Before presenting my question (which I already formulate in the title of this post) is important to establish the context of my problem:
Definition: The $p$-Wasserstein metric $W_{p}(\mu,\nu)$ between $\mu,\nu\in\mathcal{P}_{p}(\Xi)$ is defined by $$W_{p}^{p}(\mu,\nu):=\min_{\Pi\in\mathcal{P}(\Xi\times\Xi)}\left\{\int_{\Xi\times\Xi}d^{p}(\xi,\zeta)\Pi(d\xi,d\zeta)\: :\: \Pi(\cdot \times\Xi)=\mu(\cdot),\: \Pi(\Xi\times\cdot)=\nu(\cdot)\right\}$$ where $$\mathcal{P}_{p}(\Xi):=\left\{\mu\in\mathcal{P}(\Xi)\: :\: \int_{\Xi}d^{p}(\xi,\zeta_{0})\mu(d\xi) < \infty\ \mbox{for some }\zeta_{0}\in\Xi\right\}$$ where $d$ is a metric on $\Xi$.
The $p$-Wasserstein metric (with $p\geq 1$) is also defined for any measure in $\mathcal{P}(\Xi)$ the space of all the measures of probability, the only difference is that in that set it can take values as infinite.
We consider $\Xi=\mathbb{R}^{m}$ and $\xi$ a random vector with support in $\Xi$ and distribution $\mathbb{P}$, let $\widehat{\xi}_{1},\ldots,\widehat{\xi}_{N} $ be a sample of $\xi$, then we consider the empirical distributión given by $$\widehat{\mathbb{P}}_{N}:=\sum_{i=1}^{N}\delta_{\widehat{\xi}_{i}}.$$ Given $x\in\mathbb{R}^{m}$ we consider the random variable $\zeta^{x}:=\langle x,\xi\rangle$, let $\mathbb{P}^{x}$ the distribution of $\zeta^{x}$, note that as $\widehat{\xi}_{1},\ldots,\widehat{\xi}_{N} $ is a sample of $\xi$, then $\widehat{\zeta}^{x}_{1},\ldots,\widehat{\zeta}^{x}_{N}$ given by $\widehat{\zeta}^{x}_{i}:=\langle x,\widehat{\xi}_{i}\rangle$ is a sample of $\zeta^{x}$, therefore, we have $\widehat{P}^{x}_{N}$ the empirical distribution given by $$\widehat{\mathbb{P}}_{N}^{x}:=\sum_{i=1}^{N}\delta_{\widehat{\zeta}_{i}^{x}}.$$ From now on we consider the $2$-Wasserstein metric with $d$ as euclidean distance, that is $d(x,y)=\left\|x-y\right\|$ where $\left\|\cdot\right\|$ is the euclidean norm.
The question: What is the relationship between $W_2^2(\widehat{\mathbb{P}}_{N},\mathbb{P})$ and $W^{2}_{2}(\widehat{\mathbb{P}}_{N}^{x},\mathbb{P}^{x})$?
My attepmt: I think that $$ W^{2}_{2}(\widehat{\mathbb{P}}_{N}^{x},\mathbb{P}^{x}) \leq \left\|x\right\|^{2} W_{2}^{2}\left( \widehat{\mathbb{P}}_{N} ,\mathbb{P}\right)$$ The next is my attempt to prove it, although there are steps of which I do not feel safe.
We consider $\widehat{\widehat{\xi}}_{1},\ldots,\widehat{\widehat{\xi}}_{M}$ other sample of $\xi$, then we consider $\widehat{\widehat{\mathbb{P}}}_{M}$ the empirical distribution determined by this sample. Also, we consider of sample $\widehat{\widehat{\zeta}}^{x}_{1},\ldots,\widehat{\widehat{\zeta}}^{x}_{M}$ of $\zeta^{x}$ given by $\widehat{\widehat{\zeta}}^{x}_{i}:=\left\langle x, \widehat{\widehat{\xi}}_{i}\right\rangle$ and $\widehat{\widehat{\mathbb{P}}}^{x}_{M}$ the empirical distribution determined by this sample.
We know that $ \widehat{\widehat{\mathbb{P}}}_{M} \rightarrow \mathbb{P}$ weakly, then, by Corollary 6.11 in Villani, we have $$W_{2}^{2}\left( \widehat{\mathbb{P}}_{N} ,\widehat{\widehat{\mathbb{P}}}_{M}\right)\overset{{\scriptstyle M\rightarrow \infty}}{\longrightarrow}W_{2}^{2}\left( \widehat{\mathbb{P}}_{N} ,\mathbb{P}\right). \tag{I}$$ Analogously we have $ \widehat{\widehat{\mathbb{P}}}^{x}_{M} \rightarrow \mathbb{P}^{x}$ weakly, then, by Corollary 6.11 in Villani, we have $$W_{2}^{2}\left( \widehat{\mathbb{P}}^{x}_{N} ,\widehat{\widehat{\mathbb{P}}}^{x}_{M}\right)\overset{{\scriptstyle M\rightarrow \infty}}{\longrightarrow}W_{2}^{2}\left( \widehat{\mathbb{P}}_{N}^{x} ,\mathbb{P}^{x}\right). \tag{II}$$
But note that
$$ \begin{align} W_{2}^{2}\left( \widehat{\mathbb{P}}^{x}_{N} ,\widehat{\widehat{\mathbb{P}}}^{x}_{M}\right) &= \inf\left\{\sum_{i=1}^{N}\sum_{j=1}^{M} \lambda_{i,j}\left|\widehat{\zeta}^{x}_{i}-\widehat{\widehat{\zeta}}^{x}_{j} \right|^{2} \:\left|\: \begin{array}{l} \sum_{i=1}^{N}\lambda_{i,j}=\frac{1}{M},\\ \sum_{j=1}^{M}\lambda_{i,j}=\frac{1}{N},\\ \lambda_{i,j}\geq 0, \\ i=1,\ldots,N,\\ j=1,\ldots,M \end{array} \right.\right\} \\ &= \inf\left\{\sum_{i=1}^{N}\sum_{j=1}^{M} \lambda_{i,j}\left|\left\langle x, \widehat{\xi}_{i}\right\rangle-\left\langle x, \widehat{\widehat{\xi}}_{j}\right\rangle \right|^{2} \:\left|\: \begin{array}{l} \sum_{i=1}^{N}\lambda_{i,j}=\frac{1}{M},\\ \sum_{j=1}^{M}\lambda_{i,j}=\frac{1}{N},\\ \lambda_{i,j}\geq 0, \\ i=1,\ldots,N,\\ j=1,\ldots,M \end{array} \right.\right\} \\ &\leq \inf\left\{\sum_{i=1}^{N}\sum_{j=1}^{M} \lambda_{i,j} \left\|x\right\|^{2}\left\| \widehat{\xi}_{i}- \widehat{\widehat{\xi}}_{j} \right\|^{2} \:\left| \: \begin{array}{l} \sum_{i=1}^{N}\lambda_{i,j}=\frac{1}{M},\\ \sum_{j=1}^{M}\lambda_{i,j}=\frac{1}{N},\\ \lambda_{i,j}\geq 0, \\ i=1,\ldots,N,\\ j=1,\ldots,M \end{array} \right.\right\} \:\: \begin{array}{l}\mbox{by Hölder}\\ \mbox{inequality}\end{array}\\ &= \left\|x\right\|^{2}W_{2}^{2}\left( \widehat{\mathbb{P}}_{N} ,\widehat{\widehat{\mathbb{P}}}_{M}\right). \end{align} $$ Therefore, by $(I)$ and $(II)$ we have $$W_{2}^{2}\left( \widehat{\mathbb{P}}_{N}^{x} ,\mathbb{P}^{x}\right)\leq \left\|x\right\|^{2} W_{2}^{2}\left( \widehat{\mathbb{P}}_{N} ,\mathbb{P}\right).$$
Remark: I feel that it was a very simple demonstration, for this reason I do not trust my argument, I would like someone to help me see if my reasoning has errors. In my research this result is very important, for that reason I need to know if my idea is correct.
I published my question here because I feel that here are the people trained to answer it.