# Quasi-convex function must be “partially monotonic”?

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

$$x^*\in\arg\min_{x\in C}f(x).$$

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

Is this simple result a part of an ancient theorem?

• You can easily prove it, using the definition of function convexity and the fact that $x^*$ is a minimizer. I'm hesitant to say more, lest I ruin a perfectly good homework problem. – prubin Aug 6 '20 at 20:32

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 $$w\in[0,1]$$ $$f(wx_1+(1-w)x_2)\geq \max\{f(x_1),f(x_2)\}$$.
Let $$x^* = \arg\min_{x\in C}f(x)$$ where $$C$$ is the compact support of $$f$$. Then consider $$x_1,x_2\in [x^*,\infty)$$.
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:
$$\lim_{x_2\downarrow x_1} f(wx_1+(1-w)x_2)-f(x_1)\geq \max\{0,f(x_2)-f(x_1)\}~\forall w\in[0,1].$$ Thus the right derivative is non-negative. This then holds for all $$x_1\geq x^*$$. Thus $$f$$ is weakly monotone increasing on $$[x^*,\infty)$$.
We can do likewise for $$x_1,x_2\in(-\infty,x^*]$$ using left limits and show that $$f$$ is weakly monotone decreasing on $$(-\infty,x^*]$$.
• Thank you kurtosis! I like your method but I am not sure if we are implicitly assuming that $f$ is differentiable (which is not necessarily true). – High GPA Aug 8 '20 at 13:59
• I suppose we could forego the limit existing (and hence the derivative) and just have the inequality be true ($\forall\delta>0:0<x_2-x_1\leq\delta\ldots$) and still get weak monotonicity. – kurtosis Aug 8 '20 at 19:46