About the weak compactness of a certain set.

Question

Why is the following set weakly compact in $L^1(d\mu)$? $$\left\{-\frac{|x|^2}{2}+O(l)\right\}$$ where $\mu$ is a probability measure in $\mathbb{R}^n$ with finite second order moment: $$\int_{\mathbb{R}^n}|x|^2d\mu<\infty$$ and $O(l)$ is any function bounded by some fixed $l\in \mathbb{N}$.

This can be found on page 65 of Villani's Topics in optimal transport. It must be simple.

Answer 1

Just to clarify the problem statement, we have for each fixed integer $\ell$, a sequence $(\varphi_k^{(\ell)})_{k \in \mathbb{N}}$ of functions on $\mathbb{R}^n$ satisfying the following bound:

\begin{equation} 0 \leq \varphi_k^{(\ell)}(x) + \frac{|x|^2}{2} \leq \ell \end{equation}

The claim is that $(\varphi_k^{(\ell)})_{k \in \mathbb{N}}$ is relatively compact for the weak topology $\sigma(L^1(d\mu), L^\infty(d\mu))$.

As per the comment above, we can use the Dunford-Pettis theorem (the below criterion is only applicable for $\sigma$-finite measure spaces, but we are fine here):

$(\varphi_k^{(\ell)})_{k \in \mathbb{N}}$ is relatively compact for the weak topology $\sigma(L^1(d\mu), L^\infty(d\mu))$ iff

1. $\sup_{k}\int_{\mathbb{R}^n}\varphi_k^{(\ell)}d\mu < \infty$, and
2. $\lim_{j \to \infty}\sup_{k}\int_{A_j}\varphi_k^{(\ell)}d\mu = 0$ for every ($\mu$-measurable) sequence $A_j \downarrow \emptyset$

Using the above bound, we easily see that $|\int_{\mathbb{R}^n}\varphi_k^{(\ell)}d\mu| \leq \ell + \int_{\mathbb{R}^n}\frac{|x|^2}{2}d\mu < \infty$ is uniformly bounded in $k$.

As well, we have $|\int_{A_j}\varphi_k^{(\ell)}d\mu| \leq \ell\mu(A_j) + \int_{A_j}\frac{|x|^2}{2}d\mu \to 0$ as $A_j \downarrow \emptyset$

($\int_{A_j}\frac{|x|^2}{2}d\mu \to 0$ by dominated convergence).

As 1 and 2 are satisfied, we invoke Dunford-Pettis.