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REVIEW on SPACES of LINEAR OPERATORS Throughout, We Let X, Y Be Complex Banach Spaces. Definition 1. B(X, Y ) Denotes the Vector

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REVIEW on SPACES of LINEAR OPERATORS Throughout, We Let X, Y Be Complex Banach Spaces. Definition 1. B(X, Y ) Denotes the Vector REVIEW ON SPACES OF LINEAR OPERATORS Throughout, we let X; Y be complex Banach spaces. Definition 1. B(X; Y ) denotes the vector space of bounded linear maps from X to Y . • For T 2 B(X; Y ), let kT k be smallest constant C such that kT xk ≤ Ckxk, or equivalently kT k = sup kT xk : kxk=1 Lemma 2. kT k is a norm on B(X; Y ), and makes B(X; Y ) into a Banach space. Proof. That kT k is a norm is easy, so we check completeness of B(X; Y ). 1 Suppose fTjgj=1 is a Cauchy sequence, i.e. kTi − Tjk ! 0 as i; j ! 1 : For each x 2 X, Tix is Cauchy in Y , since kTix − Tjxk ≤ kTi − Tjk kxk : Thus lim Tix ≡ T x exists by completeness of Y: i!1 Linearity of T follows immediately. And for kxk = 1, kTix − T xk = lim kTix − Tjxk ≤ sup kTi − Tjk j!1 j>i goes to 0 as i ! 1, so kTi − T k ! 0. Important cases ∗ • Y = C: we denote B(X; C) = X , the \dual space of X". It is a Banach space, and kvkX∗ = sup jv(x)j : kxk=1 • If T 2 B(X; Y ), define T ∗ 2 B(Y ∗;X∗) by the rule (T ∗v)(x) = v(T x) : Then j(T ∗v)(x)j = jv(T x)j ≤ kT k kvk kxk, so kT ∗vk ≤ kT k kvk. Thus ∗ kT kB(Y ∗;X∗) ≤ kT kB(X;Y ) : • If X = H is a Hilbert space, we saw we can identify X∗ and H by the relation v $ y where v(x) = hx; yi : Then T ∗ is defined on H by hx; T ∗yi = hT x; yi : Note the identification is a conjugate linear map between X∗ and H: cv(x) = chx; yi = hx; cy¯ i : 1 2 526/556 LECTURE NOTES • The case of greatest interest is when X = Y ; we denote B(X) ≡ B(X; X). Then B(X) is an algebra under composition, and is easy to see that kTSk ≤ kT k kSk : Thus B(X) is a Banach algebra with identity. Power series in B(X). P1 k Suppose that k=0 ak z is a power series with radius of convergence R > 0, where −1 1=k R = lim sup jakj : If T 2 B(X) and kT k < R, then the sum 1 X k ak T k=0 P1 k is dominated in the norm by k=0 jakj kT k < 1, and so converges in B(X) by completeness. P1 −k k Important case: For z 2 C ; z 6= 0 ; k=0 z T converges if kT k < jzj. By taking limits 1 1 1 X X X (zI − T ) z−kT k = z1−kT k − z−kT k+1 = zI ; k=0 k=0 k=1 P1 −k k and similarly k=0 z T (zI − T ) = zI. Corollary 3. If kT k < jzj, then zI − T is invertible (both left and right) and 1 X (zI − T )−1 = z−k−1T k : k=0 THE RESOLVENT SET We review the basics of the resolvent set for an operator in T 2 B(X). Define ρ(T ) ⊂ C = fz : zI − T has an inverse (both left and right) in B(X)g : −1 For complex z 2 ρ(T ), the operator RT (z) = (zI − T ) is then defined, and is a holomorphic function of z with values in the Banach space B(X). To see this, suppose that z0 2 ρ(T ), and consider (z0 − w)I − T = (z0I − T ) − wI = (z0I − T ) I − w RT (z0) 526/556 LECTURE NOTES 3 −1 If jwj < kRT (z0)k , then the operator I − wRT (z0) is invertible by the Neumann series expansion 1 −1 X n n I − wRT (z0) = w RT (z0) n=1 Since (z0I − T ) is invertible with inverse RT (z0), we get that (z0 − w)I − T is invertible, with inverse 1 1 X n n X n n+1 R(z0 − w; T ) = RT (z0) w RT (z0) = w RT (z0) n=1 n=1 Setting w = z0 − z yields that RT (z) has a power series expansion about z0 −1 with radius of convergence at least kRT (z0)k , 1 X n n n+1 RT (z) = (−1) (z − z0) RT (z0) : n=1 Alternatively, you could deduce this by formally expanding (zI − T )−1 in a power series about z0, observe that it converges provided jz − z0j ≤ −1 kRT (z0)k , and check explicitly that the power series gives an inverse for (zI − T ) for all z for which it converges. • In particular, the resolvent set of T is open. P1 n In general, given a power series n=1 z vn with coefficients vn belonging to a Banach space, the radius of convergence is 1 −1 lim sup kvnk n n!1 proven just as in the one-variable case. As seen before, if jzj < kT k−1, then (I − zT )−1 exists, and is given by the expansion 1 X znT n : n=0 This series converges, and yields an inverse for (I − zT ), provided −1 n 1 jzj < lim sup kT k n : n!1 Since z(I − zT )−1 = (z−1I − T )−1 for z 6= 0, changing z $ z−1 shows that n 1 z 2 ρ(T ) if jzj > lim sup kT k n : n!1 Conversely, suppose that the set jzj > M is contained in ρ(T ). Then the function 1 X n n −1 −1 −1 z T = (I − zT ) = z RT (z ) n=0 4 526/556 LECTURE NOTES extends holomorphically to jzj < M −1. The Cauchy integral formula over jzj = (M + )−1 shows that n −n n 1 kT k ≤ C (M + ) ; therefore lim sup kT k n ≤ M: n!1 It follows that n 1 rσ(T ) = lim sup kT k n n!1 is the smallest number such that ρ(T ) contains jzj > rσ(T ). THE SPECTRUM OF A BOUNDED LINEAR OPERATOR Definition 4. For T 2 B(X), we define the spectrum σ(T ) = C n ρ(T ). Then z 2 σ(T ) means that (zI − T ) is not invertible in B(X). (It may have a one-sided inverse, but not both.) The spectrum σ(T ) is a closed set, contained in the set jzj ≤ kT k. More precisely, we note by the above section that the spectral radius rσ(T ) of T is given by n 1 rσ(T ) = sup jzj = lim sup kT k n : z2σ(T ) n!1 The spectrum of T is divided into 3 disjoint subsets. • The point spectrum σP (T ), also called eigenvalues, are those z for which ker(zI − T ) 6= 0 : That is, T x = zx for some non-zero x. • The continuous spectrum σC (T ) are those z for which ker(zI − T ) = 0 ; and for which the range of (zI − T ) is dense in X. • The residual spectrum σR(T ) are those z for which ker(zI − T ) = 0 ; and for which the range of (zI − T ) is not dense in X. Example 1. If an operator T is an isomorphism of X onto a proper subspace of X, then 0 is in the residual spectrum of T . This is the case for the right shift operator on spaces of sequences. Example 2. For the map (T f)(x) = xf(x) defined on L2([0; 1]), we have σ(T ) = [0; 1], and the spectrum is pure continuous spectrum, σ(T ) = σC (T ). 526/556 LECTURE NOTES 5 COMPACT LINEAR OPERATORS Let B1 denote the unit ball in X. Definition 5. We say that T 2 B(X; Y ) is a compact linear operator (or compact, for short) if T (B1) is compact in Y . • The condition is that the closure of the image of the unit ball is compact; the image itself need not be closed. • Compactness of T implies that the image of any bounded set has compact closure, since the set cK is compact if K is. Lemma 6. A linear map is compact if and only if, for every bounded se- 1 1 quence fxjgj=1 ⊂ X, the sequence fT xjgj=1 ⊂ Y has a convergent subse- quence. 1 Proof. By scaling assume fxjgj=1 ⊂ B1, and use that T (B1) has compact closure in Y if and only if every sequence in it has a convergent subsequence (since Y is a complete metric space.) Examples of compact operators. Example 1. Let X = C(Ω), in the uniform topology, where Ω is a compact n subset of R . Recall by Arzela-Ascoli that a subset F ⊂ C(Ω) has compact closure if and only if: •F is bounded, and •F is uniformly equicontinuous: given > 0 there exists δ > 0 so that jf(x) − f(x)j < if jx − yj < δ, for all f 2 F. n Let K(x; y) 2 C(Ω × Ω), where Ω is a compact subset of R . R Define (T f)(x) = Ω K(x; y)f(y) dy : The T is compact as a map from C(Ω) to C(Ω). R Proof. jK(x; y)j ≤ C for some C, so jT f(x)j ≤ C Ω jf(y)j dy ≤ C m(Ω) kfku. In particular, T (B1) is a bounded set. Next, K(x; y) is uniformly continuous on Ω × Ω, so given > 0 find δ > 0 with jK(x; y) − K(x0; y)j ≤ /m(Ω) if jx − x0j ≤ δ. Then 0 0 jT f(x) − T f(x )j ≤ m(Ω)kfku · if jx − x j < δ : Thus T (B1) is uniformly equicontinuous.
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