2
$\begingroup$

I'm interested in finding $\arg\min_{x \in X} (f(x) + \lVert x\rVert_2^2)$ where $X$ is a $[0,1]^n$, $f$ is Lipschitz but non-convex and we already have a procedure to find some $x^* \in \arg\min_{x\in X^*} \lVert x\rVert_2^2$, where $X^* = \arg\min_{x \in X}(f(x))$, that is we can find minimal values of $f$, and specifically one which minimizes the norm of the argument in the set that minimizes $f$.

$f(x)$ is some polynomial in $x_i$ which is linear with respect to each $x_i$ and with coefficients either $1$ or $-1$, for example $x_1 x_2 - x_1 x_3$

Using prox-linear algorithms I can find local minima of the sum, is there any way to use global minima of $f$ to improve this?

$\endgroup$
2
  • $\begingroup$ In general no because the global minima of $f$ can occur far away from the origin and they all can be lifted up by $\|x\|^2$ too high to compete for being (close to) the minima of the sum, so you'll have to investigate a completely new region to minimize. However, if one of your global minima for $f$ is reasonably close to the origin, it can be used at least to restrict the search region to a small ball. If is hard to say more without knowing the function. $\endgroup$ Commented Feb 4, 2021 at 15:01
  • $\begingroup$ @fedja I suspect you're sadly right. Question edited anyway describing the space to which my horrible function belongs. $\endgroup$ Commented Feb 4, 2021 at 17:36

0

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.