I am currently working on unbalanced optimal transport, where the Hellinger (or sometimes Fisher-Rao) distance
$$
H^2(\rho,\mu)=\int\_{\Omega}\left|\sqrt{\frac{d\rho}{d\lambda}}-\sqrt{\frac{d\mu}{d\lambda}}\right|^2 d\lambda
$$
shows up. Here $\Omega\subset R^d$ is a (possibly unbounded) smooth domain, $\rho,\mu$ are nonnegative Borel measures with finite mass on $\Omega$, $\lambda$ is any reference (nonnegative, Borel) measure on $\Omega$ such that $\rho,\mu$ are simultaneously absolutely continuous with respect to $\lambda$, and $\frac{d\rho}{d\lambda},\frac{d\mu}{d\lambda}$ denotes the corresponding Radon-Nikodym derivatives. Note that by 1-homogeneity this definition does not depend on the choice of the reference measure $\lambda$. This distance is ususally defined for probability measures (i-e with fixed unit mass $|\rho|=|\mu|=1$) but still gives a distance on the set $\mathcal M^+(\Omega)$ of nonnegative Borel measures with finite mass (thus possibly $|\rho|\neq |\mu|$).



> 
> **Question**: is it true that $H$ is lower semi-continuous with respect to the weak-$\ast$ convergence of measures,
>  $$
> \rho^n\overset{\ast}{\rightharpoonup}\rho,\mu^n\overset{\ast}{\rightharpoonup}\mu
> \quad\Rightarrow\quad
> H(\rho,\mu)\leq \liminf \,H(\rho^n,\mu^n)???
> $$
> 
> 
> 


Let me remind that the weak-$\ast$ convergence of measures is defined by duality with continuous compactly supported functions $\phi\in \mathcal C^\infty\_c(\Omega)$. Also, I would be happy if this were true for the (stronger) weak $L^1(\Omega,dx)$ convergence for absoluetly continuous measures (with respect to, say, Lebesgue's measure $dx$) instead of weak-$\ast$ convergence of measures.



Lower semi-continuity of the Hellinger-Fisher-Rao distance
The lower semicontinuity of the Hellinger-Rao distance follows from the following more general lower semicontinuity result (see e.g. Buttazzo, *Semicontinuity, relaxation and integral representation in the calculus of variations*, Thm. 3.4.1):


If $f\colon \mathbb{R}^n\to [0,\infty]$ is a proper lower semicontinuous convex positively $1$-homogeneous function, then the functional
$$
F\colon \mathcal{M}(\Omega)^n\to [0,\infty],\,\mu\mapsto \int\_\Omega f\left(\frac{d\mu^1}{d\lambda},\dots,\frac{d\mu^n}{d\lambda}\right)\,d\lambda
$$
is sequentially weak\* lower semicontinuous. As in the question, $\lambda$ is any scalar measure on $\Omega$ such that $\mu^1,\dots,\mu^n\ll\lambda$.


Let me sketch the proof. First note that $f$ is the supremum of a sequence of affine-linear functions $f\_i$. By monotone convergence,
$$
\int\_\Omega \sup\_{1\leq i\leq N}f\_i^+\left(\frac{d\mu}{d\lambda}\right)\,d\lambda\to F(\mu),\;N\to \infty.
$$
For each $N\in\mathbb{N}$ we can decompose $\Omega$ into disjoint subsets $E\_j$ such that $f\_j^+=\sup\_{1\leq i\leq N}f\_i^+$. Thus
$$
F(\mu)=\sup\left\lbrace\sum\_{i\in I}\int\_{E\_i}f\_i^+\left(\frac{d\mu}{d\lambda}\right)\,d\lambda\,\Bigg|\, I\subset \mathbb{N}\text{ finite}, E\_i\cap E\_j=\emptyset\text{ for }i\neq j\right\rbrace.
$$
Using the regularity of the measures, one can restrict to disjoint open sets $E\_i$ in the supremum above. For $U\subset \Omega$ open one has
$$
\int\_U f\_i^+\left(\frac{d\mu}{d\lambda}\right)\,d\lambda=\sup\left\lbrace \int\_\Omega f\_i\left(\frac{d\mu}{d\lambda}\right)\phi\,d\lambda\,\Bigg|\,\phi\in C\_c^\infty(U),\,0\leq \phi\leq 1\right\rbrace.
$$
Since $f\_i$ is affine-linear, the functional in the supremum on the right side is sequentially weak\* continuous. Therefore $F$ is sequentially weak\* lower semicontinuous as supremum of sequentially weak\* continuous maps.


Remark: I treated the reference measure as fixed. This is justified since for a sequence $(\mu\_k)$ one can always take something like
$$
\lambda=\sum\_{k=1}^\infty\sum\_{l=1}^n |\mu\_k^l|.
$$
For this reason the argument only works for *sequential* semicontinuity.


