Let $f$ be a smooth probability distribution on the unit square $S$ such that $f(x)>0$ on $S$. Let $\{g_i\}$ be a sequence of smooth probability distributions such that $g_i(x)>0$ on $S$ as well and such that the Wasserstein distance $W_1(f,g_i)$ approaches zero. Is it true that $$\iint_S \sqrt{f(x)} dx \geq \limsup_{i\to\infty} \iint_S \sqrt{g_i(x)}dx$$ assuming both quantities exist? The other direction of the inequality is definitely not true because we could construct $\{g_i\}$ that is almost discrete, which would have very tiny values of $\iint_S \sqrt{g_i(x)}dx$.

## 1 Answer

Yes. Also, both quantities are guaranteed to exist: the left hand side because $f$ is smooth and bounded, and the right hand side because a lim sup always exists, and it's finite because the LHS is finite and the inequality is true.

Here is a sketch for the proof of (a slightly generalized version of) the inequality: suppose $f: \mathbb{R}^d \to \mathbb{R}$ is a uniformly continuous density and let $\delta$ be its modulus of continuity. For $\epsilon > 0$, let $$ \eta_\epsilon(x) = \frac{1}{|B_\epsilon|} \chi_{B_\epsilon}(x), $$ where $B_\epsilon$ is the Euclidean ball of radius $\epsilon$ in $\mathbb{R}^d$.

**Lemma**
If $\|f - \eta_\epsilon * g\|_\infty \ge 4 \delta(\epsilon)$ then
$W_1(f, g) \ge C_d \delta(\epsilon)^2 \epsilon^{d+1}$, where $C_d$ is a constant depending only on $d$.

**Proof** (sketch): Choose a point $x$ such that $|f(x) - (\eta_\epsilon * g)(x)| \ge 3 \delta(\epsilon)$. Suppose that $f(x) \ge (\eta_\epsilon * g)(x) + 3 \delta$ (the other direction is similar). Consider a ball $B$ of radius $\epsilon$
around $x$: by the definition of $\delta$, $f(y) \ge f(x) - \delta \ge (\eta_\epsilon * g)(x) + 2 \delta$ for all $y \in B$. On the other hand, $(\eta_\epsilon * g)(x)$ is the average value of $g$ on $B$. Using this, you can find $\epsilon' < \epsilon$ so that $f$ has $\delta$ more mass on $B(x, \epsilon')$ than $g$ has on $B(x, \epsilon)$. To move $g$ onto $f$, you will have to move at least $\delta$ mass from outside of $B(x, \epsilon)$ into $B(x, \epsilon')$, which costs at least $\delta (\epsilon - \epsilon')$ in $W_1$ distance. Then you need to compute how small $\epsilon'$ can be: you'll get
$\epsilon - \epsilon' \asymp \delta \epsilon^{d+1}$.

Since $W_1(f, g_i) \to 0$, the lemma implies that $$ \lim_{\epsilon \to 0} \lim_{i \to \infty} \|f - \eta_\epsilon * g_i\|_\infty \to 0. $$ Since $f$ and $\eta_\epsilon * g_i$ are both bounded, for every continuous $\phi: \mathbb{R} \to \mathbb{R}$ we have $$ \lim_{\epsilon \to 0} \lim_{i \to \infty} \int \phi(\eta_\epsilon * g) = \int \phi(f). $$ Finally, if $\phi$ is a convex function (like the square root) then convolution with $\eta_\epsilon$ is an averaging operation and so Jensen's inequality implies $\phi(\eta_\epsilon * g) \ge \eta_\epsilon * \phi(g)$. So $$ \int \phi(\eta_\epsilon * g_i) \ge \int \eta_\epsilon * \phi(g_i) = \int \phi(g_i) $$ for every $\epsilon$ and every $i$.