Quasiconcavity of $f(x) = \min\{k | \sum_{i=1}^k |x_i| > 1\}$ with $f(x) = \infty$ if $\sum_{i=1}^n {|x_i| \leq 1}$

Question

For $x \in \mathbb{R}^n$, we define $f(x) = \min\{k | \sum_{i=1}^k |x_i| > 1\}$ with $f(x) = \infty$ if $\sum_{i=1}^n {|x_i| \leq 1}$. How to use sublevel set to show the quasiconcavity of $f$?

Answer 1

To show quasi concavity of $f$, we must show that the set $$C_{\alpha}=\{x:f(x)\geq\alpha\},$$ is convex for any $\alpha$. For this, we need to show that, if $y$ and $z$ are in $C_{\alpha}$, all convex combinations $\beta y+(1-\beta)z$ are in $C_{\alpha}$, where $\beta\in(0,1)$.

Pick any such $w=\beta y+(1-\beta)z$. Let $K$ be any integer smaller than $\alpha$. Then, by triangle inequality, $$\sum_{i=1}^{K}|w_i|\leq\beta\sum_{i=1}^{K}|y_i|+(1-\beta)\sum_{i=1}^{K}|z_i|.$$ Since $y$ and $z$ are in $C_{\alpha}$, $\sum_{i=1}^{K}|y_i|\leq 1$ and $\sum_{i=1}^{K}|z_i|\leq 1$ (otherwise they cannot be members of $C_{\alpha}$). Hence, $$\sum_{i=1}^{K}|w_i|\leq \beta*1+(1-\beta)*1\leq 1.$$ Hence, $f(w)\geq\alpha$, and hence $w\in C_{\alpha}$. Since $\beta$ was arbitrary, the result follows.