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Let $f: A \to \mathbb{R}$ be a function and $A\subset \mathbb{R}^d$. It looks like I can ensure existence and uniqueness of a maximizer of $f$ if I let $f$ be strictly concave and upper semicontinuous with bounded super-level sets. I wonder why this is true and moreover what conditions should I place on $A$. I would like to avoid $A$ being bounded. I wonder if I can let $A$ be an open or closed rectangle.

user12703198
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  • It is not true, let $A=(0,1)$ and $f(x) = x$, then $f$ has no maximiser. – copper.hat Jul 29 '23 at 18:28
  • Thanks. What if $A$ is closed e.g. $A=[3,+\infty)$? Is the statement true if I let $A$ be a closed rectangle? Can I let $A$ be a closed convex set? – user12703198 Jul 29 '23 at 18:51
  • No, try $f(x) = 1-e^{-x}$ on $x \ge 0$. You need to try some simple examples yourself. – copper.hat Jul 29 '23 at 18:56
  • Does the logarithm $\log : \left[ 3, +\infty \right) \to \mathbb{R}$ which is strictly concave on a closed connected set have a maximizer? Edit: Oh, my example was unbounded, but copper.hat's works. – Jeppe Stig Nielsen Jul 29 '23 at 18:56
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    Neither $f(x)=1-e^{-x}$ has a maximizer on $[0,+\infty)$ nor the $\log$ function on $[3,+\infty)$ but they do not have bounded superlevel sets. – user12703198 Jul 29 '23 at 19:02
  • You are right. Wikipedia states: By Weierstrass's theorem, the boundness of some non-empty sublevel set and the lower-semicontinuity of the function implies that a function attains its minimum. – Jeppe Stig Nielsen Jul 29 '23 at 19:10
  • Thank you. I believe that is referred to functions defined on the whole $\mathbb{R}^d$. But if a function is defined only on a subset $A$ of $\mathbb{R}^d$, what conditions should I place on $A$? – user12703198 Jul 29 '23 at 19:16
  • It should be enough that $A$ is closed, I believe. If you can pick just one nonempty superlevel set which is bounded, then it is clear that the maximum cannot occur outside the closure (which will be compact and inside $A$) of that set? – Jeppe Stig Nielsen Jul 29 '23 at 19:20
  • ($A$ would automatically be a convex set, otherwise it will not make sense to assume $f$ is strictly concave.) – Jeppe Stig Nielsen Jul 29 '23 at 19:27
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    Thank you. To my understanding, we need to exploit that the superlevel set is also convex because $f$ is concave (to ensure uniqueness of the maximizer by strict concavity) and that the superlevel set is closed because $f$ is upper semicontinuous. So probably, as you suggest, we only need to let $A$ be closed and convex. – user12703198 Jul 29 '23 at 19:32
  • Sorry, I misread, I missed the super-level sets part. If a super level set is non empty and bounded then you are essentially dealing with the compact case. – copper.hat Jul 29 '23 at 19:42

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