The spectrum of a self-adjoint operator is a compact subset of

Jordan Bell
April 3, 2014

In these notes I prove that the spectrum of a bounded linear operator from a Hilbert space to itself is a nonempty compact subset of , and that if the operator is self-adjoint then the spectrum is contained in . To show that the spectrum is nonempty I prove various facts about resolvents.

1 Adjoints

1.1 Operator norm

Let H be a Hilbert space with inner product ,:H×H, and define I:HH by Ix=x, xH.. For vH, let v=v,v, and if T:HH is a bounded linear map, let


namely, the operator norm of T.

1.2 Definition of adjoint

The Riesz representation theorem states that if ϕ:H is a bounded linear map then there is a unique vϕH such that


for all xH. Let T:HH be a bounded linear map, and for yH, define ϕy:H by


ϕy:H is a bounded linear map, so by the Riesz representation theorem there is a unique vy such that


for all xH. Define T*:HH by


T*y is well-defined because of the uniqueness in the Riesz representation theorem. For all x,yH,


We call T*:HH the adjoint of T:HH.

1.3 Adjoint is linear

For y1,y2H, we have for all xH that

x,T*(y1+y2) = Tx,y1+y2
= Tx,y1+Tx,y2
= x,T*y1+T*y2.

Hence for all xH,


In particular this is true for x=T*(y1+y2)-T*y1-T*y2, so by the nondegeneracy of , we get


We similarly obtain for all λ and all yH that


Hence T*:HH is a linear map.

1.4 Adjoint is bounded

For x,yH, by the Cauchy-Schwarz inequality we have


so ϕyvy, i.e. the operator norm of ϕy is less than or equal to the norm of vy. If vy0, then vyvy=1 and


It follows that


Then for yH, by the Cauchy-Schwarz inequality and because T is bounded we have

T*y = vy
= ϕy
= supx1ϕy(x)
= supx1|Tx,y|

Therefore T* is bounded. Thus if T:HH is a bounded linear map then its adjoint T*:HH is a bounded linear map.

1.5 Adjoint is involution

Because T*:HH is a bounded linear map, it has an adjoint T**:HH, and T** is itself a bounded linear map. For all x,yH,

Tx,y = x,T*y
= T*y,x¯
= y,T**x¯
= T**x,y.

Hence for all x,yH,


This is true in particular for y=Tx-T**x, so by the nondegeneracy of , we obtain


Thus for any bounded linear map T:HH, T**=T. In words, if T is a bounded linear map from a Hilbert space to itself, then the adjoint of its adjoint is itself. We have shown already that T*T. Hence also T=T**T*, so


If T*=T, we say that T is self-adjoint.

2 Bounded linear operators

Let (H) be the set of bounded linear maps HH. With the operator norm, one checks that (H) is a Banach space. We define a product on (H) by T1T2=T1T2, and thus (H) is an algebra. We have


and thus (H) is a Banach algebra.11 1 The adjoint map *:(H)(H) satisfies, for λ and T1,T2(H), T**=T,(T1+T2)*=T1*+T2*,(λT)*=λ¯T*,T*T=T2. Thus (H) is a C*-algebra. I(H), so we say that (H) is unital. Let sa(H) be the set of all T(H) that are self-adjoint.

Theorem 1.

If TB(H), then T is self-adjoint if and only if Tx,xR for all xH.


If Tsa(H), then for all xH,


so Tx,x.

If T(H) and Tx,x for all xH, then


so, putting A=T-T*, for all xH we have


Thus, for all x,yH we have


and combining these three equations,


But A*=-A, so we get



Ax,y-Ax,y¯=0. (1)

As well, for all x,yH we have



Ax,y+Ax,y¯=0. (2)

By (1) and (2), for all x,yH we have


and thus A=0, i.e. T=T*. ∎

Using the above characterization of bounded self-adjoint operators, we can prove that a limit of bounded self-adjoint operators is itself a bounded self-adjoint operator.

Theorem 2.

sa(H) is a closed subset of B(H).


If Tnsa(H) and TnT(H), then for xH we have


hence Tsa(H). ∎

If Tsa(H) and Tx,x0 for all xH, we say that T is positive. Let +(H) be the set of all positive Tsa(H). For S,Tsa(H), if


we write ST. Thus, we can talk about one self-adjoint operator being greater than or equal to another self-adjoint operator. ST is equivalent to


for all xH.

3 A condition for invertibility

Theorem 3.

If TB(H) and there is some α>0 such that αITT* and αIT*T, then T-1B(H).


By αIT*T, we have for all xH,


so Txαx. This implies that T is injective. By αITT*, we have for all xH,


so T*xαx, and hence T* is injective. Let TxnyH. Then,


Since Txn converges it is a Cauchy sequence, and from the above inequality it follows that xn is a Cauchy sequence, hence there is some xH with xnx. As T is continuous, y=TxT(H), showing that T(H) is a closed subset of H. But it is a fact that if T(H) then the closure of T(H) is equal to (kerT*).22 2 It is straightforward to show that if v is in the closure of T(H) and wkerT* then v,w=0. It is less straightforward to show the opposite inclusion. Thus, as we have shown that T* is injective,


i.e. T is surjective. Hence T:HH is bijective. It is a fact that if T(H) is bijective then T-1(H), completing the proof.33 3 T-1:HH is linear. The open mapping theorem states that if X and Y are Banach spaces and S:XY is a bounded linear map that is surjective, then S is an open map, i.e., if U is an open subset of X then S(U) is an open subset of Y. Here, T(H) and T is bijective, and so by the open mapping theorem T is open, from which it follows that T-1:HH is continuous, and so bounded (a linear map between normed vector spaces is continuous if and only if it is bounded).

4 Spectrum

For T(H), we define the spectrum σ(T) of T to be the set of all λ such T-λI is not bijective, and we define the resolvent set of T to be ρ(T)=σ(T). To say that λρ(T) is to say that T-λI is a bijection, and if T-λI is a bijection it follows from the open mapping theorem that its inverse function is an element of (H): the inverse of a linear bijection is itself linear, but the inverse of a continuous bijection need not itself be continuous, which is where we use the open mapping theorem.

We prove that the spectrum of a bounded self-adjoint operator is real.

Theorem 4.

If TBsa(H), then σ(T)R.


If λ, λ=a+ib, b0, and X=T-λI, then

XX* = (T-λI)(T-λI)*
= (T-(a+ib)I)(T-(a-ib)I)
= T2-(a-ib)T-(a+ib)T+(a2+b2)I
= (a2+b2)I-2aT+T2
= b2I+(aI-T)2
= b2I+(aI-T)(aI-T)*

X*X=XX*b2I and b>0, so by Theorem 3, X=T-λI has an inverse (T-λI)-1(H), showing λσ(T). ∎

5 The spectrum of a bounded linear map is bounded

If λρ(T) then we define Rλ=(T-λI)-1(H), called the resolvent of T.

Theorem 5.

If TB(H) and |λ|>T then λρ(T).


Define Rλ,N(H) by


As T|λ|<1, the geometric series n=0Tn|λ|n converges, from which it follows that Rλ,N is a Cauchy sequence in (H) and so converges to some Sλ(H). We have

Sλ(T-λI)-I Sλ(T-λI)-Rλ,N(T-λI)
= Sλ-Rλ,NT-λI+-TN+1λN+1

which tends to 0 as N. Therefore Sλ(T-λI)=I. And,

(T-λI)Sλ-I (T-λI)Sλ-(T-λI)Rλ,N

whence (T-λI)Sλ=I, showing that


Thus, if |λ|>T then λρ(T). ∎

The above theorem shows that σ(T) is a bounded set: it is contained in the closed disc |λ|T. Moreover, if |λ|>T then we have an explicit expression for the resolvent Rλ:


6 The spectrum of a bounded linear map is closed

Theorem 6.

If TB(H), then ρ(T) is an open subset of C.


If λρ(T), let |μ-λ|<Rλ-1, and define Rμ,N(H) by


Because |μ-λ|<Rλ-1, Rμ,N is a Cauchy sequence in (H) and converges to some Sμ(H). We have, as Rλ=(T-λI)-1,

Sμ(T-μI)-I Sμ(T-μI)-Rμ,N(T-μI)
= Sμ-Rμ,NT-μI
= Sμ-Rμ,NT-μI+-(μ-λ)N+1RλN+1
= Sμ-Rμ,NT-μI+|μ-λ|N+1RλN+1,

which tends to 0 as N. Therefore Sμ(T-μI)=I. One checks likewise that (T-μI)Sμ=I, and hence that


showing that μρ(T). ∎

As σ(T) is bounded and closed, it is a compact set in . Moreover, if λσ(T) and |μ-λ|<Rλ-1, then


7 The spectrum of a bounded linear map is nonempty

Theorem 7.

If TB(H) is self-adjoint, then σ(T).


Suppose by contradiction that σ(T)=.44 4 For each v,wH we are going to construct a bounded entire function depending on v and w, which by Liouville’s theorem must be constant, and it will turn out to be 0. This will lead to a contradiction. If λ,μ, then

(T-λI)(Rλ-Rμ)(T-μI) = (I-(T-λI)Rμ)(T-μI)
= T-μI-(T-λI)
= (λ-μ)I,


Rλ-Rμ=(λ-μ)RλRμ, (3)

the resolvent identity. Thus


and together with Rμ-RλRμ-Rλ we get


If |λ-μ|12Rλ-1, then


whence, for |λ-μ|12Rλ-1,


Therefore, λRλ is a continuous function (H). From this and (3) it follows that for each λ,55 5 There are no complications that appear if we do complex analysis on functions from to a complex Banach algebra rather than on functions from to . Thus this statement is that λRλ is a holomorphic function (H).


Let v,wH and define fv,w: by


For λ,


Thus fv,w is an entire function. For |λ|>T, Rλ=-1λn=0Tnλn, so, for r=T|λ|,

Rλ = 1|λ|n=0Tnλ
= 1|λ|11-r
= 1|λ|11-T|λ|
= 1|λ|-T.

Hence, for |λ|>T,

|fv,w(λ)| = |Rλv,w|

from which it follows that fv,w is bounded and that lim|λ|fv,w(λ)=0. Therefore by Liouville’s theorem, fv,w(λ)=0 for all λ. Let’s recap: for all v,wH and for all λ, Rλv,w=0. Switching the order of the universal quantifiers, for all λ and for all v,wH we have Rλv,w=0, which implies that for all λ we have Rλ=0. But by assumption Rλ is invertible, so this is a contradiction. Hence σ(T) is nonempty. ∎