Point, Continuous, and Residual Spectrum

Point, continuous, and residual spectrum distinguish different failures of the equation $(A-\lambda)\psi=f$. The classification uses injectivity and the range of an operator. It must not be confused with the absolutely continuous and singular continuous parts of a self-adjoint operator’s spectral measures.

Helpful background. Spectra of Operators defines the resolvent set and approximate eigenvectors. Adjoints of Unbounded Operators supplies the range-kernel identity.

A disjoint classification by solvability

Let $A$ be closed and densely defined. This page uses the following disjoint convention:

• $\lambda\in\sigma_{\mathrm p}(A)$ if $A-\lambda$ is not injective.
• $\lambda\in\sigma_{\mathrm c}(A)$ if $A-\lambda$ is injective and its range is dense but is not all of $\mathcal H$.
• $\lambda\in\sigma_{\mathrm r}(A)$ if $A-\lambda$ is injective and its range is not dense.

These sets partition the spectrum. Some texts allow the residual set to overlap the point spectrum by omitting injectivity in its definition; comparing statements requires checking that convention.

In the continuous case the inverse exists on the range but is unbounded. Otherwise it would extend continuously to all of $\mathcal H$, and closedness of $A$ would make the range all of $\mathcal H$. Thus the issue is unstable inversion on a dense set of forcing terms.

Residual spectrum means that some directions cannot even be approximated by outputs. The identity

$$\operatorname{Ran}(A-\lambda)^\perp =\ker(A^*-\overline\lambda)$$

makes that obstruction visible as an adjoint eigenvector.

Why self-adjoint operators have no residual spectrum

For self-adjoint $A$, nonreal numbers belong to the resolvent set. At real $\lambda$,

$$\operatorname{Ran}(A-\lambda)^\perp=\ker(A-\lambda).$$

If $\lambda$ is not an eigenvalue, the kernel vanishes, so the range is dense. Hence every self-adjoint spectral point is either point or continuous spectrum under the disjoint convention.

This argument does not imply that the range is closed. At a continuous spectral point it must fail to be closed, since it is dense and proper. Approximate eigenvectors quantify that failure.

Nor does the absence of residual spectrum hold for an arbitrary symmetric operator. Its adjoint can have a larger domain, giving nonzero adjoint kernels even though the original operator has no corresponding eigenvectors.

Three explicit examples

For a diagonal operator $Ae_n=n e_n$, each positive integer belongs to the point spectrum. The eigenvectors are normalizable and satisfy the equation in the operator domain.

For $M_x$ on $L^2(0,1)$, zero is a continuous spectral value. Multiplication by $x$ is injective because its zero set has measure zero. The range is dense: a vector orthogonal to every $xf$ must satisfy $xg=0$ almost everywhere and therefore vanish. But the constant function $1$ is not in the range, since its putative preimage $1/x$ is not square-integrable. The inverse therefore fails exactly in the continuous-spectrum way.

For residual spectrum, use the unilateral right shift on $\ell^2(\mathbb N_0)$:

$$S(c_0,c_1,c_2,\ldots)=(0,c_0,c_1,\ldots).$$

It is bounded and injective. Its range is the closed proper subspace of sequences whose first coordinate vanishes, so $0\in\sigma_{\mathrm r}(S)$. Its adjoint is the left shift and has eigenvector $e_0$ with eigenvalue zero, matching the range-kernel identity. This shift is not self-adjoint, so the example does not contradict the preceding theorem.

Spectral measures answer a different question

For a self-adjoint operator, each vector defines a measure $\mu_\psi(\Delta)=\langle\psi,E_A(\Delta)\psi\rangle$. Such measures split into pure point, absolutely continuous, and singular continuous parts.

Absolutely continuous means admitting a density relative to Lebesgue measure. Singular continuous means having no atoms while being concentrated on a Lebesgue-null set. Both are continuous measure types, and both can occur in the continuous spectral subspace.

The set $\sigma_{\mathrm c}(A)$ from the range classification need not equal the support of that continuous subspace. For example, let $Ae_n=n^{-1}e_n$. Every vector’s spectral measure is a sum of atoms at $1/n$, so the operator has pure point spectral type. Nevertheless zero belongs to $\sigma_{\mathrm c}(A)$: it is a spectral accumulation point but not an eigenvalue.

A reader should therefore ask whether “continuous spectrum” denotes a set of spectral parameters or a spectral-measure subspace. The distinction matters in scattering, where transport and long-time decay depend on more than the absence of eigenvectors.

Exercises

Show directly that the range of $Ae_n=n^{-1}e_n$ is dense but proper.

Solution

Every finite sequence is in the range, so it is dense. The vector with coordinates $y_n=1/n$ lies in $\ell^2$, but solving $Ax=y$ would require $x_n=1$, which is not in $\ell^2$. Thus the range is proper. Since the kernel is zero, zero is continuous spectrum in the range classification.

For the right shift, determine whether its adjoint has point spectrum inside the unit disk.

Solution

For $|\lambda|<1$, the vector $(1,\lambda,\lambda^2,\ldots)$ is square-summable and is an eigenvector of the left shift $S^*$ with eigenvalue $\lambda$. Therefore $\operatorname{Ran}(S-\overline\lambda)$ is not dense. The right shift has no eigenvectors, so these conjugate parameters belong to its residual spectrum.

References

• K. Schmüdgen, Unbounded Self-adjoint Operators on Hilbert Space, Graduate Texts in Mathematics 265, Springer, 2012, doi:10.1007/978-94-007-4753-1 — spectral classification and spectral measures.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., American Mathematical Society, 2014, author’s book page — spectral subspaces and distinctions among spectral types.