Spectrum of operator defined by composition with linear function

Question

For any $b\in\mathbb{R}^d$ and $B\in\mathbb{R}^{d\times d}$ symmetric positive definite, define the functional operator $\mathcal{A}f(x)=f(Bx+b)$ ($f$ real-valued). We have that $\mathcal{A}$ is linear; compact (I think); normal, since $$\mathcal{A}^*f(x)=\frac{1}{\mathrm{det}(B)} f(B^{-1}x-B^{-1}b)$$ and hence $\mathcal{A}^*\mathcal{A}=\mathcal{A}\mathcal{A}^*=\frac{1}{\mathrm{det}(B)} I$; but not self-adjoint.

I have not defined the underlying Hilbert space yet; this is a fluid aspect of my problem definition. The only function $f$ I need to apply this to for now is $e^{-\|x\|^2/2}$.

The question is, can $\mathcal{A}$ be expressed in terms of some simple orthonormal sequence of operators? Again, the Hilbert space is a free parameter.

Possible, but rather undesirable, simplification -- take $B=I$.

I've perused Conway and Rudin, but being very new to functional analysis, so far even establishing conditions [non-]existence escapes me.

Thanks!

Answer 1

Take $H=L^2(\mathbb{R}^n, e^{-|x|^2}dx)$ as your Hilbert space, the $L^2$ space with weight $e^{-|x|^2}$, so that polynomials are in the space (Hermite Polynomials are an ONB for this space).

Your map $(B, b) \mapsto \widehat{(B, b)}$ defines a group homomorphism from the group $T(n)$ of transformations $x \mapsto Bx+b$ into the group $\mathrm{GL}(H)$ of bounded linear isomorphisms of $H$. Now the differential of your map at the identity assigns to an element $x \mapsto Mx + b$ in $\mathfrak{t}(n)$ (i.e. $M \in \mathrm{End}(\mathbb{R}^n)$, $b \in \mathbb{R}^n$) the unbounded operator (infinitesimal generator) $$ \widehat{(M, b)}f(x) = Df(x) \cdot(Mx + b).$$ I didn't think too long about this, but the generalized eigenfunctions of this operator should simply be certain polynomials. For example, if $c$ is a left eigenvector of $B$ with eigenvalue $\lambda \neq 0$ and we set $d = \langle c,b\rangle/\lambda$, then $f(x) = \langle c, x \rangle + d$ is and eigenvector to the eigenvalue $\lambda$. I think that if $M$ is diagonalizable, then $\widehat{(M, b)}$ has a discrete spectrum $\lambda_1, \lambda_2, \dots$ and a complete eigendecomposition, with each eigenfunction being a polynomial.

If now $B = e^M$, then $\widehat{(B, b)}$ has the same eigenfunctions with eigenvalus $e^{\lambda_1}, e^{\lambda_2}, \dots$.

By the way: A functional is usually a function from a space to the scalars, so you do not define a functional but an operator. This confused me for a bit.

\Edit: Assume $B = e^M$. Then $(M, b)$ is an element of the Lie algebra of $T(n)$. Let $U_\varepsilon = \exp(\varepsilon M, \varepsilon b)$ (Note this is the exponential in the Lie group $T(n)$, which is given on the matrix part by the usual matrix exponential, but on the translation part, $\exp(0, b)(x) = x + b$. $U_\varepsilon$ is given by the Baker-Campbell-Hausdorff formula in general, but its Taylor expansion is $$\exp(\varepsilon M, \varepsilon b)(x) = x + \varepsilon M + \varepsilon b + O(\varepsilon^2).$$ Hence if $f$ is somewhat regular (say a polynomial at first), $$\hat{U}_\varepsilon f(x) = f(x + \varepsilon M + \varepsilon b) + O(\varepsilon^2) = f(x) + \varepsilon Df(x)(M + b) + O(\varepsilon^2).$$ Now forming $$f = \lim_{\varepsilon \rightarrow 0} \frac{1}{\varepsilon} (U_\varepsilon f - f)$$ gives what I said. Notice that the limit is actually in $L^2$ with the given weight (if $f$ is a polynomial, the space of which is dense).

Answer 2

This operator is not compact --- as you point out, it is invertible.

You say that you have not defined the underlying Hilbert space, but your adjoint calculation is the adjoint as an operator on $L^2({\bf R}^d)$ where ${\bf R}^d$ is equipped with Lebesgue measure.

Probably you need to explain more about your motivation in order for us to be able to help ...

Answer 3

This is really a comment, not an answer but I am not empowered. I am taking the liberty of reinterpreting your question (always a dangerous course) since it has already been pointed out that your operator is not compact. I think that the proper framework for your question is that of $L^2$ on a space which has symmetries induced by a group action, say transation on the real line and the rotations of the sphere, both of which are hinted at in your query. Here, although the operators you allude to are not compact, one does get an orthonormal system of eigenvectors, provided that the group is compact, as is the case of the rotation group, but not the translation one. The corresponding eigenvectors for the sphere are the spherical functions which have played a central role in mathematical physics since their introduction by Thomson (aka Lord Kelvin) and Tait. The non-compactness in the second case leads to the fact that the "eigenvectors" do not lie in the Hilbert space (but in that of the tempered distributions). There is a considerable literature on this subject at various levels of sophistication (mathematical or physical).