4
$\begingroup$

Given a general function $f:\Bbb Z\to\Bbb R$ is there a simple way to verify whether $f(x)$ is pseudo-convex or not?

$\endgroup$

1 Answer 1

2
$\begingroup$

For a continuous function, all you need to do is prove that it's (i) non-convex, and (ii) monotonic. (i) can be shown using the eigenvalues of the hessian matrix, and (ii) using the gradient.

However, in your case your domain is $\mathbb{Z}$, therefore derivatives are generally not defined, and neither is the concept of (pseudo)convexity.

You can show whether the relaxed $f:\mathbb{R}\rightarrow \mathbb{R}$ is pseudo-convex, but the concept is not defined in the integral domain. In a way, this directly answers your question: $f:\mathbb{Z}\rightarrow \mathbb{R}$ can not be pseudo-convex.

$\endgroup$
2
  • $\begingroup$ If the function's domain is the integers, is there even necessarily a Hessian? $\endgroup$ Apr 15, 2021 at 16:56
  • $\begingroup$ @LarrySnyder610 Good point, I updated the answer. $\endgroup$ Apr 15, 2021 at 17:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.