOINT OF AN OPERATOR 51 3. Define, on the space H := L2 [0, ∞[, the operator D(A1 ) := {f ∈ E; f 00 ∈ L2 and f (0) = f 0 (0) = 0}, A1 f (t) = −f 00 (t). By the same way as in the first example, y ∈ D(A∗ ) implies that y 00 ∈ L2 , then Z ∞ < A1 f, g > = (−f 00 )ḡ dt 0 Z ∞ = −[f 0 ḡ]∞ + f 0 (t)g¯0 (t) dt 0 cel-00376389, version 1 - 17 Apr 2009 0 00 = [−f 0 ḡ + f g¯0 ]∞ 0 + < f, −g > . Therefore we see that < A1 f, f >≥ 0, and that y ∗ = A∗ y = −y 00 and D(A∗ ) = {y ∈ L2 ; y 00 ∈ L2 }. So A1 ⊂ A∗1 and A1 is symmetric but not self-adjoint. 4. Consider the same operation on the same space, L2 [0, ∞[, D(A2 ) := {f ∈ H; f 00 ∈ L2 and f (0) = 0}, A2 f (t) = −f 00 (t). Obviously A1 ⊂ A2 and then A∗2 ⊂ A∗1 . Repeating the same calculation as above we get < Af, g >=< f, g ∗ >= −f 0 (0)g(0)+ < f, −g 00 >, and necessarily g(0) = 0 (otherwise consider a sequence fn → 0 in L2 with fn0 (0) = 1 to get a contradiction). This shows that A2 is self-adjoint (extension of A1 ). Theorem 5.2.2 If A is a symmetric operator and ImA = H then A is self-adjoint. Proof. We know that A ⊂ A∗ . Now let y ∈ D(A∗ ) and set y ∗ = A∗ y. Since ImA = H there is x ∈ D(A) such that A∗ y = y ∗ = Ax. For every z ∈ D(A) we have < Az, y >=< z, A∗ y >=< z, y ∗ >=< z, Ax >=< Az, x >, thus y = x and so A = A∗ . ¤ 52 CHAPTER 5. UNBOUNDED SELF-ADJOINT OPERATORS Theorem 5.2.3 Let A be a bounded self-adjoint operator. Assume that kerA = {0} then A−1 is also self adjoint. cel-00376389, version 1 - 17 Apr 2009 Proof. First let’s show that A−1 is densely defined: If not, D(A−1 ) = ImA 6= H, then by proposition 5.2.1.6, there is y0 ∈ kerA∗ , y0 6= 0, which ∗ is not possible since A = A∗ . Now < A−1 x, y >=< x, A−1 y >, setting z = A−1 x, x = Az we get < z, y >=< x, y ∗ >=< Az, y ∗ >=< z, A∗ y ∗ > and so ∗ y = A∗ y ∗ = Ay ∗ or y ∗ = A−1 y, but y ∗ = A−1 y. Thus y ∈ ImA = D(A−1 ), ∗ and A−1 y = A−1 y. Thus A−1 is self-adjoint. ¤ Remark. This theorem 5.2.3 gives us many examples of unbounded selfadjoint operators. Start with any self-adjoint compact operator A with kerA = {0}. Then A−1 is an unbounded self-adjoint operator. 5.3 The L∞ -spectral theorem In this section we will show a theorem, known as spectral theorem, for bounded self-adjoint or unitary operators, stating that each self-adjoint or unitary operator is unitary equivalent to a real multiplication operator. Thus, self-adjoint and real multiplication operators are effectively the same things. It is frequent to regard an arbitrary self-adjoint operator as being a real multiplication operator. We will start by defining multiplication operator: Let (X, µ) a measured space and f ∈ L∞ (X, µ). Definition 5.3.1 The multiplication operator Af : L2 (X, µ) −→ L2 (X, µ) is defined by Af (g) := f g. It is easy to see that Af is a linear bounded operator on L2 , and kAf k ≤ kf k∞ . Moreover, A∗f = Af¯, hence Af is normal. If in addition f is realvalued then Af is self-adjoint and if |f | = 1 a.e. then A∗f Af = A|f |2 = IdL2 . Proposition 5.3.1 σ(Af ) = Ress (f ) = {λ ∈ C; ∀ε > 0, the set of x ∈ X, |f (x) − λ| < ε is not µ-negligeable}. Proof. Let λ ∈ C\Ress (f ). If there is ε > 0, such that the set of x ∈ X, |f (x)−λ| < ε is µ-negligeable, denoting by h the function h(x := (f (x)− λ)−1 for f (x) 6= λ and 0 if not. |h(x)| < ε−1 for µ-a.e. x and h(x)(λ−f (x)) = 1. Thus h ∈ L∞ (X, µ) and Ah (Af − λ) = (Af − λ)Ah = IdL2 . Now if λ ∈ Ress (f ) then for every ε, the set Aε := {x ∈ X, |f (x) − λ| < ε} 5.3. THE L∞ -SPECTRAL THEOREM 53 is not µ-negligeable, consider a function χ ∈ L2 (X, µ), kχk2 = 1, χ = 0 outside Aε . Then |(Af − λ)χ| ≤ ε|χ|, hence k(Af − λ)χk ≤ ε. Thus Af − λ is not bijective. ¤ Proposition 5.3.2 For all g ∈ RAf we have g(Af ) = Ag(f ) . If Af is selfadjoint or unitary, then for all g ∈ C(σ(Af ) then g(Af ) = Ag(f ) . cel-00376389, version 1 - 17 Apr 2009 Proof. The mapping g 7−→ Ag(f ) is linear morphism of ring, so we get the first point using uniqueness in proposition 3.1.1. The second point could be obtained by applying theorem ??. ¤ In the following, H is a Hilbert space and T a bounded self-adjoint or unitary operator. Lemma 5.3.1 Let x ∈ H. 1. There exists a finite measure R µx on σ(T ) such that, for all f ∈ C(σ(T )) we have < f (T )x, x >= σ(T ) f (t) dµx (t). 2. Denote by φx : C(σ(T )) −→ H the linear mapping defined by φx (f ) := f (T )x, and w: C(σ(T )) −→ L2 (σ(T ), µx ) the mapping that to a continuous function associate its class in L2 . There exists an isometry ψx : L2 (σ(T ), µx ) −→ H such that ψx ◦ w = φx . Moreover, ψx (1) = x, ψx (Az ) = T φx and ψx (Az̄ ) = T ∗ φx . Proof. 1. The linear form Φx : f 7−→< f (T )x, x > is positive on C(σ(T )): Indeed, if f is positive, then f (T ) is a positive operator by theorems ?? and 3.2.3 (f (T ) is self-adjoint and σ(f (T )) = f (σ(T )) ⊂ R+ ). There exists R then a unique measure µx on σ(T ) such that < f (T )x, x >= Φx (f ) = σ(A) f (t) dµx (t). 2. Let f ∈ C(σ(T )). We have kφx (f )k2 =< f (T )x, f (T )x >=< f¯(T )f (T )x, x > R =< [f f¯](T )x, x >= σ |f (t)|2 dµ(t) = kw(f )k2 . On the space C(σ(T )) endowed with semi-norm kφx (f )k, w is a linear isometric of dense image, there exists then ψx ∈ L(L2 (σ(T ), µx ), H) such that φx = ψx ◦ w. For all f ∈ C(σ(T )) we have kψx (w(f ))k = kφx (f )k = kw(f )k, then by density of the image of w we have kψx (g)k = kgk for all g ∈ L2 (σ(T ), µx ): ψx is isometric. ψx (1) = φx (1) = Ix = x. Finally, for all f, g ∈ C(σ(T )), we have ψx (Af (w(g)))ψx (w(f g)) = φx (f g) = f (T )g(T )x = f (T )φx (g) = f (T )ψx (w(g)). Again, by the density of the image of w, we get for all f ∈ C(σ(T )) and all g ∈ L2 (σ(T ), µx ), we have ψx (Af (g)) = f (T )ψx (g). Take f = z and f = z̄ to terminate. ¤ 54 CHAPTER 5. UNBOUNDED SELF-ADJOINT OPERATORS Lemma 5.3.2 Let x ∈ H and denote by Ex the image of ψx . 1. If y ∈ Ex⊥ then Ey ⊂ Ex⊥ . 2. There exists a subset D ⊂ H such that for all x, y ∈ D, x 6= y, Ex ⊥ Ey and ⊕x∈D Ex = H. cel-00376389, version 1 - 17 Apr 2009 Proof. 1. If y ∈ Ex⊥ , then for all f ∈ σ(T ) and all g ∈ L2 (σ(T ), µ), we have < ψx (g), f (T )y >=< f¯(T )ψx (g), y >=< ψx (Af¯g), y >= 0. Since {f (T )y; f ∈ C(σ(T ))} is dense in Ey , we get the result. 2. Denote by G the set of subsets D of H\{0} such that for all x, y ∈ D, Ex ⊥ Ey . Endowed with inclusion G is inductive. It is easy to show that G admits a maximal element D that is our candidate. ¤ Theorem 5.3.1 Let H be a separable Hilbert space and T ∈ L(H) a selfadjoint or unitary bounded operator. There exists a measured space (X, µ), a function f ∈ L∞ (X, µ) and an isomorphism ψ: L2 (X, µ) −→ H such that T = ψAf ψ ∗ Proof. Let x ∈ H and D as in the last lemma. For all y ∈ H, we have y ∈ Ey and since H is separable, D is countable. Rearrange D to be a discrete set and set X := σ(T ) × D. If g is a function on X, denote for y ∈ D, gy the function t 7−→ g(t, y). Denote by Cc (X) the set of continuous functions of compact support on X, i.e. g ∈ Cc (X) if all except finite number of the functions gy are null. Denote by XZ Φ(g) := gy (t) dµy (t). y∈D σ(T ) Since Φ is a positive linear form on Cc (X), there exists a unique measure µ on X such that, for all lg ∈ Cc (X) we have Z XZ g(x) dµ(x) = Φ(g) = gy (t) dµy (t). X y∈D σ(T ) P Denote by φ: Cc (X) −→ H the mapping defined by φ(g) := y∈D gy (T )y and w: Cc (X) −→ L2 (X, µ) the function class. PFor g ∈ Cc (X) we have gy (T )y ∈ Ey . by orthogonality we get kφ(g)k2 = y∈D < gy (T )y, gy (T )y > R R P =< ḡy (T )gy (T )y, y >= y∈D σ(T ) |gy (t)|2 dµy (t) = X |g(x)|2 dµ(x) = kw(g)k2 . On the space Cc (X) endowed with semi-norm kφ(f )k, w is a linear isometric of dense image, there exists then ψ ∈ L(L2 (X, µ), H) such that φ = ψ ◦ w. For all f ∈ C(σ(T )) we have kψ(w(f ))k = kφ(f )k = kw(f )k, then by density of the image of w we have kψ(g)k = kgk for all g ∈ L2 (X, µ): ψ is isometric. Let’s show that ψ is surjective. Let y ∈ D and h ∈ C(σ(T )). Set g(t, y) := 5.4. THE L2 -SPECTRAL THEOREM 55 cel-00376389, version 1 - 17 Apr 2009 h(t) and g(t, x) = 0 for x ∈ D, x 6= y. We have ψ(w(g)) = φ(g) = h(T )y. Hence the image of ψ contains all h(T )y, y ∈ D, h ∈ C(σ(T )). Since ψ is isometric, its image is closed, hence contains all Ey . Thus ψ is surjective. Denote by P1 the first projection on X. For all g ∈ Cc (X) and all y ∈ D we have (P1 g)y = zgy P , hence (P1 g)y (T ) = T gy (T ). So ψ(AP1 w(g)) = ψ(w(P1 g)) = φ(P1 g) = y∈D (P1 g)y (T )y = T φ(g) = T ψ(w(g)). By density of the image of w, we deduce that for all g ∈ L2 (X, µ), ψ(Af (g)) = T ψ(g). This implies that ψAf = T ψ hence T = ψAf ψ −1 = ψAf ψ ∗ . ¤ Using this theorem, one can define a symbolic calculus from C(σ(T )) into L(H): for g ∈ C(σ(T )), wet g(T ) := ψAg◦f ψ ∗ . Notice that this symbolic calculus could be extended to the B(σ(T )) the vector space of bounded borelean functions on σ(T ). 5.4 The L2 -spectral theorem In this section we consider a particular self-adjoint operator which appears to be a very particular (and simple) example, but which will be central to the description and application of the spectral theorem. This will be seen by the main theorem of the next section. In the last section we have defined multiplication operator for a bounded function, which gives a bounded operator. In this section we will define the multiplication operator for L2 -functions, which gives unbounded operator. The proofs are roughly the same, so they are omitted. Let (X, µ) be a measured space. Define H := L2 (X, µ) the space of all measurable functions of square integrable, with the classical identification between two functions if ever they are equal almost everywhere. Fix a measurable real-valued function a that is bounded on every bounded subset of X. Let D be the set of all functions f ∈ H such that Z [1 + a(x)2 ]|f (x)|2 dµ < ∞, X and define the operator Aa with domain D by Aa f (x) := a(x)f (x), the multiplication operator. Lemma 5.4.1 The operator (Aa , D) is self-adjoint. Define the essential range of a, Ress (a), the set of all λ ∈ R such that for all ε > 0 the measure of the set {x ∈ X; |a(x) − λ| < ε} is zero. 56 CHAPTER 5. UNBOUNDED SELF-ADJOINT OPERATORS Lemma 5.4.2 σ(Aa ) = Ress (a), and if λ 6∈ σ(A) then £ ¤ (λ − Aa )−1 f (x) = [λ − a(x)]−1 f (x) for all f ∈ H and all x ∈ X, and k(λ − Aa )−1 k = 1 . dist(λ, σ(Aa )) cel-00376389, version 1 - 17 Apr 2009 Now we can generalize the results of the last section to the case of unbounded self-adjoint operators. Theorem 5.4.1 (Spectral theorem) Let H be a Hilbert space and T a densely defined self-adjoint operator on H. Then 1. σ(T ) ⊂ R. 2. The operator U := (i − T )(i + T )−1 is a unitary operator in L(H). 3. There exists a measured space (X, µ) a measurable function f : X → R and an isomorphism ψ: L2 (X, µ) → H of Hilbert spaces such that T = ψAa ψ ∗ . Proof. 1. Let λ ∈ C\R. Denote by b its imaginary part. For all x ∈ D(T ) we have < T x, x >=< x, T x > hence < T x, x >∈ R and the imaginary part of < (λ − T )x, x > is then bkxk2 . Thus |b|kxk2 ≤ | < (λ − T )x, x > | ≤ k(λ − T )xkkxk and so k(λ − T )xk ≥ |b|kxk. Thus, for all (x, y) ∈ G(λ − T ), we have kyk ≥ |b|kxk, hence (1 + b2 )kyk2 ≥ b2 (kxk2 + kyk2 ). By proposition , the mapping (x, y) 7−→ y from G(λ − T ) into H is injective of closed image. Since (λ − T )∗ = λ̄ − T is also injective, we deduce, by proposition 5.2.1.6, that the image of λ − T is dense. 2. Since Im((i + T )−1 ) = D(i − T ), D(U ) = H and U is bijective by 1. Now for x ∈ D(T ), we have k(i − T )xk2 = kT xk2 + kxk2 − i < x, T x > +i < T x, x >= kT xk2 + kxk2 = k(i + T )xk2 . For y ∈ H, set x = (i + T )−1 y, we have kU yk = k(i − T )xk = k(i + T )xk = kyk. Thus U is isometric. 3. Let y ∈ H and set x := (i + T )−1 y, we have U y = (i − T )x = 2ix − (i + T )x = 2ix − y. Thus x = 2(U y + y)/i and then (i + T )−1 = 2(U + Id)/i and T = i(U + Id)−1 − i. By theorem 5.3.1, there exists a measured space (X, µ), a function g: X −→ C, |g| = 1 measurable and an isomorphism ψ: L2 (X, µ) −→ H such that U = ψAg ψ ∗ . Since U − Id is injective, Ag−1 is injective, and so the set {x ∈ X; g(x) = 1} is µ-negligeable. Then (U − Id)−1 = (ψAg−1 ψ ∗ )−1 = (ψ ∗ )−1 Ag ψ −1 ψAg ψ ∗ = U . Thus T = ψAf ψ ∗ where f := 2(g − 1)−1 /i − i = −i(g + 1)(g − 1)−1 . ¤ As in the bounded case, one can define a symbolic calculus on B(σ(T )) the space of bounded borelean functions on σ(T ). 5.5. STONE’S THEOREM 5.5 57 Stone’s theorem Definition 5.5.1 Let E be a Banach space. We call a one parameter C0 group any family of linear bounded operators (G(t))t∈R ⊂ L(E) verifying 1. G(0) = IdE . 2. G(t + s) = G(t)G(s), for all t, s ∈ R. 3. For all x ∈ E, the mapping t 7−→ G(t)x is continuous. The operator defined by G(t)x − x exists}, t→0 t cel-00376389, version 1 - 17 Apr 2009 D(A) := {x ∈ E; lim Ax := lim t→0 G(t)x − x t is called generator of the C0 -group. Let H be a Hilbert space. A C0 -group is called unitary C0 -group if each operator is unitary. Theorem 5.5.1 Let H be a separable Hilbert space. Let (A, D(A)) be a densely defined operator. The following are equivalent: (i) iA generates a unitary C0 -group. (ii) A is self-adjoint. Proof. (i)=⇒(ii). We have G∗ (t) = G(t)−1 = G(−t). Let’s show that A ⊂ A∗ . Indeed, let x, y ∈ D(A), we have À ¿ À ¿ G(t)x − x G∗ (t)y − y , y = −i lim x, < Ax, y > = −i lim t→0 t→0 t t ¿ À −1 G (t)y − y = −i lim x, t→0 t ¿ À G(−t)y − y = −i lim x, = −i < x, −iAy >, t→0 t thus x ∈ D(A∗ ) and < Ax, y >=< A∗ x, y >. A = A∗ : Let x ∈ D(A), y ∈ D(A∗ ) we have ¿ À ¿ À G(t)x − x G∗ (t)y − y ∗ < x, A y > = < Ax, y >= −i lim , y = −i lim x, t→0 t→0 t t À ¿ −1 G (t)y − y = −i lim x, t→0 t ¿ À G(−t)y − y = −i lim x, =< x, Ay >, t→0 t 58 CHAPTER 5. UNBOUNDED SELF-ADJOINT OPERATORS Therefore y ∈ D(A) and hence A = A∗ . (ii)=⇒(i). Since A admits a L∞ (σ(B)) symbolic calculus. Denote by Φ this symbolic calculus and define, for all t ∈ R, G(t) := Φ(et ) = et (A), where et (s) := exp(ist). Since σ(A) ⊂ R, et is bounded. Using properties of the symbolic calculus, it is easy to verify that (G(t)) is a unitary group generated iA. ¤ 5.6 Laplace operator on bounded open domain of cel-00376389, version 1 - 17 Apr 2009 RN Let Ω be an open of RN and H = L2 (Ω) as a real Hilbert space. Define the operator ∆0 on H by D(∆0 ) := {u ∈ H01 (Ω), ∆u ∈ L2 (Ω)}, ∆0 u = ∆u, u ∈ D(∆0 ). Then we have Proposition 5.6.1 (∆0 , D(∆0 )) is negative self adjoint operator. Proof. Since D(Ω) ⊂ D(∆0 ), D(∆0 ) is dense in H. Let u ∈ D(∆0 ) ⊂ H01 (Ω), by Green’s formula, we have Z Z < ∆0 u, u >= ∆u · u dx = − |∇u|2 dx Ω Ω so ∆0 is negative. By a similar calculation, one can see that ∆0 is symmetric. In order to use theorem 5.2.2, let’s show that Im∆0 = H. In fact we will show that 0 6∈ σ(∆0 ). For this, and using Lax-Milgram lemma, for all f ∈ H, there exists u ∈ H01 (Ω) such that, for all v ∈ H01 (Ω) Z Z (λuv + ∇u · ∇v) dx = f v, for all λ > −λ0 , λ0 being one over the Poincar constant. Which gives (by Green) that in the distribution sens λu − ∆u = f. Thus (u ∈ H01 ) ∆u = u − f ∈ L2 , i.e. u ∈ D(∆0 ). In other terms σ(∆0 ) ⊂ ] − ∞, −λ0 [. ¤ Corollary 5.6.1 i∆0 generates a unitary C0 -group. 5.6. LAPLACE OPERATOR ON BOUNDED OPEN DOMAIN OF RN 59 Remark 5.6.1 If the boundary of Ω is bounded and is of class C 2 , then D(∆0 ) = H 2 (Ω) ∩ H01 (Ω) with equivalent norm. In order to determine the eigenvalues of Laplace operator, notice first that if u is an eigenvalue then there is λ (≤ −λ0 ) such that ∆0 u = λu. (5.3) cel-00376389, version 1 - 17 Apr 2009 This means that u ∈ D(∆0 ). But asking u to be in H01 is sufficient, since in this case, ∆u ∈ L2 . Therefore it is sufficient to solve (5.3) in H01 . We start by the following direct application of theorem 5.2.3. Corollary 5.6.2 (−∆0 )−1 : L2 (Ω) 7−→ L2 (Ω) is a positive bounded self-adjoint operator. Corollary 5.6.3 (−∆0 )−1 : H01 (Ω) 7−→ H01 (Ω) is a positive compact selfadjoint operator. Proof. Remainder to show that this operator is compact. Denoting by A this operator then A = −∆−1 0 ◦ J, where J: u 7−→ u is the canonical injection from H01 into L2 . Since J is compact (Rellich theorem) and using proposition 4.1.1 A is compact. ¤ We terminate by Theorem 5.6.1 The set of eigenvalues of Laplace operator with Dirichlet condition ∆0 on Ω is a strictly decreasing sequence that tends to −∞. Each eigen-space is of finite dimension. Denote by (µn ) the sequence of eigenvalues of −∆0 , repeated each with its multiplicity. Then there exists a Hilbert basis (un ) of H01 (Ω) such that for all n, we have ∆0 un = µn un Remark 5.6.2 By the same argument above each un ∈ H0∞ (Ω), hence C ∞ and so un is an ordinary solution of the equation ∆un = µn un . View publication stats