A translation operator on measures

Question

We consider $P$ the space of probability measures on $\mathbb{R}^{d}$ endowed with the topology of weak convergence.(note that this topology makes $P$ polish space)

For all $y \in \mathbb{R}^{d} $ we define the "translation by $y$ operator" as the operator $\tau_y : P \rightarrow P$ such that , for all $\mu \in P$, $\int \limits_{x \in \mathbb{R}^{d} } f(x) d\tau_y \mu (x) = \int \limits_{x \in \mathbb{R}^{d} } f(x+y) d\mu (x) $, for all measurable and bounded functions $f:\mathbb{R}^{d} \rightarrow \mathbb{R}$.

why is it called "translation operator" ? could someone give me an example that illustrates the choice of this name ?

Also, how can we prove continuity ?

Answer 1

It is easy to see why this operator is called a translation operator once you understand what the measure $\tau_y \mu$ actually is. By taking $f$ in the definition to be $1_A$ for some measurable set $A \subseteq \mathbb{R}^d$, we get $$\tau_y \mu(A) = \int_{\mathbb{R}^d} 1_A(x+y) d \mu(x) = \int_{\mathbb{R}^d} 1_{A-y}(x) d \mu(x) = \mu(A-y)$$ and hence $\tau_y \mu$ is really the translation of the measure $\mu$ by $y$ in the natural sense.

To show continuity, it is convenient to work with the sequential characterisation. So we want to show that for every bounded, continuous function $f$ and sequence of probability measures such that $\mu_n \to \mu$ in the topology of weak convergence, we have that $$\int_{\mathbb{R}^d} f d \tau_y \mu_n \to \int_{\mathbb{R}^d} f d \tau_y \mu$$ The result is then immediate by writing out the definition of integration against $d \tau_y \nu$ for a measure $\nu$ on both sides, since if $f$ is a continuous and bounded function then so is $f_y(\cdot) := f(\cdot + y)$.

Answer 2

Let $P^p(\mathbb{R}^{d})$ denote the wasserstein space of order p. As a special case of the previous operator we define $T:P^p(\mathbb{R}^{d}) \rightarrow P^p(\mathbb{R}^{d}) $ by $T\mu = \tau_{-\xi} \, \mu$, where $\xi = \int xd\mu (x) $ the mean value of $\mu$.

I want to show continuity of this operator.

So let $(\mu_n)_n \subset M^p$ such that $\mu_n \rightarrow \mu$ in Wasserstein topology. This means that $\mu_n \rightarrow \mu$ weakly and that \begin{equation*} \lim \limits_{R \rightarrow \infty} \limsup \limits_{n\rightarrow\infty} \int \limits_{|x|\geqslant R} |x|^p d \mu_n (x) = 0 \,\, (*) \end{equation*} First, we show that $T\mu_n \rightarrow T\mu$ weakly. So we want \begin{equation*} \int \limits_{x \in \mathbb{R}^{d}} f(x - \xi_{\mu_n} )d\mu_n(x) \rightarrow \int \limits_{x \in \mathbb{R}^{d}} f(x - \xi_{\mu} )d\mu(x) \end{equation*} for $f$ continuous and bounded function. But my problem here is that the first integral depends on a sequence of functions, so the weak convergence $\mu_n \rightarrow \mu$ doesn't seem applicable.

Secondly, we must show that \begin{equation*} \lim \limits_{R\rightarrow\infty} \limsup \limits_{n\rightarrow\infty} \int \limits_{|x- \xi_{\mu_n}|\geqslant R} |x-\xi_{\mu_n}|^p d\mu_n(x) = 0 \end{equation*} which seems to follow from (*) but I can't prove how.