Are $\min_{x,y} f(x,y)$ and $\min_{x} \min_{y} f(x,y)$ equivalent?

Question

Are the following two equations equivalent?

The first equation is $$ \min_{x,y} f(x,y). $$

The second equation is $$ \min_x g(x) \quad \text{where} \quad g(x) = \min_y f(x,y). $$ The second equation can be considered as $$\min_x \left(\min_y f(x,y)\right).$$

I tried to find some counter example, but I failed. If you know the answer, I would really appreciate it if you share it, and I would be even more grateful if you provide counterexamples. If the two are equivalent, please let me know and share any relevant sources or references related to the proof.

Answer 1

The issue is really about the existence of sections minima, if the function is well behaved then you can use the answer from robjohn in this post:

When $\min_{x \in X,y \in Y} f(x,y) = \min_{x \in X} \min_{y \in Y} f(x,y)$?

But a global minimum may exist while for sections only an infimum may exist, except maybe for one section where the min is reached.

E.g. $f(x,y)=|x|e^{-y^2}$ only the $x=0$ section reaches the minimum.

Answer 2

Let $f$ be a real function with eah argument $x$ and $y$ is an element of $X$ and $Y$, respectively ; $f:X\times Y\to\mathbb R$

For each $x\in X$, there exists $y_x\in Y$ such that $\min_yf(x,y)=f(x,y_x)=g(x)$. And there exists $x_0\in X$ such that $\min_xg(x)=g(x_0)$. So we have $$\min_x\min_yf(x,y)=f(x_0,y_{x_0}).$$ That $$\min_{x,y}f(x,y)\le \min_x\min_yf(x,y)$$ is then trivial since $$\min_{x,y}f(x,y)\le f(x',y')$$ for every $x'\in X$ and $y'\in Y$.

To prove the reverse inequality, let $x'\in X$ and $y'\in Y$. Then,

$$f(x',y')\ge\min_{y}f(x',y)\ge\min_x\min_yf(x,y).$$ Taking mininum for all $x'$ and $y'$ in $A$, we get $$\min_{x,y}f(x,y)\ge\min_x\min_yf(x,y).$$