Functional analysis II Prof. Dr. Thomas Sørensen winter term 2014/15

Transcription

1 Functional analysis II Prof. Dr. Thomas Sørensen winter term 2014/15 Marcel Schaub February 4,

2 Contents 0 Motivation and repetition 3 1 Spectral Theory for compact operators Spectral Theorem for compact operators Consequences of the spectral Theorem for compact operators The finite-dimensional case Spectral Theory for normal compact operators on Hilbert spaces The Spectral Theorem for normal compact operators on a Hilbert space Spectral Theory for bounded, self-adjoint operators on a Hilbert space The Continuous Functional Calculus Properties of the Continuous Functional Calculus Outlook on the Measurable Functional Calculus A recall on Measure Theory and Integration Theory Statements about b (D) The Measurable Functional Calculus Projection Valued Measures Characterization of the Measurable Functional Calculus Spectral Theorem for self-adjoint bounded operators Unbounded operators Motivation Unbounded operators and their adjoint Closed operators Essential self-adjointness Existence of self-adjoint extensions The spectrum of an unbounded operator Spectral Theory for unbounded self-adjoint operators Spetral Theorem Multiplication operator version Spectral Theorem Spectral measure version Consequences of the Spectral Theorems Outlook on other approaches to the Spectral Theorem (for bounded/unbounded self-adjoint operators Another outlook: Spectral theory of bounded operatos; Generalisation to Banach algebras Banach algebras setup (B(X )) Adding a unit to a unit-less algebra Functional Calculus Outlook on the Functional Calculus and Spectral decompositions: Usefullness?!? (Partial) Differential Operators

3 0 Motivation and repetition Recall A Banach space is a vector space (mostly over ) with a norm such that (s.t.) the metric space (X, d) with d(x, y) := x y is a complete metric space. A Hilbert space is a Banach space where the norm is given by a scalar product (inner product),, and x = x, x. Example. L 2 () with f, g := f (x)g(x) dx is a Hilbert space (the Hilbert space). So is l 2 () = {x = (x 1, x 2,...) i=1 x i 2 < }. But the norm p (p 2) with 1/p x p = x i p i=1 turnes l p = {x x p < } into a Banach space, but not a Hilbert space (p 2). The space B(X, Y ) is the set of bounded (i.e. continuous) linear operators from X to Y, i.e. T B(X, Y ) iff T : X Y is linear and bounded ( C > 0 : T x C x for every x X (this definition needs only normed spaces not completeness). Set B(X ) := B(X, X ). If Y = (if X is a -vector space (v.s.); if X is some -vector space then Y = ) we call the elements in B(X, ) linear bounded functionals and write X := B(X, ), X is called the dual (or, the dual vector space) of X. Note: For a Hilbert space, H = H (Riesz representation theorem) in the sense that φ : H H, y y, such that φ is an isometry (bijection, conserves norm). However: φ is not linear it is conjugate linear, i.e. φ(αx + y) = αφ(x)+φ(y), α, x, y H. One has X X (X is the bi-dual) in the sense that there is a canonical embedding ι : X X that embeds X isometrically in X (ι is injective, but not always surjective). If X = X (in the sense that ι is surjective), then X is called reflexive (so, all Hilbert spaces are reflexive). So are l p, L p (1 < p < ). Definition 0.1. Let T B(X ) (X -Banach space, X 0). We define the resolvent set of T by ρ(t) = {x N(T λi) = {0} and R(T λi) = X } Here N(S) = {x X S x = 0} is the kernel and R(S) = { y X x X = S x = y} is the range of the operator S B(X ). Also I : X X, x x is the identity on X (I = I X ). We define the spectrum of T B(X ) by σ(t) := \ ρ(t) The spectrum can be split in to 3 types (3 subsets). The point spectrum ( Punkt-Spektrum ) σ p (T) = {λ σ(t) N(T λi) {0}} σ(t) 3

4 The continuous spectrum ( Stetiges Spektrum ) σ c (T) = {λ σ(t) N(T λi) = {0}, R(T λi) X, R(T λi) = X } The rest spectrum/residual spectrum: σ r (T) = {λ σ(t) N(T λi) = {0}, R(T λi) X } Remark 0.2. (1) Note that λ ρ(t) iff T λi : X X is bijective. This is equivalent to the existence of R λ (T) := (T λi) 1! B(X ) R λ (T) is also continuous and called the resolvent of T at λ. (2) λ σ p (T) iff x 0, x X : T x = λx (or (T λi)x = 0). Then λ is called an eigenvalue, and x an eigenvector. However, when X is some space of functions, e.g. L p (), we often call x an eigenfunction. N(T λi) is the eigenspace of T belonging to the eigenvalue λ σ p (T). It is a T-invariant (linear) subspace of X. If Y X is a linear subspace, then Y is called T-invariant iff T(Y ) Y. Proposition 0.3. Let T B(X ), then ρ(t) is an open subset of (hence σ(t) is closed) and the resolvent map (or, function) ρ(t) λ R λ (T) B(X ) is complex analytic map with R λ (T) 1 dist(λ, σ(t)). Here, complex analytic means: λ 0 ρ(t) r 0 (λ 0 ) > 0 and c j (λ 0 ) B(X ): R λ (T) = c j (λ λ 0 ) j j=0 for λ B r0 (λ 0 ) = {µ λ 0 µ < r 0 } with the series convergent in the (operator) norm on B(X ). The aim of the course is (mainly) to study the spectrum (and its properties) for various classes/types of operators and prove theorems in analogy to one on diagonalisation of symmetric/self adjoint/hermitian matrices in linear algebra. We would like to see how/why this is useful. The theory here (in infinite dimensional case) is, however, much, much more rich. Definition 0.4. The compact (linear) operators from X to Y are defined by K(X, Y ) = {T B(X, Y ) T(B 1 (0)) Y is compact} B(X, Y ) Remarks 0.5. (1) As for B(X ), we write K(X ) if X = Y. (2) If Y is Banach, then T(B 1 (0) is compact can be replaced with T(B 1 (0)) is pre-compact (3) That is, T K(X, Y ) iff T maps bounded sequences (in X ) into sequences in Y which have a convergent subsequence. (Note: T B(X, Y ) = T maps bounded sequences into bounded sequences.) (4) Example: I = [0, 1], k [I I], f (I) =: X. For x I set (T k f )(x) := 1 0 k(x, y)f (y) dy 4

5 and let X be equipped with f = f = sup x I f (x). Then T k : X X is a bounded, linear operator which is compact (i.e. T k K(X ) B(X )). We shall see other examples and further properties of compact operators soon. Definition 0.6 (Banach space adjoint). Let T B(X, Y ) and define for y Y (y : Y ) and x X : (T y )(x) := y (T x) Then T : Y X, y T y is linear and bounded (i.e. T B(Y, X ). It is called the (Banach space) adjoint of T (or, the adjoint operator of T). Furthermore: T = T (the first norm is in B(Y, X ) and the second one in B(X, Y )). In fact, : B(X, Y ) B(Y, X ) T T is an isometric, linear embedding (i.e. injective). NB. In the above definition there are in fact things to prove exercise! Linearity: Let α, u, v Y, x X, then: T (αu + v )(x) = (αu + v )(T x) = αu (T x) + v (T x) So T (αu + v ) = αt u + v. = α(t u )(x) + (T v )(x) = (αt u + T v )(x) Boundedness comes by T y (x) = y (T x) y T x and then T y T y (x) = sup y T = T T = sup y x X x y Y y T For T = T see Theorem 4.18 in [FA1]. is isometric by the above and linear by 0.8 (1). Definition 0.7 (Hilbert space adjoint). Let H be a Hilbert space 1 and let φ : H H be the map y y,, identifying H and its dual H, and let T B(H) (T : H H) (T : H H ). Then T = φ 1 T φ : H H is called the Hilbert space adjoint of T B(H). It satisfies: T is called selfadjoint (s.a.) iff 2 T = T. T x, y = x, T y x, y H We shall see other examples and further properties of the adjoint and self adjoint operator(s) soon. 1 We could do this for T B(H 1, H 2 ), H 1, H 2 do different Hilbert spaces. We need φ H1, φ H2 then. 2 Note that T B(H). For unbounded operators (typical examples occur in quantum mechanics or partial differential equations) there is a more complicated definition of (self)adjoint. We will see this later. 5

6 Program of the course (mainly...) Spectral theory for compact operators (in X ). Spectral theory for bounded, self adjoint operators (in H). Unbounded operators (in particular, symmetric operators and quatratic forms). Spectral theory for unbounded self adjoint operators. Lemma 0.8 (Algebraic properties of adjoints). Let X, Y be Banach spaces, let H be a Hilbert space. Let T, T 1, T 2 B(X, Y ), S, S 1, S 2 B(H) and α. Then the following statements hold: (1) (αt 1 + T 2 ) = αt 1 + T 2 and (αs 1 + S 2 ) = αs 1 + S 2. (2) (I X ) = I X (I X : X X, I X : X X ). (3) (T 2 T 1 ) = T 1 T 2 and (S 1S 2 ) = S 2 S 1 (4) For ι X : X X, ι Y : Y Y the canonical embedding: T ι X = ι Y T : X Y and S = S. Proof. Simple algebraic manipulations from the definitions. Exercise. Proposition 0.9. Let X, Y be Banach spaces and T B(X, Y ). Then T 1 B(Y, X ) exists iff (T ) 1 B(X, Y ) exists and, in this case, (T ) 1 = (T 1 ). If X = Y = H is a Hilbert space, then (T 1 ) = (T ) 1. Remark. For T B(X ) we have: T 1 B(X ) exists iff (T 0 I) 1 B(X ) exists iff 0 ρ(t). I.e. invertability of T is a spectral question related to 0 ρ(t) or 0 σ(t)?. The proposition 0.9 says (for T B(X )): This is therefore a property of the spectrum. The proof uses [FA1]: 0 σ(t) 0 σ(t ) Theorem (Open mapping theorem). Let X, Y be Banach spaces. Let T B(X, Y ) be onto. Then T is open. Proof of Proposition 0.9. Note that T : X Y, T : Y X, T 1 : Y X, (T 1 ) : X Y, (T ) 1 : X Y. ( ) If T is invertible, then (by 0.8 (2) + (3)) and I X = (I X ) = (T 1 T) = T (T 1 ) I Y = (I Y ) = (T T 1 ) = (T 1 ) T Hence: (T is invertible, and) (T ) 1 = (T 1 ) B(X, Y ). 6

7 ( ) Assume now T is invertible. By the above, T : X Y is invertible in particular it is a bijection and continuous. Therefore, by the open mapping theorem, it maps closed subsets in closed subsets. Recall that by 0.8 T ι X = ι Y T (ι X, ι Y are the canonical embeddings) and that ι X, ι Y are isometries, hence: R(ι Y T) = R(T ι X ) = T (R(ι X )) Y is closed in Y because: Take a convergent sequence in R(ι X ) Y. Then it s Cauchy and also Cauchy in X (ι X isometry). Therefore it is convergent with limit in X. Then (ι X isometry) the limit in X of the sequence in R(ι X ) is in R(ι X ). It follows that R(T) = ι 1 Y (R(ι Y T)) is closed in Y. Since T is injective, it follows that {0} = N(T ). By Lemma 0.10 we get {0} = N(T ) = Annil(R(T)). It now follows from the proposition 0.11 (a consequence of the Hahn-Banach Theorem) that Y = R(T) which is equal to R(T) since R(T) is closed, hence T is surjective. Since T is injective (since T = (T ) is invertible) we have {0} = N(T ι X ) = N(ι Y T) = N(T). So T is injective. Hence T 1 exists (and is linear) and is automatically bounded (i.e. continuous by the open mapping theorem), i.e. T 1 B(X ). Lemma Let W be a Banach space. For a subspace Z W let Annil(Z) = {w W w (z) = 0 z Z} W be the set of bounded linear functionals on W which vanish on all of Z W. For T B(X, Y ) we have N(T ) = Annil(R(T)) Proof. Let y N(T ), i.e. T y = 0 ( X ) iff (T y )(x) = 0 x X (definition of a map being 0) iff y (T x) = 0 x X (definition of T ) iff y Annil(R(T)). Proposition Let W be a normed space, Z W a closed subspace and w 0 / Z. Then there exists w W such that w 0 on Z, w = 1 and w (w 0 ) = dist(w 0, Z) > 0 The proof uses [FA1]: Theorem (Hahn-Banach; complex version). Let X be a -vector space. Let p: X with p(αx + β y) α p(x) + β p(y) for every x, y X and α, β with α + β = 1. Let Y X be a subspace and λ: Y linear with λ(x) p(x) for every x Y. Then there exists a linear extention Λ: X of λ (i.e. Λ Y = λ ) with Λ(x) p(x) x X. Proof of Set U := span(y {w 0 }) and define f : U as f (w 0 ) := dist(w 0, Z) > 0 because Z is closed and f (z) = 0 z Z. For some u U, we decompose as u = λx 0 + z for some λ and z Z and extend f linearly on U by setting f (u) := λ f (w 0 )+ f (z) = λ f (w 0 ). We have f (u) = f (λw 0 ) + f (z) = dist(w 0, Z) = inf y Z w 0 y λw 0 z = u 7

8 i.e. f (u) u for every u U. By Hahn-Banach with p = we obtain a linear extension l of f on W with l(w) w w W, so l 1. By closeness of Z there is some (y n ) n with w 0 y n n dist(w 0, Z) and then dist(x 0, Z) = l(w 0 ) l(y n ) l w 0 y n n l dist(w 0, Z) hence l 1 yielding l = 1. Proposition 0.12 (Neumann-series). Let X be a Banach space, T B(X ), with T < 1. Then (I T) 1 B(X ) and (I T) 1 = n=0 in B(X ), i.e. convergence in the operatornorm in B(X ). Proof. See [FA1], Lemma T n ( ) Remark operators) (i) The series in ( ) is called the Neumann-series (for T) (when applied for (ii) Of course (T 1 I) 1 = (T I) 1 = (I T) 1 = Hence: T < 1 implies 1 ρ(t) ( spectral property ). T n B(X ) (iii) Perturbation result : The identity I : X X is invertible. If one adds something not too big (a small perturbation, i.e. add T ) then the result (I T) is also invertible ( perturbing I a little preserves invertability ). Corollary Let X, Y be Banach spaces. Then the subset (of B(X, Y )) of invertible operators in B(X, Y ) is an open subset (in B(X, Y )). More precisely: For S, T B(X, Y ): If T is invertible and S T < T 1 1, then S is invertible. Note: Again, a perturbative result : If S is close to T and T is invertible, then so is S. Remark Functions of an operator : (1) For T B(X ), we have the obvious definition T 2 = T T = T T, T n+1 := T T n, T 1 = T, T 0 := I (already used), and T 1 is the inverse of T (if exists!) and (exercise) then T n := (T 1 ) n (Note: T k T m = T k+m ). (2) More generally: If p: is a polynomial, then, if p(z) = a 0 + a 1 z + + a n z n = we can define p(t) B(X ) the obvious way, by p(t) := n a i T i B(X ) i=0 n=0 n a i z i Exercise: For p, q polynomials as above: p(t)q(t) = (pq)(t). i=0 8

9 (3) We have already seen other functions of T, namely the resolvent of T at λ: R λ (T) = (T λi) 1 B(X ) and, if T < 1, the Neumann-series: (I T) 1 = T n B(X ) n=0 i.e. f (T) for f : \ {1}, z 1 1 z = (1 z) 1. λ ρ(t) (4) More generally: Let f (z) = n=0 a nz n be a powerseries in (being or ) with convergence radius r > 0. Let X be a Banach space and T B(X ). Then, by the same type of argument that goes into proving Prop 0.12 (see [FA1]), n=0 a nt n converges (in B(X ), i.e. in operatornorm) for all T B(X ) which satisfy T < r (r = 1 in Prop 0.12). NB. In fact, both here, and in 0.12, it is enough (exercise) to have lim sup T m 1 /m < r m Conclusion: For T B(X ) we can define f (T) B(X ) for functions f which are powerseries (analytic functions). Hence a part of spectral theory is the goal of enlarging the class of functions f for which we can make sense of f (T) (possibly at the price of not making sence for all T B(X ), as above for example ( T < r needed). Why do so?!? - We shall see later what this is good for. Also, we shall study the relationship between spectra of T and f (T). Examples (1) For all T B(X ) we define the exponential function by exp(t) := e T 1 := n! T n B(X ) (in ODE for matrices...). For S, T B(X ): If ST = TS then e T+S = e T e S (otherwise may fail). (2) For T B(X ), define A(s) := e st B(X ) for s ( s A(s) B(X ); A (, B(X )) then n=0 d A(s) = TA(s) = A(s)T ds A(0) = I (i.e. A = TA, A(0) = I. Note the analogy to f (t) = e at, f = a f, f (0) = 1). (3) For T B(X ) with I T < 1 let ln(t) = log(t) = 1 n (I T)n B(X ) (4) For T B(X ) with T < 1 and s, s < 1 define A(s) = log(i st) (as in (3)). Then A (( 1, 1), B(X )) with d ds A(s) =! T(I st) 1 = (I st) 1 T and (!) exp(a(s)) = exp(log(i st)) = I st = (exp log)(i st). NB. This says f (g(t)) = (f g)(t). We already saw this for f (x) = x 1, g(x) = x n. n=1 9

10 1 Spectral Theory for compact operators For a while we will be interested in the point spectrum σ p (T) for operators T B(X ) (or, for some of them...), i.e. we look at the eigenvalue problem for T. Given y X, λ, x X we look for all solutions to (T λi)x = T x λx = y ( ) If λ ρ(t) (so that (T λi) 1 B(X ) exists), then ( ) has a unique solution x 0 (for all y X, x 0 = x 0 (y)) given by x 0 := (T λi) 1 y. If, on the other hand, λ σ p (T) then a solution x 0 to ( ) if such exists is not unique anymore. If x N(T λi) ( {0}, when λ σ p (T)) then (T λi)(x + x 0 ) = (T λi)x + (T λi)x 0 = 0 + y = y So the number of degrees of freedom for solutions x equals (!) the dimension of N(T λi). On the other hand, for a solution x 0 to ( ) to exist we must have that y R(T λi). This can be thought of as a constraint. If we had/have a scalar product in X : y has to be orthogonal to the orthogonal complement (U := R(T λi) ) of R(T λi). So y R(T λi) iff y U iff y, u i = 0 for all u i {u j } j ONB for U. I.e. the number of u js in the basis is equal to the number of constraints on y. An important class of operators are those when both the number of degrees of freedom for x (= dim N(T λi)) as well as the number of constraints on y are finite. Definition 1.1. An operator A B(X, Y ) is called a Fredholm operator iff (i) dim N(A) < (ii) R(A) Y is closed (iii) co dim R(A) < The index of a Fredholm operator is then defined as: Remark 1.2. co dim R(A) < means ind(a) = dim N(A) co dim R(A) < Y = R(A) Y 0 (#) with Y 0 Y a linear subspace with dim Y 0 < where the claim is that in this case, co dim R(A) := dim Y 0 is independent of the choice of Y 0 such that (#) holds. I.e. if also Y 1 Y is a linear subspace with dim Y 1 < and Y = R(A) Y 1 holds, then dim Y 0 = dim Y 1 holds aswell 1. A large and important class of Fredholm operators (the most important class?!?) is when X = Y and A is a compact perturbation of the identity I: 1 In a Hilbert space we can replace this with Y = R(A) R(A). We see that R(A) is closed is important. 10

11 Theorem 1.3. Let X be a Banach space and T K(X ). Then A := I T is a Fredholm operator with index zero. In particular (1) dim N(A) < (2) R(A) is closed. (3) N(A) = {0} implies R(A) = X. NB. close to finite dimensions!!! (4) co dim R(A) dim N(A) (5) dim N(A) co dim R(A) NB. The information in (1) - (5) is redundant, but as we refer to them later, we list all the points. Proof. (1) Note that Ax = 0 is equivalent to x = T x. So B 1 (0) N(A) T(B 1 (0)) which is pre-compact in X since T K(X ). Hence N(A) is finite-dimensional. n (2) Let x R(A) then there is a sequence (y n ) n R(A), with y n x. I.e. there is n a sequence (x n ) n X with Ax n = y n R(A) such that Ax n x. Claim: We can assume, with d n := dist(x n, N(A)), that x n 2 d n. True because if not, we change x n : Choose a n N(A) with x n a n dist(x n, N(A)) + dist(x n, N(A)) = 2 dist(x n, N(A)) 2d n NB. This is just the property of inf being the biggest lower bound. The right number could have been any b > 0. We chose it to be dist(x n, N(A)). Let then x n := x n a n then A x n = Ax n and x n 2 dist(x n, N(A)) = 2 dist( x n, N(A)) (since a n N(A)). Then the sequence (d n ) n is bounded, because assume for contradiction that not, i.e. d n for a subsequence (called the same!). Let z n := x n n, d n then Az n = Ax n n 0. Now z d n 2, because x n 2d n, so (z n ) n is bounded and T n n is compact, so there is a subsequence (called the same) such that Tz n z for some z. Hence (A = I T), z n = Az n + Tz n n z ( ) and, A is continuous, so (see ( )). So z N(A). Hence Az = lim n Az n = 0 z n z dist(z n, N(A)) xn = dist, N(A) d n = 1 d n dist(x n, N(A)) = 1 11

12 which is a contradiction to ( ). Hence, (d n ) n is a bounded sequence. So, since x n 2d n n 1, it follows that (x n ) n is bounded. Since T K(X ), there exists a subsequence n (called the same) such that T x n u for some u. Hence R(A) x n y n = Ax n = A(A + T)x n = A(Ax n + T x n ) n A(x + u) So x = A(x + u) (uniqueness of limit). So x R(A) and R(A) is closed. (3) Assume that N(A) = {0} and that, for contradiction, there exists an x X \ R(A). Claim: Then A n x R(A n ) \ R(A n+1 ) ( ) for every n 1. True because, if not, then there exists n, such that A n x R(A n+1 ), then A n x = A n+1 y for some y, so A(A n 1 (x Ay)) = A n (x Ay) = 0 i.e. A n 1 (x Ay) = 0, since N(A) = {0} and A n 1 x = A n y. By induction x Ay = 0. But then x = Ay R(A), contradiction (we chose x X \ R(A)). R(A n+1 ) is closed 2. Recall A n x R(A n ) \ R(A n+1 ) for every n 1. So then there exists a n+1 R(A n+1 ) 0 < A n x a n+1 2 dist(a n x, R(A n+1 )) ( ) Let now For y R(A n+1 ) we have x n := An x a n+1 A n x a n+1 R(An ) R(A n+1 ) {}}{ x n y = An (a n+1 + A n x a n+1 y) A n x a n+1 dist(an x, R(A n+1 ) A n x a n (by ( )) So dist(x n, R(A n+1 )) 1 2. Now, for m > n we have R(Am ) R(A n ): T x n T x m = x n (Ax n + x m Ax m ) 1 }{{} 2 R(A n+1 ) ( ) But x n = 1 by definition, for all n, so (x n ) n is bounded. Since T is compact, (T x n ) n should at least have one convergent subsequence, but ( ) says it does not. Contradiction! Hence X = R(A), so A is surjective. (4) By (1) we have that n := dim N(A) < is finite. Let {x 1,..., x n } be a basis for N(A) and assume for contradiction that co dim R(A) > dim N(A). Then there exist linearly independent vectors { y 1,..., y n } X such that span{ y 1,..., y n } R(A) X ( ) 2 This is by (2), A n+1 is still Fredholm according to ( ) in the proof of Theorem 1.4 (3) 12

13 Then, by Theorem 1.5 there exists {x 1,..., x n } X such that x l (x k) = δ kl, k, l = 1,..., n. For x X define n T x := T x + x k (x)y k This defines a compact operator T K(X ). Let à = I T then N(Ã) = {0}. True because if Ãx = 0, then n 0 = Ãx = (I T)x x }{{} k (x)y k k=1 =Ax R(A) }{{} span{ y 1,...,y n } and by ( ) it follows that Ax = 0 and x k (x) = 0 for k = 1,..., n (since { y 1,..., y n } is linearly independent). But x N(A) and then x = n k=1 α k x k for some α k. Hence k=1 n 0 = x l (x) = α k x l (x k) = α l k=1 l = 1,..., n So x = 0. Hence N(Ã) = {0}. Now use (3) on Ã, then R(Ã) = X. Since Ãx l = y l for l = 1,..., n and n à x + x l (x)x l = Ax x X l=1 so X = R(Ã) span{ y 1,..., y n } R(A). This contradicts ( ). Hence co dim R(A) dim N(A). (5) Let m := co dim R(A) and n := dim N(A). Then by (4) m n. We first reduce to the case m = 0. As in (4), choose x 1,..., x n N(A) and x 1,..., x n X and y 1,..., y m X with X = span{ y 1,..., y m } R(A) As in the proof of (4) it follows that the map T x := T x + m x k (x)y k K(X ) k=1 is compact and à = I T is surjective with N(Ã) = span{x i m < i n}. It then remains to prove that N(Ã) = {0} (i.e. dim N(Ã) = 0). This means the statement is reduced to be proved for the case m = 0. In the case m = 0 we have R(A) = X. Assume, for contradiction, that there exists x 1 N(A) \ {0}. Since A is surjective, we can inductively choose a sequence x k X, k 2 with Ax k = x k 1. Then x k N(A k ) \ N(A k 1 ). By the Riesz-Lemma there exists z k N(A k ) with z k = 1 and dist(z k, N(A k 1 )) 1. Then for 2 l < k Tz k Tz l = z k (Az k + z l Az l ) 1 }{{} 2 N(A k 1 ) So (as earlier...) the sequence (Tz k ) k does not have any convergent subsequence. However, z k = 1 is a contradiction to the compactness of T. This result is important in itself, but also essential in proving the following: 13

14 1.1 Spectral Theorem for compact operators Theorem 1.4 (Spectral Theorem for compact operators, Riesz-Schauder). Let X be a Banach space. For every compact T K(X ) one has (1) σ(t)\{0} consists of countably (finite or infinitely) many eigenvalues, with 0 as only possible accumulation point. If σ(t) contains infinitely many points, then it follows (= σ(t), since σ(t) closed). NB. 0 might not be an eigenvalue! (2) For λ σ(t) \ {0} one has σ(t) = σ p (T) {0} 1 n λ := max{n N((T λi) n 1 ) N((T λi) n )} < n λ is the order (or index 3...) of λ (as an eigenvalue) and dim(n(t λi)) is the multiplicity of λ. (3) (Riesz decomposition). For λ σ(t) \ {0} one has X = N((T λi) n λ ) R((T λi) n λ ) ( ) Both subspaces are closed, T-invariant and N((T λi) n λ) is finite-dimensional ( generalised eigenspace ). (4) For T R((T λi) n λ ) : R((T λi) n λ) R((T λi) n λ) we have σ(t R((T λi) n λ ) ) = σ(t)\{λ}. (5) Let, for λ σ(t) \ {0}, E λ : X X be the projection on N((T λi) n λ) according to ( ), then E µ E λ = δ µλ E λ for λ, µ σ(t) \ {0}. Proof. (1) Let λ / σ p (T), λ 0. Then N(I T/λ) = N(λI T) = {0} since λ / σ p (T). So, since T/λ is compact, it follows from Theorem 1.3 (3) that R(T λi) = R(λI T) = R(I T/λ) = X So λ ρ(t) (i.e. λ / σ(t)). In other words σ(t) \ {0} σ p (T) ( ) If σ(t) \ {0} is not finite, choose λ n σ(t) \ {0}, n with λ n λ m for n m, and eigenvectors e n 0 corresponding to λ n (Te n = λ n e n, see ( )) and define X n := span{e 1,..., e n } (dim X n n < ). Claim: Then the e n s are linearly independent (in particular, dim X n = n). True because: By induction in n. n = 1 is trivial. Assume the claim is true for n 1 3 This is no good choice since we use index for Fredholm operators. 14

15 (i.e. {e 1,..., e n 1 } is linearly independent). Assume that e n = n 1 k=1 α ke k (for n ) and α k. Then n 1 0 = Te n λ n e n = T α k e k λ n k=1 n 1 k=1 n 1 n 1 = α k (Te k λ n e k ) = α k (λ k e k λ n e k ) k=1 k=1 n 1 n 1 = α k (λ k λ n )e k =: β k e k k=1 k=1 α k e k where β k := α k (λ k λ n ). Hence, β k = 0 for all k, so α k = 0 for every k (since λ n λ k k!). So e n = n 1 k=1 α ke k = 0. Hence, dim X n = n and X n 1 X n is a proper subspace (meaning X n 1 X n ). Also, X n 1 is closed (since finite-dimensional). Then 4 there exists an x n X n, with x n = 1 and dist(x n, X n 1 ) 1/2. Note that all the X k s are T-invariant, for let x X k, then x = k i=1 β ie i for some β i. Then T x = k β i Te i = i=1 k (β i λ i )e i X k Also, since x n X n, X n 1 X n, and X n = span{e 1,..., e n } we have x n = α n e n + x n for some α n, x n X n 1. Then So, for m < n T i=1 T x n λ n x n = α n Te n + T x n α n λ n e n λ n x n xn λ n Hence, the sequence = α n (Te n λ n e }{{} n ) + T x n λ n x n = T x n λ n x n X n 1 =0 T xm λ m = x n + 1 λ n (T x n λ n x n ) 1 T xk λ k T x m λ 1 m 2 } {{ } X n 1 by calc just done m<n k has no convergent subsequences (i.e. no accumulation points). Since T is compact, it maps bounded sequences into sequences with convergent subsequences. So, the sequence xn cannot have any bounded subsequence. So ( x n = 1) we have that λ n n 1 λ n = x n/λ n n that is: λ n n 0. This proves that 0 is the only possible accumulation point for σ(t) \ {0}. In particular σ(t) \ B r (0) is finite for all r > 0, so σ(t) \ {0} is countable. 4 This is Lemma 2.20 in [FA1]: If X is normed, Y X is a closed, proper subspace and θ (0, 1) then x θ X with x θ = 1 and θ dist(x 0, Y ) 1. Note: If X is an inner product space (Hilbert for example), one can find x θ such that dist(x 0, Y ) = 1, take it in the orthogonal complement! 15

16 (2) Let A = λi T = (T λi) Then N(A n 1 ) N(A n ) for all n (always!). Assume (for contradiction) that N(A n 1 ) N(A n ) for all n 1. Note: N(S) = S 1 ({0}) is always a closed subspace when S is bounded. So N(A n 1 ) is a closed, proper subspace of N(A n ) so we an choose x n N(A n ) such that x n = 1, dist(x n, N(A n 1 )) 1. Then for m < n 2 (T = λi A) T xn T x m = λxn (Ax n + λx m Ax m ) = λ x n 1 λ (Ax n + λx m Ax m ) Claim: Ax n + λx m Ax m N(A n 1 ). In this case, also 1 λ (Ax n + λx m Ax m ) N(A n 1 ) so T x n T x m λ dist(x n, N(A n 1 )) λ 2 > 0. To see the claim, we have x m N(A m ), m < n (by construction) and N(A k 1 ) N(A k ). So x m N(A n 1 ) (n > m). Hence λx m N(A n 1 ) and A n 1 (Ax m ) = A(A n 1 x m ) = A 0 = 0, so Ax m N(A n 1 ), similarly: A n 1 (Ax n ) = A n x n = 0 since x n N(A n ), i.e. Ax n + λx m Ax m N(A n 1 ). But x n = 1 n, so (x n ) n is bounded and (T x n ) n has no convergent subsequence which is a contradiction to T being compact. Hence there exists an n such that N(A n 1 ) = N(A n ). Set n λ := n. Then for m > n, x N(A m ): A m n x N(A n ) = N(A n 1 ) (since A n (A m n x) = A m x = 0). Then A n 1 A m n x = 0, so A m 1 x = 0 and x N(A m 1 ). So N(A m ) = N(A m 1 ). By induction it follows N(A m ) = N(A n ) = N(A n 1 ) for all m n. Hence n λ < and since λ σ(t) \ {0} σ p (T) we have so n λ 1. N(A) = N(λI T) = N(T λi) {0} (3) Let again A := λi T and λ 0, since λ σ(t) \ {0}, so by Theorem 1.3 A = λ(i T/λ) is a Fredholm operator. We claim that N(A n λ) R(A n λ) X, i.e. the two subspaces form a direct sum. Their intersection is {0}. True because if x N(A n λ) R(A n λ) then A n λ x = 0 and x = A n λ y for some y X. Then A 2n λ y = A n λa n λ x = A n λ x = 0. So y N(A 2n λ) = N(A n λ) (by (2)), so x = A n λ y = 0, i.e. N(A n λ) R(A n λ) = {0}. Note that 5 n λ A n λ = (λi T) n λ = λ n λ I + k=1 nλ k λ n λ k ( T) k Note that if B, C B(X ), B or C compact, then BC is compact 6, so ( T) k is compact. Hence n λ nλ k=1 k λ n λ k ( T) k is in K(X ). So A n λ = λ n λ(i T), T K(X ), i.e. by Theorem 1.3, A n λ is Fredholm. 5 Binomial Formula (B + C) n = n k=0 n k B k C n k if BC = CB, B, C B(X ) 6 By exercise sheet 1 ( ) 16

17 NB. Note that there is nothing particular about n λ here. This would have worked with every n. Therefore, by 1.3 (4) and 1.3 (1) we have co dim R(N n λ ) dim N(A n λ ) < so it follows that X = N(A n λ) R(A n λ). Note that T commutes with A (AT = TA) and so TA n λ = A n λ T, therefore both N(A n λ) and R(A n λ) are T-invariant. (4) From (3) we have that R(A n λ) = R((λI T) n λ) is T-invariant. Let T λ be the restriction of T to R(A n λ). Then T λ is still compact and by Theorem 1.3 (2) R(A n λ) is a closed subspace of X, since A n λ is Fredholm (see above). Hence, R(A n λ) is a Banach space. Also we have 7 N(λI T λ ) = N(A) R(A n λ ) = {0} So λi T λ is injective and by 1.3 (3) it follows that λi T λ is onto. So we have So λ ρ(t λ ). R(λI T λ ) = R(A n λ ) It remains to prove σ(t λ ) \ {λ} = σ(t) \ {λ}. Let µ \ {λ}. Recall that N(A n λ) is also T-invariant, hence so is under µi T (so was R(A n λ) already used). We claim: µi T is injective on N(A n λ). Assume x N(µI T), then (µi T)x = 0, i.e. (λ µ)x = Ax. If A m x = 0 for some m 1, then (λ µ)a m 1 x = A m 1 (λ µ)x = A m 1 (Ax) = A m x = 0 Hence (λ µ), A m 1 x = 0. By induction we get that x = A 0 x = 0. So N(µI T) N(A m ) = {0} for all m 1. In particular, for m = n λ : N(µI T) N(A n λ) = {0}. So (µi T) N(A n λ ) is injective. But N(A n λ) is finite dimensional (by 1.3), so µi T is a bijection on N(A n λ). Hence, wheter or not µi T is invertible (with bounded inverse...) on X is equivalent to whether or not it is invertible on R(A n λ). That is µ ρ(t) µ ρ(t λ ). Summing up: By splitting off (off X ) the finite dimensional characteristic subspace N(A n λ) belonging to the eigenvalue λ, we get a rest operator T λ (on R(A n λ)) for which σ(t λ ) = σ(t) \ {λ}. (5) Let λ, µ σ(t) \ {0}, λ µ and let A λ := λi T, A µ := µi T. Let x N(A n µ µ ) X = N(A n λ λ ) R(An λ λ ). Then there exist unique elements z N(An λ λ ), y R(An λ λ ) such that x = z + y. Both N(A n λ λ ) and R(An λ λ ) are T-invariant, so also A µ-invariant, hence A n µ µ -invariant. So, since x N(A n µ µ ) so, by X = N(A n λ λ ) R(An λ) we get λ 0 = A n µ µ x = A n µ µ z 7 Since N(A n λ) R(A n λ) = {0} and N(A) N(A n λ). N(A n λ λ ) + A A n µ µ z = 0 = A n λ µ y n µ µ y R(A n λ λ ) 17

18 Recall that A µ is a bijection on N(A n λ λ ), so An µ µ is also a bijection on N(A n λ). Hence z = 0. λ It follows that x = y R(A n λ). So λ N(A n µ µ ) R(A n λ λ ) ( ) Now E µ : X X is the projection on N(A n µ µ ) (ditto for λ). So R(E µ ) = N(A n µ µ ) R(E λ ) = N(A n λ λ ) So ( ) says: R(E µ ) N(E λ ). Interchanching the roles of λ and µ, we get R(E λ ) N(E µ ) hence (check!) E µ E λ = δ λµ E λ for µ, λ σ(t) \ {0}. Theorem 1.5. Let X be a normed space and E an n-dimensional subspace with basis {e 1,..., e n }. Let Y X be a closed subspace with Y E = {0}. Then (1) There exists e 1,..., e n X with e j = 0 on Y and e i (e j) = δ i j for i, j = 1,..., n (2) There exists a projection P B(X ) on E, with Y N(P). Proof. (1) Let Y j := span{e k k j} Y. Then, by Lemma 1.6, Y j X is a closed subspace. Also e j / Y j. Hence, by Lemma 1.7, there is e j X with e j 0 on Y j and e j (e j) = 1. In particular e j 0 on Y and e j (e k) = 0 for all k j, i.e. e j (e k) = δ jk. (2) Let P x := n e j (x)e j E P continuous because it has finite rank and identically 0 on Y. j=1 Lemma 1.6. Let X be a normed space. For all finite-dimensional subspaces E X and Y X closed with Y E = {0}, then the direct sum Y E X is also a closed subspace. Proof. Let X be the space X with the equivalence relation x 1 x 2 : x 1 x 2 Y Then X with x X := dist(x, Y ) is a normed vector space (since Y is closed). Since Y E = {0} the dimension of E, as a subspace of X is the same as its dimension as a subspace of X. Let k (y n ) n Y and (z n ) n E with y k + z k x X. Then z k x X = dist(z k x, Y ) k 0 since (z k x) + y k k 0 and y k Y. Hence (z k ) k E is a Cauchy sequence in ( X, X ). Since E is complete (it is a finite dimensional subspace of X ), so there is z E such that k z k z X 0. I.e. x = z X by uniqueness of limits, i.e. x z =: y Y by the definition of X. That is, x = z + y E Y. Lemma 1.7. Let X be a normed space and Y X a closed proper subspace and x 0 / Y. Then there is x X such that x 0 on Y, x = 1 and x (x 0 ) = dist(x 0, Y ). 18

19 Proof. On Y 0 := Y span{x 0 }, define y 0 (y + αx) := α dist(x 0, Y ) for y Y, α. Then y 0 : Y 0 is linear and y 0 on Y. For y Y, 0 α, then Hence dist(x 0, Y ) x 0 y α y 0 (y + αx 0) α x 0 y α = αx0 + y so y 0 Y with y 0 1. To prove y 0 = 1. Since Y is closed, we have dist(x 0, Y ) > 0, so for every ɛ > 0 there is some y ɛ Y such that Then x 0 y ɛ (1 + ɛ) dist(x 0, Y ) y 0 (x 0 y ɛ ) = dist(x 0, Y ) ɛ x 0 y ɛ Now x 0 y ɛ 0, so y 0 1 ɛ 0 1, so y 1+ɛ 0 = Consequences of the spectral Theorem for compact operators Proposition 1.8 (Fredholm Alternative). Let X be a Banach space. For T K(X ) and λ 0 one has that either the equation T x λx = y has a unique solution x for any y X or that the equation T x λx = 0 has non-trivial solutions. NB. In this case Either or is mutually-exclusive (not ). Proof. Follows from σ(t) \ {0} σ p (T). NB. Compare with finite-dimensional case! Proposition 1.9. Let X be a Banach space and T K(X ) and λ σ(t)\{0}. Then the resolvent map ρ(t) µ R µ (T) = (T µi) 1 has an isolated pole of order n λ at λ (see 1.4 (2)). That is, the map ρ(t) µ (µ λ) n λ R µ (T) B(X ) can be continued at the point λ to an analytic map (i.e. with a convergent power series representation, with values in B(X )) and its value at µ = λ is different from the 0-operator. Proof. See e.g. Alt, Lineare Funktionalanalysis, 6. Auflage, p The finite-dimensional case Proposition 1.10 (finite-dimensional case). Let X be a finite-dimensional -vector space, and T : X X linear. Then there exist pairwise different λ 1,..., λ m, 1 m dim X such that σ(t) = σ p (T) = {λ 1,..., λ m } and orders n λj (of λ j ) with the properties in 1.4 (2) - (5). That is, X = N((T λ 1 I) n λ 1 ) N((T λ m I) n λm ) 19

20 Proof. Equip X with your favorite norm, then X is Banach and T K(X ). So T µ = T µi for any µ is also compact since I K(X ). Now apply 1.4 on (for example) T 0 and T 1. Proposition 1.11 (Jordan Normal Form). Let X be a finite-dimensional -vector space and T K(X ), λ σ p (T). For A = T λi one has (1) For n = 1,..., n λ there exist subspaces E n N(A n ) \ N(A n 1 ) such that n λ k 1 N(A n λ ) = N k N k := A l (E k ) k=1 (2) The subspaces N k (k = 1,..., n λ ) are T-invariant and d k = dim A l (E k ) is independent of l {0,..., k 1} (3) If {e k,j j = 1,..., d k } E k are a basis for E k then l=0 {A l e k,j 0 l < k n λj, 1 j d k } is a basis for N(A n λ) and with x = α k, j,l A l e k,j y = β k,j,l A l e k,j k,j,l the equation T x = y is equivalent to λ λ λ λ k,j,l α k,j,0.... α k, j,k 1 = β k, j,0.... β k,j,k 1 (that is, the matrix representing T in this basis has Jordan Normal Form). Proof. If E is a subspace with N(A n 1 ) E N(A n ) then N(A n l 1 ) A l (E) N(A n l ) for all 0 l < n and A l is injective on E, since if x E with A l x = 0 then A n 1 x = 0 since l n 1, hence x N(A n 1 ) E = {0}. Now using this inductively, choose E n for n = n λ, n λ 1,..., 2, 1 such that from which the claim follows. N(A n ) = N(A n 1 ) n λ n l=0 A l (E n+l ) Theorem 1.12 (Schauder). T K(X, Y ) if and only if T K(Y, Y ) Proof. See [FA1] Theorem 5.30 (b) 20

21 1.4 Spectral Theory for normal compact operators on Hilbert spaces The last part of this chapter on spectral theory of compact operators is to investigate the case of compact, normal operators on a Hilbert space. We shall see: The Riesz-Schauder decomposition (Theorem 1.4) X = N((T λi) n λ ) R((T λi) n λ ) in this case results in an orthonormal decomposition of the whole space X (H when Hilbert) in eigenspaces. This is the diagonalisation in the infinite-dimensional case. Throughout let H be a Hilbert space. Definition T B(H) is called a normal operator iff T T = T T (i.e. [T, T ] = T T T T = 0 B(H)). Lemma Let T B(H). T is normal iff T x = T x for every x H. Proof. ( ) This is done by the computation T x 2 = T x, T x = x, T T x = x, T T x = T x, T x = T x 2 ( ) From polarisation 1 x + y 2 x y 2 = Re x, y 4 x, y H follows Re T x, T y = Re T x, T y. If =, then replace y by i y to get 0 = T x, T y T x, T y = T T x T T x, y = (T T T T )x, y Theorem Let X be a Banach space and T B(X ). Then σ(t) is a compact, non-empty subset of (if X 0!) and sup λ = lim T m 1 /m T λ σ(t) m We call r(t) := sup λ σ(t) the spectral radius of T. NB. The sup is in fact a max and σ(t) B T (0). We shall need some complex analysis for the proof. Theorem (Cauchy s Integral Theorem). Let Ω open and simply connected and assume f : Ω is analytic. Let γ: [a, b] Ω by any closed path, i.e. γ(a) = γ(b). Then γ f (z) dz = 0 The same holds when f : Ω X where X is a -Banach space (!!). Only the -linear structure and completeness are needed! Obviously we need to generalize the Riemann-integral to functions g : [a, b] X easy using middle sums. 21

22 Proof. By Proposition 0.3, σ(t) is open, so σ(t) is closed. By Proposition 0.12, the operator I T/λ is invertible in B(X ) if T < λ and in this case R λ (T) = 1 λ (I T/λ) 1 = 1 T n 1 = λ λ λ n+1 T n Hence r(t) T (since λ > T = λ ρ(t)). So σ(t) is bounded and so (by Heine- Borel) σ(t) is compact 8 ). Note that m 1 T m λ m I = (T λi)p m (T) = p m (T)(T λi) p m (T) = λ m 1 i T i (!) Hence if λ σ(t) then 9 λ m σ(t m ). So λ m T m (just proved!) and λ T m 1 /m for all m. Hence r(t) lim inf m T m 1 /m. We now show also n=0 r(t)! lim sup T m 1 /m m Then it follows that ( T m 1 /m ) m is convergent, and r(t) = lim m T m 1 /m. Recall that Proposition 0.3 says that ρ(t) is open and that the map λ R λ (T) is complex analytic, i.e. for all λ 0 \ B r (0) (since σ(t) B r (0) we have ρ(t) \ B r (0) with r r(t)), there exist c j (λ 0 ) B(X ), δ(λ 0 ) > 0 such that R λ (T) = n=0 i=1 c j (λ 0 )(λ λ 0 ) j λ B r (λ 0 ) j=0 Hence, as a consequence of the Cauchy Integral Theorem: For and 0 j, and s > r(t) = r the (curve) integral 1 λ j R λ (T)dλ 2πi B s (0) is independent of s. To see this we deform the curve and use Cauchy Integral Theorem (not done here). Hence 1 λ j R λ (T)dλ = 1 λ j n 1 T n dλ 2πi 2πi B s (0) B s (0) We curve B s (0) as γ: [0, 2π), θ se iθ with γ (θ) = ise iθ since the convergence is uniform = 1 2π = 1 2π n=0 n=0 s j n 2π e iθ(j n) dθ T n 0 s j n 2πδ j,n T n = T j n=0 8 This holds for any bounded operator not only compact operators! 9 This relates σ(f (T)) and σ(t) with f (x) = x m! 22

23 Hence, for any j 0 and s > r(t) T j = 1 λ j R 2π λ (T)dλ = 1 2π 1 2π 2π 0 B s (0) 2π 0 (se iθ ) j R se iθ (T)ise iθ dθ 1 2π (se iθ ) j R se iθ (T)ise iθ dθ s j+1 1 sup R λ (T) dθ λ =s 2π = s j+1 sup R λ (T) λ =s So for any s > r(t) and any sequence j we have T j 1 /j s s sup R λ (T) λ =s 1/j j s 0 Hence, lim sup T j 1 /j s s > r(t) j So lim sup j T j 1 /j r(t). Assume that σ(t) =, then for j = 0 and s 0 i.e. I = 0, so X = {0}. I = T 0 s sup R λ (T) s 0 0 λ =1 Proposition Let H be a -Hilbert space, H {0} and T B(H) be normal. Then sup λ = T λ σ(t) Proof. Let T 0. Since lim m T m 1 /m T it suffices to prove that T m T m for m 0 (then in fact T m = T m ). For m = 0, 1 this is clear. For m 1 and x H: T m x 2 = T m x, T m x = T T m x, T m 1 x T T m x T m 1 x and since T y = T y for (in particular) y = T m x, by Lemma 1.14, we have T m+1 T m 1 x 2 Hence T m 2 T m+1 T m 1. If we assume, by induction T m T m (so, in fact, equality!), then T m+1 T m 2 T m 1 = T 2m (m 1) = T m+1 Definition Let T B(H) be self adjoint. T is called positive semi-definite, T 0, iff x, T x 0 x H 23

24 Proposition Let H be a -Hilbert space, T B(H). (i) If T = T, then σ(t) [ T, T ] If even T K(H), then T or T is an eigenvalue of T (or both). (ii) If T = T and T is positive semi-definite, then If also T K(H) then T is an eigenvalue. σ(t) [0, T ] Proof. Let U := { x, T x x = 1, x H}. Since x, T x = T x, x = x, T x we have U. Let λ U c. We have to show that T λi is invertible. Let 0 x ker(t λi). Then for y H 0 = y, (T λi)x = y, T x λ y, x Inserting y := x 2 x yields λ = x, T x, hence λ U. So T λi is injective. Let A := T λi. Since U is closed, we have dist(λ, U) ɛ > 0. Then R(A) is closed. True because, for x H with x = 1, we have ɛ x, T x λ and so for every 0 x H: x, x ɛ x, T x λ x, x = x, (T λ)x Let (T λ)x n be a Cauchy-sequence in R(A) converging to z H. Then x n x m 2 = x n x m, x n x m 1 ɛ x n x m, (T λ(x n x m ) 1 ɛ x n x m (T λ)(x n x m ) So x n x m 1 ɛ (T λ)(x n x m ) Hence, (x n ) n H is a Cauchy-sequence and therefore x n x for some x H and since A is continuous, we have Ax = z, so z R(A). Now assume for contradiction that A is not surjective. Then there exists some non-zero y R(A) and for all x R(A) it follows that 0 = Ax, y = x, Ay Inserting x = Ay gives Ay = 0. Hence A is not injective. Summing up, we have U and if λ U c, then A = T λi is invertible. Hence σ(t) inf x H, x =1 x, T x, sup x H, x =1 x, T x ( ) 24

25 By 1.15, we have σ(t) B T (0), hence σ(t) [ T, T ]. In fact sup λ = r(t) = T λ σ(t) by If T K(H), then σ(t) \ {0} consists of isolated eigenvalues, so sup λ = max λ = λ 0 λ σ(t) λ σ(t) where λ 0 σ p (T), hence λ 0 = ± T. Again by ( ) σ(t) 0, sup x H, x =1 x, T x So σ(t) [0, T ]. If T K(H), we also have that T is an eigenvalue (all non-zeroeigenvalues are non-negative, so λ 0 = T above. Note (to 1.18 (2)). In particular: If T K(H), T = T and T is positive semi-definite, then λ 0 := sup λ σ(t) λ = sup x, T x = x 0, T x 0 = λ0 x 0 2 = λ 0 x H, x =1 I.e. maximising x, T x under the constraint x = 1 gives the largest eigenvalue, i.e λ 0 x, T x, x H, x = 1. This we can use to compute a lower bound to λ 0 by choosing some x and compute. Afterwards we repeat this, for λ 1 λ 0 (next eigenvalue) by restricting T to span{x 0 } =: H. H is invariant, since for x H: T x, x0 = x, T x0 = λ0 x, x0 = 0 So T x H. This can be used to compute the λ s and their eigenspaces. Recall the following (Problem 6, Sheet 2, T : l 2 l 2 translated to general setting) Example Let H be a -Hilbert space, {e j } j N, N an orthonormal system and {λ k } k N with λ k r < for k N. Then T x := λ j e j, x e j defines a normal, bounded operator. Also T is compact iff 10 lim k λ k = 0. j N Proof. Let N = and N = {n 1, n 2,...} be an increasing enumeration of N. Let T be compact. We know that e nj 0 as j. Hence, since T is compact, we have Te n j j 0 and 0 k Te nk, Te nk = λ i e i, e nk e i, λ j e j, e nk e j = i N j N 2 λ i λ j e nk, e i e j, e nk e i, e j = λ nk }{{}}{{} =δ i,nk =δ j,nk i,j N k So λ k 0. On the other hand if λ k 0 let T k x := k j=1 λ n j e n j, x e nj, then T x T k x = λ nj e nj, x e n j λ nj e nj, x e nj x j=k+1 So T is a limit of finite-rank operators. j=k+1 j=k+1 The following Theorem shows that every compact normal operator has this form. 10 If N is finite, T is compact. λ n j k 0 25

26 1.4.1 The Spectral Theorem for normal compact operators on a Hilbert space Theorem 1.20 (Spectral Theorem for normal compact operators on a Hilbert space). Let H be a -Hilbert space, T K(H), T 0 and T T = T T. Then (1) There exists an orthonormal system {e j } j N with N and λ k \ {0} such that Te k = λ k e k k N and σ(t) \ {0} = {λ k k N}, i.e. the λ k s are the eigenvalues of T different from 0 with eigenvectors e k. Note. Here (with this notation) λ k s for different k s may be equal. If N is infinite, then λ k k 0. (2) For all k N we have n λk = 1. (3) H = N(T) span{e k k N} Note. This is the orthogonal direct sum: N(T) span{e k k N} (4) We have T x = λ k e k, x e k T is diagonal in {e k } k N k N x H Proof. Apply first the spectral Theorem for compact operators 1.4 on T: σ(t) \ {0} consists of eigenvalues λ k, k N with λ k 0 if N is infinite 11. Moreover, the space N k := N(T λ k I) is finite dimensional (see 1.4). Let N 0 := N(T), λ 0 := 0. Note that N(T λ k I) = N T λ k I k N {0} ( ) which follows from T x = T x for x H by Claim: N k N l for k, l N {0}, k l. True, because λ l x k, x l = x k, T x l ( ) = λ k x k, x l = λ k x k, x l Hence x k, x l = 0 and H = N k =: W k N {0} ( ) True because let y Y := W. By the projection Theorem we have H = Y W (W closed). By ( ), we have for x N k, k N {0}: x, T y = T x, y = λ k x, y = λ k x, y = 0 Hence T y N k for k N {0}, so, since T is continuous, T y Y. Hence, Y is a T-invariant closed subspace of H. We look at T 0 := T Y : Y Y. Then T 0 K(Y ) and T 0 is normal. In the case Y {0}, then since max λ =! T 0 λ σ(t) 11 In this ennumeration, the λ k s are pairwise different. 26

27 there exists λ σ(t 0 ) such that λ = T 0. If T 0 0 then T 0 0, so λ 0 and λ is by 1.4 (on T 0 ) an eigenvalue for T 0 : Y Y. I.e. there is some non-zero u Y such that T 0 u = λu, i.e. Tu = λu, so λ is an eigenvalue for T. So u N k Y {0} for some k N. This leads to u, u = 0 since Y N k which is a contradiction to u 0. Hence T 0 = 0, so Y N(T), but Y N 0 = {0} so Y = 0. Let E k for k N {0} be the orthogonal projection on N k (Problem 5 on Sheet 2), then, from ( ) x = E k x x H So k N {0} T x = T(E k x) = λ k E k x (E k x N k ) ( ) k N {0} k N Now the representation of T follows. Let d k := dim N k < and choose an orthonormal basis: {e k1,..., e kdk }. Then d k E k x = e k j, x e k j j=1 From the representation ( ) of T follows that dk T x = λ k e k j, x e k j λ k e k, x e k k N i=1 (re-number the λ k s and e k j s). From the representation of T also follows that True because for x N((T λ k I) 2 ): 0 = (T λ k I) 2 x ( ) = N k = N(T λ k I) = N((T λ k I) 2 ) j N {0} So E j x = 0 for j k, hence x = E k x N k. k (λ k λ j ) 2 E j x (E 2 j = E j ) Recall that T T = T T and T K(H). Let E k be the orthogonal projection on the eigenspace N(T λ k I) = N k, belonging to the eigenvalue λ k ( λ k λ l, k l), that is The spectral Theorem says d k E k x = e k j, x e k j j=1 T x = λ k E k x k=1 x H I.e. the series on the right converges pointwise to T x at any x H. In fact, we have more: 27

28 Corollary 1.21 (Spectral Theorem for normal, compact operators Projection version). Let H be a -Hilbert space, T K(H), T 0 and and T T = T T, then T = with convergence in operator norm on B(H). λ k E k k=1 Proof. The calculation with the notation as above gives M T λ k E k = λ k E k k=1 and the operator k>m λ k E k is normal and compact, so its norm is its biggest eigenvalue: k>m = sup λ k M 0 k>m Next, we shall use the spectral Theorem to compute square roots of positive semi-definite operators 12. Theorem Let T K(H), T = T, T 0. Then there exists a unique, self-adjoint operator S 0 with S 2 = T. We write S := T 1 /2 = T. Proof. Write from the Spectral Theorem 1.20 T x = λ k e k, x e k Since T 0, we have λ k Let S x := λ k e k, x e k k k Then, since (λ k 0, k ) = ( λ k 0, k ) it follows from Example 1.19 that S is compact, that S is self-adjoint and positive semi-definite. Also we have S 2 x = T x, x H. Let s prove uniqueness. Assume R K(H) satisfies R 0, R 2 = T, R = R. Write, as in Corollary 1.21 R = ν k F k k=1 (convergence in operator norm, ν k eigenvalues of R, F k orthogonal projection on the eigenspaces). Then T = ν 2 k F k k=1 since F k F l = δ kl F k and ν k 0 (since R 0, 1.18). Hence, the ν 2 are the eigenvalues of T, k and the F k s are the eigenprojections of T on the eigenspaces. Hence, since the convergence is pointwise unconditionally, we have R = S. 12 I.e. f (T), for f (x) = x. 28

29 Let T : H 1 H 2 be a compact operator between two Hilbert spaces, then T T : H 1 H 1 is positive semi-definite, self-adjoint and compact. Its unique square root is denoted by T := (T T) 1 /2. Theorem 1.23 (Polar decomposition). For T K(H 1, H 2 ) there exists an operator U B(H 1, H 2 ) with T = U T so that U (N(U)) is an isometry and hence a unitary operator between N(U) and R(U). U is uniquely determined by the demand N(U) = N(T). Proof. T is selfadjoint (all square roots are, by Definition). So T x 2 = x, T 2 x = x, T T x = T x, T x = T x 2 ( ) Hence: U( T x) := T x defines an isometric operator from R( T ) to R(T), which can by (uniquely) extended 13 to U : R( T ) R(T) (same notation: U). Let U x = 0 for x (R( T )) = N( T ) (since N(S) = (R(S )), N(S ) = R(S) ). This proves the existence of U (such that T = U( T )). The uniqueness follows from N( T ) = N(T) (see ( ). Remarks. (1) An operator with the properties of U is called a partial isometry. (2) T = U T reminds of λ = e it λ ( U U = UU = 1 ). However, in general, S + T S + T does not hold (for example). Here A B : B A 0. Theorem 1.24 (Singular Value Decomposition SVD). For any compact T K(H 1, H 2 ) there exist orthonormal systems {e i } i N H 1, { f j } j N H 2 and numbers s 1 s 2 s 3 0 with s k k 0 such that T x = s k e k, x f k x H 1 k N The numbers s 2 k are the eigenvalues of T T (counted with multiplicity). The s k s are called the singular values of T. Proof. Write T = U T as above, and use the Spectral Theorem for normal compact operators 1.20 on T : T x = s k e k, x e k k N Then the s k s and the e k s have the properties in the Theorem. Let f k := U(e k ) H 2. Since U is an isometry on R( T ), which, by the polarization identity, means, it conserves the scalar product, it maps {e k } k N into an orthonormal system { f k } k N. 13 This is Theorem 2.32 in [FA1]. 29