Decomposition of the spectrum of a bounded linear operator

Jordan Bell
April 3, 2014

1 Definitions

Let H be a complex Hilbert space. If λ, we also write λ to denote λidHB(H).

For TB(H), the spectrum σ(T) of T is the set of those λ such that the map T-λ is not bijective.11 1 σ(T) is defined to be the set of λ such that vTv-λv is not a bijection. It is a fact that if TB(H) and vTv-λv is a bijection then it is an element of B(H). That it is linear can be proved quickly. The fact that it is bounded is proved using the open-mapping theorem, which states that a surjective bounded linear map from one Banach space to another is an open map, from which it follows that a bijective bounded linear map from one Banach space to another has a bounded inverse. It happens that it is useful for some purposes to write σ(T) as a union of three particular disjoint subsets of itself.

  • The point spectrum σpoint(T) is the set of those λ such that T-λ is not injective. Equivalently, λσpoint(T) if λ is an eigenvalue of T.22 2 The point spectrum is often called the discrete spectrum. From the definition by itself it is not apparent what σpoint(T) has to do either with points or discreteness.

  • The continuous spectrum σcont(T) is the set of those λ such that T-λ is injective, has dense image, and is not surjective.

  • The residual spectrum σres(T) is the set of those λ such that T-λ is injective and does not have dense image.

It is apparent that the sets σpoint(T), σcont(T), and σres(T) are disjoint, and that


If TB(H) then σ(T), but any of the above three sets may be empty; they merely can’t all be empty for a given operator.

2 Residual spectrum

If TB(H) is a normal operator then σres(T)=.33 3 TB(H) is normal if T*T=TT*; in particular, a self-adjoint operator is normal. Equivalently, TB(H) is normal if and only if Tv=T*v for all vH. We prove this. Suppose that T-λ is injective. We have to show that im (T-λ) is dense in H, and thus that λσres(T). (λ might be in σcont(T) or might not be in σ(T); we merely want to show that it is not in σres(T).) We have

H=im (T-λ)¯(im (T-λ)).

Let w(im (T-λ)); we have to show that w=0. For all vH,


so for all vH we have v,(T-λ)*w=0 and therefore (T-λ)*w=0, so wker(T-λ)*=ker(T-λ).44 4 If S is normal then kerS=kerS*. Proof: If vkerS then Sv,Sv=0, hence S*Sv,v=0, hence SS*v,v=0, hence S*v,S*v=0, hence S*v=0, hence vkerS*. As T-λ is injective, w=0, completing the proof.

3 Point spectrum

If AB(H) is normal then it is straightforward to show that kerA=kerA*. Also, if TB(H) is normal then for any z, T-z is normal. Thus, ker(T-z)={0} if and only if ker((T-z)*)={0}. That is, λσpoint(T) if and only if λ¯σpoint(T*). For X we define X*={z¯:zX}. We have shown that if TB(H) is normal then


4 Continuous spectrum

If λσcont(T), then im (T-λ) is dense in H. Also,

(T-λ)-1:im (T-λ)H

is a surjective linear map (the inverse of a linear map is itself a linear map) that is not continuous. For im (T-λ) is dense in H, so if (T-λ)-1 were continuous then it would have a unique extension to a continuous, hence bounded, map HH. Using this and the fact that T-λ is not surjective will give a contradiction.

5 Approximate point spectrum

Let λ(σpoint(T)σres(T)). I claim that λσcont(T) if and only if λ is in the approximate point spectrum of T, the set of those λ such that there is is a sequence vnH with vn=1 and (T-λ)vn0 as n. It is a fact that for TB(H), T is invertible if and only if T(H) is dense in H and there is some α>0 such that Tvαv for all vH.55 5 In words, TB(H) is invertible if and only if it has dense image and is bounded below. This result is proved in Paul Halmos, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, p. 38, §21, Theorem 3. If λσcont(T), then T-λ is not invertible but the image of T-λ is dense in H, then it must therefore be that T-λ is not bounded below. That is, there is no α>0 such that for every wH we have (T-λ)wαw. Then for each n there is some wH such that (T-λ)wn<1nwn. Let vn=wnwn. We have vn=1 and (T-λ)vn<1n, showing that λσap(T).

On the other hand, if λσap(T) then there is a sequence vnH such that (T-λ)vn0 as n. Then for any α>0, there is some vn such that (T-λ)vn<α=αvn, so T-λ is not invertible and hence λσ(T). Since we assumed that λ(σpoint(T)σres(T)), we then have λσcont(T).

Halmos shows in Problem 62 of his Hilbert Space Problem Book that σap(T) is a closed subset of , and proves in Problem 63 that σ(T)σap(T): the boundary of the spectrum of T is contained in the approximate point spectrum of T.

6 Normal operators

We showed earlier that if TB(H) is normal then σpoint(T)*=σpoint(T*), where, for X, X*={z¯:zX}. This is one reason why it can be helpful to know that an operator is normal. Using this we can show something more about normal operators. Let TB(H) be normal and suppose that λ,μσpoint(T) are distinct. Then there are nonzero v,wH with Tv=λv and Tw=μw. Using T*w=μ¯w we get

λv,w = λv,w
= Tv,w
= v,T*w
= v,μ¯w
= μv,w.

As λμ, this means that v,w=0. In words, if TB(H) is normal, then its eigenspaces are mutually orthogonal.

7 Compact operators

Let K(H) be the closure in B(H) of the set of finite-rank operators. We call the elements of K(H) compact operators. The following are equivalent ways to state that an operator is compact.

  • TB(H) is compact if and only if for every bounded subset S of H, the closure of the image T(S) is compact.

  • TB(H) is compact if and only if for every sequence vnH with vn=1, T(vn) has a convergent subsequence.

  • Let B be the closed unit ball in H, and let B be a topological space with the weak topology: a net vαB converges weakly to vB if for all wH we have vα,wv,w. If H is separable, then the weak topology on B is metrizable and thus can be characterized using merely sequences instead of nets.66 6 This is proved in Paul Halmos, Hilbert Space Problem Book, Problem 18. TB(H) is compact if and only if the restriction of T to B is continuous BH, where B has the weak topology and H has the norm topology.

If KK(H) and V is a closed subspace of V such that K(V)V, then the restriction of K to V is an element of K(V).

B(H) is a C*-algebra, and K(H) is a C*-subalgebra of B(H). If TK(H) and SB(H), then


Hence K(H) is an ideal of the C*-algebra B(H).77 7 The C*-algebra B(H)/K(H) is called the Calkin algebra of H. TB(H) is called a Fredholm operator if T+K(H) is an invertible element of the Calkin algebra. In particular, if TK(H) then idH-T is a Fredholm operator. TB(H) is a Fredholm operator if and only if the following three conditions holds: im T is closed in H, kerT is finite dimensional, and kerT* is finite dimensional. This equivalence is called Atkinson’s theorem. The index of a Fredholm operator is dimkerT-dimkerT*. If TK(H), then idH-T has index 0.

Useful facts about compact operators are proved in Yuri A. Abramovich and Charalambos D. Aliprantis, An Invitation to Operator Theory, p. 272, §7.1.

8 Fredholm alternative

The Fredholm alternative states that if KK(H), λ0, and ker(K-λ)={0} then (K-λ)-1B(H).88 8 We are following Paul Halmos, Hilbert Space Problem Book, p. 293, Problem 140. Equivalently, if KK(H), λ0, and λσpoint(K) then λσ(K). Equivalently, if KK(H), then


The above forms are the ones that we want to use. The following is the one that we want to prove, which is equivalent because a nonzero multiple of a compact operator is compact: If KK(H) and ker(idH-K)={0} then (idH-K)-1B(H).

We prove two standalone lemmas that we then use to prove the Fredholm alternative.

  • Let KK(H) let A=idH-KB(H), and suppose that A(H)=H. Define Kn=ker(An). We have K1K2. Assume by contradiction that K1{0}. Then there is some nonzero f1K1. As A(H)=H, there is some f2H with Af2=f1; but A2f2=Af1=0 and Af2=f10, so f2K2K1. Let fn+1Kn+1Kn with Afn+1=fn. Therefore K1,K2, are a strictly increasing sequence of subspaces of H. Using Gram-Schmidt, there is an orthonormal sequence e1,e2, with enKn for all n; we caution that we do not necessarily have Aen+1=en. As Aen+1Kn, Aen+1,en+1, giving


    Each en is an element of the closed unit ball B, and en0 weakly (this is the case for any orthonormal sequence in H, basis or not, and is proved using Bessel’s inequality). Since K is compact, it is continuous BH where B has the weak topology and H has the norm topology; but en0, K(0)=0, and Ken1, so Ken does not converge to 0 in H, a contradiction. Therefore K1={0}, that is, kerA={0}.

  • Let KK(H) and let A=idH-KB(H). Suppose by contradiction that A is not bounded below on (kerA). So for every α>0 there is some w(kerA) such that Awαw. Then for all n there is some wn(kerA) with Awn<1nwn. Let vn=wnwn(kerA). Then


    As K is compact, there is some subsequence va(n) such that Kva(n) converges to some vH. AvnAv and Avn0, so vkerA. On the other hand, because vn=Avn+Kvnv and (kerA) is closed, we have v(kerA), so vkerA(kerA)={0}. But as vn=1 for each n, we have v=1, a contradiction. Therefore, A is bounded below on (kerA).

If KK(H) and A=idH-K, then by the second of the two lemmas, we have that A is bounded below on (kerA): there is some α>0 such that Avαv for all v(kerA). If


with wnwH, then there are vn(kerA) such that Avn=wn. As wn converges, for all ϵ>0 there is some N such that if n,mN then wn-wmϵ, so A(vn-vm)=Avn-Avmϵ. But




so vn is a Cauchy sequence and hence converges, say to v. vn(kerA), which is closed, so v(kerA). Then wn=AvnAvA(H). Therefore, if KK(H) and A=idH-K, then A(H) is closed in H.

Let KK(H) and A=idH-K, and suppose that kerA={0}. By the above paragraph, A(H) is closed in H. K*K(H) and A*=idH-K*,99 9 The adjoint of a compact operator is itself a compact operator. This is true even for Banach spaces, and a proof of this is given by Paul Garrett in his note Compact operators on Banach spaces: Fredholm-Riesz. A bounded linear map K:XY, where X and Y are Banach spaces, is said to be compact if for every bounded subset S of X, the closure of K(S) in Y is compact. so by the above paragraph we also get that A*(H) is closed in H. It is a fact that if TB(H) then kerT*=(T(H)), so using this with T=A* we get


taking orthogonal complements and using the fact that the double orthogonal complement of a subspace is its closure and that A*(H) is closed, we obtain


Since kerA={0}, we have A*(H)=H. Then we can apply the first of the two lemmas: as A*=idH-K*, K*K(H), and A*(H)=H, we have kerA*={0}. We now apply the second of the two lemmas: A* is bounded below on (kerA*)=H. Using the fact that if TB(H) is bounded below and has dense image then T-1B(H), we get (A*)-1B(H) (A*(H)=H so A* certainly has dense image). Taking adjoints commutes with taking inverses, so A-1B(H). This completes the proof of the Fredholm alternative.

9 Compact self-adjoint operators

It is a fact that if TB(H) is self-adjoint then1010 10 This is proved in Anthony W. Knapp, Advanced Real Analysis, p.  37, Proposition 2.2.


Let TB(H) be compact and self-adjoint and T0. Since T is self-adjoint, Tv,v, so either T=supv=1Tv,v or T=-infv=1Tv,v. Say the first is the case. Let vn=1 and Tvn,vnT as n. Then, as T=T*,

Tvn-Tvn,Tvn-Tvn = Tvn,Tvn-Tvn,Tvn-Tvn,Tvn
= Tvn2-2TTvn,vn+T2vn2
= 2T2-2TTvn,vn.

Thus Tvn-Tvn0 as n, that is, Tvn-Tvn0. On the other hand, as vn=1 for each n, there is some subsequence va(n) such that Tva(n) converges, say to v. Together with Tvn-Tvn0 this gives Tva(n)v as n, from which we get v=1T>0. Thus v0. And


which means that Tσpoint(T). Likewise, in the case T=-infv=1Tv,v we get -Tσpoint(T).

10 Multiplication operators

Let (X,Σ,μ) be a σ-finite measure space. L2(X) is a Hilbert space1111 11 Whether L2(X) is separable depends on the measure space (X,μ). Let 𝒮 be the set of all measurable subsets of X with finite measure, and let ρ(A,B)=μ(ABAB), the measure of the symmetric difference of A and B. One shows that 𝒮 is a pseudometric space with pseudometric ρ. It is a fact that L2(X) is separable if and only if 𝒮 is separable; cf. Paul Halmos, Measure Theory, p. 177, §42. For this to be the case, it suffices that X is σ-finite and that its σ-algebra is countably generated. with inner product


where g*(x)=g(x)¯.

A multiplication operator on L2(X) is an operator Mϕ:L2(X)L2(X), ϕL(X), of the form




where ϕ is the essential supremum of ϕ(x) for xX, we have Mϕϕ and so MϕB(H). L(X) is a C*-algebra, and so is B(L2(X)). If X is σ-finite then I claim that


is an injective homomorphism of C*-algebras. It is straightforward to show that this map is a homomorphism of C*-algebras, and this does not use the assumption that X is σ-finite. For our benefit, we shall show that ϕMϕ is injective. If ϕ0, then ϕ>0, so for


we have 0<μ(E). Because (X,μ) is σ-finite, there is some subset F of E with 0<μ(F)<. As f=ϕ*χFL2(X), we have

Mϕf = Fϕ(x)ϕ(x)¯𝑑μ(x)
= 14ϕ2μ(F)
> 0,

so Mϕ0. Generally, an injective homomorphism of C*-algebras is an isometry, so Mϕ=ϕ.

As Mϕϕ*=MϕMϕ*=MϕMϕ* and Mϕϕ*=Mϕ*ϕ, we have MϕMϕ*=Mϕ*Mϕ, namely, a multiplication operator is a normal operator. Since residual spectrum of a normal operator is empty, the residual spectrum of a multiplication operator is empty.

For ϕL(X), we define the essential range of ϕ to be the set

{z:if ϵ>0 then μ({xX:|f(x)-z|<ϵ})>0}.

Equivalently, the essential range of ϕ is the set of those z such that for all ϵ>0,


in words, those z such that the inverse image of every ϵ-disc about z has positive measure. Equivalently, the essential range of ϕ is the intersection of all closed subsets K of such that for almost all xX, ϕ(x)K. It is a fact that if ϕL(X) then the essential range of ϕ is a compact subset of .

Let ϕL(X). If λ is not in the essential range of ϕ, then there is some ϵ>0 such that μ(ϕ-1(Dϵ(λ)))=0, which means that for almost all xX we have |ϕ(x)-λ|ϵ. Define ψ(x)=1ϕ(x)-λ. For almost all xX,


hence ψL(X). Then


so Mϕ-λ is invertible. But Mϕ-λ=Mϕ-λ, so Mϕ-λ is invertible and hence λσ(Mϕ).

If λ is in the essential range of ϕ, then for each n we have 0<μ(ϕ-1(D1/n(λ))); since Σ is σ-finite, for each n there is a subset En of ϕ-1(D1/n(λ)) with 0<μ(En)<, and so χEnL2(X). We have, since |ϕ(x)-λ|<1n for xEn,

(Mϕ-λ)χEn2 = X|(ϕ(x)-λ)χEn(x)|2𝑑μ(x)
= En|ϕ(x)-λ|2𝑑μ(x)
= 1n2XχEn𝑑μ(x)
= 1n2χEn2,

so for each n,


It follows that Mϕ-λ is not invertible, as it is not bounded below. Therefore λσ(Mϕ). Therefore the essential range of ϕL(X) is equal to the spectrum of MϕB(L2(X)).

We say that ϕL(X) is invertible if there is some ψL(X) such that ϕ(x)ψ(x)=1 for almost all xX. (It would not make sense to demand that ϕ(x)ψ(x)=1 for all xX.) For ϕL(X) to be invertible, it is necessary and sufficient that there is some α>0 such that |ϕ(x)|α for almost all xX (lest its inverse not have an essential supremum).

If λ is not just an element of the essential range ϕ but is an isolated element of the essential range, then we can say more than just that λσ(Mϕ). In this case, there is some ϵ>0 such that the intersection of Dϵ(λ) and the essential range of ϕ is equal to the singleton {λ}. Let E be a subset of ϕ-1(Dϵ(λ)) with 0<μ(E)<. For almost all xX, ϕ(x) is an element of the essential range of ϕ, hence for almost all xE we have ϕ(x)=λ. Therefore, for almost all xX we have


Hence (Mϕ-λ)χE=0, and as μ(E)>0 we have χE0. Therefore, if λ is an isolated element of the essential range of ϕ then λσpoint(Mϕ).1212 12 If λ is an isolated element of the essential range of ϕ then one finds that the inverse image of the singleton {λ} has positive measure. I would be surprised if this were not the origin of the term point spectrum. Being isolated corresponds to being discrete.

11 Functional calculus

Let TB(H) be self-adjoint. The spectrum σ(T) is a compact subset of , and one checks that the set C(σ(T)) of continuous functions σ(T) is a C*-algebra, with norm f=supλσ(T)|f(λ)|.

Let [x] be the set of polynomials with complex coefficients. For TB(H) self-adjoint and p(x)=k=0nakxk[x], we define




Tp(T) is a homomorphism of C*-algebras, where1313 13 If we had not stipulated that T be self-adjoint then we would have to define the conjugation of polynomials as conjugation of polynomial functions: for p defined by p(z)=k=0nakzk, then p* is defined by p*(z)=k=0nak¯(z¯)k.


It is a fact that1414 14 See Paul Halmos, Hilbert Space Problem Book, p. 62, Problem 97.


where for M we define p(M)={p(z):zM}. It is also a fact that


this is proved using the result that the norm of a normal operator T is equal to its spectral radius, which is given by the two following expressions that one proves are equal:


The above is used to define f(T) for any continuous function f:σ(T). This map C(σ(T))B(H) is called the continuous functional calculus. It is an isometric homomorphism of C*-algebras. The continuous functional calculus can be used to prove things about the spectrum of self-adjoint operators that do not obviously have to do with making sense of continuous functions applied to these operators.

Let TB(H) be self-adjoint and let λσ(T) be an isolated point in σ(T). I will show that that λσpoint(T). Since λ is isolated in σ(T), the function f:σ(T) defined by


is continuous. Since f is continuous, f(T)B(H), and because f=f*, f(T) is self-adjoint. Let P=f(T). As P=f=1, P0. Define gC(σ(T)) by g(x)=(x-λ)f(x). Then g(x)=0 for all xσ(T), so g(T)=0, i.e.


Hence im Pker(T-λ). As P0, there is some vH with Pv0. Then (T-λ)Pv=0, Pv0, so λσpoint(T).

12 Spectral measures

It is a fact that if f0 then f(T)0, where, for a self-adjoint operator T, T0 means Tv,v0 for all vH. For TB(H) self-adjoint and vH, using the continuous functional calculus talked about in the previous section we define ϕ:C(σ(T)) by


ϕ is a positive linear functional: if f is real-valued and f(x)0 for all xσ(T), then ϕ(f)0. σ(T) is indeed a locally compact Hausdorff space and since σ(T) is compact the continuous functions of compact support on σ(T) are precisely the continuous functions on σ(T), so we satisfy the conditions of the Riesz-Markov theorem. Thus there exists a unique regular Borel measure μ on the Borel σ-algebra of σ(T) such that, for all fC(σ(T)),


Lebesgue’s decomposition theorem states that



  • μac is absolutely continuous with respect to Lebesuge measure: if A is a measurable subset of σ(T) and its Lebesgue measure is 0, then μac(A)=0.

  • μsing and Lebesgue measure are mutually singular,1515 15 There are disjoint measurable sets A and B with AB=σ(T) such that μsing(A)=0 and the Lebesgue measure of B is 0. and if λσ(T) then μsing({λ})=0.

  • There is a countable subset J of σ(T) such that


We define Hac to be the set of those vH such that μ is equal to the absolutely continuous part of its Lebesgue decomposition, i.e. the other two parts are 0, and we define Hsing and Hpp likewise. (Note that we first took vH and then defined μ using v.) One proves that Hac, Hsing and Hpp are closed subspaces of H and that they are invariant under T, and defines the absolutely continuous spectrum of T to be the spectrum of the restriction of T to Hac; the singular spectrum of T to be the spectrum of the restriction of T to Hsing; and the pure point spectrum of T to be the spectrum of the restriction of T to Hpp. It is a fact that1616 16 See Reed and Simon, Methods of Modern Mathematical Physics. I: Functional Analysis, revised and enlarged ed., p. 231, §VII.2.


but these three sets might not be disjoint.