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?



Distance between empirical measures and thickened version
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\*}


