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.