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 finitedimensional linear algebra. Suppose that is an ndimensional 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 finitedimensional). 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 welldefined, 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 welldefined, 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 (σalgebras, Measurable Sets and Functions). Let be a set, and let Ω be a collection of subsets of. Then Ω is a σalgebra 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 extendedreal values, i.e., f(x) is allowed to take the values ±. An extendedrealvalued 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 complexvalued function f : C is called a measurable function if its real and imaginary parts are measurable (realvalued) functions. Definition 1.11 (Measure Space). Let be a set and Ω a σalgebra 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 σfinite. For example, Lebesgue measure on R n is σfinite. It is often useful to allow measures to take negative values. Definition 1.12 (Signed Measure). Let be a set and Ω a σalgebra 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 seminorm 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 seminorm 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 σfinite 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 welldefined, 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 σfinite, 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 σfinite, 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 σfinite 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 σfinite 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 squareintegrable, 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 σfinite 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 welldefined 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 σfinite measure space, and Assume that k is a measurable function on which satisfies the mixednorm 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 σfinite). 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 lefthand 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 1periodically 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 rightshift 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 selfadjoint or Hermitian if A = A. (b) We say that A is normal if AA = A A. Example A real n n matrix A is selfadjoint if and only if it is symmetric, i.e., if A = A T. A complex n n matrix A is selfadjoint if and only if it is Hermitian, i.e., if A = A H. Exercise Show that every selfadjoint operator is normal. Show that every unitary operator is normal, but that a unitary operator need not be selfadjoint. For H = C n, find examples of matrices that are not normal. Are the left and rightshift operators on l 2 (N) normal? Exercise (a) Show that if A, B B(H) are selfadjoint, then AB is selfadjoint if and only if AB = BA. (b) Give an example of selfadjoint operators A, B such that AB is not selfadjoint. (c) Show that if A, B B(H) are selfadjoint then A + A, AA, A A, A + B, ABA, and BAB are all selfadjoint. What about A A or A B? Show that AA A A is selfadjoint. 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 selfadjoint 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 selfadjoint. (c) Determine a necessary and sufficient condition on the kernel k so that the integral operator L defined in (1.23) is selfadjoint. The following result gives a useful condition for telling when an operator on a complex Hilbert space is selfadjoint. Proposition Let H be a complex Hilbert space (i.e., F = C), and let A B(H) be given. Then: A is selfadjoint 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 nonselfadjoint operator provides a counterexample. For example, if H = R n then any nonsymmetric matrix A is a counterexample. The next result provides a useful way of calculating the operator norm of a selfadjoint operator. Proposition If A B(H) is selfadjoint, 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 selfadjoint. In the case of statement (b), we can conclude that A is selfadjoint 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 selfadjoint, 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 infinitedimensional 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 selfadjoint. (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 righthand 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 εballs centered at points of E. Choose any f 1 E. Since E cannot be covered by a single εball, 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 εballs, 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 εballs. 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 infinitedimensional, 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 infinitedimensional 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 (FiniteRank Operators). Recall that the rank of an operator T : H K is the dimension of range(t ). We say that T is a finiterank 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 finiterank}, A linear, finiterank 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 finiterank 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 finitedimensional space range(t ). All finitedimensional 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 finiterank operators T n that converge to T in operator norm. Corollary Suppose that T n B(H, K) are finiterank 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 finiterank. (Exercise: Show that L is not finiterank 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 selfadjoint if and only if every λ n is real.
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