Functional Calculus of Unbounded Operator (revise at 10th August) Hayato Arai (荒井駿) 321801019 1 Introduction In some of exercises and scattering theory, functional calculus plays an essential role. That is because I proof functional calculus of unbounded operator version for my subject. Functional calculus says that any self adjoint operator (even if on infinity dimension or un- bounded) can calculus like complex function. Therefore we regard spectral decomposition as a kind of integration, represent unbounded operator (such as differential operator or Hamiltonian) as a unitary group which is well-known as Stone's theorem, and so on. 2 Proof of functional calculus First, we check the notation. H= 6 (0) is Hilbert space and B(H) is a bounded linear operators on H. C(H) is densely defined closed linear operators on H which have a domain. σ(S) is the spectral set of S and ρ(S) is the resolvent set of S. If λ 2 ρ(S), then λ − S has inverse and this inverse −1 ~ denoted by (λ − S) or Rλ. σ(S) is defined by onepoint conpactification when σ(S) is unbounded, 1 if σS is unbounded, σ(~S) is defined by σ(S). We write e(s) ≡ 1 and r (s) = ; λ 2 ρ(S) for λ λ − s some S. At first, we show following proposition Proposition 2.1. Let S : H!H be densely defined and symmetric. The following are eauivalent: 1. S = S∗ 2. σ(S) ⊂ R 3. i 2 ρ(S). Proof. Suppose 1 and suppose λ = µ+iν with µ, ν real and ν =6 0. When x 2 D(S), j((λ−S)x; (λ− S)x)j ≥ jλj2jjxjj2 ≥ jνj2jjxjj2. Therefore λ − S is 1-1 and has closed range sinse S = S∗ is closed. Again R(λ − S)? = N((λ − S)∗) = N(λ − S) = (0) with N(S) be a kernel of S. Thus λ − S is onto, so λ 2 ρ(S). 2 ) 3 is clear. Suppose 3. Then N(S∗ + i) = R(S∗ − i)? = (0). This means that S∗ + i, which is an extension of the onto operator S + i, is 1-1. This is only possible if S∗ + i = S + i, which implies S∗ = S. 1 Next, we show the theorem which should be called the "continuous" functional calculus. −1 Theorem 2.2. Let S : H!H be a self adjoint operator. The map e 7! I; rλ 7! (λ − S) has a unique extension and this extension is an isometric algebra isomorphism f 7! f(S): C(~σ(S)) !B. Moreover, this extension preserves adjoint and order, i.e. f(S)∗ = f ∗(S), f(S) ≥ 0 if and only if f ≥ 0, on σ(S). Proof. Let F be the algebra of functions on C generated by e and rλ. When such map exists and ··· let p be a polynomial in n-variables, then the function f defined by f(s) = p(rλ1 (s); ; rλn (s)) ··· F correspond to the operator f(S) = p(Rλ1 ; ;Rλn ). Conversely, this will extend the map to ··· ··· when it is well-defined, i.e. p(rλ1 (s); ; rλn (s)) = 0 imply p(Rλ1 ; ;Rλn ) = 0. ≡ ≡ To show this, we induce on n. For n = 1, p(rλ1 ) 0 imply p is zero polynomial, so p(Rλ1 ) 0. ≤ ··· ≡ Next, suppose the assumption is true in n m, and suppose p(rλ1 (s); ; rλm+1 (s)) 0. Since 1 each rλj = 0 at , the constant term of p is zero. Now, 1 − − (λm+1 s)rλj (s) = (λm+1 s) λj − s 1 1 = λm+1 + 1 − λj λj − s λj − s − = (λm+1 λj)rλj + 1; − ··· ··· so (λm+1 s)p(rλ1 (s); ; rλm+1 (s)) = q(rλ1 (s); ; rλm (s)), with q is polynomial in m-variables because p has no constant term. Similarly, − − (λm+1 S)Rλj = (λm+1 λj)Rλj + I; − ··· ··· so (λm+1 S)p(Rλ1 ; ;Rλm+1 ) = q(Rλ1 ; ;Rλm ) and this is equal to 0 by the assumption of − − induction. Finally, p has no constant terms, each Rλj has range D(S), and λm+1 Sis 1 1, so ··· ≡ these impies p(Rλ1 ; ;Rλm+1 ) 0. Thus there is a unique extension to an algebra homomorphism f 7! f(S): F!B(H). Next, we show that for f 2 F,σ(f(S)) = f(~σ(S)). To show this, note that F consists precisely of all bounded rational functions which are holomorphic on σ(S). Suppose x 62 f(~σ(S)). Then put (x − f(s))−1 = h(s) 2 F and by homomophismness, (x − f(S))h(S) = e(S) = I = h(S)(x − f(S)): Thus x 62 σ(f(S)). Conversely if λ 2 σ(S) and x = f(λ), then s − f(s) = (λ − s)g(s), where g(s) 2 F and g(1) = 0byboundness. Therefore, using the fractions decomposition of g we get g(S)(λ − S) ⊂ (λ − S)g(S) = s − f(S): Since λ 2 σ(S), λ − S is either not 1-1 or not onto, so the same is true of x 2 σ(f(S)). Thus f(~σ(S)) ⊂ σ(f(S)). Next, we show isometry and uniqueness. Recall that operator norm is specified its spectral. we get jjf(S)jj = supff(λ); λ 2 σ(~S)g = jjfjj; f 2 F 2 Therefore this map is an isometry of F, considered as a subalgebra of C(~σ(S)), into B(H). On ⊂ R ∗ F σ(S) , rλ = rλ. Therefore is a closed complex conjugation. The Stone-Weierstrass theorem show that F is dense in C(~σ(S)), so f 7! f(S)has a unique extension to an algebra isometry. − ∗ −1 − −1 ∗ ∗ ∗ Finally, we show the other relations. (λ S ) = ((λ S) ) imply rλ(S) = rλ = rλ(S) . Thus f ∗(S) = f(S)∗; 8f 2 F. This relation is carried to C(~σ(S)). suppose f ≥ 0; f 2 C(~σ(S)),, then there exists g 2 C(~σ(S)); g ≥ 0, such that f = g2. Then f(S) = g(S)2 = g(S)∗g(S) ≥ 0. Conversely if f(S) ≥ 0, then f(~σ(S)) = σ(f(S)) ⊂ 0; 1), (Recall that put a self adjoint operator S, then σ(S) ⊂ [α; β], where α = inff(Sx; x); jjxjj = 1g; β = supf(Sx; x); jjxjj = 1g.) Thus f ≥ 0. Corollary 2.3. S is a self adjoint operator. Then S is bounded if and only if σ(S) is bounded. Proof. If S bounded, σ(S) ⊂ [α; β], where α = inff(Sx; x); jjxjj = 1g; β = supf(Sx; x); jjxjj = 1g is bounded. Conversely suppose σ(S) is bounded. Let f(s) ≡ s. Then f 2 C(σ(S)). Moreover, (f − λ)rλ = e,λ 2 ρ(S), so (f(S) − λ)Rλ = Rλ(f(S) − λ) = I. Therefore f(S) − λ = S − λ, so S = f(S) 2 B(H). Now, we have functional calculus which have only continuous functions, so we extends this calculus for some general functions. Definition 2.4. First, let σ be a closed subset of R, and let A0(σ) be the algebra continuous complex valued function on σ which is 0 at 1. Next, let A+(σ) be the set of bounded functions on σ which are pointwise limits of increasing sequence of non-negative functions in A0. Finally, let A1 be the algebra generated by A1. If σ is a bounded, A0 equal to algebra of continuous function C(σ). All bounded continuous non- negative functions are written as pointwise limits of increasing sequence of non-negative functions, so A+ contains all of them. Similarly, A1 contained all bounded continuous function on σ. Write fn % f is real valued and ffn(s)gn is increase sequence, all s 2 σ. Definition 2.5. fSg ⊂ B(H) is called full if _R(S) = H −1 Note that if S is self adjoint, f(λ − S) gλ is full. Then, next lemma is essential to extend functional calculus. Lemma 2.6. Let σ is a closed subset in R, and let A0 : A0(σ) !B(H) be an isometric algebra isomorphism which preserve adjoints. Then A0 has a unique extension to an algebra homomorphism A1 : A1 !B(H) with property that if ffng ⊂ A+ and fn % f 2 A+, then A1(fn) ! A1(f). jj jj ≤ j j Moreover, the extention A1 also preserving adjoint and order, and A1(f) sups2σ f(s) . The image of A0 is full if and only if A1(e) = I. Proof. First, the same argument in theorem 2.1 shows that A0 preserves order. Now, we will define A1(f) which has uniqueness step by step, and then proof the other relation. Suppose 0 ≤ fn % f, where fn 2 A0, f is bounded. Then fA0(fn)g is a bounded increase sequence of operators in B(H), so A0(fn) ! S 2 B(H). Suppose also gn 2 A0; 0 ≤ gn % f. Then A0(gn) ! T 2 B(H). Let fn ^ gm(s) := minffn(s); gm(s)g. Then fn ^ gm % fn as m ! 1. These functions are continuous and 0 at 1 (if σ is unbounded), so the convergence is uniform. Thus T = lim A0(gm) ≥ lim A0(fn ^ gm) = A0(fn), all n, so T ≥ S. Also we get S ≥ T . Therefore the 3 limitS = T =: A1(f) is independent of the approximating sequence. If f 2 A0 \A+, set fn ≡ f. Then A1(f) = lim A0(fn) = A0(f). Suppose fn; gn 2 A0, 0 ≤ fn % f 2 A+, 0 ≤ gn % g 2 A+.