Does convergence in law to absolutely continuous limit imply convergence in convex distance? Let $(X_n)$ be a sequence of $\mathbb{R}^d$-valued random variables converging in distribution to some limiting random variable $X$ whose CDF is absolutely continuous with respect to the Lebesgue measure.
Does it follow that $X_n$ converges to $X$ in convex distance, i.e. that
$$\sup_{h} \lvert \operatorname{E}(h(X)) - \operatorname{E}(h(X_n)) \rvert \to 0,$$
where the supremum is taken over all indicator functions of measurable convex subsets of $\mathbb{R}^d$, if necessary assuming (absolute) continuity of the CDFs of the $X_n$ as well?
Remark 1:
For $d=1$, the implication is true and can be proven by Polya's theorem (convergence in law of real valued random variables towards a limit with continuous CDF implies uniform convergence of the CDF). Is it still true for $d \geq 2$?
Remark 2:
If absolute continuity is replaced by continuity the conclusion is false, see here
 A: What is essential here is that the distribution of $X$ assigns little mass to sets which are essentially $(d-1)$-dimensional.

The standard approach to problems of this kind is to estimate
$$ \operatorname{P}(X_n \in K) - \operatorname{P}(X \in K) $$
from above by
$$ \operatorname{E}(g(X_n)) - \operatorname{E}(f(X)) , $$
and from below by
$$ \operatorname{E}(f(X_n)) - \operatorname{E}(g(X)) , $$
where $0 \leqslant f \leqslant \mathbb{1}_K \leqslant g \leqslant 1$, and $f$ and $g$ are continuous. In $k$-th step we choose $f$ and $g$ in such a way that $\operatorname{E}(g(x) - f(x)) < \tfrac{1}{2 k}$. Convergence of $\operatorname{E}(f(X_n))$ to $\operatorname{E}(f(X))$ and convergence of $\operatorname{E}(g(X_n))$ to $\operatorname{E}(g(X))$ imply that
$$ -\tfrac{1}{k} \leqslant \operatorname{P}(X_n \in K) - \operatorname{P}(X \in K) \leqslant \tfrac{1}{k} $$
for all $n$ large enough.
This works as expected, i.e. leads to convergence of $\operatorname{P}(X_n \in K)$ to $\operatorname{P}(X \in K)$, if (and only if) $\operatorname{P}(X \in \partial K) = 0$: then (and only then) it is possible to choose $f$ and $g$ with the desired property.

Now in order to get uniform convergence for a class of sets $K$ — here the class of convex subsets of $\mathbb{R}^d$ — in every step $k$ we should choose $f$ and $g$ from a fixed finite-dimensional set of functions (which, of course, can well depend on $k$). One solution for the class of convex $K$ is as follows.
By tightness, there is $R > 1$ such that $\operatorname{P}(|X_n| > R) < \tfrac{1}{4 k}$ uniformly in $n$. We will choose (small) $\delta \in (0, 1)$ at a later stage. We cover the ball $\overline{B}(0, R)$ using $d$-dimensional cubes $Q_j$ with edge length $\delta$ and vertices at lattice points $(\delta \mathbb{Z})^d$. To be specific, suppose that the cubes $Q_j$ are open sets. Let $h_j = \mathbb{1}_{Q_j}$ be the indicator of $Q_j$. Then
$$h_1 + \ldots + h_J = 1 \quad \text{a.e. on $\overline{B}(0, R)$}$$
(more precisely: everywhere on $\overline{B}(0, R)$, except possibly on the faces of cubes $Q_j$). We add $h_0 = 1 - \sum_{j = 1}^J h_j$ to this collection. Observe that $h_0 = 0$ a.e. on the complement of $\overline{B}(0, R)$, and hence $\operatorname{E}(h_0(X)) \leqslant \operatorname{P}(|X_n| > R) < \tfrac{1}{4 k}$. Furthermore, by the first part of this answer, we already know that $\operatorname{E}(h_j(X_n)) = \operatorname{P}(X_n \in Q_j)$ converges to $\operatorname{E}(h_j(X)) = \operatorname{P}(X \in Q_j)$ as $n \to \infty$ for every $j = 0, 1, \ldots, J$ (because the distribution of $X$ does not charge the boundaries of $Q_j$).
Given a convex set $K$, we define $f$ to be the sum of all $h_j$ corresponding to cubes $Q_j$ contained in $K$, and $g$ to be the sum of all $h_j$ corresponding to cubes $Q_j$ which intersect $K$. Clearly, $$0 \leqslant f \leqslant \mathbb{1}_K \leqslant g \leqslant 1 \qquad \text{a.e.}$$ As in the first part of the proof,
$$ \operatorname{E}(f(X_n)) - \operatorname{E}(g(X)) \leqslant \operatorname{P}(X_n \in K) - \operatorname{P}(X \in K) \leqslant \operatorname{E}(g(X_n)) - \operatorname{E}(f(X)) . $$
For $n$ large enough we have
$$ \sum_{j = 0}^J |\operatorname{E}(h_j(X_n)) - \operatorname{E}(h_j(X))| \leqslant \tfrac{1}{2 k} , $$
and so
$$ |\operatorname{E}(f(X_n)) - \operatorname{E}(f(X))| \leqslant \tfrac{1}{2 k} , \qquad |\operatorname{E}(g(X_n)) - \operatorname{E}(g(X))| \leqslant \tfrac{1}{2 k} .$$
Thus,
$$ -\tfrac{1}{2k} - \operatorname{E}(g(X) - f(X)) \leqslant \operatorname{P}(X_n \in K) - \operatorname{P}(X \in K) \leqslant \tfrac{1}{2k} + \operatorname{E}(g(X) - f(X)) $$
for $n$ large enough, uniformly with respect to $K$. It remains to choose $\delta > 0$ such that $\operatorname{E}(g(X) - f(X)) < \tfrac{1}{2k}$ uniformly with respect to $K$; once this is proved, we have
$$ -\tfrac{1}{k} \leqslant \operatorname{P}(X_n \in K) - \operatorname{P}(X \in K) \leqslant \tfrac{1}{k} $$
for $n$ large enough, uniformly with respect to $K$, as desired.
By definition, $g - f$ is the sum of some number of functions $h_j$ with $j \geqslant 1$ — say, $m$ of them — and possibly $h_0$. Recall that $\operatorname{E}(h_0(X)) \leqslant \operatorname{P}(|X_n| > R) < \tfrac{1}{4 k}$. It follows that
$$ \operatorname{E}(g(X) - f(X)) \leqslant \tfrac{1}{4 k} + \sup \{ \operatorname{P}(X \in A) : \text{$A$ is a sum of $m$ cubes $Q_j$} \} . \tag{$\heartsuit$} $$
We now estimate the size of $m$.

Lemma. For a convex $K$, the number $m$ defined above is bounded by a constant times $(R / \delta)^{d - 1}$.

Proof: 
Suppose that $Q_j$ intersects $K$, but it is not contained in $K$. Consider any point $z$ of $K \cap Q_j$, and the supporting hyperplane $$\pi = \{x : \langle x - z, \vec{u} \rangle = 0\}$$ of $K$ at that point. We choose $\vec{u}$ in such a way that $K$ is contained in $\pi^- = \{x : \langle x - z, \vec{u} \rangle \leqslant 0\}$. If the boundary of $K$ is smooth at $z$, then $\vec{u}$ is simply the outward normal vector to the boundary of $K$ at $z$.
To simplify notation, assume that $\vec{u}$ has all coordinates non-negative. Choose two opposite vertices $x_1, x_2$ of $Q_j$ in such a way that $\vec{v} = x_2 - x_1 = (\delta, \ldots, \delta)$. Then the coordinates of $x_2 - z$ are all positive. It follows that for every $n = 1, 2, \ldots$, all coordinates of $(x_1 + n \vec{v}) - z = (x_2 - z) + (n - 1) \vec{v}$ are non-negative, and therefore the translated cubes $Q_j + n \vec{v}$ all lie in $\pi^+ = \{x : \langle x - z, \vec{u} \rangle \geqslant 0\}$. In particular, all these cubes are disjoint with $K$.
In the general case, when the coordinates of $\vec{u}$ have arbitrary signs, we obtain a similar result, but with $\vec{v} = (\pm \delta, \ldots, \pm \delta)$ for some choice of signs. It follows that with each $Q_j$ intersecting $K$ but not contained in $K$ we can associate the directed line $x_2 + \mathbb{R} \vec{v}$, and this this line uniquely determines $Q_j$: it is the last cube $Q$ with two vertices on this line that intersects $K$ (with "last" referring to the direction of the line).
It remains to observe that the number of lines with the above property is bounded by $2^d$ (the number of possible vectors $\vec{v}$) times the number of points in the projection of $(\delta \mathbb{Z})^d \cap \overline{B}(0, R)$ onto the hyperplane perpendicular to $\vec{v}$. The latter is bounded by a constant times $(R / \delta)^{d - 1}$, and the proof is complete. $\square$
(The above proof includes simplification due to Iosif Pinelis.)
Since the Lebesgue measure of $Q_j$ is equal to $\delta^d$, the measure of $A$ in ($\heartsuit$) is bounded by $m \delta^d \leqslant C R^{d - 1} \delta$ for some constant $C$. Furthermore, since the distribution of $X$ is absolutely continuous, we can find $\delta > 0$ small enough, so that $\operatorname{P}(X \in A) < \tfrac{1}{4 k}$ for every set $A$ with measure at most $C R^{d - 1} \delta$ (recall that $R$ was chosen before we fixed $\delta$). By ($\heartsuit$), we find that
$$ \operatorname{E}(g(X) - f(X)) \leqslant \tfrac{1}{4 k} + \sup \{ \operatorname{P}(X \in A) : |A| \le C R^{d - 1} \delta\} \leqslant \tfrac{1}{2 k} , $$
uniformly with respect to $K$.
A: $\newcommand{\R}{\mathbb{R}}
\newcommand{\ep}{\varepsilon}
\newcommand{\p}{\partial}
\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}$
This is to try to provide a simplification and detalization of the answer by Mateusz Kwaśnicki. 
Suppose that the distribution of $X$ is absolutely continuous (with respect to the Lebesgue measure) and $X_n\to X$ in distribution. We are going to show that then $X_n\to X$ in in convex distance, that is, 
$$\sup_K|\mu_n(K)-\mu(K)|\overset{\text{(?)}}\to0$$
(as $n\to\infty$), where $\mu_n$ and $\mu$ are the distributions of $X_n$ and $X$, respectively, and $\sup_K$ is taken over all measurable convex sets in $\R^d$. 
Take any real $\ep>0$. Then there is some real $R>0$ such that $P(X\notin Q_R)\le\ep$, where $Q_R:=(-R/2,R/2]^d$, a left-open $d$-cube. Since $X_n\to X$ in distribution and $P(X\in\p Q_R)=0$, there is some natural $n_\ep$ such that for all natural $n\ge n_\ep$ we have 
$$P(X_n\notin Q_R)\le2\ep.$$
Take a natural $N$ and partition the left-open $d$-cube $Q_R$ naturally into $N^d$ left-open $d$-cubes $q_j$ each with edge length $\de:=R/N$, where $j\in J:=[N^d]:=\{1,\dots,N^d\}$. 
Using again the conditions that $X_n\to X$ in distribution and $\mu$ is absolutely continuous (so that $\mu(\p q_j)=0$ for all $j\in J$), and increasing $n_\ep$ is needed, we may assume that for all natural $n\ge n_\ep$
$$\De:=\sum_{j\in J}|\mu_n(q_j)-\mu(q_j)|\le\ep.$$
Take now any measurable convex set $K$ in $\R^d$. Then 
$$|\mu_n(K)-\mu(K)|\le|\mu_n(K\cap Q_R)-\mu(K\cap Q_R)| \\ 
+|\mu_n(K\setminus Q_R)-\mu(K\setminus Q_R)|
$$
and 
$$|\mu_n(K\setminus Q_R)-\mu(K\setminus Q_R)|\le
P(X_n\notin Q_R)+P(X\notin Q_R)\le3\ep. 
$$
So, without loss of generality (wlog) $K\subseteq Q_R$. Let 
$$J_<:=J_{<,K}:=\{j\in J\colon q_j\subseteq K^\circ\},$$
$$J_\le:=J_{\le,K}:=\{j\in J\colon q_j\cap \bar K\ne\emptyset\},$$
$$J_=:=J_{=,K}:=\{j\in J\colon q_j\cap\p K\ne\emptyset\},$$
where $K^\circ$ is the interior of $K$ and $\bar K$ is the closure of $K$. 
The key to the whole thing is 

Lemma. $|\bigcup_{j\in J_=}q_j|\le2d(d+2)R^{d-1}\de$, where $|\cdot|$ is the Lebesgue measure.  

This lemma will be proved at the end of this answer. Using the absolute continuity of the distribution of $X$, we can take $N$ so large that for any Borel subset $B$ of $\R^d$ we have the implication
$$|B|\le2d(d+2)R^{d-1}\de\implies \mu(B)\le\ep.$$ 
Using now the lemma, for $n\ge n_\ep$ we have 
$$\mu_n(K)-\mu(K)
\le\sum_{j\in J_\le}\mu_n(q_j)-\sum_{j\in J_<}\mu(q_j) \\ 
\le\sum_{j\in J_\le}|\mu_n(q_j)-\mu(q_j)|
+\mu
\Big(\bigcup_{j\in J_=}q_j\Big)\le\De+\ep\le2\ep. 
$$
Similarly, 
$$\mu(K)-\mu_n(K)
\le\sum_{j\in J_\le}\mu(q_j)-\sum_{j\in J_<}\mu_n(q_j) \\ 
\le\sum_{j\in J<}|\mu(q_j)-\mu_n(q_j)|
+\mu
\Big(\bigcup_{j\in J_=}q_j\Big)\le\De+\ep\le2\ep. 
$$
So, $|\mu_n(K)-\mu(K)|\le2\ep$. That is, the desired result is proved modulo the lemma. 

Proof of the lemma. Since $K$ is convex, for any $x\in\p K$ there is some unit vector $\nu(x)$ such that $\nu(x)\cdot(y-x)\le0$ for all $y\in K$ (the support half-space thing), where $\cdot$ denotes the dot product. For each $j\in[d]$, let 
$$S_j^+:=\{x\in\p K\colon\nu(x)_j\ge1/\sqrt d\},\quad 
S_j^-:=\{x\in\p K\colon\nu(x)_j\le-1/\sqrt d\},$$
$$J_{=,j}^+:=\{j\in J\colon q_j\cap S_j^+\ne\emptyset\},\quad 
J_{=,j}^-:=\{j\in J\colon q_j\cap S_j^-\ne\emptyset\},$$
where  $v_j$ is the $j$th coordinate of a vector $v\in\R^d$. 
Note that 
$\bigcup_{j\in[d]}(S_j^+\cup S_j^-)=\p K$ and hence 
$\bigcup_{j\in[d]}(J_{=,j}^+\cup J_{=,j}^-)=J_=$, so that 
$$\Big|\bigcup_{j\in J_=}q_j\Big|
\le
\sum_{j\in[d]}\Big(\Big|\bigcup_{j\in J_{=,j}^+}q_j\Big|+\Big|\bigcup_{j\in J_{=,j}^-}q_j\Big|\Big)
\le\de^d
\sum_{j\in[d]}(|J_{=,j}^+|+|J_{=,j}^-|),\tag{*}
$$
where now $|J_{=,j}^\pm|$ denotes the cardinality of $J_{=,j}^\pm$. 
Now comes the key step in the proof of the lemma: 
Take any $x$ and $y$ in $S_d^+$ such that $x_d\le y_d$. We have the support "inequality" $\nu(x)\cdot(y-x)\le0$, which implies 
$$\frac{y_d-x_d}{\sqrt d}\le\nu(x)_d(y_d-x_d)\le\sum_{j=1}^{d-1}\nu(x)_j(x_j-y_j)
\le|P_{d-1}x-P_{d-1}y|, 
$$
where $P_{d-1}x:=(x_1,\dots,x_{d-1})$. So, we get the crucial Lipschitz condition 
$$|y_d-x_d|\le\sqrt d\,|P_{d-1}x-P_{d-1}y| \tag{**}$$ 
for all $x$ and $y$ in $S_d^+$. 
Partition the left-open $(d-1)$-cube $P_{d-1}Q_R$ naturally into $N^{d-1}$ left-open $(d-1)$-cubes $c_i$ each with edge length $\de=R/N$, where $i\in I:=[N^{d-1}]$. 
For each $i\in I$, let 
$$ J_{=,d,i}^+:=\{j\in J_{=,d}^+\colon P_{d-1}q_j=c_i\},\quad
s_i:=\bigcup_{j\in J_{=,d,i}}q_j,  
$$ 
so that $s_i$ is the "stack" of all the $d$-cubes $q_j$ with $j\in J_{=,d}^+$ that $P_{d-1}$ projects onto the same $(d-1)$-cube $c_i$. Let $r_i$ be the cardinality of the set $J_{=,d,i}$, that is, 
the number of the $d$-cubes $q_j$ in the stack $s_i$. Then for some two points $x$ and $y$ in $s_i\cap S_d^+$ we have $|y_d-x_d|\ge(r_i-2)\de$, whence, in view of (**), 
$$\sqrt d\,\sqrt{d-1}\,\de\ge\sqrt d\,|P_{d-1}x-P_{d-1}y|\ge|y_d-x_d|\ge(r_i-2)\de,$$ 
so that $r_i\le d+2$. So, 
$$|J_{=,d}^+|=\sum_{i\in I}r_i\le\sum_{i\in I}(d+2)=(d+2)N^{d-1}=(d+2)(R/\de)^{d-1}.$$
Similarly, $|J_{=,j}^\pm|\le(d+2)(R/\de)^{d-1}$ for all $j\in[d]$. Now the lemma follows from (*). 
