0
$\begingroup$

I am self-studying "Introduction to nonlinear optimization" by Amir Beck, and after studying chapter two which is called "Optimality Conditions for Unconstrained Optimization" I came across the following problem:

Find a function $f: \mathbb{R}^2 \to \mathbb{R}$ which is not coercive and satisfies that for any $\alpha \in \mathbb{R}$: $$\lim_{|x_1| \to \infty} {f(x_1, \alpha x_1)} = \lim_{|x_2| \to \infty} {f(\alpha x_2, x_2)} = \infty$$

My thought process for solving this problem was:

The function has to be symmetric with respect to variables since when we replace $x_1$ with $x_2$ it acts the same.

Then I said to myself since the limit requires both variables to approach $\infty$ at the same time, maybe I should find a function that goes to $-\infty$ if one of the variables is constant and the other one approaches $\infty$. But then I realized this case is equivalent to setting $\alpha = 0$.

Then I became pretty sure that I should find a function such that when variables approach $\infty$ on a different trajectory than $x_2 = \alpha x_1$ for example on $x_2 = x_1^2$, it goes to $-\infty$. But I have no idea how to find such a function.

Is my thought process correct? If not, what is wrong with it? If it is correct, then what is the next step? I really appreciate it if you can help me.

$\endgroup$

2 Answers 2

1
$\begingroup$

The function has to go to infinity along any ray, but the rate is not specified. Neither do you have any continuity constraints. Thus, you may define a function for every half-ray through the origin (by which is meant a set of the form $$ \{\lambda x : \|x\| = 1\} $$), but with slopes that, though always positive, become arbitrarily small.

$\endgroup$
1
  • $\begingroup$ thanks for your help. $\endgroup$
    – john
    Apr 1, 2023 at 13:20
0
$\begingroup$

as @Cloudscape guided me, I found the function: $$f(x_1, x_2) = \frac{\lvert x_1 \rvert * \lvert x_2 \rvert} {\lvert x_1 \rvert + \lvert x_2 \rvert}$$

which has the mentioned characteristics.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .