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Let $f : \mathbf R^n \to \mathbf R$ be a convex function with $f(0)=0$ and $\displaystyle\lim_{t\to\infty}f(tx)=\infty$ for any $x\in\mathbf R-\{0\}$. Is $$A:=\{x \mid f(x)\le 1 \}$$ bounded?

I was thinking about projecting $A$ onto the unit spherical surface, but did not know how to proceed.

Hans
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    From Rockafellar's "Convex Analysis", Theorem 8.4: "A non-empty closed convex set $C$ in $\mathbb{R}^n$ is bounded if and only if its recession cone $0^+C$ consists of the zero vector alone". – copper.hat Aug 19 '18 at 20:17
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    @copper.hat: Thank you for pointing out this is a classical result. – Hans Aug 19 '18 at 20:40

2 Answers2

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We already know that $A$ is convex as the sub-level sets of a convex function are convex. If $A$ is unbounded, then for every $x\in A$, there is some nonzero vector $h_{x}$ such that $x+th_{x}\in A$ for all $t\geq 0$. Since $f(0)=0$, we can find $v:=h_{0}$ so $th_{0}\in A$ for all $t\geq 0$. Thus, $f(th_{0})\leq 1$ for all $t\geq 0$. This contradicts that $\lim_{t\to\infty} f(th_{0}) = \infty$.

Wraith1995
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  • Just a comment, as shown in the link, it's important that $A$ is closed. –  Aug 19 '18 at 18:18
  • @Balgani: It is sufficient in this case to have $0 \in \operatorname{ri} A$, and since $0 \in A^\circ$ that is satisfied. – copper.hat Aug 19 '18 at 20:13
  • +1. Nice answer. I hope you do not mind me accepting mathcounterexample.net's answer as the proof of his for the existence of ray is simpler than the accepted answer of the linked question. – Hans Aug 19 '18 at 20:47
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$A$ is indeed bounded

We can prove it by contradiction, using the result that for finite dimensional real vector spaces an unbounded convex subset contains a ray. See my website for the proof.

$A$ being a level set of $f$ is convex and not empty as $0 \in A$. $0$ is even an interior point of $A$ as $f$ is continuous. According to the result above, $A$ contains a ray $[0, \infty v )$ with $v \in A$. By definition of $A$, the value of $f$ on this ray is less or equal to $1$. In contradiction with the hypothesis $\displaystyle\lim_{t\to\infty}f(tv)=\infty$.

mathcounterexamples.net
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  • How does this differ from the answer below? – copper.hat Aug 19 '18 at 19:59
  • @copper.hat: I suppose the difference is that the proof here of "an unbounded closed convex subset of a finite dimensional real vector space contains a ray" is simpler than the accepted answer in the linked question in the answer of Wraith1995. – Hans Aug 19 '18 at 20:44