Distance between empirical measures and thickened version

Question

Let $\mathcal{H}$ be a separable Hilbert space and let $x_1,...,x_n$ be points in $\mathcal{H}$. Let $\varepsilon >0 $ be given and consider the measures $$ \mu := \frac1{n}\,\sum_{i=1}^n\, \delta_{x_i} \mbox{ and } \mu^{\varepsilon} := \frac1{n}\,\sum_{i=1}^n\, G(x_i,\varepsilon\, T). $$ Here $T$ is any trace-class operator on $\mathcal{H}$ and $G(x,\Sigma) $ denotes the Gaussian measure on $\mathcal{H}$ with mean $x$ and covariance operator $\Sigma$.

Is there an upper bound on the 2-Wasserstein distance between $\mu$ and $\mu^{\varepsilon}$ that depends only on $\varepsilon$ and tends to 0 as $\varepsilon$ does?

Answer 1

The claim follows by a synchronous coupling: $X = x_I$ and $X^{\epsilon} = x_I + \sqrt{\epsilon} \sum_{j=1}^{\infty} \sqrt{\lambda_j} \rho_j e_j$ where $I \sim \operatorname{Uniform}(\{1, \dots, n \})$; $e_j$ are the eigenfunctions of $T$; $\lambda_j$ are the corresponding eigenvalues; and $\{ \rho_j \} \overset{i.i.d}{\sim} \mathcal{N}(0,1)$. Indeed, \begin{align*} \mathcal{W}_2(\mu,\mu^{\epsilon})^2 \le E\left[ |X - X^{\epsilon} |^2 \right] = \epsilon E\left[| \sum_{j=1}^{\infty} \sqrt{\lambda_j} \rho_j e_j|^2 \right] = \epsilon \operatorname{trace}(T) \;. \end{align*}