Do eigenvalues of the Hilbert-Schmidt integral operator determine the underlying measure up to translation, reflection and rotation?

Details: Suppose we have a measure $\mu$ on a Euclidean space $X=\mathbb R^n$ and a kernel $\kappa: X \times X \rightarrow \mathbb R$ (which is symmetric and every Gram matrix $G$ defined by $G_{i,j} = \kappa(x_i, x_j)$ from a finite set $\{x_1, \dots x_m\} \subset X$ is positive semidefinite). Assume that $\kappa$ is given by distance function: $\kappa(x,y) = \kappa_0(\|x-y\|)$, for example $\kappa(x,y) = e^{-\gamma \|x-y\|^2}$. Suppose we are given a class of measures $\mathcal S$ on $X=\mathbb R^n$, for example those induced by continuous probability density functions that integrate to 1.

We then define the Hilbert-Schmidt integral operator: $$K_{\mu}: \phi \mapsto \int_X \kappa(x,-) \phi(x) \text{d} \mu (x)$$ The operator is both compact and self-adjoint, and thus admits an orthogonal decomposition and real eigenvalues by the spectral theorem. Let's define its vector of eigenvalues by: $$\vec \lambda(\mu) = (\lambda_1, \lambda_2, \cdots) \text{ where } \lambda_i \phi_i = K_{\mu} \phi_i \text{ and } \lambda_1 \ge \lambda_2 \ge \cdots $$

We immediately observe that $\vec \lambda(\mu) = \vec \lambda (\rho \cdot \mu)$ where $\rho$ is an isometry of $X = \mathbb R^n$ and $(\rho \cdot \mu)(U) := \mu(\rho(U))$. This is because $K_{\rho \cdot \mu}(\rho \cdot \phi) = K_\mu(\phi)$. The question is now whether the converse holds: do we have $\vec \lambda(\mu) = \vec \lambda(\nu) \implies \exists \rho: \mu = \rho \cdot \nu$?

The kernel $\kappa$ is characteristic iff the map $\Phi: \mu \mapsto K_\mu(\textbf{1}) = \mathbb{E}_{x \sim \mu} [\kappa(x,-)]$ is injective. The Gaussian (or RBF) kernel $\kappa(x,y) = e^{-\gamma\|x-y\|^2}$ is an example of a characteristic kernel. Thus, the map $\mu \mapsto K_\mu$ is all the more injective. Therefore, the Hilbert-Schmidt integral operator determines the underlying measure (but isn't invariant with respect to isometry).

My question then concerns how much information we can remove from the operator while telling apart underlying measures up to the isometries of a Euclidean space. Namely, can we do without the eigenvectors?

  • $\begingroup$ When you say that the map $\Phi:\mu \mapsto K_\mu({\bf 1})$ is injective, what's the precise domain of $\Phi$? Are you allowing all Borel probability measures on $X={\mathbb R}^n$? $\endgroup$ – Yemon Choi Oct 14 at 16:19
  • 1
    $\begingroup$ I was being imprecise, but I think it's certainly injective for measures induced by continuous probability density functions. $\endgroup$ – Finn Lim Oct 14 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.