Generalized eigenvectors of arbitrary Hilbert spaces

Question

Per wikipedia:

If $\lambda$ is an eigenvalue of $T$, then the operator $T - \lambda I$ is not one-to-one, and therefore the inverse $(T - \lambda I)$ is not defined. However, the inverse statement is not true: the operator $T - \lambda I$ may not have an inverse, even if $\lambda$ is not an eigenvalue. Thus the spectrum of an operator always contains all its eigenvalues, but is not limited to them.

The article goes on to give an example from $\ell^2(\mathbb{Z})$ of an operator with such a value in its spectrum (namely, the right-shift operator).

I know that we can employ some machinery to "recover" eigenvectors corresponding to these values. For example, for $L_2(\mathbb{R})$, we can construct the Gelfand triple $H^s(\mathbb{R}) \subseteq L_2(\mathbb{R}) \subseteq H^{-s}(\mathbb{R})$ and recover, for instance, the "generalized eigenvectors" of the momentum operator $-i\frac{d}{dx}$.

In that case, it is easy to interpret the generalized eigenvectors - they are sinusoids. But what about the earlier case of $\ell^2(\mathbb{Z})$? Here, I have no real intuition, which suggests that I'm missing something fundamental in my understanding.

Is there anything that can be said about what these "generalized eigenvectors" look like, in general? Is there any theory characterizing what "sort" of operators on a general Hilbert space require this sort of additional machinery to recover their eigenbases, and why?

Edit: It occurs to me upon reflection that the generalized eigenvectors of the right-shift operator on $\ell^2(\mathbb{Z})$ must be those sequences that grow or decay geometrically. This bears a notable similarity to the above case of $L_2(\mathbb{R})$, though again I don't feel like I quite grasp the nature of the relationship.

Answer 1

I'm not sure about a general approach, but here's a rigged Hilbert space that yields "generalized eigenfunctions" for the shift operators.

For $0 < \alpha < 1$, define $$ \Phi_{\alpha} = \left\{f \in \ell^2(\Bbb Z) : \sum_{n \in \Bbb Z} \alpha^{|n|} |f(n)|^2 < \infty \right\}. $$ For any $\lambda \in \Bbb C$ with $|\lambda| = 1$, let $f_\lambda \in \Phi_\alpha^*$ be defined by $$ f_{\lambda}(n) = \lambda^{-n}, \quad n \in \Bbb Z. $$ If $T$ is the bilateral shift operator (i.e. $T(f)(n) = f(n-1)$), then we find that $T^*(\Phi_{\alpha}) \subseteq \Phi_{\alpha}$ and that for any $\lambda \in \Bbb C$ with $|\lambda| = 1$ and any $g \in \Phi_\alpha$, we have $$ \langle T(f_\lambda),g\rangle_{\ell^2(\Bbb Z)} := \langle f_\lambda,T^*(g)\rangle_{\ell^2(\Bbb Z)} = \lambda \cdot \langle f_\lambda,g\rangle_{\ell^2(\Bbb Z)}. $$ So, the functions $f_\lambda$ form a (complete) set of generalized eigenfunctions to the shift operator.

Answer 2

The right shift operator $S : \ell^2(\mathbb{Z})\rightarrow\ell^2(\mathbb{Z})$ is defined by $$ S(\alpha_0,\alpha_1,\alpha_2,\alpha_3,\cdots)=(0,\alpha_0,\alpha_1,\alpha_2,\alpha_3,\cdots). $$ You can view this as multiplication by $z$ on the Hardy space $H^2(\mathbb{D})$ on the unit disk in $\mathbb{C}$, which consists of power series functions $p(z)=\sum_{n=0}^{\infty}p_n z^n$ with $\sum_{n=0}^{\infty}|p_n|^2 < \infty$. The adjoint $S^*$ is given by its action on power series functions: $$ S^*p=\frac{p(z)-p(0)}{z}. $$ $S^*$ maps $(\alpha_0,\alpha_1,\alpha_2,\cdots)$ to $(\alpha_1,\alpha_2,\alpha_3,\cdots)$. This is the so-called left shift operator. Note that $S^*Sp=p$, but $SS^*q\ne q$ unless $q(0)=0$. $S$ is not invertible, even though it has a left inverse. $S-0I$ does not have an inverse, even though $\mathcal{N}(S)=\{0\}$.