I have been struggling with a definition of spectrum that I find in some papers and I cannot seem to relate it to functional analysis.

To exemplify what I mean, consider the momentum operator $L=i\partial_x$ densely defined from $H^1(\mathbb{R})$ to $L^2(\mathbb{R})$. I know that this is a self-adjoint operator, the spectrum is the entire real line and the residual spectrum is empty. I also know that for self-adjoint operators we have a characterization of the spectrum in terms of discrete and continuous:

  • The discrete spectrum is the set of isolated eigenvalues $\lambda \in \mathbb{C}$ with finite algebraic multiplicity
  • The continuous spectrum is the set of $\lambda \in \mathbb{C}$ such that there exists a Weyl sequence.

For this particular operator:

  1. The solutions of the spectral problem $Lu=\lambda u$ are given by $u(x) = Ke^{-i\lambda x}$, where $K$ is an integration constant, and there is no complex $\lambda$ (for $K\neq 0$) such that $u$ is square integrable, so the discrete spectrum is empty.
  2. For $\lambda \in \mathbb{R}$ and the same $u$ of 1., we build a Weyl sequence to conclude that the continuous spectrum is the entire line.

This is fine and I completely understand it.

Now, let us consider a different definition for the continuous spectrum, see this paper by Pelinovsky (page 4, right after equation (3.3)):

Definition: the continuous (Lax) spectrum is the set of $\lambda \in \mathbb{C}$ such that the solutions of $Lu=\lambda u$ are $L^{\infty}(\mathbb{R})$.

If we use this definition on 2., we obtain the following:

2'. The solutions of the spectral problem (see $u$ in 1.) are in $L^{\infty}(\mathbb{R})$ if and only if $\lambda \in \mathbb{R}$ so the continuous (Lax) spectrum is the entire real line.

Conclusion: continuous spectrum = continuous (Lax) spectrum.


  1. Is this a coincidence or is there a relation between the two definitions considering self-adjoint operators?
  2. Can I characterize the spectrum in terms of bounded 'eigenvectors'?


  1. Every mathematician I have ever asked this question says that the answer is in Quantum Mechanics (the same type of relation will appear for self-adjoint Schrödinger operators). Physicists then tell me it is because the wavefunctions are always square integrable with norm equals 1 (the particle needs to be somewhere). It makes sense to me, but mathematically it seems off because this would mean the spectrum for the momentum operator is empty. But then the conversation continues about how plane waves are not physically relevant and, well, I am too ignorant in Physics to really understand more.
  2. Similarly to what is done for the continuous spectrum of the momentum operator, I have tried to show that if $u\in L^{\infty}(\mathbb{R})$ then

$$g_k(x) = \frac{1}{\sqrt{k}\Vert f\Vert_{L^2}\Vert u\Vert_{L^{\infty}}} f\left(\frac{x}{k}\right) u,$$ for an arbitrary fixed nonzero compactly supported function $f$, is a Weyl sequence. Well, it did not work because I cannot guarantee that the norm is 1. Conversely, I have no idea if is possible to build a bounded 'eigenvector' from a Weyl sequence.



You must log in to answer this question.

Browse other questions tagged .