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Home , Compact operator, Equicontinuity

REVIEW ON SPACES OF LINEAR OPERATORS

Throughout, we let X,Y be complex Banach spaces. Definition 1. B(X,Y ) denotes the of bounded linear maps from X to Y .

• For T ∈ 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 on B(X,Y ), and makes B(X,Y ) into a .

Proof. That kT k is a norm is easy, so we check completeness of B(X,Y ). ∞ Suppose {Tj}j=1 is a Cauchy , i.e. kTi − Tjk → 0 as i, j → ∞ . For each x ∈ X, Tix is Cauchy in Y , since kTix − Tjxk ≤ kTi − Tjk kxk . Thus

lim Tix ≡ T x exists by completeness of Y. i→∞ Linearity of T follows immediately. And for kxk = 1,

kTix − T xk = lim kTix − Tjxk ≤ sup kTi − Tjk j→∞ j>i goes to 0 as i → ∞, so kTi − T k → 0.  Important cases ∗ • Y = C: we denote B(X,C) = X , the “ of X”. It is a Banach space, and kvkX∗ = sup |v(x)| . kxk=1

• If T ∈ B(X,Y ), define T ∗ ∈ B(Y ∗,X∗) by the rule (T ∗v)(x) = v(T x) . Then |(T ∗v)(x)| = |v(T x)| ≤ kT k kvk kxk, so kT ∗vk ≤ kT k kvk. Thus ∗ kT kB(Y ∗,X∗) ≤ kT kB(X,Y ) . • If X = H is a , 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 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 with identity.

Power series in B(X). P∞ k Suppose that k=0 ak z is a power series with radius of convergence R > 0, where −1 1/k R = lim sup |ak| . If T ∈ B(X) and kT k < R, then the sum ∞ X k ak T k=0 P∞ k is dominated in the norm by k=0 |ak| kT k < ∞, and so converges in B(X) by completeness.

P∞ −k k Important case: For z ∈ C , z 6= 0 , k=0 z T converges if kT k < |z|. By taking limits ∞ ∞ ∞ X X X (zI − T ) z−kT k = z1−kT k − z−kT k+1 = zI , k=0 k=0 k=1 P∞ −k k and similarly k=0 z T (zI − T ) = zI.

Corollary 3. If kT k < |z|, then zI − T is invertible (both left and right) and ∞ 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 ∈ B(X). Define

ρ(T ) ⊂ C = {z : zI − T has an inverse (both left and right) in B(X)} .

−1 For complex z ∈ ρ(T ), the operator RT (z) = (zI − T ) is then defined, and is a of z with values in the Banach space B(X).

To see this, suppose that z0 ∈ ρ(T ), and consider  (z0 − w)I − T = (z0I − T ) − wI = (z0I − T ) I − w RT (z0) 526/556 LECTURE NOTES 3

−1 If |w| < kRT (z0)k , then the operator I − wRT (z0) is invertible by the Neumann series expansion ∞ −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 ∞ ∞ 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 , ∞ 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 |z − z0| ≤ −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. P∞ 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→∞ proven just as in the one-variable case. As seen before, if |z| < kT k−1, then (I − zT )−1 exists, and is given by the expansion ∞ X znT n . n=0 This series converges, and yields an inverse for (I − zT ), provided

−1  n 1  |z| < lim sup kT k n . n→∞ Since z(I − zT )−1 = (z−1I − T )−1 for z 6= 0, changing z ↔ z−1 shows that

n 1 z ∈ ρ(T ) if |z| > lim sup kT k n . n→∞

Conversely, suppose that the set |z| > M is contained in ρ(T ). Then the function ∞ X n n −1 −1 −1 z T = (I − zT ) = z RT (z ) n=0 4 526/556 LECTURE NOTES extends holomorphically to |z| < M −1. The Cauchy integral formula over |z| = (M + )−1 shows that n −n n 1 kT k ≤ C (M + ) , therefore lim sup kT k n ≤ M. n→∞ It follows that n 1 rσ(T ) = lim sup kT k n n→∞ is the smallest number such that ρ(T ) contains |z| > rσ(T ).

THE SPECTRUM OF A BOUNDED LINEAR OPERATOR

Definition 4. For T ∈ B(X), we define the spectrum σ(T ) = C \ ρ(T ).

Then z ∈ 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 |z| ≤ kT k. More precisely, we note by the above section that the rσ(T ) of T is given by n 1 rσ(T ) = sup |z| = lim sup kT k n . z∈σ(T ) n→∞

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 . 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 ∈ B(X,Y ) is a compact linear operator (or compact, for short) if T (B1) is compact in Y .

• The condition is that the 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 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- ∞ ∞ quence {xj}j=1 ⊂ X, the sequence {T xj}j=1 ⊂ Y has a convergent subse- quence.

∞ Proof. By scaling assume {xj}j=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 .) 

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 |f(x) − f(x)| <  if |x − y| < δ, for all f ∈ F.

n Let K(x, y) ∈ 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. |K(x, y)| ≤ C for some C, so |T f(x)| ≤ C Ω |f(y)| 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 |K(x, y) − K(x0, y)| ≤ /m(Ω) if |x − x0| ≤ δ. Then 0 0 |T f(x) − T f(x )| ≤ m(Ω)kfku ·  if |x − x | < δ .

Thus T (B1) is uniformly equicontinuous. 

Example 2. Suppose that T is an operator of finite , that is, TX lies in a finite dimensional subspace Z ⊂ Y . Since any two norms on a finite dimensional space are equivalent, it follows that Z is metric equivalent to Euclidean space (via any choice of basis.) In particular, Z is complete, hence closed in Y . 6 526/556 LECTURE NOTES

Then T (B1) is a closed, bounded subset of the finite diminensional space Z, hence compact by Heine-Borel theorem.  • Any map of finite rank can be written in the form n X T x = vj(x) yj j=1 ∗ where vj ∈ X and yj ∈ Y .

Theorem 7. The collection of compact linear operators from X to Y is a closed subspace of B(X,Y ).

Proof. That S + cT is compact if S and T are compact follows by choosing, ∞ for a sequence {xj}j=1 ⊂ B1, a subsequencex ˜j such that both Sx˜j and T x˜j converge. Then (S + cT )˜xj converges. ∞ Now suppose {Tj}j=1 is a sequence of compact linear operators, and that kTj − T k → 0. We need to show that T is compact, i.e. that T xj has ∞ a convergent subsequence if {xj}j=1 ⊂ B1, which we do by a standard diagonal argument: ∞ By compactness of T1, given {xj}j=1 ⊂ B1, find a subsequence {x1,j} such that limj→∞ T1x1,j = y1.

From {x1,j}, extract a subsequence {x2,j} such that limj→∞ T2x2,j = y2.

Repeat, to get successive subsequences {xn,j} such that limj→∞ Tnxn,j = yn.

Letx ˜j = xj,j so that limj→∞ Tnx˜j = yn for all n. Note that yn is a Cauchy sequence in Y , since

kyn − ymk = lim kTnx˜j − Tmx˜jk ≤ kTm − Tnk . j→∞

Let y = limn→∞ yn; we check that limj→∞ T x˜j = y. For this:

kT x˜j − yk ≤ kT x˜j − Tmx˜jk + kTmx˜j − ymk + kym − yk . Take m large enough that the first and last term are less than /3; then for this m take j large so the middle term is less than /3. 

2 P∞ Example 3. Hilbert-Schmidt operators on ` (N): (Ax)i = j=1 Aij xj, P∞ P∞ 2 2 where i=1 j=1 |Aij| = kAkHS is finite. By Cauchy-Schwartz, ∞ ∞ 2 ∞ ∞ ∞ X X X X 2 X 2 2 2 Aij xj ≤ |Aij| |xj| = kAk kxk . HS i=1 j=1 i=1 j=1 j=1 A similar calculation shows that the finite rank truncations of A converge in the to A, so A is compact. Theorem 8. Suppose that T : X → Y is a between Ba- nach spaces. Then T ∗ : Y ∗ → X∗ is compact, where X∗ and Y ∗ are the Banach spaces dual to X and Y . 526/556 LECTURE NOTES 7

∗ ∗ ∗ Proof. Let {yj } be a sequence in Y with kyj k ≤ 1. We need to show that ∗ ∗ ∗ ∗ there is a subsequence of T yj which converges to some x in X . A sequence converges in X∗ if and only if it converges uniformly when considered as ∗ ∗ ∗ continuous functions on the unit ball B1,X in X. But (T yj )(x) = yj (T x), ∗ so we need a subsequence of yj to converge uniformly as functions on the image T (B1,X ) ⊂ Y , which holds if it converges uniformly on T (B1,X ).

The set T (B1,X ) is compact, so by the Arzela-Ascoli theorem it is sufficient ∗ to prove that the family of functions {yj } is uniformly bounded and equicon- ∗ tinuous on T (B1,X ). The uniform boundedness follows since kyj k ≤ 1 and T (B1,X ) is bounded in Y . is a simple consequency of linear- ity: for general y, y˜ ∈ Y ∗ ∗ ∗ ∗ kyj (y) − yj (˜y)k = kyj (y − y˜)k ≤ kyj k ky − y˜k ≤ ky − y˜k . ∗ So, the set of functions yj are equicontinuous as functions on Y , and hence on the subset T (B1,X ). 

SPECTRAL THEORY FOR COMPACT LINEAR OPERATORS

Lemma 9. Suppose that T is a compact operator, and that z 6= 0 is not in the point spectrum of T . Then (zI − T ) is bounded below, in that k(zI − T )xk ≥ c kxk for some c > 0 , for all x ∈ X.

Proof. Suppose k(zI − T )xnk → 0 where kxnk = 1 . Pass to a subsequence −1 to assume by compactness that T xn → y. Then xn → x = z y, so kxk = 1 and T x = zx. 

• This says that, for z 6= 0, then either z is an eigenvalue, or (zI − T ) maps X in an invertible manner onto a closed subspace of X. If the range is dense, then it must be all of X, so z∈ / σ(T ). In particular, z 6= 0 cannot be in the continuous spectrum of T . We will prove that, in fact, (zI − T ) must be onto if z is not an eigenvalue, so z 6= 0 cannot be in the residual spectrum either. • It is possible for compact T that 0 is in the continuous or residual spectrum of T . An example with 0 ∈ σC (T ): 1 1 T (x1, x2, x3,...) = (x1, 2 x2, 3 x3,...) and an example with 0 ∈ σR(T ): 1 1 T (x1, x2, x3,...) = (0, x1, 2 x2, 3 x3,...)

Lemma 10. If z 6= 0 is an eigenvalue of the compact operator T , then ker(zI − T ) is finite dimensional. ∞ If {zj}j=1 is an infinite set of distinct nonzero eigenvalues of T , then zj → 0. 8 526/556 LECTURE NOTES

Proof. Let Nz = ker(zI − T ). Then Nz is a closed subspace of X, and the −1 identity operator I = z T restricted to Nz is compact. Thus the unit ball of Nz is compact, which can only happen if dim(Nz) < ∞ . We prove the second part for a Hilbert space; the Banach space case is on the homework. Given an infinite sequence of zj, let En be the span of Nzj for 1 ≤ j ≤ n. Then En is a strictly increasing family of finite dimensional subspaces of X, and TEn ⊆ En.

Suppose {zj} is an infinite sequence of distinct eigenvalues, and |zj| ≥ c > 0. Choose vn ∈ En with kvnk = 1 and vn ⊥ En−1. Then T vn −znvn ∈ En−1, so dist(T vn, T vj) ≥ |zn| ≥ c if n > j. Since T vn has a convergent subsequence, we reach a contradiction.  Corollary 11. The set of nonzero eigenvalues of a compact operator T form a finite or countable set. For any c > 0, there are only finitely many eigenvalues with |z| ≥ c.

• We now show that, if z 6= 0 is not an eigenvalue, then it cannot be in the spectrum. We have already seen in Lemma 9 that (zI − T ) maps X in an invertible manner onto a closed subspace of X, so we need to show that (zI − T ) is onto if it is 1-1. • Suppose that (zI − T ) is not onto. Its range is a proper closed subspace of X, so there is some nonzero v ∈ X∗ with v(zI − T )x = 0 for all x, that is, (zI − T )∗v = 0 . So T ∗v = zv (Banach space dual) or T ∗v =zv ¯ (Hilbert space dual). • If (zI − T ) is 1-1, then (zI − T )∗ is onto. To see this, given v ∈ X∗ define a linear functional w on the range of (zI − T ) by w(zI − T )x) = v(x). By Hahn-Banach w extends to an element of X∗, and (zI − T )∗w = v.

Theorem 12 (The ). Supose that T : X → X is a compact operator on a Banach space X, and z 6= 0. Then zI − T is onto if and only if it is 1-1. In particular, if z is not an eigenvalue of T , and z 6= 0, then zI − T is an isomorphism of X onto X.

Proof. We will prove that (zI −T ) onto implies (zI −T ) is 1-1. This implies the other direction as well: since T ∗ is compact, then (zI − T )∗ onto implies (zI −T )∗ is 1-1. By the preceeding bullet points we conclude that if (zI −T ) is 1-1 then it is onto.

Suppose that (zI − T ) is onto but not 1-1. Then there exists x1 6= 0 with (zI − T )x1 = 0. Since (zI − T ) is onto, we can successively choose xj such j−1 j that (zI − T )xj+1 = xj. Then (zI − T ) xj = x1 6= 0, but (zI − T ) xj = 0. j j j If we let Nz denote the of (zI − T ) , then Nz is a strictly increasing j−1 j family of vector subspaces, Nz ( Nz . 526/556 LECTURE NOTES 9

j j−1 j j By definition (zI − T )Nz = Nz , and TNz ⊆ Nz , since T commutes with (zI − T ). j We next note that each Nz is finite dimensional. This is because j−i j j X j j−1 i j ˜ (zI − T ) = z I − i z T ≡ z I − T i=1 j j where T˜ is a compact operator. Then Nz is the z eigenspace of T˜, hence is finite dimensional. j We now derive a contradiction. Successively choose vectors yj ∈ Nz such j−1 that kyjk = 1 and dist(yj,Nz ) = 1. Suppose that j ≥ i. Then j−1 kT yj − T yik = kzyj − (zI − T )yj − T yik ≥ dist(zyj,Nz ) ≥ |z| .

Thus, we have an infinite sequence {yj} with kyjk = 1, but T yj has no convergent subsequence, contradicting compactness of T . 

Collecting together the above results, we conclude: If T is a compact operator on a Banach space, then σ(T )\{0} consists of a finite or countable set of eigenvalues {zj}, each with finite dimensional eigenspace. For c > 0, there are only a finite number of zj with |zj| > c. The point 0 may or may not belong to σ(T ), and may be of any spectral type.

There is a stronger version of the Fredholm alternative which states that, for all z 6= 0, the codimension of the range of (zI − T ) is equal to the dimension of the kernel of (zI − T ). Theorem 12 states that this is true if z is not an eigenvalue. To prove it is true for general z, we first make some observations.

• For all z 6= 0, the range of (zI − T ) is closed. Lemma 9 establishes this if z is not an eigenvalue. If z is an eigenvalue, with eigenspace Nz, choose 0 0 a closed complement Nz to Nz, so (zI − T )X = (zI − T )Nz. One can find 0 0 such Nz since dim(Nz) < ∞ . Then (zI − T ) is 1-1 on the Banach space Nz, 0 and by Lemma 9 is an isomorphism of Nz onto a closed subspace of X.

∗ • Let {v1, . . . , vm} be a basis for ker(zI −T ) . Since the range of (zI −T ) is closed, it is equal to the set of y such that vi(y) = 0 , i = 1 . . . , m . Indeed, one can take {y1, . . . , ym} such that vi(yj) = δij, then X = range(zI − T ) ⊕  ∗ span(y1, . . . , ym). So codim range(zI − T ) = dim ker(zI − T ) < ∞ .

• Similarly, let {x1, . . . , xn} be a basis for Nz = ker(zI − T ). Then one can 0 0 find {w1, . . . , wn} such that wi(xj) = δij, and wi vanish on Nz, where Nz is ∗ ∗ a closed complement to Nz. Then X = range(zI − T ) ⊕ span(w1, . . . , wn), and dimker(zI − T ) = codimrange(zI − T )∗ . 10 526/556 LECTURE NOTES

Theorem 13 (The Fredholm alternative). Suppose that T : X → X is a compact operator on a Banach space X, and z 6= 0 is an eigenvalue of T . Then zI − T has closed range of finite codimension, and dimker(zI − T ) = codimrange(zI − T ) .

Proof. We have already seen that range(zI −T ) is closed. We will show that dimker(zI − T ) ≤ codimrange(zI − T ) . Applying the same inequality for (zI − T )∗ and using the above points shows they are equal.

With vectors {x1, . . . , xn}, etc., chosen as above, assume that n > m and consider the map m X x → (zI − T ) x − wj(x) yj . j=1 0 This map is onto, since Nz is mapped onto span(y1, . . . ym), and Nz is mapped onto range(zI − T ). The map is of the form (zI − T˜) with T˜ compact, since finite rank operators are compact. If n > m then xn belongs to the kernel of this map, contradicting Theorem 12.