Basis of eigenfunctions in Banach space

Question

I have a question about proving the existence of a basis of eigenfunctions.

Assume we have a compact operator $L:\mathcal{B}\rightarrow\mathcal{B}$, where $\mathcal{B}$ is a Banach space of analytic even maps defined in a neighborhood of $[-1,1]$.

Since the operator is compact, I know that the spectrum is discrete. But what do I need to do in order to show that there exists a basis of eigenfunctions, i.e. $\exists f_{j}(x)\in\mathcal{B}$ such that $Lf_{j}(x)=\lambda_{j}f_{j}(x)$ and $\lbrace f_{j}(x)\rbrace$ span the whole space?

Any help will be appreciated! (this is not homework)

Answer 1

Compact operators on Banach spaces can in general be upper-triangularized---generalizing the Jordan canonical form. This means that the space is Schauder-spanned by the generalized eigenvectors. So what you would need to show is that, given any non-zero $\lambda \in \sigma(L)$, the finite dimensional subspace $\mbox{ker}(L - \lambda)$ is spanned by eigenvectors, rather than generalized eigenvectors.

Answer 2

See if you can extend your operator to some $L^2$ space containing your functions. Maybe your operator is even self adjoint and still compact. Use the result for compact self adjoint operators on Hilbert spaces to find a basis of eigenvectors in that $L^2$ space. Show that in fact your eigenfunctions are already in your original space. Show that any functions can be uniformly approximated by a combination of your functions.

Obs: This approach is used in one of the proofs of the theorem of Peter and Weyl.

Answer 3

I think you are thinking about it backwards. Every Banach space is a vector space. Every vector space has a basis.

If your eigenfunctions are: (1) linearly independent under the inner product (2) have the same dimension as the space.

They must be an equivalent basis.

However, neither (1) nor (2) are necessary conditions of eigenfunctions as far as I know. Its perfectly possible to have eigen functions which are not linearly independent. Its also possible to have an operator with linearly independent functions but not have enough levels in your spectrum to span the whole space. In this case, you have a set which is equivalent to some subset.

Establishing (1) and (2) is usually not difficult for physical systems.

My background is physics, so excuse the lack of formalism.