Functional Calculus of Unbounded Operator (Revise at 10Th August)

Home , Functional calculus, Unbounded operator

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 unbounded) 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 ̸= (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 ∈ 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 ∈ 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 ∈ ρ(S). Proof. Suppose 1 and suppose λ = µ+iν with µ, ν real and ν ≠ 0. When x ∈ D(S), |((λ−S)x, (λ− S)x)| ≥ |λ|2||x||2 ≥ |ν|2||x||2. 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 ∈ 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 ∞ 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 ∈ 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 ̸∈ f(˜σ(S)). Then put (x − f(s))−1 = h(s) ∈ F and by homomophismness,

(x − f(S))h(S) = e(S) = I = h(S)(x − f(S)).

Thus x ̸∈ σ(f(S)). Conversely if λ ∈ σ(S) and x = f(λ), then s − f(s) = (λ − s)g(s), where g(s) ∈ F and g(∞) = 0byboundness. Therefore, using the fractions decomposition of g we get

g(S)(λ − S) ⊂ (λ − S)g(S) = s − f(S).

Since λ ∈ σ(S), λ − S is either not 1-1 or not onto, so the same is true of x ∈ 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 ||f(S)|| = sup{f(λ); λ ∈ σ(˜S)} = ||f||, f ∈ 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)∗, ∀f ∈ F. This relation is carried to C(˜σ(S)). suppose f ≥ 0, f ∈ C(˜σ(S)),, then there exists g ∈ 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, ∞), (Recall that put a self adjoint operator S, then σ(S) ⊂ [α, β], where α = inf{(Sx, x); ||x|| = 1}, β = sup{(Sx, x); ||x|| = 1}.) 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 = inf{(Sx, x); ||x|| = 1}, β = sup{(Sx, x); ||x|| = 1} is bounded. Conversely suppose σ(S) is bounded. Let f(s) ≡ s. Then f ∈ C(σ(S)). Moreover, (f − λ)rλ = e,λ ∈ ρ(S), so (f(S) − λ)Rλ = Rλ(f(S) − λ) = I. Therefore f(S) − λ = S − λ, so S = f(S) ∈ 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 ∞. 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 {fn(s)}n is increase sequence, all s ∈ σ. Deﬁnition 2.5. {S} ⊂ B(H) is called full if ∨R(S) = H

−1 Note that if S is self adjoint, {(λ − S) }λ 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 {fn} ⊂ A+ and fn ↗ f ∈ A+, then A1(fn) → A1(f). || || ≤ | | Moreover, the extention A1 also preserving adjoint and order, and A1(f) sups∈σ 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 ∈ A0, f is bounded. Then {A0(fn)} is a bounded increase sequence of operators in B(H), so A0(fn) → S ∈ B(H). Suppose also gn ∈ A0, 0 ≤ gn ↗ f. Then A0(gn) → T ∈ B(H). Let fn ∧ gm(s) := min{fn(s), gm(s)}. Then fn ∧ gm ↗ fn as m → ∞. These functions are continuous and 0 at ∞ (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 ∈ A0 ∩ A+, set fn ≡ f. Then A1(f) = lim A0(fn) = A0(f). Suppose fn, gn ∈ A0, 0 ≤ fn ↗ f ∈ A+, 0 ≤ gn ↗ g ∈ A+. Then fn + gn ↗ f + g, fngn ↗ fg, so A1(f + g) = A1(f) + A1(g), A1(fg) = A1(f)A1(g). Similarly, if f ∈ A+, α ≥ 0, then A1(αf) = αA1(f). If f ≤ g, f, g ∈ A+, then we put the sequence fn ≤ gn. Thus A1(f) = lim A0(fn) ≤ lim A0(gn) = A1(g). Now, since A+ is closed under addition, multiplication, and multiplication by non-negative numbers, A1 contains precisely of all functions of the form f = f1f2 + if3 − if4, fj ∈ A+. Also if f = g1−g2+ig3−ig4, gj ∈ A+, then f1+g2 = g1+f2, f3+g4 = g3+f4. Thus by the algebraic property of A1, A1(f1)−A1(f2)+iA1(f3)−iA1(f4) = A1(g1)−A1(g2)+iA1(g3)−iA1(g4). Therefore A1 has a unique linear extension to A1, and extension is an algebra homomorphism.Moreover, this argument ∗ ∗ also shows A1(f ) = A1(f) . If f, g ∈ A1 and f ≥ g, then f = f1 − f2, g = g1 − g2, fj, gj ∈ A+, and f1 + g2 ≥ g1 + f2, so A1(f1 + g2) ≥ A1(g1 + f2). Therefore A1(f) ≥ A1(g). fn ∈ A+ andfn ↗ f ∈ A+, take sequences {fn,m} ⊂ A0 ∩ A1 with fn,m ↗ fn. Let gn(s) := max{f1,n(s), ··· fn,n(s)}. Then gn ∈ A0 ∩ A+ andgn ≤ fn, gn ↗ f. Therefore A1(gn) ≤ A1(fn) ≤ ∗ 2 A1(f), and since A1(gn) → A1(f) we have A1(fn) → A1(f). If |f(s)| ≤ c, then (f f)(s) ≤ c , so ∗ 2 2 2 2 A1(f) A1(f) ≤ c A1(e) ≤ c I, so ||A1(f)|| ≤ c . Finally, we discuss about fullness. if {A0(f)} is full, then since A1(e)A1(f) = A1(ef) = A1(f), all f, we have A1(e) = I. Conversely, let H1 = ∨R(A0(f)) be a subspace of H. Then each A1(f) has range in H1. Therefore since A1(e) = I, H1 = H, this means fullness.

Finally, we extend functional calculus to the locally continuous function, so the image of functional calculus is extended to the unbounded densely deﬁned closed operators.

Deﬁnition 2.7. Let eN be the characteristic function of (−N,N) and the restriction of en to σ (it N is in A+) also denoted by e . Let A be the algebra of functions on σ which are locally in A1, i.e. N N N those fsuch that e f ∈ A1, all N > 0. Write f := e f.

Theorem 2.8. Let σ be a closed subset of R and let A0 : A0(σ) → B(H) be an isometric algebra isomorphism onto a full subalgebra of B(H) of which preserve adjoints. There is an unique extension A : A(σ) → C(H), which has the properties: 1. A(αf) = αA(f), α ≠ 0; 2. A(f) + A(g) ⊂ A(f + g); 3. A(f)A(g) ⊂ A(fg);

4. A(f ∗) = A(f)∗; ⇒ || || ≤ | | 5. f is bounded A(f) supσ f(s) ; 6. f ≥ 0 ⇒ (A(f)x, x) ≥ 0, all x ∈ D(A(f));

7. fn ∈ A+, fn ↗ f ⇒ (A(fn)x, x) → (A(f)x, x) all x ∈ D(A(f)); 8. if s ∈ D(A(f)), then A(f N )x → A(f)x as N → ∞.

4 Proof. Lemma 2.6 give us a unique extension A1 : A1 → B(H). Given f ∈ A, let D(A(f)) = { || N || ∞} ∈ || N ||2 | N |2 x ; supN A1(f )x < . Suppose x D(A(f)). A1(f )x = (A1( f )x, x) is bounded and non-decreasing with N, hence convergent as N → ∞. For M < N,(f ∗)M f N = (f ∗)M f M , so

N M 2 N 2 M 2 ||A1(f )x − A1(f )x|| = ||A1(f )x|| − ||A1(f )|| → 0 (M,N → ∞). Let A(f)x be this limit. This is independent of choice of internal. By this extension, A(f) is linear. N N Since e ↗ e, A1(e )A1(f) = A1(f). Therefore if f ∈ A1, so that D(A(f)) = H, we have N N A(f) = lim A1(f ) = lim A1(e )A1(f) = A1(f).

N N In general, if f is bounded then the f are uniformly bounded; by the lemma ||A1(f )|| ≤ sup |f N (s)| ≤ sup |f(s)|. Therefore A(f) ∈ B(H), ||A(f)|| ≤ sup |f(s)|. Each EN := A(eN ) is projection whose range is clearly contained in D(A(f)) for any f ∈ A. Since EN → I, ∨EN (H) = H. Therefore each A(f) is densely deﬁned. Suppose y ∈ D(A(f ∗)), x ∈ D(A(f)). Then (y, A(f)x) = lim(y, A(f N )x) = lim(A(f ∗)N x, y) = (A(f ∗)y, x). Conversely, suppose y ∈ D(A(f)∗) and A(f)∗y = z. Then (EN y, A(f N )x) = (y, A(f)EN x) = (z, EN x) = (EN z, x). Thus A((f)∗N )y = EN z → z, so y ∈ D(A(f ∗)∗) and A(f ∗)y = z. This proves 4 and it follows that A(f) = A(f ∗)∗ is closed. So far we have deﬁned A, shown that A(f) ∈ C(H) and proved 4 and 7. Before show the other relations, let us show uniquness. If B is another extension of A0, it also extends A1. By 7, B(f) ⊂ A(f). Taking adjoint, B(f) = B(f ∗)∗ ⊃ A(f ∗)∗ = A(f). Therefore A(f) = B(f). Now, we show the other relations. 1 is trivial, Suppose f, g ∈ A.(f + g)N = f N + gN , (fg)N = N N N f g and ||A1((f + g) x)|| = ||A1(f)x + A1(g)x||. Thus D(A(f)) ∩ D(A(g)) ⊂ D(A(f + g)). For x ∈ D(A(f)) ∩ D(A(g)),

N A(f + g)x = lim(A1((f + g) x) = A(f)x + A(g)x. If x ∈ D(A(g)) and A(g)x ∈ D(A(f)), then ||A((fg)N )x|| = ||A(f N )A(g)x|| is bounded. Thus N M N D(A(f)A(g)) ⊂ D(A(fg)) and since A1(f )A1(g ) = A1((fg) ), N < M, N A(f)A(g)x = lim A1(f )A(g)x N N = lim A1(f )(lim A1(g )x) N = lim A1((fg) x) = A(fg)x.

N If f ≥ 0, then f ≥ 0, so 6 follows fram that A1 preserves order. Finally, suppose fn ∈ A+, fn ↗ f ∈ A. Then for x ∈ D(A(f)), N (A(fn) x, x) ≤ (A(fn)x, x) ≤ (A(f)x.x).

5 N → N N → By the Lemma 2.6, A(fn ) A(f ), and since (A(f )x, x) (A(f)x, x), we get 7.

Corollary 2.9. S is a self adjoint operator and σ isσ(S). Let A0 : A0(σ) → B(H) be given by A0(f) = f(S). Let A be the extension of theorem, and f(s) ≡ s. Then A(f) = S. Proof. Let h(s) = (s − i)−1. Then hf = 1 + if, so (S − i)−1A(f) ⊂ I + i(S − i)−1. Multiplication on the left by S − i gives A(f) ⊂ S. Then A(f) = A(f)∗ ⊃ S∗ = S, so A(f) = S.

Conbining Theorem 2.2, 2.8, and Corollary 2.9, we have a map from A(σ(S)) to C(H), which we shall denoted by f 7→ f(S). This is an abstract construction of Functional Calculus. Next, we see explicit construction of this functional calculus.

Proposition 2.10. Let {Et} be a spectral measure with support∫ σ and deﬁne et(s) := 1, s ≤ t, et(s) := 0, s > t. The map et 7→ Et has extension, written f 7→ f(s)dEs : A(σ) → C(H) and it satisfy the same consitions of theorem 2.8.

B B Proof. If B is a half-open interval (a, b], let e be the characteristic function. Then e = eb−ea. Let E(B) = Eb − Ea, and let E(∅) = 0. If B1,B2 are two such intervals, then the fact that Et increases with t implies E(B1 ∩ B2) = E(B1)E(B2), while E(B1 ∪ B2) = E(B1) + E(B2) − E(B1 ∩ B2), if B B1 ∪ B2 is again an interval. It follows that the map e → E(B) has a unique linear extension B which is an algebra∑ homomorphism from the algebra of functoins generated by the e into B(H). Bj Furthermore if f = αje where the Bj are disjoint and the sum is ﬁnitem then for x ∈ H, ∑ ∑ 2 2 2 || αjE(Bj)x|| = |αj| ||E(Bj)x|| ∑ 2 ≤ sup{|αj| ; R(Bj) ≠ 0} ||E(Bj)x|| ≤ sup{|f(s)|}||x||2. σ {| |} | | ̸ Conversely, supσ ∑f(s) = αk for some k such that E(Bk) = 0. Take x in∑ the range of E(Bk)with 2 2 2 ||x|| = 1. Then || αjE(Bj)x|| = ||αkx|| = |αk| . Thus the map f → αjE(Bj) has a unique extension to an algebra isometry into B(H) from the algebra A∗(σ)which is the closure in the sup norm on σ of the algebra generated by the eB. Moreover, this map is seen to preserve adjoint. Since A0(σ) ⊂ A∗(σ) we may restrict to A0(σ) and apply Theorem 2.2 to get a mapping A1 : A1(σ) → B(H). We assert that A1(et) = Et. Given t ∈ R, choose a sequence {fn} of 1 continuous functions with 0 ≤ f ≤ e, f ↗,f (s) = 0, s ≤ t, f (s) = 1, t + ≤ s ≤ t + n, and n n n n n 1 f (s) = 0, s ≥ t + n + 1. Then f ↗ 1 − e . Let B = (t + , t + n], and let F := A (f ). Our n n t n n n 1 n construction gives E(Bn) ≤ Fn ≤ I − Et. However, E(Bn) → I − Et, so A1(1 − et) = lim A1(fn) = I − Et, or A1(et) = Et. Thus A/1 extends the map et → Et; moreover Et ↗ I as t → ∞ implies that A1(A0) is full. The assertion now follows from Theorem 2.2.

Proposition 2.11. Let {Et} be a spectral measure and let ∫ its Ut := e dEt.

6 Then {U } is strongly continuous unitary group. t ∫ ∫ ist Proof. Let ut(s) := e . Ut = ut dEs is in the image of extension map f 7→ f(s) dEs in ∗ Proposition 2.10. Therefore Ut satisﬁes corresponding properties of ut. ut ut = 1 shows each Ut is unitary, and us+t = utus shows Us+t = UsUt. Now we have already ∨R(Eb − Ea) = H. Put x ∈ R(Eb − Ea), then ut(eb − ea) → us(eb − ea) uniformly as t → s. Thus Utx → Usx everywhere.

1 Proposition 2.12. Let {U } be a strongly continuous unitary group in H and let S := lim → (U x− t t 0 it t x). Then S is exists and self adjoint.

Proof. The existence is following meaning. {Ut} is unitary group, so the domain D(S) is defined as the set of all x such that limt→0(Utx − x) is exists, and letSx = y. This is well-defined.∫ ∞ ∞ | | Now, we proof self-adjointness. Suppose ϕ is a continuous function on [0, ) with 0 ϕ(s) ds < ∞. Define Tϕx, x ∈ H, as ∫ ∞ Tϕx = ϕ(s)Usxds. 0 Since Utxis continuous about t and has norm ||x||, this show the existence of the improper Riemann ′ integral, and so Tϕ ∈ B(H). Suppose also that the derivation ϕ is continuous and absolutely integrable on [0, ∞). For t > 0, ∫ 1 1 ∞ (U − I)T x = ϕ(s)[U x − U x]ds it t ϕ it s+t s ∫ 0 ∫ ∞ 1 1 t = [ϕ(s − t) − ϕ(s)]Usxds + ϕ(s)Usxds. t it it 0

Then we think about limit this fomula, and since Usx → x as s → 0 from Proposition 2.11, ∫ ∫ 1 ∞ 1 1 t lim (U − I)T x = lim [ϕ(s − t) − ϕ(s)]U xds + ϕ(s)U xds → t ϕ → s s t 0 it t 0 t it it 0 = iTϕ′ + iϕ(0)I, ∫ ∞ ′ { } so we get STϕ = iTϕ + iϕ(0)I. If ϕn is a sequence of functions as above with 0 ϕn(s)ds = 1 1,ϕ (s) = 0 for s ≥ , then T x → as n → ∞. Thus D(S) is dense. n n ϕn 1 1 Since ( (U − I)x, y) = (y, (U − I)x), S is symmetric, so we should show only that i ∈ ρ(S) it t it t −s from proposition 2.1. Take ϕ(s) = ie and consider STϕ = iTϕ′ + iϕ(0)I. Deﬁne R+ as ∫ ∞ −s R+x = −i e Usxds. 0 Then ∫ ∞ −s (S + i)R+x = (S + i) − i e Usxds 0 ∫ ∞ −s = −iSTϕx + e Usxds 0 = iTϕ′ x + iϕ(0)Ix − iTϕ′ s = Ix,

7 so (S + i)R+ = I and the domain is everywhere. Since UtUs = UsUt from Proposition2.12, D(S) is invariant under Ut and UtS ⊂ SUt. It follows that R+(S + i) ⊂ (S + i)R+ = I Therefore −i ∈ ρ(S) −1 −1 and R+ = (S + i) . Similarly, i ∈ ρ(s) and (S − i) = R−, where ∫ ∞ −s R−x = i e U−ssds. 0 Thus S is self adjoint.

Theorem 2.13. There are 1-1 correspondence between self adjoint operators S, spectral measure {Et} and unitary groups {Ut} given by: ∫

Et = et(S),S = s dEs;

1 Ut = ut(S),S = lim (Us − I); ∫ s→0 is ist Ut = e dEs.

Proof. We show ﬁrst correspondence. If S is self adjoint and Et = et(∫S), then by uniqueness of the extensions to A(R) and the corollary of theorem∫ 2.8, S = f(s) = f(∫s)dEs, wheref(s) = s. −1 Conversely, if {Et} is a spectral measure and∫S = sdEs, them (λ−S) = rλ(s)dEs, λ ̸∈ R; this follows from the properties∫ of the map g 7→ g(s)dEs. Since the {rλ} generate a dense subalgebra of A0(σ), we get∫g(S) = g(S)dEs for every g ∈ A0(σ) and therefore for every g ∈ A1(σ). In particular, Et = et(S)dEs = et(S). 1 Now, we show second correspondence. If U = u (S) and T = lim (U − I), let {E } be the t t it t t 1 spectral measure associated with S. Now (u − e)(e − e ) → f(e − e ) unifomly as ϵ → 0, where iϵ ϵ b a b a f(s) = s. Therefore the range of Eb − Ea, a < b, is in D(T ) and T = S on this range. For any x ∈ D(S) → 0, xN := (EN − E−N )x → x and T xN = SxN → Sx as N → ∞. Therefore S ⊂ T . 1 However, S = S∗ ⊂ T ∗ = T , so S = T . Conversely, let {U } be a unitary group, S = lim (U − I), t it t 1 and V = u (S). Then we have also S = lim (V − I). If x, y ∈ D(S), then U x, V y ∈ D(S), all t. t t it t t t Therefore the scalar functionϕ(t) = (Utx, Vty) is diﬀerentiable, and ′ ϕ (t) = (iSUtx, Vxy) + (Utx, iSVty) = 0. ≡ ≡ ∗ Thus (Utx, Vty) (U0x, V0y) (x, y). Since D(S) is dence, this implies Vt Ut = I, or Vt = Ut, i.e. {Ut} and {Vt} are equivalent. ∫ its Finally, we show third correspondence. IfUt = e dEs where {Et} is a spectral measure. Let ∫ 1 S = s dE . Then we know U = u (S), and consequently S = lim (U − I). Then we know s t t it t 1 U = u (S), and consequently S = lim (U −I). Therefore we have already {U } and {E } uniquely t t it t t t from S.

8 This is the∫ explicit construction of functional calculus. For example, we want to calculous f(S), we calculus f(s)dEs. Moreover, this construction reveal the representation of S as unitary group. ′ 1 For example, we have S0f := if on C , and S is closure of S0. Then S is self adjoint unbounded operator. From theorem 2.12, there exists unitary group{Ut} which correspond to S. {Ut} must satisfy the relation Ut = ut(S), so we get Utf(s) = f(s − t) by simple calculus of Taylor’s formula. Therefore Ut is a shift operator.

3 Reference

[1] The lecture notes of this class [2] R. Beals, ”topics in operator theory”, The University of Chicago Press Chicago and London, (1971).