CHAPTER 3 Hilbert spaces There are really three `types' of Hilbert spaces (over C): The finite dimen- sional ones, essentially just Cn; for different integer values of n; with which you are pretty familiar, and two infinite dimensional types corresponding to being separa- ble (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 con- tents when complete and 1. pre-Hilbert spaces mention problems 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) hλ1v1 + λ2v2; wi = λ1hv1; wi + λ2hv2; wi; hw; vi = hv; wi for any v1; v2; v and w 2 H and λ1; λ2 2 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; λ1w1 + λ2w2i = λ1hv; w1i + λ2hv; w2i 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 8 v 2 H =) u = 0: Lemma 3.1. If H is a pre-Hilbert space with Hermitian inner product h; i then 1 (3.5) kuk = hu; ui 2 is a norm on H: 69 70 3. HILBERT SPACES Proof. The first condition on a norm follows from (3.2). Absolute homogene- ity follows from (3.1) since (3.6) kλuk2 = hλu, λui = jλj2kuk2: 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, (3.8) jhu; vij ≤ kukkvk 8 u; v 2 H holds in any pre-Hilbert space. Proof. This inequality is trivial if either u or v vanishes. For any non-zero u; v 2 H and s 2 R positivity of the norm shows that (3.9) 0 ≤ ku + svk2 = kuk2 + 2s Rehu; vi + s2kvk2: 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 (3.10) 2skvk2 + 2 Rehu; vi = 0: Substituting this into (3.9) gives (3.11) kuk2 − (Rehu; vi)2=kvk2 ≥ 0 =) j Rehu; vij ≤ kukkvk which is what we want except that it is only the real part. However, we know that, for some z 2 C with jzj = 1; Rehzu; vi = Re zhu; vi = jhu; vij 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 − ujk ! 0 and kv − vjk ! 0 so from (3.12) jhu; vi − huj; vjij ≤ jhu; vi − hu; vjij + jhu; vji − huj; vjij ≤ kukkv − vjk + ku − ujkkvjk continuity follows. Corollary 3.2. The Cauchy-Scwharz inequality is optimal in the sense that (3.13) kuk = sup jhu; vij: v2H;kvk≤1 I really will leave this one to you. 3. ORTHONORMAL SEQUENCES 71 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 n 0 X 0 (3.14) hz; z i = zjzj 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) 1 X 2 (3.15) ha; bi = ajbj; a; b 2 l : 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 2 1 2 1 (3.16) L (R) = fu 2 Lloc(R); juj 2 L (R)g=N where N is the space of null functions. The inner product is Z (3.17) hu; vi = uv: Note that we showed that if u; v 2 L2(R) then uv 2 L1(R): We also showed that if R juj2 = 0 then u = 0 almost everywhere, i.e. u 2 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 2 H; (finite or infinite) is said to be orthonormal if keik = 1 for all i and hei; eji = 0 for all i 6= j: Proposition 3.1 (Bessel's inequality). If ei; i 2 N; is an orthonormal sequence in a pre-Hilbert space H; then X 2 2 (3.19) jhu; eiij ≤ kuk 8 u 2 H: i Proof. Start with the finite case, i = 1; : : : ; N: Then, for any u 2 H set N X (3.20) v = hu; eiiei: i=1 This is supposed to be `the projection of u onto the span of the ei'. Anyway, computing away we see that N X (3.21) hv; eji = hu; eiihei; eji = hu; eji i=1 72 3. HILBERT SPACES using orthonormality. Thus, u − v ? ej for all j so u − v ? v and hence (3.22) 0 = hu − v; vi = hu; vi − kvk2: Thus kvk2 = jhu; vij 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) N 2 X 2 2 kvk = jhu; eiij ≤ kuk 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, feig; (finite or infinite) in a pre- Hilbert space is said to be maximal if (3.23) u 2 H; hu; eii = 0 8 i =) u = 0: Theorem 3.1. Every separable pre-Hilbert space contains a maximal orthonor- mal sequence. Proof. Take a countable dense subset { which can be arranged as a sequence fvjg and the existence of which is the definition of separability { and orthonormalize it. First if v1 6= 0 set ei = v1=kv1k: 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 n X w (3.25) w = v − hv ; e ie 6= 0 so e = n+1 n+1 j j m+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 (3.26) H 3 u ? ej 8 j =) u = 0: 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 2 X 2 X 2 2 (3.27) kwkk = jhwk; elij = jhu − wk; elij ≤ ku − wkk l l 6. ISOMORPHISM TO l2 73 where hu; eli = 0 for all l has been used.