Spectral family and resolution of the identity

Posted on 2026/06/25 by arkajad

This post is (hopefully) the last one in a short series on real and complex Hilbert spaces. After this post we will return to the study of real and complex spaces with indefinite metric, Krein spaces and their geometry.

In the previous post, Square root lemma, we constructed the positive square root of any bounded positive self-adjoint operator on a real or complex Hilbert space H. In the present post we will use that square-root construction to associate, with every self-adjoint operator, a family of projection operators, its spectral family or resolution of the identity.

Throughout the post, H denotes a real or complex Hilbert space, and L(H) denotes the algebra of bounded linear operators on H.

1. The absolute value of an operator

Let A\in L(H) be self-adjoint. Then A^2 is self-adjoint and positive. We define

(1)   \begin{equation*} |A| = \sqrt{A^2}. \end{equation*}

From Theorem 1 in Square root lemma we know that |A| commutes with every bounded linear or antilinear operator that commutes with A^2, and hence in particular with every such operator that commutes with A. We will use this “absolute value” in the construction of the resolution of the identity associated with A.

Remark 1. [The finite-dimensional case] In the finite-dimensional case we can always choose an orthonormal basis in which the matrix representing A is diagonal, with diagonal entries equal to the eigenvalues of A. In such a representation, taking powers of A corresponds to taking powers of the diagonal entries. Therefore taking the square root of a positive operator is the same as taking square roots of its non-negative diagonal eigenvalues, and taking the absolute value |A| is simply taking absolute values of the eigenvalues.

Let A=A^*\in L(H). For any real \lambda we set

(2)   \begin{equation*} A_\lambda := A - \lambda I, \end{equation*}

and

(3)   \begin{equation*} A_\lambda^+ := \frac{1}{2}\bigl(A_\lambda + |A_\lambda|\bigr). \end{equation*}

We define \mathcal{N}_\lambda to be the kernel of A_\lambda^+:

(4)   \begin{equation*} \mathcal{N}_\lambda := \{x\in H : A_\lambda^+ x = 0\}. \end{equation*}

2. Resolution of the identity: the construction

Definition 1. For any real \lambda we define E_\lambda to be the orthogonal projection onto the closed subspace \mathcal{N}_\lambda:

    \[ E_\lambda : H \to H, \qquad \operatorname{Ran}(E_\lambda) = \mathcal{N}_\lambda, \quad \ker(E_\lambda) = \mathcal{N}_\lambda^\perp. \]

We call the family \{E_\lambda : \lambda\in \mathbb{R}\} the resolution of the identity associated with A.

Remark 2. The factor \tfrac{1}{2} in the definition of A_\lambda^+ is not essential. It is included mainly to make the formula look symmetric. The kernels of A_\lambda + |A_\lambda| and of A_\lambda^+ coincide.

The next proposition records a simple but important structural property: the resolution of the identity commutes with every operator that commutes with A.

Proposition 1. With the assumptions and notation as above, E_\lambda commutes with any bounded linear or antilinear operator C that commutes with A. In particular,

(5)   \begin{equation*} E_\lambda E_\mu = E_\mu E_\lambda \end{equation*}

for all \lambda,\mu\in \mathbb{R}.

Proof. Suppose C is a bounded linear operator commuting with A, i.e.\ CA = AC. Then C commutes with A_\lambda = A - \lambda I, and by the construction of |A_\lambda| and A_\lambda^+ it also commutes with |A_\lambda| and A_\lambda^+.

Since A_\lambda^+=A_\lambda^{+*}, from C A_\lambda^+ = A_\lambda^+ C we obtain

    \[ A_\lambda^+ C^* = C^* A_\lambda^+. \]

Thus C^* also commutes with A_\lambda^+.

Let x\in \mathcal{N}_\lambda, so A_\lambda^+ x = 0. Then

    \[ A_\lambda^+ (C x) = C \bigl(A_\lambda^+ x\bigr) = C\cdot 0 = 0, \]

so C x\in \mathcal{N}_\lambda. Hence

    \[ C\mathcal{N}_\lambda \subseteq \mathcal{N}_\lambda. \]

Similarly, from A_\lambda^+ C^* = C^* A_\lambda^+ we obtain

    \[ C^* \mathcal{N}_\lambda \subseteq \mathcal{N}_\lambda. \]

We also have C\mathcal{N}_\lambda^\perp \subseteq \mathcal{N}_\lambda^\perp. Indeed, if x\in \mathcal{N}_\lambda^\perp, i.e.\ (x,y) = 0 for all y\in \mathcal{N}_\lambda, then for every y\in\mathcal{N}_\lambda we have

    \[ (C x, y) = (x, C^* y) = 0, \]

since C^* y\in \mathcal{N}_\lambda. Thus C x \in \mathcal{N}_\lambda^\perp.

Therefore both \mathcal{N}_\lambda and its orthogonal complement \mathcal{N}_\lambda^\perp are invariant under C. It follows that C commutes with the orthogonal projection E_\lambda onto \mathcal{N}_\lambda, i.e.

    \[ C E_\lambda = E_\lambda C \]

on the whole space H. The same argument works for C antilinear.

In particular, taking C=E_\mu we see that E_\lambda E_\mu = E_\mu E_\lambda for all \lambda,\mu\in \mathbb{R}.

\blacksquare

Recall the simple but useful fact: the product of two commuting orthogonal projections is again an orthogonal projection. If P projects onto a closed subspace M, and Q projects onto a closed subspace N, and if PQ=QP, then PQ projects onto M\cap N. We use this repeatedly below.

Lemma 1.  With the assumptions and notation as above, the following operator inequalities hold:

(6)   \begin{equation*} A_\lambda E_\lambda \leq 0, \end{equation*}

(7)   \begin{equation*} A_\lambda (I - E_\lambda) \geq 0. \end{equation*}

Proof. From the definition of E_\lambda we have

    \[ A_\lambda^+ E_\lambda = 0, \]

that is,

    \[ \bigl(A_\lambda + |A_\lambda|\bigr) E_\lambda = 0. \]

Therefore

    \[ A_\lambda E_\lambda = - |A_\lambda| E_\lambda. \]

Since |A_\lambda| is positive and commutes with E_\lambda, we have

    \[ -|A_\lambda| E_\lambda = -E_\lambda |A_\lambda| E_\lambda \leq 0, \]

which gives 6.

For (7) we observe first that

    \[ \bigl(A_\lambda + |A_\lambda|\bigr)\bigl(A_\lambda - |A_\lambda|\bigr) = A_\lambda^2 - |A_\lambda|^2 = 0. \]

Thus the range of the operator A_\lambda - |A_\lambda| is contained in the kernel of A_\lambda + |A_\lambda|, which is \mathcal{N}_\lambda. Hence

    \[ E_\lambda(A_\lambda - |A_\lambda|) = A_\lambda - |A_\lambda|, \]

and therefore

    \[ A_\lambda(I - E_\lambda) = |A_\lambda|(I - E_\lambda). \]

Since |A_\lambda is positive and commutes with the projection I - E_\lambda, we obtain

    \[ A_\lambda(I - E_\lambda) = |A_\lambda|(I - E_\lambda) \geq 0. \]

This proves (7).

\blacksquare

 

Proposition 2. With the assumptions and notation as above, if \lambda<\mu then

(8)   \begin{equation*} E_\lambda E_\mu = E_\lambda. \end{equation*}

Equivalently, \lambda<\mu implies E_\lambda \leq E_\mu.

Proof.  From(6) we have

(9)   \begin{equation*} A_\lambda E_\lambda (I - E_\mu) \leq 0. \end{equation*}

On the other hand, for A_\mu = A - \mu I we again have

    \[ \bigl(A_\mu + |A_\mu|\bigr)\bigl(A_\mu - |A_\mu|\bigr) = A_\mu^2 - |A_\mu|^2 = 0. \]

Therefore the range of A_\mu - |A_\mu| lies in the kernel \mathcal{N}_\mu of A_\mu + |A_\mu|. Hence

    \[ E_\mu(A_\mu - |A_\mu|) = A_\mu - |A_\mu|, \]

or

    \[ A_\mu (I - E_\mu) = |A_\mu|(I - E_\mu). \]

Since I - E_\mu is a projection commuting with |A_\mu|, it follows that

(10)   \begin{equation*} A_\mu (I - E_\mu) \geq 0. \end{equation*}

Multiplying this inequality on the left by E_\lambda (which commutes with A_\mu and E_\mu) we obtain

(11)   \begin{equation*} A_\mu E_\lambda (I - E_\mu) \geq 0. \end{equation*}

Now substitute A_\lambda = A - \lambda I and A_\mu = A - \mu I into (9) and (11), and subtract (9) from (11). The terms containing A cancel, and we obtain

(12)   \begin{equation*} (\lambda - \mu) E_\lambda (I - E_\mu) \geq 0. \end{equation*}

If \lambda<\mu, then \lambda - \mu < 0, and E_\lambda(I - E_\mu) is a positive operator (in fact a projection). The inequality (12) can therefore hold only if

    \[ E_\lambda(I - E_\mu) = 0, \]

that is,

    \[ E_\lambda E_\mu = E_\lambda. \]

This proves (8). The equivalence with E_\lambda\leq E_\mu is the usual order relation for projections.

\blacksquare

 

Exercise 1. Prove that (8) is equivalent to the statement E_\lambda \leq E_\mu in the usual operator order.

Proposition 3. The function \lambda\mapsto E_\lambda is right-continuous in the strong operator topology:

(13)   \begin{equation*} E_{\lambda+} := s\!\lim_{\mu\to \lambda+} E_\mu = E_\lambda. \end{equation*}

Proof.  For \lambda<\mu define

(14)   \begin{equation*} E_{\lambda,\mu} := E_\mu - E_\lambda. \end{equation*}

By Proposition~??, E_\lambda \leq E_\mu, so E_{\lambda,\mu} is an orthogonal projection. In particular, E_{\lambda,\mu} is self-adjoint and positive, and \|E_{\lambda,\mu}\|\le 1.

By the standard argument for bounded monotone families of self-adjoint operators (using monotonicity of the scalar functions \langle E_{\lambda,\mu} x, x\rangle and the polarization identity), there exist strong limits

(15)   \begin{align*} \Delta^+(\lambda) &:= s\!\lim_{\mu\to \lambda+} E_{\lambda,\mu}, \\ \Delta^-(\mu) &:= s\!\lim_{\lambda\to \mu-} E_{\lambda,\mu}, \end{align*}

which are orthogonal projections (possibly the zero projection).

We will now show that \Delta^+(\lambda)=0. This will prove right-continuity of \lambda\mapsto E_\lambda.

First, we consider the consequence of (8). Multiplying \(14) by E_\lambda on the left and using (8) we obtain

    \[ E_\lambda E_{\lambda,\mu} = E_\lambda(E_\mu - E_\lambda) = E_\lambda E_\mu - E_\lambda^2 = E_\lambda - E_\lambda = 0. \]

Taking the strong limit as \mu\to \lambda+ we have

(16)   \begin{equation*} E_\lambda \Delta^+(\lambda) = 0. \end{equation*}

Next we use the inequalities (6) and (7). Inequality (6) holds for every \lambda, therefore in particular for \lambda=\mu:

    \[ (A - \mu I) E_\mu \leq 0. \]

We can multiply this inequality on the right by the commuting projection E_{\lambda,\mu} without changing the inequality. Using E_\mu E_{\lambda,\mu} = E_{\lambda,\mu}, we obtain

    \[ (A - \mu I) E_{\lambda,\mu} \leq 0, \]

or

(17)   \begin{equation*} A E_{\lambda,\mu} \leq \mu E_{\lambda,\mu}. \end{equation*}

Similarly, multiply (7),

    \[ (A_\lambda)(I - E_\lambda) = (A - \lambda I)(I - E_\lambda) \geq 0, \]

by the commuting projection E_\mu on the right. Using E_\mu(I - E_\lambda) = E_{\lambda,\mu}, we obtain

    \[ (A - \lambda I) E_{\lambda,\mu} \geq 0, \]

that is,

(18)   \begin{equation*} A E_{\lambda,\mu} \geq \lambda E_{\lambda,\mu}. \end{equation*}

Combining (19) and (18) we obtain

(19)   \begin{equation*} \lambda E_{\lambda,\mu} \leq A E_{\lambda,\mu} \leq \mu E_{\lambda,\mu}. \end{equation*}

Now take the strong limit as \mu\to \lambda+. Since E_{\lambda,\mu}\to \Delta^+(\lambda) strongly and A is bounded and self-adjoint, we get

    \[ \lambda \Delta^+(\lambda) \leq A \Delta^+(\lambda) \leq \lambda \Delta^+(\lambda), \]

hence

    \[ A \Delta^+(\lambda) = \lambda \Delta^+(\lambda), \]

or

    \[ (A - \lambda I)\Delta^+(\lambda) = 0. \]

By the definition of A_\lambda^+ we then have

    \[ A_\lambda^+ \Delta^+(\lambda) = \frac{1}{2}\bigl(A_\lambda + |A_\lambda|\bigr) \Delta^+(\lambda) = 0, \]

so every vector in the range of \Delta^+(\lambda) lies in \mathcal{N}_\lambda. Therefore

    \[ E_\lambda \Delta^+(\lambda) = \Delta^+(\lambda). \]

Comparing this with (17), we obtain

    \[ \Delta^+(\lambda) = E_\lambda \Delta^+(\lambda) = 0. \]

Thus \Delta^+(\lambda)=0, which means precisely that

    \[ s\!\lim_{\mu\to \lambda+} E_{\lambda,\mu} = 0, \]

or equivalently,

    \[ s\!\lim_{\mu\to \lambda+} E_\mu = E_\lambda. \]

This completes the proof.

\blacksquare

 

We recall (Proposition 3 in Positive operators II that

(20)   \begin{equation*} \|A\| = \sup_{\|x\|=1} |(x,Ax)|. \end{equation*}

Denoting

(21)   \begin{equation*} m = \inf_{\|x\|=1} (x,Ax), \qquad M = \sup_{\|x\|=1} (x,Ax), \end{equation*}

we can write

(22)   \begin{equation*} \|A\| = \max(|m|,|M|). \end{equation*}

The extremal values m and M give a natural interval on the real line outside of which the spectral family is trivial.

Proposition 4. With the assumptions and notation as above we have

(23)   \begin{align*} E_\lambda &= 0 \quad \text{for all } \lambda < m, \\ E_\lambda &= I \quad \text{for all } \lambda > M. \end{align*}

Proof. We sketch the argument for the second statement; the first one is similar and uses the inequality A_\lambda E_\lambda\le 0.

Fix \lambda>M. Suppose, for contradiction, that there exists x\in H with \|x\|=1 and E_\lambda x = 0. Then x\in \ker(E_\lambda) = \mathcal{N}_\lambda^\perp, so x is orthogonal to the kernel of A_\lambda^+. Using (7) we have

    \[ (x, A_\lambda x) = (x, (A_\lambda (I - E_\lambda))x) \ge 0, \]

because (I - E_\lambda)x = x.

Thus

    \[ (x,Ax) - \lambda \ge 0, \]

or

    \[ (x,Ax) \ge \lambda. \]

But by the definition of M we have (x,Ax)\le M. Hence \lambda\le M, which contradicts the assumption \lambda > M. Therefore no such x exists and E_\lambda x\ne 0 for all unit x, which forces E_\lambda = I.

A similar argument, using A_\lambda E_\lambda\le 0, shows that for \lambda<m we must have E_\lambda = 0.
\blacksquare

When I started working on this post, I expected it to be the final one in the series on Hilbert spaces. The material turned out to be richer than I had anticipated, so there will be one more post, devoted to spectral integrals.

In writing this post I followed the exposition in the three monographs listed in the references. Many other textbooks and lecture notes treat this topic, but these three are closest to my own perspective: their methods apply equally well to real Hilbert spaces, even when this is not stated explicitly.

References
[1] Kreyzig, E., Introduction to Functional Analysis with Applications, Wiley, 1978.
[2] Lusternik, L.A., Sobolev, V.J., Elements of Functional Analysis, Wiley,
[3] Schmüdgen, K., Unbounded Self-adjoint Operators on Hilbert Space, Springer, 2012.

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