6

Definition

$f : \mathbb{R}^n \to \mathbb{R}$ is radially unbounded if $\|x\| \to \infty$ implies $f(x) \to \infty$

Question

Which conditions on $f$ imply that if the one-dimensional function $t \to f(t d)$ 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?

Rodrigo de Azevedo
  • 23,788
Alex Shtoff
  • 208
  • Is there more context for this? I think some more information might get you an answer. – 3-in-441 Jan 08 '17 at 22:50
  • No additional context. This is what I am trying to prove. – Alex Shtoff Jan 10 '17 at 08:21
  • @Alex: Your two definitions are equivalent, it's just that you seem not to see this. In fact, what you call definition is NOT a definition, while the statement with $t$ and $d$ is just the explanation of the notation $\lim \limits _{| x | \to \infty} f(x)$ (which, in itself, does not mean anything, even if you believe the opposite). This problem is a false problem, it is confusion caused by the improper use of some notations. – Alex M. Jan 11 '17 at 13:37

1 Answers1

3

$\newcommand{\Reals}{\mathbf{R}}$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.)

Andrew D. Hwang
  • 82,708
  • If I understand correctly your idea, you seem to use the notation $| (x,y) | \to \infty$ as a synonym for $(x,y) \to \infty$ (i.e. "the limit at $\infty$"). This allows you to consider the limit of $f$ along $y=x^2$, which is $0$. Are you sure that this is the OP's intended use of the notation $| (x,y) | \to \infty$? To me, the fact that he uses the adjective "radially" when writing $| x | \to \infty$ suggests that he uses this notation to indicate convergence along lines, without realizing it (read his "definition" again). This also explains my comment below the question. – Alex M. Jan 11 '17 at 22:56
  • @AlexM.: As I read it, the OP has two conditions, "radially unbounded" (i.e., $f \to \infty$ as $|x| \to \infty$), and "radially unbounded along every ray" (i.e., for every unit vector $d$, $f(td) \to \infty$ as $t \to \infty$), and asks whether they are equivalent for $C^{1}$ functions. The polynomial function $f(x, y) = (y - x^{2})^{2}$ satisfies the second condition (for each $\theta$, the right-hand expression goes to $\infty$ as $r \to \infty$) but not the first (because $f$ vanishes on the parabola). [...] – Andrew D. Hwang Jan 12 '17 at 00:15
  • I see your point that the term "radially unbounded" may be inapt, but the mathematical intent seems clear. Presumably OP will clarify as needed when they return. :) – Andrew D. Hwang Jan 12 '17 at 00:16
  • AFAIK, notation $||x|| \rightarrow \infty ~\Rightarrow~ f(x) \rightarrow \infty$ which is used in the literature means that for every sequence ${x_n}{n=1}^{\infty}$ which satisfies $\lim{n \rightarrow \infty} ||x_n|| = \infty$ we have $\lim_{n \rightarrow \infty} f(x_n) = \infty$. @AndrewD.Hwang indeed provided a counter-example for the conjecture, since the sequence can go along any path, and not only rays. – Alex Shtoff Jan 12 '17 at 07:35