radially unbounded functions and 1D characterization. Definition
$f: \mathbb{R}^n \rightarrow \mathbb{R}$ is radially unbounded if $||x|| \rightarrow \infty$ implies $f(x) \rightarrow \infty$
Question
Which conditions on $f$ imply that if the one-dimensional function $t \rightarrow f(t d)$  is radially unbounded for any $d \in \mathbb{R}^n$ such that $||d||_2 = 1$, then $f$ is radially unbounded?
I have a counter-example, which is a discontinous $f$. I was not able to find any counterexample for the case when $f$ is continuously differentiable, but I was also unable to prove the statement for this case.
Are there any known theorems on the subject?
 A: $\newcommand{\Reals}{\mathbf{R}}$The polynomial function
$$
f(x, y) = (y - x^{2})^{2}
  = r^{2}(\sin^{2}\theta - 2r\sin\theta \cos^{2}\theta + r^{2} \cos^{4}\theta)
$$
goes to infinity along every line through the origin, but does not go to infinity as $\|(x, y)\| \to \infty$. (Examples of the same type with even more pathological behavior, and in arbitrarily many variables, are easy to construct along the same (ahem) lines.)

If $f$ is continuous, you can define a function $m:[0, \infty) \to S^{n-1}$ by
$$
m(r) = \min_{\|x\| = r} f(x),
$$
and consider the set $V$ (for valley) of points $x$ in $\Reals^{n}$ at which "the minimum is achieved", i.e., $m(\|x\|) = f(x)$. Loosely, if there exists a ray asymptotic to $V$ at $\infty$, then the fact that $f$ is radially unbounded along this ray guarantees that $f \to \infty$ as $\|x\| \to \infty$.
(I haven't thought carefully about how one might define "asymptotically" in a suitable sense for your problem as stated, but compactifying $\Reals^{n}$ to the sphere $S^{n}$ and looking for lines through infinity and tangent to the closure of $V$ looks promising.)
