Conversely, we assume that $\forall \bb t\in\R^d,\, \, \phi_{{\bb X}^{(n)}}(\bb t)\to \phi_{\bb X}(\bb t)$ and show that for any continuous function $g$ that is zero outside a bounded and closed set, we have $\E(g({\bb X}^{(n)}))\to \E(g(\bb X))$. Using the Portmanteau theorem, this implies that ${\bb X}^{(n)}\tood \bb X$. Since $g$ is continuous on a compact set, it is uniformly continuous and we can select for all $\epsilon>0$ a $\delta>0$ such that $\|\bb x-\bb y\| < \delta$ implies $|g(\bb x)-g(\bb y)| < \epsilon$.

Denoting by $\bb Z$ a $N(\bb 0,\sigma^2 I)$ random vector that is independent of $\bb X$ and the sequence ${\bb X}^{(n)}$, we have \begin{align*} & |\E(g({\bb X}^{(n)})) - \E(g(\bb X))| \\ &\quad = |\E(g({\bb X}^{(n)})) - \E(g(\bb X)) + \E(g({\bb X}^{(n)}+\bb Z)) \\ &\quad \quad - \E(g({\bb X}^{(n)}+\bb Z)) + \E(g({\bb X}+\bb Z)) - \E(g({\bb X}+\bb Z)) |\\ &\quad \leq |\E(g({\bb X}^{(n)})) - \E(g({\bb X}^{(n)}+\bb Z)) | + |\E(g({\bb X}^{(n)}+\bb Z)) -\E(g({\bb X}+\bb Z))|\\ &\quad \quad + | \E(g({\bb X}+\bb Z)) - \E(g(\bb X))|. \end{align*} The first term above is bounded by $2\epsilon$ since for $\sigma$ sufficiently small \begin{align*} |\E(g({\bb X}^{(n)})) - \E(g({\bb X}^{(n)}+\bb Z)) | &\leq \E(|g({\bb X}^{(n)})) - \E(g({\bb X}^{(n)}+\bb Z))| I(\|\bb Z\|\leq\delta)\\ & \quad + \E(|g({\bb X}^{(n)})) - \E(g({\bb X}^{(n)}+\bb Z)) | I(\|\bb Z\|>\delta)\\ &\leq \E(\epsilon) + 2 \left(\sup_{\bb w}|g(\bb w)|\right) \P(\|\bb Z\|>\delta)\\ &\leq 2\epsilon. \end{align*} The third term above is also bounded by $2\epsilon$ due to a similar argument. It remains to show that the second term converges to zero: $\E(g({\bb X}^{(n)}+\bb Z))\to \E (g(\bb X+\bb Z))$. We will then have that $|\E(g({\bb X}^{(n)})) - \E(g(\bb X))|\to 0$, implying that $\E(g({\bb X}^{(n)})) \to \E(g(\bb X)$ (for all continuous functions $g$ that are zero outside a bounded and closed set), which together with the Portmanteau theorem implies ${\bb X}^{(n)}\tood \bb X$.

We show below that $\E(g({\bb X}^{(n)}+\bb Z))\to \E (g(\bb X+\bb Z))$. We have \begin{align} \tag{*} \E(g({\bb X}^{(n)}+\bb Z)) &= \frac{1}{(\sqrt{2\pi}\sigma)^d} \iint g(\bb x+\bb z) \exp(-{\bb z}^{\top}\bb z/(2\sigma^2)) \, d\bb z d F_{{\bb X}^{(n)}} \\ &= \frac{1}{(\sqrt{2\pi}\sigma)^d} \iint g(\bb u) \exp(-(\bb u-\bb x)^{\top}(\bb u-\bb x)/(2\sigma^2)) \, d\bb u d F_{{\bb X}^{(n)}} \nonumber\\ &= \frac{1}{(\sqrt{2\pi}\sigma)^d} \iint g(\bb u) \prod_{j=1}^d \exp\left(-\frac{(u_j-x_j)^2}{2\sigma^2}\right) \, d\bb u d F_{{\bb X}^{(n)}}\nonumber \\ &= \frac{1}{(\sqrt{2\pi}\sigma)^d} \iint g(\bb u) \prod_{j=1}^d \frac{\sigma}{\sqrt{2\pi}} \int \exp\left(it_j(u_j-x_j)-\sigma^2t_j^2/2\right) \\ & \qquad \qquad\qquad \qquad\qquad \qquad \qquad \qquad \,dt_j d\bb u d F_{{\bb X}^{(n)}}\nonumber \\ &= \frac{1}{(2\pi)^d} \iiint g(\bb u) \exp\left(i{\bb t}^{\top} (\bb u-\bb x)-\sigma^2 {\bb t}^{\top} \bb t/2 \right) \,d\bb t d\bb u d F_{{\bb X}^{(n)}}\nonumber \\ &= \frac{1}{(2\pi)^d} \iint g(\bb u) \exp\left(i{\bb t}^{\top} \bb u-\sigma^2 {\bb t}^{\top} \bb t/2 \right) \phi_{{\bb X}^{(n)}} (-\bb t) \, d \bb td\bb u, \nonumber \end{align} where $\bb u=\bb x+\bb z$. Note that we used Lemma 8.7.1 in the fourth equality and Proposition 8.7.2 in the last equality.

Since $g$ is continuous and non-zero only on a closed and bounded set, $g(\bb u)$ may be made into a distribution by adding a constant to it and dividing by a constant. This implies that $\E(g({\bb X}^{(n)}+\bb Z))$ may be considered as an expectation over a two random vectors $\bb U$ having density $c(g(\bb u)+b)$ and $\bb T$ have a Gaussian density. The argument of that expectation is the bounded function $\exp\left(i{\bb t}^{\top} \bb u\right) \phi_{{\bb X}^{(n)}} (-\bb t)$, and so by the dominated convergence theorem for random variables (Proposition 8.3.1) \begin{multline*} \frac{1}{(2\pi)^d} \iint g(\bb u) \exp\left(i{\bb t}^{\top} \bb u-\sigma^2 {\bb t}^{\top} \bb t/2 \right) \phi_{{\bb X}^{(n)}} (-\bb t) \, d \bb td\bb u \\ \to \frac{1}{(2\pi)^d} \iint g(\bb u) \exp\left(i{\bb t}^{\top} \bb u-\sigma^2 {\bb t}^{\top} \bb t/2 \right) \phi_{\bb X} (-\bb t) \, d \bb td\bb u. \end{multline*} Repeating the derivation in Equation (*) with $\bb X$ substituting ${\bb X}^{(n)}$ we see that \begin{multline*} \E(g({\bb X}^{(n)}+\bb Z)) = \frac{1}{(2\pi)^d} \iint g(\bb u) \exp\left(i{\bb t}^{\top} \bb u-\sigma^2 {\bb t}^{\top} \bb t/2 \right) \phi_{\bb X} (-\bb t) \, d \bb td\bb u, \end{multline*} implying that $\E(g({\bb X}^{(n)}+\bb Z))\to \E(g({\bb X} +\bb Z))$.