Hilbert Spaces

Hilbert spaces There are really three ‘types’ of Hilbert spaces (over C). The finite dimensional ones, essentially just Cn , for different integer values of n, with which you are pretty familiar, and two infinite dimensional types corresponding to being separable (having a countable dense subset) or not. As we shall see, there is really only one separable infinite-dimensional Hilbert space (no doubt you realize that the Cn are separable) and that is what we are mostly interested in. Nevertheless we try to state results in general and then give proofs (usually they are the nicest ones) which work in the non-separable cases too. I will first discuss the definition of pre-Hilbert and Hilbert spaces and prove Cauchy’s inequality and the parallelogram law. This material can be found in many other places, so the discussion here will be kept succinct. One nice source is the book of G.F. Simmons, “Introduction to topology and modern analysis” [5]. I like it – but I think it is long out of print. RBM:Add description of contents when complete and 1. pre-Hilbert spaces A pre-Hilbert space, H, is a vector space (usually over the complex numbers but there is a real version as well) with a Hermitian inner product h, i : H × H −→ C, (3.1) hλ1 v1 + λ2 v2 , wi = λ1 hv1 , wi + λ2 hv2 , wi, hw, vi = hv, wi for any v1 , v2 , v and w ∈ H and λ1 , λ2 ∈ C which is positive-definite (3.2) hv, vi ≥ 0, hv, vi = 0 =⇒ v = 0. Note that the reality of hv, vi follows from the ‘Hermitian symmetry’ condition in (3.1), the positivity is an additional assumption as is the positive-definiteness. The combination of the two conditions in (3.1) implies ‘anti-linearity’ in the second variable (3.3) hv, λ1 w1 + λ2 w2 i = λ1 hv, w1 i + λ2 hv, w2 i which is used without comment below. The notion of ‘definiteness’ for such an Hermitian inner product exists without the need for positivity – it just means (3.4) hu, vi = 0 ∀ v ∈ H =⇒ u = 0. Lemma 3.1. If H is a pre-Hilbert space with Hermitian inner product h, i then (3.5) is a norm on H. 1 kuk = hu, ui 2 mention problems Proof. The first condition on a norm follows from (3.2). Absolute homogeneity follows from (3.1) since kλuk2 = hλu, λui = |λ|2 kuk2 . (3.6) So, it is only the triangle inequality we need. This follows from the next lemma, which is the Cauchy-Schwarz inequality in this setting – (3.8). Indeed, using the ‘sesqui-linearity’ to expand out the norm (3.7) ku + vk2 = hu + v, u + vi = kuk2 + hu, vi + hv, ui + kvk2 ≤ kuk2 + 2kukkvk + kvk2 = (kuk + kvk)2 .  Lemma 3.2. The Cauchy-Schwarz inequality, |hu, vi| ≤ kukkvk ∀ u, v ∈ H (3.8) holds in any pre-Hilbert space. Proof. This inequality is trivial if either u or v vanishes. For any non-zero u, v ∈ H and s ∈ R positivity of the norm shows that 0 ≤ ku + svk2 = kuk2 + 2s Rehu, vi + s2 kvk2 . (3.9) This quadratic polynomial in s is non-zero for s large so can have only a single minimum at which point the derivative vanishes, i.e. it is where 2skvk2 + 2 Rehu, vi = 0. (3.10) Substituting this into (3.9) gives (3.11) kuk2 − (Rehu, vi)2 /kvk2 ≥ 0 =⇒ | Rehu, vi| ≤ kukkvk which is what we want except that it is only the real part. However, we know that, for some z ∈ C with |z| = 1, Rehzu, vi = Re zhu, vi = |hu, vi| and applying (3.11) with u replaced by zu gives (3.8).  Corollary 3.1. The inner product is continuous on the metric space (i.e. with respect to the norm) H × H. Proof. Corollaries really aren’t supposed to require proof! If (uj , vj ) → (u, v) then, by definition ku − uj k → 0 and kv − vj k → 0 so from (3.12) |hu, vi − huj , vj i| ≤ |hu, vi − hu, vj i| + |hu, vj i − huj , vj i| ≤ kukkv − vj k + ku − uj kkvj k continuity follows.  Corollary 3.2. The Cauchy-Scwharz inequality is optimal in the sense that (3.13) kuk = sup v∈H;kvk≤1 I really will leave this one to you. |hu, vi|. 2. Hilbert spaces Definition 3.1. A Hilbert space H is a pre-Hilbert space which is complete with respect to the norm induced by the inner product. As examples we know that Cn with the usual inner product hz, z 0 i = (3.14) n X zj zj0 j=1 is a Hilbert space – since any finite dimensional normed space is complete. The example we had from the beginning of the course is l2 with the extension of (3.14) (3.15) ha, bi = ∞ X aj bj , a, b ∈ l2 . j=1 Completeness was shown earlier. The whole outing into Lebesgue integration was so that we could have the ‘standard example’ at our disposal, namely (3.16) L2 (R) = {u ∈ L1loc (R); |u|2 ∈ L1 (R)}/N where N is the space of null functions. The inner product is Z (3.17) hu, vi = uv. 2 1 Note R that we showed that if u, v ∈ L (R) then uv ∈ L (R). We also showed that if |u|2 = 0 then u = 0 almost everywhere, i.e. u ∈ N , which is the definiteness of the inner product (3.17). It is fair to say that we went to some trouble to prove the completeness of this norm, so L2 (R) is indeed a Hilbert space. 3. Orthonormal sequences Two elements of a pre-Hilbert space H are said to be orthogonal if (3.18) hu, vi = 0 which can be written u ⊥ v. A sequence of elements ei ∈ H, (finite or infinite) is said to be orthonormal if kei k = 1 for all i and hei , ej i = 0 for all i 6= j. Proposition 3.1 (Bessel’s inequality). If ei , i ∈ N, is an orthonormal sequence in a pre-Hilbert space H, then X (3.19) |hu, ei i|2 ≤ kuk2 ∀ u ∈ H. i Proof. Start with the finite case, i = 1, . . . , N. Then, for any u ∈ H set N X v= hu, ei iei . (3.20) i=1 This is supposed to be ‘the projection of u onto the span of the ei ’. Anyway, computing away we see that (3.21) hv, ej i = N X hu, ei ihei , ej i = hu, ej i i=1 using orthonormality. Thus, u − v ⊥ ej for all j so u − v ⊥ v and hence 0 = hu − v, vi = hu, vi − kvk2 . (3.22) Thus kvk2 = |hu, vi| and applying the Cauchy-Schwarz inequality we conclude that kvk2 ≤ kvkkuk so either v = 0 or kvk ≤ kuk. Expanding out the norm (and observing that all cross-terms vanish) kvk2 = N X |hu, ei i|2 ≤ kuk2 i=1 which is (3.19). In case the sequence is infinite this argument applies to any finite subsequence, ei , i = 1, . . . , N since it just uses orthonormality, so (3.19) follows by taking the supremum over N.  4. Gram-Schmidt procedure Definition 3.2. An orthonormal sequence, {ei }, (finite or infinite) in a preHilbert space is said to be maximal if u ∈ H, hu, ei i = 0 ∀ i =⇒ u = 0. (3.23) Theorem 3.1. Every separable pre-Hilbert space contains a maximal orthonormal sequence. Proof. Take a countable dense subset – which can be arranged as a sequence {vj } and the existence of which is the definition of separability – and orthonormalize it. First if v1 6= 0 set ei = v1 /kv1 k. Proceeding by induction we can suppose we have found, for a given integer n, elements ei , i = 1, . . . , m, where m ≤ n, which are orthonormal and such that the linear span (3.24) sp(e1 , . . . , em ) = sp(v1 , . . . , vn ). We certainly have this for n = 1. To show the inductive step observe that if vn+1 is in the span(s) in (3.24) then the same ei ’s work for n + 1. So we may as well assume that the next element, vn+1 is not in the span in (3.24). It follows that (3.25) w = vn+1 − n X w hvn+1 , ej iej 6= 0 so em+1 = kwk j=1 makes sense. By construction it is orthogonal to all the earlier ei ’s so adding em+1 gives the equality of the spans for n + 1. Thus we may continue indefinitely, since in fact the only way the dense set could be finite is if we were dealing with the space with one element, 0, in the first place. There are only two possibilities, either we get a finite set of ei ’s or an infinite sequence. In either case this must be a maximal orthonormal sequence. That is, we claim H 3 u ⊥ ej ∀ j =⇒ u = 0. (3.26) This uses the density of the vj ’s. There must exist a sequence wk where each wk is a vj , such that wk → u in H, assumed to satisfy (3.26). Now, each vj , and hence each wk , is a finite linear combination of el ’s so, by Bessel’s inequality X X (3.27) kwk k2 = |hwk , el i|2 = |hu − wk , el i|2 ≤ ku − wk k2 l l where hu, el i = 0 for all l has been used. Thus kwk k → 0 and hence u = 0.  Although a non-complete but separable pre-Hilbert space has maximal orthonormal sets, these are not much use without completeness. 5. Orthonormal bases Definition 3.3. In view of the following result, a maximal orthonormal sequence in a separable Hilbert space will be called an orthonormal basis; it is often called a ‘complete orthonormal basis’ but the ‘complete’ is really redundant. This notion of basis is not quite the same as in the finite dimensional case (although it is a legitimate extension of it). There are other, quite different, notions of a basis in infinite dimensions. See for instance ‘Hamel basis’ which arises in some settings – it is discussed briefly in §1.12 and can be used to show the existence of a noncontinuous functional on a Banach space. Theorem 3.2. If {ei } is an orthonormal basis (a maximal orthonormal sequence) in a Hilbert space then for any element u ∈ H the ‘Fourier-Bessel series’ converges to u : ∞ X (3.28) u= hu, ei iei . i=1 In particular a Hilbert space with an orthonormal basis is separable! Proof. The sequence of partial sums of the Fourier-Bessel series (3.29) uN = N X hu, ei iei i=1 is Cauchy. Indeed, if m < m0 then 0 (3.30) 2 kum0 − um k = m X |hu, ei i|2 ≤ i=m+1 X |hu, ei i|2 i>m which is small for large m by Bessel’s inequality. Since we are now assuming completeness, um → w in H. However, hum , ei i = hu, ei i as soon as m > i and |hw − un , ei i| ≤ kw − un k so in fact (3.31) hw, ei i = lim hum , ei i = hu, ei i m→∞ for each i. Thus u − w is orthogonal to all the ei so by the assumed completeness of the orthonormal basis must vanish. Thus indeed (3.28) holds.  6. Isomorphism to l2 A finite dimensional Hilbert space is isomorphic to Cn with its standard inner product. Similarly from the result above Proposition 3.2. Any infinite-dimensional separable Hilbert space (over the complex numbers) is isomorphic to l2 , that is there exists a linear map (3.32) T : H −→ l2 which is 1-1, onto and satisfies hT u, T vil2 = hu, viH and kT ukl2 = kukH for all u, v ∈ H. Proof. Choose an orthonormal basis – which exists by the discussion above – and set T u = {hu, ej i}∞ j=1 . (3.33) This maps H into l2 by Bessel’s inequality. Moreover, it is linear since the entries in the sequence are linear in u. It is 1-1 since T u = 0 implies hu, ej i = 0 for all j implies u = 0 by the assumed completeness of the orthonormal basis. It is surjective 2 since if {cj }∞ j=1 ∈ l then (3.34) u= ∞ X cj ej j=1 converges in H. This is the same argument as above – the sequence of partial sums is Cauchy since if n > m, k (3.35) n X j=m+1 cj ej k2H = n X |cj |2 . j=m+1 Again by continuity of the inner product, T u = {cj } so T is surjective. The equality of the norms follows from equality of the inner products and the latter follows by computation for finite linear combinations of the ej and then in general by continuity.  7. Parallelogram law What exactly is the difference between a general Banach space and a Hilbert space? It is of course the existence of the inner product defining the norm. In fact it is possible to formulate this condition intrinsically in terms of the norm itself. Proposition 3.3. In any pre-Hilbert space the parallelogram law holds – (3.36) kv + wk2 + kv − wk2 = 2kvk2 + 2kwk2 , ∀ v, w ∈ H. Proof. Just expand out using the inner product (3.37) kv + wk2 = kvk2 + hv, wi + hw, vi + kwk2 and the same for kv − wk2 and see the cancellation.  Proposition 3.4. Any normed space where the norm satisfies the parallelogram law, (3.36), is a pre-Hilbert space in the sense that (3.38) hv, wi =
1 kv + wk2 − kv − wk2 + ikv + iwk2 − ikv − iwk2 4 is a positive-definite Hermitian inner product which reproduces the norm. Proof. A problem below.  So, when we use the parallelogram law and completeness we are using the essence of the Hilbert space. 8. Convex sets and length minimizer The following result does not need the hypothesis of separability of the Hilbert space and allows us to prove the subsequent results – especially Riesz’ theorem – in full generality. Proposition 3.5. If C ⊂ H is a subset of a Hilbert space which is (1) Non-empty (2) Closed (3) Convex, in the sense that v1 , v2 ∈ C implies 12 (v1 + v2 ) ∈ C then there exists a unique element v ∈ C closest to the origin, i.e. such that (3.39) kvk = inf kuk. u∈C Proof. By definition of the infimum of a non-empty set of real numbers which is bounded below (in this case by 0) there must exist a sequence {vn } in C such that kvn k → d = inf u∈C kuk. We show that vn converges and that the limit is the point we want. The parallelogram law can be written (3.40) kvn − vm k2 = 2kvn k2 + 2kvm k2 − 4k(vn + vm )/2k2 . Since kvn k → d, given  > 0 if N is large enough then n > N implies 2kvn k2 < 2d2 + 2 /2. By convexity, (vn + vm )/2 ∈ C so k(vn + vm )/2k2 ≥ d2 . Combining these estimates gives (3.41) n, m > N =⇒ kvn − vm k2 ≤ 4d2 + 2 − 4d2 = 2 so {vn } is Cauchy. Since H is complete, vn → v ∈ C, since C is closed. Moreover, the distance is continuous so kvk = limn→∞ kvn k = d. Thus v exists and uniqueness follows again from the parallelogram law. If v and v 0 are two points in C with kvk = kv 0 k = d then (v + v 0 )/2 ∈ C so (3.42) kv − v 0 k2 = 2kvk2 + 2kv 0 k2 − 4k(v + v 0 )/2k2 ≤ 0 =⇒ v = v 0 . Alternatively you can just observe that we have actually shown above that any sequence in C such that kvn k → d converges, so this is true for the alternating sequence of v and v 0 .  9. Orthocomplements and projections Proposition 3.6. If W ⊂ H is a linear subspace of a Hilbert space then (3.43) W ⊥ = {u ∈ H; hu, wi = 0 ∀ w ∈ W } is a closed linear subspace and W ∩ W ⊥ = {0}. If W is also closed then (3.44) H = W ⊕ W⊥ meaning that any u ∈ H has a unique decomposition u = w + w⊥ where w ∈ W and w⊥ ∈ W ⊥ . Proof. That W ⊥ defined by (3.43) is a linear subspace follows from the linearity of the condition defining it. If u ∈ W ⊥ and u ∈ W then u ⊥ u by the definition so hu, ui = kuk2 = 0 and u = 0; thus W ∩ W ⊥ = {0}. Since the map H 3 u −→ hu, wi ∈ C is continuous for each w ∈ H its null space, the inverse image of 0, is closed. Thus \ (3.45) W⊥ = {u ∈ H; hu, wi = 0} w∈W is closed. Now, suppose W is closed. If W = H then W ⊥ = {0} and there is nothing to show. So consider u ∈ H, u ∈ / W and set (3.46) C = u + W = {u0 ∈ H; u0 = u + w, w ∈ W }. Then C is closed, since a sequence in it is of the form u0n = u + wn where wn is a sequence in W and u0n converges if and only if wn converges. Also, C is non-empty, since u ∈ C and it is convex since u0 = u + w0 and u00 = u + w00 in C implies (u0 + u00 )/2 = u + (w0 + w00 )/2 ∈ C. Thus the length minimization result above applies and there exists a unique v ∈ C such that kvk = inf u0 ∈C ku0 k. The claim is that this v is orthogonal to W – draw a picture in two real dimensions! To see this consider an aritrary point w ∈ W and λ ∈ C then v + λw ∈ C and (3.47) kv + λwk2 = kvk2 + 2 Re(λhv, wi) + |λ|2 kwk2 . Choose λ = teiθ where t is real and the phase is chosen so that eiθ hv, wi = |hv, wi| ≥ 0. Then the fact that kvk is minimal means that (3.48) kvk2 + 2t|hv, wi)| + t2 kwk2 ≥ kvk2 =⇒ t(2|hv, wi| + tkwk2 ) ≥ 0 ∀ t ∈ R =⇒ |hv, wi| = 0 which is what we wanted to show. Thus indeed, given u ∈ H \ W we have constructed v ∈ W ⊥ such that u = v + w, w ∈ W. This is (3.44) with the uniqueness of the decomposition already shown since it reduces to 0 having only the decomposition 0 + 0 and this in turn is W ∩ W ⊥ = {0}.  Since the construction in the preceding proof associates a unique element in W, a closed linear subspace, to each u ∈ H, it defines a map (3.49) ΠW : H −→ W. This map is linear, by the uniqueness since if ui = vi + wi , wi ∈ W, hvi , wi i = 0 are the decompositions of two elements then (3.50) λ1 u1 + λ2 u2 = (λ1 v1 + λ2 v2 ) + (λ1 w1 + λ2 w2 ) must be the corresponding decomposition. Moreover ΠW w = w for any w ∈ W and kuk2 = kvk2 + kwk2 , Pythagoras’ Theorem, shows that (3.51) Π2W = ΠW , kΠW uk ≤ kuk =⇒ kΠW k ≤ 1. Thus, projection onto W is an operator of norm 1 (unless W = {0}) equal to its own square. Such an operator is called a projection or sometimes an idempotent (which sounds fancier). Finite-dimensional subspaces are always closed by the Heine-Borel theorem. Lemma 3.3. If {ej } is any finite or countable orthonormal set in a Hilbert space then the orthogonal projection onto the closure of the span of these elements is X (3.52) Pu = hu, ek iek . Proof. We know that the series in (3.52) converges and defines a bounded linear operator of norm at most one by Bessel’s inequality. Clearly P 2 = P by the same argument. If W is the closure of the span then (u−P u) ⊥ W since (u−P u) ⊥ ek for each k and the inner product is continuous. Thus u = (u − P u) + P u is the orthogonal decomposition with respect to W.  Lemma 3.4. If W ⊂ H is a linear subspace of a Hilbert space which contains the orthocomplement of a finite dimensional space then W is closed and W ⊥ is finite-dimensional. Proof. If U ⊂ W is a closed subspace with finite-dimensional orthocomplement then each of the N elements, vi , of a basis of (Id −ΠU )W is the image of some wi ∈ W. Since U is the null space of Id −ΠU it follows that any element of W can be written uniquely in the form (3.53) w =u+ N X ci vi , u = ΠU w ∈ U, ci = hw, vi i. i=1 Then if φn is a sequence in W which converges in H it follows that ΠU φn converges in U and hφn , vi i converges and hence the limit is in W.  Note that the existence of a non-continuous linear functional H −→ C is equivalent to the existence of a non-closed subspace of H with a one-dimensional complement. Namely the null space of a non-continuous linear functional cannot be closed, since from this continuity follows, but it does have a one-dimensional complement (not orthocomplement!) Question 1. Does there exist a non-continuous linear functional on an infinitedimensional Hilbert space? 1 10. Riesz’ theorem The most important application of the convexity result above is to prove Riesz’ representation theorem (for Hilbert space, there is another one to do with measures). Theorem 3.3. If H is a Hilbert space then for any continuous linear functional T : H −→ C there exists a unique element φ ∈ H such that (3.54) T (u) = hu, φi ∀ u ∈ H. Proof. If T is the zero functional then φ = 0 gives (3.54). Otherwise there exists some u0 ∈ H such that T (u0 ) 6= 0 and then there is some u ∈ H, namely u = u0 /T (u0 ) will work, such that T (u) = 1. Thus (3.55) C = {u ∈ H; T (u) = 1} = T −1 ({1}) 6= ∅. The continuity of T implies that C is closed, as the inverse image of a closed set under a continuous map. Moreover C is convex since (3.56) T ((u + u0 )/2) = (T (u) + T (u0 ))/2. Thus, by Proposition 3.5, there exists an element v ∈ C of minimal length. Notice that C = {v + w; w ∈ N } where N = T −1 ({0}) is the null space of T. Thus, as in Proposition 3.6 above, v is orthogonal to N. In this case it is the unique element orthogonal to N with T (v) = 1. 1The existence of such a functional requires some form of the Axiom of Choice (maybe a little weaker in the separable case). You are free to believe that all linear functionals are continuous but you will make your life difficult this way. Now, for any u ∈ H, (3.57) u − T (u)v satisfies T (u − T (u)v) = T (u) − T (u)T (v) = 0 =⇒ u = w + T (u)v, w ∈ N. Then, hu, vi = T (u)kvk2 since hw, vi = 0. Thus if φ = v/kvk2 then (3.58) u = w + hu, φiv =⇒ T (u) = hu, φiT (v) = hu, φi.  11. Adjoints of bounded operators As an application of Riesz’ Theorem we can see that to any bounded linear operator on a Hilbert space (3.59) A : H −→ H, kAuk ≤ Ckuk ∀ u ∈ H there corresponds a unique adjoint operator. This has profound consequences for the theory of operators on a Hilbert space, as we shall see. Proposition 3.7. For any bounded linear operator A : H −→ H on a Hilbert space there is a unique bounded linear operator A∗ : H −→ H such that (3.60) hAu, viH = hu, A∗ viH ∀ u, v ∈ H and kAk = kA∗ k. Proof. To see the existence of A∗ v we need to work out what A∗ v ∈ H should be for each fixed v ∈ H. So, fix v in the desired identity (3.60), which is to say consider H 3 u −→ hAu, vi ∈ C. (3.61) This is a linear map and it is clearly bounded, since (3.62) |hAu, vi| ≤ kAukkvk ≤ (kAkkvk)kuk. Thus it is a continuous linear functional on H which depends on v. In fact it is just the composite of two continuous linear maps (3.63) u7−→Au H −→ H w7−→hw,vi −→ C. By Riesz’ theorem there is a unique element in H, which we can denote A∗ v (since it only depends on v) such that (3.64) hAu, vi = hu, A∗ vi ∀ u ∈ H. This defines the map A∗ : H −→ H but we need to check that it is linear and continuous. Linearity follows from the uniqueness part of Riesz’ theorem. Thus if v1 , v2 ∈ H and c1 , c2 ∈ C then (3.65) hAu, c1 v1 + c2 v2 i = c1 hAu, v1 i + c2 hAu, v2 i = c1 hu, A∗ v1 i + c2 hu, A∗ v2 i = hu, c1 A∗ v2 + c2 A∗ v2 i where we have used the definitions of A∗ v1 and A∗ v2 – by uniqueness we must have A∗ (c1 v1 + c2 v2 ) = c1 A∗ v1 + c2 A∗ v2 . Using the optimality of Cauchy’s inequality (3.66) kA∗ vk = sup |hu, A∗ vi| = sup |hAu, vi| ≤ kAkkvk. kuk=1 kuk=1 This shows that A∗ is bounded and that (3.67) kA∗ k ≤ kAk. The defining identity (3.60) also shows that (A∗ )∗ = A so the reverse equality in (3.67) also holds and therefore (3.68) kA∗ k = kAk.  One useful property of the adjoint operator is that (3.69) Nul(A∗ ) = (Ran(A))⊥ . Indeed w ∈ (Ran(A))⊥ means precisely that hw, Avi = 0 for all v ∈ H which translates to (3.70) w ∈ (Ran(A))⊥ ⇐⇒ hA∗ w, vi = 0 ⇐⇒ A∗ w = 0. Note that in the finite dimensional case (3.69) is equivalent to Ran(A) = (Nul(A∗ ))⊥ but in the infinite dimensional case Ran(A) is often not closed in which case this cannot be true and you can only be sure that (3.71) Ran(A) = (Nul(A∗ ))⊥ . 12. Compactness and equi-small tails A compact subset in a general metric space is one with the property that any sequence in it has a convergent subsequence, with its limit in the set. You will recall, with pleasure no doubt, the equivalence of this condition to the (more general since it makes good sense in an arbitrary topological space) covering condition, that any open cover of the set has a finite subcover. So, in a Hilbert space the notion of a compact set is already fixed. We want to characterize it, actually in two closely related ways. In any metric space a compact set is both closed and bounded, so this must be true in a Hilbert space. The Heine-Borel theorem gives a converse to this, for Rn or Cn (and hence in any finite-dimensional normed space) any closed and bounded set is compact. Also recall that the convergence of a sequence in Cn is equivalent to the convergence of the n sequences given by its components and this is what is used to pass first from R to C and then to Cn . All of this fails in infinite dimensions and we need some condition in addition to being bounded and closed for a set to be compact. To see where this might come from, observe that Lemma 3.5. In any metric space the set, S, consisting of the points of a convergent sequence, together with its limit, is compact. Proof. We show that S is compact by checking that any sequence in S has a convergent subsequence. To see this, observe that a sequence {tj } in S either has a subsequence converging to the limit, s, of the original sequence or it does not. So we only need consider the latter case, but this means that, for some  > 0, d(tj , s) > ; but then tj takes values in a finite set, since S \ B(s, ) is finite – hence some value is repeated infinitely often and there is a convergent subsequence.  Lemma 3.6. The image of a convergent sequence in a Hilbert space is a set with equi-small tails with respect to any orthonormal sequence, i.e. if ek is an othonormal sequence and un → u is a convergent sequence then given  > 0 there exists N such that X (3.72) |hun , ek i|2 < 2 ∀ n. k>N Proof. Bessel’s inequality shows that for any u ∈ H, X (3.73) |hu, ek i|2 ≤ kuk2 . k The convergence of this series means that (3.72) can be arranged for any single element un or the limit u by choosing N large enough, thus given  > 0 we can choose N 0 so that X (3.74) |hu, ek i|2 < 2 /2. k>N 0 Consider the closure of the subspace spanned by the ek with k > N. The orthogonal projection onto this space (see Lemma 3.3) is X (3.75) PN u = hu, ek iek . k>N Then the convergence un → u implies the convergence in norm kPN un k → kPN uk, so X (3.76) kPN un k2 = |hun , ek i|2 < 2 , n > n0 . k>N So, we have arranged (3.72) for n > n0 for some N. This estimate remains valid if N is increased – since the tails get smaller – and we may arrange it for n ≤ n0 by choosing N large enough. Thus indeed (3.72) holds for all n if N is chosen large enough.  This suggests one useful characterization of compact sets in a separable Hilbert space since the equi-smallness of the tails, as in (3.72), for all u ∈ K just means that the Fourier-Bessel series converges uniformly. Proposition 3.8. A set K ⊂ H in a separable Hilbert space is compact if and only if it is bounded, closed and the Fourier-Bessel sequence with respect to any (one) complete orthonormal basis converges uniformly on it. Proof. We already know that a compact set in a metric space is closed and bounded. Suppose the equi-smallness of tails condition fails with respect to some orthonormal basis ek . This means that for some  > 0 and all p there is an element up ∈ K, such that X (3.77) |hup , ek i|2 ≥ 2 . k>p Consider the subsequence {up } generated this way. No subsequence of it can have equi-small tails (recalling that the tail decreases with p). Thus, by Lemma 3.6, it cannot have a convergent subsequence, so K cannot be compact if the equismallness condition fails. Thus we have proved the equi-smallness of tails condition to be necessary for the compactness of a closed, bounded set. It remains to show that it is sufficient. So, suppose K is closed, bounded and satisfies the equi-small tails condition with respect to an orthonormal basis ek and {un } is a sequence in K. We only need show that {un } has a Cauchy subsequence, since this will converge (H being complete) and the limit will be in K (since it is closed). Consider each of the sequences of coefficients hun , ek ) in C. Here k is fixed. This sequence is bounded: |hun , ek i| ≤ kun k ≤ C (3.78) by the boundedness of K. So, by the Heine-Borel theorem, there is a subsequence unl such that hunl , ek i converges as l → ∞. We can apply this argument for each k = 1, 2, . . . . First extract a subsequence {un,1 } of {un } so that the sequence hun,1 , e1 i converges. Then extract a subsequence un,2 of un,1 so that hun,2 , e2 i also converges. Then continue inductively. Now pass to the ‘diagonal’ subsequence vn of {un } which has kth entry the kth term, uk,k in the kth subsequence. It is ‘eventually’ a subsequence of each of the subsequences previously constructed – meaning it coincides with a subsequence from some point onward (namely the kth term onward for the kth subsquence). Thus, for this subsequence each of the hvn , ek i converges. Consider the identity (the orthonormal set ek is complete by assumption) for the difference X X kvn − vn+l k2 = |hvn − vn+l , ek i|2 + |hvn − vn+l , ek i|2 k≤N (3.79) ≤ X k>N 2 |hvn − vn+l , ek i| + 2 k≤N X |hvn , ek i|2 + 2 k>N X |hvn+l , ek i|2 k>N where the parallelogram law on C has been used. To make this sum less than 2 we may choose N so large that the last two terms are less than 2 /2 and this may be done for all n and l by the equi-smallness of the tails. Now, choose n so large that each of the terms in the first sum is less than 2 /2N, for all l > 0 using the Cauchy condition on each of the finite number of sequence hvn , ek i. Thus, {vn } is a Cauchy subsequence of {un } and hence as already noted convergent in K. Thus K is indeed compact.  This criterion for compactness is useful but is too closely tied to the existence of an orthonormal basis to be easily applicable. However the condition can be restated in a way that holds even in the non-separable case (and of course in the finite-dimensional case, where it is trivial). Proposition 3.9. A subset K ⊂ H of a Hilbert space is compact if and only if it is closed and bounded and for every  > 0 there is a finite-dimensional subspace W ⊂ H such that (3.80) sup inf ku − wk < . u∈K w∈W So we see that the extra condition needed is ‘finite-dimensional approximability’. Proof. Before proceeding to the proof consider (3.80). Since W is finitedimensional we know it is closed and hence the discussion in §9 above applies. In particular u = w + w⊥ with w ∈ W and w⊥ ⊥ W where (3.81) inf ku − wk = kw⊥ k. w∈W This can be restated in the form (3.82) sup k(Id −ΠW )uk <  u∈K where ΠW is the orthogonal projection onto W (so Id −ΠW is the orthogonal projection onto W ⊥ ). Now, let us first assume that H is separable, so we already have a condition for compactness in Proposition 3.8. Then if K is compact we can consider an orthonormal basis of H and the finite-dimensional spaces WN spanned by the first N elements in the basis with ΠN the orthogonal projection onto it. Then k(Id −ΠN )uk is precisely the length of the ‘tail’ of u with respect to the basis. So indeed, by Proposition 3.8, given  > 0 there is an N such that k(Id −ΠN )uk < /2 for all u ∈ K and hence (3.82) holds for W = WN . Now suppose that K ⊂ H and for each  > 0 we can find a finite dimensional subspace W such that (3.82) holds. Take a sequence {un } in K. The sequence ΠW un ∈ W is bounded in a finite-dimensional space so has a convergent subsequence. Now, for each j ∈ N there is a finite-dimensional subspace Wj (not necessarily corresponding to an orthonormal basis) so that (3.82) holds for  = 1/j. Proceeding as above, we can find successive subsequence of un such that the image under Πj in Wj converges for each j. Passing to the diagonal subsequence unl it follows that Πj uni converges for each j since it is eventually a subsequence of the jth choice of subsequence above. Now, the triangle inequality shows that (3.83) kuni − unk k ≤ kΠj (uni − unk )kWj + k(Id −Πj )uni k + k(Id −Πj )unk k. Given  > 0 first choose j so large that the last two terms are each less than 1/j < /3 using the choice of Wj . Then if i, k > N is large enough the first term on the right in (3.83) is also less than /3 by the convergence of Πj uni . Thus uni is Cauchy in H and hence converges and it follows that K is compact. This converse argument does not require the separability of H so to complete the proof we only need to show the necessity of (3.81) in the non-separable case. Thus suppose K is compact. Then K itself is separable – has a countable dense subset – using the finite covering property (for each p > 0 there are finitely many balls of radius 1/p which cover K so take the set consisting of all the centers for all p). It follows that the closure of the span of K, the finite linear combinations of elements of K, is a separable Hilbert subspace of H which contains K. Thus any compact subset of a non-separable Hilbert space is contained in a separable Hilbert subspace and hence (3.80) holds.  13. Finite rank operators Now, we need to starting thinking a little more seriously about operators on a Hilbert space, remember that an operator is just a continuous linear map T : H −→ H and the space of them (a Banach space) is denoted B(H) (rather than the more cumbersome B(H, H) which is needed when the domain and target spaces are different). Definition 3.4. An operator T ∈ B(H) is of finite rank if its range has finite dimension (and that dimension is called the rank of T ); the set of finite rank operators will be denoted R(H). Why not F(H)? Because we want to use this for the Fredholm operators. Clearly the sum of two operators of finite rank has finite rank, since the range is contained in the sum of the ranges (but is often smaller): (T1 + T2 )u ∈ Ran(T1 ) + Ran(T2 ) ∀ u ∈ H. (3.84) Since the range of a constant multiple of T is contained in the range of T it follows that the finite rank operators form a linear subspace of B(H). What does a finite rank operator look like? It really looks like a matrix. Lemma 3.7. If T : H −→ H has finite rank then there is a finite orthonormal set {ek }L k=1 in H and constants cij ∈ C such that (3.85) Tu = L X cij hu, ej iei . i,j=1 Proof. By definition, the range of T, R = T (H) is a finite dimensional subspace. So, it has a basis which we can diagonalize in H to get an orthonormal basis, ei , i = 1, . . . , p. Now, since this is a basis of the range, T u can be expanded relative to it for any u ∈ H : p X (3.86) Tu = hT u, ei iei . i=1 On the other hand, the map u −→ hT u, ei i is a continuous linear functional on H, so hT u, ei i = hu, vi i for some vi ∈ H; notice in fact that vi = T ∗ ei . This means the formula (3.86) becomes (3.87) Tu = p X hu, vi iei . i=1 Now, the Gram-Schmidt procedure can be applied to orthonormalize the sequence e1 , . . . , ep , v1 . . . , vp resulting in e1 , . . . , eL . This means that each vi is a linear combination which we can write as L X (3.88) vi = cij ej . j=1 Inserting this into (3.87) gives (3.85) (where the constants for i > p are zero).  It is clear that (3.89) B ∈ B(H) and T ∈ R(H) then BT ∈ R(H). Indeed, the range of BT is the range of B restricted to the range of T and this is certainly finite dimensional since it is spanned by the image of a basis of Ran(T ). Similalry T B ∈ R(H) since the range of T B is contained in the range of T. Thus we have in fact proved most of Proposition 3.10. The finite rank operators form a ∗-closed two-sided ideal in B(H), which is to say a linear subspace such that (3.90) B1 , B2 ∈ B(H), T ∈ R(H) =⇒ B1 T B2 , T ∗ ∈ R(H). Proof. It is only left to show that T ∗ is of finite rank if T is, but this is an immediate consequence of Lemma 3.7 since if T is given by (3.85) then (3.91) T ∗u = N X cij hu, ei iej i,j=1 is also of finite rank.  Lemma 3.8 (Row rank=Colum rank). For any finite rank operator on a Hilbert space, the dimension of the range of T is equal to the dimension of the range of T ∗ . Proof. From the formula (3.87) for a finite rank operator, it follows that the vi , i = 1, . . . , p must be linearly independent – since the ei form a basis for the range and a linear relation between the vi would show the range had dimension less than p. Thus in fact the null space of T is precisely the orthocomplement of the span of the vi – the space of vectors orthogonal to each vi . Since hT u, wi = p X hu, vi ihei , wi =⇒ i=1 (3.92) p X hw, T ui = hvi , uihw, ei i =⇒ i=1 p X T w= hw, ei ivi ∗ i=1 the range of T ∗ is the span of the vi , so is also of dimension p.  14. Compact operators Definition 3.5. An element K ∈ B(H), the bounded operators on a separable Hilbert space, is said to be compact (the old terminology was ‘totally bounded’ or ‘completely continuous’) if the image of the unit ball is precompact, i.e. has compact closure – that is if the closure of K{u ∈ H; kukH ≤ 1} is compact in H. Notice that in a metric space, to say that a set has compact closure is the same as saying it is contained in a compact set; such a set is said to be precompact. Proposition 3.11. An operator K ∈ B(H), bounded on a separable Hilbert space, is compact if and only if it is the limit of a norm-convergent sequence of finite rank operators. Proof. So, we need to show that a compact operator is the limit of a convergent sequence of finite rank operators. To do this we use the characterizations of compact subsets of a separable Hilbert space discussed earlier. Namely, if {ei } is an orthonormal basis of H then a subset I ⊂ H is compact if and only if it is closed and bounded and has equi-small tails with respect to {ei }, meaning given  > 0 there exits N such that X (3.93) |hv, ei i|2 < 2 ∀ v ∈ I. i>N Now we shall apply this to the set K(B(0, 1)) where we assume that K is compact (as an operator, don’t be confused by the double usage, in the end it turns out to be constructive) – so this set is contained in a compact set. Hence (3.93) applies to it. Namely this means that for any  > 0 there exists n such that X (3.94) |hKu, ei i|2 < 2 ∀ u ∈ H, kukH ≤ 1. i>n For each n consider the first part of these sequences and define X (3.95) Kn u = hKu, ei iei . k≤n This is clearly a linear operator and has finite rank – since its range is contained in the span of the first n elements of {ei }. Since this is an orthonormal basis, X (3.96) kKu − Kn uk2H = |hKu, ei i|2 i>n Thus (3.94) shows that kKu − Kn ukH ≤ . Now, increasing n makes kKu − Kn uk smaller, so given  > 0 there exists n such that for all N ≥ n, kK − KN kB = sup kKu − Kn ukH ≤ . (3.97) kuk≤1 Thus indeed, Kn → K in norm and we have shown that the compact operators are contained in the norm closure of the finite rank operators. For the converse we assume that Tn → K is a norm convergent sequence in B(H) where each of the Tn is of finite rank – of course we know nothing about the rank except that it is finite. We want to conclude that K is compact, so we need to show that K(B(0, 1)) is precompact. It is certainly bounded, by the norm of K. Let Wn = Tn H be the range of Tn . By definition it is a finite dimensional subspace and hence closed. Let Πn be the orthogonal projection onto Wn , so Id −Πn is projection onto Wn⊥ . Thus the composite (Id −Πn )Tn = 0 and hence (3.98) (Id −Πn )K = (Id −Πn )(K − Tn ) =⇒ k(Id −Πn )Kk → 0 as n → ∞. So, for any  > 0 there exists n such that (3.99) sup inf kKu − wk ≤ sup k(Id −Πn )Kuk <  u∈B(0,1) w∈Wn kuk≤1 and it follows from Proposition 3.9 that K(B(0, 1)) is precompact and hence K is compact.  Proposition 3.12. For any separable Hilbert space, the compact operators form a closed and ∗-closed two-sided ideal in B(H). Proof. In any metric space (applied to B(H)) the closure of a set is closed, so the compact operators are closed being the closure of the finite rank operators. Similarly the fact that it is closed under passage to adjoints follows from the same fact for finite rank operators. The ideal properties also follow from the corresponding properties for the finite rank operators, or we can prove them directly anyway. Namely if B is bounded and T is compact then for some c > 0 (namely 1/kBk unless it is zero) cB maps B(0, 1) into itself. Thus cT B = T cB is compact since the image of the unit ball under it is contained in the image of the unit ball under T ; hence T B is also compact. Similarly BT is compact since B is continuous and then (3.100) BT (B(0, 1)) ⊂ B(T (B(0, 1))) is compact since it is the image under a continuous map of a compact set.  15. Weak convergence It is convenient to formalize the idea that a sequence be bounded and that each of the hun , ek i, the sequence of coefficients of some particular Fourier-Bessel series, should converge. Definition 3.6. A sequence, {un }, in a Hilbert space, H, is said to converge weakly to an element u ∈ H if it is bounded in norm and huj , vi → hu, vi converges in C for each v ∈ H. This relationship is written (3.101) un * u. In fact as we shall see below, the assumption that kun k is bounded and that u exists are both unnecessary. That is, a sequence converges weakly if and only if hun , vi converges in C for each v ∈ H. Conversely, there is no harm in assuming it is bounded and that the ‘weak limit’ u ∈ H exists. Note that the weak limit is unique since if u and u0 both have this property then hu − u0 , vi = limn→∞ hun , vi − limn→∞ hun , vi = 0 for all v ∈ H and setting v = u − u0 it follows that u = u0 . Lemma 3.9. A (strongly) convergent sequence is weakly convergent with the same limit. Proof. This is the continuity of the inner product. If un → u then |hun , vi − hu, vi| ≤ kun − ukkvk → 0 (3.102) for each v ∈ H shows weak convergence.  Lemma 3.10. For a bounded sequence in a separable Hilbert space, weak convergence is equivalent to component convergence with respect to an orthonormal basis. Proof. Let ek be an orthonormal basis. Then if un is weakly convergent it follows immediately that hun , ek i → hu, ek i converges for each k. Conversely, suppose this is true for a bounded sequence, just that hun , ek i → ck in C for each k. The norm boundedness and Bessel’s inequality show that X X (3.103) |ck |2 = lim |hun , ek i|2 ≤ sup kun k2 k≤p n→∞ k≤p n for all p. Thus in fact {ck } ∈ l2 and hence X (3.104) u= ck ek ∈ H k by the completeness of H. Clearly hun , ek i → hu, ek i for each k. It remains to show that hun , vi → hu, vi for all v ∈ H. This is certainly true for any finite linear combination of the ek and for a general v we can write (3.105) hun , vi − hu, vi = hun , vp i − hu, vp i + hun , v − vp i − hu, v − vp i =⇒ |hun , vi − hu, vi| ≤ |hun , vp i − hu, vp i| + 2Ckv − vp k where vp = P hv, ek iek is a finite part of the Fourier-Bessel series for v and C is a k≤p bound for kun k. Now the convergence vp → v implies that the last term in (3.105) can be made small by choosing p large, independent of n. Then the second last term can be made small by choosing n large since vp is a finite linear combination of the ek . Thus indeed, hun , vi → hu, vi for all v ∈ H and it follows that un converges weakly to u.  Proposition 3.13. Any bounded sequence {un } in a separable Hilbert space has a weakly convergent subsequence. This can be thought of as different extension to infinite dimensions of the HeineBorel theorem. As opposed to the characterization of compact sets above, which involves adding the extra condition of finite-dimensional approximability, here we weaken the notion of convergence. Proof. Choose an orthonormal basis {ek } and apply the procedure in the proof of Proposition 3.8 to it. Thus, we may extract successive subsequence along the kth of which hunp , ek i → ck ∈ C. Passing to the diagonal subsequence, vn , which is eventually a subsequence of each of these ensures that hvn , ek i → ck for each k. Now apply the preceeding Lemma to conclude that this subsequence converges weakly.  Lemma 3.11. For a weakly convergent sequence un * u kuk ≤ lim inf kun k (3.106) and a weakly convergent sequence converges strongly if and only if the weak limit satisfies kuk = limn→∞ kun k. Proof. Choose an orthonormal basis ek and observe that X X |hun , ek i|2 . (3.107) |hu, ek i|2 = lim n→∞ k≤p k≤p 2 The sum on the right is bounded by kun k independently of p so X (3.108) |hu, ek i|2 ≤ lim inf kun k2 n k≤p by the definition of lim inf . Then let p → ∞ to conclude that (3.109) kuk2 ≤ lim inf kun k2 n from which (3.106) follows. Now, suppose un * u then (3.110) ku − un k2 = kuk2 − 2 Rehu, un i + kun k2 . Weak convergence implies that the middle term converges to −2kuk2 so if the last term converges to kuk2 then u → un .  Observe that for any A ∈ B(H), if un * u then Aun * Au using the existence of the adjoint:(3.111) hAun , vi = hun , A∗ vi → hu, A∗ vi = hAu, vi ∀ v ∈ H. Lemma 3.12. An operator K ∈ B(H) is compact if and only if the image Kun of any weakly convergent sequence {un } in H is strongly, i.e. norm, convergent. This is the origin of the old name ‘completely continuous’ for compact operators, since they turn even weakly convergent into strongly convergent sequences. Proof. First suppose that un * u is a weakly convergent sequence in H and that K is compact. We know that kun k < C is bounded so the sequence Kun is contained in CK(B(0, 1)) and hence in a compact set (clearly if D is compact then so is cD for any constant c.) Thus, any subsequence of Kun has a convergent subsequence and the limit is necessarily Ku since Kun * Ku. But the condition on a sequence in a metric space that every subsequence of it has a subsequence which converges to a fixed limit implies convergence. (If you don’t remember this, reconstruct the proof: To say a sequence vn does not converge to v is to say that for some  > 0 there is a subsequence along which d(vnk , v) ≥ . This is impossible given the subsequence of subsequence condition converging to the fixed limit v.) Conversely, suppose that K has this property of turning weakly convergent into strongly convergent sequences. We want to show that K(B(0, 1)) has compact closure. This just means that any sequence in K(B(0, 1)) has a (strongly) convergent subsequence – where we do not have to worry about whether the limit is in the set or not. Such a sequence is of the form Kun where un is a sequence in B(0, 1). However we know that we can pass to a subsequence which converges weakly, unj * u. Then, by the assumption of the Lemma, Kunj → Ku converges strongly. Thus un does indeed have a convergent subsequence and hence K(B(0, 1)) must have compact closure.  As noted above, it is not really necessary to assume that a weakly convergent sequence in a Hilbert space is bounded, provided one has the Uniform Boundedness Principle, Theorem 1.3, at the ready. Proposition 3.14. If un ∈ H is a sequence in a Hilbert space and for all v∈H (3.112) hun , vi → F (v) converges in C then kun kH is bounded and there exists w ∈ H such that un * w. Proof. Apply the Uniform Boundedness Theorem to the continuous functionals (3.113) Tn (u) = hu, un i, Tn : H −→ C where we reverse the order to make them linear rather than anti-linear. Thus, each set |Tn (u)| is bounded in C since it is convergent. It follows from the Uniform Boundedness Principle that there is a bound (3.114) kTn k ≤ C. However, this norm as a functional is just kTn k = kun kH so the original sequence must be bounded in H. Define T : H −→ C as the limit for each u : (3.115) T (u) = lim Tn (u) = lim hu, un i. n→∞ n→∞ This exists for each u by hypothesis. It is a linear map and from (3.114) it is bounded, kT k ≤ C. Thus by the Riesz Representation theorem, there exists w ∈ H such that (3.116) T (u) = hu, w) ∀ u ∈ H. Thus hun , ui → hw, ui for all u ∈ H so un * w as claimed.  16. The algebra B(H) Recall the basic properties of the Banach space, and algebra, of bounded operators B(H) on a separable Hilbert space H. In particular that it is a Banach space with respect to the norm kAk = (3.117) sup kAukH kukH =1 and that the norm satisfies kABk ≤ kAkkBk (3.118) as follows from the fact that kABuk ≤ kAkkBuk ≤ kAkkBkkuk. Consider the set of invertible elements: (3.119) GL(H) = {A ∈ B(H); ∃ B ∈ B(H), BA = AB = Id}. Note that this is equivalent to saying A is 1-1 and onto in view of the Open Mapping Theorem, Theorem 1.4. This set is open, to see this consider a neighbourhood of the identity. Lemma 3.13. If A ∈ B(H) and kAk < 1 then Id −A ∈ GL(H). (3.120) Proof. This follows from the convergence of the Neumann series. If kAk < 1 then kAj k ≤ kAkj , from (3.118), and it follows that ∞ X Aj (3.121) B= j=0 (where A0 = Id by definition) is absolutely summable in B(H) since ∞ P kAj k con- j=0 verges. Since B(H) is a Banach space, the sum converges. Moreover by the continuity of the product with respect to the norm (3.122) AB = A lim n→∞ n X Aj = lim j=0 n→∞ n+1 X Aj = B − Id j=1 and similalry BA = B − Id . Thus (Id −A)B = B(Id −A) = Id shows that B is a (and hence the) 2-sided inverse of Id −A.  Proposition 3.15. The invertible elements form an open subset GL(H) ⊂ B(H). Proof. Suppose G ∈ GL(H), meaning it has a two-sided (and unique) inverse G−1 ∈ B(H) : (3.123) G−1 G = GG−1 = Id . Then we wish to show that B(G; ) ⊂ GL(H) for some  > 0. In fact we shall see that we can take  = kG−1 k−1 . To show that G + B is invertible set (3.124) E = −G−1 B =⇒ G + B = G(Id +G−1 B) = G(Id −E) From Lemma 3.13 we know that (3.125) kBk < 1/kG−1 k =⇒ kG−1 Bk < 1 =⇒ Id −E is invertible. Then (Id −E)−1 G−1 satisfies (3.126) (Id −E)−1 G−1 (G + B) = (Id −E)−1 (Id −E) = Id . Moreover E 0 = −BG−1 also satisfies kE 0 k ≤ kBkkG−1 k < 1 and (3.127) (G + B)G−1 (Id −E 0 )−1 = (Id −E 0 )(Id −E 0 )−1 = Id . Thus G + B has both a ‘left’ and a ‘right’ inverse. The associativity of the operator product (that A(BC) = (AB)C) then shows that (3.128) G−1 (Id −E 0 )−1 = (Id −E)−1 G−1 (G+B)G−1 (Id −E 0 )−1 = (Id −E)−1 G−1 so the left and right inverses are equal and hence G + B is invertible.  Thus GL(H) ⊂ B(H), the set of invertible elements, is open. It is also a group −1 – since the inverse of G1 G2 if G1 , G2 ∈ GL(H) is G−1 2 G1 . This group of invertible elements has a smaller subgroup, U(H), the unitary group, defined by (3.129) U(H) = {U ∈ GL(H); U −1 = U ∗ }. The unitary group consists of the linear isometric isomorphisms of H onto itself – thus (3.130) hU u, U vi = hu, vi, kU uk = kuk ∀ u, v ∈ H, U ∈ U(H). This is an important object and we will use it a little bit later on. The groups GL(H) and U(H) for a separable Hilbert space may seem very similar to the familiar groups of invertible and unitary n × n matrices, GL(n) and U(n), but this is somewhat deceptive. For one thing they are much bigger. In fact there are other important qualitative differences. One important fact that you should know, and there is a proof towards the end of this chapter, is that both GL(H) and U(H) are contractible as metric spaces – they have no significant topology. This is to be constrasted with the GL(n) and U(n) which have a lot of topology, and are not at all simple spaces – especially for large n. One upshot of this is that U(H) does not look much like the limit of the U(n) as n → ∞. In fact there is another group which is essentially the large n limit of the U(n), namely (3.131) U−∞ (H) = {Id +K ∈ U(H); K ∈ K(H)}. It does have lots of interesting (and useful) topology. Another important fact that we will discuss below is that GL(H) is not dense in B(H), in contrast to the finite dimensional case. In other words there are operators which are not invertible and cannot be made invertible by small perturbations. 17. Spectrum of an operator Another direct application of Lemma 3.13, the convergence of the Neumann series, is that if A ∈ B(H) and λ ∈ C has |λ| > kAk then kλ−1 Ak < 1 so (Id −λ−1 A)−1 exists and satisfies (3.132) (λ Id −A)λ−1 (Id −λ−1 A)−1 = Id = λ−1 (Id −λ−1 A)−1 (λ − A). Thus, λ Id −A ∈ GL(H), which we usually abbreviate to λ − A, has inverse (λ − A)−1 = λ−1 (Id −λ−1 A)−1 . The set of λ for which this operator is invertible is called the resolvent set and we have shown Res(A) = {λ ∈ C; (λ Id −A) ∈ GL(H)} ⊂ C (3.133) {|λ| > kAk} ⊂ Res(A). From the discussion above, it is an open, and non-empty, set on which (A−λ)−1 , called the resolvent of A, is defined. The complement of the resolvent set is called the spectrum of A (3.134) Spec(A) = {λ ∈ C; λ Id −A ∈ / GL(H)} ⊂ {λ ∈ C; |λ| ≤ kAk}. As follows from the discussion above it is a compact set – in fact it cannot be empty. This is quite easy to see if you know a little complex analysis since it follows from Liouville’s Theorem. One way to show that λ ∈ Spec(A) is to check that λ − A is not injective, since then it cannot be invertible. This means precisely that λ is an eigenvalue of A : ∃ 0 6= u ∈ H s.t. Au = λu. (3.135) However, you should strongly resist the temptation to think that the spectrum is the set of eigenvalues of A, this is sometimes but by no means always true. The other way to show that λ ∈ Spec(A) is to prove that λ − A is not surjective. Note that by the Open Mapping Theorem if λ − A is both surjective and injective then λ ∈ Res(A). For a finite rank operator the spectrum does consist of the set of eigenvalues. For a bounded self-adjoint operator we can say more quite a bit more. Proposition 3.16. If A : H −→ H is a bounded operator on a Hilbert space and A∗ = A then A − λ Id is invertible for all λ ∈ C \ [−kAk, kAk] and conversely at least one of A − kAk Id and A + kAk Id is not invertible. The proof of this depends on a different characterization of the norm in the self-adjoint case. Lemma 3.14. If A∗ = A ∈ B(H) then kAk = sup |hAu, ui|. (3.136) kuk=1 Proof. Certainly, |hAu, ui| ≤ kAkkuk2 so the right side can only be smaller than or equal to the left. Set a = sup |hAu, ui| ≤ kAk. kuk=1 Then for any u, v ∈ H, |hAu, vi| = hAeiθ u, vi for some θ ∈ [0, 2π), so we can arrange that hAu, vi = |hAu0 , vi| is non-negative and ku0 k = 1 = kuk = kvk. Dropping the primes and computing using the polarization identity (3.137) 4hAu, vi = hA(u + v), u + vi − hA(u − v), u − vi + ihA(u + iv), u + ivi − ihA(u − iv), u − ivi. By the reality of the left side we can drop the last two terms and use the bound |hAw, wi| ≤ akwkw on the first two to see that (3.138) 4hAu, vi ≤ a(ku + vk2 + ku − vk2 ) = 2a(kuk2 + kvk2 ) = 4a Thus, kAk = supkuk=kvk=1 |hAu, vi| ≤ a and hence kAk = a.  This suggests an improvement on the last part of the statement of Proposition 3.16, namely If A∗ = A ∈ B(H) then a− , a+ ∈ Spec(A) and Spec(A) ⊂ [a− , a+ ] (3.139) where a− = inf hAu, ui, a+ = sup hAu, ui. kuk=1 kuk=1 Observe that Lemma 3.14 shows that kAk = max(a+ , −a− ). Proof of Proposition 3.16. First we show that if A∗ = A then Spec(A) ⊂ R. Thus we need to show that if λ = s + it where t 6= 0 then A − λ is invertible. Now A − λ = (A − s) − it and A − s is bounded and selfadjoint, so it is enough to consider the special case that λ = it. Then for any u ∈ H, (3.140) Imh(A − it)u, ui = −tkuk2 . So, certainly A − it is injective, since (A − it)u = 0 implies u = 0 if t 6= 0. The adjoint of A − it is A + it so the adjoint is injective too. It follows from (3.71) that the range of A − it is dense in H. By this density of the range, if w ∈ H there exists a sequence un ∈ H with wn = (A − it)un → w. So again we find that (3.141) | Imh(A − it)hun − um ), (un − um )i| = |t|kun − um k2 = | Imh(wn − wm ), (un − um )i| ≤ kwn − wm kkun − um k 1 =⇒ kun − um k ≤ kwn − wm k. |t| Since wn → w it is a Cauchy sequence and hence un is Cauchy so by completeness, un → u and hence (A − it)u = w. Thus A − it is 1-1 and onto and kA−1 k ≤ 1/|t|. So we have shown that Spec(A) ⊂ R. We already know that Spec(A) ⊂ {z ∈ C; |z| ≤ kAk} so finally then we need to show that one of A ± kAk Id is NOT invertible. This follows from (3.136). Indeed, by the definition of sup there is a sequence un ∈ H with kun k = 1 such that either hAun , un i → kAk or hAun , un i → −kAk. Assume we are in the first case, so hAun , un i → kAk. Then (3.142) k(A − kAk)un k2 = kAun k2 − 2kAkhAun , un i + kAk2 kun k2 ≤ 2kAk2 − 2kAkhAun , un i → 0. Since the sequence is positive it follows that k(A − kAk)un k → 0. This means that A − kAk Id is not invertible, since if it had a bounded inverse B then 1 = kun k ≤ kBkk(A − kAk)un k which is impossible. In the other case it follows similarly that A + kAk is not invertible, or one can replace A by −A and use the same argument. So one of A ± kAk is not invertible.  Only slight modifications of this proof are needed to give (3.139) which we restate in a slightly different form. Lemma 3.15. If A = A∗ ∈ B(H) then (3.143) Spec(A) ⊂ [α−, α+ ] ⇐⇒ α− ≤ hAu, ui ≤ α+ ∀ u ∈ H, kuk = 1. Proof. Take a± to be defined as in (3.139) then set b = (a+ − a− )/2 and consider B = A − b Id which is self-adjoint and clearly satisfies (3.144) sup |hBu, ui| = b kuk=1 Thus kBk = b and Spec(B) ⊂ [−b, b] and the argument in the proof above shows that both end-points are in the spectrum. It follows that (3.145) {a− } ∪ {a+ } ⊂ Spec(A) ⊂ [a− , a+ ] from which the statement follows.  In particular if A = A∗ then (3.146) Spec(A) ⊂ [0, ∞) ⇐⇒ hAu, ui ≥ 0. 18. Spectral theorem for compact self-adjoint operators One of the important differences between a general bounded self-adjoint operator and a compact self-adjoint operator is that the latter has eigenvalues and eigenvectors – lots of them. Theorem 3.4. If A ∈ K(H) is a self-adjoint, compact operator on a separable Hilbert space, so A∗ = A, then H has an orthonormal basis consisting of eigenvectors of A, uj such that (3.147) Auj = λj uj , λj ∈ R \ {0}, combining an orthonormal basis for the possibly infinite-dimensional (closed) null space and eigenvectors with non-zero eigenvalues which can be arranged into a sequence such that |λj | is non-increasing and λj → 0 as j → ∞ (in case Nul(A)⊥ is finite dimensional, this sequence is finite). The operator A maps Nul(A)⊥ into itself so it may be clearer to first split off the null space and then look at the operator acting on Nul(A)⊥ which has an orthonormal basis of eigenvectors with non-vanishing eigenvalues. Before going to the proof, let’s notice some useful conclusions. One is that we have ‘Fredholm’s alternative’ in this case. Corollary 3.3. If A ∈ K(H) is a compact self-adjoint operator on a separable Hilbert space then the equation (3.148) u − Au = f either has a unique solution for each f ∈ H or else there is a non-trivial finite dimensional space of solutions to (3.149) u − Au = 0 and then (3.148) has a solution if and only if f is orthogonal to all these solutions. Proof. This is just saying that the null space of Id −A is a complement to the range – which is closed. So, either Id −A is invertible or if not then the range is precisely the orthocomplement of Nul(Id −A). You might say there is not much alternative from this point of view, since it just says the range is always the orthocomplement of the null space.  Let me separate off the heart of the argument from the bookkeeping. Lemma 3.16. If A ∈ K(H) is a self-adjoint compact operator on a separable (possibly finite-dimensional) Hilbert space then (3.150) F (u) = hAu, ui, F : {u ∈ H; kuk = 1} −→ R is a continuous function on the unit sphere which attains its supremum and infimum where sup |F (u)| = kAk. (3.151) kuk=1 Furthermore, if the maximum or minimum of F (u) is non-zero it is attained at an eivenvector of A with this extremal value as eigenvalue. Proof. Since |F (u)| is the function considered in (3.136), (3.151) is a direct consequence of Lemma 3.14. Moreover, continuity of F follows from continuity of A and of the inner product so (3.152) |F (u)−F (u0 )| ≤ |hAu, ui−hAu, u0 i|+|hAu, u0 i−hAu0 , u0 i| ≤ 2kAkku−u0 k since both u and u0 have norm one. If we were in finite dimensions this almost finishes the proof, since the sphere is then compact and a continuous function on a compact set attains its supremum and infimum. In the general case we need to use the compactness of A. Certainly F is bounded, |F (u)| ≤ sup |hAu, ui| ≤ kAk. (3.153) kuk=1 − Thus, there is a sequence such that F (u+ n ) → sup F and another un such that − F (un ) → inf F. The properties of weak convergence mean that we can pass to a ± weakly convergent subsequence in each case, and so assume that u± n * u converges ± weakly; then ku k ≤ 1 by the properties of weak convergence. The compactness of ± A means that Au± n → Au converges strongly, i.e. in norm. But then we can write u+ n ± ± ± ± ± ± ± (3.154) |F (u± n ) − F (u )| ≤ |hA(un − u ), un i| + |hAu , un − u i| ± ± ± ± ± ± ± = |hA(u± n − u ), un i| + |hu , A(un − u )i| ≤ 2kAun − Au k to deduce that F (u± ) = lim F (u± n ) are respectively the supremum and infimum of F. Thus indeed, as in the finite dimensional case, the supremum and infimum are attained, and hence are the max and min. Note that this is not typically true if A is not compact as well as self-adjoint. Now, suppose that Λ+ = sup F > 0. Then for any v ∈ H with v ⊥ u+ and kvk = 1, the curve Lv : (−π, π) 3 θ 7−→ cos θu+ + sin θv (3.155) lies in the unit sphere. Expanding out (3.156) F (Lv (θ)) = hALv (θ), Lv (θ)i = cos2 θF (u+ ) + sin(2θ) Re(Au+ , v) + sin2 (θ)F (v) we know that this function must take its maximum at θ = 0. The derivative there (it is certainly continuously differentiable on (−π, π)) is 2 RehAu+ , vi which must therefore vanish. The same is true for iv in place of v so in fact (3.157) hAu+ , vi = 0 ∀ v ⊥ u+ , kvk = 1. Taking the span of these v’s it follows that hAu+ , vi = 0 for all v ⊥ u+ so Au+ must be a multiple of u+ itself. Inserting this into the definition of F it follows that Au+ = Λ+ u+ is an eigenvector with eigenvalue Λ+ = sup F. The same argument applies to inf F if it is negative, for instance by replacing A by −A. This completes the proof of the Lemma.  Proof of Theorem 3.4. First consider the Hilbert space H0 = Nul(A)⊥ ⊂ H. Then, as noted above, A maps H0 into itself, since (3.158) hAu, vi = hu, Avi = 0 ∀ u ∈ H0 , v ∈ Nul(A) =⇒ Au ∈ H0 . Moreover, A0 , which is A restricted to H0 , is again a compact self-adjoint operator – where the compactness follows from the fact that A(B(0, 1)) for B(0, 1) ⊂ H0 is smaller than (actually of course equal to) the whole image of the unit ball. Thus we can apply the Lemma above to A0 , with quadratic form F0 , and find an eigenvector. Let’s agree to take the one associated to sup F0 unless sup F0 < − inf F0 in which case we take one associated to the inf . Now, what can go wrong here? Nothing except if F0 ≡ 0. However in that case we know from Lemma 3.14 that kAk = 0 so A = 0. So, we now know that we can find an eigenvector with non-zero eigenvalue unless A ≡ 0 which would implies Nul(A) = H. Now we proceed by induction. Suppose we have found N mutually orthogonal eigenvectors ej for A all with norm 1 and eigenvectors λj – an orthonormal set of eigenvectors and all in H0 . Then we consider (3.159) HN = {u ∈ H0 = Nul(A)⊥ ; hu, ej i = 0, j = 1, . . . , N }. From the argument above, A maps HN into itself, since (3.160) hAu, ej i = hu, Aej i = λj hu, ej i = 0 if u ∈ HN =⇒ Au ∈ HN . Moreover this restricted operator is self-adjoint and compact on HN as before so we can again find an eigenvector, with eigenvalue either the max of min of the new F for HN . This process will not stop uness F ≡ 0 at some stage, but then A ≡ 0 on HN and since HN ⊥ Nul(A) which implies HN = {0} so H0 must have been finite dimensional. Thus, either H0 is finite dimensional or we can grind out an infinite orthonormal sequence ei of eigenvectors of A in H0 with the corresponding sequence of eigenvalues such that |λi | is non-increasing – since the successive FN ’s are restrictions of the previous ones the max and min are getting closer to (or at least no further from) 0. So we need to rule out the possibility that there is an infinite orthonormal sequence of eigenfunctions ej with corresponding eigenvalues λj where inf j |λj | = a > 0. Such a sequence cannot exist since ej * 0 so by the compactness of A, Aej → 0 (in norm) but kAej k ≥ a which is a contradiction. Thus if null(A)⊥ is not finite dimensional then the sequence of eigenvalues constructed above must converge to 0. Finally then, we need to check that this orthonormal sequence of eigenvectors constitutes an orthonormal basis of H0 . If not, then we can form the closure of the span of the ei we have constructed, H0 , and its orthocomplement in H0 – which would have to be non-trivial. However, as before F restricts to this space to be F 0 for the restriction of A0 to it, which is again a compact self-adjoint operator. So, if F 0 is not identically zero we can again construct an eigenfunction, with nonzero eigenvalue, which contradicts the fact the we are always choosing a largest eigenvalue, in absolute value at least. Thus in fact F 0 ≡ 0 so A0 ≡ 0 and the eigenvectors form and orthonormal basis of Nul(A)⊥ . This completes the proof of the theorem.  19. Functional Calculus As we have seen, the non-zero eigenvalues of a compact self-adjoint operator A form the image of a sequence in [−kAk, kAk] either converging to zero or finite. If ej is an orthonormal sequence of eigenfunctions which spans Nul(A)⊥ with associated eigenvalues λi then X (3.161) A= λi Pi , Pi u = hu, ei iei i being the projection onto the span Cei . Since Pi Pj = 0 if i 6= j and Pi2 = Pi it follows inductively that the positive powers of A are given by similar sums converging in B(H) : X (3.162) Ak = λki Pi , Pi u = hu, ei iei , k ∈ N. i There is a similar formula for the identity of course, except we need to remember that the null space of A then appears (and the series does not usually converge in the norm topology on B(H)) : X (3.163) Id = Pi + PN , N = Nul(A). i The sum (3.163) can be interpreted in terms of a strong limit of operators, meaning that the result converges when applied term by term to an element of H, so X (3.164) u= Pi u + PN u, ∀ u ∈ H i which is a form of the Fourier-Bessel series. Combining these formulæ we see that for any polynomial p(z) X (3.165) p(A) = p(λi )Pi + p(0)PN i converges strongly, and in norm provided p(0) = 0. In fact we can do this more generally, by choosing f ∈ C([−kAk, kAk) and defining an operator by X (3.166) f (A) ∈ B(H), f (A)u = f (λi )(u, ei )ei i This series converges in the norm topology provided f (0) = 0 so to a compact operator and if f is real it is self-adjoint. You can easily check that, always for A = A∗ compact here, this formula defines a bounded linear map (3.167) C([−kAk, kAk] −→ B(H) which has nice properties. Most importantly (3.168) (f g)(A) = f (A)g(A), (f (A))∗ = f¯(A) so it takes the product of two continuous functions to the product of the operators. We will proceed to show that such a map exists for any bounded self-adjoint operator. Even though it may not have eigenfunctions – or even if it does, it might not have an orthonormal basis of eigenvectors. Even so, it is still possible to define f (A) for a continous function defined on [a− , a+ ] if Spec(A) ⊂ [a− , a+ ]. (In fact it only has to be defined on the compact set Spec(A) which might be quite a lot smaller). This is an effective extension of the spectral theorem to the case of non-compact self-adjoint operators. How does one define f (A)? Well, it is easy enough in case f is a polynomial, since then we can simply substitute An in place of z n . If we factorize the polynomial this is the same as setting (3.169) f (z) = c(z−z1 )(z−z2 ) . . . (z−zN ) =⇒ f (A) = c(A−z1 )(A−z2 ) . . . (A−zN ) and this is equivalent to (3.166) in case A is also compact. Notice that the result does not depend on the order of the factors or anything like that. To pass to the case of a general continuous function we need to estimate the norm in the polynomial case. Proposition 3.17. If A = A∗ ∈ B(H) is a bounded self-adjoint operator on a Hilbert space then for any polynomial with real coefficients (3.170) kf (A)k ≤ |f (z)|, Spec(A) ⊂ [a− , a+ ]. sup z∈[a− ,a+ ] Proof. For a polynomial we have defined f (A) by (3.169). We can drop the constant c since it will just contribute a factor of |c| to both sides of (3.170). Now, recall from Lemma 3.14 that for a self-adjoint operator the norm can be realized as (3.171) kf (A)k = sup{|t|; t ∈ Spec(f (A))}. That is, we need to think about when f (A) − t is invertible. However, f (z) − t is another polynomial (with leading term z N because we normalized the leading coefficient to be 1). Thus it can also be factorized: f (z) − t = N Y (z − ζj (t)), j=1 (3.172) f (A) − t = N Y (A − ζj (t)) j=1 where the ζj ∈ C are the roots (which might be complex even though the polynomial is real). Written in this way we can see that (3.173) (f (A) − t)−1 = N Y (A − ζj (t))−1 if ζj (t) ∈ / Spec(A) ∀ j. j=1 Indeed the converse is also true, i.e. the inverse exists if and only if all the A − ζj (t) are invertible, but in any case we see that (3.174) Spec(f (A)) ⊂ {t ∈ C; ζj (t) ∈ Spec(A), for some j = 1, . . . , N } since if t is not in the right side then f (A) − t is invertible. Now this can be restated as (3.175) Spec(f (A)) ⊂ f (Spec(A)) since t ∈ / f (Spec(A)) means f (z) 6= t for z ∈ Spec(A) which means that there is no root of f (z) = t in Spec(A) and hence (3.174) shows that t ∈ / Spec(f (A)). In fact it is easy to see that there is equality in (3.175). Then (3.170) follows from (3.171), the norm is the sup of |z|, for z ∈ Spec(f (A)) so kf (A)k ≤ sup |f (t)|. t∈Spec(A)  This allows one to pass by continuity to f in the uniform closure of the polynomials, which by the Stone-Weierstrass theorem is the whole of C([a− , a+ ]). Theorem 3.5. If A = A∗ ∈ B(H) for a Hilbert space H then the map defined on polynomials, through (3.169) extends by continuity to a bounded linear map (3.176) C([a− , a+ ]) −→ B(H) if Spec(A) ⊂ [a− , a+ ], Spec(f (A)) ⊂ f ([a− , a+ ]). Proof. By the Stone-Weierstrass theorem polynomials are dense in continous functions on any compact interval, in the supremum norm.  Remark 3.1. You should check the properties of this map, which also follow by continuity, especially that (3.168) holds in this more general context. In particular, f (A) is self-adjoint if f ∈ C([a− , a+ ]) is real-valued and is non-negative if f ≥ 0 on Spec(A). 20. Spectral projection I have not discussed this in lectures but it is natural at this point to push a little further towards the full spectral theorem for bounded self-adjoint operators. If A ∈ B(H) is self-adjoint, and [a− , a+ ] ⊃ Spec(A), we have defined f (A) ∈ B(H) for A ∈ C([a− , a+ ]) real-valued and hence, for each u ∈ H, (3.177) C([a− , a+ ]) 3 f 7−→ hf (A)u, ui ∈ R. Thinking back to the treatment of the Lebesgue integral, you can think of this as a replacement for the Riemann integral and ask whether it can be extended further, to functions which are not necessarily continuous. In fact (3.177) is essentially given by a Riemann-Stieltjes integral and this suggests finding the increasing function which defines it. Of course we have the rather large issue that this depends on a vector in Hilbert space as well – clearly we want to allow this to vary too. One direct approach is to try to define the ‘integral’ of the characteristic function (−∞, a] for fixed a ∈ R. To do this is consider (3.178) Qa (u) = inf{hf (A)u, ui; f ∈ C([a− , a+ ]), f (t) ≥ 0, f (t) ≥ 1 on [a− , a]}. Since f ≥ 0 we know that hf (A)u, ui ≥ 0 so the infimum exists and is non-negative. In fact there must exist a sequence fn such that (3.179) Qa (u) = limhfn (A)u, ui, fn ∈ C([a− , a+ ]), fn ≥ 0, fn (t) ≥ 1, a− ≤ t ≤ a, where the sequence fn could depend on u. Consider an obvious choice for fn given what we did earlier, namely   a− ≤ t ≤ a 1 (3.180) fn (t) = 1 − (t − a)/n a ≤ t ≤ a + 1/n   0 t > a + 1/n. Certainly (3.181) Qa (u) ≤ limhfn (A)u, ui where the limit exists since the sequence is decreasing. Lemma 3.17. For any a ∈ [a− , a+ ], (3.182) Qa (u) = limhfn (A)u, ui. Proof. For any given f as in (3.178), and any  > 0 there exists n such that f (t) ≥ 1/(1 + ) in a ≤ t ≤ a + 1/n, by continuity. This means that (1 + )f ≥ gn and hence hf (A)u, ui ≥ (1 + )−1 hfn (A)u, ui from which (3.182) follows, given (3.181).  Thus in fact one sequence gives the infimum for all u. Now, use the polarization identity to define 1 (3.183) Qa (u, v) = (Qa (u + v) − Qa (u − v) + iQa (u + iv) − iQa (u − iv)) . 4 The corresponding identity holds for hfn (A)u, vi so in fact Qa (u, v) = lim hfn (A)u, vi. (3.184) n→∞ It follows that Qa (u, v) is a sesquilinear form, linear in the first variable and antilinear in the second. Moreover the fn (A) are uniformly bounded in B(H) (with norm 1 in fact) so |Qa (u, v)| ≤ Ckukkvk. (3.185) Now, using the linearity in v of Qa (u, v) and the Riesz Representation theorem it follows that for each u ∈ H there exists a unique Qa u ∈ H such that Qa (u, v) = hQa u, vi, ∀ v ∈ H, kQa uk ≤ kuk. (3.186) From the uniqueness, H 3 u 7−→ Qa u is linear so (3.186) shows that it is a bounded linear operator. Thus we have proved most of Proposition 3.18. For each a ∈ [a− , a+ ] ⊃ Spec(A) there is a uniquely defined operator Qa ∈ B(H) such that Qa (u) = hQa u, ui (3.187) recovers (3.182) and (3.188) Q∗a = Qa = Q2a is a projection satisfying Qa Qb = Qb Qa = Qb if b ≤ a, [Qa , f (A)] = 0 ∀ f ∈ C([a− , a+ ]). This operator, or really the whole family Qa , is called the spectral projection of A. Proof. We have already shown the existence of Qa ∈ B(H) with the property (3.187) and since we defined it directly from Qa (u) it is unique. Self-adjointness follows from the reality of Qa (u) ≥ 0 since hQa u, vi = hu, Qa vi then follows from (3.186). From (3.184) it follows that hQa u, vi = lim hfn (A)u, vi =⇒ (3.189) n→∞ hQa u, f (A)vi = lim hfn (A)u, f (A)vi = hQa f (A)u, vi n→∞ since f (A) commutes with fn (A) for any continuous f. This proves the statement in (3.188). Since fn fm ≤ fn is admissible in the definition of Qa in (3.178) (3.190) hQa u, vi = lim h(fn fm )(A)u, vi = lim hfn (A)u, fm (A)vi = hQa (A)u, fm (A)vi n→∞ n→∞ and now letting m → ∞ shows that Q2a = Qa . A similar argument shows the first identity in (3.188).  Returning to the original thought that (3.177) represents a Riemann-Stieltjes integral for each u we see that collectively what we have is a map [a− , a+ ] 3 a 7−→ Qa ∈ B(H) (3.191) taking values in the self-adjoint projections and increasing in the sense of (3.188). A little more application allows one to recover the functional calculus as an integral which can be written Z (3.192) f (A) = f (t)dQt [a− ,a+ ] which does indeed reduce to a Riemann-Stieltjes integral for each u : Z (3.193) hf (A)u, ui = f (t)dhQt u, ui. [a− ,a+ ] This, meaning (3.192), is the spectral resolution of the self-adjoint operator A, replacing (and reducing to) the decomposition as a sum in the compact case X f (λj )Pj (3.194) f (A) = n where the Pj are the orthogonal projections onto the eigenspaces for λj . 21. Polar Decomposition One nice application of the functional calculus for self-adjoint operators is to get the polar decomposition of a general bounded operator. 1 Lemma 3.18. If A ∈ B(H) then E = (A∗ A) 2 , defined by the functional calculus, is a non-negative self-adjoint operator. Proof. That E exists as a self-adjoint operator satisfying E 2 = A∗ A follows directly from Theorem 3.5 and positivity follows as in Remark 3.1.  Proposition 3.19. Any bounded operator A can be written as a product (3.195) 1 A = U (A∗ A) 2 , U ∈ B(H), U ∗ U = Id −ΠNul(A) , U U ∗ = ΠRan(A) . 1 Proof. Set E = (A∗ A) 2 . We want to define U and we can see from the first condition, A = U E, that (3.196) U (w) = Av, if w = Ev. This makes sense since Ev = 0 implies hEv, Evi = 0 and hence hA∗ Av, vi = 0 so kAvk = 0 and Av = 0. So let us define ( Av if w ∈ Ran(E), w = Ev (3.197) U (w) = 0 if w ∈ (Ran(E))⊥ . So U is defined on a dense subspace of H, Ran(E) ⊕ (Ran(E))⊥ which may not be closed if Ran(E) is not closed. It follows that (3.198) U (w1 + w2 ) = U (w1 ) = Av1 =⇒ kU (w1 + w2 )k2 = |hAv1 , Av1 i|2 = hE 2 v1 , v1 i = kEv1 k2 = kw1 k2 ≤ kw1 + w2 k2 if w1 = Ev, w2 ∈ (Ran E)⊥ . Thus U is bounded on the dense subspace on which it is defined, so has a unique continuous extension to a bounded operator U ∈ B(H). From the definition of U the first, factorization, condition in (3.195) holds. From the definition U vanishes on Ran(E)⊥ . We can now check that the continuous extension is a bijection (3.199) U : Ran(E) −→ Ran(A). Indeed, if w ∈ Ran(E) then kwk = kU wk from (3.198) so (3.199) is injective. The same identity shows that the range of U in (3.199) is closed since if U wn converges, kwn − wm k = kU (wn − wm )k shows that the sequence wn is Cauchy and hence converges; the range is therefore Ran(A). This same identity, kU wk = kwk, for w ∈ Ran(E), implies that (3.200) hU w, U w0 i = hw, w0 i, w, w0 ∈ Ran(E). This follows from the polarization identity (3.201) 4hU w, U w0 i = kU (w + w0 )k2 − kU (w − w0 )k2 + ikU (w + iw0 )k2 − ikU (w − iw0 )k2 = kw + w0 k2 − kw − w0 k2 + ikw + iw0 k2 − ikw − iw0 k2 = 4hw, w0 i The adjoint U ∗ of U has range contained in the orthocomplement of the null space of U, so in Ran(E), and null space precisely Ran(A)⊥ so defines a linear map from Ran(A) to Ran(E). As such it follows from (3.201) that (3.202) U ∗ U = Id on Ran(E) =⇒ U ∗ = U −1 on Ran(A) since U is a bijection it follows that U ∗ is the two-sided inverse of U as a map in (3.199). The remainder of (3.195) follows from this, so completing the proof of the Proposition.  A bounded linear operator with the properties of U above, that there are two decompositions of H = H1 ⊕ H2 = H3 ⊕ H4 into orthogonal closed subspaces, such that U = 0 on H2 and U : H1 −→ H3 is a bijection with kU wk = kwk for all w ∈ H1 is called a partial isometry. So the polar decomposition writes a general bounded operator as product A = U E where U is a partial isometry from Ran(E) 1 onto Ran(A) and E = (A∗ A) 2 . If A is injective then U is actually unitary. 1 Exercise 1. Show that in the same sense, A = F V where F = (AA∗ ) 2 and V is a partial isometry from Ran(A∗ ) to Ran F . 22. Compact perturbations of the identity I have generally not had a chance to discuss most of the material in this section, or the next, in the lectures. Compact operators are, as we know, ‘small’ in the sense that they are norm limits of finite rank operators. Accepting this, then you will want to say that an operator such as Id −K, K ∈ K(H) (3.203) is ‘big’. We are quite interested in this operator because of spectral theory. To say that λ ∈ C is an eigenvalue of K is to say that there is a non-trivial solution of Ku − λu = 0 (3.204) where non-trivial means other than than the solution u = 0 which always exists. If λ is an eigenvalue of K then certainly λ ∈ Spec(K), since λ−K cannot be invertible. For general operators the converse is not correct, but for compact operators it is. Lemma 3.19. If K ∈ B(H) is a compact operator then λ ∈ C \ {0} is an eigenvalue of K if and only if λ ∈ Spec(K). Proof. Since we can divide by λ we may replace K by λ−1 K and consider the special case λ = 1. Now, if K is actually finite rank the result is straightforward. By Lemma 3.7 we can choose a basis so that (3.85) holds. Let the span of the ei be W – since it is finite dimensional it is closed. Then Id −K acts rather simply – decomposing H = W ⊕ W ⊥ , u = w + w0 (Id −K)(w + w0 ) = w + (IdW −K 0 )w0 , K 0 : W −→ W (3.205) being a matrix with respect to the basis. It follows that 1 is an eigenvalue of K if and only if 1 is an eigenvalue of K 0 as an operator on the finite-dimensional space W. A matrix, such as IdW −K 0 , is invertible if and only if it is injective, or equivalently surjective. So, the same is true for Id −K. In the general case we use the approximability of K by finite rank operators. Thus, we can choose a finite rank operator F such that kK − F k < 1/2. Thus, (Id −K + F )−1 = Id −B is invertible. Then we can write (3.206) Id −K = Id −(K − F ) − F = (Id −(K − F ))(Id −L), L = (Id −B)F. Thus, Id −K is invertible if and only if Id −L is invertible. Thus, if Id −K is not invertible then Id −L is not invertible and hence has null space and from (3.206) it follows that Id −K has non-trivial null space, i.e. K has 1 as an eigenvalue.  A little more generally:Proposition 3.20. If K ∈ K(H) is a compact operator on a separable Hilbert space then null(Id −K) = {u ∈ H; (IdK )u = 0} is finite dimensional Ran(Id −K) = {v ∈ H; ∃u ∈ H, v = (Id −K)u} is closed and (3.207) Ran(Id −K)⊥ = {w ∈ H; (w, Ku) = 0 ∀ u ∈ H} is finite dimensional and moreover (3.208)
dim (null(Id −K)) = dim Ran(Id −K)⊥ . Proof of Proposition 3.20. First let’s check this in the case of a finite rank operator K = T. Then (3.209) Nul(Id −T ) = {u ∈ H; u = T u} ⊂ Ran(T ). A subspace of a finite dimensional space is certainly finite dimensional, so this proves the first condition in the finite rank case. Similarly, still assuming that T is finite rank consider the range (3.210) Ran(Id −T ) = {v ∈ H; v = (Id −T )u for some u ∈ H}. Consider the subspace {u ∈ H; T u = 0}. We know that this this is closed, since T is certainly continuous. On the other hand from (3.210), (3.211) Ran(Id −T ) ⊃ Nul(T ). Now, Nul(T ) is closed and has finite codimension – it’s orthocomplement is spanned by a finite set which maps to span the image. As shown in Lemma 3.4 it follows from this that Ran(Id −T ) itself is closed with finite dimensional complement. This takes care of the case that K = T has finite rank! What about the general case where K is compact? If K is compact then there exists B ∈ B(H) and T of finite rank such that 1 (3.212) K = B + T, kBk < . 2 Now, consider the null space of Id −K and use (3.212) to write (3.213) Id −K = (Id −B) − T = (Id −B)(Id −T 0 ), T 0 = (Id −B)−1 T. Here we have used the convergence of the Neumann series, so (Id −B)−1 does exist. Now, T 0 is of finite rank, by the ideal property, so (3.214) Nul(Id −K) = Nul(Id −T 0 ) is finite dimensional. Here of course we use the fact that (Id −K)u = 0 is equivalent to (Id −T 0 )u = 0 since Id −B is invertible. So, this is the first condition in (3.207). Similarly, to examine the second we do the same thing but the other way around and write (3.215) Id −K = (Id −B) − T = (Id −T 00 )(Id −B), T 00 = T (Id −B)−1 . Now, T 00 is again of finite rank and (3.216) Ran(Id −K) = Ran(Id −T 00 ) is closed and of finite codimension. What about (3.208)? This time let’s first check first that it is enough to consider the finite rank case. For a compact operator we have written (Id −K) = G(Id −T ) (3.217) where G = Id −B with kBk < want to see is that (3.218) 1 2 is invertible and T is of finite rank. So what we dim Nul(Id −K) = dim Nul(Id −T ) = dim Nul(Id −K ∗ ). However, Id −K ∗ = (Id −T ∗ )G∗ and G∗ is also invertible, so (3.219) dim Nul(Id −K ∗ ) = dim Nul(Id −T ∗ ) and hence it is enough to check that dim Nul(Id −T ) = dim Nul(Id −T ∗ ) – which is to say the same thing for finite rank operators. Now, for a finite rank operator, written out as (3.85), we can look at the vector space W spanned by all the fi ’s and all the ei ’s together – note that there is nothing to stop there being dependence relations among the combination although separately they are independent. Now, T : W −→ W as is immediately clear and T ∗v = (3.220) N X (v, fi )ei i=1 0 so T : W −→ W too. In fact T w = 0 and T ∗ w0 = 0 if w0 ∈ W ⊥ since then (w0 , ei ) = 0 and (w0 , fi ) = 0 for all i. It follows that if we write R : W ←→ W for the linear map on this finite dimensional space which is equal to Id −T acting on it, then R∗ is given by Id −T ∗ acting on W and we use the Hilbert space structure on W induced as a subspace of H. So, what we have just shown is that (3.221) (Id −T )u = 0 ⇐⇒ u ∈ W and Ru = 0, (Id −T ∗ )u = 0 ⇐⇒ u ∈ W and R∗ u = 0. Thus we really are reduced to the finite-dimensional theorem dim Nul(R) = dim Nul(R∗ ) on W. (3.222) You no doubt know this result. It follows by observing that in this case, everything is now in W, Ran(W ) = Nul(R∗ )⊥ and in finite dimensions (3.223) dim Nul(R) + dim Ran(R) = dim W = dim Ran(W ) + dim Nul(R∗ ).  23. Hilbert-Schmidt, Trace and Schatten ideals As well as the finite rank and compact operators there are other important ideals. Since these results are not exploited in the subsequence sections, the many proofs are relegated to exercises. First consider the Hilbert-Schmidt operators. The definition is based on Lemma 3.20. For a separable Hilbert space, H, if A ∈ B(H) then once the sum for any one orthonormal basis {ei } X (3.224) kAk2HS = kAei k2 i is finite it is finite for any other orthonormal basis and is independent of the choice of basis. It is straightforward to show that the operators of finite rank satisfy (3.224); this is basically Bessel’s inequality. Proof. This is Problem XX. Starting from (3.224) for some orthonormal basis ei , consider any other orthonormal basis fj . Using the completeness, expand using Bessel’s identity X X (3.225) kAei k2 = |hAei , fj i|2 = |hei , A∗ fj i|2 . j j This converges absolutely, so the convergence of (3.224) implies the convergence of the double sum, which can then be rearranged to give X XX XX X (3.226) kAei k2 = |hei , A∗ fj i|2 = |hei , A∗ fj i|2 = kA∗ fj k2 i i j j i j where Bessel’s identity is used again. Thus the sum for A∗ with respect to the new basis is finite. Applying this argument again shows that the sum is independent of the basis, and the same for the adjoint.  Proposition 3.21. The operators for which (3.224) is finite form a 2-sided ∗ -ideal HS(H) ⊂ B(H), contained in the ideal of compact operators, it is a Hilbert space and the norm satisfies ! 21 X 2 , kAkB ≤ kAkHS = kAei k (3.227) i kADkHS ≤ kAkHS kDkB , A ∈ HS(H), D ∈ B(H). The inner product is hA, BiHS = X hAei , Bei iH , A, B ∈ HS(H). i For a compact operator the polar decomposition can be given a more explicit form and we can use this to give another characterization of the Hilbert-Schmidt operators. Proposition 3.22. If A ∈ K(H) then there exist orthonormal bases ei of Nul(A)⊥ and fj of Nul(A∗ )⊥ such that X Au = si hu, ei ifi i 1 where the si are the non-zero eigenvalues of (A∗ A) 2 repeated with multiplicity. The si are called the characteristic values of A. Proof. First take a basis ei of eigenvectors of A∗ A restricted to Nul(A)⊥ = Nul(A∗ A)⊥ with eigenvalues s2i > 0, so the si are the non-zero eigenvalues of 1 |A| = (A∗ A) 2 . Then A = U |A| with U a unitary operator from Ran(|A|) = Nul(A)⊥ to Ran(A) so (3.22) follows by taking fi = U ei .  Extending the ei to an orthonormal basis of H it follows that ! 21 X 2 kAkHS = si = ks∗ kl2 . i So to say that A is Hilbert-Schmidt is to say that the sequence of its characteristic values is in l2 (with the caveat that the sequence might be finite). One reason that the Hilbert-Schmidt operators are of interest is their relation to the ideal of operators ‘of trace class’, T (H). Definition 3.7. The space T (H) ⊂ B(H) for a separable Hilbert space consists of those operators A for which X |hAei , fi i| < ∞ (3.228) kAkTr = sup i where the supremum is over pairs of orthonormal sequences {ei } and {fi }. Proposition 3.23. The trace class operators form an ideal, T (H) ⊂ HS(H), which is a Banach space with respect to the norm (3.228) which satisfies 1 1 2 kAkB ≤ kAkTr , kAkHS ≤ kAkB2 kAkTr ; (3.229) the following two conditions are equivalent to A ∈ T (H) : (1) The operator defined by the functional calculus, 1 1 |A| 2 = (A∗ A) 4 ∈ HS(H). (3.230) (2) There are operators Bi , Bi0 ∈ HS(H) such that (3.231) A= N X Bi0 Bi . i=1 Proof. Note first that T (H) is a linear space and that k · kTr is a norm on 1 it. Now suppose A ∈ T (H) and consider its polar decomposition A = U (A∗ A) 2 . 1 Here U is a partial isometry mapping Ran(A∗ A) 2 to Ran(A) and vanishing on 1 ⊥ 1 Ran(A∗ A) 2 . Consider an orthonormal basis {ei } of Ran(A∗ A) 2 . This is an orthonormal sequence in H as is fi = U ei . Inserting these into (3.228) shows that X X 1 1 1 (3.232) |hU (A∗ A) 2 ei , fi i| = |h(A∗ A) 4 ei , (A∗ A) 4 ei i| < ∞ i i where we use the fact that U ∗ fi = U ∗ U ei = ei . Since the closure of the range of 1 1 (A∗ A) 4 is the same as the closure of the range of (A∗ A) 2 it follows from (3.232) 1 that (3.230) holds (since adding an orthonormal basis of Ran((A∗ A) 4 )⊥ does not increase the sum). Next assume that (3.230) holds for A ∈ B(H). Then the polar decomposition 1 1 can be written A = (U (A∗ A) 4 )(A∗ A) 4 showing that A is the product of two Hilbert-Schmidt operators, so in particular of the form (3.231). Now assume that A is of the form (3.231), so is a sum of products of HilbertSchmidt operators. The linearity of T (H) means it suffices to assume that A = BB 0 where B, B 0 ∈ HS(H). Then, (3.233) |hAe, fi i| = |hB 0 ei , B ∗ fi i| ≤ kB 0 ei kH kB ∗ fi kH . Taking a finite sum and applying Cauchy-Schwartz inequality (3.234) N X i=1 |hAe, fi i| ≤ ( N X N X 1 1 kB 0 ei k2 ) 2 ( kB ∗ fi k2 ) 2 . i=1 i=1 If the sequences are orthonormal the right side is bounded by the product of the Hilbert-Schmidt norms so (3.235) kBB 0 kTr ≤ kBkHS kB 0 kHS and A = BB 0 ∈ T (H). The first inequality in (3.229) follows the choice of single unit vectors u and v as orthonormal sequences, so (3.236) |hAu, vi| ≤ kAkTr =⇒ kAk ≤ kAkTr . The completeness of T (H) with respect to the trace norm follows standard arguments which can be summarized as follows (1) If An is Cauchy in T (H) then by the equality just established, it is Cauchy in B(H) and so converges in norm to A ∈ B(H). (2) A Cauchy sequence is bounded, so there is a constant C = supn kAm kTr such that for any N, any orthonormal sequences ei , fi , N X (3.237) |hAn ei , fi i| ≤ C. i=1 Passing to the limit An → A in the finite sum gives the same bound with An replaced by A and then allowing N → ∞ shows that A ∈ T (H). Similarly the Cauchy condition means that for  > 0 there exists M such that for all N, and any orthonormal sequences ei , fi (3.238) m.n > M =⇒ N X |h(An − Am )ei , fi i| ≤ . i=1 Passing first to the limit m → ∞ in the finite sum and then N → ∞ shows that n > M =⇒ kAn − AkTr ≤  and so An → A in the trace norm.  Proposition 3.24. The trace functional T (H) 3 A 7−→ Tr(A) = (3.239) X hAei , ei i i is a continuous linear functional (with respect to the trace norm) which is independent of the choice of orthonormal basis {ei } and which satsifies (3.240) Tr(AB − BA) = 0 if A ∈ T (H), B ∈ B(H) or A, B ∈ HS(H). Proof. The complex number Tr(AB − BA) depends linearly on A and, separately, on B. The ideals are ∗-closed so decomposing A = (A + A∗ )/2 + i(A − A∗ )/2i and similarly for B shows that it suffices to assume that A and B are self-adjoint. If A ∈ T (H) we can choose use an orthonormal basis of eigenvectors for it to evaluate the trace. Then if Aei = λi ei X (3.241) Tr(AB − BA) = (hBei , Aei i − hAei , Bei i) i = X (λi hBei , ei i − λi hei , Bei i) = 0. i The case that A, B ∈ HS(H) is similar.  This is the fundamental property of the trace functional, that it vanishes on commutators where one of the elements is of trace class and the other is bounded. Two other important properties are that Lemma 3.21. (3.242) (1) If A, B ∈ HS(H) then hA, BiHS = Tr(A∗ B) (2) If T = T ∗ ∈ K(H) then T ∈ T (H) if and only if the sequence of non-zero eigenvalues λj of T (repeated with multiplicity) is in l1 and X X (3.243) Tr(T ) = λj , kT kTr = |λj |. j j In fact the second result extends to Lidskii’s theorem: If T ∈ Tr(H) then the spectrum outside 0 is discrete, so countable, and each λi P point is an eigenvalue of finite algebraic multiplicity ki and then Tr(T ) = ki λi converges in l1 . The i algebraic multiplicity is the limit as k → ∞ of the dimension of the null space of (T − λi )k . The standard proof of this is not elementary. Next we turn to the more general Schatten classes. Definition 3.8. An operator A ∈ K(H) is ‘of Schatten class,’ A ∈ Scp (H), p ∈ [1, ∞) if and only if |A|p ∈ T (H), i.e. ! p1 (3.244) kT kScp = X spi <∞ i where si are the non-zero characteristic values of A repeated with multiplicity. So T (H) = Sc1 (H), HS(H) = Sc2 (H). Of course the notation is suggestive, but we need to be a bit careful in proving the results which are implied by the notation! Proposition 3.25. Each of the Schatten classes is a two-sided ∗-ideal in B(H) which is a Banach space with respect to the norm (3.244); the norm is also given by X (3.245) kT kpScp = sup |hT ei , fi i|p i with the supremum over orthonormal sequences, with finiteness implying that T ∈ Scp (H). If q is the conjugate index to p ∈ (0, ∞) then (3.246) A ∈ Scq (H), B ∈ Scp (H) =⇒ AB ∈ T (H), kABkTr ≤ kAkScq kBkScp and conversely, if A ∈ B(H) then A ∈ Scp (H) if and only if AB ∈ T (H) for all B ∈ Scq (H) and kAkScp = sup kABkTr . kBkScq =1 Proof. The alternate realization of the Schatten norm in (3.245) is particularly useful since whilst it is clear from the definition that cT ∈ Scp (H) if T ∈ Scp (H) and c ∈ C, it is not otherwise immediately clear that the space is linear (or that the triangle inequality holds). From the definition (3.244), that if T is self-adjoint then T ∈ Scp (H) if and only if X (3.247) sup |hT fi , fi i|p = kT kpScp < ∞ i with the supremum over orthonormal sequences. To see this let ej be an orthonomal basis of eigenvectors for T. Then expanding in the Fourier-Bessel series (3.248) hT fi , fi i = X λj |hfi , ej i|2 ≤ X j ≤( |λj ||hfi , ej i|2/p |hfi , ej i|2/q j X X 1 X 1 1 |λj | hfi , ej i| ) p ( |hfi , ej i|2 ) q = ( |λj |p hfi , ej i|2 ) p p 2 j j j by Hölder’s inequality, so X X X X (3.249) |hT fi , fi i|p ≤ |λj |p hfi , ej i|2 = |λj |p = kT kpScp . i j i j This proves (3.247) when T = T ∗ ∈ K(H). Now consider (3.245). Let PN be the orthogonal projection onto the span of the eigenspaces corresponding to the the N largest eigenvalues of |T |. Then we replace T by TN = T PN ; certainly T PN → T in norm. Since TN has finite rank, both Nul(TN ) and Nul(TN∗ ) are infinite dimensional so we can write the polar decomposition as TN = UN AN , AN = PN |T |PN and take UN to be unitary (rather than a partial isometry) by extending it by an isometric isomorphism Nul(TN )⊥ −→ Nul(TN∗ )⊥ . Then using Cauchy-Schwartz inequality and then (3.247) for AN , (3.250) X |hTN ei , fi i|p = i X 1 1 ∗ 2 2 |hAN ei , AN UN fi i|p i X X 1 1 1 1 ∗ 2 2 ei k2p ) 2 ( UN fi k2p ) 2 ≤( kAN kAN i i X X 1 1 ∗ |hAN UN fi , U ∗ fi i|p ) 2 ≤( |hAN ei , ei i|p ) 2 | i i ≤ kAN kpScp = kTN kpScp ≤ kT kpScp . As usual dropping to a finite sum on the left we can pass to the limit as N → ∞ and obtain a uniform bound on any finite sum for T from which (3.245) follows. At this point we know that if A ∈ Scp (H) and U1 , U2 are unitary then U1 AU2 ∈ Scp (H) and kU1 AU2 kScp = kAkScp . From (3.245) it follows directly that Scp (H) is linear, that the triangle inequality holds, so that k · kScp is a norm, and Scp (H) is complete and that it is ∗-closed. Now, if A ∈ Scq (H) and B ∈ Scp (H) for conjugate indices p, q ∈ (1, ∞) choose a finite rank orthogonal projection P and consider ABP which is of finite rank, and hence of trace class. We can compute its trace with respect to any orthonormal basis. Choose an orthonormal basis ei of the range of P AP and fi so that the polar decomposition of P AP becomes P AP fi = si ei =⇒ P A∗ P ei = si fi where the si are the characteristic values of P A. With finite sums X (3.251) | Tr(P AP BP )| = | hP AP 2 BP ei , ei i| =| N X X hP BP ei , P A∗ P ei i| ≤ si |hP BP ei , fi i| i=1 X q 1 X 1 |hP BP ei , fi i|p ) p ≤ kP AP kScq kP BP kScp ≤( sj ) q ( by Hölder’s inequality. Now |P BP | = P |B|P (and similarly for A) and from minimax arguments discussed earlier it follows that, sj (|P BP |) ≤ sj (|B|) for all j. So we see that (3.252) | Tr(P AP BP )| ≤ kAkScq kBkScp . Fixing B this is true for any A, so A can be replaced by U A with U unitary, in such a way that AP B = |AP B|. We also know that kU AkScq = kAkScq and since P |AP B|P is positive and (3.253) Tr(P AP BP ) = Tr(P |AP B|P ) = kP |AP B|P kTr ≤ kAkScq kBkScp . Taking an increasing sequence of projections PN , it follows that PN |APN B|PN → |AB| in trace norm and that (3.246) holds. The proof of optimality in this ‘non-commutative Hölder inequality’ is left as an exercise. That Scp (H) is an ideal then follows from the fact that T (H) is an ideal.  24. Fredholm operators Definition 3.9. A bounded operator F ∈ B(H) on a Hilbert space is said to be Fredholm, written F ∈ F(H), if it has the three properties in (3.207) – its null space is finite dimensional, its range is closed and the orthocomplement of its range is finite dimensional. In view of Proposition 3.20, if K ∈ K(H) then Id −K ∈ F(H). For general Fredholm operators the row-rank=colum-rank result (3.208) does not hold. Indeed the difference of these two integers, called the index of the operator,
(3.254) ind(F ) = dim (null(F )) − dim Ran(F )⊥ is a very important number with lots of interesting properties and uses. Notice that the last two conditions in (3.207) are really independent since the orthocomplement of a subspace is the same as the orthocomplement of its closure. There is for instance a bounded operator on a separable Hilbert space with trivial null space and dense range which is not closed. How could this be? Think for instance of the operator on L2 (0, 1) which is multiplication by the function x. This is assuredly bounded and an element of the null space would have to satisfy xu(x) = 0 almost everywhere, and hence vanish almost everywhere. Moreover the density of the L2 functions vanishing in x <  for some (non-fixed)  > 0 shows that the range is dense. However this operator is not invertible and not Fredholm. On the other hand we do know that a subspace with finite codimension is closed, so we can replace the last two conditions in Definition 3.9 by saying that the range of the operator has finite codimension. I have not done this directly since it is a little too easy to fall into the trap of thinking that it is enough to check that the closure of the range has finite codimension; it isn’t! Before looking at general Fredholm operators let’s check that, in the case of operators of the form Id −K, with K compact the third conclusion in (3.207) really follows from the first. This is a general fact which I mentioned, at least, earlier but let me pause to prove it. Proposition 3.26. If B ∈ B(H) is a bounded operator on a Hilbert space and B ∗ is its adjoint then (3.255) Ran(B)⊥ = (Ran(B))⊥ = {v ∈ H; (v, w) = 0 ∀ w ∈ Ran(B)} = Nul(B ∗ ). Proof. The definition of the orthocomplement of Ran(B) shows immediately that (3.256) v ∈ (Ran(B))⊥ ⇐⇒ (v, w) = 0 ∀ w ∈ Ran(B) ⇐⇒ (v, Bu) = 0 ∀ u ∈ H ⇐⇒ (B ∗ v, u) = 0 ∀ u ∈ H ⇐⇒ B ∗ v = 0 ⇐⇒ v ∈ Nul(B ∗ ). On the other hand we have already observed that V ⊥ = (V )⊥ for any subspace – since the right side is certainly contained in the left and (u, v) = 0 for all v ∈ V implies that (u, w) = 0 for all w ∈ V by using the continuity of the inner product to pass to the limit of a sequence vn → w.  There is a more ‘analytic’ way of characterizing Fredholm operators, rather than Definition 3.9. Lemma 3.22. An operator F ∈ B(H) is Fredholm, F ∈ F(H), if and only if it has a generalized inverse P satisfying P F = Id −ΠNul(F ) (3.257) F P = Id −ΠRan(F )⊥ with the two projections of finite rank. Proof. If (3.257) holds then F must be Fredholm, since its null space is finite dimensional, from the second identity the range of F must contain the range of Id −ΠNul(F )⊥ and hence it must be closed and of finite codimension. Conversely, suppose that F ∈ F(H). We can divide H into two pieces in two ways as H = Nul(F ) ⊕ Nul(F )⊥ and H = Ran(F )⊥ ⊕ Ran(F ) where in each case the first summand is finite-dimensional. Then F defines four maps, from each of the two first summands to each of the two second ones but only one of these is non-zero and so F corresponds to a bounded linear map F̃ : Nul(F )⊥ −→ Ran(F ). These are two Hilbert spaces with a bounded linear bijection between them, so the inverse map, P̃ : Ran(F ) −→ Nul(F )⊥ is bounded by the Open Mapping Theorem and we can define (3.258) Then (3.257) follows directly. P = P̃ ◦ ΠNul(F )⊥ .  What we want to show is that the Fredholm operators form an open set in B(H) and that the index is locally constant. To do this we show that a weaker version of (3.257) also implies that F is Fredholm. Lemma 3.23. An operator F ∈ F(H) is Fredholm if and only if it has a parametrix Q ∈ B(H) in the sense that QF = Id −EL (3.259) F Q = Id −ER with EL and ER of finite rank. Moreover any two such parametrices differ by a finite rank operator. The term ‘parametrix’ refers to an inverse modulo an ideal. Here we are looking at the ideal of finite rank operators. In fact this is equivalent to the existence of an inverse modulo compact operators. One direction is obvious – since finite rank operators are compact – the other is covered by one of the problems. Notice that the parametrix Q is itself Fredholm, since reversing the two equations shows that F is a parameterix for Q. Similarly it follows that if F is Fredholm then so is F ∗ and that the product of two Fredholm operators is Fredholm. Proof. If F is Fredholm then Q = P certainly is a parameterix in this sense. Conversely suppose that Q as in (3.259) exists. Then Nul(Id −EL ) is finite dimensional – from (3.207) for instance. However, from the first identity Nul(F ) ⊂ Nul(QF ) = Nul(Id −EL ) so Nul(F ) is finite dimensional too. Similarly, the second identity shows that Ran(F ) ⊃ Ran(F Q) = Ran(Id −ER ) and the last space is closed and of finite codimension, hence so is the first. Thus the existence of such a parameterix Q implies that F is Fredholm. 0 Now if Q and Q0 both satisfy (3.259) with finite rank error terms ER and EL0 0 for Q then (Q0 − Q)F = EL − EL0 (3.260) is of finite rank. Applying the generalized inverse, P, of F on the right shows that the difference (3.261) (Q0 − Q) = (EL − EL0 )P + (Q0 − Q)ΠNul(F ) is indeed of finite rank.  Observe that (3.260) can be reversed. If F is Fredholm, so has a parametrix Q then all the operators Q + E where E is of finite rank are also parametrices. It is also the case that if F is Fredholm and K is compact then F + K is Fredholm. Indeed, if you go through the proof above replacing ‘finite rank’ by ‘compact’ you can check this. Thus an operator is Fredholm if and only if it has invertible image in the Calkin algebra, B(H)/K(H). Now recall that finite-rank operators are of trace class, that the trace is welldefined and that the trace of a commutator where one factor is bounded and the other trace class vanishes. Using this we show Lemma 3.24. If Q and F satisfy (3.259) then (3.262) ind(F ) = Tr(EL ) − Tr(ER ). Proof. We certainly know that (3.262) holds in the special case that Q = P is the generalized inverse of F, since then EL = ΠNul(F ) and ER = ΠRan(F )⊥ and the traces are the dimensions of these spaces. Now, if Q is a parameterix as in (3.259) consider the straight line of operators Qt = (1 − t)P + tQ. Using the two sets of identities for the generalized inverse and paramaterix (3.263) Qt F = (1 − t)P F + tQF = Id −(1 − t)ΠNul(F ) − tEL , F Qt = (1 − t)F P + tF Q = Id −(1 − t)ΠRan(F )⊥ − tER . Thus Qt is a curve of parameterices and what we need to show is that (3.264) J(t) = Tr((1 − t)ΠNul(F ) + tEL ) − Tr((1 − t)ΠRan(F )⊥ + tER ) is constant. This is a linear function of t as is Qt . We can differentiate (3.263) with respect to t and see that (3.265) d d ((1 − t)ΠNul(F ) + tEL ) − ((1 − t)ΠRan(F )⊥ + tER ) = [Q − P, F ] dt dt =⇒ J 0 (t) = 0 since it is the trace of the commutator of a bounded and a finite rank operator (using the last part of Lemma 3.23).  Proposition 3.27. The Fredholm operators form an open set in B(H) on which the index is locally constant. Proof. We need to show that if F is Fredholm then there exists  > 0 such that F + B is Fredholm if kBk < . Set B 0 = ΠRan(F ) BΠNul(F )⊥ then kB 0 k ≤ kBk and B−B 0 is finite rank. If F̃ is the operator constructed in the proof of Lemma 3.22 then F̃ + B 0 is invertible as an operator from Nul(F )⊥ to Ran(F ) if  > 0 is small. The inverse, PB0 , extended as 0 to Nul(F ) as P is defined in that proof, satisfies (3.266) PB0 (F + B) = Id −ΠNul(F ) + PB0 (B − B 0 ), (F + B)PB0 = Id −ΠRan(F )⊥ + (B − B 0 )PB0 and so is a parametrix for F + B. Thus the set of Fredholm operators is open. The index of F +B is given by the difference of the trace of the finite rank error terms in the second and first lines here. It depends continuously on B in kBk <  so, being integer-valued, is constant.  This shows in particular there is an open subset of B(H) which contains no invertible operators, in strong contrast to the finite dimensional case. In fact even the Fredholm operators do not form a dense subset of B(H). One such open subset consists of the semi-Fredholm operators – those with closed range and with either null space or complement of range finite-dimensional. Why is the index important? For one thing it actually labels the components of the Fredholm operators – two Fredholm operators can be connected by a curve of Fredholms if and only if they have the same index. One of the main applications of the index is quite trivial to see – if the index of a Fredholm operator is positive then the operator must have non-trivial null space. This is a remarkably powerful method for showing that certain sorts (‘elliptic’ for one) of equations have nontrivial solutions. 25. Kuiper’s theorem For finite dimensional spaces, such as CN , the group of invertible operators – in this case matrices and denoted typically GL(N ) – is a particularly important example of a Lie group. One reason it is important is that it carries a good deal of ‘topological’ structure. In particular – if you have done a little topology – its fundamental group is not trivial, in fact it is isomorphic to Z. This corresponds to the fact that a continuous closed curve c : S −→ GL(N ) is contractible if and only if its winding number is zero – the effective number of times that the determinant goes around the origin in C. There is a lot more topology than this and it is actually quite complicated. Perhaps surprisingly, the corresponding group of the invertible bounded operators on a separable (complex) infinite-dimensional Hilbert space is contractible. This is Kuiper’s theorem, and means that this group, GL(H), has no ‘topology’ at all, no holes in any dimension and for topological purposes it is like a big open ball. The proof is not really hard, but it is not exactly obvious either. It depends on an earlier idea, ‘Eilenberg’s swindle’ - it is an unusual name for a theorem - which shows how the infinite-dimensionality is exploited. As you can guess, this is sort of amusing (if you have the right attitude . . . ). The proof I give here is due to B. Mityagin, [3]. Let’s denote by GL(H) this group. In view of the open mapping theorem we know that (3.267) GL(H) = {A ∈ B(H); A is injective and surjective}. Contractibility means precisely that there is a continuous map γ : [0, 1] × GL(H) −→ GL(H) s.t. (3.268) γ(0, A) = A, γ(1, A) = Id, ∀ A ∈ GL(H). Continuity here means for the metric space [0, 1] × GL(H) where the metric comes from the norms on R and B(H). I will only show ‘weak contractibility’ of GL(H). This has nothing to do with weak convergence, rather just means that we only look for an homotopy over compact sets. As a warm-up exercise, let us show that the group GL(H) is contractible to the unitary subgroup (3.269) U(H) = {U ∈ GL(H); U −1 = U ∗ }. These are the isometric isomorphisms. Proposition 3.28. There is a continuous map (3.270) Γ : [0, 1] × GL(H) −→ GL(H) s.t. Γ(0, A) = A, Γ(1, A) ∈ U(H) ∀ A ∈ GL(H). Proof. This is a consequence of the functional calculus, giving the ‘polar decomposition’ of invertible (and more generally bounded) operators. Namely, if A ∈ GL(H) then AA∗ ∈ GL(H) is self-adjoint. Its spectrum is then contained in an interval [a, b], where 0 < a ≤ b = kAk2 . It follows from what we showed earlier 1 that R = (AA∗ ) 2 is a well-defined bounded self-adjoint operator and R2 = AA∗ . Moreover, R is invertible and the operator UA = R−1 A ∈ U(H). Certainly it is bounded and UA∗ = A∗ R−1 so UA∗ UA = A∗ R−2 A = Id since R−2 = (AA∗ )−1 = (A∗ )−1 A−1 . Thus UA∗ is a right inverse of UA , and (since UA is a bijection) is the unique inverse so UA ∈ U(H). So we have shown A = RUA then (3.271) Γ(s, A) = (s Id +(1 − s)R)UA , s ∈ [0, 1] satisfies (3.270). There is however the issue of continuity of this map. Continuity in s is clear enough but we also need to show that the map (3.272) 1 GL(H) 3 A 7−→ (A∗ A) 2 ∈ GL(H), defining R, is continuous. Certainly the map A − 7 → A∗ A is (norm) continuous. Suppose An → A in GL(H) then given  there is a polynomial p such that 1 k(B ∗ B) 2 − p(B ∗ B)k ≤ /3 for B = A, An ∀ n. (3.273) On the other hand (A∗n An )k → (A∗ A)k for any k so 1 1 (3.274) k(A∗ A) 2 − (A∗n An ) 2 k ≤ 1 1 k(A∗ A) 2 − p(A∗ A)k + kp(A∗ A) − p(A∗n An )k + k(A∗n An ) 2 − p(A∗n An )k → 0.  So, for any compact subset X ⊂ GL(H) we seek a continuous map (3.275) γ : [0, 1] × X −→ GL(H) s.t. γ(0, A) = A, γ(1, A) = Id, ∀ A ∈ X, note that this is not contractibility of X, but of X in GL(H). In fact, to carry out the construction without having to worry about too many things at once, just consider (path) connectedness of GL(H) meaning that there is a continuous map as in (3.275) where X = {A} just consists of one point – so the map is just γ : [0, 1] −→ GL(H) such that γ(0) = A, γ(1) = Id . The construction of γ is in three stages (1) Creating a gap (2) Rotating to a trivial factor (3) Eilenberg’s swindle. Lemma 3.25 (Creating a gap). If A ∈ B(H) and  > 0 is given there is a decomposition H = HK ⊕ HL ⊕ HO into three closed mutually orthogonal infinitedimensional subspaces such that if QI is the orthogonal projections onto HI for I = K, L, O then kQL BQK k < . (3.276) Proof. Choose an orthonormal basis ej , j ∈ N, of H. The subspaces Hi will be determined by a corresponding decomposition (3.277) N = K ∪ L ∪ O, K ∩ L = K ∩ O = L ∩ O = ∅. Thus HI has orthonormal basis ek , k ∈ I, I = K, L, O. To ensure (3.276) we choose the decomposition (3.277) so that all three sets are infinite and so that (3.278) |(el , Bek )| < 2−l−1  ∀ l ∈ L, k ∈ K. Once we have this, then for u ∈ H, QK u ∈ HK can be expanded to P (Qk u, ek )ek k∈K and expanding in HL similarly, X XX QL BQK u = (BQK u, el )el = (Bek , el )(QK u, ek )el l∈L k∈L k∈K ! (3.279) =⇒ kQL BQK uk2 ≤ X k∈K |(Qk u, ek )|2 X |(Bek , el )|2 l∈L 1 X 1 ≤ 2 |(Qk u, ek )|2 ≤ 2 kuk2 2 2 k∈K giving (3.276). The absolute convergence of the series following from (3.278). Thus, it remains to find a decomposition (3.277) for which (3.278) holds. This follows from Bessel’s inequality. First choose 1 ∈ K then (Be1 , el ) → 0 as l → ∞ so |(Be1 , el1 )| < /4 for l1 large enough and we will take l1 > 2k1 . Then we use induction on N, choosing K(N ), L(N ) and O(N ) with K(N ) = {k1 = 1 < k2 < . . . , kN }, L(N ) = {l1 < l2 < · · · < lN }, lr > 2kr , kr > lr−1 for 1 < r ≤ N and O(N ) = {1, . . . , lN } \ (K(N ) ∪ L(N )). Now, choose kN +1 > lN such that |(el , BekN +1 )| < 2−l−N , for all l ∈ L(N ), and then lN +1 > 2kN +1 such that |(elN +1 , Bk )| < e−N −1−k  for k ∈ K(N +1) = K(N )∪ {kN +1 } and the inductive hypothesis follows with L(N + 1) = N (N ) ∪ {lN +1 }.  Given a fixed operator A ∈ GL(H) Lemma 3.25 can be applied with  = kA−1 k−1 . It then follows, from the convergence of the Neumann series, that the curve (3.280) A(s) = A − sQL AQK , s ∈ [0, 1] lies in GL(H) and has endpoint satisfying (3.281) QL BQK = 0, B = A(1), QL QK = 0 = QK QL , QK = Q2K , QL = Q2L where all three projections, QL , QK and Id −QK − QL have infinite rank. These three projections given an identification of H = H ⊕ H ⊕ H and so replace the bounded operators by 3 × 3 matrices with entries which are bounded operators on H. The condition (3.281) means that       B11 B12 B13 1 0 0 0 0 0 B22 B23  , QK = 0 0 0 , QL = 0 1 0 . (3.282) B= 0 B31 B32 B33 0 0 0 0 0 0 So, now we have a ‘little hole’. Under the conditions (3.281) consider (3.283) P = BQK B −1 (Id −QL ). The condition QL BQK = 0 and the definition show that QL P = 0 = P QL . Moreover, P 2 = BQK B −1 (Id −QL )BQK B −1 (Id −QL ) = BQK B −1 BQK B −1 (Id −QL ) = P. So, P is a projection which acts on the range of Id −QL ; from its definition, the range of P is contained in the range of BQK . Since P BQK = BQK B −1 (Id −QL )BQK = BQK it follows that P is a projection onto the range of BQK . The next part of the proof can be thought of as a result on 3 × 3 matrices but applied to a decomposition of Hilbert space. First, observe a little result on rotations. Lemma 3.26. If P and Q are projections on a Hilbert space with P Q = QP = 0 and M = M P = QM restricts to an isomorphism from the range of P to the range of Q with ‘inverse’ M 0 = M 0 Q = P M 0 (so M 0 M = P and M M 0 = Q) (3.284) [−π/2, π/2] 3 θ 7−→ R(θ) = cos θP + sin θM − sin θM 0 + cos θQ + (Id −P − Q) is a path in the space of invertible operators such that R(0)P = P, R(π/2)P = M 0 P. (3.285) Proof. Computing directly, R(θ)R(−θ) = Id from which the invertibility follows as does (3.285).  We have shown above that the projection P has range equal to the range of BQK ; apply Lemma 3.26 with M = S(BQK )−1 P where S is a fixed isomorphism of the range of QK to the range of QL . Then (3.286) L1 (θ) = R(θ)B has L1 (0) = B, L(π/2) = B 0 with B 0 QK = QL SQK an isomorphism onto the range of Q. Next apply Lemma 3.26 again but for the projections QK and QL with the isomorphism S, giving R0 (θ) = cos θQK + sin θS − sin θS 0 + cos θQL + QO . (3.287) Then the curve of invertibles L2 (θ) = R0 (θ − θ0 )B 0 has L(0) = B 0 , L(π/2) = B 00 , B 00 QK = QK . So, we have succeed by succesive homotopies through invertible elements in arriving at an operator   Id E (3.288) B 00 = 0 F where we are looking at the decomposition of H = H ⊕ H according to the projections QK and Id −QK . The invertibility of this is equivalent to the invertibility of F and the homotopy   Id (1 − s)E (3.289) B 00 (s) = 0 F connects it to (3.290)     Id 0 Id −(1 − s)EF −1 00 −1 L= , (B (s)) = 0 F 0 F −1 through invertibles. The final step is ‘Eilenberg’s swindle’. In (3.290) we arrived at a family of operators on H ⊕ H. Reversing the factors we can consider   F 0 (3.291) . 0 Id Eilenberg’s idea is to connect this to the identity by an explicit curve which ‘only uses F ’ and so is uniform in parameters. So for the moment just take F to be a fixed unitary operator on H. We use several isomorphism involving H and l2 (H) which are isomorphic of course, as separable Hilbert spaces. First consider the two simple ‘rotations’ on H ⊕H     Id cos t F sin t F cos t F sin t (3.292) , . −F −1 sin t Id cos t −F −1 sin t cos tF −1 These are both unitary norm-continuous curves, the first starts at the identity and is off-diagonal at t = π/2 and equal to the second at that point. So by reversing the second and going back we connect     Id 0 F 0 (3.293) Id = to . 0 Id 0 F −1 Now, we can also identify H ⊕ H with l2 (H ⊕ H). So an element of this space is an l2 sequence with values in H ⊕ H and the identity just acts as the identity on each 2 × 2 block. We can perform the to-and-fro rotation in (3.292) in each block. That this is actually a norm-continuous curve acting on l2 (H ⊕ H) is a consequence of the fact that it is ‘the same’ in each block and so it is actually a sequence of operators, each on H ⊕ H, which are continuous, and uniformly so with respect to the index i corresponding to a sequence in l2 ; so the whole operator is continuous. This connects Id to the second matrix (3.293) acting in each block of l2 (H ⊕H). So, here is one part of the swindle, we can reorder the space so it becomes l2 (H) where now the operator is diagonal but with alternating entries (3.294) Diag(F −1 , F, F −1 , F, . . . ) on l2 (H). This requires just a unitary isomorphism corresponding to relabelling the basis elements. Now, go back to the operator (3.291) and look at the lower left identity element acting on H. We can identify the H in this spot with l2 (H) and then we have a curve linking this entry to (3.294). For the whole operator this gives a norm-continuous curve connecting   F 0 (3.295) to Diag(F, F −1 , F, F −1 , F, . . . ) on H ⊕ l2 (H) 0 Id just adding the first entry. But now we reverse the procedure using F −1 in place of F so the end-point in (3.295) is connected to the identity on l2 (H) = H! The fact that this construction only uses F itself and 2 × 2 matrices means that it works uniformly when F depends continously on parameters in a compact set. So we have constructed a curve as desired in (3.275) and hence we have proved:Theorem 3.6 (Kuiper). For any compact subset X ⊂ GL(H) there is a retraction γ as in (3.275). Note that it follows from a result of Milnor (on CW complexes) that in this case contractibility follows from weak contractibility. If you are topologically inclined you might like to look up some applications of Kuiper’s Theorem - for instance that the projective unitary group is a classifying space for two dimensional integral cohomology, an Eilenberg-MacLane space.