Functional Analysis – Does Weak Convergence Imply Convergence in Probability?

functional-analysisprobability theoryweak-convergence

I'm striving to understand a proof from a paper.

Denote the symbol $\rightsquigarrow$ as a weak convergence, and $l^{\infty}(\Omega)$ as the space of bounded functions on a metric space $(\Omega, d)$.

Let $M_n$ be a sequence of random objects of $l^{\infty}(\Omega)$, which converges weakly to $M$, i.e. $M_n \rightsquigarrow M$.

Is it true that $\lVert M_n – M \rVert _{l^{\infty} (\Omega)}$ converges to zero in probability?

I know that convergence in probability implies convergence in distribution, but the converse is not always true.

If you need a reference, please read the last paragraph of https://arxiv.org/pdf/1608.03012.pdf. (Alexander Petersen and Hans-Georg Müller, Fréchet Regression for Random Objects with Euclidean Predictors)

Thank you very much.

Caution : $\Omega$ is not a probability space in this question, instead it is a metric space.

P.S. I just highlighted the part. Thank you. Theorem 1.3.6 in van der Vaart(1996) is about the continuous mapping theorem.

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Best Answer

Recall that $l^\infty(\Omega)$ is a normed space endowed with the norm $\|f\|_{\Omega} = \sup_{\omega \in \Omega}|f(\omega)|$.

Fix some bounded function $g:\Omega \to \mathbb R$. Define the following map $\Psi$ from $l^\infty(\Omega)$ to $\mathbb R$: $$\Psi:f\mapsto \sup_{w\in \Omega}|f(\omega)-g(\omega)|.$$

The following estimate holds for every $f_1,f_2\in l^\infty(\Omega)$ and $\omega \in \Omega$: $$|f_1(\omega)-g(\omega)|\leq \|f_1-f_2\|_{\Omega} + \Psi(f_2),$$ hence $\Psi$ is $1$-Lipschitz, thus continuous.

It is known that $M_n \rightsquigarrow M$. Set $g = M$ and then by the continuous mapping theorem (Theorem 1.3.6 in van der Vaart and Wellner), $$\Psi(M_n) \rightsquigarrow \Psi(M),$$ which rewrites $$\sup_{w\in \Omega}|M_n(\omega)-M(\omega)|\rightsquigarrow 0. $$