Continuity of $\arg\min$

Question

Let $f:\mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}$ be a continuous function.

Let $A = \{(x,\arg\min_x f(x,y)) | x \in \mathbb{R}^n\}$.

Is there necessarily a continuous function $g: \mathbb{R}^n \to \mathbb{R}^m$ that satisfies: $(x,g(x))\in A$ for all $x$?

What about taking $A = \{(x,y) | x \in \mathbb{R}^n, f(x,y) \leq \min_x f(x,y)+\epsilon\}$ instead?

Answer 1

The answer is no (also for the question with $\epsilon$).

Here is a counterexample:

Consider $m=n=1$ and $$ f(x,y) = (x^2-1)^2 + yx. $$

• For $y>0$ there is a unique minimizer to $\min_x f(x,y)$ which is smaller than $-1$.
• For $y<0$ there is a unique minimizer to $\min_x f(x,y)$ which is larger than $1$.
• For $y=0$ there are two minimizers of $\min_x f(x,y)$ which are $1$ and $-1$.

Minimizing up to $\epsilon$ does not help…

I may add that even convexity of $f$ does no help (consider the convex envelope of the $f$ above).