Let $\Omega\subset \mathbb R^N$ open bounded, smooth boundary be given. Define $$ F(u,v):=\int_\Omega |\nabla u|^2v^2dx+\int_\Omega(|\nabla v|^2+(1-v)^2)dx, $$ and two sets $\mathcal U:=\{u\in H^1(\Omega),\,T[u]=T[w]\}$, where $w\in H^1(\Omega)$ is a fixed function. Also, $\mathcal V:=\{v\in H^1(\Omega),\,\,0\leq v\leq 1\}$.
My question: does it hold that ? $$ \operatorname{argmin}_{v\in\mathcal V}\operatorname{argmin}_{u\in\mathcal U} F(u,v) = \operatorname{argmin}_{u\in\mathcal U} \operatorname{argmin}_{v\in\mathcal V}F(u,v) $$ Clearly the $\inf$ exists on the both hand side, but I am not sure whether they are equal...
If one of them exists, the other does, too, and both are equal. This follows by using just general properties of infima. Both sides equal $$ \inf_{u,v} F(u,v) $$ since clearly, given $u$, we have $$ \inf_{u,v} F(u,v) \le \inf_v F(u,v) $$ so, $$ \inf_{u,v} F(u,v) \le \inf_u \inf_v F(u,v) $$ On the other side, given $\epsilon > 0$, choose $u',v'$ with $F(u',v') \le \inf_{u,v} F(u,v) + \epsilon$. Then $$ \inf_v F(u',v) \le F(u',v') $$ and hence $$ \inf_u \inf_v F(u,v) \le F(u',v') \le \inf_{u,v} F(u,v) + \epsilon $$ As $\epsilon$ was arbitrary, $$ \inf_{u,v} F(u,v) = \inf_u \inf_v F(u,v)$$ As the left hand side is symmetric in $u,v$, we have analogously, $$ \inf_{u,v} F(u,v) = \inf_v \inf_u F(u,v)$$