Finding minimizer from different order 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?
 A: 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.
