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Functional Analysis Lecture Notes Chapter 2. Operators on Hilbert Spaces

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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 X, Y be normed linear spaces, and let L: X Y be a linear operator. ! (a) L is continuous at a point f X if f f in X implies Lf Lf in Y . 2 n ! n ! (b) L is continuous if it is continuous at every point, i.e., if fn f in X implies Lfn Lf in Y for every f. ! ! (c) L is bounded if there exists a finite K 0 such that ≥ f X; Lf K f : 8 2 k k ≤ k k Note that Lf is the norm of Lf in Y , while f is the norm of f in X. k k k k (d) The operator norm of L is L = sup Lf : k k kfk=1 k k (e) We let (X; Y ) denote the set of all bounded linear operators mapping X into Y , i.e., B (X; Y ) = L: X Y : L is bounded and linear : B f ! g If X = Y = X then we write (X) = (X; X). B B (f) If Y = F then we say that L is a functional. The set of all bounded linear functionals on X is the dual space of X, and is denoted X0 = (X; F) = L: X F : L is bounded and linear : B f ! g 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 (X; Y ) of all bounded linear operators from X to Y , and we have the composition propertyBthat if L (X; Y ) and K (Y; Z), then KL (X; Z), with KL K L . 2 B 2 B 2 B k k ≤ k k k k Date: February 20, 2006. These notes closely follow and expand on the text by John B. Conway, \A Course in Functional Analysis," Second Edition, Springer, 1990. 1 2 CHRISTOPHER HEIL Exercise 1.2. Suppose that L: X Y is a bounded map of a Banach space X into a Banach space Y . Prove that if there !exists a c > 0 such that Lf c f for every f X, then range(L) is a closed subspace of Y . k k ≥ k k 2 n n Exercise 1.3. Let Cb(R ) be the set of all bounded, continuous functions f : R F. Let n n ! C0(R ) be the set of all continuous functions f : R F such that limjxj!1 f(x) = 0 (i.e., for every " > 0 there exists a compact set K such that! f(x) < " for all x = K). Prove that these are closed subspaces of L1(Rn) (under the L1j-norm;j note that for2a continuous function we have f 1 = sup f(x) ). Define δ : C (Rnk) k F by j j b ! δ(f) = f(0): n 0 Prove that δ is a bounded linear functional on Cb(R ), i.e., δ (Cb) , and find δ . This linear functional is the delta distribution (see also Exercise 1.262below). k k Example 1.4. In finite dimensions, all linear operators are given by matrices, this is just standard finite-dimensional linear algebra. Suppose that X 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 = x ; : : : ; x for X and a basis = y ; : : : ; y for Y . If x X, BX f 1 ng BY f 1 mg 2 then x = c1x1 + + cnxn for a unique choice of scalars ci. Define the coordinates of x with respect to the basis· · · to be BX c1 . Cn [x]BX = . : 2 3 2 cn The vector x is completely determined by its4 coordinates,5 and conversely each vector in Cn is the coordinates of a unique x X. The mapping x [x]BX is a linear mapping of X onto Cn 2Cm 7! . We similarly define [y]BY for vectors y Y . Let A: X Y be a linear map2 (it is automatically2 bounded since X is finite-dimensional). Then A transforms! vectors x X into vectors Ax Y . The vector x is determined by its 2 2 coordinates [x]BX 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]BX and [Ax] are related by multiplication by an m n matrix determined by A. We call this BY × matrix the standard matrix of A with respect to X and Y , and denote it by [A]BX ;BY . That is, the standard matrix should satisfy B B [Ax] = [A] [x] ; x X: BY BX ;BY BX 2 We claim that the standard matrix is the matrix whose columns are the coordinates of the vectors Axk, i.e., [A]B ;B = [Ax1] [Axn]B : X Y 2 BY · · · Y 3 4 5 CHAPTER 2. OPERATORS ON HILBERT SPACES 3 To see this, choose any x X and let x = c1x1 + + cnxn be its unique representation with respect to the basis 2 . Then · · · BX c1 . [A]BX ;BY [x]BX = [Ax1]B [Axn]BY . 2 Y · · · 3 2 3 cn = c4 [Ax ] + + c [Ax ]5 4 5 1 1 BY · · · n n BY = [c Ax + + c Ax ] 1 1 · · · n n BY = [A(c x + + c x )] 1 1 · · · n n BY = [Ax]BY : Exercise 1.5. Extend the idea of the preceding example to show that that any linear mapping L: `2(N) `2(N) (and more generally, L: H K with H, K separable) can be realized in terms of!multiplication by an (infinite but coun!table) matrix. Exercise 1.6. Let A be an m n complex matrix, which we view as a linear transformation A: Cn Cm. The operator×norm of A depends on the choice of norm for Cn and Cm. Compute!an explicit formula for A , in terms of the entries of A, when the norm on Cn and Cm is taken to be the `1 norm. kThenk do the same for the `1 norm. Compare your results to the version of Schur's Lemma given in Theorem 1.23. The following example is one that we will return to many times. Example 1.7. Let en n2N be an orthonormal basis for a separable Hilbert space H. Then we know that every ff gH can be written 2 1 f = f; e e : h ni n n=1 X Fix any sequence of scalars λ = (λn)n2N, and formally define 1 Lf = λ f; e e : (1.1) n h ni n n=1 X 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. 2 Note that if H = ` (N) and e N is the standard basis, then L is given by the formula f ngn2 Lx = (λ x ; λ x ; : : : ); x = (x ; x ; : : : ) `2(N): 1 1 2 2 1 2 2 We will show the following (the `1-norm of the sequence λ is λ = sup λ ). k k1 n j nj (a) The series defining Lf in (1.1) converges for each f H if and only if λ `1. In this case L is a bounded linear mapping of H into itself,2 and L = λ .2 k k k k1 4 CHRISTOPHER HEIL (b) If λ = `1, then L defines an unbounded linear mapping from the domain 2 1 domain(L) = f H : λ f; e 2 < (1.2) 2 j n h nij 1 n=1 n X o (which is dense in H) into H. Proof. (a) Suppose that λ `1, i.e., λ is a bounded sequence. Then for any f we have 2 1 1 λ f; e 2 λ 2 f; e 2 = λ 2 f 2 < ; j n h nij ≤ k k1 h nij k k1 k k 1 n=1 n=1 X X so the series defining Lf converges (because en is an orthonormal sequence). Moreover, the preceding calculation also shows that Lff 2g= 1 λ f; e 2 λ 2 f 2, so we k k n=1 j n h nij ≤ k k1 k k see that L λ 1. On the other hand, by orthonormality we have Len = λnen (i.e., each k k ≤ k k P en is an eigenvector for L with eigenvalue λn). Since en = 1 and Len = λn en = λn we conclude that k k k k j j k k j j L = sup Lf sup Len = sup λn = λ 1: k k kfk=1 k k ≥ n2N k k n2N j j k k The converse direction will be covered by the proof of part (b). (b) Suppose that λ = `1, i.e., λ is not a bounded sequence. Then we can find a subsequence 2 1 N (λnk )k2 such that λnk k for each k. Let cnk = k and define all other cn to be zero. Then 2 1 j j ≥ c = 2 < , so f = c e converges (and c = f; e ). But the formal series n j nj k k 1 n n n n h ni Lf = λncnen does not converge, because P n P P 1 1 1 P k2 c λ 2 = c λ 2 = : j n nj j nk nk j ≥ k2 1 n=1 X Xk=1 Xk=1 In fact, the series defining Lf in (1.1) only converges for those f which lie in the domain defined in (1.2).
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