Let $x\_1,\dots,x\_n\in X$ some Polish space $X$ and let $\Delta$ be the probability simplex in $\mathbb{R}^n$. Consider the map sending every $(w\_1,\dots,w\_n)\in\Delta$ to the finitely supported measure $\sum\_{k=1}^n w\_k\delta\_{x\_k}$. This map is clearly continuous with respect to the Wasserstein distance, but is it also Lipschitz?



Modulus of continuity of parameterizing Wasserstein
$\newcommand\De\Delta\newcommand\de\delta$Yes, this map is Lipschitz. Indeed, the map is
\begin{equation\*}
 \De\ni w=(w\_1,\dots,w\_n)\mapsto\mu\_w:=\sum\_{k=1}^n w\_k\de\_{x\_k}. \tag{1}
\end{equation\*}
Let $d$ denote the metric on $X$, and then let
\begin{equation\*}
 D:=\max\_{i,j\in[n]}d(x\_i,x\_j),
\end{equation\*}
the diameter of the set $\{x\_1,\dots,x\_n\}$, where $[n]:=\{1,\dots,n\}$.


Take any $v=(v\_1,\dots,v\_n)$ and $w=(w\_1,\dots,w\_n)$ in $\De$ and let
\begin{equation\*}
 h:=\max\_{j\in[n]}|v\_j-w\_j|.
\end{equation\*}
Consider the following $(n-1)$-step transportation plan of transporting the probability measure $\mu\_v$ to $\mu\_w$ (or vice versa).


For the first step of the plan, transport $\mu\_v$ to $\mu\_{v^{(1)}}$, where
\begin{equation\*}
 v^{(1)}:=\Big(w\_1,v\_2+\frac{v\_1-w\_1}{n-1},\dots,v\_n+\frac{v\_n-w\_n}{n-1}\Big),
\end{equation\*}
assuming without loss of generality that $v\_1\ge w\_1$. Then (i) the first coordinate of $v^{(1)}$ is the same as that of $w$, (ii) $v^{(1)}\in\De$, (iii)
\begin{equation\*}
 \max\_{2\le j\le n}|v^{(1)}\_j-w\_j|\le\max\_{2\le j\le n}|v\_j-w\_j|+\frac{v\_1-w\_1}{n-1}\le h+\frac h{n-1}=\frac n{n-1}\,h, 
\end{equation\*}
and (iv)
\begin{equation\*}
 W(\mu\_v,\mu\_{v^{(1)}})\le\sum\_{j=2}^n \frac{v\_1-w\_1}{n-1}\,d(x\_1,x\_j)
 \le(v\_1-w\_1)D\le hD,
\end{equation\*}
where $W$ denotes the Wasserstein distance.


In the remaining $n-2$ steps of the plan, similarly and consecutively equalizing the remaining $n-1$ coordinates of the initially given vectors $v=(v\_1,\dots,v\_n)$ and $w=(w\_1,\dots,w\_n)$, we see that
\begin{equation\*}
 W(\mu\_v,\mu\_w)\le hD+\frac n{n-1}\,hD+\frac n{n-2}\,hD+\cdots+\frac n1\,hD
 \le Lh,
\end{equation\*}
where $L:=n(1+\ln n)D$. So, the map (1) is Lipschitz.


