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 ?


1 Answer 1


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.)

  • $\begingroup$ 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 ? $\endgroup$
    – vggls
    Feb 27, 2019 at 21:25
  • $\begingroup$ @vl.ath Oops, you are right -- I missed that. I'm not sure what to do-- maybe try taking the Fourier transform? $\endgroup$
    – Elle Najt
    Feb 27, 2019 at 21:55
  • $\begingroup$ ok i will try that.. do you know if the statement is correct in general ? $\endgroup$
    – vggls
    Feb 27, 2019 at 22:45
  • 1
    $\begingroup$ @vl.ath It appears to be answered here: math.stackexchange.com/questions/1552171/… $\endgroup$
    – Elle Najt
    Feb 27, 2019 at 22:58
  • $\begingroup$ 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. $\endgroup$
    – vggls
    Feb 28, 2019 at 9:20

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