Let's say you have a covariance matrix $\sigma$ and a vector of expected returns $\mu$. Basically $\sigma$ is a symmetric matrix with positive eigenvalues and we can safely assume that $\mu$ is just a non null vector of $\mathbb{R}^n$ not collinear with $\mathbb{1}^{\top}=(1,1,...,1)^{\top}$. We introduce the $n-$simplex $X:=\left\{x \in \mathbb{R}_+^n | \mathbb{1}^{\top}x=1\right\}$ and we also define the function : $f:\begin{aligned}\mathbb{R}^n &\to \mathbb{R}^2 \\ x &\mapsto \left( x^{\top} \sigma x , -\mu^{\top}x \right)^{\top}\end{aligned}$
My question is : is $f(X)$ a convex set ? It is straightforward to show that the images of X by each component are convex by using for example their continuity and the compactness of $X$, but that doesn't really help me to show ( or disprove ) that $f(X)$ is convex.
The reason I'm trying to prove that is that it is said in this paper http://www.sciencedirect.com/science/article/pii/S0377221713008515 that the fact that $f(X)$ is a convex set is "straightforward", and so either I'm passing by something really obvious or they made a mistake.
Let $n=2, \sigma = I, \mu = (-1,0)^T$, Then $X = \{(t,1-t)\}_{t \in [0,1]}$ and $f(X) = \{ (t^2+(1-t)^2, t) \}_{t \in [0,1]}$, which is not convex (it is the graph of $t \mapsto t^2+(1-t)^2$ on $[0,1]$).