# Continuity of $\arg\min$

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?

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

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).

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