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