Proof is quasiconvex function

Question

For $x=(x_1, \cdots, x_n) \in \mathbb{R}^n,$ and

$f(x) = \left\{ \begin{array}{ll} \text{max}{\{i : x_i \neq 0\}} &, \mbox{if $x \neq 0$};\\ 0 &, \mbox{if $x = 0$}.\end{array} \right. $

Proof $f$ is quasiconvex function.

Where

Let $C\subset\mathbb{R}^n$ be a nonempty convex set. A function $f:C→\mathbb{R}$ is called quasiconvex.

$f(\lambda u+(1-\lambda)v)\leq\max\{f(u), f(v)\}, \quad\forall u,v\in C, \forall\lambda\in(0,1)$

But if

$f(\lambda u+(1-\lambda)v)=max\{i:\lambda u_i+(1-\lambda)v_i \neq 0 \}$

as you get to

$\leq max\{max\{i:u_i \neq o\},max\{i:v_i\neq 0\}\}$

Thanks!

Answer 1

If $k>max\{f(u),f(v)\}$ then $u_k=0$ and $v_k=0$. Hence $(\lambda u+(1-\lambda )v)_k=0$ and the result follows immediately.