Probability density from Hilbert-Schmidt integral operator

Question

The Hilbert-Schmidt integral operator determines the underlying measure, if a universal kernel is used. Now, 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 $\rho \cdot \mu := \mu \circ \rho^{-1}$. This is because $\rho^{-1} \cdot K_{\rho \cdot \mu}(\rho \cdot \phi) = K_\mu(\phi)$ if we define $\rho \cdot \phi = \phi \circ \rho^{-1}$: $$\left( \rho^{-1} \cdot K_{\rho \cdot \mu}(\rho \cdot \phi)\right) (y) = \int_X \kappa(x, \rho y) \phi(\rho^{-1} x) d (\rho \cdot \mu)(x) \\ = \int_X \kappa(\rho x, \rho y) \phi(\rho^{-1} \rho x) d (\rho \cdot \mu)(\rho x) = K_\mu(\phi)(y)$$ so that $$K_\mu \phi = \lambda \phi \implies K_{\rho \cdot \mu} (\rho \cdot \phi) = \rho \cdot (K_\mu \phi) = \rho \cdot (\lambda \phi) = \lambda (\rho \cdot \phi) $$ 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?

Answer 1

...do we have $\vec \lambda(\mu) = \vec \lambda(\nu) \implies \exists > \rho: \mu = \rho \cdot \nu$?

For measures that are convex sums of point masses, the answer seems to yes,

Consider the case where $\mu = \delta_{x_0}$, point mass at $x_0$. Then the operator is given by $$ (K_{\delta_{x_0}} \phi)(y) = \kappa(x_0, y) \phi(x_0), $$ where we restrict to $\phi \in C_c(\mathbb{R}^n)$ so that pointwise evaluation makes sense then extend to $L^2$ by denseness of $C_c(\mathbb{R}^n)$ in $L^2$.

Evidently, $K_{\delta_{x_0}}$ is a rank-one operator whose range is the linear span of $\kappa(x_0, \cdot)$. The eigenvalue equation $$ \kappa(x_0, y) \phi(x_0) = \lambda \phi(y) $$ tells us that $\lambda = \kappa(x_0, x_0) = 1$. So $K_{\delta_{x_0}}$ is a rank-one projection onto the the linear span of $\kappa(x_0, \cdot)$.

(If you restrict the question to point masses, then it is true: $K_{\delta_{x_0}}$ and $K_{\delta_{x_1}}$ have the same spectrum $\{1,0,0,\cdots\}$, and $\delta_{x_0}$ can be mapped to $\delta_{x_1}$ by translation.)

Now consider a convex sum $\alpha \delta_{x_1} + (1-\alpha)\delta_{x_2}$, where $0 < \alpha < 1$. By definition, $$ K_{\alpha \delta_{x_1} + (1-\alpha)\delta_{x_2}} = \alpha K_{\delta_{x_1}} + (1- \alpha) K_{\delta_{x_2}}, $$ which is a sum of two rank-one projections. In what follows, things are not as clean as one would like, because the two rank-one projections in the sum do not commute---but we do know the sum is a rank-two self-adjoint operator. In particular, it has two non-zero eigenvalues.

The eigenvalue equation $$ \alpha \kappa(x_1, y) \phi(x_1) + (1 -\alpha )\kappa(x_2, y) \phi(x_2) = \lambda \phi(y) $$ has two non-zero solutions: \begin{align} \lambda_1 &= \alpha + (1-\alpha) \kappa(x_2, x_1) \frac{\phi(x_2)}{\phi(x_1)}, \\ \lambda_2 &= \alpha \kappa(x_2, x_1) \frac{\phi(x_1)}{\phi(x_2)} + (1-\alpha). \end{align} Inspecting these equations show that, for any $\phi$ in the eigenspace corresponding to $\lambda_1$, $\frac{\phi(x_2)}{\phi(x_1)} = \kappa(x_2, x_1)$. Similarly, for any $\phi$ in the eigenspace corresponding to $\lambda_2$, $\frac{\phi(x_1)}{\phi(x_2)} = \kappa(x_2, x_1)$. So we have the spectrum of $K_{\alpha \delta_{x_0} + (1-\alpha)\delta_{x_1}}$: \begin{align} \lambda_1 &= \alpha + (1-\alpha) \kappa(x_2, x_1)^2, \\ \lambda_2 &= \alpha \kappa(x_2, x_1)^2 + (1-\alpha). \quad (*) \end{align}

In other words, the two non-zero eigenvalues are convex sums of $1$ and $\kappa(x_2, x_1)^2$ with weights $\alpha$ and $1-\alpha$.

Let $(\lambda_1, \lambda_2)$ and $(\lambda_1' = \lambda_2')$ be the nonzero eigenvalues of HS operators corresponding to measures of the above type. If $\lambda_1 = \lambda_1'$ and $\lambda_2 = \lambda_2'$, the explicit solution $(*)$ implies that $$ \kappa(x_2, x_1) = \kappa(x_2', x_1'), \mbox{ and } \alpha = \alpha'. $$ Therefore there is an isometry mapping $\alpha \delta_{x_1} + (1-\alpha)\delta_{x_2}$ to $\alpha' \delta_{x_1'} + (1-\alpha')\delta_{x_2'}$. My guess is that this argument extends to general finite convex sums of point masses, i.e. the explicit expression of the eigenvalues should characterize the support of the measure up to isometry.

For general measures, I do not know, although the above argument is somewhat suggestive. Perhaps one can use the fact that finite convex sums of point masses are weak-* dense in the family of Borel probability measures and approximate.