I am studying functional analysis and we proved in class that the eigenvalues $\mu_1\geq \mu_2\geq \cdots$ of a compact and self-adjoint operator defined on some Hilbert space converge to zero, i.e., $\lim_{n\rightarrow \infty}\mu_n=0$. Does this hold true for other types of operators (non-compact, non-self adjoint) as well? I'm actually solving the Sturm-Liouville eigenvalue problem and noticed that the eigenvalues actually never converge to zero.
Do Eigenvalues of an operator always converge to zero
functional-analysis
Related Solutions
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.
Best Answer
No. The identity operator has only one eigen value, namely $1$.