Let a nonnegative function $f(x,y)$: $\mathbb R^2\to \mathbb R$ be second order continuous differentiable. We also know that $f$ is not convex in its two arguments, but only separately in each of them.
It is clear that $$ \inf_{(x,y)\in\mathbb R^2} f(x,y)\geq0. $$ My question: do we have $$ \inf_{y\in\mathbb R}\inf_{x\in\mathbb R}f(x,y)=\inf_{x\in\mathbb R}\inf_{y\in\mathbb R}f(x,y) $$ hold?
Let $\mathcal{X}$ and $\mathcal{Y}$ be abstract sets and consider any real-valued function $f:\mathcal{X}\times \mathcal{Y}\rightarrow\mathbb{R}$. For any $(x,y) \in \mathcal{X}\times \mathcal{Y}$ we get: \begin{align} f(x,y) &\geq \inf_{a \in \mathcal{X}} f(a,y) \\ &\geq \inf_{b \in \mathcal{Y}}\left[ \inf_{a \in \mathcal{X}} f(a,b)\right] \end{align} Thus, taking the $\inf$ of both sides over $(x,y) \in \mathcal{X}\times \mathcal{Y}$ gives: $$ \inf_{(x,y) \in\mathcal{X}\times \mathcal{Y}} f(x,y) \geq \inf_{b \in \mathcal{Y}}\left[ \inf_{a \in \mathcal{X}} f(a,b)\right] \quad (Eq 1)$$ On the other hand, a similar argument also shows: $$ \inf_{(x,y) \in \mathcal{X}\times \mathcal{Y}} f(x,y) \leq \inf_{b \in \mathcal{Y}}\left[\inf_{a \in \mathcal{X}} f(a,b)\right] \quad (Eq 2) $$ Combining (Eq 1) and (Eq 2) gives: $$ \inf_{b \in \mathcal{Y}} \left[\inf_{a \in \mathcal{X}} f(a,b)\right] = \inf_{(x,y) \in\mathcal{X}\times\mathcal{Y}} f(x,y) $$ The same argument holds for the reversed order of infs, so: $$ \boxed{ \inf_{a \in \mathcal{X}}\left[\inf_{b \in \mathcal{Y}} f(a,b)\right] = \inf_{b \in \mathcal{Y}} \left[\inf_{a \in \mathcal{X}} f(a,b)\right] = \inf_{(x,y) \in\mathcal{X}\times\mathcal{Y}} f(x,y) } $$
Things get more interesting if we use $\inf$ and $\sup$: A similar argument gives $$ \sup_{y \in \mathcal{Y}} \left[ \inf_{x \in \mathcal{X}} f(x,y)\right] \leq \inf_{x \in \mathcal{X}} \left[\sup_{y \in \mathcal{Y}} f(x,y)\right] $$ This is weak duality. The interesting thing is that the reverse inequality does not necessarily hold. When it holds, it is called strong duality.