30 RICHARD B. MELROSE5. Hilbert spaceWe have shown that Lp(X, µ) is a Banach space – a complete normedspace. I shall next discuss the class of Hilbert spaces, a special class ofBanach spaces, of which L2(X, µ) is a standard example, in which thenorm arises from an inner product, just as it does in Euclidean space.An inner product on a vector space V over C (one can do the realcase too, not much changes) is a sesquilinear formV × V → Cwritten (u, v), if u, v ∈ V . The ‘sesqui-’ part is just linearity in the firstvariable(5.1) (a1u1+ a2u2, v) = a1(u1, v) + a2(u2, v),anti-linearly in the second(5.2) (u, a1v1+ a2v2) = a1(u, v1) + a2(u, v2)and the conjugacy condition(5.3) (u, v) = (v, u) .Notice that (5.2) follows from (5.1) and (5.3). If we assume in additionthe positivity condition8(5.4) (u, u) ≥ 0 , (u, u) = 0 ⇒ u = 0 ,then(5.5) kuk = (u, u)1/2is a norm on V , as we shall see.Suppose that u, v ∈ V have kuk = kvk = 1. Then (u, v) = eiθ|(u, v)|for some θ ∈ R. By choice of θ, e−iθ(u, v) = |(u, v)| is real, so expandingout using linearity for s ∈ R,0 ≤ (e−iθu − sv , e−iθu − sv)= kuk2− 2s Re e−iθ(u, v) + s2kvk2= 1 − 2s|(u, v)| + s2.The minimum of this occurs when s = |(u, v)| and this is negativeunless |(u, v)| ≤ 1. Using linearity, and checking the trivial cases u =or v = 0 shows that(5.6) |(u, v)| ≤ kuk kvk, ∀ u, v ∈ V .This is called Schwarz’9inequality.8Notice that (u, u) is real by (5.3).9No ‘t’ in this Schwarz.LECTURE NOTES FOR 18.155, FALL 2004 31Using Schwarz’ inequalityku + vk2= kuk2+ (u, v) + (v, u) + kvk2≤ (kuk + kvk)2=⇒ ku + vk ≤ kuk + kvk ∀ u, v ∈ Vwhich is the triangle inequality.Definition 5.1. A Hilbert space is a vector space V with an innerproduct satisfying (5.1) - (5.4) which is complete as a normed space(i.e., is a Banach space).Thus we have already shown L2(X, µ) to be a Hilbert space for anypositive measure µ. The inner product is(5.7) (f, g) =ZXfg dµ ,since then (5.3) gives kfk2.Another important identity valid in any inner product spaces is theparallelogram law:(5.8) ku + vk2+ ku − vk2= 2kuk2+ 2kvk2.This can be used to prove the basic ‘existence theorem’ in Hilbert spacetheory.Lemma 5.2. Let C ⊂ H, in a Hilbert space, be closed and convex (i.e.,su + (1 − s)v ∈ C if u, v ∈ C and 0 < s < 1). Then C contains aunique element of smallest norm.Proof. We can certainly choose a sequence un∈ C such thatkunk → δ = inf {kvk ; v ∈ C} .By the parallelogram law,kun− umk2= 2kunk2+ 2kumk2− kun+ umk2≤ 2(kunk2+ kumk2) − 4δ2where we use the fact that (un+um)/2 ∈ C so must have norm at leastδ. Thus {un} is a Cauchy sequence, hence convergent by the assumedcompleteness of H. Thus lim un= u ∈ C (since it is assumed closed)and by the triangle inequality|kunk − kuk| ≤ kun− uk → 0So kuk = δ. Uniqueness of u follows again from the parallelogram lawwhich shows that if ku0k = δ thenku − u0k ≤ 2δ2− 4k(u + u0)/2k2≤ 0 .32 RICHARD B. MELROSE The fundamental fact about a Hilbert space is that each elementv ∈ H defines a continuous linear functional byH 3 u 7−→ (u, v) ∈ Cand conversely every continuous linear functional arises this way. Thisis also called the Riesz representation theorem.Proposition 5.3. If L : H → C is a continuous linear functional ona Hilbert space then this is a unique element v ∈ H such that(5.9) Lu = (u, v) ∀ u ∈ H ,Proof. Consider the linear spaceM = {u ∈ H ; Lu = 0}the null space of L, a continuous linear functional on H. By the as-sumed continuity, M is closed. We can suppose that L is not identicallyzero (since then v = 0 in (5.9)). Thus there exists w /∈ M. Considerw + M = {v ∈ H ; v = w + u , u ∈ M} .This is a closed convex subset of H. Applying Lemma 5.2 it has aunique smallest element, v ∈ w + M. Since v minimizes the norm onw + M,kv + suk2= kvk2+ 2 Re(su, v) + ksk2kuk2is stationary at s = 0. Thus Re(u, v) = 0 ∀ u ∈ M, and the sameargument with s replaced by is shows that (v, u) = 0 ∀ u ∈ M.Now v ∈ w + M, so Lv = Lw 6= 0. Consider the element w0=w/Lw ∈ H. Since Lw0= 1, for any u ∈ HL(u − (Lu)w0) = Lu − Lu = 0 .It follows that u − (Lu)w0∈ M so if w00= w0/kw0k2(u, w00) = ((Lu)w0, w00) = Lu(w0, w0)kw0k2= Lu .The uniqueness of v follows from the positivity of the norm. Corollary 5.4. For any positive measure µ, any continuous linearfunctionalL : L2(X, µ) → Cis of the formLf =ZXfg dµ , g ∈ L2(X, µ) .LECTURE NOTES FOR 18.155, FALL 2004 33Notice the apparent power of ‘abstract reasoning’ here! Although weseem to have constructed g out of nowhere, its existence follows fromthe completeness of L2(X, µ), but it is very convenient to express theargument abstractly for a general Hilbert