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Let $f(\cdot,\cdot):\Omega\to\mathbb{R}$ be a continuous function with $\Omega$ a compact subset of $\mathbb{R}^n\times\mathbb{R}^m$.

From these assumptions, $f(x,\cdot)$ must possess a maximum for each $x\in\mathcal{X}$, where $\mathcal{X} = \{x\in\mathbb{R}^n: (x,y)\in\Omega \text{ for some } y\in\mathbb{R}^m\}$.

In general, however, $f(x,\cdot)$ may contain multiple global maximizers.

My question is -- are there any (mild) conditions that we can impose on $f$ to ensure that a function $F:\mathcal{X}\to\mathcal{Y}$ $$F(x) \in \operatorname{argmax} f(x,\cdot),$$ where $\mathcal{Y} = \{y\in\mathbb{R}^m: (x,y)\in\Omega \text{ for some } x\in\mathbb{R}^n\}$, is continuous?

In other words, $F(\cdot)$ is such that $$f(x,F(x)) = \max_{y\in\Omega_x} f(x,y),$$ where $\Omega_x = \{y\in\mathbb{R}^m: (x,y)\in\Omega\}$.

And I mean other than the "trivial" case of $f(x,\cdot)$ having exactly one global maximizer for each $x$ and thus $F(x)$ being uniquely defined.

EDIT: Follow-up question: Under the conditions that I stated (or other similar mild ones), where $F$ is non-unique, must one of them be continuous?

EDIT 2: Scratch the previous follow-up question. I guess the better question is: are there any mild conditions on $f$ (and/or $\Omega$) that ensure that at least one particular $F$ (when the argmax may not be uniquely defined) must be continuous?

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    $\begingroup$ Consider $f(x,y)=\sin x+yx$ on $[0,3\pi]\times[-1,1]$. A function cannot be much nicer than this. $\endgroup$
    – Hagen von Eitzen
    Sep 3, 2018 at 6:32
  • $\begingroup$ I'm not sure I understand what you mean. For that particular function, $f(x,\cdot)$ has a unique maximizer $y=1$. Therefore $F(x) = 1$, which is obviously continuous. What I'm wondering is if there are conditions to guarantee continuity of argmax, especially when it is not uniquely defined. $\endgroup$
    – Orlando
    Sep 3, 2018 at 6:40
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    $\begingroup$ There is a large collection of theorems called 'selection theorems'. There are also books on this topic. Perhaps you will find some answers by searching for them. $\endgroup$
    – Kavi Rama Murthy
    Sep 3, 2018 at 7:21
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    $\begingroup$ If there are mild conditions, it probably won't be easy to identify them. The maximum theorem only guarantees upper hemicontinuity of the $\text{argmax}$ correspondence, but the selection theorems I know of require lower hemicontinuity. $\endgroup$
    – Theoretical Economist
    Sep 4, 2018 at 14:11
  • $\begingroup$ Thank you! It's not what I would have preferred, but it gives me a place to start. $\endgroup$
    – Orlando
    Sep 4, 2018 at 17:13

1 Answer 1

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No. Let $\Omega=[-1,1]\times[-1,1]$ and $f$ be given by $f(x,y)=x\cdot y$.

Let $F:[-1, 1]\to [-1,1]$ be any function such that $F(x)\in\arg\max_{[-1,1]}f(x,\cdot)$. Then we must have $F(x)=-1$ for $x<0$ and $F(x)=1$ for $x>1$. So $F$ must have a discontinuity at $0$.

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  • $\begingroup$ Right, I know of that example. I wasn't asking if $F$ must always be continuous, which I definitely know isn't true. I was asking if there are any mild conditions (i.e. conditions that are not ridiculously specific) that guarantee F to be continuous. I'm especially interested in conditions that would allow me to guarantee $F$ to be continuous without demanding argmax to be uniquely defined (say by requiring $f(x,\cdot)$ to be strictly concave). $\endgroup$
    – Orlando
    Sep 3, 2018 at 14:49
  • $\begingroup$ How does this example not answer the question "Under the conditions that I stated (or other similar mild ones), where F is non-unique, must one of them be continuous?" $\endgroup$
    – Michael Greinecker
    Sep 3, 2018 at 14:56
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    $\begingroup$ Fair enough -- it does answer the follow-up question, but not the main question itself. Which is, again, under what conditions can we guarantee $F$ to be continuous? $\endgroup$
    – Orlando
    Sep 3, 2018 at 15:46

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