1 Definition and Basic Properties of Compa Operator

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1 1 Definition and Basic Properties of Compa Operator 1.1 Let X be a infinite dimensional Banach space. Show that if A C(X ), A does not have bounded inverse. Proof. Denote the unit ball of X by B and the unit sphere S. Suppose that {x n } S, then x n A 1 Ax n for all n. Because {Ax n } has a convergent subsequence, we know that x n has a convergent subsequence too, which implies that S is sequentially compa. is contradi s with eorem , which states that a normed linear space is finite dimensional iff the unit sphere is sequentially compa. 1.2 Let X be a Banach space and A L (X ) satisfy Ax a x for all x X, where a is a positive constant. Prove that A C(X ) iff X is finite-dimensional. Proof. It suffices to show that A C(X ) iff every bounded set in X is sequentially compa. `If': Let {x n } be a bounded sequence, thus Ax n is bounded since A is bounded, thus it has a convergent subsequence and A is therefore compa. `Only if': Let B be a bounded set and {x n } B. We can find a convergent subsequence in {Ax n }, say Ax nk. Note that Ax n a x n, we know that x nk is a Cauchy sequence thus convergent (as X is complete). 1.3 Let X and Y be Banach spaces, A L (X, Y ), K C(X, Y ) and R(A) R(K). Show that A C(X, Y ). Proof. Let K : X / ker K X be the canonical map, then K is also a compa operator, since B + ker K is the unit ball in X / ker K, where B is the unit ball in X. Note that K is continuous, thus K 1 is a closed map, and D(K 1 ) = R(K) R(A), hence K 1 A : X X / ker K is a closed map, and its domain is the entire X, thus from the closed graph theorem that K 1 A is continuous, whence it follows that A = K(K 1 A) is compa. 1.4 Let H be a Hilbert space and A : H H is a compa operator. Suppose that x n x 0 and y n y 0. Show that (x n, Ay n ) (x 0, Ay 0 ). Proof. We have that (x n, Ay n ) (x 0, Ay 0 ) (x n, Ay n Ay 0 ) + (x n x 0, Ay 0 ). Since x n x 0, it is clear that {x n } is bounded, say by M, and the second term goes to 0. Since y n y 0 and A is compa (thus completely continuous), we have that Ay n Ay 0. Notice that (x n, Ay n Ay 0 ) x n Ay n Ay 0 M Ay n Ay 0, thus the first term also goes to Let X, Y be Banach spaces and A L (X, Y ). Suppose that R(A) is closed and infinite-dimensional. Show that A / C(X, Y ). Proof. Suppose that A is compa. Note that R(A) is a Banach space, and there exist a bounded set B which is not sequentially compa, since R(A) is infinite dimensional. Take {y n } B such that it has no convergent subsequence. Consider X / ker A and A : X / ker A R(A) is the induced natural map, which is bije ive. Let [x n ] = A 1 (y n ), we can choose x n [x n ] such that Ax n = y n and x n 2 [x n 2 A 1 y n, thus {x n } is bounded. We meet a contradi ion. erefore A can not be compa. 1.6 Let w n K with w n 0. Show that the map defined as is a compa operator on l p (p 1). T : {ξ n } {w n ξ n } 1

2 Proof. It is clear that T L (l 2 ) since {w n } is bounded. Let T n be a linear operator defined on l p as T n : (ξ 1,..., ξ n, ξ n+1,... ) (w 1 ξ 1,..., w n ξ n, 0, 0,... ). Since dim T n (l 2 ) <, T n has finite-rank. It is also bounded, thus compa. Given ϵ > 0, there exists N such that w n < ϵ for all n > N. It then follows that T n x T x ϵ x, thus T n T ϵ, and T n T 0. Because C(l 2 ) is closed, T is compa. 1.7 Let R n be a measurable set and f be a bounded measurable fun ion on. Prove that F : x(t) f(t)x(t) is a compa operator on L 2 () iff f = 0 almost everywhere on. Proof. `If': Trivial. `Only if': Assume that mq > 0. If f(x) > 0 on a set A with ma > 0, we can find a compa set C A with mc > 0. en f is bounded below on C, say, f(x) c > 0 for all x C. We can find {x n } L 2 (C) such that x n 2 = 1 while x n 0 (for instance, take an orthonormal basis). Since F is completely continuous, we have F x 2 = f(t) 2 x n (t) 2 0. On the other hand, contradi ion. F x 2 C f(t) 2 x n (t) 2 c 2 C x n (t) 2 = c 2, 1.8 Let R n be a measurable set and K L 2 ( ). Show that A : u(x) K(x, y)u(y)dy, u L 2 () is a compa operator on L 2 (). Proof. It is clear that L 2 () is separable, hence there exists an orthonormal basis {u i } L 2 (). en where K(x, y) = K i (y) = for almost all y. e Parseval identity gives that and thus K i (y)u i (x), K(x, y) 2 dx = K(x, y) 2 dxdy = We now define the following operator of rank N A N u = K(x, y)u i (x). K i (y) 2 K N (x, y)f(y)dy, K i (y) 2 dy. (1) 2

3 where K N (x, y) = N K i (y)u i (x). By Cauchy-Schwarz inequality, (A A N ) 2 K(x, y) K N (x, y) 2 dxdy = = K(x, y) 2 dxdy 2 N N K(x, y) K i (y)u i (x)dxdy + K(x, y) 2 dxdy K i (y) 2 dy 0 as N. Hence A N A and A is therefore compa. K i (y) 2 dy 1.9 Let H be a Hilbert space, A C(H), {e n } is an orthonormal set in H. Show that lim n (Ae n, e n ) = 0. Proof. It can be proved that e n 0 (See proof to Exercise ), thus the conclusion follows from Exercise Let X be a Banach space, A C(H), X 0 is a closed subspace of X such that A(X 0 ) X 0. Prove that the map T : [x] [Ax] is a compa operator on X /X 0. Proof. It can be proved that B +ker A is the unit ball in X / ker A, where B is the unit ball in X. Let {[x n ]} be a bounded sequence, we can find {x n } such that x n 2 [x n ], thus {x n } is bounded, and {Ax n } has a convergent subsequence, thus {T [x n ]} = {[Ax n ]} has also a convergent subsequence. T is compa Let X, Y, Z be Banach spaces, X Y Z, if the embedding map from X to Y is compa and from Y to Z continuous. Prove that for any ϵ > 0, there exists c(ϵ) > 0 such that x Y ϵ x X + c(ϵ) x Z, x X. Proof. Prove by contradi ion. Suppose that there exists ϵ 0, for all n there exists x n X such that x n Y > ϵ 0 x n X + n x n Z. Let y n = x n / x n, then it holds that y n Y > ϵ 0 + n y n Z. Since the embedding map X Y is compa and y n = 1 for all n, we know that y n Y is bounded thus y n Z 0. Also we know that {y n } has a convergent subsequence in Y, say y nk y in Y as k. en z nk y in Z as the embedding map from Y to Z is continuous, and therefore y must be 0. But y nk ϵ, we reach a contradi ion. 2 Riesz-Fredholm eory 2.1 Let X be a Banach space and M X is a closed linear subspace with codim M = n. Show that there exists linearly independent set {ϕ k } n k=1 X such that M = n N(ϕ k ). k=1 Proof. Let {e i + M} (i = 1,..., n) be a basis of X /M, D = {e 1,..., e n }, D i = span{d \ {e i }, M}. en we have that e i / D i and we can a bounded linear fun ional ϕ i such that ϕ i (D i ) = 0 and ϕ i (e i ) = 1. It is easy to verify that {ϕ i } is linearly independent, and M = n N(ϕ i). 3

4 2.2 Let X, Y be Banach space and T L (X, Y ) is surje ive. Define T : X /N(T ) Y as Show that T is a linear homeomorphism. T [x] = T x, x [x], [x] X /N(T ); Proof. It is clear that T is well-defined and linear. For each [x] we can find x [x] such that x 2 [x]. us T [x] = T x T x 2 T [x] and T is continuous. Also T is a bije ion, hence it is a homeomorphism. 2.3 Let X be a Banach space, M, N 1, N 2 be closed linear subspaces of X. Suppose that show that N 1 is homeomorphic to N 2. M N 1 = X = M M 2, Proof. It suffices to show that N 1 and N 2 are both homeomorphic to X /M. Define F : N 1 X /M as F (x) = x + M, and it is easy to verify that F is well-defined. Since X = M N 1, F is bije ive. Besides, it holds that F (x) = x + M = inf m M x + m x, whence we know that F is continuous thus a homeomorphism. 2.4 Let A C(X ), T = I A, show that (1) x X /N(T ), x 0 [x], such that x 0 = [x] ; (2) Suppose that y X such that T x = y has at least one solution, show that one of the solutions has the minimum norm. Proof. (1) Since [x] = inf z N(T ) x + z, we can choose z n N(T ) for each n such that x + z n < [x] + 1 n, so {x + z n} is bounded. Since A is compa, we have that {Ax + Az n } = {Ax + z n } has a convergent subsequence, say Ax+z nk z, thus z nk z Ax. It follows that x+z nk T x+z, combining with x + z n [x] we have that T x + z = [x]. We verify that T x + z [x], or, T x + z x N(T ): T (T x + z x) = T (z Ax) = lim T (z nk ) = 0. (2) Suppose that x is a solution to T x = y, then the set of all the solutions is exa ly [x ]. From (1) we know that there exists x 0 [x ], thus a solution to T x = y, with the minimum norm [x ]. 2.5 Let A C(X ) and T = I A. Show that (1) N(T k ) is finite dimensional; and (2) R(T k ) is closed for all k N. Proof. T k = (I A) k = I A k, where A k is compa, as a result of Proposition 4.1.2(2) and (6). 2.6 Let M be a closed linear subspace on Banach space B. Call a bounded linear operator P : X M with P 2 = P a proje ion operator on M. Show that (1) If M is finite dimensional then a proje ion operator on M do exist; (2) If P is a proje ion operator on M then I P is a proje ion operator on R(I P ) from X ; (3) If P is a proje ion operator on M then X = M N, where N = R(I P ); 4

5 (4) If A C(X ) and T = I A, then it holds that N(T ) X /N(T ) = X = R(T ) X /R(T ) in the sense of isomorphism both algebraical and topological. Proof. (1) Let e 1,..., e n be a normal basis of M. From Hahn-Banach eorem, there exists continuous linear fun ionals f 1,..., f n such that f k (e j ) = δ kj. en define P x = f k (x)e k, and it is easy to verify that P is bounded and satisfies that P 2 = P. (2) e conclusion follows from that (I P ) 2 = I 2P + P 2 = I 2P + P = I P. (3) It is clear that X = M +R(I P ) and we shall show that M R(I P ) = 0. Let x M R(I P ), then there exists y such that x = (I P )y, thus P x = P (I P )y = (P P 2 )y = 0. Since x M, it holds that x = P x, whence we obtain that x = 0. (4) Since N(T ) is finite dimensional, from (1) there exists a proje ion operator P on N(T ), and from (3) it suffices to show that X /N(T ) is isomorphic to R(I P ). Let F : X /N(T ) R(I P ) be defined as F ([x]) = (I P )x. It is clear that F is well-defined, bije ive and linear (algebraically isomorphic). For all [x] X /N(T ) there exists x [x] such that x 2 [x], so F ([x]) = (I P )x I P x 2 I P [x], and thus F is continuous (topologically homeomorphic). erefore we obtain that N(T ) X /N(T ) = X. Since codim R(T ) = dim N(T ), we know that X /R(T ) and N(T ) are isomorphic both algebraically and topologically. And it is obvious that R(T ) is isomorphic to X /N(T ), since the map y = T x [x] is an isomorphism. us we also have that X /R(T ) R(T ) = X. 3 Spe rum eory of Compa Operators (Riesz-Schauder eory) (X denotes Banach space in this se ion) 3.1 Given sequence of numbers {a n } and define operator A on l 2 as (1) Show that A L (l 2 ) iff {a n } is bounded; A : (x 1, x 2,... ) (a 1 x 1, a 2 x 2,... ). (2) If A L (l 2 ) find σ(a) and the types of the spe ral points. Proof. (1) `If': Suppose that a n M, then Ax M x. `Only if': If {a n } is not bounded, then there exists n 1 < n 2 < such that a nk > k. For each m, Take x m = (ξ 1, ξ 2,... ) where ξ nm = 1 and ξ j = 0 for all of the rest indices j. It is clear that x l 2 and x m = 1. We compute Ax m > m 1 2 as m, which contradi s with the continuity of A. erefore {a n } is bounded. (2) Since A l 2, we know that {a n } is bounded, say, by M. If λ = a i for some i, then (λi A)x = 0 has nonzero solutions and λ σ p (A). Now assume that λ a i for all i, then (λi A) 1 exists, sending (x 1, x 2,... ) to ( x1 x λ a 1, 2 λ a 2,... ). 1 If λ is not a limit point of {a i }, then λ a i is bounded away from 0, so ( x 1 x λ a 1, 2 λ a 2,... ) l 2 whenever (x 1, x 2,... ) L 2 and R(λI A) = l 2, thus λ σ(a). Now let λ a i be a limit point of {a i }, suppose that a nk λ < 1 k, where {a n k } are pairwise distin. Consider x = (x 1, x 2,... ) with x nk = (λ a nk )/ k and x i = 0 for i n k, then x 2 i = (λ a nk ) 2 /k < 1/k 3 <. However, x 2 i /(λ a i) 2 = 1/k =, hence x R(λI A). Note that any x with finitely many non-zero components is in R(λI A), we know that R(λI A) is dense in l 2. erefore, λ σ c (A). We conclude that σ(a) = {a i } with σ p (A) = {a i } and σ c (A) the rest spe ral points. 5

6 3.2 In C[0, 1] consider the operator T : x(t) (1) Show that T is a compa operator; t 0 x(s)ds, x(t) C[0, 1]. (2) Find σ(t ) and a nontrivial closed invariant subspace of T. Proof. (1) It suffices to show that T (B 1 ) is sequentially compa, or, uniformly bounded and equi-continuous. First we have that T x 1 0 x(s)ds x implying that T (B1 ) is uniformly bounded. Besides it holds that (T x)(s ) (T x)(s ) = s x(s)ds s x s s implying that T (B 1 ) is equicontinuous. (2) First of all, T = 1 implies that σ(t ) is contained in the closed disc. Since C[0, 1] is infinitedimensional, we know that 0 σ(t ). Any other spe ral point must be eigenvalue, that is, if λ 0 belongs to σ(t ), then T x = λx has non-zero solution. But T x = λx has only zero solution, hence σ(t ) = {0}. An invariant space of T is C 1 [0, 1]. 3.3 Let A C(X ). Prove that x Ax = 0 has only zero solution iff x Ax = y has solution for all y X. Proof. `Only if': is is eorem `If': Let T = I A, then dim N(T ) = codim R(T ) = 0, thus N(T ) = {0}. 3.4 Let T L (X ) and there exists m N such that Show that p(t ) = q(t ) m. X = N(T m ) R(T m ). Proof. Let x N(T m+1 ). We have that T m x R(T m ) N(T m ), yielding that T m x = 0 and x N(T m ). So N(T m+1 ) N(T m ), thus p(t ) m. Now we show that q(t ) = p(t ). First we show that q(t ) p(t ). For simplicity, we use notations p and q instead of p(t ) and q(t ) respe ively. (1) Proof of p q. We have that T (R(T q )) = R(T q+1 ) = R(T q ), thus for y R(T q ), we have x R(T q ) such that T x = y. We claim that if T x = 0 for some x R(T q ) then x must be zero. If not, there exists x 1 R(T q )\{0} such that T x 1 = 0, then there exists x 2 0 such that T x 2 = x 1. Continuing this process, we obtain {x n } such that 0 x 1 = T x 2 = = T n 1 x n, but 0 = T x 1 = T n x n. us x n / N(T n 1 ) and x n N(T n ) for all n, which contradi s with p <. Now we show that N(T q+1 ) = N(T q ), which would imply that p q. It suffices to show that N(T q+1 ) N(T q ). Let x N(T q+1 ). Since T q x R(T q ) and T (T q x) = 0, we must have that T q x = 0 and x N(T q ). (2) Proof of p q. is is obviously true for q = 0. Assume that q > 0. It suffices to show that N(T q 1 ) N(T q ). Let y R(T q 1 ) \ R(T q ). en there exists x such that y = T q 1 x, and there also exists z such that T y = T q+1 z since T y R(T q ) = R(T q+1 ). us T q 1 (x T z) = y T q z 0 because y / R(T q ). So x T z does not belong to N(T q 1 ). And it is obvious that it belongs to N(T q ), which establishes that N(T q 1 ) N(T q ). 3.5 Let A, B L (X ) and AB = BA. Prove that (1) R(A) and N(A) are invariant subspaces of B; 6

7 (2) R(B n ) and N(B n ) are invariant subspaces of B for all n N. Proof. (1) Let y R(A), then y = Ax for some x. It holds that By = BAx = A(Bx) R(A), thus R(A) is an invariant subspace of B. Let y N(A), then A(By) = B(Ay) = 0, indicating that By N(A). Hence N(A) is an invariant subspace of B. (2) e conclusion follows from B(B n x) = B n (Bx) R(B n ) and B n (By) = B(B n y) = 0 for y N(B n ). 3.6 Let A L (X ) and M is a finite-dimensional invariant subspace of A. Show that (1) e a ion of A on M can be described by a matrix; (2) At least one eigenve or of A is in M. Proof. Trivial, as A M can be viewed as a linear transformation over M (which is finite dimensional). 3.7 Let x 0 X and f X satisfy f, x 0 = 1. Let A = x 0 f and T = I A. Find p(t ). Proof. We have that Ax = f, x x 0 and A 2 = A. us N(T ) = N(T 2 ). If dim X > 1 then N(T ) X so p(t ) = 1; otherwise N(T ) = X, so p(t ) = 0. 4 Hilbert-Schmidt eorem (H denotes complex Hilbert space in this se ion) 4.1 Let A L (H), show that A + A, AA and A A are all symmetric and AA = A A = A 2. Proof. It is trivial to prove that A + A, AA and A A are symmetric. With respe to norm, we have AA = sup x =1 (AA x, x) = sup x =1 (A x, A x) = sup x =1 A x 2 = A 2 = A 2. Similarly we have A A = A Let A L (H) satisfying (Ax, x) 0 for all x H and (Ax, x) = 0 iff x = 0. Show that Ax 2 A (Ax, x), x H. Proof. It is not hard to show that the following generalized Cauchy's Inequality holds. (Au, v) 2 (Au, u)(av, v). Let u = x and v = Ax, we have (Au, Au) 2 simplifies to our desired result. (Ax, x)(a 2 x, Ax) (Ax, x) A Ax 2, which 4.3 Let A be a symmetric compa operator on H, and m(a) = inf (Ax, x), x =1 M(A) = sup (Ax, x) x =1 Prove that (1) If m(a) 0 then m(a) σ p (A); (2) If M(A) 0 then M(A) σ p (A); 7

8 Proof. Consider A α = A+αI, then the spe rum is translated by α, so m(a α ) = m(a)+α and M(A α ) = M(A) + α. For α < 0 small enough, m(a α ) < M(A α ) < 0. Suppose that (A α x n, x n ) m(a α ) with x n = 1. From Proposition 4.4.5(5), it holds that A α = m(a α ). Note that A α x n m(a α )x n 2 = A α 2 2m(A α )(A α x n, x n ) + m(a α ) 2 0 as n, it follows that m(a α ) is in the spe rum of A α. Hence m(a) is in σ(a), and A is compa, thus if m(a) 0 it must be in σ p (A). Similarly consider A + αi for α > 0 enough, it yields that M(A) σ p (A) if M(A) Let A be a symmetric compa operator, show that (1) If A is non-zero then it has at least one non-zero eigenvalue; (2) If M is an non-trivial invariant subspace then M contains some eigenve or of A. Proof. (1) It follows dire ly from eorem (2) Assume M is closed, then A M is compa and symmetric. Since M is nontrivial, A M is non-zero, and therefore has an eigenvalue on M. 4.5 Show that P L (H) is an orthogonal proje or if and only if (1) P is symmetric, i.e., P = P ; (2) P is idempotent, i.e., P 2 = P. Proof. `Only if': Trivial. `If': Let M = {x : P x = x}, then M is a linear subspace of H. Since P is continuous, it follows that M is closed. If P x = y then P y = P 2 x = P x = y, which means that M is the range of P. Now notice that (P y, x P x) = (y, P x P P x) = (y, P x P 2 x) = 0, so R(P ) R(I P ). Also x = P x+(x P x), we see that P x is an orthogonal proje or onto M. 4.6 Show that P L (H) is an orthogonal proje or if and only if (P x, x) = P x 2 for all x H. Proof. `Only if': Suppose that P is a proje or, then x = y + z with y R(P ) and z R(P ). en (P x, x) = (y, y + z) = (y, y) + (y, z) = (y, y) = P x 2. `If': From Proposition (1), we know that P is symmetric, hence P 2 P is symmetric. Notice that (P 2 x P x, x) = (P 2 x, x) (P x, x) = (P x, P x) (P x, x) = 0, it follows from Proposition (5) that P 2 P = 0. Hence P is an orthogonal proje or by the previous problem. 4.7 Let A L (H), it is called positive operator if (Ax, x) 0 for all x H. Show that (1) All positive operators are symmetric; (2) e eigenvalues of a positive operators are non-negative. Proof. (1) Proposition 4.4.5(1). (2) From Proposition 4.4.5(2), we need only to consider λi A for real λ. Notice that for negative λ it holds that (λi A)x 2 = ((λi A)x, (λi A)x) = λ 2 x 2 2λ(Ax, x) + Ax 2 λ 2 x 2 and thus λi A is inje ive for negative λ. Hence σ(a) [0, ). 8

9 4.8 Let L and M be two closed linear subspace of H. Show that L M iff P M P L is positive. Proof. `Only if': Suppose that x = x L + y L = x M + y M where x A A and y A A. Decompose x M as x M = u + v, where u L and v L. Notice that y M L hence x = u + (v + y M ) is an orthogonal decomposition along L, and from the uniqueness of decomposition it must hold that x L = u. Since L M, it holds that x M x L M and thus (P M x P L x, x) = (x M x L, x M + y M ) = (x M x L, x M ) = (v, u + v) = (v, v) 0. `If': Suppose that x L then P L x = x. Suppose that x = x M + y M where x M M and y M M. en 0 (P M x P L x, x) = (P M x x, x) = (x M x, x) = (y M, x M + y M ) = (y M + y M ) 0, hence y M = 0 and x = x M, hence x M, and L M. 4.9 Let (a ij ) satisfy i,j=1 a ij 2 <. Define in l 2 where y i = j=1 a ijx j. Show that (1) A is compa ; A : x = {x 1, x 2,... } y = {y 1, y 2,... } (2) If a ij = a ji then A is a symmetric compa operator. Proof. (1) Using Cauchy-Schwarz inequality it is easy to verify that A L (l 2 ). Define A N L(l 2 ) as A : x = {x 1, x 2,... } y = {y 1,..., y N, 0,... } then A N is a finite-rank operator and thus is compa. And it follows from Cauchy-Schwarz inequality that A N A 0, hence A is compa. (2) Suppose that z = (z 1, z 2,... ). It is not hard to show that N N a ij x j z i a ij x j z i j=1 as N using Cauchy-Schwarz inequality. Also, j=1 i=j j=i 2 j=1 0 N N N N a ij x i = a ji z i x j x j a ij z i, which can be proved in the same way. We have established Ax, z = x, Az, hence A is symmetric Let A be a symmetric operator on H and there exists an orthonormal basis in which every ve or is an eigenve or of A. Suppose that (1) dim N(λI A) < ( λ σ p (A) \ {0}) (2) For any ϵ > 0, the set σ p (A) \ [ ϵ, ϵ] is finite. Show that A is a compa operator on H. Proof. From the two assumptions we know that A has countably many eigenvalues (including the multiplicity). List them in the decreasing order of absolute value (with multiplicity) as λ 1 λ 2. Since A is symmetric, we know that the basis contains a basis of N(λI A) for all eigenvalue λ 0. en Ax = (x, e i)e i (no need to consider the eigenve ors associated with eigenvalue 0), where e i is an eigenve or associated with λ n. Define A N = N λ n(x, e i )e i then A N is of finite rank and thus compa. By the Remark 1 of eorem 4.4.7, A A N 0 and thus A is compa. 9