Reformulating Equality of Infinima Assume I have two functions both having an infinum $f_i:X_i \to \mathbb{R}$ with $i \in \{1,2\}$.
Does the following hold for $i, \bar{i}\in \{1,2\}$ with $i \neq\bar{i}$? 
$$\inf_{x_1 \in X_1}(f_1(x_1)) = \inf_{x_2 \in X_2}(f_2(x_2)) \iff \forall x_i \in X_i. \exists x_{\bar{i}} \in X_{\bar{i}} \text{ such that } f_i(x_i) \geq f_{\bar{i}}(x_{\bar{i}})$$
As I am not too good in analysis I have no idea how to proceed... I appreciate any suggestions.
 A: The statement is true. By the assumption, there exist $\bar x, \bar y$ such that $f_1(\bar x) = \inf_{x \in X_1}(f_1(x))$  and $f_2(\bar y) =\inf_{y \in X_2}(f_2(y))$.
Suppose we have 
$$
\inf_{x \in X_1}(f_1(x)) = I = \inf_{y \in X_2}(f_2(y)),
$$
then for any $x\in X_1$, we must have $f_1(x)\ge I = f_2(\bar y)$. Similarly, for any $y\in X_2$, we have $f_2(y)\ge I = f_2(\bar x)$.
Conversely, suppose that 
$$
\forall x_i \in X_i. \exists x_{\bar{i}} \in X_{\bar{i}} \text{ such that } f_i(x_i) \geq f_{\bar{i}}(x_{\bar{i}}),
$$
if we take $x_1= \bar x$, then there exists $x_2\in X_2$ such that
$$
\inf_{x \in X_1}(f_1(x)) = f_1(\bar x) \ge f_2(x_2)
$$
so $\inf_{x \in X_1}(f_1(x))  \ge \inf_{y \in X_2}(f_2(y))$. We do the same for the other direction (i.e. take $x_2 = \bar y$) to get $\inf_{x \in X_1}(f_1(x)) \le \inf_{y \in X_2}(f_2(y))$. This shows that 
$$
\inf_{x \in X_1}(f_1(x)) = \inf_{y \in X_2}(f_2(y))
$$
as required.

Remark: The statement is false if we don't assume that the minima of both $f_1,f_2$ are attained.
Indeed, we can take $X_1=[0,1], X_2 = (0,\infty)$ and consider $f_1:[0,1]\to\Bbb R$ and $f_2:(0,\infty)\to\Bbb R$ defined by 
$$
f_1(x)=x,\quad f_2(x') = \frac 1{x'}.
$$
It is not hard to see that 
$$
\inf_{x \in [0,1]}f_1(x) = 0 = \inf_{x' \in (0,\infty)}f_2(x').
$$
However, for $x=0\in X_1$ we have
$$
f_1(0) = 0
$$
but there is no $x'\in X_2$ such that $0\ge f_2(x')$.
