Let $$(M,g,m_0)$$ be a pointed-Hadamard manifold with Riemmanian distance function $$d_g$$, $$(X,\Sigma,\mu)$$ be a finite measure space. We use $$L^2(\mu;M,m_0)$$ to denote the metric space consisting of all equivalence classes of measurable functions $$g:X\rightarrow M$$ with finite the distance $$D(m_0,g)$$ from the constant function $$m_0$$ where $$D$$ is the metric: $$D(g,h):=\int_{m \in M} d_g^2(g(m),h(m))d\mu(m) = \int_{m \in M} \|\operatorname{exp}_{g(m)}^{-1}(h(m))\|^2d\mu(m).$$
Fix some non-constant $$f \in L^2(\mu;M,m_0)$$.
Let $$c\in M$$ and let $$C\subseteq L^2(\mu;M,m_0)$$ be proper and non-empty. Under what joint conditions on $$C$$ and $$c$$ can we guarantee that $$\operatorname{argmin}_{g \in C} D(f,g) = \operatorname{argmin}_{g \in C} \int_{m \in M} \|\exp_{c}^{-1}(f(m))-\operatorname{exp}_{c}^{-1}(g(m))\|^2d\mu(m) .$$
• What do you mean by $\exp_c$? Also, is $g$ supposed to be $h$? Jan 4, 2021 at 17:56
• @MoisheKohan I mean the Riemannian exponential at $c$''s inverse function (which is globally defined) Jan 4, 2021 at 18:44
• @Topology_Catologist unclear: $c$ is s not in $M$. Jan 4, 2021 at 22:35
• @MoisheKohan Thanks for noticing the misprint. Indeed $c\in M$ is correct. Jan 5, 2021 at 8:31