Why are the inner products of the eigenfunctions of an operator with a discrete eigenvalue spectrum guaranteed to exist? I was reading through a textbook, and the statement was made that the inner products are guaranteed to exist if the eigenvalue spectrum of the operator is discrete. I have come across no support for this claim, and the basis for this claim was not immediately apparent after quite a long time of consideration on my behalf. Furthermore, while trying to deduce the answer, I was lead to the complement of my first question: why does a continuous eigenvalue spectrum of an operator lead to non-normalizable eigenfunctions (i.e. the inner products don't exist)?
 A: The spectrum of a bounded linear operator $A$ is by definition the set of numbers $\lambda$ where $A-\lambda$ is not invertible. In the finite dimensional case, this means $A-\lambda$ is neither injective nor surjective, and the former statement is just a fancy way of saying that there exists an eigenvector; as this eigenvector is by definition a part of the Hilbert space, it is in partiular normalizable.
However, in the infinite-dimensional case, injective and surjective are no longer equivalent and the spectrum decomposes into the point spectrum, the continuous spectrum and the residual spectrum.
The point spectrum is the part of the spectrum where the map is not injective (it might not be hard to prove that it is indeed discrete - too lazy to investigate right now). The residual spectrum is empty for normal operators (and thus in particular selfadjoint operators), which leaves the continuous spectrum where the map is injective (and thus there are no eigenvectors) but not surjective. The concept of eigenvector just doesn't make sense for the coninuous spectrum in this formalism.
The situation gets muddled (or, depending on your point of view, un-muddled) in the formalism of rigged Hilbert spaces, which is the proper framework for the treatment of unbounded operators: Rigged Hilbert spaces allow us to introduce 'eigenvectors' which aren't part of our Hilbert space. In particular, in case of the Hilbert space $L^2$, they can be non-normalizable functions (eg plane waves) or not functions at all (eg delta distributions).
A: Here is a partial answer the question (v1). OP would have to provide more mathematically precise details and assumptions in order to guarantee that an eigenvector (for an operator $H$ with discrete spectrum) has finite norm, i.e. is normalizable. For instance, within the framework of rigged Hilbert spaces.
Counterexample: Domain $D:=C^{\infty}(\mathbb{R})$=infinitely often differentiable complex-valued functions on the real line $\mathbb{R}$. 
Norm: 
$$|| f||^2 ~:=~ \int_{\mathbb{R}} \! dx~|f(x)|^2~. $$
Let the operator $H:=0$ to be the zero-operator, taking all functions $f$ to the zero-function $0$. (Let us mention that the pair $(H,D)$ can be modified to become a selfadjoint operator, but skip the details here.) The spectrum is discrete 
$${\rm spec}(H)=\{0\}.$$ 
Any function $f\in D\backslash\{0\}$ is an eigenfunction for $H$ with eigenvalue $0$, but the norm $|| f||^2$  is not necessarily finite. 
A: For a complete explanation look at a functional analysis book.
But for discrete points of the spectrum one can explicitly write down the projection operator onto the corresponding eigenspace :
The resolvent of the operator has poles at the discrete points of the spectrum and the residue of such a pole is the projection operator onto the corresponding eigenspace , see http://en.wikipedia.org/wiki/Resolvent_formalism .
A: The complimentary question is the real question here. If you have a continuous set of vectors there could be a discontinuity, in which case the inner product would depend on how you approach the discontinuity. In that case we can say the inner product does not exist. This isn't a problem with a discrete set of vectors. 
