# Quasi-convex function must be "partially monotonic"?

Operations Research Asked by High GPA on December 13, 2020

$$f(x)$$ is quasi-convex,

$$x^*inargmin_{xin C}f(x).$$

How to prove that, for any $$ain C$$, $$f(x)$$ is weakly monotonic in the direction of $$(x^*-a)$$?

Is this simple result a part of an ancient theorem?

By the definition of quasiconvex: $$f(x)$$ with compact support $$C$$ is quasiconvex if for two points in the domain $$x_1,x_2$$ and $$win[0,1]$$ $$f(wx_1+(1-w)x_2)geq max{f(x_1),f(x_2)}$$.

Let $$x^* = argmin_{xin C}f(x)$$ where $$C$$ is the compact support of $$f$$. Then consider $$x_1,x_2in [x^*,infty)$$.

Choose $$x_2>x_1$$. By the definition of quasiconvexity, the secant segment from $$(x_1,f(x_1))$$ to $$(x_2,f(x_2))$$ lies below or at the maximum of the segment endpoints $${f(x_1),f(x_2)}$$. Since $$x^*$$ is a global minimizer, we can choose $$x_1=x^*$$ which implies the right limit inequality:

$$lim_{x_2downarrow x_1} f(wx_1+(1-w)x_2)-f(x_1)geq max{0,f(x_2)-f(x_1)}~forall win[0,1].$$ Thus the right derivative is non-negative. This then holds for all $$x_1geq x^*$$. Thus $$f$$ is weakly monotone increasing on $$[x^*,infty)$$.

We can do likewise for $$x_1,x_2in(-infty,x^*]$$ using left limits and show that $$f$$ is weakly monotone decreasing on $$(-infty,x^*]$$.

Correct answer by kurtosis on December 13, 2020

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