Spectral Mapping Theorem

Spectral Mapping Theorem Dan Sievewright Definition of a Hilbert space Let H be a Hilbert space over C. 1) H is a vector space over C. 2) H has an inner product, : H H C with the following properties. a) Linear in first component: λx 1 + x 2, y = λ x 1, y + x 2, y for all λ C, x 1, x 2, y H. b) Conjugate symmetric: x, y = y, x, for all x, y H. c) Positive definite: x, x 0 for all x H and x, x = 0 if and only if x = 0. 3) H is complete with respect to the norm x = x, x. Definition of complete A sequence {x n } n N in a normed space is Cauchy if for any ɛ > 0, there exists N N such that if n, m N, then x n x m < ɛ. A complete space is one where all Cauchy sequences converge. Examples: 1) H = C n with dot product: x y = n k=1 x ky k 2) H = l 2 is the set of square summable sequences, k=1 x k 2. The inner product is {x n }, {y n } = x n y n. In order to understand the spectral mapping theorem, we need to introduce the spaces L(H) and C(K). n=1 Definition of a L(H) L(H) consists of the continuous linear transformations T : H H. L(H) is a vector space over C and its multiplication is composition. The multiplicative identity is the identity operator I defined by Ix = x for all x H. An operator T L(H) is invertible if there exists S H such that T S = ST = I. For T L(H), we define the norm of T by T = sup T x. x =1 It can be shown that T < if and only if T is continuous. Here are some facts about T. 1) This does indeed define a norm on L(H) and L(H) is complete with respect to this norm. 2) T x T x for all x H. 1

3) ST S T for all S, T L(H). Definition of a C(K) Let K be a compact set. The set of continuous functions f : K C is called C(K). Just like L(H), this is a vector space over C. There is also a multiplication, but it is defined pointwise as opposed to composition, (fg)(x) := f(x)g(x). Hence the multiplicative identity is the constant function, 1(x) = 1 for all x K. A function f C(K) will be invertible if there exists g C(K) such that fg = gf = 1. This is equivalent to f(x) 0 for all x K since the inverse of f would be f 1 (x) = 1 f(x). Note that f 1 is used to denote the multiplicative inverse, not inverse under composition. For f C(K), we define the norm of f by f = max x K f(x). Since continuous functions are bounded, f <. This does indeed define a norm on C(K) and C(K) is complete with respect to this norm. Definitions of adjoint and normal operators Let T L(H). The adjoint of T is T defined by T x, y = x, T y. Using the Riesz representation theorem, one can show that this defines a unique operator T L(H). An operator T L(H) is normal if T T = T T. Example: For T L(C n ), with matrix T ij, the adjoint T has the matrix (T ) ij = T ji. Here are some facts about the adjoint. 1) (T ) = T. 2) (ST ) = T S. 3) T = T. Subalgebras of L(H) We will call a set A L(H) a subalgebra if it satisfies the following properties. 1) A is a closed subspace of L(H). 2) If T A, then T A. 3) If S, T A, then ST A. 4) I A. For an operator T L(H), let A[T ] denote the smallest subalgebra of L(H) containing T. Theorem Let T L(H) and let C = {subalgebras A L(H) : T A}. Then A[T ] = A CA. 2

Examples: 1) L(H) is a subalgebra of L(H). 2) A[0] = {λi : λ C}. Spectrum Let T L(H). The spectrum of T is σ(t ) = {λ C : T λi is not invertible}. Theorem Let T L(H). Then σ(t ) is a nonempty, compact set. Examples: 1) For T L(C n ), the spectrum of T is the set of eigenvalues of T. 2) Let S L(l 2 ) be defined by S(x 1, x 2, x 3, ) = (0, x 1, x 2, x 3, ). S is not invertible so 0 σ(s) but the only solution to Sx = 0x is x = 0. Hence σ(s) consists of more than just the eigenvalues. It can actually be shown that σ(s) = {λ : λ 1}. The Continuous Functional Calculus Let T L(H) be normal. Then there exists an isomorphism Φ : A[T ] C(σ(T )). What does isomorphism mean in this context? 1) Both A[T ] and C(σ(T )) are vector spaces over C so we request that Φ is an invertible linear transformation. 2) We also want to preserve the topological structure from the norms, so Φ(A) = A for all A A[T ]. 3) To preserve the multiplicative structure, Φ(AB) = Φ(A)Φ(B) for all A, B A[T ]. 4) For A A[T ], we also ask that Φ(A )(z) = Φ(A)(z) for all z σ(t ). How does this isomorphism work? If f(z) C(σ(T )) is analytic in a neighborhood of σ(t ), we can write f(z) = a n z n. n=0 Under the isomorphism, we have that Φ 1 (f) = a n T n. n=0 3

This series will converge in A[T ]. Also, if p is a polynomial in z and z, p(z, z) = a ij z i z j, then Φ 1 (p) = a ij T i (T ) j. These formulas give justification to saying that f(t ) := Φ 1 (f). With this definition, we can finally state the spectral mapping theorem. Spectral Mapping Theorem Let T L(H) be normal and let f : σ(t ) C be a continuous function. Then σ(f(t )) = f(σ(t )). Next week in Analysis Seminar First, we will show how the spectral mapping theorem follows from the continuous functional calculus. Why should we study normal operators? Rotations, reflections, and projections are all examples of normal operators with obvious physical applications. For those studying differential equations, the operator which sends f to the Fourier transform tranform of f, ˆf(t) = f(x)e 2πixt dx is a normal operator (the adjoint is the inverse transform). There are several special types of normal operators. We will talk about self adjoint, positive, and unitary operators. We call T self adjoint if T = T. Self adjoint operators must be normal, T T = T T = T T. We will show that if T is normal, then T is self adjoint if and only if σ(t ) R. We call T unitary if T T = T T = I. From this definition, we know T must be normal. We can then show that if T is normal, then T is unitary if and only if σ(t ) {λ C : λ = 1}. We call T positive if T x, x 0 for all x H. It is much more difficult to see that T is normal, but this follows from the second polarization identity (this proves a stronger statement, T is self adjoint). If T is normal, then T is positive if and only if σ(t ) [0, ). We will show that backwards direction of this theorem and give a description of how one would prove the forwards direction. Using this information, we can prove that if T is self adjoint, then e T is a positive operator. The remaining time next week will be devoted to the proof of the continuous functional calculus. We need quite a few lemmas and a couple more definitions before we reach this point though. Lemma Let T L(H) be normal. Then A[T ] is the norm closure of the set of operators of the form a ij T i (T ) j. 4

Spectral radius The spectrum of T L(H) is a nonempty compact set, so we can define the spectral radius of T, We will prove the following two theorems. r(t ) = max λ σ(t ) λ. Theorem For T L(H), we have that r(t ) = lim T n 1/n. Theorem If T L(H) is normal, then r(t ) = T. Multiplicative linear functionals A multiplicative linear functional ϕ : A[T ] C must be linear, continuous, and satisfy the conditions, ϕ(ab) = ϕ(a)ϕ(b) for all A, B A[T ], and ϕ(i) = 1. The set of multiplicative linear functionals on A[T ] is denoted M A[T ] and called the maximal ideal space (we will explain why). We will introduce the topology on M A[T ] and explain why it is compact. The Gelfand transform The Gelfand transform is the function Γ : A[T ] C(σ(T )) defined by Γ(A)(ϕ) = ϕ(a). The Gelfand transform has many nice properties when T is normal. Theorem Let T L(H) be normal and let A A[T ]. Then Γ(A) is invertible if and only if A is invertible. Also, σ(a) = range(γ(a)) and r(a) = Γ(A). Using this theorem, we can now prove the following very important results. Theorem Let T L(H) be normal. Since range of Γ(T ) is σ(t ), this allows us to define a homeomorphism : M A[T ] σ(t ) by (ϕ) = Γ(T )(ϕ) = ϕ(t ). Theorem Let T L(H) be normal. Then Γ : A[T ] C(M A[T ] ) is an isomorphism. Finally, these two theorems combine to prove the Continuous Functional Calculus where the isomorphism, Φ, is defined by Φ(A)(z) = Γ(A)( 1 (z)) 5