Discrete eigenvalues and their eigenfunctions

[ralqs]

What's the proof that eigenfunctions of discrete eigenvalues are in Hilbert Space?

 

[dextercioby]

This is included in the definition of an <eigenvalue>. The <eigenfunctions> are elements of a Hilbert space. The set of all eigenvalues is called the <point spectrum> of an operator. What you call <discrete eigenvalue> is probably a point in the <discrete spectrum>, which is a subset of the <point spectrum>, again by definition:

"The discrete spectrum σ_d(A) is the
set of all eigenvalues which are discrete points of the spectrum and whose
corresponding eigenspace is finite dimensional. The complement of the discrete
spectrum is called the essential spectrum σ_ess(A) = σ(A)\σ_d(A)" (*)

for A a linear operator in a Hilbert space. *Quote form G. Teschl's <Mathematical Methods for Quantum Mechanics>, page 145.

 

[ralqs]

This is included in the definition of an <eigenvalue>. The <eigenfunctions> are elements of a Hilbert space.

I don't think it's a matter of definition. If the eigenvalues are continuous, then the eigenfunctions are not necessarily part of the Hilbert Space. e.g. eigenfunctions of the operator $\frac{1}{i} \frac{d}{dx}$ are $A e^{ikx}$ (k = eigenvalue), which isn't in the Hilbert Space.

 

[dextercioby]

I don't understand what you're denying. It's true that in the literature there's no common definition for some of the concepts, but apparently everyone agrees that

<Let A:D(A)$\subset$H $\rightarrow$ Ran(A)$\subset$H be a linear operator in a Hilbert space. The complex number $\lambda$ is called eigenvalue of the operator A, iff there's at least one vector $\psi$≠ 0 in D(A) satisfying

$$Aψ=λψ$$. Such a vector is called eigenvector of the operator A and is assumed by definition to be a member of the Hilbert space> is a definition for eigenvector/eigenvalue.

The statement <The eigenvalues are continuous> is inaccurate, for eigenvalues are numbers which can't be continuous. I think you meant <If the spectral values are elements of the continuous spectrum of an operator A> which is whole different thing.

Your second assertion could be taken as a counterexample to a statement such as: For a linear self-adjoint operator in a Hilbert space*, its spectral equation always has solutions in the Hilbert space.

As for your original question, I believe my previous post above pretty much settles it.

*The derivative operator (momentum operator) is self-adjoint on a dense everywhere subset of $L^2 (\mathbb{R}, dx)$.

 

[ralqs]

If the operator is in Hilbert space, then it's a tautology. But an operator like $-i \frac{d}{dx}$ doesn't have to be defined on a Hilbert space alone; it can act on functions outside Hilbert space, like $e^{ikx}$. But apparently if the spectrum of such an operator is discrete, its eigenfunctions are in Hilbert space. I'm looking for a proof of this proposition.

 

[dextercioby]

I would be very interested to learn a way to evade the tautology, namely that you can't prove something which is assumed in a definition (in this case, the definition of an eigenvalue/eigenvector).