The Gini-coefficient is defined as $$ G(x) = \sum_{i = 1}^n \left( \frac{i}{n} - \sum_{j=1}^{i} \frac{x_{(j)}}{\mathbb{1}^{T}x}\right), $$ where $x_{i} $ is nonnegative numbers with positive sum. $x_{(j)}$ denotes the j-th smallest number among $\{ x_1,x_2, ..,x_n \}$.
Show that the Gini coefficient is quasiconvex.
My approach:
Rewrite the sum as
$$ G(x) = \sum_{i = 1}^n \frac{i}{n} - \sum_{i=1}^{n}\sum_{j=1}^{i} \frac{x_{(j)}}{\mathbb{1}^{T}x}. $$
We can rewrite this as
$$ b - \frac{c^Tx}{\mathbb{1}^Tx},$$
where $b = \sum_{i = 1}^n \frac{i}{n}$ and $c_i$ is the number of times $x_i$ is counted. Noting that G(x) is invariant to permutations we can assume that $x_1 \leq x_2,...,\leq x_{n-1} \leq x_n$ and then $c = (n,n-1,...,1)^T.$
Looking at the $\alpha$-sublevel set we have
$$S_{\alpha} = \{ x \in \mathbb{R}^{n}_{+}, \mathbb{1}^{T}x > 0: G(x) \leq \alpha \} =$$ $$ \{ x \in \mathbb{R}^{n}_{+}, \mathbb{1}^{T}x > 0: b - \frac{c^Tx}{\mathbb{1}^{T}x} \leq \alpha \} = $$ $$ \{ x \in \mathbb{R}^{n}_{+}, \mathbb{1}^{T}x > 0: ((b-\alpha)\mathbb{1} -c)^Tx \leq 0 \}. $$
which is an intersection of three convex sets, hence convex.
Is this correct?
Let $f_i(x) = \sum_{j=1}^i x_{(j)}$. This is a concave function of $x$. To prove it, define $$S_{(i)}\triangleq \{s\in\{0,1\}^n\,|\,\textstyle\sum_j s_j = i\}$$ In other words, $S_{(i)}$ is the set of all $\{0,1\}$ vectors with exactly $i$ ones. Then $$f_i(x) = \inf_{s\in S_{(i)}} s^Tx.$$ Because $f_i(x)$ is the pointwise infimum of affine functions, it is concave. (See, for instance, $\S3.2.3$ of Convex Optimization by Boyd & Vandenberghe.)
Now we can write $G(x)$ as follows: $$G(x) = \sum_{i=1}^n \frac{i}{n}-\frac{f_i(x)}{1^Tx}$$ And the sublevel set inequality becomes $$ G(x) \leq \alpha \quad \Longleftrightarrow\quad \sum_{i=1}^n \frac{i}{n}(1^Tx)-f_i(x) \leq \alpha (1^Tx)$$ The left-hand side is convex, the right-hand side is affine, so the inequality describes a convex set. Hence the sublevel set is convex, and $G(x)$ is quasiconvex.