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.
On
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 $\lim_{t\to\infty} f(th_{0}) = \infty$.
$A$ 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 $\displaystyle\lim_{t\to\infty}f(tv)=\infty$.