0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ What are your conditions ensuring the existence of at least one minimum? The answer may depend on this. $\endgroup$ – András Bátkai Jun 21 at 7:48
  • $\begingroup$ Also, have you looked for "selection theorems"? Maybe there is one you can use. $\endgroup$ – András Bátkai Jun 21 at 7:48
4
$\begingroup$

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).

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.