# Generalized eigenvectors of arbitrary Hilbert spaces

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.

• My intuition is that when you have a rigged Hilbert space $S\subset H\subset S^*$ and operator $T:S\to S$, then you define $T$ acting on $S^*$ as $(T\psi)(u) = \psi(T^*u)$ and get something like this? For any $\lambda\in\operatorname{spec}(T)$, there is a generalized eigenvector $\phi\in S^*$ such that $T\phi = \lambda\phi$. Leaving this as a comment since if true it might be helpful, I'll try to verify later this week if it actually is the case.
– Neal
Commented Jun 8, 2021 at 16:55
• @Neal what you are saying is exactly what one finds in Reed and Simon Volume 1 p.134-135 and example with the Fourier transform Thm IX.2 p.5 of the volume 2 with $T$ the Fourier transform, $S$ the Schwartz space. I just want to point out that one must have $T^*$ coincide with $T$ on $S$ which for the Fourier transform is sometimes known as "exchange lemma". What I wish to know more about are the possible extension when one knows that an (unbounded) operator is defined on a dense subset... Commented Jun 9, 2021 at 16:11
• @Noix07 If one has a densely-defined bounded-below symmetric operator $T$, I think one could complete the domain of $T$ wrt the norm $(u,v) + (Tu,v)$ (call it $H_T$) to get a rigged Hilbert space $H_T\subset H\subset H_T^*$ and I would guess the same machinery ought to work
– Neal
Commented Jun 10, 2021 at 12:15
• @Neal Noted. In fact I don't know... I'm interested in this kind of subject. What you say remind me of the "Friedrichs" extension. But again I don't know Commented Jun 10, 2021 at 12:20
• @Noix07 that is exactly what I'm thinking of. The Friedrichs extension form domain and its dual should give a Gelfand triple. I will see if I have time in the next few days to puzzle thru some details.
– Neal
Commented Jun 10, 2021 at 13:48

## 2 Answers

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.

• Interesting! This makes a fair bit of sense. This raises the question of the case of the continuous shift operator acting on $L_2(\mathbb{R})$. I'm trying to apply similar reasoning, but keep tripping up over the apparent degeneracy of the eigenvalue $1$ (for any shift operator with shift $a$, any sinusoid whose period divides $a$ is invariant under said shift, and so is an eigenvector of the shift operator with eigenvalue 1). Other eigenvalues occur as normal, but what do we make of the degeneracy? Commented Jun 9, 2021 at 12:13
• @user3716267 I'm not sure. I suppose I don't see anything "wrong" with having degenerate eigenvalues. Are you sure that the other eigenvalues occur as normal? For instance, with $a = 1$, sinusoids with period $2$ and $2/3$ would both be eigenvectors associated with eigenvalue $-1$. Commented Jun 9, 2021 at 12:57
• Yes, indeed - all of the eigenvalues are degenerate in this case, with a family of integer-indexed eigenfunctions. What implications does this have for interpretation of, say, the Gelfand transform? I've always liked thinking of continuous functions on the spectrum as "weightings" of the eigenvectors, with which they can be integrated to recover a given operator (basically, a naive interpretation as an eigenbasis). How does degeneracy interact with this? Each eigenvalue now has multiple eigenfunctions. Commented Jun 9, 2021 at 12:59
• @user I would suspect that the eigenvectors are "stuck" together when the transformation is acted upon by a continuous function. That is, all eigenvectors associated with the same eigenvalue will be acted upon in the same way. Commented Jun 9, 2021 at 13:10
• That seems intuitive, but I can't convince myself intuitively that it converges...I suspect I'll probably have to actually define the projection-valued measure on the spectrum and work through the details at this point. Commented Jun 9, 2021 at 13:18

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 $$\mathcal{N}(S)=\{0\}$$.