OPERATOR THEORY - PART 2/3. Contents

Transcription

1 OPERATOR THEORY - PART 2/3 T.K.SUBRAHMONIAN MOOTHATHU Contents 1. Introduction to C -algebras 1 2. Continuous functional calculus for normal elements 4 3. Borel functional calculus for normal operators 8 4. Spectral theorem for normal operators What is the dual notion of surjectivity? Introduction to C -algebras C -algebras are modeled after Γ(X), where X is a Hilbert space. We will define self-adjoint, unitary, and normal elements in a C -algebra. Results about them apply to the corresponding type of operators on Hilbert spaces. Definition: Let Γ be a Banach algebra. An involution of Γ is a map : Γ Γ that is conjugatelinear (i.e., (αs + βt) = αs + βt ) and enjoys the properties (st) = t s and t = t. We call t the adjoint of t Γ. We say Γ is a C -algebra if Γ is a Banach algebra with an involution that satisfies t t = t 2 for every t Γ (this condition is known as the C -condition). Examples: (i) C is a unital, commutative C -algebra with involution z z. (ii) Γ(C k ) is a unital C -algebra where the involution is taking the conjugate-transpose of a matrix. (ii ) More generally, if X is a Hilbert space, then Γ(X) is a unital C -algebra with involution T T, where T is the Hilbert-adjoint of T given by T x, y = x, T y for every x, y X. If dim(x) > 1, then Γ(X) is non-commutative. (iii) If K is compact Hausdorff, then C(K, C) is a unital, commutative C -algebra with involution f f. There are more examples, but we ignore them at present. Exercise-17: Let Γ be a unital C -algebra. Then, (i) t = t for every t Γ, (ii) t t is a homeomorphism of Γ with 0 = 0 and e = e. (iii) t Inv(Γ) iff t Inv(Γ); and (t ) 1 = (t 1 ). (iv) σ(t ) = {α : α σ(t)}. [Hint: (i) s 2 = s s s s for s = t, t. (ii) s t = (s t) = s t and e = (e ) = (ee ) = ee = e. (iv) Apply part (iii) to s = αe t.] Definition: Let Γ be a unital C -algebra and t Γ. We say (i) t is normal if t t = tt, (ii) t is self-adjoint if t = t, and (iii) t is unitary if t Inv(Γ) and t 1 = t (i.e., if t t = e = tt ). 1

2 2 T.K.SUBRAHMONIAN MOOTHATHU Exercise-18: Let Γ be a unital C -algebra. (i) If t Γ is either self-adjoint or unitary, then t is normal; but there are no other implications among the three concepts. (ii) If t Γ is unitary, then t = 1, t is unitary, and t = 1. (iii) Any t Γ can be uniquely written as t = t 1 + it 2 with t 1, t 2 Γ self-adjoint. (iv) If t Γ is self-adjoint, then r(t) = t. [Hint: (ii) 1 = e = t t = t 2. (iii) Take t 1 = (t + t )/2 and t 2 = (t t )/2i. (iv) t 2 = t t = t 2 and hence t 2n = t 2n. Also, by considering the subsequence (2 n ) in the spectral radius formula, r(t) = lim n t 2n 1/2n.] [126] Let Γ be a unital C -algebra and t Γ. (i) If t is unitary, then σ(t) S 1 := {z C : z = 1}. (ii) If t is self-adjoint, then σ(t) R. Proof. (i) If α σ(t), then 1/α σ(t 1 ) = σ(t ). So α t = 1 and 1/α t = 1. (ii) Let f : C C be f(z) = exp(iz). Then f is holomorphic, and observe f(z) S 1 iff z R. Consider f[t] Γ defined via holomorphic functional calculus. Since the involution is continuous and conjugate-linear, (f[t]) = (exp[it]) = (lim k k n=0 (it)n /n!) = lim k k n=0 ((it)n /n!) = lim k k n=0 ( it)n /n! = exp[ it] = (f[t]) 1 and hence f[t] is unitary. By the Spectral mapping theorem and part (i) above, f(σ(t)) = σ(f[t]) S 1, and use the initial observation. Definition: A subalgebra Π of a unital C -algebra Γ with t C -subalgebra of Γ. Π for every t Π is called a The following should be contrasted with [124]. Also see later that [127] is a corollary of [129]. [127] [Independence of the spectrum] Let Γ be a unital C -algebra, t Γ, and let Π be a closed C -subalgebra of Γ with {e, t} Π. Then σ(t, Π) = σ(t, Γ). Proof. We have to prove ρ(t, Γ) ρ(t, Π). Consider α ρ(t, Γ), let s = αe t, and u Γ be such that su = us = e. Then u s = s u = e also. We need to show u Π. It suffices to show u u Π, for then u = eu = (s u )u = s (u u) Π. Verify by direct multiplication that u u is the inverse of ss in Γ. Next, note that ss is self-adjoint and therefore σ(ss, Π) = σ(ss, Γ) by [126](ii) and [124](v). Since ss Inv(Γ), we have 0 / σ(ss, Γ) = σ(ss, Π), and thus ss Inv(Π). This implies u u Π by the uniqueness of the inverse of ss in Γ. Seminar topic: [Stone-Weierstrass theorem] Let K be a nonempty compact Hausdorff space and let Π be a closed C -subalgebra of C(K, C) containing constant functions and separating points of K. Then Π = C(K, C). [see Theorem 7.33 in Rudin, Principles of Mathematical Analysis.] Remark: Certain properties of C may be transferred to the C -algebra C(K, C) (where K is a nonempty compact Hausdorff space), and then transferred to suitable commuting portions of an abstract unital C -algebra Γ by the help of Gelfand mapping. The following table indicates some of the similarities (what has been already done, and a few results which are yet to be established).

3 OPERATOR THEORY - PART 2/3 3 C C(K, C) Γ z f t z z f f t t C \ {0} nonvanishing f Inv(Γ) (1 z) 1 = n=0 zn 1/(1 f) = n=0 f n (e t) 1 = n=0 tn (when < 1) R C(K, R) {self-adjoint elements} {0, 1} {f : f = χ A } {self-adjoint idempotents (projections)} S 1 = {z 0 : z 1 = z} {f : f 1} {unitary elements} [0, ) {f : f(k) [0, )} Γ + := {positive elements} z = x + iy f = f 1 + if 2 ; f j (K) R t = t 1 + it 2 with t 1, t 2 self-adjoint 0 z = z exp(iθ) nonvanishing f = f f invertible t = su; s Γ +, u unitary f x 0 x = y 2, y 0 f 0 f = g 2, g 0 t Γ + t = s 2, s Γ + x R, x = x + x f(k) R, f = f + f t self-adjoint, t = t + t x, y 0 x + y 0 f, g 0 f + g 0 s, t Γ + s + t Γ + zz 0 ff 0 t t Γ + Definition: A homomorphism F : Γ 1 Γ 2 between two unital C -algebras with (F (t)) = F (t ) for every t Γ 1 is called a -homomorphism. [128] [Gelfand-Naimark theorem (unital case)] Let Γ be a commutative, unital C -algebra and M be its Gelfand space. Then the Gelfand mapping E : Γ C(M, C) given by E t (φ) = φ(t) for t Γ and φ M is a unital, isometric, -isomorphism of C -algebras. Proof. We already know E is a homomorphism. We now verify E is a -homomorphism by showing φ(t ) = φ(t) for t Γ and φ M. If t Γ is self-adjoint, then by [121] and [126], we have {φ(t) : φ M} = σ(t) R and we are done. In the general case, write t = t 1 + it 2 with t 1, t 2 self-adjoint and see that φ(t ) = φ(t 1 it 2 ) = φ(t 1 ) iφ(t 2 ) = φ(t 1 ) + iφ(t 2 ) = φ(t 1 + it 2 ) = φ(t). Since E is a -homomorphism, E t E t = E t t. Also note that t t is self-adjoint so that r(t t) = t t by Exercise-18(iv). Therefore, using the C condition and [121], we have E t 2 = E t E t = E t t = sup{ φ(t t) : φ M} = r(t t) = t t = t 2, and this shows E is an isometry (and in particular injective). Since E is an isometric -homomorphism, E(Γ) is a closed C -subalgebra of C(M, C). Also we know from [123] that E(Γ) separates points of Γ. If α C, then E αe α (in particular, E e 1). Thus E(Γ) = C(M, C) by Stone-Weierstrass theorem. Remark: Let Γ be a commutative, unital C -algebra. Then by [123] and [128] we see that the radical of Γ, namely φ M ker(φ), is {0} (in technical terms, Γ is semisimple).

4 4 T.K.SUBRAHMONIAN MOOTHATHU Remark: Let Γ be a commutative, non-unital C -algebra. To prove the compactness of M in the unital case, we used the fact that φ = 1 for every φ M which in turn depended on the fact φ(e) = 1. In the non-unital case, we can say only that φ 1 for every φ M and that M is a locally compact Hausdorff space. There is a process by which we can adjoin a unit to Γ to obtain a commutative, unital C -algebra Γ where Γ = Γ C as a set. If M is the Gelfand space of Γ, then Γ = C( M, C) by [128]. Let ψ M be the projection of Γ to C. Then M can be identified with the one-point compactification M {ψ} of M, and hence C(M, C) = {f C( M, C) : f(ψ) = 0}. But the latter space can be identified with C 0 (M, C) := {f C(M, C) : for every ɛ > 0, f < ɛ outside a compact subset of M} (called the space of continuous functions vanishing at infinity). Filling in the details of this outline, the following can be proved. [128 ] [Gelfand-Naimark theorem] Let Γ be a commutative, C -algebra and M be its Gelfand space. Then the Gelfand mapping E : Γ C 0 (M, C) given by E t (φ) = φ(t) for t Γ and φ M is a norm-decreasing (i.e., E t t ) -isomorphism of C -algebras. Next, we may ask what can be said if Γ is not commutative. The answer is elegant, but proof is difficult; so we give only an outline. The actual detailed proof also requires some concepts mentioned later (for instance, positive elements); hence placing this theorem here is not so correct. [129] [Gelfand-Naimark-Segal construction/theorem] If Π is a C -algebra, then there exists a Hilbert space X such that Π is isometrically -isomorphic to a closed C -subalgebra of Γ(X). Proof. (Outline) Step-1 : Consider φ Π satisfying φ(t t) 0 for every t Π (such a φ is called a positive functional) and let Λ = {t Π : φ(t t) = 0}. Then Λ is a closed left-ideal of Π and we can define an inner product on Π/Λ by the rule s + Λ, t + Λ = φ(t s). Fix t Π, and define a bounded linear operator L : Π/Λ Π/Λ as L(s+Λ) = ts+λ. Let Y be the Hilbert space obtained by taking the completion of the inner product space Π/Λ. Then we can extend L to an element L Γ(Y ). To show the dependence, let us now write Y φ for Y and L φ, t for L. Step-2 : Let Ω = {φ Γ : φ is positive and φ = 1}. Let X be the Hilbert space defined as the Hilbertian sum of Y φ s, φ Ω. That is, X = {(y φ ) φ Ω Y φ : φ Ω y φ 2 < } with the inner product (y φ ), (z φ ) = φ Ω y φ, z φ. The required embedding F : Π Γ(X) is given as t F t : X X, where F t ((y φ )) = (L φ, t (y φ )) for t Π and φ Ω. 2. Continuous functional calculus for normal elements Definition: By a polynomial in z and z we mean a function or expression of the form n i,j=0 α ijz i z j where α ij C. When Γ is a unital C -algebra and t Γ is normal, by a polynomial in t and t we mean an expression of the form n i,j=0 α ijt i (t ) j Γ where α ij C.

5 OPERATOR THEORY - PART 2/3 5 Remark: (i) If K C is compact, then {polynomials in z and z} is dense in C(K, C) by Stone-Weierstrass theorem since the collection is a C -subalgebra of C(K, C) containing all constant functions and separates points of K. (ii) Let Γ be a unital C -algebra, t Γ be normal and Π be the closed C -subalgebra generated by {e, t}. Then Π = {polynomials in t and t }. [130] [Continuous functional calculus] Let Γ be a unital C -algebra, let t Γ be normal, let Π be the closed C -subalgebra of Γ generated by {e, t}, and let M be the Gelfand space of Π. Then, (i) There is a unique unital, isometric -isomorphism F : C(σ(t), C) Π that extends the holomorphic functional calculus in the following sense: if f C(σ(t), C) has a holomorphic extension to a neighborhood of σ(t), then F (f) = f[t]. Now write f[t] for F (f) Π for any f C(σ(t), C). (ii) For f C(σ(t), C) and s Π, we have f[t] = s iff f(φ(t)) = φ(s) for every φ M. Consequently, f(σ(t)) = σ(f[t]) [Spectral mapping theorem]. (iii) If f C(σ(t), C) and g C(σ(f[t]), C), then (g f)[t] = g[f[t]]. (iv) If s Γ commutes with t and t, then f[t]s = sf[t] for every f C(σ(t), C). Proof. (i) The main idea is simple: we have to verify C(σ(t), C) = C(M, C) = Π. Recall from [127] that σ(t, Π) = σ(t, Γ) so that we may just write σ(t). The Gelfand space M is defined for Π since Π commutative: Π is generated by the commutative set {e, t, t }. Let E : Π C(M, C), E s (φ) = φ(s), be the Gelfand mapping. By [128], E is a unital, isometric -isomorphism. By the definition of Gelfand topology and by [121](ii) we know E t : M σ(t), E t (φ) = φ(t), is a continuous surjection. If φ, ψ M are such that φ(t) = ψ(t), then φ(t ) = φ(t) = ψ(t) = ψ(t ) since E t = E t. By linearity and the multiplicative property, φ(s) = ψ(s) whenever s is a polynomial in t and t. Since polynomials in t and t are dense in Π, we conclude φ = ψ by continuity. Thus E t is a continuous bijection, and hence a homeomorphism since the spaces are compact Hausdorff. Therefore, F 1 : C(σ(t), C) C(M, C) defined as F 1 (f) = f E t, or F 1 (f)(φ) = f(φ(t)), is a unital, isometric -isomorphism. The required F : C(σ(t), C) Π is now taken as F = E 1 F 1. To see F extends the holomorphic functional calculus, it is enough to check F (p) = p[t] for polynomial p (then the equality extends to holomorphic functions via rational holomorphic functions; see the Remark after [115]). By linearity, it suffices to consider z k. If f(z) = z k, then by definition, F (f)(φ) = f(φ(t)) = φ(t) k = φ(t k ) = E t k(φ) so that F (f) = t k, and we are through. If f(z) = n i,j=1 α ijz i z j, then F (f) = n i,j=1 α ijt i (t ) j by what is stated in the above paragraph and by the fact that F is a -homomorphism. The continuity of F now imposes uniqueness on F since polynomials in z and z are dense in C(σ(t), C). (ii) f[t] = F (f) = s F 1 (f) = E s f(φ(t)) = φ(s) for every φ M. (iii) Let s = f[t] Π, let Π be the closed C -subalgebra generated by {e, s}, and note u := g[s] Π Π. If M is the Gelfand space of Π, then φ := φ Π M for φ M. Therefore by (ii), for φ M we get φ(u) = φ(u) = g( φ(s)) = g(φ(s)) = g(f(φ(t))) and hence (g f)[t] = u again by (ii).

6 6 T.K.SUBRAHMONIAN MOOTHATHU (iv) s {t, t } c and f[t] Π {t, t } cc. Example: The continuous functional calculus takes a simple form for diagonal operators. Let X be a Hilbert space with orthonormal basis {e n : n N} and T Γ(X) be the diagonal operator given by the bounded sequence (β n ) in C. That is, T x = n=1 β nα n e n if x = n=1 α ne n. Then T is the diagonal operator given by the sequence (β n ). In particular, T is normal. Recall from [101] that σ(t ) = {β n : n N}, and consider the linear map F : C(σ(T ), C) Γ(X), where F (f) Γ(X) is the diagonal operator given by the sequence (f(β n )). We have F (f) = sup n N f(β n ) = f. Moreover, if f is a polynomial in z and z, it may be seen that F (f) = f[t ]. Since polynomials in z and z are dense in C(σ(T ), C), it follows by continuity that F (f) = f[t ] for every f C(σ(T ), C). In other words, f[t ]x = n=1 f(β n)α n e n if x = n=1 α ne n for every f C(σ(T ), C). Example: Let (Y, µ) be a σ-finite measure space, g L C (Y, µ), let L2 = L 2 C (Y, µ), and M g Γ(L 2 ) be the multiplication operator M g h(y) = g(y)h(y) for h L 2 and y Y. Imitating the argument given above for diagonal operators, verify that f[m g ] = M f g for every f C(σ(M g ), C). Example: Consider T Γ(C 2 ) given by T e 1 = e 2 and T e 2 = 0. Polynomials p(z) = z and q(z) = z 2 agree on σ(t ) = {0}, but T 0 = T 2. So we cannot define continuous functional calculus for T. A consequence of [130] is that normal elements of C -algebras share many properties of f C(K, C) where K C is compact. The details can be seen below. Definition: Let Γ be a unital C -algebra and let Γ + = {t Γ : t = t and σ(t) [0, )}. Elements of Γ + are called positive elements of Γ, and we use the notation t 0 to mean t Γ +. Example: Let K be compact Hausdorff and Γ = C(K, C). For f Γ, note the following: (i) f = f and σ(f) = f(k). Hence r(f) = f and Γ + = {f Γ : f(k) [0, )}. (ii) f is self-adjoint iff σ(f) = f(k) R. (iii) f is a self-adjoint idempotent iff σ(f) = f(k) {0, 1}. (iv) f is unitary iff σ(f) = f(k) S 1. (v) If f is self-adjoint, then we can write f = f + f uniquely with f +, f Γ + and f + f = 0. (vi) If f Γ +, then for every n N, there is unique g Γ + such that g n = f. (vii) If f is self-adjoint and f 1, then there are unitary f 1, f 2 Γ with f = (f 1 + f 2 )/2. In fact, f 1 = f + i 1 f 2 and f 2 = f i 1 f 2 (this comes from the following: if x [ 1, 1] and if we define z 1 = x + i 1 x 2, z 2 = x i 1 x 2, then z 1, z 2 S 1 and x = (z 1 + z 2 )/2). Translating these properties by the help of [130], we have the following: [131] Let Γ be a unital C -algebra and let t Γ be normal. Then, (i) r(t) = t. (ii) t is self-adjoint iff σ(t) R. (iii) t is a self-adjoint idempotent iff σ(t) {0, 1}.

7 OPERATOR THEORY - PART 2/3 7 (iv) t is unitary iff σ(t) S 1. (v) If t is self-adjoint, then t = t + t for unique t +, t Γ + satisfying t + t = 0 = t t +. (vi) If t Γ +, then for every n N, there is unique s Γ + such that s n = t. (vii) If t is self-adjoint and t 1, then we can write t = (u 1 + u 2 )/2 with u 1, u 2 Γ unitary. Proof. Consider f C(σ(t), C) given by f(z) = z, use [130] and the example above. This will establish (i)-(iv), (vii), and the existence part of (v) and (vi). It remains to show uniqueness in (v) and (vi). Let Π be the closed C -subalgebra generated by {e, t}. Note that Π = {p[t] : p is a polynomial} since t = t. Also note t +, t from (v) and s from (vi) belong to Π. (v) [Uniqueness] Suppose t = t 1 t 2 is another decomposition, where t 1, t 2 Γ + and t 1 t 2 = 0 = t 2 t 1. Both t 1 and t 2 commute with t, hence with p[t], and hence with every element of Π. Considering the closed C -subalgebra Π generated by the commutative set {e, t +, t, t 1, t 2 } and using the uniqueness in C( M, C) via [128] (where M is the Gelfand space of Π), we get t 1 = t + and t 2 = t. (vi) [Uniqueness] If u Γ + is with u n = t, then as in (v), check that u commutes with every element of Π and in particular with s. Now consider the closed C -subalgebra of Γ generated by the commutative set {e, s, u} and as before apply [128] to see u = s. Remark: If Γ is a unital C -algebra, then Γ = span(γ + ) and Γ = span{t Γ : t is unitary} by [131](iv) and [131](vi) since any t Γ can be written as t = t 1 + it 2 with t 1, t 2 self-adjoint. Remark: Mainly because of the equality r(t) = t for normal elements, the following fact is true: if X is a Hilbert space, then T σ(t ) and T r(t ) are continuous when restricted to the collection of normal operators, see Problem 105 in Halmos, A Hilbert Space Problem Book. Exercise-19: Let Γ be a unital C -algebra. Then, (i) Γ + is a closed cone (here cone means s + t, αt Γ + for every s, t Γ + and every real α 0). (ii) Γ + = {t t : t Γ}. (iii) e + t t Inv(Γ) for every t Γ. [Hint: Use [128] and borrow some properties of C(M, C); but some extra work is needed. we will indicate partial solution below. For the remaining arguments see results 3.6 and 3.7 in Conway, A Course in Functional Analysis (also see p of Rudin, Functional Analysis). (i) Suppose t 1, t 2 Γ +. Assume t i 1 so that σ(t i ) [0, 1] and hence σ(e t i ) [0, 1] by spectral mapping theorem. Thus s i := e t i Γ + and s i = r(s i ) 1. By self-adjointness and the estimate e (t 1 + t 2 )/2 = (1/2) s 1 + s 2 1, we have σ(e (t 1 + t 2 )/2) [ 1, 1] so that σ( t 1 + t 2 ) [0, 2] and thus σ(t 1 + t 2 ) [0, 4], again by spectral mapping theorem. So t 1 + t 2 Γ +. 2 (ii) If s Γ +, choose t Γ + with t 2 = s, and then s = tt = t t. The converse is slightly difficult. (iii) Follows from (ii) since σ(e + t t) = 1 + σ(t t) by spectral mapping theorem.]

8 8 T.K.SUBRAHMONIAN MOOTHATHU Remark: Let X be a Hilbert space. Usually, T Γ(X) is called a positive operator if T x, x 0 for every x X. With respect to this definition, statements in Exercise-19 are almost trivial (verify). Later, after introducing the notion of approximate eigenvalues, we will see in [139] that the definition of a positive operator given here agrees with that of a positive element in Γ(X). The following corresponds to writing z C \ {0} as z = z w with z > 0 and w S 1. [132] [Polar decomposition] Let Γ be a unital C -algebra and t Inv(Γ). Then there exist s Γ + and unitary u Γ such that t = su. This decomposition is unique and s = (tt ) 1/2. If t is also normal, then {t, s, u} is a commutative set so that t = us also. Proof. Note tt = (t ) t Γ + by Exercise-19. Let s = (tt ) 1/2 Γ + whose existence is guaranteed by [131]. Since t is invertible, tt is invertible (with inverse (t ) 1 t 1 ) and hence 0 / σ(tt ). Consequently, 0 / σ(s) and thus s Inv(Γ). Taking u = s 1 t, it may be directly verified that u u = uu = e by using the fact s 2 = tt. If t = s 1 u 1 is another decomposition with s 1 Γ + and u 1 unitary, then tt = s 1 u 1 u 1 s 1 = s 2 1. By the uniqueness of square root asserted by [131], we get s 1 = s, and then u 1 = u. If t is also normal, the closed C -subalgebra Π generated by {e, t, t } is commutative. Since tt Π +, we have s = (tt ) 1/2 Π + Π. Therefore, t commutes with s, and hence with s 1. It follows that u commutes with both t and s. Another proof when t is normal: Take s = f 1 [t], u = f 2 [t], where f 1 (z) = z, f 2 (z) = z/ z. Remark: In [132], su us in general. However, if we start with s = (t t) 1/2 and put u = ts 1, then the decomposition of t can take the form t = us. 3. Borel functional calculus for normal operators Our aim here is to extend the continuous functional calculus. Some preparations first. Definition and Remark: If K is a compact metric space, let B(K) denote the Borel σ-algebra on K. Let B (K, C) = {f : K C : f is Borel and bounded}, which is a commutative, unital C -algebra with respect to the sup-norm and the involution f := f. Also, C(K, C) B (K, C) as a closed C -subalgebra. By a complex Borel measure on K, we mean a function µ : B(K) C satisfying µ[ ] = 0 and µ[ n=1 A n] = n=1 µ[a n] whenever A 1, A 2,... B(K) are disjoint. We define µ := sup n=1 µ[a n], where the supremum is taken over all Borel partitions K = n=1 A n. We have µ < by Theorem 6.4 in Rudin, Real and Complex Analysis. Any complex Borel measure µ on K has a decomposition µ = µ 1 + iµ 2 = (µ + 1 µ 1 ) + i(µ+ 2 µ 2 ), where µ+ 1, µ 1, µ+ 2, µ 2 0 (i.e., they are usual measures). This decomposition helps to define fdµ by linearity. Exercise-20: Let K be a compact metric space. Then the collection of simple functions (i.e., functions of the form n j=1 α jχ Aj, where α j C and A j B(K)) is dense in B (K, C). [Hint:

9 OPERATOR THEORY - PART 2/3 9 Given f B (K, C) and ɛ > 0, consider a Borel partition f(k) = n j=1 D j with diam[d j ] < ɛ and consider n j=1 α jχ Aj, where α j D j and A j = f 1 (D j ).] Seminar topic: [Riesz representation theorem of functionals] Let K be a compact metric space and φ : C(K, C) C be a bounded linear functional. Then there is a unique complex Borel measure µ on K such that φ(f) = K fdµ for every f C(K, C). Moreover, µ = φ. [see Theorem 6.19 in Rudin, Real and Complex Analysis; also see Theorem 2.14 for the real version with φ positive.] We need some information about sesquilinear forms. Definition: Let X be a Hilbert space. A map ψ : X 2 C is a sesquilinear form if ψ is linear in the first variable and conjugate-linear in the second variable. For example, if T Γ(X), then (x, y) T x, y is sesquilinear. We say a sesquilinear form ψ : X 2 C is bounded if there is c > 0 such that ψ(x, y) c x y for every x, y X; and the infemum of such c is taken as ψ. Exercise-21: Let X be a Hilbert space. (i) Let S, T Γ(X). If Sx, y = T x, y for every x, y X, then S = T. (ii) [Polarization identity] If ψ : X 2 C is sesquilinear, then ψ(x, y) = p=0 ψ(x + ip y, x + i p y) for every x, y X. (iii) Let S, T Γ(X). If Sx, x = T x, x for every x X, then S = T. (iv) [Reisz representation theorem of sesquilinear forms] If ψ : X 2 C is a bounded sesquilinear form, there is unique T Γ(X) such that ψ(x, y) = T x, y for every x, y X. Also, T = ψ. [Hint: (i) Note that (S T )x, y = 0 and take y = (S T )x. (ii) Verify directly. (iii) Use (ii) and (i) by considering ψ 1 (x, y) = Sx, y, ψ 2 (x, y) = T x, y. (iv) For a fixed x, the functional y ψ(x, y) is bounded linear, so there is z X with ψ(x, y) = y, z. Take T x = z.] For a normal operator T of a Hilbert space, below we extend the continuous functional calculus to Borel functional calculus by defining f[t ] for every f B (σ(t ), C). The idea of the proof is as follows: for f C(σ(T ), C), we will show f[t ]x, y = σ(t ) fdµ x,y for a suitable measure µ x,y, and then we will define f[t ] for f B (σ(t ), C) by the same expression, f[t ]x, y = σ(t ) fdµ x,y. [133] [Borel functional calculus] Let X be a Hilbert space, T Γ(X) be normal, let K = σ(t ) and F : C(K, C) Γ(X) be the continuous functional calculus of T. Then, (i)-(x) below are true. (i) For every x, y X, there is a unique complex Borel measure µ x,y on K such that F (f)x, y = K fdµ x,y for every f C(K, C). Moreover, µ x,y x y and µ x,y = µ y,x. Proof. Fixing x, y X, define φ x,y : C(K, C) C as φ x,y (f) = F (f)x, y. Then φ x,y is linear. Also, φ x,y (f) f x y since F is an isometry so that φ x,y x y. By Riesz representation of functionals, there is a unique complex Borel measure µ x,y on K such that φ x,y (f) = K fdµ x,y for every f C(K, C), and µ x,y = φ x,y x y. If f C(K, R), then F (f) = F (f) = F (f)

10 10 T.K.SUBRAHMONIAN MOOTHATHU so that K fdµ x,y = F (f)x, y = x, F (f)y = F (f)y, x = K fdµ y,x = K fdµ y,x. By linearity, K fdµ x,y = K fdµ y,x for any f = f 1 +if 2 C(K, C). So µ x,y = µ y,x by the uniqueness of µ x,y. (ii) If g C(K, C) +, then µ F (g)x,x 0. In particular (by taking g 1), µ x,x 0. Proof. Fix x X. If f C(K, C) +, then fg C(K, C) + so that F (fg) Γ(X) +. By [131], there is S Γ(X) + with S 2 = F (fg). Hence 0 Sx 2 = S 2 x, x = F (fg)x, x = F (f)f (g)x, x = K fdµ F (g)x,x, implying µ F (g)x,x 0 by the uniqueness part in the real version of Riesz representation theorem about positive functionals. (iii) For f B (K, C), ψ f : X 2 C given by ψ f (x, y) = K fdµ x,y is sesquilinear, and ψ f f. Proof. For f C(K, C), observe K fdµ x 1 +x 2,y = F (f)(x 1 + x 2 ), y = F (f)x 1, y + F (f)x 2, y = K fdµ x 1,y + K fdµ x 2,y = K fd(µ x 1,y + µ x2,y) so that µ x1 +x 2,y = µ x1,y + µ x2,y by the uniqueness in Riesz theorem. By similar arguments, we see x µ x,y is linear and y µ x,y is conjugate-linear since the expression F (f)x, y is linear in x and conjugate linear in y. Therefore, (x, y) µ x,y [A] = K χ Adµ x,y is sesquilinear for any A B(K). Since simple functions are dense in B (K, C), by linearity and approximation ψ f is sesquilinear. And ψ f (x, y) f µ x,y f x y. (iv) There is a norm-decreasing -homomorphism F : B (K, C) Γ(X) such that F (f)x, y = K fdµ x,y for every f B (K, C) and every x, y X. Then F is an extension of the continuous functional calculus F by Exercise-21 so that we write f[t ] for F (f) henceforth. Proof. For f B (K, C), by Exercise-21(iv), there is unique F (f) Γ(X) such that ψ f (x, y) = F (f)x, y for every x, y X and F (f) = ψ f f. Linearity of F : B (K, C) Γ(X) is easy to verify. Since µ x,x 0, for any f B (K, R) we have F (f)x, x = F (f)x, x = K fdµ x,x = K fdµ x,x = F (f)x, x = x, F (f) x = F (f) x, x. By linearity, F (f)x, x = F (f) x, x for every f = f 1 + if 2 B (K, C) and every x X. Thus F (f) = F (f) by Exercise-21(iii). It remains to show F (fg) = F (f) F (g). We will establish this for every f B (K, C) and g C(K, C), for then the same technique can be repeated. Writing g = g 1 + ig 2 = (g + 1 g 1 ) + i(g + 2 g 2 ), it suffices to consider g C(K, C)+ by the linearity of F. In view of Exercise-21(iii), it suffices to show F (fg)x, x = F (f) F (g)x, x, or equivalently K fgdµ x,x = K fdµ F (g)x, x for every f B (K, C) and x X. By the linearity of F and the density of simple functions in B (K, C), it suffices to prove K χ Agdµ x,x = K χ Adµ F (g)x, x for every A B(K). Fix x X and write µ 1 = µ x,x, µ 2 = µ F (g)x,x. Note that µ 1, µ 2 0 by (ii). What we know is the following: since the continuous functional calculus F is multiplicative, F (fg)x, x = F (f)f (g)x, x and consequently, K fgdµ 1 = K fdµ 2 for any f C(K, C). If A K is compact, by Urysohn s lemma we may pick continuous functions f n : K [0, 1] such that (f n ) χ A pointwise; and hence (f n g) χ A g pointwise. By Lebesgue dominated convergence theorem, K χ Agdµ 1 = lim n K f ngdµ 1 = lim n K f ndµ 2 = K χ Adµ 2. If A B(K) is

11 OPERATOR THEORY - PART 2/3 11 arbitrary, choose an increasing sequence (A n ) of compact subsets of A with lim n µ j [A n ] = µ j [A] for j = 1, 2 by the regularity of the two Borel measures. This for j = 2 means lim n K χ A n dµ 2 = K χ Adµ 2. Letting A = n=1 A n and applying Lebesgue dominated convergence theorem, we also have K χ Agdµ 1 = K χ A gdµ 1 = lim n K χ A n gdµ 1 since (χ An g) χ A g pointwise (and is dominated by g). But K χ A n gdµ 1 = K χ A n dµ 2 by what is proved above, since A n is compact. We conclude that K χ Agdµ 1 = K χ Adµ 2 for every A B(K), completing the proof. (v) For A B(K), let P A = χ A [T ] Γ(X). Then P A is a self-adjoint idempotent (i.e., a projection). Moreover, for A, B B(K), we have the following: A B P A P B = P B P A = P A P A (X) P B (X); and A B = P A P B = P B P A = 0 P A (X) P B (X). Proof. To see P A is a self-adjoint idempotent, observe χ A = χ A = χ A χ A and use the fact T f[t ] is a -homomorphism. The remaining two assertions are also proved similarly, where the implication appearing in the beginning cannot be reversed since T f[t ] may not be injective. (vi) If U K is nonempty and relatively open in K, then χ U [T ] 0. Proof. We may choose f C(K, C) + with 0 < f χ U as follows: consider a nonempty compact set A U, and put f(z) = dist(z, K \U)/[dist(z, A)+dist(z, K \U)]. As the continuous functional calculus is an isometric -isomorphism to its image, f[t ] Γ(X) + \ {0}. By [131], there is S Γ(X) + \ {0} with S 2 = f[t ]. Choosing x X with Sx 0, we get 0 < Sx 2 = Sx, Sx = f[t ]x, x = K fdµ x,x χ U dµ x,x = χ U [T ]x, x since f χ U and µ x,x 0. Thus χ U [T ] 0. (vii) If S Γ(X) commutes with both T and T, then Sf[T ] = f[t ]S for every f B (K, C). In other words, f[t ] {T, T } cc for every f B (K, C). Proof. S commutes with T and T Sf[T ] = f[t ]S for every f C(K, C) by [130](iv) K fdµ x,s y = f[t ]x, S y = Sf[T ]x, y = f[t ]Sx, y = K fdµ Sx,y for every f C(K, C) µ x,s y = µ Sx,y by the uniqueness in Riesz theorem Sf[T ]x, y = f[t ]x, S y = K fdµ x,s y = K fdµ Sx,y = f[t ]Sx, y for every f B (K, C) Sf[T ] = f[t ]S for every f B (K, C) by Exercise-21. (viii) σ(f[t ]) f(σ(t )) for every f B (K, C). Proof. Recall that K denotes σ(t ). If g B (K, C) is invertible and if h B (K, C) is such that gh = hg = 1, then g[t ]h[t ] = h[t ]g[t ] = T and thus g[t ] Inv(G(X)). This observation implies σ(f[t ]) σ(f). Imitating the argument given for members of C(K, C), it is easy to check f(k) σ(f) f(k) and hence σ(f) = f(k) as σ(f) must be compact. (ix) If f B (K, C) and g C(f(K), C), then (g f)[t ] = g[f[t ]].

12 12 T.K.SUBRAHMONIAN MOOTHATHU Proof. Observe that σ(f[t ]) f(k) and so g[f[t ]] is defined. If g(z) = n i,j=1 α i,jz i z j, then (g f)(z) = n i,j=1 α i,jf(z) i f(z) j and hence (g f)[t ] = n i,j=1 α i,jf[t ] i (f[t ] ) j = g[f[t ]]. The general assertion follows since polynomials in z and z are dense in C(f(K), C) and since the Borel functional calculus is continuous. (x) Let f, f n B (K, C). Suppose sup n N f n < and (f n ) f pointwise. Then (f n [T ]x) f[t ]x for every x X (note: the conclusion is weaker than saying (f n [T ]) f[t ] in Γ(X)). Proof. Fix x X. By considering real and imaginary parts separately, assume f, f n are realvalued. Replacing f n with f f n, also assume f = 0. If c = sup n N f n, then f n 2 c 2 for every n N, and therefore K f ndµ 2 x,x 0 by Lebesgue dominated convergence theorem. Since f n [T ] is self-adjoint, we get f n [T ]x = f n [T ]x, f n [T ]x = fn[t 2 ]x, x = K f ndµ 2 x,x 0. Remark: (i) Since ff = ff, we have f[t ]f[t ] = f[t ] f[t ] (i.e., f[t ] is normal) for every f B (σ(t ), C). (ii) The continuous/borel functional calculus cannot be extended beyond the class of normal operators as a -homomorphism since the commutativity zz = zz implies T T = T T. Remark: Some differences between the continuous functional calculus F and the Borel functional calculus F : (i) F is an isometry and hence injective; but F is only norm-decreasing and may not be injective (this is why [133](vi) is non-trivial). The reason for non-injectivity can be vaguely stated as follows: if f, g B (K, C) are distinct members which agree µ x,y -almost everywhere, then F (f)x, y = F (g)x, y. (ii) The image of F is contained in the closed C -subalgebra generated by {I, T }; in the case of F, we can only say the image is contained in {T, T } cc. (iii) Spectral mapping theorem holds for F, but we only have an inclusion statement [133](viii) for F. Remark: If T Γ(X) is normal, Fuglede s theorem says {T, T } c = {T } c ; hence {T, T } cc = {T } cc. Exercise-22: Let X be a Hilbert space. Then, (i) If T Γ(X) is unitary, then there is self-adjoint S Γ(X) with exp[is] = T. (ii) If T Γ(X) is normal, then for every n N, there is normal S Γ(X) with S n = T. (iii) Inv(Γ(X)) is path connected. [Hint: (i) Note σ(t ) S 1. Let f : S 1 [0, 2π) be z = e iw w and g : [0, 2π] S 1 be g(w) = e iw. Then f B (S 1, C), g is continuous, and (g f)(z) = z. Apply [133](ix) and take S = f[t ]. (ii) re iθ r 1/n e iθ/n belongs to B (σ(t ), C). (iii) Write T Inv(Γ(X)) as T = T 1 T 2 with T 1 Γ(X) + and T 2 unitary, by [132]. By [117] and (i), there are S 1, S 2 Γ(X) so that exp[s 1 ] = T 1 and exp[is 2 ] = T 2. The path w exp[ws 1 ]exp[iws 2 ] joins I and T in Inv(Γ(X)).] Remark: When X is a Hilbert space, a deep theorem of Stone says the following: if w T w is a continuous group homomorphism from (R, +) to the multiplicative group of unitary elements in Γ(X), then there is self-adjoint S Γ(X) such that T w = exp[iws] for every w R.

13 OPERATOR THEORY - PART 2/3 13 Exercise-23: [An explanation for the name point spectrum] Let X be a Hilbert space and T Γ(X) be normal. If α σ(t ) is an isolated point of σ(t ), then α σ p (T ) (i.e., α is an eigenvalue of T ). [Hint: Since U := {α} is open in σ(t ), there is x X \ {0} with χ U [T ]x = x by [133](vii). Since (α z)χ U (z) 0, we have 0 = (αi T )χ U [T ]x = (αi T )x.] 4. Spectral theorem for normal operators The most important result about a normal operator is spectral theorem. There are a few interrelated results which go by the name spectral theorem (sometimes, all these results together is called spectral theorem). We know from Linear Algebra that if T is a normal matrix with distinct eigenvalues α 1,..., α k, and if P 1,..., P k are the projections to the corresponding (mutually orthogonal) eigenspaces, then T = k j=1 α jp j. In [133], we came very close to the following version of the spectral theorem which generalizes the finite dimensional case mentioned above: if T Γ(X) is normal, then there is a unique spectral measure/projection-valued measure/resolution of identity, say P, for T such that T = σ(t ) z dp. This version follows quickly from [133] once we clarify what P is. We leave it as a reading assignment. Below, we will derive another version of spectral theorem that says normal operators are unitarily equivalent to multiplication operators. Example: Recall the multiplication operator: let (Y, µ) be a σ-finite measure space, let g L (Y, µ), and M g : L 2 (Y, µ) L 2 (Y, µ) be M g h = gh. We observe that (M g ) = M g and it follows M g is normal. Also note M g = g. Borel functional calculus takes a simple form for the multiplication operator; f[m g ] = M f g for every f B (σ(m g ), C) ( we know this for f C(σ(M g ), C); next consider f = χ A and imitate some arguments from the proof of [133](iv)). The main concept used in the proof of spectral theorem is that of a cyclic vector. Definition: Let X be a Hilbert space and T Γ(X) be normal. If x X, temporarily we denote by X[x] the smallest closed vector subspace containing x and invariant under both T and T. Note that X[x] = {g[t ]x : g is a polynomial in z and z}. We say x X is a cyclic vector for {T, T } for a closed vector subspace Z X if Z = X[x]. If this happens, observe that Z is invariant under both T and T, and therefore Z is invariant under both T and T. Exercise-24: Let X {0} be a separable Hilbert space and T Γ(X). Then we can write X as a finite or countable Hilbertian sum X = { n x n : x n X n and n x n 2 < }, where (i) each X n is a non-zero, closed vector subspace of X such that both X n and X n (ii) X n X m for n m, (iii) for each n, there is a cyclic vector for {T, T } for X n. are T -invariant, [Hint: Let F = {A X \ {0} : X[a] X[b] for every a b in A}, ordered by inclusion. Union of members of any chain is an upper bound for the chain, so by Zorn s lemma, F has a maximal

14 14 T.K.SUBRAHMONIAN MOOTHATHU element, say A. If the Hilbertian sum Z of X[a] s (a A) is X, pick b Z \ {0} and consider A {b} F to contradict the maximality of A. Finally, A is countable since X is separable.] From Linear Algebra we know that if T Γ(C k ) is normal, then there is unitary U Γ(C k ) such that U 1 T U is diagonal. The spectral theorem below can be thought of as a generalization of this. [134] [Spectral theorem] Let X be a separable Hilbert space and T Γ(X) be normal. Then there exist a σ-finite measure space (Y, µ), an element g L (Y, µ), and a unitary operator U : L 2 (Y, µ) X such that T = UM g U 1, where M g : L 2 (Y, µ) L 2 (Y, µ) is M g f = gf. Proof. First suppose there is a cyclic vector x X for {T, T } in X. let µ = µ x,x Let K = σ(t ) and 0 be the finite Borel measure on K obtained in [133] by Riesz representation theorem. That is, f[t ]x, x = K fdµ for every f C(K, C). Now, C(K, C) is a dense subspace of L 2 (K, µ) with respect to the L 2 -norm ( using the regularity of the measure and Urysohn s lemma, see that a characteristic function can be approximated by continuous functions). Define U : C(K, C) X as U(f) = f[t ]x. Then U is linear and the computation U(f) 2 = f[t ]x, f[t ]x = (f[t ]) f[t ]x, x = ff[t ]x, x = K f 2 dµ = f 2 2 shows U is an isometry. Since x is a cyclic vector for X, the image of U is dense in X. Therefore, U extends to a unique surjective linear isometry U : L 2 (K, µ) X. And it is known from Functional Analysis that any surjective linear isometry is unitary. Let g L (K, µ) be g(z) = z. For f C(K, C), we have UM g f = U(gf) = (gf)[t ]x = g[t ]f[t ]x = T f[t ]x = T U(f), and hence this holds also for every f L 2 (K, µ) since C(K, C) is dense in L 2 (K, µ). In the general case, write X = { n x n : x n X n and n x n 2 < } according to Exercise-24. Let T n = T Xn and K n = σ(t n ). By the first case, there exist a finite Borel measure µ n 0 on K n, a unitary operator U n : L 2 (K n, µ n ) X n and g n L (K n, µ n ) such that T n = U n M gn U 1 n. Replacing K n with K n {n}, assume K n s are pairwise disjoint. Let Y = n K n and µ be the Borel measure defined on Y by the condition µ[a] = n µ n[a K n ] for A B(Y ). Then, we have the Hilbertian sum L 2 (Y, µ) = { n f n : f n L 2 (K n, µ n ) and n f n 2 < }. Define g : Y C as g Kn = g n. Since g n = M gn = T n T, we have g sup n g n T and thus g L (Y, µ). Let U : L 2 (Y, µ) X be U( n f n) = n U n(f n ). It may be seen that U is a surjective linear isometry (hence unitary) and UM g = T U. Remark: Examining the above proof, we see: (i) Y is a σ-compact metric space. (ii) g n s are continuous, and the image of g n is contained in K n B(0, T ) so that g is continuous and bounded by T. (iii) In the first part of the proof, if φ x,x is the functional f f[t ]x, x, then µ = µ x,x = φ x,x = x 2. Therefore, replacing x with αx for some α 0, we can have any positive value for µ. So in the second part of the proof, we can achieve n µ n = 1. Hence:

15 OPERATOR THEORY - PART 2/3 15 [134 ] [Spectral theorem - slightly improved version] Let X be a separable Hilbert space and T Γ(X) be normal. Then there exist a σ-compact metric space Y, a probability Borel measure µ on Y, a bounded continuous function g : Y C, and a unitary operator U : L 2 (Y, µ) X such that T = UM g U 1, where M g : L 2 (Y, µ) L 2 (Y, µ) is M g f = gf. Remark: For j = 1, 2, let X j be Hilbert spaces, T j Γ(X j ) be normal, and U : X 1 X 2 be unitary with UT 1 = T 2 U. Since U 1 = U, we also have UT1 = U(U T2 U) = T 2U. It follows that Ug[T 1 ] = g[t 2 ]U whenever g is a polynomial in z and z. Let K denote the compact set σ(t 1 ) = σ(t 2 ). Since polynomials in z and z are dense in C(K, C), we get Uf[T 1 ] = f[t 2 ]U for every f C(K, C). Now as in the proof of [133](vii), we may see that µ x,u y = µ Ux,y for every x X 1 and y X 2, and then Uf[T 1 ] = f[t 2 ]U for every f B (K, C). 5. What is the dual notion of surjectivity? Part of this section is a preparation to discuss approximate eigenvalues in the next section. Definition: Let X be a Banach space and T Γ(X). The dual (transpose) operator T Γ(X ) is given by T φ(x) = φ(t x) for φ X and x X. Remark: Properties of T may be related to the dual properties of T and vice versa. For certain properties, this relation will be unidirectional only, since X may not be reflexive. Remark: We have T T T T and hence equality throughout, where the second inequality is by the isometric embedding x E x of X in X and the observation T E x = E T x. Consequently, the map T T is a linear isometry from Γ(X) to Γ(X ); if X is reflexive, this map is also surjective (and hence an isomorphism). Now suppose X is a Hilbert space and consider the Hilbert adjoint operator T. The map T T is not linear, but it is a conjugate-linear isometric surjection from Γ(X) to itself. Moreover, T is related to T. Riesz representation theorem gives a map J : X X defined by J(φ) = y iff φ(x) = x, y for every x X. This J is a conjugate-linear isometric surjection, and T = JT J 1. Hence, results comparing T and T can often be imitated (with some extra work) to obtain analogues or weaker results comparing T and T. Definition: Let X be a Banach space. We say T Γ(X) is bounded below if there is c > 0 such that x c T x for every x X (equivalently, for every unit vector x X). The following implications are easy: T is invertible T is bounded below T is injective. Remark: In Linear Algebra, the dual notion of surjectivity is injectivity: if V 1, V 2 are finite dimensional vector spaces, then a linear map T : V 1 V 2 is surjective iff T : V2 V 1 is injective (and vice versa). We will see that in the Banach space setting (where the dimension is in general infinite), the dual notion of surjectivity is being bounded below (stronger than injectivity), and the dual notion of injectivity is having dense range (weaker than surjectivity) for bounded linear operators.

16 16 T.K.SUBRAHMONIAN MOOTHATHU Exercise-25: Let X be a Banach space and T Γ(X). (i) T has dense range T is injective. (ii) T has dense range T is injective. The converse holds if X is reflexive. (iii) Let X is a Hilbert space. Then ker(t ) = T (X) and ker(t ) = T (X). Hence, T has dense range T is injective, and vice versa. [Hint: Do (iii) first to grasp the idea: y ker(t ) T y = 0 T x, y = x, T y = 0 for every x X y T (X). For (i), use the following corollary of Hahn-Banach: for a vector subspace Y X, we have Y X iff there is φ X \ {0} such that φ Y 0. Similar idea for (ii) also.] Exercise-26: If X is a Banach space and T Γ(X), then the following are equivalent: (i) T is not bounded below. (ii) There are x n X such that x n = 1 for every n N and lim n T x n = 0. (iii) There are S n Γ(X) such that S n = 1 for every n N and lim n T S n = 0. [Hint: (i) (ii): There are z n X with z n > n T z n and take x n = z n / z n. (ii) (iii): Choose φ X with φ = 1 and put S n x = φ(x)x n. (iii) (i): Consider unit vectors z n X with S n z n 1/2 and look at x n = S n z n.] Remark: Let X be a Banach space. For T Γ(X), the property of having a closed range is important, for then X/ker(T ) = T (X) as Banach spaces via x+ker(t ) T x. Therefore, conditions necessary and/or sufficient for T to have a closed range are also important. Exercise-27: If X is a Banach space and T Γ(X). Then, (i) T has closed range there is c > 0 such that y T (X), x T 1 (y) with x c y. (ii) T is bounded below T is injective and has closed range. [Hint: (i) : Let S be the inverse of the isomorphism T : X/ker(T ) T (X) and take c > S. : If y T (X), choose y n T (X) such that y n+1 y n < 1/2 n and (y n ) y. Let y 0 = 0 = x 0. Having chosen x n, pick z n+1 T 1 (y n+1 y n ) with z n+1 c y n+1 y n and put x n+1 = z n+1 +x n. Then x n+1 x n < c/2 n, implying (x n ) is Cauchy. If (x n ) x, then T x = y. (ii) Use simpler versions of the above argument. For instance, to show T (X) is closed in, note that if (T x n ) y, then (x n ) must be Cauchy since x n x m c T x n T x m ; and for, note that T : X T (X) is an isomorphism of Banach spaces.] Hahn-Banach theorem is useful to show suitable disjoint objects can be separated by functionals. In [135] below, we present one simple result of this type which is needed in the proof of [136]. Definition: A nonempty subset Y of a Banach space X is said to be balanced if αy Y for every α C with α 1. Note that if T Γ(X) and ɛ > 0, then T (B(0, ɛ)), T (B(0, ɛ)) are balanced. [135] Let X be a Banach space, and Y X be a closed, convex, and balanced set. If z X \ Y, then there is φ X such that φ(y) 1 for every y Y and φ(z) (1, ).

17 OPERATOR THEORY - PART 2/3 17 Proof. Let λ = dist(z, Y ) > 0 and let U = {x X : dist(x, Y ) < λ/2}. Then Y U, z / U, and U is a convex open balanced set. Note that for every x X, there is c > 0 such that x U. Define c p : X [0, ) as p(x) = inf{c > 0 : x U}. Clearly, p(x) 1 for every x U. We see p(z) > 1 c by observing the following: if 0 < c 1, then z is a convex combination of 0 and z c ; moreover, 0 U and z / U so that z / U since U is convex. We claim that c (i) p(αx) = α p(x) for every x X and α C, and (ii) p(x + y) p(x) + p(y) for every x, y X. Statement (i) is evident since U is balanced. For (ii), consider x, y X and ɛ > 0. Enough to show p(x + y) p(x) + p(y) + 2ɛ. Choose c 1, c 2 such that x, y U, 0 < c 1 < p(x) + ɛ, and c 1 c 2 0 < c 2 < p(y) + ɛ. Now by convexity, x + y c 1 = ( ) x c 2 + ( ) y U and hence p(x + y) c 1 + c 2 p(x) + p(y) + 2ɛ. c 1 + c 2 c 1 + c 2 c 1 c 1 + c 2 c 2 Define a linear functional φ : span{z} C as φ(αz) = αp(z), note φ p, and use Hahn-Banach theorem for complex vector spaces to get a linear extension φ : X C with φ(x) p(x) for every x X. Hahn-Banach theorem does not say that the extended linear functional φ is bounded, but we can argue it as follows in our special case. Choosing r > 0 with B(0, r) U, we see that rx U for every x X \ {0}, and therefore φ(x) p(x) 2 x /r, or φ 2/r. Finally, 2 x observe that φ(y) p(y) 1 for every y Y, and φ(z) = p(z) (1, ). [136] Let X be a Banach space and T Γ(X). Then, (i) T is surjective T is bounded below. (ii) T is bounded below T is surjective. (iii) T is invertible T is invertible. Applying this to αi T, we have σ(t ) = σ(t ). (iv) T has closed range T has closed range. Proof. (i) : The surjectivity of T implies T is injective. Now by Exercise-27, it suffices to show the range of T is closed. Consider ψ T (X ) and let (φ n ) be a sequence in X so that (T φ n ) ψ. Let c > 0 be a constant given by Exercise-27(i). Fix y X and choose x X such that x c y and T x = y. We have φ n (y) φ m (y) = φ n (T x) φ m (T x) = T φ n (x) T φ m (x) T φ n T φ m x c T φ n T φ m y and thus φ n φ m c T φ n T φ m implying that (φ n ) is Cauchy. If φ = lim n φ n (where the limit exists since X is complete), then T φ = ψ. (ii) : Let Y = T (X). By hypothesis, T : X Y is an isomorphism of Banach spaces and let S Γ(Y, X) be its inverse. Given ψ X, we have ψ S Y so that there is φ X with φ Y = ψ S by Hahn-Banach theorem. And T φ(x) = φ(t x) = ψ(st x) = ψ(x) so that T φ = ψ. The two implication proved so far, give the other two implications in (i) and (ii) also, provided X is reflexive, since we may identify T with T. If X is not reflexive, some extra work is needed.

18 18 T.K.SUBRAHMONIAN MOOTHATHU (i) : Let c > 0 be such that φ c T φ for every φ X. Let Y = T (B(0, 1)), and we claim B(0, 1/c) Y. If the claim is false, consider z B(0, 1/c) \ Y, and choose φ X such that φ(z) (1, ) and φ(y) 1 for every y Y by [135]. The last two assertions imply respectively that φ > c and T φ = φ T 1. This contradicts the initial statement φ c T φ. Hence our claim must be true. Therefore, B(0, r/c) T (B(0, r)) for every r > 0. Now imitate the standard proof of Open mapping theorem to show T is an open map and hence surjective. (ii) : Using (i), we obtain T is bounded below. Considering X as embedded in X, we have T = T X (to be precise, T E x = E T x ), and hence T is bounded below. (iii) If T is invertible, then by taking the dual of the identity T T 1 = I = T 1 T, we see T is invertible. Conversely assume T is invertible. Then T is bounded below and surjective. Hence T is surjective and bounded below (in particular injective) by (i), (ii), and hence invertible. (iv) For a proof, one has to argue using the concept of annihilators. assignment; see Theorem 4.14 in Rudin, Functional Analysis. We leave it as a reading Remark: Let X be a Banach space and T Γ(X). It follows from [136] and Exercise-26 that T fails to be surjective there are φ n X such that φ n = 1 for every n N and lim n (φ n T ) = 0. Simpler and more transparent arguments can be used in a Hilbert space to compare T and T. Exercise-28: Let X be a Hilbert space and T Γ(X). Then, (i) T is bounded below there is S Γ(X) such that ST = I. (ii) T is surjective there is S Γ(X) such that T S = I. (iii) T is bounded below T is surjective (and vice versa). (iv) T is invertible both T and T are bounded below both T and T are surjective. (v) T has closed range T has closed range. [Hint: (i) : Define S(y 1 + y 2 ) = T 1 y 1 for y 1 + y 2 T (X) T (X) = X. (ii) : Let Z = (ker(t )) and T 1 = T Z. Define Sy = T 1 1 y. Use (i) and (ii) to deduce (iii), and then (iv). To prove in (v), write X = Y Y = Z Z, where Y = ker(t ) and Z = T (X). We know T (X) Y by Exercise-25(iii), and it suffices to show equality. If T 1 = T Y, then T 1 : Y Z is an isomorphism. Let S = T 1 1 : Z Y so that T 1 S = I Y. If x = x 1 + x 2 Y Y and y Y, then verify that x, T S y = T x 2, S y = x, T 1 S y = x, y and thus T (S y) = y.] Exercise-29: Let X be a Banach space. Then, Γ b := {T Γ(X) : T is bounded below} and Γ s := {T Γ(X) : T is surjective} are open in Γ(X). [Hint: Let T Γ b, and c > 0 be such that x c T x. Check that if T S < 1 2c, then x 2c Sx and hence S Γ b. To see Γ s is open, consider the dual, note that T S = T S, and use [136] and the openness of Γ b.] *****