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!



Spectrum of operator defined by composition with linear function
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).


