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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?

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  • $\begingroup$ What are your conditions ensuring the existence of at least one minimum? The answer may depend on this. $\endgroup$ Jun 21, 2019 at 7:48
  • $\begingroup$ Also, have you looked for "selection theorems"? Maybe there is one you can use. $\endgroup$ Jun 21, 2019 at 7:48

1 Answer 1

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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).

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  • $\begingroup$ I hate it (or love it?) if there are quick and easy counterexamples... $\endgroup$ Jun 21, 2019 at 8:16
  • $\begingroup$ I love quick and easy counterexamples! $\endgroup$
    – Dirk
    Jun 21, 2019 at 8:46

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