Portmanteaus Theorem and Lipschitz functions I was looking at some basics in the theory of convergence of distributions, e.g. Portmanteaus Thm. One part of it asserts that it suffices to check $\mathbb{E}[f(X_n)]\to\mathbb{E}[f(X)]$ for bounded Lipschitz-continuous $f$ instead of $f\in\mathcal{C}_b(E)$ to establish $X_n\Rightarrow X$ on $E$. The proof always cirlces with other parts of the Thm. (the limsup/liminf inequalities and convergence along sets having borders which are $\mathbb{P}$-null sets). 
Now I ask myself if there is a "direct" proof of this fact ($E=\mathbb{R}$ suffices).
Thx a lot!
 A: The following proof shows that it suffices to have $\mathbb{E}f(X_n) \to \mathbb{E}f(X)$ for bounded $C^{\infty}$ functions.


*

*It suffices to show that $\mathbb{E}f(X_n) \to \mathbb{E}f(X)$ for uniformly continuous bounded functions. Indeed: Suppose that $f$ is bounded and continuous. Choose a cut-off function $\chi_R \in C_c$ such that $\chi|_{B(0,R)}=1$, $0 \leq \chi_R \leq 1$. Then $$\begin{align*} &\quad |\mathbb{E}f(X_n)-\mathbb{E}f(X)| \\&\leq \|f\|_{\infty} \bigg( \mathbb{E}(1-\chi_R(X_n))+ \mathbb{E}(1-\chi_R(X)) \bigg) + |\mathbb{E}(f \chi_R(X_n))-\mathbb{E}(f \chi_R(X))| \\ &= \|f\|_{\infty} (2- \mathbb{E}\chi_R(X_n)-\mathbb{E}\chi_R(X)) + |\mathbb{E}(f \chi_R(X_n))-\mathbb{E}(f \chi_R(X))|. \end{align*}$$ By assumption, $$\mathbb{E}\chi_R(X_n) \to \mathbb{E}(\chi_R(X)) \qquad \text{and} \quad |\mathbb{E}(f \chi_R(X_n))-\mathbb{E}f \chi_R(X)| \to 0$$ as $n  \to \infty$. Consequently, $$\liminf_{n \to \infty} |\mathbb{E}(f(X_n))-\mathbb{E}f(X)| \leq 2 \|f\|_{\infty} \mathbb{E}(1-\chi_R(X)).$$ Letting $R \to \infty$ finishes the proof.

*Let $f$ be a uniformly continuous bounded function and denote by $$p_t(x) := \frac{1}{\sqrt{2\pi t}} \exp \left(- \frac{x^2}{2t} \right)$$ the heat kernel. Then it is not difficult to see that the convolution $f \ast p_t$ satisfies $$\|f-f \ast p_t\|_{\infty} \to 0$$ as $t \to 0$. Moreover, $f \ast p_t$ is differentiable (with bounded derivative) and bounded.

*It remains to prove that $\mathbb{E}f(X_n) \to \mathbb{E}f(X)$ holds for uniformly continuous bounded functions. To this end, we define $f_t := f \ast p_t$. By step 2, $f_t$ is bounded and Lipschitz-continuous and for $t$ sufficiently small we have $\|f-f_t\|_{\infty} \leq \varepsilon$. Using a standard $\varepsilon/3$-argument, we find that $$\mathbb{E}f(X_n) \to \mathbb{E}f(X).$$


Remark: In fact, the weak convergence is equivalent to $$\mathbb{E}e^{\imath \, \xi X_n} \to \mathbb{E}e^{\imath \, \xi X} \quad \text{for all} \, \xi,$$
i.e. it suffices to have convergence for $f$ of the form $f(x)=e^{\imath \, x \xi}$.
