OPERATOR THEORY - PART 3/3. Contents

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1 OPERATOR THEORY - PART 3/3 T.K.SUBRAHMONIAN MOOTHATHU Contents 1. Approximate eigenvalues, and another polar decomposition 1 2. The spectrum of a compact operator 5 3. Index theory and compact operators 8 4. Hilbert-Schmidt and trace class operators 1 5. Integral operators are Hilbert-Schmidt operators Invariant subspace problem and the spectrum Invariant subspace problem and compact operators Approximate eigenvalues, and another polar decomposition Definition: Let X be a Banach space and T Γ(X). The approximate point spectrum σ a (T ) of T is σ a (T ) = {α C : αi T is not bounded below}. Note that α σ a (T ) iff there is a sequence (x n ) of unit vectors in X with lim n (αi T )x n =. Any α σ a (T ) is called an approximate eigenvalue of T. Clearly, σ p (T ) σ a (T ) σ(t ), and all the three coincide when dim(x) <. [137] Let X be a Banach space and T Γ(X). Then σ a (T ) is closed and σ(t ) σ a (T ). In particular, σ a (T ) ( σ(t ) always as σ(t ) is a nonempty compact subset of C). Proof. Closedness of σ a (T ) is a consequence of Exercise-29. To prove σ(t ) σ a (T ), consider α σ(t ) and let ɛ >. It suffices to find unit vector x X with (αi T )x < 2ɛ. Let β ρ(t ) B(α, ɛ). Writing R(β) = (βi T ) 1, by [11] we have 1/ R(β) dist(β, σ(t )) β α < ɛ. Choose unit vector y X with R(β)y > 1/ɛ and put x = R(β)y/ R(β)y. Then x = 1, and (αi T )x (α β)x + (βi T )x α β x + y / R(β)y < ɛ+ɛ = 2ɛ. Remark: From [137], it is immediate that if T is a self-adjoint operator of a Hilbert space, then σ(t ) = σ(t ) σ a (T ) since σ(t ) R. We can improve this: [138] Let X be a Hilbert space and T Γ(X) be normal. Then σ(t ) = σ a (T ). First proof. Let α σ(t ) and ɛ >. Enough to find unit vector x X with (αi T )x ɛ. If U = B(α, ɛ) σ(t ), then the projection χ U [T ] by [133](vi), and so there is a unit vector x X with χ U [T ]x = x. Consider f B (σ(t ), C) defined as f(z) = (α z)χ U (z) and note 1

2 2 T.K.SUBRAHMONIAN MOOTHATHU f(z) < ɛ. Since the Borel functional calculus is a norm decreasing homomorphism, we obtain ɛ f f[t ] f[t ]x = (αi T )χ U [T ]x = (αi T )x. Second proof (more elementary): Suppose α / σ a (T ). Then S := αi T is bounded below. Since S is normal, Sx 2 = S Sx, x = SS x, x = S x 2, or Sx = S x for every x X, and therefore S is also bounded below. Now Exercise-28(iv) implies S is invertible. Thus α / σ(t ). Part (iii) of the following shows our definition of a positive element in a C -algebra agrees with the usual definition of a positive operator. [139] Let X be a Hilbert space and T Γ(X). Then, (i) If α σ(t ) \ σ a (T ), then α σ p (T ). (ii) If α σ(t ), then there are unit vectors x n X such that lim n T x n, x n = α. (iii) T Γ(X) + (i.e., T = T and σ(t ) [, )) T x, x for every x X. Proof. (i) Since αi T is bounded below, αi T is surjective by Exercise-28, and hence αi T must fail to be injective since α σ(t ). (ii) If α σ a (T ), choose unit vectors x n X with lim n (αi T )x n =, and observe that α T x n, x n = α x n, x n T x n, x n = (αi T )x n, x n (αi T )x n x n. If α σ(t ) \ σ a (T ), take unit vector x ker(αi T ) by part (i) and put x n = x for every n N. (iii) : Consider S Γ(X) + with S 2 = T. : σ(t ) [, ) by (ii) and hence T = T also. Remark: Let X be a Hilbert space and T Γ(X). The numerical range of T is W (T ) := { T x, x : x = 1}. Result [139](i) says that σ(t ) W (T ). In fact, W (T ) is convex and W (T ) contains the convex hull of σ(t ), with equality when T is normal; see Chapter 22 in Halmos, A Hilbert Space Problem Book. Knowing the numerical range helps to locate the spectrum. [14] Let X be a Hilbert space and T Γ(X). (i) If T is an isometry, then σ a (T ) S 1. Moreover, either σ(t ) S 1 or σ(t ) = B(, 1). (ii) Let T be self-adjoint and define α = inf{ T x, x : x X and x = 1}, β = sup{ T x, x : x X and x = 1}. Then, {α, β} σ(t ) [α, β] and T = max{ α, β }. Proof. (i) Since T = 1, we have σ(t ) B(, 1). If α B(, 1), then we see αi T is bounded below as follows: x = T x (αi T )x + α x, or x (1 α ) 1 (αi T )x. The second assertion follows by the geometry of a subset of B(, 1) since σ(t ) σ a (T ). (ii) We have σ(t ) [α, β] by [139](ii). If T is replaced by T + δi for some δ >, then α, β become α + δ, β + δ, and moreover σ(t + δi) = σ(t ) + δ. Therefore, replacing T with T + δi for some δ >, we may assume T Γ(X) +. Consider S Γ(X) + with S 2 = T. Since S, T are self-adjoint, we get β = sup{ Sx, Sx : x X and x = 1} = S 2 = S 2 = T = r(t ) by Exercise-18(iv), and thus β σ(t ) since σ(t ) [, ). Similarly, α σ( T ) so that α σ(t ).

3 OPERATOR THEORY - PART 3/3 3 For Hilbert space operators, below we extend [132] by including non-invertible operators also. Definition: Let X be a Hilbert space, T Γ(X), and let Y = ker(t ). We say T is a partial isometry if T Y : Y X is an isometry (if Y = {} here, then T becomes an isometry). We call Y the initial space of T. Note that T 1 and T (X) = T (Y ) is closed if T is a partial isometry. An example of a partial isometry which is not an isometry is the left shift T : l 2 l 2 given by (x, x 1, x 2,...) (x 1, x 2,...) whose initial space is span{e }. Exercise-3: Let X be a Hilbert space, T Γ(X), and let Y = ker(t ), Z = T (X). Then the following are equivalent: (i) T is a partial isometry with initial space Y, and T (X) = Z. (ii) T T is a projection onto Y. (iii) T T is a projection onto Z. (iv) T is a partial isometry with initial space Z, and T (X) = Y. [Hint: (i) (ii): Clearly Y ker(t T ). Since T T Γ(X) + and T T = T 2 1, we deduce I T T Γ(X) + by spectral mapping theorem. Let S = (I T T ) 1/2 Γ(X) +. For x Y, Sx 2 = (I T T )x, x = x, x T x 2 = so that Sx =, or (I T T )x = S 2 x =. (ii) (iii): Z = ker(t ) ker(t T ). If x Y, then T T (T x) = T (T T x) = T x. Also T (Y ) = T (X) is dense in Z. Next, (ii) (i) is easy, and the other implications follow by replacing T with T.] Definition: If X is a Hilbert space and T Γ(X), then the absolute value of T is the operator defined as T := (T T ) 1/2 Γ(X) +. We observe that T = (T T ) 1/2 = (T T ) 1/2 Γ(X) +. Remark: Let X be a Hilbert space and T Γ(X). Note T x 2 = T x, T x = T 2 x, x = T T x, x = T x 2, and hence ker( T ) = ker(t ). Using T = T again, we also see T (X) = ker( T ) = ker( T ) = ker(t ) = T (X), and therefore T (X) = ker(t ) = T (X). [141] [Polar decomposition] Let X be a Hilbert space, T Γ(X), Y = ker(t ), and Z = T (X). Then, there exist unique partial isometries T 1, T 2 Γ(X) with initial space Y and range Z such that T = T 1 T = T T 2. Moreover, T1 T = T and T T 2 = T. Proof. As noted above, T x 2 = T x 2. Hence if we define T 1 on T (X) as T 1 ( T x) = T x, then T 1 is a well-defined linear isometry. Then, T 1 extends uniquely to an isometry on T (X) = Y. Also put T 1 on Y so that T 1 Γ(X) becomes a partial isometry. It is clear that T 1 (X) = Z. Similarly, we can write T = T 3 T, where T 3 Γ(X) is a partial isometry with initial space Z and range Y. Take T 2 = T 3 which is a partial isometry with initial space Y and range Z by Exercise-3, and note T = T = T T3 = T T 2. To prove the uniqueness part, consider partial isometries S 1, S 2 playing the roles of T 1, T 2 respectively. From S 1 T x = T x = T 1 T x, we deduce S 1 = T 1 on their common initial space Y = T (X). Hence S 1 = T 1 ; and similarly, S 2 = T 2. By Exercise-3, T1 T 1 is a projection onto Y = T (X), and therefore T1 T = T 1 T 1 T = T. Similarly, T T 2 = T T 2 T 2 = T because T 2 T 2 is a projection onto Z = T (X).

4 4 T.K.SUBRAHMONIAN MOOTHATHU Example: Let (w n ) be a bounded sequence in C \ {} and let T : l 2 l 2 be the weighted left shift (x, x 1, x 2,...) (w 1 x 1, w 2 x 2,...). Then T : l 2 l 2 is the weighted right shift (x, x 1,...) (, w 1 x, w 2 x 1,...); to confirm this, verify the defining property T x, y = x, T y. Following [141], it may be seen that T has the polar decompositions T = T 1 T = T T 2, where T = (T T ) 1/2 : l 2 l 2 is (x, x 1, x 2,...) (, w 1 x 1, w 2 x 2,...), T = (T T ) 1/2 : l 2 l 2 is (x, x 1, x 2,...) ( w 1 x, w 2 x 1,...), and T 1 = T 2 : l 2 l 2 is the partial isometry (x, x 1, x 2,...) ( w 1 1 w 1 x 1, w 2 1 w 2 x 2,...). We conclude this section by determining the spectrum and its parts for the shift. [142] Consider the Hilbert space X = l 2 (Z). Let T Γ(X) be the left shift, [T x] n = x n+1. Then, (i) T is unitary and T is the right shift given by [T x] n = x n 1. (ii) If β S 1, there is unitary U Γ(X) such that βt = U 1 T U. Consequently, σ(βt ) = σ(t ). (iii) σ p (T ) = = σ p (T ) and σ a (T ) = σ(t ) = σ a (T ) = σ(t ) = S 1. Proof. (ii) Let U Γ(X) be Ux = (β n x n ). Then U 1 x = (β n x n ) = (β n x n ) = U x and thus U is unitary. Also, [U 1 T Ux] n = β n [T Ux] n = β n [Ux] n+1 = β n β n+1 x n+1 = βx n+1 = [βt x] n. (iii) Verify directly that σ p (T ) = = σ p (T ). Now, σ(t ) S 1 since T is unitary. Also we know σ(t ). Let α σ(t ). If β S 1, then β σ(βα 1 T ) by spectral mapping theorem; and σ(βα 1 T ) = σ(t ) by (ii) since βα 1 S 1. Thus σ(t ) = S 1. Since σ(t ) σ a (T ), we conclude σ a (T ) = S 1. Finally, σ(t ) = {α : α σ(t )} = S 1, and then σ a (T ) = S 1 as argued above. [143] Let X = l 2 and T Γ(X) be the left shift (x, x 1, x 2,...) (x 1, x 2,...). Then, (i) T is the right shift (x, x 1,...) (, x, x 1,...) which is an isometry. (ii) T is not normal (i.e., T T T T ). In fact, σ(t T ) = σ(i) = {1} = σ(t T ) since σ(t T ). (iii) σ p (T ) = B(, 1), σ p (T ) =, σ a (T ) = σ(t ) = σ(t ) = B(, 1), and σ a (T ) = S 1. Proof. (iii) Note σ(t ) B(, 1) since T = 1. We have σ p (T ) since T e =. If < α < 1, then x = (α n ) X \ {} and T x = αx so that α σ p (T ). But, if α = 1, then for x X, the equality T x = αx implies x n = α n x ; and now the fact n= x n 2 < forces x =. This proves σ p (T ) = B(, 1). From σ p (T ) σ a (T ) σ(t ) B(, 1), it now follows by the closedness of σ a (T ) that σ a (T ) = σ(t ) = B(, 1). Therefore, σ(t ) = {α : α σ(t )} = B(, 1). Easy to see σ p (T ) =. Since T is an isometry, σ a (T ) S 1 by [14](i); and S 1 = σ(t ) σ a (T ) also. Remark: For a non-normal operator, the approximate point spectrum may or may not coincide with the spectrum. The operators T and T in [143] illustrate both the possibilities. Remark: To know about the spectrum and numerical range of weighted shifts, see p of the book Topics in Operator Theory, edited by C.Pearcy. Remark: Let T : l 2 l 2 be the left shift (x, x 1, x 2,...) (x 1, x 2,...). We claim T has no square root. Let if possible, S Γ(l 2 ) be such that S 2 = T. Since {} = ker(s) ker(t ) and since

5 OPERATOR THEORY - PART 3/3 5 ker(t ) = span{e } is one-dimensional, we must have ker(s) = span{e } = ker(t ). If x l 2 \ {} is such that Sx = e, then x ker(s 2 ) = ker(t ) = ker(s) implying e =, a contradiction. 2. The spectrum of a compact operator Definition: Let X be a Banach space. We say T Γ(X) is a compact operator if T (A) is compact for every bounded subset A X (equivalently, if (T x n ) has a convergent subsequence for any bounded sequence (x n ) in X). Let Γ K (X) = {T Γ(X) : T is a compact operator}. We assume the reader is familiar with the following basic facts from Functional Analysis. [144] Let X be a Banach space. Then Γ K (X) is a closed, two-sided ideal in Γ(X). This means, (i) Γ K (X) is a closed vector subspace of Γ(X). (ii) ST, T S Γ K (X) for every S Γ(X) and T Γ K (X). Moreover, (iii) I Γ K (X) Γ K (X) = Γ(X) dim(x) <. (iv) If T Γ(X), then T Γ K (X) T Γ K (X ) (similarly, T Γ K (X) T Γ K (X) when X is a Hilbert space). (v) If T Γ K (X) and T (X) is closed, then dim(t (X)) <. (vi) If T Γ K (X), then T (X) is separable. (vii) If X is a Hilbert space and T Γ K (X), then there is a sequence (T n ) of finite rank operators (i.e., dim(t n (X)) < for each n) in Γ(X) such that T T n. Remark: Perhaps the only difficult statements above are (iv) and (v). To establish (v), by Open mapping theorem first note that there is ɛ > such that {y T (X) : y < ɛ} T (B(, 1)). Then the closed ball B(, ɛ) in T (X) is compact, and it follows that dim(t (X)) <. proof of (iv) requires the use of Arzela-Ascoli theorem (see Theorem 4.19 in Rudin, Functional Analysis). However, when X is a Hilbert space, it is relatively easy to establish (iv) via (vii) as follows. Since T (X) = n=1 T (B(, n)) and since each T (B(, n)) is separable by the compactness of its closure, we see T (X) is separable. If {e 1, e 2,...} is an orthonormal basis for T (X), and if P n : X span{e 1,..., e n } is the orthogonal projection, then T n := P n T is of finite rank and (T n ) T. Now, T n = T P n is compact since it is of finite rank, and (T n) T so that T is compact by (i). This argument cannot be imitated in Banach spaces since a compact operator may not be the limit of a sequence of finite rank operators in a general Banach space. Remark: Let X be a Hilbert space and T Γ(X). Then T Γ K (X) iff T = (T T ) 1/2 Γ K (X) by the polar decomposition [141], and [144](ii). Remark: If X is an infinite dimensional Banach space and T Γ K (X), then T is not surjective because of [144](v), and hence σ(t ). becomes compact by [144](ii), a contradiction to [144](iii). The Another argument is: if / σ(t ), then I = T T 1

6 6 T.K.SUBRAHMONIAN MOOTHATHU Definition: Let X be a Banach space and Y, Z X be vector subspaces (may not be closed). (i) The codimension of Y, denoted as codim(y ), is defined as the dimension of the quotient vector space X/Y. (ii) The expression X = Y Z means X = Y + Z and Y Z = {}. Exercise-31: Let X be a Banach space and Y X be a vector subspace. (i) If dim(y ) <, then there is a closed vector subspace Z X with X = Y Z. (ii) If codim(y ) <, then there is a finite dimensional vector subspace Z X with X = Y Z. [Hint: (i) Let {y 1,..., y n } be a basis of Y, and choose φ j Y such that φ j (y i ) = 1 or according to whether i = j or i j. Extend by Hahn-Banach to get φ j X and take Z = n j=1 ker( φ j ). Check that Y Z = {} and any x X is x = y+z where y = n j=1 φ j (x)y j Y and z = x y Z. (ii) Let {x 1 + Y,..., x n + Y } be a basis for X/Y and take Z = span{x 1,..., x n }.] Remark: If X is an infinite dimensional Banach space, then a vector subspace of finite codimension need not be closed in X. Let Y 1 X be a dense vector subspace, let B be a Hamel basis for Y 1, and C X \ Y 1 be so that B C is a Hamel basis for X. Fixing x C, we see Y := span(b C \ {x}) is a dense vector subspace of codimension one in X, but Y is not closed in X since Y X. General Principle: If X is a Banach space and T Γ K (X), then I T can be thought of as a small perturbation of I; hence we may anticipate I T to behave almost like I. Exercise-32: Let X be a Banach space and T Γ K (X). Then, (i) dim(ker(i T )) <. (ii) If 1 σ a (T ), then 1 σ p (T ). (iii) (I T )(X) is closed. [Hint: (i) If (x n ) is a sequence in the closed unit ball D of ker(i T ), then a subsequence (x nk ) = (T x nk ) converges by the compactness of T, implying D is compact. (ii) There is (x n ) in X with x n = 1 and lim n (I T )x n =. If (T x nk ) y by the compactness of T, then (x nk ) y. Now y = 1 and T y = y. (iii) We may write X = ker(i T ) Z by part(i) and Exercise-31. Check that (I T ) Z is bounded below as in (ii). Note (I T )(Z) = (I T )(X) and apply Exercise-24.] [145] Let X be an infinite dimensional Banach space and T Γ K (X). Then, either (i) σ(t ) is a finite set containing, or (ii) σ(t ) = {} {α n : n N}, where (α n ) is a sequence in C \ {} with lim n α n =. Moreover, σ(t ) \ {} σ p (T ) always, and dim(ker(αi T )) < for every α σ(t ) \ {}. Proof. Claim-1 : σ(t ) \ {} σ p (T ) σ p (T ), where T Γ(X ) is the dual operator. To prove the claim, we consider a non-zero α / σ p (T ) σ p (T ), and then show α / σ(t ). Replacing T with α 1 T, we may assume α = 1. That is, suppose 1 / σ p (T ) σ p (T ). Then I T and I T are injective. The range of I T is closed by Exercise-32, and dense by the injectivity of I T, by Exercise-25. Thus I T is also surjective and hence invertible in Γ(X). Therefore 1 / σ(t ).

7 OPERATOR THEORY - PART 3/3 7 Claim-2 : For any δ >, the set {α σ p (T ) : α δ} is finite. Otherwise choose infinitely many distinct members α n σ p (T ) such that α n δ for every n N. Let x n ker(α n I T ) be unit vectors and put X n = span{x 1,..., x n }. We may verify that X n 1 X n since {x 1,..., x n } is linearly independent; T (X n ) X n, and (α n I T )(X n ) X n 1 where we take X = {}. Since X n 1 X n, by Riesz lemma there is y n X n \X n 1 such that y n = 1 and y n x 1/2 for every x X n 1. Now (αn 1 y n ) is a bounded sequence since αn 1 y n 1/δ. On the other hand, if m < n, then T (αm 1 y m ) T (αn 1 y n ) = y n z 1/2 since z := (α n I T )αn 1 y n + T (αm 1 y m ) X n 1. This contradicts the compactness of T, and completes the proof of Claim-2. Since T is compact by [144](iv), note that Claim-2 applies to T also. Combining with Claim-1, we conclude that for any δ >, the set {α σ(t ) : α δ} is finite. Hence (i) or (ii) holds. Now consider α σ(t ) \ {}. By what is already proved, α is an isolated point of σ(t ). Therefore α σ(t ) σ a (T ) by [135]. Hence 1 σ a (α 1 T ) σ p (α 1 T ) by Exercise-32(ii), and thus α σ p (T ). Moreover, dim(ker(αi T )) = dim(ker(i α 1 T )) < by Exercise-32(i). Remark: If (α n ) is a sequence in C\{} converging to, and if we define a bounded linear operator T : l 2 l 2 by the condition T e n = α n e n, then T is compact and σ(t ) = {} {α n : n }. We know from Functional Analysis that the Volterra operator defined below on C([, 1], C) is compact, thanks to Arzela-Ascoli Theorem. We may consider Volterra operator on L 2 ([, 1], µ) also, where µ is the Lebesgue measure; this will be taken up a little later. [146] [Spectrum of Volterra operator]. Let T : C([, 1], C) C([, 1], C) be the Volterra operator given by (T f)(y) = y f(x)dx. Then, T n 1/n! for every n N, σ(t ) = {}, and σ p (T ) =. Proof. For f C([, 1], C) and x [, 1], observe that (T n f)(x ) = x (T n 1 f)(x 1 )dx 1 = x If f 1, then f(x n ) 1 so that we obtain the estimate x1 (T n 2 f)(x 2 )dx 2 dx 1 = x x n 1 f(x n )dx n dx 1. (T n f)(x ) x x n 1 1 dx n dx 1 = x x n 2 x n 1 dx n 1 dx 1 = x x n 3 (x 2 n 2 /2)dx n 2 dx 1 = x x n 4 (x 3 n 3 /3!)dx n 3 dx 1 = = x n /n! 1/n!. Thus T n 1/n!. It follows that r(t ) = lim n T n 1/n = and therefore σ(t ) = {}. Let f C([, 1], C) be such that T f =. If f(y) for some y [, 1], then by the continuity of f, we can find [a, b] [, 1] such that b a f(x)dx. But b a f(x)dx = (T f)(b) (T f)(a) = =, a contradiction. Hence f and therefore / σ p (T ). 3. Index theory and compact operators Definition: Let X be a Banach space. We say T Γ(X) is a Fredholm operator if dim(ker(t )) < and codim(t (X)) <. The index of T is defined as ind(t ) = dim(ker(t )) codim(t (X)) Z, when T is Fredholm. Similar definition applies to T Γ(X, Y ) when X, Y are Banach spaces.

8 8 T.K.SUBRAHMONIAN MOOTHATHU Remark: If X is a Banach space and T Γ(X) is Fredholm, then T (X) is closed. Proof : Using Exercise-31, write X = ker(t ) Y = T (X) Z, where Y, Z X are vector subspaces with Y closed and dim(z) <. Note that S : Y Z X defined as S(y, z) = T y + z is a bounded linear bijection, and hence an isomorphism of Banach spaces. Thus T (X) = S(Y {}) is closed. Example: Let T be the left shift (x, x 1,...) (x 1, x 2,...) and S be the right shift (x, x 1,...) (, x, x 1,...) on l 2. Then T n, S n are Fredholm, ind(t n ) = n, and ind(s n ) = n for every n. Remark: Intuitively, T Γ(X) is Fredholm if T is almost invertible in Γ(X); this is made precise in [149] later. According to Peter Lax [p.35, Functional Analysis], there is not much historical justification for calling operators of finite index as Fredholm, but we will stick to the common usage. [147] [Homomorphism property] Let X, Y, Z be Banach spaces and T Γ(X, Y ), S Γ(Y, Z) be Fredholm. Then ST Γ(X, Z) is Fredholm and ind(st ) = ind(s) + ind(t ). Proof. Case-1 : Suppose T is injective and S is surjective. By Exercise-31, write Y as direct sum of closed vector subspaces as Y = ker(s) W = F T (X), where dim(ker(s)) < and dim(f ) <. Now, ker(st ) = T 1 (T (X) ker(s)) so that by the injectivity of T we have dim(ker(st )) = dim(t (X) ker(s)) dim(ker(s)) <. Next, Z = S(Y ) = S(W ) = S(T (X) W ) S(F W ) = (ST )(X) S(F W ) and thus codim(st ) = dim(s(f W )) = dim(f W ) < since S W : W Z is a bijection. We have shown ST is Fredholm. Now compute ind(st ) = dim(t (X) ker(s)) dim(f W ) = dim(ker(s)) dim(f ) = ind(s) + ind(t ), where we have added and subtracted dim(f ker(s)) in the calculation process. Case-2 : In the general case, write X = ker(t ) X 1 and Z = S(Y ) Z 1 where X 1 is a closed vector subspace and Z 1 is finite dimensional. Let T 1 Γ(X 1, Y ) be T 1 = T X1 and S 1 Γ(Y, S(Y )) be S 1 y = Sy. Since T 1 is injective and S 1 is surjective, S 1 T 1 is Fredoholm with ind(s 1 T 1 ) = ind(s 1 )+ind(t 1 ) by Case-1. To complete the proof, observe that ind(t ) = ind(t 1 )+dim(ker(t )), ind(s) = ind(s 1 ) dim(z 1 ), and ind(st ) = ind(s 1 T 1 ) + dim(ker(t )) dim(z 1 ). Exercise-33: Let X be a Banach space, T Γ K (X), and let S = I T. (i) If X n = ker(s n ), then X 1 X 2, and there is k N such that X n = X k for every n k. (ii) If Y n = S n (X), then Y 1 Y 2, and there is k N such that Y n = Y k for every n k. [Hint: Note that X n, Y n are T -invariant, S(X n+1 ) X n and S(Y n ) Y n+1. Also, X n is clearly closed, and Y n is closed by Exercise-32 and induction. Now, for each of (i) and (ii), imitate the argument based on Riesz lemma given for the proof of Claim-2 in [145].] If X is a Hilbert space and S Γ(X) is with closed range, then we know X is the orthogonal direct sum X = S(X) ker(s ) so that codim(s) = dim(ker(s )). Similarly, we have: Exercise-34: Let X be a Banach space, T Γ K (X), and S = I T. Then, codim(s(x)) = dim(ker(s )) <. [Hint: Since T is compact, dim(ker(s )) < by Exercise-32. Let Y =

9 OPERATOR THEORY - PART 3/3 9 X/S(X) and q : X Y be the quotient map. Then q : Y X is injective and q (Y ) ker(s ). Also, if φ ker(s ) and if we put ψ(x + S(X)) := φ(x), then ψ Y and q (ψ) = φ, showing q (Y ) = ker(s ). Thus codim(s(x)) = dim(y ) = dim(y ) = dim(ker(s )).] [148] Let X be a Banach space, T Γ K (X), α C \ {}, and S = αi T. Then, (i) S is Fredholm with index zero. (ii) [Fredholm alternative] S is injective iff S is surjective. (iii) dim(ker(s)) = codim(s(x)) = dim(ker(s )) = codim(s (X )). Moreover, if X is a Hilbert space, then dim(ker(s)) = codim(s(x)) = dim(ker(s )) = codim(s (X)). Proof. Replacing α 1 T with T, we may assume α = 1 for all the statements. (i) We have S := I T is Fredholm by Exercise-32 and Exercise-34. Hence S n is Fredholm and ind(s n ) = n ind(s) for every n N by [147]. Using Exercise-33, choose k N such that ker(s n ) = ker(s k ) and S n (X) = S k (X) for every n k. Then ind(s n ) = ind(s k ), or n ind(s) = k ind(s) for every n k. This implies ind(s) =. And (ii) is a corollary of (i). (iii) Combine Exercise-34, and part (i) above applied to both S and S (respectively, S and S ). Remark: Result [148](i) can be read as ind(αi T ) = ind(αi), i.e., a compact perturbation does not change the index of αi when α. This can be generalized, see the theory below. Definition: Let X be a Banach space and S Γ(X). We say S is pseudo-invertible in Γ(X) if there is S 1 Γ(X) such that both I SS 1 and I S 1 S are compact operators. If this happens, S 1 is called a pseudo-inverse of S. Note that a pseudo-inverse is in general not unique. [149] Let X be a Banach space and S Γ(X). Then the following are equivalent: (i) S is Fredholm. (ii) There is Fredholm S 1 Γ(X) such that I SS 1 and I S 1 S are finite rank operators. (iii) S is pseudo-invertible in Γ(X). (iv) There are S 2, S 3 Γ(X) such that I S 2 S and I SS 3 are compact operators. Proof. (i) (ii): Write X = ker(s) W = S(X) Z, where W, Z are vector subspaces with W closed and dim(z) <, by Exercise-31. Note that S := S W : W S(X) is an isomorphism of Banach spaces. Let S 1 Γ(X) be S 1 (y + z) = S 1 (y) W for y + z S(X) Z. Then ker(s 1 ) = Z and S 1 (X) = W so that S 1 is Fredholm. It may be seen that I SS 1, I S 1 S are the finite rank projections to Z and ker(s) respectively. Next, (ii) (iii) and (iii) (iv) are trivial. (iv) (i): Let T 2 = I S 2 S, T 3 = I SS 3. Then S 2 S = I T 2 and SS 3 = I T 3 are Fredholm by [148]. Since ker(s) ker(s 2 S) and S(X) SS 3 (X), we get S is Fredholm. Remark: Let X be a Banach space and S Γ(X) be Fredholm. If S 1 is a Fredholm, pseudoinverse of S given by [149], then by [148] and [147] we have = ind(i (I SS 1 )) = ind(ss 1 ) = ind(s) + ind(s 1 ), and therefore ind(s) = ind(s 1 ).

10 1 T.K.SUBRAHMONIAN MOOTHATHU [15] [Stability of index] Let X be a Banach space and S Γ(X) be Fredholm. Then, (i) S + T is Fredholm and ind(s + T ) = ind(s) for every T Γ K (X). (ii) ɛ > such that T Γ(X) with T < ɛ, S + T is Fredholm and ind(s + T ) = ind(s). That is, F := {Fredholm operators on X} is open in Γ(X) and index : F Z is continuous. Proof. (i) To see S + T is Fredholm, we use [149]: if S 1 Γ(X) is a Fredholm pseudo-inverse of S, then using the fact that Γ K (X) is an ideal, it may be checked that S 1 is also a pseudo-inverse of S + T. Now, ind(s) = ind(s 1 ) = ind(s + T ) by the Remark above. (ii) If dim(x) <, then every member of Γ(X) is Fredholm with index zero (verify). So we may assume dim(x) =. Let S 1 Γ(X) be a Fredholm pseudo-inverse of S, chosen by [149], and let T, T 1 Γ K (X) be so that SS 1 = I T and S 1 S = I T 1. Since dim(x) =, the identity I is not compact, and hence S 1. We take ɛ = 1/ S 1, and consider T Γ(X) with T < ɛ. Note that I + S 1 T, I + T S 1 are invertible in Γ(X) by [14](ii) since S 1 T < 1, T S 1 < 1. Now S 1 (S + T ) = I +S 1 T T and hence (I +S 1 T ) 1 S 1 (S+T ) = I T 2, where T 2 = (I S 1 T ) 1 T Γ K (X). If we set S 2 = (I +S 1 T ) 1 S 1, then S 2 is Fredholm with ind(s 2 ) = ind(s 1 ), and I S 2 (S +T ) = T 2 Γ K (X). Similarly, (S +T )S 1 = I +T S 1 T 1 so that if we put S 3 = S 1 (I +T S 1 ) 1, then S 3 is Fredholm with ind(s 3 ) = ind(s 1 ) and I (S + T )S 3 Γ K (X). Therefore, S + T is Fredholm by [149]. Moreover, ind(s + T ) = ind(s 2 ) = ind(s 3 ) = ind(s 1 ) = ind(s) as argued in the Remark above. 4. Hilbert-Schmidt and trace class operators Let X be an infinite dimensional separable Hilbert space throughout this section. Note that l is a commutative Banach algebra with respect to the product (x n )(y n ) := (x n y n ). We are about to see the following analogy: Commutative Noncommutative l c = {(x n ) : lim n x n = } l 2 l 1 c = {(x n ) : x n = for all large n} c l 1 l 2 c l Γ(X) Γ K (X) = {compact operators} Γ 2 (X) = {Hilbert-Schmidt operators} Γ 1 (X) = {trace class operators} Γ F (X) = {finite-rank operators} Γ F (X) Γ 1 (X) Γ 2 (X) Γ K (X) Γ(X) (l 2 ) = l 2, c = l1, (l 1 ) = l Γ 2 (X) = Γ 2 (X), Γ K (X) = Γ 1 (X), Γ 1 (X) = Γ(X) Consider T Γ K (X). Then T is compact and positive. By the spectral theorem for compact self-adjoint operators from Functional Analysis, X has an orthonormal basis {e n : n N} consisting of eigenvectors of T. Let β n be the eigenvalue corresponding to e n, i.e., T e n = β n e n. We observe that β n for every n N since T ; (β n ) and any positive β n is repeated at most finitely many times, by [145] applied to T. If β n > only for finitely many n N, let (α n ) = (β n ). If

11 OPERATOR THEORY - PART 3/3 11 β n > for infinitely many n N, then arrange such positive β n s as a decreasing sequence, counted with multiplicity, and label the sequence as (α n ). Definition: Let T Γ K (X) and let (α n ) be the nonnegative sequence of eigenvalues of T arranged as described above. We call (α n ) the singular value sequence of T. For 1 p <, the Schatten p- class is defined as Γ p (X) = {T Γ K (X) : the singular value sequence of T belongs to l p }. Shortly, Γ 2 (X) and Γ 1 (X) will be defined in another equivalent fashion. [151] [Singular value decomposition of a compact operator] Let T Γ K (X) \ {} and let (α n ) be the singular value sequence of T. (i) Suppose α n > for every n N. Then T Γ K (F )\Γ F (X), and there are orthonormal sequences (e n ) n=1, (v n) n=1 in X such that T x = n=1 α n x, e n v n for every x X. (ii) Suppose α n = for all large n N. Then T Γ F (X) and orthonormal sequences (e n ) k n=1, (v n ) k n=1 in X such that T (X) = span{v 1,..., v k } and T x = k n=1 α n x, e n v n for every x X. Proof. Let T = T 1 T be the polar decomposition as in [141], and Y = ker(t ) = ker( T ). (i) Let B be an orthonormal basis of X consisting of eigenvectors of T. If z B, then either T z = so that z Y or T z = α n z for some n N. Let e n B be such that T e n = α n e n. Then (e n ) is orthonormal and Y = span{e n : n N}. Next observe (v n ) := (T 1 e n ) is an orthonormal sequence in X since T 1 is an isometry on Y. Any x X can be written as x = y + n=1 x, e n e n for some y Y. Hence T x = T 1 T x = + n=1 x, e n T 1 T e n = n=1 x, e n T 1 α n e n = n=1 α n x, e n v n. Moreover, T / Γ F (X) because v n = T 1 e n = T 1 T (e n /α n ) = T (e n /α n ) T (X) for every n N. (ii) an orthonormal basis {e n : n N} of X with T e n = α n e n n N. If k is the maximum with α n > for 1 n k, then Y = span{e 1,..., e k }. Now imitate the proof of (i). Remark: Let (β n ) be a bounded sequence in C, and let (e n ), (v n ) be two orthonormal sequences in X. Define T : X X as T x = n=1 β n x, e n v n. Then, it may be verified that T Γ(X) with T = sup{ β n : n N} and T y = n=1 β n y, v n e n. Remark: If T x = n α n x, e n v n is the singular value decomposition of T Γ K (X), then for any bounded function f : σ( T ) C we can define f[t ] Γ(X) as f[t ]x = n f(α n) x, e n v n. Exercise-35: Let T Γ(X), and let {e n : n N}, {v k : k N} be orthonormal bases for X. Then, (i) n=1 T e n 2 = n=1 k=1 T e n, v k 2 = k=1 n=1 e n, T v k 2 = k=1 T v k 2 [, ]. (ii) n=1 T e n, e n = k=1 T v k, v k [, ]. [Hint: For (ii), note that T e n, e n = T 1/2 e n 2 and apply part (i) to T 1/2.] Remark: In view of Exercise-35, from now onwards, when we write n=1 T e n 2 or n=1 T e n, e n, it will be understood that the value is independent of the particular choice of the orthonormal basis.

12 12 T.K.SUBRAHMONIAN MOOTHATHU Remark: If T Γ(C k ) is considered as a matrix, and if this matrix is considered as an element k of C k2, then we may define another norm on Γ(C k ) as T 2 := k i=1 j=1 T ij 2. Moving onto infinite dimension, if we consider the infinite matrix of an operator T Γ(X) with respect to an orthonormal basis {e n : n N} of X, then the (m, n)th entry of the matrix is T e n, e m so that n=1 T e n 2 = n=1 m=1 T e n, e m 2 is the sum of squares of the absolute values of the entries of the matrix. Hence we may attempt to define T 2 := ( n=1 T e n 2 ) 1/2 provided the sum is finite. Also note n=1 T e n, e n (if the series converges in C) is the trace of the matrix. Definition: Let {e n : n N} be an orthonormal basis for X. Let Γ 2 (X) = {T Γ(X) : T 2 2 := n=1 T e n 2 < }, and Γ 1 (X) = {T Γ(X) : T 1 := n=1 T e n, e n < }. Any T Γ 2 (X) is called a Hilbert-Schmidt operator, any T Γ 1 (X) is called a trace class operator, T 2 is the Hilbert-Schmidt norm, and T 1 is the trace norm of T (we will see that they are indeed norms). Remarks and observations: (i) Γ F (X) Γ j (X) for j = 1, 2. (ii) T Γ 2 (X) iff T T Γ 1 (X) since T e n 2 = T T e n, e n and T T. (iii) If Λ {Γ 1 (X), Γ 2 (X), Γ 1 (X), Γ 2 (X)}, then T Λ iff T Λ ( for the case Λ = Γ 2 (X), note that T e n 2 = T e n 2 ; other cases are easy). (iv) Let Y = ker(t ) = ker( T ). To check whether T Γ j (X) for j = 1, 2, it suffices to ask n T e n, e n <?, n T e n 2 <? for an orthonormal basis {e n : n = 1, 2,...} of Y. (v) In the definition of Γ 1 (X) we have used T instead of T since the convergence of n=1 T e n, e n does not imply the convergence of k=1 T v k, v k if another orthonormal basis {v k : k N} is considered; see Exercise-2 on page 92 of Conway, A Course in Operator Theory. Results about Γ 2 (X) and Γ 1 (X) are interrelated, and a little care is needed in choosing the order of the results for the proofs. The following treatment is roughly based on Conway, A Course in Operator Theory. An ideal Λ of Γ(X) is called a -ideal if T Λ whenever T Λ. Note that any left -ideal Λ of Γ(X) is always a two-sided -ideal since T S = (S T ) for T Λ and S Γ(X). [152] (i) Γ 2 (X) is a two-sided -ideal of Γ(X). (ii) T T 2 = T 2 and max{ ST 2, T S 2 } S T 2 for every T Γ 2 (X) and S Γ(X). (iii) If T Γ 2 (X) and ɛ >, S Γ F (X) with T S S T 2 < ɛ. Hence Γ 2 (X) Γ K (X). Proof. (i) Easy to see Γ 2 (X) is a left ideal, and it is closed under involution by Exercise-35(i). (ii) For ɛ >, choose a unit vector x X with T x > T ɛ, and then choose an orthonormal basis {e n : n N} for X with e 1 = x. Then T 2 = ( n=1 T e n 2 ) 1/2 T e 1 > T ɛ, and therefore T 2 T. Exercise-35(i) gives T 2 = T 2. Since ST e n 2 S 2 T e n 2, we get ST 2 S T 2. Then T S = S T S T 2 = S T 2 also. (iii) Imitate the proof that c is dense in l 2. Suppose n>k T e n 2 < ɛ 2, and define S Γ F (X) as S = T on Y := span{e 1,..., e k }, and S = on Y. Also note Γ F (X) = Γ K (X) by [144](vii).

13 OPERATOR THEORY - PART 3/3 13 If (x n ), (y n ) l 2, then (x n y n ) l 1 by Cauchy-Schwarz. The following result is of the same spirit. [153] For T Γ(X), the following are equivalent: (i) T is a trace class operator. (ii) T 1/2 = (T T ) 1/4 is a Hilbert-Schmidt operator. (iii) T is the product of two Hilbert-Schmidt operators. (iv) n T e n, v n < for any two orthonormal sequences (e n ), (v n ) in X. Proof. Let T = T 1 T be the polar decomposition given by [141] so that T1 T = T. (i) (ii): T 1/2 e n 2 = T e n, e n. (ii) (iii): T = (T 1 T 1/2 ) T 1/2 and use [152](i). (iii) (iv): If T = S 1 S 2 with S j Γ 2 (X), then S 1 Γ 2(X) also by [152](i). Hence n=1 T e n, v n = n S 2e n, S 1 v n n S 2e n S 1 e n ( n S 2e n 2 ) 1/2 ( n S 1 v n 2 ) 1/2 <. (iv) (i): Let Y = ker(t ) = ker( T ) and {e n : n = 1, 2,...} be an orthonormal basis for Y. Since T 1 is an isometry on Y by [141], we get (v n ) := (T 1 e n ) is an orthonormal sequence and hence n T e n, e n = n T 1 T e n, e n = n T e n, T 1 e n = n T e n, v n <. Exercise-36: (i) Γ 1 (X) is a two-sided -ideal in Γ(X) with Γ 1 (X) Γ 2 (X). (ii) If (α n ) is the singular value sequence of T Γ K (X), then we have T 2 2 = n=1 α2 n and T 1 = n=1 α n. Consequently, Γ 2 (X) = Γ 2 (X) and Γ 1 (X) = Γ 1 (X). [Hint: Write Γ 1, Γ 2, Γ K, Γ for simplicity. (i) By [153](iv), Γ 1 is a vector space, and ΓΓ 1 Γ = Γ(Γ 2 Γ 2 )Γ = (ΓΓ 2 )(Γ 2 Γ) = Γ 2 Γ 2 = Γ 1. If T Γ 1 and if T = T 1 T = T T 2 as in [141], then T Γ 1 and T = T T 2 = T 1 T T 2 Γ 1, and thus T Γ 1. Also, Γ 1 = Γ 2 Γ 2 Γ 2. (ii) We know Γ j Γ K for j = 1, 2. For T Γ K, let (e n ) be an orthonormal sequence corresponding to (α n ), and note α 2 n = T e n 2 = T e n 2 and α n = T e n, e n.] [154] Let {e n : n N} be an orthonormal basis for X. Then, (i) trace : Γ 1 (X) C defined as trace(t ) = n=1 T e n, e n is linear, is independent of the choice of {e n : n N}, and trace(t ) = trace(t ). (ii) The inner product S, T 2 := n=1 Se n, T e n = n=1 T Se n, e n = trace(t S) makes Γ 2 (X) into a separable Hilbert space. The corresponding norm is precisely the Hilbert-Schmidt norm 2 defined earlier. Moreover, T, S 2 = S, T 2 for every S, T Γ 2 (X). (iii) trace(st ) = trace(t S) for every T Γ 1 (X) and S Γ(X); and also for every S, T Γ 2 (X). (iv) T 1 = T and max{ ST 1, T S 1 } S T 1 for every T Γ 1 (X) and S Γ(X). (v) (Γ 1 (X), 1 ) is a separable Banach space, and trace (Γ 1 (X), 1 ) with trace = 1. Proof. (i) Trace is well-defined since the series converges absolutely by [153](iv). Linearity and the property trace(t ) = trace(t ) are clear. If {v k : k N} is another orthonormal basis,

14 14 T.K.SUBRAHMONIAN MOOTHATHU then the absolute convergence from [153](iv) allows an interchange of the order of summation so that we see n=1 T e n, e n = n=1 k=1 T e n, v k v k, e n = n=1 k=1 T e n, v k v k, e n = k=1 n=1 e n, T v k v k, e n = k=1 v k, n=1 T v k, e n e n = k=1 v k, T v k = k=1 T v k, v k. (ii) The equality S, T 2 = trace(t S) ensures by (i) that, 2 is independent of the choice of an orthonormal basis. Now consider the Hilbertian sum Y := {(y n ) X N : n=1 y n 2 < } of infinitely many copies of X, and let J : Γ 2 (X) Y be T (T e n ). Clearly J is injective. If (y n ) Y and if we define T : X X as T x = n=1 α ny n for x = n=1 α ne n, then T is linear and T x 2 (y n ) 2 ( n=1 α n 2 ) = (y n ) 2 x 2, showing T is bounded. Moreover, T Γ 2 (X) and J(T ) = (y n ). Thus J is a bijection, and the inner product defined on Γ 2 (X) is just the one from Y transferred via J. Since Y is a Hilbert space, Γ 2 (X) also becomes a Hilbert space. The separability of Γ 2 (X) can be deduced using [152](iii). In a Hilbert space, the inner product can be recovered from the norm, see p.134 of Kreyszig, Introductory Functional Analysis with Applications. Since the inner product S, T := T, S 2 also induces the norm 2, we get T, S 2 = S, T 2. (iii) Fix S Γ(X). Since T trace(st ), T trace(t S) are linear and since T can be written as T = T 1 +it 2 with T 1, T 2 self-adjoint, we may assume T Γ 1 (X) under consideration is self-adjoint. By the spectral theorem for compact self-adjoint operators from Functional Analysis, there exist an orthonormal basis {v k : k N} and {α k : k N} R such that T v k = α k v k for every k N. Hence trace(st ) = k=1 ST v k, v k = k=1 α k Sv k, v k = k=1 Sv k, T v k = k=1 T Sv k, v k = trace(t S). If S, T Γ 2 (X), part (ii) gives trace(st ) = T, S 2 = S, T 2 = trace(t S). (iv) & (v): For proofs (based on polar decomposition) of most of the assertions, see pages of Conway, A Course in Operator Theory. We just present a sample. If T = T 1 T is the polar decomposition, put S = T 1/2 Γ 2 (X). We have trace(t ) = trace(t 1 SS) = S, (T 1 S) 2 S 2 (T 1 S) 2. But (T 1 S) 2 = T 1 S 2 T 1 S 2 by [152](ii) and hence trace(t ) S 2 2 = T 1/2 2 2 = T 1. Also, if {e n : n N} is an orthonormal basis and T Γ(X) is the orthogonal projection to span{e 1 }, then trace(t ) = 1 = T 1. Thus trace = 1. Remark: If T Γ 1 (X) is positive, then Exercise-36(ii) says trace(t ) is the sum of eigenvalues of T, repeated according to multiplicity, just as in Linear Algebra. Lidskii s theorem gives the same conclusion without assuming T ; see Chapter 3 of P.Lax, Functional Analysis, for a proof. Definition: For x, y X, let x y : X X be z z, y x. As per this notation, the singular value decomposition in [151] becomes T = n α nv n e n. Exercise-37: (i) x y Γ(X) is a finite rank operator with ker(x y) = {y} and range = span{x}. (ii) trace(x y) = x, y, (x y) = y x, and T (x y) = T x y for every T Γ(X). (iii) If T Γ F (X), then there are finitely many x j, y j X such that T = k j=1 x j y j.

15 OPERATOR THEORY - PART 3/3 15 [Hint: (ii) Assume x and choose an orthonormal basis {e n : n N} with e 1 = x/ x. Denoting S = x y, we compute trace(s) = Se 1, e 1 + = x/ x, y x, x/ x = x, y. (iii) If {x 1,..., x k } is an orthonormal basis for T (X), then T z = k j=1 T z, x j x j = k j=1 z, T x j x j. Or use [151].] The result [155] below is analogous to: (l 2 ) = l 2, c = l1 and (l 1 ) = l. In the statement below, note that the spaces Γ K (X), Γ(X) are equipped with the operator norm, Γ 2 (X) with the Hilbert-Schmidt norm 2, and Γ 1 (X) with the trace norm 1. [155] Γ 2 (X) = Γ 2 (X), Γ K (X) = Γ 1 (X) and Γ 1 (X) = Γ(X). In all the three cases, the required isometric isomorphism is T trace( T ) from right to left. Proof. We leave the proof as a reading assignment: see Theorems 19.1 and 19.2 in Conway, A Course in Operator Theory. However, we indicate how the surjectivity is established. If φ Γ 2 (X), then by Riesz representation theorem, there is T Γ 2 (X) such that φ(s) = S, T 2 = trace(t S) = trace(st ). If φ Γ K (X) or φ Γ 1 (X), then ψ : X 2 C defined as ψ(x, y) = φ(x y) is a bounded sesquilinear form with ψ φ x y. Hence by Exercise-21(iv) and Exercise-37, there is unique T Γ(X) with φ(x y) = ψ(x, y) = T x, y = trace(t x y) = trace(t (x y)) = trace((x y)t ). By linearity and Exercise-37(iii), φ(s) = trace(st ) for every S Γ F (X). If φ Γ K (X), one shows also that T Γ 1 (X). Then one obtains φ(s) = trace(st ) for every S Γ K (X) or S Γ 1 (X) using the denseness of Γ F (X) and continuity. 5. Integral operators are Hilbert-Schmidt operators If T Γ(C k ) is given by a matrix A, then (T v) i = k j=1 A ijv j for v C k. Generalizing this, we have: if A : [, 1] 2 C is continuous, then the integral operator T A : C([, 1], C) C([, 1], C) is given by (T A f)x = 1 A(x, y)f(y)dy. We may consider this in a measure theoretic setting also. Definition: Let (Y, µ) be a σ-finite measure space, let ν denote the product measure on Y Y, and A L 2 (Y Y, ν). Then T A : L 2 (Y, µ) L 2 (Y, µ) defined as (T A f)x = Y A(x, y)f(y)dµ(y) = A(x, ), f is called an integral operator. If Y = [, 1] with the Lebesgue measure µ, and if A(x, y) = 1 for y x and A(x, y) = for y > x, then T A is the Volterra operator, (T A f)x = x f(y)dµ(y). Remark: Since (Y, µ) is σ-finite and A L 2 (Y Y, ν), we see A(x, ) L 2 (Y, µ) for µ-almost every x Y by Fubini s theorem (see page 37 of Royden, Real Analysis) so that A(x, ), f is meaningfully defined. Moreover, T A f 2 = Y A(x, ), f 2 dµ(x) X A(x, ) 2 f 2 dµ(x) = A 2 f 2 = A 2 f 2 and consequently, T A is a bounded linear operator with T A A. Remark: Integral operators behave almost like matrices. The following can be verified: (i) T A = T B, where B(x, y) = A(y, x). In particular, T A is self-adjoint iff A is real-valued and A(x, y) = A(y, x) for almost every x, y. (ii) T A T B = T C, where C(x, y) = Y A(x, z)b(z, y)dµ(z).

16 16 T.K.SUBRAHMONIAN MOOTHATHU Example: Let T be the Volterra operator on L 2 ([, 1], µ). (i) By the Remark above, (T g)y = 1 y g(x)dµ(x), and (T T f)x = 1 y y=x z= f(z)dµ(z)dµ(y). (ii) The facts σ(t ) = {} and T > imply that T is not normal. (iii) Since σ p (T ) =, T is injective, and the fact ker( T ) = ker(t ) = {} implies T so that T has a positive eigenvalue. In particular, σ( T ) { α : α σ(t )}. This warns us that the notation T has to be handled cautiously in general. (iv) It is known that the Volterra operator is not of trace class. But it is a Hilbert-Schmidt operator by the result below. [156] Let (Y, µ) be a σ-finite measure space and ν be the product measure on Y Y. Then, (i) If A L 2 (Y Y, ν), then T A Γ 2 (L 2 (Y, µ)) and T A 2 = A. (ii) If T Γ 2 (L 2 (Y, µ)), then there is A L 2 (Y Y, ν) such that T = T A. Proof. We will use Fubini s theorem throughout to manipulate the integrals. (i) If {f n : n N}, {g m : m N} are orthonormal bases for L 2 (Y, µ), and if h nm (x, y) := f n (x)g m (y), then it may be seen that {h nm : n, m N} is an orthonormal basis for L 2 (Y Y, ν). In particular, {f nm : n, m N} is an orthonormal basis for L 2 (Y Y, ν), where f nm (x, y) := f m (x)f n (y). And T A f n, f m = Y Y A(x, y)f n(y)f m (x)dν = Y Y A(x, y)f nm(x, y)d(ν) = A, f nm. T A 2 2 = n=1 T Af n 2 = n,m=1 T Af n, f m 2 = n,m=1 A, f nm 2 = A 2. (ii) Let T f = n α n f, g n h n be the singular value decomposition given by [151]. Hence This series converges in the L 2 -norm, but may not be pointwise convergent anywhere. If pointwise convergence holds, (T f)x = ( ) ( ) α n f(y)g n (y)dµ(y) h n (x) = α n h n (x)g n (y) f(y)dµ(y), n Y Y n suggesting A(x, y) := n α nh n (x)g n (y). Indeed, A 2 = n,k α nα k h n, h k g k, g n = n α2 n < by Exercise-36 so that A L 2 (Y Y, ν). Finally T = T A by Exercise-21 since T f, g = n α n f, g n h n, g = Y Y n α nf(y)g n (y)h n (x) g(x)dν = Y (T Af)(x) g(x) dµ = T A f, g. 6. Invariant subspace problem and the spectrum Definition: Let X be a Banach space and T Γ(X). The lattice and hyper-lattice of T are respectively defined as Lat(T ) = {Y X : Y is a closed vector subspace and T (Y ) Y }, and Lat H (T ) = {Y Lat(T ) : S(Y ) Y for every S {T } c }, where {T } c = {S Γ(X) : ST = T S}. Clearly, {{}, X} Lat H (T ) Lat(T ). Remark: The term lattice is justified by the following observations: (i) {{}, X} Lat(T ), (ii) if Y 1, Y 2 Lat(T ), then Y 1 Y 2 Lat(T ), (iii) if Y 1, Y 2 Lat(T ), then the closure of the span of Y 1 Y 2 belongs to Lat(T ). And similar statements for Lat H (T ). Definition: Let X be a Banach space and T Γ(X). We say T has non-trivial lattice if Lat(T ) {{}, X}, and we say T has non-trivial hyper-lattice if Lat H (T ) {{}, X}. Example: Let α C and X be a separable Hilbert space with dim(x) 2. Clearly Lat(αI) consists of all closed vector subspaces of X, and in particular non-trivial. But Lat H (αi) is trivial.

17 OPERATOR THEORY - PART 3/3 17 Proof : Let Y X be a non-trivial closed vector subspace and S Γ(X) = {αi} c be the orthogonal projection onto Y {}. Then S(Y ) is not contained in Y and thus Y / Lat H (αi). Example: Let T be the Volterra operator on C([, 1], C) (or on L 2 ([, 1], µ), (T f)x = x f(y)dy. For any b [, 1], let Y b := {f C([, 1], C) : f [,b] }. Clearly Y b Lat(T ), and it is known that Lat(T ) = {Y b : b [, 1]} {C([, 1], C)}. Remark: When T is the right shift on l 2, Lat(T ) is described by the work of Beurling in terms of the Hardy space and inner functions (we do not discuss this here). Invariant subspace problem: Let X be a Banach space with dim(x) 2 and T Γ(X). Is it true that T has a non-trivial lattice? In other words, does there exist a closed T -invariant vector subspace Y X different from {} and X? Remark: It is known via a complicated construction that there is T Γ(l 1 ) with Lat(T ) trivial (work of P.Enflo, and modifications by C.J.Read). However, at present (212) it is an open problem whether such an example can exist on a reflexive Banach space, in particular on a Hilbert space. We discuss below some of the sufficient conditions for T to have a non-trivial (hyper-)lattice. [157] Let X be a Banach space with dim(x) 2 and T Γ(X) \ {αi : α C}. Then, (i) If X is non-separable, then Lat(T ) is non-trivial. (ii) If T is not injective, or if T (X) X, then Lat H (T ) is non-trivial. (iii) If σ p (T ) (in particular, if dim(x) < ), or if σ a (T ) σ(t ), then Lat H (T ) is non-trivial. (iv) If σ p (T ), or if X is Hilbert space and σ p (T ), then Lat H (T ) is non-trivial. Proof. (i) Let x X \ {} and let O T (x) = {x, T x, T 2 x, T 3 x,...} be the orbit of x. Then Y := span[o T (x)] is a closed vector subspace of X different from {}. We have Y X since Y is separable, and T (Y ) Y since O T (x) is T -invariant. (This works even if T = αi). (ii) By assumption T. Now, ker(t ) and T (X) are S-invariant for every S {T } c. (iii) If α σ p (T ), then ker(αi T ) is a non-trivial member of Lat H (T ). If α σ(t ) \ σ a (T ), then αi T is bounded below, and the closed range of αi T cannot be the whole of X (else αi T would be invertible). Thus (αi T )(X) is a non-trivial member of Lat H (T ). (iv) If α σ p (T ), then (αi T )(X) is not dense by Exercise-25. So (αi T )(X) Lat H (T ) is non-trivial. Similarly, if α σ p (T ), then (αi T )(X) Lat H (T ) is non-trivial. The result below says in particular that if σ(t ) is not connected, then Lat H (T ) is non-trivial. [158] [Riesz decomposition theorem] Let X be a Banach space, T Γ(X) and suppose there are nonempty closed subsets K 1, K 2 C with σ(t ) = K 1 K 2 and K 1 K 2 =. Then, there are non-trivial members Y 1, Y 2 Lat H (T ) such that X = Y 1 Y2 and σ(t Yi ) = K i for i = 1, 2.

18 18 T.K.SUBRAHMONIAN MOOTHATHU Proof. Let U 1, U 2 C be disjoint open subsets with K i U i, and put U = U 1 U 2. Let f i H(U) be f i = χ Ui for i = 1, 2. Note that f 2 i = f i, f i f j = for i j, and f 1 + f 2 = 1 U. By holomorphic functional calculus, P i := f i [T ] Γ(X), Pi 2 = P i, P i P j = for i j and P 1 + P 2 = I. Setting Y i = P i (X), we may verify that Y i s are closed vector subspaces and X = Y 1 Y2. By a Remark after [115], Sf[T ] = f[t ]S for every f H(U) and S {T } c, and in particular SP i = P i S for every S {T } c. Consequently, Y i Lat H (T ). Since f i is not identically zero on σ(t ), we have P i by [116](i), and it follows that Y i s are non-trivial. (ii) Write T i = T Yi. Fix α C. Let S 1 Γ(Y 1 ) be S 1 y 1 = αy 1 T 1 y 1 and S Γ(X) be S(y 1 + y 2 ) = S 1 y 1 + y 2. Then α / σ(t 1 ) S 1 Inv(Γ(Y 1 )) S Inv(Γ(X)). Let f H(U) be f(z) = (α z)f 1 (z) + f 2 (z), where f 1, f 2 are as in (i). Observe that f[t ] = (αi T )P 1 + P 2 = S. Therefore by [116](ii), S Inv(Γ(X)) f has no zeroes in σ(t ) α / K 1. Thus σ(t 1 ) = K 1. [159] Let X be a Hilbert space with dim(x) 2, and T Γ(X) \ {αi : α C} be normal. Then, Lat H (T ) is non-trivial. More precisely, σ(t ) 2, and for every Borel partition σ(t ) = A 1 A 2 with A i, there are non-trivial members Y 1, Y 2 Lat H (T ) such that Y 1 Y 2, X = Y 1 Y2, and σ(t Yi ) = A i for i = 1, 2. Proof. Suppose σ(t ) = {α}. Then σ(αi T ) = {}, implying αi T = r(αi T ) = since αi T is normal. By this reasoning, we conclude that σ(t ) 2 if T αi for every α C. Now consider a non-trivial Borel partition σ(t ) = A 1 A 2. Since χ Ai B (σ(t ), C), we may obtain P i = χ Ai [T ] Γ(X) by Borel functional calculus and put Y i = P i (X). We have Y 1 Y 2 and X = Y 1 Y2 by [133](v). For the remaining assertions, imitate the proof of [158]. Next we present a result ([16] below) due to Abramovich, Aliprantis and Burkinshaw. Definition: Let X be a Banach space possessing a Schauder basis B = {e j : j N} and let {φ j : j N} X be the corresponding collection of coordinate functionals given by φ j (e i ) = 1 if i = j, and = otherwise. Note that x = j=1 φ j(x)e j for every x X. The positive cone of X with respect to B is defined as X + B = {x X : φ j(x) for every j N}. This cone defines a partial order on X by the condition that x y iff y x X + B. We say T Γ(X) and φ X are positive with respect to B if T (X + B ) X+ B and φ(x+ B ) [, ) respectively. Remark: If T Γ(X) and φ X are positive with respect to B, and x y are elements of X, then note that T x T y and φ(x) φ(y). Exercise-38: Let X be a Banach space having a Schauder basis B = {e j : j N}, and let S, T Γ(X) be commuting operators positive with respect to B. If there are x X + B and j N such that e j x and lim n S n x 1/n =, then φ j (T k Se j ) = for every k, where φ j is the jth coordinate functional. [Hint: Fix k and let β = φ j (T k Se j ). Use the positivity of the given