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 $q<p$, 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.)