FUNCTIONAL ANALYSIS LECTURE NOTES CHAPTER 2. OPERATORS ON HILBERT SPACES

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1 FUNCTIONAL ANALYSIS LECTURE NOTES CHAPTER 2. OPERATORS ON HILBERT SPACES CHRISTOPHER HEIL 1. Elementary Properties and Examples First recall the basic definitions regarding operators. Definition 1.1 (Continuous and Bounded Operators). Let, Y be normed linear spaces, and let L: Y be a linear operator. (a) L is continuous at a point f if f n f in implies Lf n Lf in Y. (b) L is continuous if it is continuous at every point, i.e., if f n f in implies Lf n Lf in Y for every f. (c) L is bounded if there exists a finite K 0 such that f, Lf K f. Note that Lf is the norm of Lf in Y, while f is the norm of f in. (d) The operator norm of L is L = sup Lf. f =1 (e) We let B(, Y ) denote the set of all bounded linear operators mapping into Y, i.e., B(, Y ) = {L: Y : L is bounded and linear}. If = Y = then we write B() = B(, ). (f) If Y = F then we say that L is a functional. The set of all bounded linear functionals on is the dual space of, and is denoted = B(, F) = {L: F : L is bounded and linear}. We saw in Chapter 1 that, for a linear operator, boundedness and continuity are equivalent. Further, the operator norm is a norm on the space B(, Y ) of all bounded linear operators from to Y, and we have the composition property that if L B(, Y ) and K B(Y, Z), then KL B(, Z), with KL K L. Date: February 20, These notes closely follow and expand on the text by John B. Conway, A Course in Functional Analysis, Second Edition, Springer,

2 2 CHRISTOPHER HEIL Exercise 1.2. Suppose that L: Y is a bounded map of a Banach space into a Banach space Y. Prove that if there exists a c > 0 such that Lf c f for every f, then range(l) is a closed subspace of Y. Exercise 1.3. Let C b (R n ) be the set of all bounded, continuous functions f : R n F. Let C 0 (R n ) be the set of all continuous functions f : R n F such that lim x f(x) = 0 (i.e., for every ε > 0 there exists a compact set K such that f(x) < ε for all x / K). Prove that these are closed subspaces of L (R n ) (under the L -norm; note that for a continuous function we have f = sup f(x) ). Define δ : C b (R n ) F by δ(f) = f(0). Prove that δ is a bounded linear functional on C b (R n ), i.e., δ (C b ), and find δ. This linear functional is the delta distribution (see also Exercise 1.26 below). Example 1.4. In finite dimensions, all linear operators are given by matrices, this is just standard finite-dimensional linear algebra. Suppose that is an n-dimensional complex normed vector space and Y is an m- dimensional complex normed vector space. By definition of dimension, this means that there exists a basis B = {x 1,..., x n } for and a basis B Y = {y 1,..., y m } for Y. If x, then x = c 1 x c n x n for a unique choice of scalars c i. Define the coordinates of x with respect to the basis B to be [x] B = c 1. c n C n. The vector x is completely determined by its coordinates, and conversely each vector in C n is the coordinates of a unique x. The mapping x [x] B is a linear mapping of onto C n. We similarly define [y] BY C m for vectors y Y. Let A: Y be a linear map (it is automatically bounded since is finite-dimensional). Then A transforms vectors x into vectors Ax Y. The vector x is determined by its coordinates [x] B and likewise Ax is determined by its coordinates [Ax] BY. The vectors x and Ax are related through the linear map A; we will show that the coordinate vectors [x] B and [Ax] BY are related by multiplication by an m n matrix determined by A. We call this matrix the standard matrix of A with respect to B and B Y, and denote it by [A] B,B Y. That is, the standard matrix should satisfy [Ax] BY = [A] B,B Y [x] B, x. We claim that the standard matrix is the matrix whose columns are the coordinates of the vectors Ax k, i.e., [A] B,B Y = [Ax 1 ] BY [Ax n ] BY.

3 CHAPTER 2. OPERATORS ON HILBERT SPACES 3 To see this, choose any x and let x = c 1 x c n x n be its unique representation with respect to the basis B. Then c 1 [A] B,B Y [x] B = [Ax 1 ] BY [Ax n ] BY. c n = c 1 [Ax 1 ] BY + + c n [Ax n ] BY = [c 1 Ax c n Ax n ] BY = [A(c 1 x c n x n )] BY = [Ax] BY. Exercise 1.5. Extend the idea of the preceding example to show that that any linear mapping L: l 2 (N) l 2 (N) (and more generally, L: H K with H, K separable) can be realized in terms of multiplication by an (infinite but countable) matrix. Exercise 1.6. Let A be an m n complex matrix, which we view as a linear transformation A: C n C m. The operator norm of A depends on the choice of norm for C n and C m. Compute an explicit formula for A, in terms of the entries of A, when the norm on C n and C m is taken to be the l 1 norm. Then do the same for the l norm. Compare your results to the version of Schur s Lemma given in Theorem The following example is one that we will return to many times. Example 1.7. Let {e n } n N be an orthonormal basis for a separable Hilbert space H. Then we know that every f H can be written f = f, e n e n. n=1 Fix any sequence of scalars λ = (λ n ) n N, and formally define Lf = λ n f, e n e n. (1.1) n=1 This is a formal definition because we do not know a priori that the series above will converge in other words, equation (1.1) may not make sense for every f. Note that if H = l 2 (N) and {e n } n N is the standard basis, then L is given by the formula Lx = (λ 1 x 1, λ 2 x 2,... ), x = (x 1, x 2,... ) l 2 (N). We will show the following (the l -norm of the sequence λ is λ = sup n λ n ). (a) The series defining Lf in (1.1) converges for each f H if and only if λ l. In this case L is a bounded linear mapping of H into itself, and L = λ.

4 4 CHRISTOPHER HEIL (b) If λ / l, then L defines an unbounded linear mapping from the domain { } domain(l) = f H : λ n f, e n 2 < (which is dense in H) into H. Proof. (a) Suppose that λ l, i.e., λ is a bounded sequence. Then for any f we have λ n f, e n 2 λ 2 f, e n 2 = λ 2 f 2 <, n=1 n=1 n=1 (1.2) so the series defining Lf converges (because {e n } is an orthonormal sequence). Moreover, the preceding calculation also shows that Lf 2 = n=1 λ n f, e n 2 λ 2 f 2, so we see that L λ. On the other hand, by orthonormality we have Le n = λ n e n (i.e., each e n is an eigenvector for L with eigenvalue λ n ). Since e n = 1 and Le n = λ n e n = λ n we conclude that L = sup f =1 Lf sup n N Le n = sup λ n = λ. n N The converse direction will be covered by the proof of part (b). (b) Suppose that λ / l, i.e., λ is not a bounded sequence. Then we can find a subsequence (λ nk ) k N such that λ nk k for each k. Let c nk = 1 and define all other c k n to be zero. Then n c n 2 = 1 k <, so f = k 2 n c ne n converges (and c n = f, e n ). But the formal series Lf = n λ nc n e n does not converge, because c n λ n 2 = c nk λ nk 2 k 2 =. k 2 n=1 k=1 In fact, the series defining Lf in (1.1) only converges for those f which lie in the domain defined in (1.2). That domain is dense because it contains the finite span of {e n } n N, which we know is dense in H. Further, that domain is a subspace of H (exercise), so it is an innerproduct space. The map L: domain(l) H is a well-defined, linear map, so it remains only to show that it is unbounded. This follows from the facts that e n domain(l), e n = 1, and Le n = λ n e n = λ n. k=1 Exercise 1.8. Continuing Example 1.7, suppose that λ l and set δ = inf n λ n. Prove the following. (a) L is injective if and only if λ n 0 for every n. (b) L is surjective if and only if δ > 0 (if δ = 0, use an argument similar to the one used in part (b) of Example 1.7 to show that range(l) is a proper subset of H). (c) If δ = 0 but λ n 0 for every n then range(l) is a dense but proper subspace of H. (d) Prove that L is unitary if and only if λ n = 1 for every n.

5 CHAPTER 2. OPERATORS ON HILBERT SPACES 5 In Example 1.7, we saw an unbounded operator whose domain was a dense but proper subspace of H. This situation is typical for unbounded operators, and we often write L: Y even when L is only defined on a subset of, as in the following example. Example 1.9 (Differentiation). Consider the Hilbert space H = L 2 (0, 1), and define an operator D : L 2 (0, 1) L 2 (0, 1) by Df = f. Implicitly, we mean by this that D is defined on the largest domain that makes sense, namely, domain(d) = { f L 2 (0, 1) : f is differentiable and f L 2 (0, 1) }. Note that if f domain(d), then Df is well-defined, Df L 2 (0, 1), and Df 2 <. Thus every vector in domain(d) maps to a vector in L 2 (0, 1) which necessarily has finite norm. Yet D is unbounded. For example, if we set e n (x) = e inx then e n 2 = 1, but De n (x) = e n (x) = ineinx so De n 2 = n. While each vector De n has finite norm, there is no upper bound to these norms. Since the e n are unit vectors, we conclude that D =. The following definitions recall the basic notions of measures and measure spaces. For full details, consult a book on real analysis. 1 Definition 1.10 (σ-algebras, Measurable Sets and Functions). Let be a set, and let Ω be a collection of subsets of. Then Ω is a σ-algebra if (a) Ω, (b) If E Ω then \ E Ω (i.e., Ω is closed under complements), (c) If E 1, E 2, Ω then E k Ω (i.e., Ω is closed under countable unions) The elements of Ω are called the measurable subsets of. If we choose F = R then we usually allow functions on to take extended-real values, i.e., f(x) is allowed to take the values ±. An extended-real-valued function f : [, ] is called a measurable function if {x : f(x) > a} is measurable for each a R. If we choose F = C then we require functions on to take (finite) complex values there is no complex analogue of ±. A complex-valued function f : C is called a measurable function if its real and imaginary parts are measurable (real-valued) functions. Definition 1.11 (Measure Space). Let be a set and Ω a σ-algebra of subsets of. Then a function µ on Ω is a (positive) measure if (a) 0 µ(e) + for all E Ω, (b) If E 1, E 2,... is a countable family of disjoint sets in Ω, then ( ) µ E k = µ(e k ). k=1 1 For example, R. Wheeden and A. Zygmund, Measure and Integral, Marcel Dekker, 1977, or G. Folland, Real Analysis, Second Edition, Wiley, k=1

6 6 CHRISTOPHER HEIL In this case, (, Ω, µ) is called a measure space. If µ() <, then we say that µ is a finite measure. If there exist countably many subsets E 1, E 2,... such that = E k and µ(e k ) < for all k, then we say that µ is σ-finite. For example, Lebesgue measure on R n is σ-finite. It is often useful to allow measures to take negative values. Definition 1.12 (Signed Measure). Let be a set and Ω a σ-algebra of subsets of. Then a function µ on Ω is a signed measure if (a) µ(e) + for all E Ω and µ( ) = 0, (b) If E 1, E 2,... is a countable family of disjoint sets in Ω, then ( ) µ E k = µ(e k ). Definition 1.13 (Integration). Let (, Ω, µ) be a measure space. k=1 (a) If f : [0, ] is a nonnegative, measurable function, then the integral of f over with respect to µ is { ( f dµ = f(x) dµ(x) = sup inf f(x) ) } µ(e j ), x E j where the supremum is taken over all decompositions E = E 1 E N of E as the union of a finite number of disjoint measurable sets E k (and where we take the convention that 0 = 0 = 0). then (b) If f : [, ] and we define k=1 f + (x) = max{f(x), 0}, f (x) = min{f(x), 0}, f dµ = f + dµ j f dµ, as long as this does not have the form (in that case the integral would be undefined). Since f = f + +f and f dµ always exists (either as a finite number or as ), it follows that f dµ exists and is finite f dµ <. (c) If f : C, then f dµ = Re (f) dµ + i Im (f) dµ, as long as both integrals on the right are defined and finite. There are many other equivalent definitions of the integral.

7 CHAPTER 2. OPERATORS ON HILBERT SPACES 7 Definition 1.14 (L p Spaces). Let (, Ω, µ) be a measure space, and fix 1 p <. Then L p () consists of all measurable functions f : [, ] (if we choose F = R) or f : C (if we choose F = C) such that f p p = f(x) p dµ(x) <. Then L p () is a vector space under the operations of addition of functions and multiplication of a function by a scalar. Additionally, the function p defines a semi-norm on L p (). Usually we identify functions that are equal almost everywhere (we say that f = g a.e. if µ{x : f(x) g(x)} = 0), and then becomes a norm on L p (). For p = we define L () to be the set of measurable functions that are essentially bounded, i.e., for which there exists a finite constant M such that f(x) M a.e. Then f = ess sup f(x) = inf { M 0 : f(x) M a.e. } x is a semi-norm on L (), and is a norm if we identify functions that are equal almost everywhere. For each 1 p, the space L p () is a Banach space under the above norm. Exercise 1.15 (l p Spaces). Counting measure on a set is defined by µ() = card(e) if E is a finite subset of, and µ() = if E is an infinite subset. Let Ω = P() (the set of all subsets of ), and show that (, Ω, µ) is a measure space. Show that L p (, Ω, µ) = l p (). Show that µ is σ-finite if and only if is countable. Exercise 1.16 (The Delta Measure). Let = R n and Ω = P(). Define δ(e) = 1 if 0 E and δ(e) = 0 if 0 / E. Prove that δ is a measure, and find a formula for R n f(x) dδ(x). Sometimes this integral is written informally as R n f(x) δ(x) dx, but note that δ is a measure on R n, not a function on R n (see also Exercise 1.26 below). Exercise Fix 0 g L 1 (R n ), under Lebesgue measure. Prove that µ(e) = g(x) dx E defines a finite measure on R n. With this preparation, we can give some additional examples of operators on Banach or Hilbert spaces. Example 1.18 (Multiplication Operators). Let (, Ω, µ) be a measure space, and let φ L () be a fixed measurable function. Then for any f L 2 () we have that fφ is measurable, and fφ 2 2 = f(x) φ(x) 2 dx f(x) 2 φ 2 dx = φ 2 f 2 2 <,

8 8 CHRISTOPHER HEIL so fφ L 2 (). Therefore, the multiplication operator M φ : L 2 () L 2 () given by M φ f = fφ is well-defined, and the calculation above shows that M φ f 2 φ f 2. Therefore M φ is bounded, and M φ φ. If we assume that µ is σ-finite, then we can show that M φ = φ, as follows. Choose any ε > 0. Then by definition of L -norm, the set E = {x : φ(x) > φ ε} has positive measure. Since is σ-finite, we can write = F m where each µ(f m ) <. Since E = (E F m ) is a countable union, we must have µ(e F m ) > 0 for some m. Let F = E F m, and set f = 1 χ µ(f ) 1/2 F. Then f 2 = 1, but M φ f 2 ( φ ε) f 2. Hence M φ 2 φ ε. Exercise: Find an example of a measure µ that is not σ-finite and a function φ such that M φ < φ. Exercise Let (, Ω, µ) be a measure space, and let φ be a fixed measurable function. Prove that if fφ L 2 () for every f L 2 (), then we must have φ L (). Solution. Assume φ / L (). Set E k = {x : k φ(x) < k + 1}. The E k are measurable and disjoint, and since φ is not in L () there must be infinitely many E k with positive measure. Choose any E nk, k N, all with positive measure and let E = E nk. Define 1 f(x) = k µ(e nk ), x E 1/2 n k, 0, x / E. Then but which is a contradiction. f 2 dµ = fφ 2 dµ k=1 k=1 E nk 1 k 2 µ(e nk ) = Enk k 2 k 2 µ(e nk ) = Exercise Continuing Example 1.18, do the following. k=1 k=1 1 k 2 <, 1 =, (a) Determine a necessary and sufficient condition on φ which implies that M φ : L 2 () L 2 () is injective. (b) Determine a necessary and sufficient condition on φ which implies that M φ : L 2 () L 2 () is surjective. (c) Prove that if M φ is injective but not surjective then M 1 φ : range(m φ) L 2 () is unbounded. (d) Extend from the case p = 2 to any 1 p.

9 CHAPTER 2. OPERATORS ON HILBERT SPACES 9 Example 1.21 (Integral Operators). Let (, Ω, µ) be a σ-finite measure space. An integral operator is an operator of the form Lf(x) = k(x, y) f(y) dµ(y). (1.3) This is just a formal definition, we have to provide conditions under which this makes sense, and the following two theorems will provide such conditions. The function k that determines the operator is called the kernel of the operator (not to be confused with the kernel/nullspace of the operator!). Note that an integral operator is just a generalization of matrix multiplication. For, if A is an m n matrix with entries a ij and u C n, then Au C m, and its components are given by n (Au) i = a ij u j, i = 1,..., m. j=1 Thus, the values k(x, y) are analogous to the entries a ij of the matrix A, and the values Lf(x) are analogous to the entries (Au) i. The following result shows that if the kernel is square-integrable, then the corresponding integral operator is bounded. Later we will define the notion of a Hilbert Schmidt operator. For the case of integral operators mapping L 2 () into itself, it can be shown that L is a Hilbert Schmidt operator if and only if the kernel k belongs to L 2 ( ). Theorem 1.22 (Hilbert Schmidt Integral Operators). Let (, Ω, µ) be a σ-finite measure space, and choose a kernel k L 2 ( ). That is, assume that k 2 2 = k(x, y) 2 dµ(x) dµ(y) <. Then the integral operator given by (1.3) defines a bounded mapping of L 2 () into itself, and L k 2. Proof. Although a slight abuse of the order of logic (technically we should show Lf exists before trying to compute its norm), the following calculation shows that L is well-defined and is a bounded mapping of L 2 () into itself: Lf 2 2 = Lf(x) 2 dµ(x) = ( 2 k(x, y) f(y) dµ(y) dµ(x) ) ( ) k(x, y) 2 dµ(y) f(y) 2 dµ(y) dµ(x)

10 10 CHRISTOPHER HEIL = = k 2 2 f 2 2, k(x, y) 2 dµ(y) f 2 2 dµ(x) where the inequality follows by applying Cauchy Schwarz to the inner integral. Thus L is bounded, and L k 2. The following result is one version of Schur s Lemma. There are many forms of Schur s Lemma, this is one particular special case. Exercise: Compare the hypotheses of the following result to the operator norms you calculated in Exercise 1.6. Theorem Let (, Ω, µ) be a σ-finite measure space, and Assume that k is a measurable function on which satisfies the mixed-norm conditions C 1 = ess sup x k(x, y) dµ(y) < and C 2 = ess sup y k(x, y) dµ(x) <. Then the integral operator given by (1.3) defines a bounded mapping of L 2 () into itself, and L (C 1 C 2 ) 1/2. Proof. Choose any f L 2 (). Then, by applying the Cauchy Schwarz inequality, we have Lf 2 2 = Lf(x) 2 dµ(x) = ( = C 1 C 1 2 k(x, y) f(y) dµ(y) dµ(x) k(x, y) 1/2 ( k(x, y) 1/2 f(y) ) dµ(y)) 2 dµ(x) ( ) ( ) k(x, y) dµ(y) k(x, y) f(y) 2 dµ(y) dµ(x) C 1 k(x, y) f(y) 2 dµ(y) dµ(x) = C 1 C 2 f 2 2, f(y) 2 f(y) 2 C 2 dµ(y) k(x, y) dµ(x) dµ(y) where we have used Tonelli s Theorem to interchange the order of integration (here is where we needed the fact that µ is σ-finite). Thus L is bounded and L (C 1 C 2 ) 1/2.

11 CHAPTER 2. OPERATORS ON HILBERT SPACES 11 Exercise Consider what happens in the preceding example if we take 1 p instead of p = 2. In particular, in part b, show that if C 1, C 2 < then L: L p () L p () is a bounded mapping for each 1 p (try to do p = 1 or p = first). Exercise 1.25 (Volterra Operator). Define L: L 2 [0, 1] L 2 [0, 1] by Show directly that L is bounded. k : [0, 1] 2 F defined by Lf(x) = x 0 f(y) dy. Then show that L is an integral operator with kernel k(x, y) = { 1, y x, 0, y > x. Observe that k L 2 ([0, 1] 2 ), so L is compact. This operator is called the Volterra operator. Exercise 1.26 (Convolution). Convolution is one of the most important examples of integral operators. Consider the case of Lebesgue measure on R n. Given functions f, g on R n, their convolution is the function f g defined by (f g)(x) = f(y) g(x y) dy, R n provided that the integral makes sense. Note that with g fixed, the mapping f f g is an integral operator with kernel k(x, y) = g(x y). (a) Let g L 1 (R n ) be fixed. Use Schur s Lemma (Theorem 1.23) to show that Lf = f g is a bounded mapping of L 2 (R n ) into itself. In fact, use Exercise 1.24 to prove Young s Inequality: If f L p (R n ) (1 p ) and g L 1 (R n ), then f g L p (R n ), and f g p f p g 1. In particular, L 1 (R n ) is closed under convolution. (b) Note that we cannot use the Hilbert Schmidt condition (Theorem 1.22) to prove Young s Inequality, since g(x y) 2 dx dy =, R n R n even if we assume that g L 2 (R n ). (c) Prove that convolution is commutative, i.e., that f g = g f. (d) Prove that there is no identity element in L 1 (R n ), i.e., there is no function g L 1 (R n ) such that f g = f for all f L 1 (R n ). This is not trivial it is easier to do if you make use of the Fourier transform on R n, and in particular use the Riemann Lebesgue Lemma to derive a contradiction.

12 12 CHRISTOPHER HEIL (e) Some texts do talk informally about a delta function that is an identity element for convolution, defined by the conditions {, x = 0, δ(x) = and δ(x) dx = 1, 0, x 0, R n but no such function actually exists. In particular, the function δ defined on the left-hand side of the line above is equal to zero a.e., and hence is the zero function as far as Lebesgue integration is concerned. That is, we have δ(x) dx = 0, not 1. The delta function is R n really just an informal use of the delta distribution (see Exercise 1.3) or the delta measure (see Exercise 1.16). Show that if we define the convolution of a function f with the delta measure δ to be (f δ)(x) = f(x y) dδ(y), (1.4) R n then f δ = f for all f L 1 (R n ). Note that in the informal notation of Exercise 1.16, (1.4) reads (f δ)(x) = f(x y) δ(y) dy, R n which perhaps explains the use of the term delta function. Exercise Prove that L 1 (R n ) is not closed under pointwise multiplication. That is, prove that there exist f, g L 1 (R n ) such that the pointwise product h(x) = (fg)(x) = f(x)g(x) does not belong to to L 1 (R n ). Exercise 1.28 (Convolution Continued). (a) Consider the space L p [0, 1], where we think of functions in L p [0, 1] as being extended 1-periodically to the real line. Define convolution on the circle by (f g)(x) = 1 0 f(y) g(x y) dy, where the periodicity is used to define g(x y) when x y lies outside [0, 1] (equivalently, replace x y by x y mod 1, the fractional part of x y). Prove a version of Young s Inequality for L p [0, 1]. (b) Consider the sequence space l p (Z). Define convolution on Z by (x y) n = m Z x m y n m. Prove a version of Young s Inequality for l p (Z). Prove that l 1 (Z) contains an identity element with respect to convolution, i.e., there exists a sequence in l 1 (Z) (typically denoted δ) such that δ x = x for every x l p (Z). (c) Identify the essential features needed to define convolution on more general domains, and prove a version of Young s Inequality for that setting.

13 CHAPTER 2. OPERATORS ON HILBERT SPACES 13 Exercise 1.29 (Convolution and the Fourier Transform). Let F be the Fourier transform on the circle, i.e., it is the isomorphism F : L 2 [0, 1] l 2 (Z) given by Ff = ˆf = { ˆf(n)} n Z, where ˆf(n) = f, e n = 1 0 f(x) e 2πinx dx, e n (x) = e 2πinx. (a) Prove that the Fourier transform converts convolution in to multiplication. That is, prove that if f, g L 2 [0, 1], then (f g) = ˆf ĝ, i.e., (f g) (n) = ˆf(n) ĝ(n), n Z. (b) Note that if g L 2 [0, 1], then g L 1 [0, 1], so by Young s Inequality we have that f g L 2 [0, 1]. Holding g fixed, define an operator L: L 2 [0, 1] L 2 [0, 1] by Lf = f g. Since {e n } n Z is an orthonormal basis for L 2 [0, 1], we have f = n Z ˆf(n) e n, f L 2 [0, 1]. Show that Lf = f g = n Z ĝ(n) ˆf(n) e n, f L 2 [0, 1]. Thus, in the Fourier domain, convolution acts by changing or adjusting the amount that each component or frequency e n contributes to the representation of the function in this basis: the weight ˆf(n) for frequency n is replaced by the weight ĝ(n) ˆf(n). Explain why this says that L is analogous to multiplication by a diagonal operator. In engineering parlance, convolution is also referred to as filtering. Explain why this terminology is appropriate. Compare this operator L to Example The Adjoint of an Operator Example 2.1. Note that the dot product on R n is given by x y = x T y, while the dot product on C n is x y = x T ȳ. Let A be an m n real matrix. Then x Ax defines a linear map of R n into R m, and its transpose A T satisfies x R n, y R m, Ax y = (Ax) T y = x T A T y = x (A T y). Similarly, if A is an m n complex matrix, then its Hermitian or adjoint matrix A H = A T satisfies x C n, y C m, Ax y = (Ax) T ȳ = x T A T ȳ = x (A H y). Theorem 2.2 (Adjoint). Let H and K be Hilbert spaces, and let A: H K be a bounded, linear map. Then there exists a unique bounded linear map A : K H such that x H, y K, Ax, y = x, A y.

14 14 CHRISTOPHER HEIL Proof. Fix y K. Then Lx = Ax, y is a bounded linear functional on H. By the Riesz Representation Theorem, there exists a unique vector h H such that Ax, y = Lx = x, h. Define A y = h. Verify that this map A is linear (exercise). To see that it is bounded, observe that A y = h = sup x, h x =1 = sup Ax, y x =1 sup Ax y x =1 sup A x y = A y. x =1 We conclude that A is bounded, and that A A. Finally, we must show that A is unique. Suppose that B B(K, H) also satisfied Ax, y = x, By for all x H and y K. Then for each fixed y we would have that x, By A y = 0 for every x, which implies By A y = 0. Hence B = A. Exercise 2.3 (Properties of the adjoint). (a) If A B(H, K) then (A ) = A. (b) If A, B B(H, K) and α, β F, then (αa + βb) = ᾱa + βb. (c) If A B(H 1, H 2 ) and B B(H 2, H 3 ), then (BA) = A B. (d) If A B(H) is invertible in B(H) (meaning that there exists A 1 B(H) such that AA 1 = A 1 A = I), then A is invertible in B(H) and (A 1 ) = (A ) 1. Remark 2.4. Later we will prove the Open Mapping Theorem. A remarkable consequence of this theorem is that if and Y are Banach spaces and A: Y is a bounded bijection, then A 1 : Y is automatically bounded. Proposition 2.5. If A B(H, K), then A = A = A A 1/2 = AA 1/2. Proof. In the course of proving Theorem 2.2, we already showed that A A. If f H, then Af 2 = Af, Af = A Af, f A Af f A Af f. (2.1) Hence Af A f (even if Af = 0, this is still true). Since this is true for all f we conclude that A A. Therefore A = A. Next, we have A A A A = A 2. But also, from the calculation in (2.1), we have Af 2 A Af f. Taking the supremum over all unit vectors, we obtain A 2 = sup Af 2 sup A Af f = A A. f =1 f =1

15 CHAPTER 2. OPERATORS ON HILBERT SPACES 15 Consequently A 2 = A A. The final equality follows by interchanging the roles of A and A. Exercise 2.6. Prove that if U B(H, K), then U is an isomorphism if and only if U is invertible and U 1 = U. Exercise 2.7. (a) Let λ = (λ n ) n N l (N) be given and let L be defined as in Example 1.7. Find L. (b)prove that the adjoint of the multiplication operator M φ defined in Exercise 1.18 is the multiplication operator M φ. Exercise 2.8. Let L and R be the left- and right-shift operators on l 2 (N), i.e., L(x 1, x 2,... ) = (x 2, x 3,... ) and R(x 1, x 2,... ) = (0, x 1, x 2,... ). Prove that L = R. Example 2.9. Let L be the integral operator defined in (1.3), determined by the kernel function k. Assume that k is chosen so that L: L 2 () L 2 () is bounded. The adjoint is the unique operator L : L 2 () L 2 () which satisfies Lf, g = f, L g, f, g L 2 (). To find L, let A: L 2 () L 2 () be the integral operator with kernel k(y, x), i.e., Af(x) = k(y, x) f(y) dµ(y). Then, given any f and g L 2 (), we have f, L g = Lf, g = Lf(x) g(x) dµ(x) = = = = f(y) f(y) = f, Ag. k(x, y) f(y) dµ(y) g(x)dµ(x) k(x, y) g(x) dµ(x) dµ(y) k(x, y) g(x) dµ(x) dµ(y) f(y) Ag(y) dµ(y) By uniqueness of the adjoint, we must have L = A. Exercise: Justify the interchange in the order of integration in the above calculation, i.e., provide hypotheses under which the calculations above are justified.

16 16 CHRISTOPHER HEIL Exercise Let {e n } n N be an orthonormal basis for a separable Hilbert space H. Define T : H l 2 (N) by T (f) = { f, e n } n N. Find a formula for T : l 2 (N) H. Definition Let A B(H). (a) We say that A is self-adjoint or Hermitian if A = A. (b) We say that A is normal if AA = A A. Example A real n n matrix A is self-adjoint if and only if it is symmetric, i.e., if A = A T. A complex n n matrix A is self-adjoint if and only if it is Hermitian, i.e., if A = A H. Exercise Show that every self-adjoint operator is normal. Show that every unitary operator is normal, but that a unitary operator need not be self-adjoint. For H = C n, find examples of matrices that are not normal. Are the left- and right-shift operators on l 2 (N) normal? Exercise (a) Show that if A, B B(H) are self-adjoint, then AB is self-adjoint if and only if AB = BA. (b) Give an example of self-adjoint operators A, B such that AB is not self-adjoint. (c) Show that if A, B B(H) are self-adjoint then A + A, AA, A A, A + B, ABA, and BAB are all self-adjoint. What about A A or A B? Show that AA A A is self-adjoint. Exercise (a) Let λ = (λ n ) n N l (N) be given and let L be defined as in Example 1.7. Show that L is normal, find a formula for L, and prove that L is self-adjoint if and only if each λ n is real. (b) Determine a necessary and sufficient condition on φ so that the multiplication operator M φ defined in Exercise 1.18 is self-adjoint. (c) Determine a necessary and sufficient condition on the kernel k so that the integral operator L defined in (1.23) is self-adjoint. The following result gives a useful condition for telling when an operator on a complex Hilbert space is self-adjoint. Proposition Let H be a complex Hilbert space (i.e., F = C), and let A B(H) be given. Then: A is self-adjoint Af, f R f H.

17 CHAPTER 2. OPERATORS ON HILBERT SPACES 17 Proof.. Assume A = A. Then for any f H we have Therefore Af, f is real. Af, f = f, Af = A f, f = Af, f.. Assume that Af, f is real for all f. Choose any f, g H. Then A(f + g), f + g = Af, f + Af, g + Ag, f + Ag, g. Since A(f + g), f + g, Af, f, and Ag, g are all real, we conclude that Af, g + Ag, f is real. Hence it equals its own complex conjugate, i.e., Similarly, since we see that Af, g + Ag, f = Af, g + Ag, f = g, Af + f, Ag. (2.2) A(f + ig), f + ig = Af, f i Af, g + i Ag, f + Ag, g i Af, g + i Ag, f = i Af, g + i Ag, f = i g, Af i f, Ag. Multiplying through by i yields Adding (2.2) and (2.3) together, we obtain Af, g Ag, f = g, Af + f, Ag. (2.3) 2 Af, g = 2 f, Ag = 2 A f, g. Since this is true for every f and g, we conclude that A = A. Example The preceding result is false for real Hilbert spaces. After all, if F = R then Af, f is real for every f no matter what A is. Therefore, any non-self-adjoint operator provides a counterexample. For example, if H = R n then any non-symmetric matrix A is a counterexample. The next result provides a useful way of calculating the operator norm of a self-adjoint operator. Proposition If A B(H) is self-adjoint, then A = sup Af, f. f =1 Proof. Set M = sup f =1 Af, f. By Cauchy Schwarz and the definition of operator norm, we have M = sup Af, f f =1 sup Af f f =1 sup A f f = A. f =1 To get the opposite inequality, note that if f is any nonzero vector in H then f/ f is a unit vector, so A f, f f f M. Rearranging, we see that f H, Af, f M f 2. (2.4)

18 18 CHRISTOPHER HEIL Now choose any f, g H with f = g = 1. Then, by expanding the inner products, canceling terms, and using the fact that A = A, we see that A(f + g), f + g A(f g), f g = 2 Af, g + 2 Ag, f = 2 Af, g + 2 g, Af = 4 Re Af, g. Therefore, applying (2.4) and the Parallelogram Law, we have 4 Re Af, g A(f + g), f + g + A(f g), f g M f + g 2 + M f g 2 = 2M ( f 2 + g 2) = 4M. That is, Re Af, g M for every choice of unit vectors f and g. Write Af, g = Af, g e iθ. Then e iθ g is another unit vector, so M Re Af, e iθ g = Re e iθ Af, g = Af, g. Hence Af = sup Af, g M. g =1 Since this is true for every unit vector f, we conclude that A M. The following corollary is a very useful consequence. Corollary Assume that A B(H). (a) If F = R, A = A, and Af, f = 0 for every f, then A = 0. (b) If F = C and Af, f = 0 for every f, then A = 0. Proof. Assume the hypotheses of either statement (a) or statement (b). In the case of statement (a), we have by hypothesis that A is self-adjoint. In the case of statement (b), we can conclude that A is self-adjoint because Af, f = 0 is real for every f. Hence in either case we can apply Proposition 2.18 to conclude that A = sup Af, f = 0. f =1 Lemma If A B(H), then the following statements are equivalent. (a) A is normal, i.e., AA = A A. (b) Af = A f for every f H.

19 CHAPTER 2. OPERATORS ON HILBERT SPACES 19 Proof. (b) (a). Assume that (b) holds. Then for every f we have (A A AA )f, f = A Af, f AA f, f = Af, Af A f, A f = Af 2 A f 2 = 0. Since A A AA is self-adjoint, it follows from Corollary 2.19 that A A AA = 0. (a) (b). Exercise. Corollary If A B(H) is normal, then ker(a) = ker(a ). Exercise Suppose that A B(H) is normal. Prove that A is injective if and only if range(a) is dense in H. Exercise If A B(H), then the following statements are equivalent. (a) A is an isometry, i.e., Af = f for every f H. (b) A A = I. (c) Af, Ag = f, g for every f, g H. Exercise If H = C n and A, B are n n matrices, then AB = I implies BA = I. Give a counterexample to this for an infinite-dimensional Hilbert space. Consequently, the hypothesis A A = I in the preceding result does not imply that AA = I. Exercise If A B(H), then the following statements are equivalent. (a) A A = AA = I. (b) A is unitary, i.e., it is a surjective isometry. (c) A is a normal isometry. The following result provides a very useful relationship between the range of A and the kernel of A. Theorem Let A B(H, K). (a) ker(a) = range(a ). (b) ker(a) = range(a ). (c) A is injective if and only if range(a ) is dense in H.

20 20 CHRISTOPHER HEIL Proof. (a) Assume that f ker(a) and let h range(a ), i.e., h = A g for some g K. Then since Af = 0, we have f, h = f, A g = Af, g = 0. Thus f range(a ), so ker(a) range(a ). Now assume that f range(a ). Then for any h H we have Af, h = f, A h = 0. But this implies Af = 0, so f ker(a). Thus range(a ) ker(a). (b), (c) Exercises. 3. Projections and Idempotents: Invariant and Reducing Subspaces Definition 3.1. a. If E B(H) satisfies E 2 = E then E is said to be idempotent. b. If E B(H) satisfies E 2 = E and ker(e) = range(e) then E is called a projection. Exercise 3.2. If E B(H) is an idempotent operator, then ker(e) and range(e) are closed subspaces of H. Further, ker(e) = range(i E) and range(e) = ker(i E). Lemma 3.3 (Characterization of Orthogonal Projections). Let E B(H) be a nonzero idempotent operator. Then the following statements are equivalent. (a) E is a projection. (b) E is the orthogonal projection of H onto range(e). (c) E = 1. (d) E is self-adjoint. (e) E is normal. (f) E is positive, i.e., Ef, f 0 for every f H. Proof. (e) (a). Assume that E 2 = E and E is normal. Then from Lemma 2.20 we know that Ef = E f for every f H. Hence Ef = 0 if and only if E f = 0, or in other words, ker(e) = ker(e ). But we know from Theorem 2.26 that ker(e ) = range(e). Hence we conclude that ker(e) = range(e), and therefore E is a projection. The remaining implications are exercises. Definition 3.4 (Orthogonal Direct Sum of Subspaces). Let {M i } i I be a collection of closed subspaces of H such that M i M j whenever i j. Then the orthogonal direct sum of the M i is the smallest closed subspace which contains every M i. This space is ( = span M i ). M i i I i I

21 CHAPTER 2. OPERATORS ON HILBERT SPACES 21 Exercise 3.5. Suppose that M, N are closed subspaces of H such that M N. Prove that M + N = {m + n : m M, n N} is a closed subspace of H, and that M N = M + N. Show that every vector x M N can be written uniquely as x = m + n with m M and n N. Extend by induction to finite collections of closed, pairwise orthogonal subspaces. (Unfortunately, the analogous statement is not true for infinite collections.) Exercise 3.6. Show that if A B(H, K) then H = ker(a) range(a ). Definition 3.7. Let A B(H) and M H. (a) We say that M is invariant under A if A(M) M, where A(M) = {Ax : x M}. That is, M is invariant if x M implies Ax M. Note that it need not be the case that A(M) = M. (b) We say that M is a reducing subspace for A if both M and M are invariant under A, i.e., A(M) M and A(M ) M. Proposition 3.8. Let A B(H) and M H be given. Then the following statements are equivalent. (a) M is invariant under A. (b) P AP = AP, where P = P M is the orthogonal projection of H onto M. Exercise 3.9. Define L: l 2 (Z) l 2 (Z) by L(..., x 1, x 0, x 1,... ) = (...,, x 0, x 1, x 2,... ), where on the right-hand side the entry x 1 sits in the 0th component position. That is, L slides each component one unit to the left (L is called a bilateral shift). Find a closed subspace of l 2 (Z) that is invariant but not reducing under L. Exercise Assume that M H is invariant under L B(H). invariant under L. Prove that M is

22 22 CHRISTOPHER HEIL 4. Compact Operators Definition 4.1 (Compact and Totally Bounded Sets). Let be a Banach space, and let E be given. (a) We say that E is compact if every open cover of E contains a finite subcover. That is, E is compact if whenever {U α } α I is a collection of open sets whose union contains E, then there exist finitely many α 1,..., α N such that E U α1 U αn. (b) We say that E is sequentially compact if every sequence {f n } n N of points of E contains a convergent subsequence {f nk } k N whose limit belongs to E. (c) We say that E is totally bounded if for every ε > 0 there exist finitely many points f 1,..., f N E such that E N B(f k, ε), k=1 where B(f k, ε) is the open ball of radius ε centered at f k. That is, E is totally bounded if and only there exist finitely many points f 1,..., f N E such that every element of E is within ε of some f k. In finite dimensions, a set is compact if and only if it is closed and bounded. In infinite dimensions, all compact sets are closed and bounded, but the converse fails. Instead, we have the following characterization of compact sets. (this characterization actually holds in any complete metric space). Theorem 4.2. Let E be a subset of a Banach space. Then the following statements are equivalent. (a) E is compact. (b) E is sequentially compact. (c) E is closed and totally bounded. Proof. (b) (a). 2 Assume that E is sequentially compact. Our first step will be to prove the following claim, where the diameter of a set S is defined to be diam(s) = sup{ f g : f, g S}. Claim 1. For any open cover {U α } α I of E, there exists a number δ > 0 (called a Lebesgue number for the cover) such that if S E satisfies diam(s) < δ, then there is an α I such that S U α. To prove the claim, suppose that {U α } α I was an open cover of E such that no δ with the required property existed. Then for each n N, we could find a set S n E with diam(s n ) < 1 n such that S n is not contained in any U α. Choose any f n S n. Since E is sequentially compact, there must be a subsequence {f nk } k N that converges to an element of 2 This proof is adapted from one given in J. R. Munkres, Topology, Second Edition, Prentice Hall, 2000.

23 CHAPTER 2. OPERATORS ON HILBERT SPACES 23 E, say f nk a E. But we must have a U α for some α, and since U α is open there must exist some ε > 0 such that B(a, ε) U α. Now choose k large enough that we have both 1 < ε and a f nk < ε n k 2 2. The first inequality above implies that diam(s nk ) < ε. Therefore, using this and second 2 inequality, we have S nk B(a, ε) U α, which is a contradiction. Therefore the claim is proved. Next, we will prove the following claim. Claim 2. For any ε > 0, there exist finitely many f 1,..., f N E such that E N k=1 B(f k, ε). To prove this claim, assume that there is an ε > 0 such that E cannot be covered by finitely many ε-balls centered at points of E. Choose any f 1 E. Since E cannot be covered by a single ε-ball, we have E B(f 1, ε). Hence there exists f 2 E \ B(f 1, ε), i.e., f 2 E and f 2 f 1 ε. But E cannot be covered by two ε-balls, so there must exist an f 3 E \ ( B(f 1, ε) B(f 2, ε) ). In particular, we have f 3 f 1, f 3 f 2 ε. Continuing in this way we obtain a sequence of points {f n } n N in E which has no convergent subsequence, which is a contradiction. Hence the claim is proved. Finally, we show that E is compact. Let {U α } α I be any open cover of E. Let δ be the Lebesgue number given by Claim 1, and set ε = δ. By Claim 2, there exists a covering of E 3 by finitely many ε-balls. Each ball has diameter smaller than δ, so by Claim 1 is contained in some U α. Thus we find finitely many U α that cover E. (c) (b). Assume that E is closed and totally bounded, and let {f n } n N be any sequence of points in E. Since E is covered by finitely many balls of radius 1, one of those balls must 2 contain infinitely many f n, say {f n (1) } n N. Then we have m, n N, f (1) m f (1) n < 1. Since E is covered by finitely many balls of radius 1 4 {f n (1) } n N such that m, n N, f (1) m f (1) n < 1 2. By induction we keep constructing subsequences {f n (k) all m, n N. Now consider the diagonal subsequence {f (n) n that 1 N f (m) m < ε. If m n > N, then f (m) m = f (n) k. Then f (m) m f (n) n = f (n) (2), we can find a subsequence {f n } n N of } n N such that f (k) m f n (k) < 1 for k } n N. Given ε > 0, let N be large enough is one element of the sequence {f (n) k } k N, say k f (n) n < 1 n < ε. Thus {f (n) n } n N is Cauchy and hence converges. Since E is closed, it must converge to some element of E.

24 24 CHRISTOPHER HEIL (a) (c). Exercise. Exercise 4.3. Show that if E is a totally bounded subset of a Banach space, then its closure E is compact. A set whose closure is compact is said to be precompact. Notation 4.4. We let Ball H denote the closed unit sphere in H, i.e., Ball H = Ball(H) = {f H : f 1}. Exercise 4.5. Prove that if H is infinite-dimensional, then Ball H is not compact. Definition 4.6 (Compact Operators). Let H, K be Hilbert spaces. T : H K is compact if T (Ball H ) has compact closure in K. We define B 0 (H, K) = {T : H K : T is compact}, and set B 0 (H) = B 0 (H, H). A linear operator By definition, a compact operator is linear, and we will see that all compact operators are bounded. Thus it will turn out that B 0 (H, K) B(H, K). In fact, we will see that B 0 (H, K) is a closed subspace of B(H, K). The following result gives some useful reformulations of the definition of compact operator. Proposition 4.7 (Characterizations of Compact Operators). Let T : H K be linear. Then the following statements are equivalent. (a) T is compact. (b) T (Ball H ) is totally bounded. (c) If {f n } n N is a bounded sequence in H, then {T f n } n N contains a convergent subsequence. Proof. (a) (b). This follows from Theorem 4.2 and Exercise 4.3. (a) (c). Suppose that T is compact and that {f n } n N is a bounded sequence in H. By rescaling the sequence (i.e., multiplying by an appropriate scalar), we may assume that f n Ball H for every n. Therefore T f n T (Ball H ) T (Ball H ). Since T (Ball H ) is compact, it follows from Theorem 4.2 that {T f n } n N contains a subsequence which converges to an element of T (Ball H ). (c) (a). Exercise. Proposition 4.8. If T : H K is compact, then it is bounded. That is, B 0 (H, K) B(H, K). Proof. Assume that T : H K is linear but unbounded. Then there exist vectors f n H such that f n = 1 but T f n n. Therefore every subsequence of {T f n } n N is unbounded, and hence cannot converge. Therefore T is not compact by Proposition 4.7.

25 CHAPTER 2. OPERATORS ON HILBERT SPACES 25 Exercise 4.9. Show that if H is infinite-dimensional then the identity operator on H is not compact. Hence a bounded operator need be compact in general. The following exercise shows that a compact operator maps an orthonormal sequence to a sequence that converges to the zero vector. Exercise (a) Let {h n } n N be a sequence of vectors in H, and let h H. Suppose that every subsequence of {h n } n N contains a subsequence that converges to h. Prove that h n h. Hint: Proceed by contradiction. Suppose that h n does not converge to h. Show that this implies that there is an ε > 0 and a subsequence {h nk } k N such that h h nk ε for every k. (b) Suppose that T : H K is compact, and let {e n } n N be an orthonormal sequence in H. Show that T e n 0. Hint: Choose any subsequence {f n } n N. Since T is compact, this sequence has a subsequence {g n } n N such that {T g n } n N converges, say T g n h. Prove that T g n, h 0 (use Bessel s Inequality to find a bound for the l 2 -norm of { T g n, h } n N ). Use part (a) to complete the proof. The following exercise shows that a compact operator maps weakly convergent sequences to convergent sequences. Definition Let {f n } n N be a sequence of vectors in H and let f H. We say that f n converges weakly to f, written f n w f, if g H, f n, g f, g as n. Exercise (a) Show that if f n f, then f n w f. (b) Show that if {e n } n N is an orthonormal sequence in H, then e n w 0. (c) Suppose that T B(H) is compact. Show that if f n w f, then T f n T f. Exercise Let φ L (R n ) be fixed, with φ 0. Then by Exercise 1.18 we know that the multiplication operator M φ : L 2 (R n ) L 2 (R n ) given by M φ f = fφ is bounded. Show that M φ is not compact. Hint: There must exist an ε > 0 and a set E R n with positive measure such that φ(x) ε for all x E. Exhibit a measure space (, Ω, µ) and a bounded, nonzero φ L () such that M φ is compact. Hint: Consider Exercise Exercise Porve that if T : H K is compact and injective, then T 1 : range(t ) H is unbounded.

26 26 CHRISTOPHER HEIL Theorem 4.15 (Limits of Compact Operators). B 0 (H, K) is a closed subspace of B(H, K) (under the operator norm). That is, (a) if S, T B 0 (H, K) and α, β F, then αs + βt B 0 (H, K), (b) if T n B 0 (H, K), T B(H, K), and T T n 0, then T B 0 (H, K). Proof. (a) Exercise. (b) Assume that T n are compact operators and that T n T in operator norm. By Proposition 4.7, it suffices to show that T (Ball H ) is a totally bounded subset of K. Choose any ε > 0. Then there exists an n such that T T n < ε. Now, T 3 n is compact, so T n (Ball H ) is totally bounded. Hence there exist finitely many points h 1,..., h m Ball H such that T n (Ball H ) We will show that T (Ball H ) is totally bounded by showing that T (Ball H ) m j=1 m j=1 B ( T n h j, ε 3). (4.1) B ( T n h j, ε ). (4.2) Choose any element of T (Ball H ), i.e., any point T f with f 1. Then T n f T n (Ball H ), so by (4.1) there must be some j such that T n f T n h j < ε 3. Consequently, T f T h j T f T n f + T n f T n h j + T n h j T h j < T T n f + ε 3 + T n T h j < ε ε 3 + ε 3 1 = ε. Hence (4.2) follows, so T is compact. Exercise Another way to prove Theorem 4.15 is to apply a Cantor diagonalization argument. Fill in the details in the following sketch of this argument. Suppose that {f n } n N is a bounded sequence in H. Then since T 1 is compact, there exists a subsequence {f n (1) } n N of {f n } n N such that {T 1 f n (1) } n N converges. Then since T 2 is compact, there exists a subsequence {f n (2) } n N of {f n (1) } n N such that {T 2 f n (2) } n N converges (and note that {T 1 f n (2) } n N also converges!). Continue to construct subsequences in this way, and then show that the diagonal subsequence {T f n (n) } n N converges (use the fact that there exists a k such that T T k < ε). Therefore T is compact. Theorem 4.17 (Compositions and Compact Operators). Let H 1, H 2, H 3 be Hilbert spaces. (a) If A: H 1 H 2 is bounded and T : H 2 H 3 is compact, then T A: H 1 H 3 is compact.

27 CHAPTER 2. OPERATORS ON HILBERT SPACES 27 (b) If T : H 1 H 2 is compact and A: H 2 H 3 is bounded, then AT : H 1 H 3 is compact. Proof. (b) Assume that A is bounded and T is compact. Let {f n } n N be any bounded sequence in H 1. Then since T is compact, there is a subsequence {T f nk } k N that converges in H 2. Since A is bounded, the subsequence {AT f nk } k N therefore converges in H 3. Hence AT is compact. (a) Exercise. Exercise Prove that if T B 0 (H, K), then range(t ) is a separable subspace of K. Hints: Since T (Ball H ) is compact, it is totally bounded. Hence for each n N we can find finitely many balls of radius 1/n with centers in T (Ball H ) that cover T (Ball H ). If we consider all these balls for every n, we have countably many balls that cover T (Ball H ). Show that this implies that T (Ball H ) contains a countable, dense subset. Then do the same for each ball of radius k N instead of just k = 1. Combine all of these together to get a countable dense subset of range(t ). Definition 4.19 (Finite-Rank Operators). Recall that the rank of an operator T : H K is the dimension of range(t ). We say that T is a finite-rank operator if range(t ) is finitedimensional. We set and set B 00 (H) = B 00 (H, H). B 00 (H, K) = {T B(H, K) : T is finite-rank}, A linear, finite-rank operator need not be bounded (that is why we include the assumption of boundedness in the definition of B 00 (H, K) above). However, the following result shows that if a finite-rank operator is bounded, then it is actually compact. Proposition If T : H K is bounded, linear, and has finite rank, then T is compact. Thus, B 00 (H, K) B 0 (H, K). Proof. Since T is bounded, T (Ball H ) is a bounded subset of the finite-dimensional space range(t ). All finite-dimensional spaces are closed. Hence the closure of T (Ball H ) is a closed and bounded subset of range(t ), and therefore is compact. This gives us the following very useful way to show that a general operator T is compact: try to construct a sequence of finite-rank operators T n that converge to T in operator norm. Corollary Suppose that T n B(H, K) are finite-rank operators, T B(H, K), and T n T in operator norm. Then T is compact. Exercise Show that if E B(H) is compact and idempotent, then E has finite rank.

28 28 CHRISTOPHER HEIL Example Let {e n } n N be an orthonormal basis for a separable Hilbert space H, and let λ = (λ n ) n N be a bounded sequence of scalars. Then we know from Example 1.7 that Lf = λ n f, e n e n defines a bounded operator on H. Suppose that λ n 0 as n. Define L N f = n=1 N λ n f, e n e n. n=1 Since range(l N ) span{e 1,..., e N } (must it be equality?), we have that L N is finite-rank. (Exercise: Show that L is not finite-rank if there are infinitely many λ n 0.) Further, L N is a good approximation to L, because (using the Plancherel Theorem) we have 2 (L L N )f 2 = λ n f, e n e n = n=n+1 n=n+1 It follows that L N converges to L in operator norm: ( ) lim L L N 2 lim λ n 2 N N λ n 2 f, e n 2 ( ) sup λ n 2 f, e n 2 n>n n=n+1 ( ) sup λ n 2 f 2. n>n sup n>n = lim sup λ n 2 = 0. N Since each L N is compact, we conclude that L is compact as well. Exercise Continuing the preceding example, prove the following. (a) Prove that if λ n does not converge to zero then L is not compact. Hint: We know at least some of the eigenvectors of L. (b) Prove that, with only the assumption that λ l, we have f H, L N f Lf. (4.3) That is, for each individual vector f we have Lf L N f 0, where this is the norm in H. A sequence of operators which satisfies (4.3) is said to converge strongly or in the strong operator topology (SOT). Prove that strong convergence of operators does not imply convergence in operator norm, i.e., (4.3) does not imply that L L N 0. (c) Assuming that λ l, prove that L is self-adjoint if and only if every λ n is real.

29 CHAPTER 2. OPERATORS ON HILBERT SPACES 29 We can characterize all the finite-rank operators, as follows. Proposition 4.25 (Finite-Rank Operators). Let L: H K be bounded and linear. Then the following statements are equivalent. (a) L has finite rank. (b) There exist vectors ϕ 1,..., ϕ N H and ψ 1,..., ψ N K such that N Lf = f, ϕ k ψ k, f H. (4.4) k=1 Proof. (a) (b). Since L has finite rank, we know that range(l) is a finite-dimensional subspace of K. Every finite-dimensional subspace is closed, so we can find a finite orthonormal basis {ψ k } N k=1 for range(l). Therefore, if f H then we can express Lf in terms of this orthonormal basis: N N N Lf = Lf, ψ k ψ k = f, L ψ k ψ k = f, ϕ k ψ k, where ϕ k = L ψ k. k=1 k=1 (b) (a). We have range(l) span{ψ 1,..., ψ N }. Corollary If L B(H, K) has rank 1, then there exist ϕ H and ψ K such that Lf = f, ϕ ψ, f H. In particular, if ϕ = ψ are unit vectors, then Lf = f, ϕ ϕ is the orthogonal projection of H onto span{ϕ}. Exercise Compute the adjoint of L given by (4.4). Conclude that the adjoint of a finite-rank operator is also finite-rank. k=1 Exercise Show that if A B(H) and AT = T A for every finite-rank T then A = ci for some scalar c. Exercise Use the idea of the Proposition 4.25 to show that if H is separable and L: H H is any bounded linear operator, then there exist finite-rank operators L N that converge to L in the strong operator topology, i.e., Lf L N f 0 for each individual f. However, observe that if L is not compact, then we cannot have L N L in operator norm. The following result shows that not only is the operator norm limit of a sequence of finiterank operators compact, but every compact operator can be realized as the operator norm limit of finite-rank operators. Theorem If T B(H, K), then the following statements are equivalent.

30 30 CHRISTOPHER HEIL (a) T is compact. (b) There exist finite-rank operators T n B(H, K) such that T n T. As a consequence, we have that B 00 (H, K) is a dense subspace of B 0 (H, K), i.e., B 00 (H, K) = B 0 (H, K) (closure in operator norm). Proof. (b) (a). This follows from Theorem 4.15 and the fact that all bounded finite-rank operators are compact. (a) (b). Assume that T is compact. Let R = range(t ). If R is finite-dimensional, then T is finite-rank, and we are done. So, assume that R is infinite-dimensional. By Exercise 4.18, we know that R is separable, so there exists a countable orthonormal basis {e n } n N for R. For any f H we have T f R, so T f = T f, e n e n, f H. Define T N f = n=1 N T f, e n e n, f H, n=1 and note that T N = P N T where P N is the orthogonal projection of K onto the closed subspace span{e 1,..., e N }. By definition, we have that T N converges to T in the strong operator topology, i.e., T N f T f for every f. Our goal is to show more, namely, to show that T N T in operator norm. That is, we need to show that sup f =1 T f T N f 0. Choose any ε > 0. Since T (Ball H ) is totally bounded, it is covered by finitely many ε-balls centered at points in T (Ball H ). Hence, there exist h 1,..., h m H such that T (Ball H ) m ( B T h k, ε ). 3 Since lim N T h k T N h k = 0 for k = 1,..., m, we can find an N 0 such that k=1 N > N 0, T h k T N h k < ε, k = 1,..., m. 3 Choose any f with f = 1 and any N > N 0. Then T f B ( T h k, ε 3) for some k, i.e., T f T h k < ε 3. Therefore we also have T N f T N h k = = N T (f h k ), e n e n n=1 ( N ) 1/2 T (f h k ), e n 2 n=1

31 CHAPTER 2. OPERATORS ON HILBERT SPACES 31 ( ) 1/2 T (f h k ), e n 2 = T f T h k < ε 3. n=1 Alternatively, this follows even more simply from the fact that T N f T N h k = P N T f P N T h k P N T f T h k < 1 ε 3. In any case, it follows that T f T N f T f T h k + T h k T N h k + T N h k T N f < ε 3 + ε 3 + ε 3 = ε. This is true for every unit vector, so we have T T N ε for all N > N 0. Therefore, we do indeed have T T N 0. Corollary If T B(H, K), then T is compact T is compact. Proof. Assume that T is compact. Then there exist finite-rank operators T N such that T N T. Hence TN T (why?), but each TN is finite-rank, so T is compact. The converse is symmetrical. Exercise Extend Example 4.23 as follows. Let H be a separable Hilbert space, and let {e n } n N be an orthonormal basis for H. Let λ = (λ n ) n N l (N) be given. Define Le n = λ n e n. Prove that the definition of L can be extended to all of H in such a way that L is a bounded linear operator. Prove that this operator L is compact if and only if λ n 0. The next result shows that an integral operator with a square-integrable kernel is compact. Theorem Let (, Ω, µ) be a σ-finite measure space. If k L 2 ( ), then the integral operator Lf(x) = k(x, y) f(y) dµ(y), f L 2 (), defines a compact mapping of L 2 () into itself. Further, L k 2. Proof. Note that by Theorem 1.22 we already know that L defines a bounded operator, and that L k 2. So, we need only show that L is compact. For simplicity, we will consider only the case where L 2 () is separable. In this case there exists an orthonormal basis {e n } n N for L 2 (). Define e mn (x, y) = e m (x) e n (y), x, y. Then it is easy to see that {e mn } m,n N is an orthonormal sequence in L 2 ( ), and with more work (exercise 3 ) it can be shown that that it is also complete and hence forms an orthonormal basis for L 2 ( ). Since k L 2 ( ), we therefore have k = k, e mn e mn, m=1 n=1 3 For details on this type of argument, see the Real Analysis Review handout on the instructor s webpage.

32 32 CHRISTOPHER HEIL where this series converges in the norm of L 2 ( ), and in fact it converges unconditionally. For each N N define an approximation to k by setting N N k N = k, e mn e mn. m=1 n=1 Then k N k in L 2 -norm. Now define an approximation to L by defining L N to be the integral operator with kernel k N, i.e., L N f(x) = k N (x, y) f(y) dµ(y), f L 2 (). Since k N L 2 ( ), we know that L N is bounded. Further, since the sums involved are finite, we can interchange sums and integrals in the following calculation to obtain that L N f(x) = k N (x, y) f(y) dµ(y) = = = N m=1 n=1 N m=1 n=1 N m=1 n=1 N k, e mn e mn (x, y) f(y) dµ(y) N k, e mn N k, e mn f, e n e m (x). e m (x) e n (y) f(y) dµ(y) This is an equality of functions, i.e., L N f = N m=1 N n=1 k, e mn f, e n e m a.e. In any case we have L N f span{e 1,..., e N }, so L N has finite rank. Since L N is bounded (why?), it is therefore compact. Consequently, if we can show that L N L, then we can conclude that L itself is compact. Note that L L N is simply the integral operator with kernel k k N. Since k k N L 2 ( ), we know that L L N is bounded, and that L L N k k N 2 0 as N. Hence L N L in operator norm, so L is compact. For the remainder of this section, we consider eigenvalues and eigenvectors of compact operators. Definition Let A B(H) be given. (a) A scalar λ F is an eigenvalue of A if there exists a nonzero vector f H such that Af = λf. Equivalently, λ is an eigenvalue if ker(a λi) {0}. (b) If λ F is an eigenvalue of A, then any nonzero vector in ker(a λi) is called an eigenvector of A corresponding to the eigenvalue λ, or simply a λ-eigenvector for short. Equivalently, a nonzero vector f H is a λ-eigenvector if Af = λf.

33 CHAPTER 2. OPERATORS ON HILBERT SPACES 33 (c) If λ F is an eigenvalue of A, then ker(a λi) is called the eigenspace of A corresponding to the eigenvalue λ, or simply a λ-eigenspace for short. Every nonzero vector in the λ-eigenspace is a λ-eigenvector of A. (d) The point spectrum σ p (A) of A is the set of eigenvalues of A: σ p (A) = {λ F : λ is an eigenvalue of A}. Exercise Let {e n } n N be an orthonormal basis for a separable Hilbert space H, and let λ = (λ n ) n N l (N) be fixed. Let L: H H be the bounded operator Lf = λ n f, e n e n defined in Example 1.7. (a) Show that σ p (L) = {λ n : n N}. (b) Show that if µ is one component of λ and J = {n N : λ n = µ}, then the µ- eigenspace of L is span{e n } n J. (c) Show that the eigenspaces of L corresponding to distinct eigenvalues are orthogonal. Exercise Let L B(H) be given. Prove the following. (a) If L is self-adjoint, then all eigenvalues of L are real. (b) If L is positive ( Lf, f 0 for all f), then all eigenvalues of L are nonnegative. (c) If L is positive definite ( Lf, f > 0 for all f 0), then all eigenvalues of L are strictly positive. (d) If L is unitary, then every eigenvalue λ satisfies λ = 1. Exercise Suppose that L B(H) is normal. Prove that if λ µ are distinct eigenvalues of L, then the corresponding eigenspaces are orthogonal, i.e., ker(l λi) ker(l µi). While any linear operator A: C n C n must have an eigenvalue, bounded operators on infinite-dimensional Hilbert spaces need not have any eigenvalues. Exercise (a) Prove that the Volterra operator defined in Exercise 1.25 is compact but has no eigenvalues, i.e., its point spectrum is empty. (b) Prove that the right-shift operator R on l 2 (N) has no eigenvalues. (c) Prove that every scalar λ < 1 is an eigenvalue of the left-shift operator L on l 2 (N), and find the corresponding eigenvectors. Thus, this operator has uncountably many eigenvalues. (d) Let φ(x) = x. Prove that the multiplication operator M φ : L 2 [0, 1] L 2 [0, 1], defined by M φ f(x) = xf(x), is self-adjoint but has no eigenvalues. (e) Define k(x, y) = { i, y x, i, y > x,

34 34 CHRISTOPHER HEIL and let L: L 2 [0, 1] L 2 [0, 1] be the integral operator with kernel k. Prove that L is both compact and self-adjoint. Prove that the eigenvalues of L are λ k = 2 (and only these), (2k+1)π and find the corresponding eigenvectors. Exercise 4.39 (Convolution). Fix g L 2 [0, 1], where we consider functions in L 2 [0, 1] to be 1-periodically extended to the real line. Let T be the convolution operator T f(x) = (f g)(x) = 1 0 g(x y) f(y) dy. (a) Prove that T is compact. Hint: Write T as an integral operator and show that its kernel is square-integrable. Note that the fact that [0, 1] has finite measure is important. (b) Prove that the complex exponential functions e n (x) = e 2πinx are eigenvectors of T, with corresponding eigenvalues ĝ(n) (the Fourier coefficients of g). Exercise (a) Assume that A B(H) is normal and let λ F be given. Show that A λi is normal. Use this to show that ker(a λi) = ker(a λi). Conclude that if λ is an eigenvalue of A then λ is an eigenvalue of A, and the corresponding eigenspaces are equal. Hint: Consider Corollary (b) Find an example of a non-normal operator for which the conclusion of part (a) fails. Hint: Consider a shift operator. The next result shows that the eigenspaces (if any) of a compact operator corresponding to nonzero eigenvalues must be finite-dimensional. Proposition Assume that T : H H is compact and that λ 0 is an eigenvalue of T. Then ker(t λi) is finite-dimensional. Proof. Since T is bounded, we know that ker(t λi) is a closed subspace of H. Suppose that it was infinite-dimensional. Then we could find an infinite orthonormal sequence {e n } n N in ker(t λi). Each e n is a λ-eigenvector, i.e., T e n = λe n. But then {e n } n N is a bounded sequence in H yet T e m T e n = λ e m e n = λ 2, so since λ 0 there can be no convergent subsequences of {T e n } n N, which contradicts the fact that T is compact. The following is one useful theoretical result which implies the existence of an eigenvalue of a compact operator T. It states that if inf f =1 T f λf = 0, then this infimum is actually achieved, i.e., T f λf = 0 for some unit vector f, or in other words, there exists a λ-eigenvector for T. Proposition Assume that T : H H is compact and that λ 0 is given. Then: inf T f λf = 0 = λ σ p(t ). f =1

35 CHAPTER 2. OPERATORS ON HILBERT SPACES 35 Proof. Assume that inf f =1 T f λf = 0. Then we can find unit vectors f n such that T f n λf n 0. Since T is compact, {T f n } n N has a convergent subsequence, say T f nk g H. Since λ 0 we have ( ) λfnk T f nk + T fnk f nk = 0 + g = g λ λ λ. (4.5) Since the f nk are unit vectors, we conclude that g 0. Moreover, since T is continuous it follow from (4.5) that T f nk T g/λ. But we also know that T f nk g, so we conclude that T g/λ = g, or in other words that g is a λ-eigenvector. Corollary Assume T : H H is compact and that λ 0. If λ / σ p (T ) and λ / σ p (T ), then range(t λi) is a bounded bijection of H onto itself, and (T λi) 1 is bounded. Proof. Since λ is not an eigenvalue, we know that T λi is injective. Further, it follows from the preceding proposition that inf f =1 T f λf > 0. Hence there exists a C > 0 such that T f λf C for every unit vector f, and hence f H, T f λf C f. (4.6) It follows from Exercise 1.2 than range(t λi) is a closed subspace of H. But then, since λ is not an eigenvalue of T, we have that range(t λi) = range(t λi) = ker ( (T λi) ) = ker(t λi) = {0} = H. Thus T λi is a bounded bijection. It remains to show that (T λi) 1 is bounded. Given f H we have from (4.6) that f = (T λi)(t λi) 1 f C (T λi) 1 f. Rearranging, we see that (T λi) 1 1 C <. Actually, it can be shown that if T : H H is compact, λ 0, and λ / σ p (T ), then λ / σ p (T ) follows automatically. 5. The Diagonalization of Compact Self-Adjoint Operators First let us summarize the facts that have been developed regarding the operator L introduced in Example 1.7 and studied in other examples in previous sections. Theorem 5.1. Let {e n } n N be an orthonormal basis for a separable Hilbert space H, and let λ = (λ n ) n l (N) be a bounded sequence of scalars. Define Lf = λ n f, e n e n, f H. (5.1) Then the following statements hold. n=1 (a) L is bounded, and L = λ.

36 36 CHRISTOPHER HEIL (b) L is normal, and L f = n=1 λ n f, e n e n. (c) L is self-adjoint if and only if λ n R for every n. (d) L is compact if and only if λ n 0. Exercise 5.2. Assume that λ 0 and that L is defined by (5.1). In the definition of L, combine those terms corresponding to identical λ n together. That is, let µ = (µ k ) k I be the sequence of distinct values in λ (so I is either {1,..., N} if there are only finitely many distinct values, or I = N if there are infinitely many). If we set J k = {n N : λ n = µ k }, then P k f = f, e n e n n J k is the orthogonal projection of H onto span{e n } n Jk. in (5.1) can be rewritten as Show that the operator L defined Lf = k I µ k P k f, f H, with convergence of the series in the norm of H. Show further that L = k I µ k P k, with convergence of the series in operator norm. Show that span{e n } n Jk is the µ k -eigenspace of L. Show that P j P k = P k P j = 0 for all j k N, and consequently the eigenspaces corresponding to distinct eigenvalues are orthogonal. In this section we will prove a converse result, showing that all compact, self-adjoint operators on a Hilbert space can be represented in the form of (5.1). First, however, we need to develop some useful machinery. Exercise 5.3. If λ is an eigenvalue of L B(H), then λ L. Exercise 5.4. Let A be an n n complex matrix. Define its spectral radius to be ρ(a) = max{ λ : λ is an eigenvalue of A} = max{ λ : λ σ p (A)}. By the preceding exercise, if we choose any norm on C n and let A be the corresponding operator norm, we have ρ(a) A. (a) Prove that if A is self-adjoint and we use the Euclidean (l 2 ) norm on C n, then A = ρ(a). (b) Prove that the same equality holds if A is normal. Find an example of a non-normal matrix for which ρ(a) < A.

37 CHAPTER 2. OPERATORS ON HILBERT SPACES 37 (c) Prove that if A is any n n matrix, then (still using the Euclidean norm on C n ), A = ρ(a A) 1/2. (d) (Harder). Prove that A is a fixed but arbitrary n n complex matrix and ε > 0 is given, then there exists a norm on C n such that the corresponding operator norm of A satisfies A ρ(a) + ε. Although an arbitrary compact operator need not have any eigenvalues (see Exercise 4.38), the following result shows that a compact, self-adjoint operator must have at least one eigenvalue. Proposition 5.5. If T : H H is compact and self-adjoint, then either T or T is an eigenvalue of T. Proof. Since T is self-adjoint, we know from Proposition 2.18 that T = sup T f, f. f =1 Hence, there must exist unit vectors f n such that T f n, f n T. Since T is self-adjoint, all the inner products T f n, f n are real, so we can find a subsequence that converges either to T or to T. Call this subsequence {g n } n N, and let λ be either T or T as appropriate. Then we have g n = 1 for every n, and T g n, g n λ. Hence, since both λ and T g n, g n are real, T g n λg n 2 = T g n 2 2λ T g n, g n + λ 2 g n 2 T 2 g n 2 2λ T g n, g n + λ 2 g n 2 = λ 2 2λ T g n, g n + λ 2 λ 2 2λ 2 + λ 2 = 0. It therefore follows from Proposition 4.42 that λ is an eigenvalue of T. Now we can prove that every compact, self-adjoint operator has a very simple and special form. Theorem 5.6 (Spectral Theorem for Compact Self-Adjoint Operators). Let T : H H be compact and self-adjoint. Then there exist nonzero real numbers {λ n } n J, either finitely many or λ n 0 if infinitely many, and an orthonormal basis {e n } n N of range(t ), such that T f = n J λ n f, e n e n, f H. Each λ n is an eigenvalue of T, and each e n is a corresponding eigenvector.

38 38 CHRISTOPHER HEIL Proof. Note that since T is compact, range(t ) is separable by Exercise If T = 0 then the result is trivial, so assume that T is not the zero operator. Let H 1 = H and T 1 = T. By Proposition 5.5, T 1 has an eigenvalue λ 1 which satisfies λ 1 = T 1 > 0. Let e 1 be a corresponding eigenvector, normalized to e 1 = 1. Let H 2 = {e 1 } and let T 2 = T H2 (the restriction of T to H 2 ). If T 2 = 0, then stop at this point. Otherwise, continue as follows. Since span{e 1 } is invariant under T 1 (after all, e 1 is an eigenvector), we know from Exercise 3.10 that H 2 is invariant under T1 = T 1. Exercise: Show that T 2 : H 2 H 2 is compact and self-adjoint. Therefore T 2 has an eigenvalue λ 2 such that λ 2 = T 2 > 0. Note that since T 2 is a restriction of T 1, we have λ 2 = T 2 T 1 = λ 1. Let e 2 be a corresponding eigenvector, normalized to e 2 = 1. Note that by definition of H 2, we have e 2 e 1. Further, λ 2 is an eigenvalue of T (not just T 2 ), and e 2 is the corresponding eigenvector of T. Let H 3 = {e 1, e 2 } and let T 3 = T H3. If T 3 = 0, then stop at this point. Otherwise, continue as before to construct an eigenvalue λ 3 and eigenvector e 3 (which will be orthogonal to both e 1 and e 2 ). Continuing in this process, there are two possibilities. Case 1: T N+1 = 0 for some N. In this case, since H N+1 = {e 1,..., e N }, we have H = span{e 1,..., e N } H N+1. Therefore, if f H then we can write f uniquely as N f = f, e n e n + v where v H N+1. Since T (v) = T N+1 (v) = 0, we therefore have T f = n=1 N f, e n T (e n ) + T (v) = n=1 In this case T is finite-rank and the proof is complete. N λ n f, e n e n. Case 2: T N 0 for any N. In this case we obtain countably many eigenvalues λ n and corresponding orthonormal eigenvectors e n. Since T is compact, we have by Exercise 4.10 that λ n e n = T (e n ) 0. Since e n = 1, we conclude that λ n 0. Let M = span{e n } n N. Then {e n } n N is an orthonormal basis for M, and H = M M. Hence, if f H then we can write f uniquely as f = f, e n e n + v n=1 for some v M. Therefore T f = f, e n T (e n ) + T (v) = n=1 If we show that T (v) = 0, then we are done. n=1 n=1 λ n f, e n e n + T (v).

39 CHAPTER 2. OPERATORS ON HILBERT SPACES 39 Note that since span{e 1,..., e N } M, we have v M span{e 1,..., e N } = H N. Hence T (v) = T N (v) T N v = λ N v 0 as N. Consequently T (v) = 0. Since each eigenspace corresponding to nonzero eigenvalues is finite-dimensional, we can group terms corresponding to the same eigenvalue together. Alternatively, we could write a more efficient proof of the Spectral Theorem (as Conway does), by using the same argument on the distinct eigenvalues and corresponding eigenspaces, instead of one eigenvalue and eigenvector at a time. Either way, an extension of the ideas used in the preceding result gives the following expanded form of the Spectral Theorem. Theorem 5.7 (Spectral Theorem for Compact Self-Adjoint Operators). Let T : H H be compact and self-adjoint. Then the following statements hold. (a) T has only a finite or countably infinite number of distinct eigenvalues, and each eigenvalue is real. (b) Let {µ 1, µ 2,... } = {µ k } k I be the distinct nonzero eigenvalues, where either I = {1,..., N} or I = N. Then each eigenspace is finite-dimensional. E k = ker(t λµ k ) (c) If I is infinite, then µ k 0 as k. (d) If P k is the orthogonal projection onto the eigenspace E k, then P j P k = P k P j = 0 for j k I. That is, eigenspaces corresponding to distinct eigenvalues are orthogonal. (e) We have T = k I µ k P k, where the series converges in operator norm. (f) There exist nonzero real numbers {λ n } n J, either finitely many or λ n 0 if infinitely many, and an orthonormal sequence {e n } n J such that T f = n J λ n f, e n e n. The λ n are obtained by repeating each µ k according to its multiplicity (the dimension of the eigenspace E k ). The sequence {e n } n J forms an orthonormal basis for ker(t ) = range(t ). Corollary 5.8. If T : H H is compact, self-adjoint, and injective, then H is separable. Proof. Since T = T, we have that range(t ) = ker(t ) = {0} = H. On the other hand, since T is compact we know from Exercise 4.18 that range(t ) is separable.

40 40 CHRISTOPHER HEIL Example 5.9 (Diagonalization of Self-Adjoint Matrices). Let us examine what the Spectral Theorem says in finite dimensions. Let A be a self-adjoint n n matrix (i.e., symmetric if real, and Hermitian if complex). Then the Spectral Theorem says that there exist real nonzero eigenvalues λ 1,..., λ k and corresponding orthonormal eigenvectors u 1,..., u k such that k Ax = λ j (x u j ) u j. (5.2) j=1 We can extend this representation by including the zero eigenvalues of A, as follows. From (5.2), we see that the column space, or range, of A is C(A) = range(a) = span{u 1,..., u k }. Since A is self-adjoint, its nullspace is the orthogonal complement of its column space, for N(A) = ker(a) = range(a) = C(A). Let u k+1,..., u n be an orthonormal basis for N(A), and let λ k+1 = = λ n = 0. Then u 1,..., u n is an orthonormal basis for C n with corresponding eigenvectors λ 1,..., λ n. Further, we have the following representations: n n x = (x u j ) u j and Ax = λ j (x u j ) u j, x F n. j=1 Let us rewrite this representation as follows: n Ax = λ j (x u j ) u j = = = j=1 j=1 λ 1 (x u 1 ) u 1 u n. λ n (x u n ) u 1 u n u 1 u n = UΛU H x, λ 1... λ 1... u n u n x u 1. x u n u H 1. x u H n where U is the matrix that has u 1,..., u n as columns, and Λ is the diagonal matrix with λ 1,..., λ n on the diagonal. On the one hand, this is nothing more than the diagonalization of A. However, this says much more: every self-adjoint matrix can be diagonalized (even if some eigenvalues are repeated), and furthermore, the eigenvector matrix is unitary (because it has orthonormal columns). We summarize this next as a theorem.

41 CHAPTER 2. OPERATORS ON HILBERT SPACES 41 Theorem 5.10 (Diagonalization of Self-Adjoint Matrices). Let A be an n n matrix. Then the following statements are equivalent. (a) A is self-adjoint. (b) A = UΛU where U is unitary and Λ is diagonal with real scalars on its diagonal. (c) There exist real scalars λ 1,..., λ n an orthonormal vectors u 1,..., u n such that n Ax = λ j (x u j ) u j, x F n. j=1 (d) There exists an orthonormal basis u 1,..., u n for F n consisting of eigenvectors of A with corresponding real eigenvalues λ 1,..., λ n.