Lower semi-continuity of the Hellinger-Fisher-Rao distance

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.

• I think the answer to your second question is affirmative: $H$ is convex and continuous on $L^1(\Omega, \mathrm{d}x)^2$, hence sequentially weakly lower semicontinuous. This doesn't help with the weak-$*$ part, though. – gerw Nov 24 '15 at 20:37
• Yes that works, thank you gew. Unfortunately I realized since I first posted that I really the need weak-* part. I think I have a proof, but it's much more involved and uses a dynamic representation of H through some Riemannian structure hidden in there. I'll triple check and post it if correct. – leo monsaingeon Nov 25 '15 at 0:47

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.