LECTURE NOTES FOR 18.155, FALL 2004 9916. Spectral theoremFor a bounded operator T on a Hilbert space we define the spectrumas the set(16.1) spec(T ) = {z ∈ C; T − z Id is not invertible}.Proposition 16.1. For any bounded linear operator on a Hilbert spacespec(T ) ⊂ C is a compact subset of {|z| ≤ kT k}.Proof. We show that the set C \ spec(T ) (generally called the resolventset of T ) is open and contains the complement of a sufficiently largeball. This is based on the convergence of the Neumann series. Namelyif T is bounded and kT k < 1 then(16.2) (Id −T )−1=∞Xj=0Tjconverges to a bounded operator which is a two-sided inverse of Id −T.Indeed, kTjk ≤ kT kjso the series is convergent and composing withId −T on either side gives a telescoping series reducing to the identity.Applying this result, we first see that(16.3) (T − z) = −z(Id −T/z)is invertible if |z| > kT k. Similarly, if (T − z0)−1exists for some z0∈ Cthen(16.4) (T −z) = (T −z0)−(z−z0) = (T −z0)−1(Id −(z−z0)(T −z0)−1)exists for |z − z0|k(T − z0)−1k < 1. In general it is rather difficult to precisely locate spec(T ).However for a bounded self-adjoint operator it is easier. One sign ofthis is the the norm of the operator has an alternative, simple, charac-terization. Namely(16.5) if A∗= A then supkφk=1hAφ, φi| = kAk.If a is this supermum, then clearly a ≤ kAk. To see the converse, chooseany φ, ψ ∈ H with norm 1 and then replace ψ by eiθψ with θ chosenso that hAφ, ψi is real. Then use the polarization identity to write(16.6) 4hAφ, ψi = hA(φ + ψ), (φ + ψ)i − hA(φ − ψ), (φ − ψ)i+ ihA(φ + iψ), (φ + iψ)i − ihA(φ − iψ), (φ − iψ)i.Now, by the assumed reality we may drop the last two terms and seethat(16.7) 4|hAφ, ψi| ≤ a(kφ + ψk2+ kφ − ψk2) = 2a(kφk2+ kψk2) = 4a.100 RICHARD B. MELROSEThus indeed kAk = supkφk=kψk=1|hAφ, ψi| = a.We can always subtract a real constant from A so that A0= A − tsatisfies(16.8) − infkφk=1hA0φ, φi = supkφk=1hA0φ, φi = kA0k.Then, it follows that A0± kA0k is not invertible. Indeed, there exists asequence φn, with kφnk = 1 such that h(A0− kA0k)φn, φni → 0. Thus(16.9)k(A0−kA0k)φnk2= −2hA0φn, φni+kA0φnk2+kA0k2≤ −2hA0φn, φni+2kA0k2→ 0.This shows that A0− kA0k cannot be invertible and the same argumentworks for A0+ kA0k. For the original operator A if we set(16.10) m = infkφk=1hAφ, φi M = supkφk=1hAφ, φithen we conclude that neither A − m Id nor A − M Id is invertible andkAk = max(−m, M).Proposition 16.2. If A is a bounded self-adjoint operator then, withm and M defined by (16.10),(16.11) {m} ∪ {M} ⊂ spec(A) ⊂ [m, M].Proof. We have already shown the first part, that m and M are inthe spectrum so it remains to show that A − z is invertible for allz ∈ C \ [m, M].Using the self-adjointness(16.12) Imh(A − z)φ, φi = − Im zkφk2.This implies that A − z is invertible if z ∈ C \ R. First it shows that(A − z)φ = 0 implies φ = 0, so A − z is injective. Secondly, the range isclosed. Indeed, if (A − z)φn→ ψ then applying (16.12) directly showsthat kφnk is bounded and so can be replaced by a weakly convergentsubsequence. Applying (16.12) again to φn− φmshows that the se-quence is actually Cauchy, hence convergens to φ so (A − z)φ = ψ is inthe range. Finally, the orthocomplement to this range is the null spaceof A∗− ¯z, which is also trivial, so A − z is an isomorphism and (16.12)also shows that the inverse is bounded, in fact(16.13) k(A − z)−1k ≤1| Im z|.When z ∈ R we can replace A by A0satisfying (16.8). Then we haveto show that A0− z is inverible for |z| > kAk, but that is shown in theproof of Proposition 16.1. LECTURE NOTES FOR 18.155, FALL 2004 101The basic estimate leading to the spectral theorem is:Proposition 16.3. If A is a bounded self-adjoint operator and p is areal polynomial in one variable,(16.14) p(t) =NXi=0citi, cN6= 0,then p(A) =NPi=0ciAisatisfies(16.15) kp(A)k ≤ supt∈[m,M]|p(t)|.Proof. Clearly, p(A) is a bounded self-adjoint operator. If s /∈ p([m, M])then p(A) − s is invertible. Indeed, the roots of p(t) − s must cannotlie in [m.M], since otherwise s ∈ p([m, M]). Thus, factorizing p(s) − twe have(16.16)p(t) − s = cNNYi=1(t − ti(s)), ti(s) /∈ [m, M] =⇒ (p(A) − s)−1existssince p(A) = cNPi(A − ti(s)) and each of the factors is invertible.Thus spec(p(A)) ⊂ p([m, M]), which is an interval (or a point), andfrom Proposition 16.3 we conclude that kp(A)k ≤ sup p([m, M]) whichis (16.15). Now, reinterpreting (16.15) we have a linear map(16.17) P(R) 3 p 7−→ p(A) ∈ B(H)from the real polynomials to the bounded self-adjoint operators whichis continuous with respect to the supremum norm on [m, M]. Sincepolynomials are dense in continuous functions on finite intervals, wesee that (16.17) extends by continuity to a linear map(16.18)C([m, M]) 3 f 7−→ f (A) ∈ B(H), kf(A)k ≤ kfk[m,M], f g(A) = f (A)g(A)where the multiplicativity follows by continuity together with the factthat it is true for polynomials.Now, consider any two elements φ, ψ ∈ H. Evaluating f (A) on φ andpairing with ψ gives a linear map(16.19) C([m, M]) 3 f 7−→ hf (A)φ, ψi ∈ C.This is a linear functional on C([m, M]) to which we can apply the Rieszrepresentatin theorem and conclude that it is defined by integration102 RICHARD B. MELROSEagainst a unique Radon measure µφ,ψ:(16.20) hf(A)φ, ψi =Z[m,M]fdµφ,ψ.The total mass |µφ,ψ| of this measure is the norm of the functional.Since it is a Borel measure, we can take the integral on −∞, b] for anyb ∈ R ad, with the uniqueness, this shows that we have a continuoussesquilinear map(16.21)Pb(φ, ψ) : H×H 3 (φ, ψ) 7−→Z[m,b]dµφ,ψ∈ R, |Pb(φ, ψ)| ≤ kAkkφkkψk.From the Hilbert space Riesz representation theorem it follows thatthis sesquilinear form defines, and is determined by, a bounded linearoperator(16.22) Pb(φ, ψ) = hPbφ, ψi, kPbk ≤ kAk.In fact, from the functional calculus (the multiplicativity in (16.18))we see that(16.23) P∗b= Pb, P2b= Pb, kPbk ≤ 1,so Pbis a projection.Thus the spectral theorem gives us an increasing (with b) family ofcommuting self-adjoint projections such that µφ,ψ((−∞, b]) = hPbφ, ψidetermines the Radon measure for which (16.20) holds. One can gofurther and think of Pbitself as determining a measure(16.24) µ((−∞, b]) = Pbwhich takes values in the