Two basic questions in Wasserstein spaces

We denote by $$P (\mathbb{R}^{d})$$ the space of probability measures on $$\mathbb{R}^{d}$$ and for $$p\geqslant 1$$ the Wasserstein space by $$\begin{equation*} P^p (\mathbb{R}^{d}) = \{ \mu \in P(\mathbb{R}^{d}) \,|\, \int |x|^p d\mu(x) <\infty \} \end{equation*}$$ Also, we say that $$\mu_N \rightarrow \mu$$ in $$P^p (\mathbb{R}^{d})$$ iff $$\mu_N \rightarrow \mu$$ weakly and $$\int |x|^p d\mu_N(x) \rightarrow \int |x|^p d\mu$$.

I am trying to clarify the following :

1)if $$q, then do we have that $$P^p (\mathbb{R}^{d}) \subseteq P^q (\mathbb{R}^{d})$$ ?

2)if the previous question is correct, then does convergence in $$P^p (\mathbb{R}^{d})$$ imply convergence in $$P^q (\mathbb{R}^{d})$$ ?

Intuitively, they should be correct, but I can't even prove the first one. I tried Holder but I can't finish it.

Could someone give me some help by telling me if at least the statements are correct or wrong ?

If $$p > q$$, then a probability distribution $$X$$ with a $$p$$th moment has a $$q$$th moment.

This follows by applying Jensen's inequality to the function $$\phi(x) = x^{p/q}$$.

Note that $$\phi''(x) = (p/q)(p/q - 1) x^{p/q - 1}$$, which is convex on $$[0,\infty)$$ exactly when $$p/q > 1$$.

So we can compute:

$$E [ ( |X|^q)^{p/q} ] \geq E [|X|^q]^{p/q}$$.

The same logic will tell you that if $$X_n \to 0$$ in $$L_p$$, then it also does in $$L^q$$, when $$q < p$$. (This is only true for probability distributions.)

• The first questions is ok. But I still can't finish the second. I would need convergence in the p-th moment, but what we have in the assumption is $E[|X_n|^p] \rightarrow E[|X|^p]$, which is weaker. No ? – vl.ath Feb 27 at 21:25
• @vl.ath Oops, you are right -- I missed that. I'm not sure what to do-- maybe try taking the Fourier transform? – Lorenzo Najt Feb 27 at 21:55
• ok i will try that.. do you know if the statement is correct in general ? – vl.ath Feb 27 at 22:45
• @vl.ath It appears to be answered here: math.stackexchange.com/questions/1552171/… – Lorenzo Najt Feb 27 at 22:58
• So to sum up, the weak convergence assumption is the convergence in distribution assumption AND the $E[|X_n|^p] \rightarrow E[|X|^p]$ assumption implies that $\sup \limits_n E[|X_n|^p] < \infty$. Consequently, for q<p, we get that $E[|X_n|^q] \rightarrow E[|X|^q]$ which is the desired result. – vl.ath Feb 28 at 9:20