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$f(x)$ 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?

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    $\begingroup$ 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. $\endgroup$
    – prubin
    Aug 6, 2020 at 20:32

1 Answer 1

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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 $(-\infty,x^*]$.

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  • $\begingroup$ 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). $\endgroup$
    – High GPA
    Aug 8, 2020 at 13:59
  • $\begingroup$ 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. $\endgroup$
    – kurtosis
    Aug 8, 2020 at 19:46

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