# Liminf of Pointwise Norms of a Weakly Convergent Sequence

Let $$X_1, X_2, \cdots$$ be a sequence of $$p$$-integrable $$\mathbb{R^d}$$ valued random variables. Assume that $$X_n$$ converges $$0$$ weakly, then can we say that $$r(\omega) = liminf\{ |X_1 (\omega)|, |X_2 (\omega)|, \cdots\}$$ is $$0$$ for almost every $$\omega ?$$

The classical example of a weakly convergent but not strongly convergent sequence is those of orthogonal basis, yet it is not a counterexample for the above claim. I could not prove the claim but intuitively I believe it holds, at least for $$\mathbb{R^d}$$ valued random vectors.

If needed one can assume that the sequence $$(X_n)_n$$ is bounded in $$p$$-norm, I do not feel like this is necessary.

Weak convergence together with boundedness of second moments implies convergence of expectations. By Fatou's Lemma we get $$E \lim \inf |X_n| \leq \lim \inf E|X_n|=0$$ since $$E|X_n| \to 0$$. This implies that $$\lim \inf |X_n|=0$$ almost surely.
• I don't see how weak convergence together with boundedness of second moments implies convergence of expectations. Are you using Vitaly type theorem here? For every $Y$ $q$-integrable, we have $E[Y X_n] \rightarrow E[Y \, 0] = 0,$ and there is some $W$ $p$-integrable such that $|X_n| \leq W$ for all $n.$ I dont see how we proceed. Could you provide more details please? Jan 4 at 9:52
• @vekinpirna Boundeness of second moments implies uniform integrabilty. And $Y_n \to 0$ weakly, $(Y_n)$ uniform integrable implies $EY_n \to 0$. (one quick way of proving this to replace weak convergence by almost sure convergence using Skorohod Representation Theorem), but you can do it without this theorem also). Jan 4 at 10:00
• We are only intersetd in the real random variables $|X_n|$. And $X_n \to 0$ weakly implies $|X_n| \to 0$ weakly, so the usual Skorohod Theorem applies. @vekinpirna Jan 4 at 10:15
• Is it obvious that $X_n \rightarrow 0$ weakly implies $|X_n| \rightarrow 0$ weakly? I will take my time working on that. Jan 4 at 10:20