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I have seen across the site questions regarding eigenvalues of linear maps between infinite dimensional vector spaces, and it regularly comes up that it is possible for there to be none or for every scalar to be an eigenvalue. When eigenvalues do exist, it seems that some of the familiar properties from finite dimensions carry across, such as this.

I would define an eigenvalue of a linear operator $L$ as any scalar $\lambda$ such that for some non-zero $u \in V$, the relevant vector space, $(L-\lambda I)(u)=0$.

Let's suppose that $X$ is a topological space, and I am working with some vector space of continuous functions from $X$ into $\mathbb{R}$, such as $C(X)$ or $C_b(X)$; call this vector space $V$. Let $L:V \to V$ be an endomorphism.

I want to know what conditions on $X$, $V$ and/or $L$:

  1. Imply the certain existence of eigenvalues for $L$.
  2. Imply the existence of a basis for $V$ of eigenfunctions/vectors of $L$.
  3. Ensure that there is a countable number of of eigenvalues, rather than a continuum.

I have experienced, for example, the case in Sturm-Liouville theory where we have $X \subset \mathbb{R}$ compact, $V=\{C^\infty(X):$ boundary conditions met$\}$ and $L$ a second-order self-adjoint differential operator, where all three of these conditions are satisfied simultaneously; alternatively I have seen the same but with '$X$ compact' replaced with 'for all $f \in V$, $f$ vanishes at $\infty$'.

Infinite differentiability and self-adjointness are rather strong and specific conditions in a more general context, but perhaps there are some weaker or related constraints which can be applied to more general topological spaces?

I expect local compactness or compactness may be an important conditions on $X$, for a start. There are obviously some simple maps (such as the identity and the zero map) which have eigenvalues for any $X$, $V$, but I am interested in as broad a range of maps as possible.

Edit: I would be satisfied with simultaneous conditions on $X$, $L$, $V$ which ensure at least point 1. I have seen plenty of discouraging non-examples, most taking the form of some kind of shift operator, but I reason that if some linear operators have eigenvalues, then there must be a way to characterise the class of operators which do in a way which will probably involve the underlying spaces.

Having done some more research myself, the spectral theorem applies to all normal operators, which form a broader class than self-adjoint operators, but is still not exclusive to these (see section 6 of that link in particular), so my question remains.

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  • $\begingroup$ Should I move this to MO? $\endgroup$ Aug 4, 2016 at 17:51

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There is essentially no condition on the space on which your functions live that will ensure existence of any eigenfunctions. For example, for $X=[a,b]$ and $V$ the space of real-valued (or complex-valued, hardly matters...) functions on $X$, and $L$ the multiplication-by-$x$ operator... there are no eigenfunctions at all (unless $a=b$) [EDIT: among continuous or $L^2$ or $L^p$ functions]. That is, unless the physical space is a finite set (or has a weird-enough topology so that there aren't very many continuous functions), some very simple, non-pathological operators fail to have any eigenvalues at all (EDIT: among continuous or $L^2$ or...].

Another non-pathological example that shows that it is not generally reasonable to expect eigenvalues is the Laplacian on the real line. Fourier inversion shows that everything in $L^2$ is a superposition of generalized eigenvalues [EDIT: oop, eigenfunctions] (the exponentials), but not a sum, and those eigenvalues are not in the space itself. EDIT: Fourier inversion expresses (e.g.) a Schwartz function $f$ as $$ f(x) \;=\; \int_{\mathbb R} e^{2\pi i\xi x}\;\widehat{f}(\xi)\; d\xi $$ where $\widehat{f}$ is the Fourier transform of $f$ Thus, $f$ is a superposition (=integral) of eigenfunctions (the functions $x\to e^{2\pi i\xi x}$ for $\Delta=\partial^2/\partial x^2$.

Yes, in the context of Sturm-Liouville problems (see also Fredholm alternative), the point is that the inverse of the differential operator (with boundary conditions) is a compact self-adjoint operator on a Hilbert space of functions, and the eigenvalues are in bijection by $\lambda \leftrightarrow \lambda^{-1}$, etc. The only general class of operators on infinite-dimensional spaces with a clear, simple, and happy spectral theory are compact self-adjoint, or unbounded operators which have compact self-adjoint operators as inverses/resolvents...

Thus, for example, the Laplace-Beltrami operator on a compact Riemannian manifold does provably have compact resolvent, so $L^2$ has an orthonormal basis of eigenfunctions.

EDIT: I'd also add that one almost surely wants a reasonable topology on the space of functions, related to the topology on the underlying physical space. And we'd want some sort of completeness on the space of functions, else we'd potentially lose eigenfunction/values for silly reasons.

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  • $\begingroup$ In the second paragraph I think you meant eigenfunctions/vectors (x2)? Some follow-up questions: I'm guessing that in the 1st paragraph you mean continuous functions? Since otherwise I can construct an eigenfunction for your operator. What do you mean by a superposition rather than a sum? $\endgroup$ Aug 7, 2016 at 18:54
  • $\begingroup$ Also I presume that when you say 'there's no condition on the space...' you mean that there's no condition on $X$ alone, but that isn't unexpected. Have edited question accordingly. $\endgroup$ Aug 7, 2016 at 19:01
  • $\begingroup$ Ah, yes, sorry about the typos... but, yes, ... I'll revise/correct/refine. $\endgroup$ Aug 7, 2016 at 19:33
  • $\begingroup$ Much clearer, thanks. Sorry to be insistent, but you still give mostly negative examples. I want to know what conditions I need to impose to avoid these examples causing problems. It seems clear that I need to consider only bounded or even compact operators, but do I have to go as far as self-adjoint? What if I don't have an inner product to work with? $\endgroup$ Aug 7, 2016 at 21:41
  • $\begingroup$ "Bounded" (=continuous) operators typically have no eigenvalues, either. The compactness is crucial, as is self-adjointness. E.g, the Volterra operator $Tf(x)=\int_0^xf(t)\;dt$ is compact, but has no eigenvalues/eigenvectors. If you want/need to avoid Hilbert space structures, the "Fredholm alternative" is about the best general fact. $\endgroup$ Aug 7, 2016 at 21:47

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