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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?

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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? $\endgroup$
    – LarrySnyder610
    Apr 15 at 16:56
  • $\begingroup$ @LarrySnyder610 Good point, I updated the answer. $\endgroup$ Apr 15 at 17:00

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