Uniform convergence in distribution

Question

Consider a sequence of stochastic processes, $X_n(x)$ and a limiting process $X(x)$. For a fixed $x$, if $\mathbb{P}(X_n(x) \leq y)$ converges to $\mathbb{P}(X(x) \leq y)$ for continuity points of $F_{X(x)}(y) = \mathbb{P}(X(x) \leq y) $, then $\mathbb{E}[f(X_n(x)]$ converges to $\mathbb{E}[f(X(x)]$ for any $f$ which is continuous and bounded.

Suppose I have uniform convergence in distribution then does $\mathbb{E}[f(X_n(x)]$ converge to $\mathbb{E}[f(X(x)]$ uniformly? That is, is the following result true:

$$\lim_{n \rightarrow \infty } \sup_{x,y} \Biggl| \mathbb{P}(X(x) \leq y) - \mathbb{P}(X_n(x) \leq y) \Biggr| = 0$$ $$ \implies$$ $$\lim_{n \rightarrow \infty } \sup_{x} \Biggl| \mathbb{E}f(X(x)) - \mathbb{E}f(X_n(x)) \Biggr| = 0.$$ for $f$ continuous and bounded.

Answer 1

It seems true. First assume that $f$ is non-negative and write $$\mathbb E\left[f(X_n(x))\right]=\int_0^{\lVert f\rVert_\infty} \mathbb P\left(f(X_n(x))\gt y\right)\mathrm dy,$$ and a similar bound for $X(x)$: then \begin{align} \left\lvert\mathbb E\left[f(X_n(x))\right]-\mathbb E\left[f(X(x))\right]\right\rvert&=\left|\int_0^{\lVert f\rVert_\infty} \mathbb P\left(f(X_n(x))\gt y\right)\mathrm dy-\int_0^{\lVert f\rVert_\infty} \mathbb P\left(f(X(x))\gt y\right)\mathrm dy\right| \\ &=\left|\int_0^{\lVert f\rVert_\infty} \mathbb P\left(f(X_n(x))\leqslant y\right)\mathrm dy-\int_0^{\lVert f\rVert_\infty} \mathbb P\left(f(X(x))\leqslant y\right)\mathrm dy\right|\\& \leqslant \lVert f\rVert_\infty\cdot \sup_{x,y} \Biggl| \mathbb{P}(X(x) \leqslant y) - \mathbb{P}(X_n(x) \leqslant y) \Biggr|. \end{align} Taking limit as $n \rightarrow \infty$, we get the desired result.
For the general case, decompose $f$ as the difference of two non-negative bounded and continuous functions.

Answer 2

Proof for the case when $f$ has compact support and the limiting distribution is continuous $ \forall x$:
Since $f$ has compact support (say over $[0,K]$) and is continuous, it is also uniformly continuous. Therefore given any $\epsilon$ there exists $m$ points $u_0 = 0 < u_1 < \cdots < u_m = K$, such that $$\sup_{u_i< y < u_{i+1}} |f(y) - f(u_i)| < \epsilon,$$ for all $i=1,2,\cdots, m$.

We have \begin{equation} \begin{split} E[f(X_n(x)] &= \int_\limits{0}^{K} f(u) d\mathbb{P}_{n,x}(u) \\ &\approx_{\epsilon} \sum\limits_{i=1}^{m-1} f(u_i) \mathbb{P}_{n,x}((u_i,u_{i+1}]), \end{split}\end{equation} where $\approx_{\epsilon}$ indicates that the RHS and LHS are $\epsilon$-close to each other. Similarly, \begin{equation} \begin{split} E[f(X(x)] \approx_{\epsilon} \sum\limits_{i=1}^{m-1} f(u_i) \mathbb{P}_{x}((u_i,u_{i+1}]). \end{split}\end{equation} Therefore \begin{equation} \begin{split} |\mathbb{E}f(X_n(x) - \mathbb{E}f(X(x))| &\approx_{2\epsilon} \sum\limits_{i=1}^{m} f(u_i) |\mathbb{P}_{n,x}((u_i,u_{i+1}]) - \mathbb{P}_{x}((u_i,u_{i+1}])|, \\ &\leq \parallel f \parallel_{\infty} \sum\limits_{i=1}^{m} |\mathbb{P}_{n,x}((u_i,u_{i+1}]) - \mathbb{P}_{x}((u_i,u_{i+1}])|.\end{split}\end{equation} The limiting distribution being continuous implies that as $n \rightarrow \infty$, $\mathbb{P}_{n,x}((u_i,u_{i+1}])$ converges to $\mathbb{P}_{x}((u_i,u_{i+1}])$. Also since the convergence is uniform over $x$, we have the desired result.

Note: The assumption that the limiting distribution be continuous helps us and lets us choose $u_i$s independently of $x$, else we have to choose the $u_i$s such that $\mathbb{P}_{n,x}((u_i,u_{i+1}])$ converges to $\mathbb{P}_{x}((u_i,u_{i+1}])$ for all $x$.