Prove or disprove the convexity of a quotient

Question

It seems that the function $$q(x,y)=\frac{\langle x, y \rangle^2}{\|x\|^2}$$ is convex over $\Omega\times \mathbb{R}^n_+$ where $\Omega$ is the unit simplex defined by $$\Omega = \{x\in \mathbb{R}^n: \sum_{i=1}^n x_i = 1, x\geq 0\}.$$ Can anyone help to prove it or disprove it? Thanks a lot!

Answer 1

I may figure out a disproof as follows:

Let $n=2$. Given two points in $\Omega\times \mathbb{R}^2_+$ as

P1: $x=(1,0)\in \Omega$, $y=(0,1)\geq 0$.

P2: $x=(0,1)\in \Omega$, $y=(1,0)\geq 0$.

Clearly, $q(P1)=q(P2)=0$.

Now considering the segment [P1,P2]. We have

$$[P1,P2] = \{\lambda P1 + (1-\lambda)P2: \lambda\in [0,1]\} = \{(\underbrace{(\lambda,1-\lambda)}_{x},\underbrace{(1-\lambda,\lambda)}_{y}): \lambda \in [0,1]\}.$$ Then, taking a point $P\in [P1,P2]$ such that $P=\lambda P1+ (1-\lambda) P2$ with $\lambda \in (0,1)$, we have $$q(P) = \frac{\langle x, y \rangle^2}{\|x\|^2} = \frac{\langle(\lambda,1-\lambda),(1-\lambda,\lambda)\rangle^2}{\|(\lambda,1-\lambda)\|^2} = \frac{4\lambda^2(1-\lambda)^2}{\lambda^2+(1-\lambda)^2}.$$ But $\lambda q(P1) + (1-\lambda) q(P2) = 0$. Hence $$q(P) = q(\lambda P1+ (1-\lambda) P2) > \lambda q(P1) + (1-\lambda) q(P2), \forall \lambda \in (0,1),$$ which contradicts the definition of the convexity.

BTW: we can disprove as well the convexity of $q$ over $\Omega\times \mathbb{R}^n_{++}$ in a similar way. Let $$P(\lambda)=(\underbrace{(\lambda,1-\lambda)}_{x},\underbrace{(1-\lambda,\lambda)}_{y}).$$ Then the two points $P(0.2)$ and $P(0.8)$ belong to $\Omega \times \mathbb{R}_{++}^2$ and $q(P(0.2)) = q(P(0.8))$. However, the mid-point $$P(0.5) = P(0.2)/2+P(0.8)/2$$ verifies $$q(P(0.5))>q(P(0.2))/2 + q(P(0.8))/2 = q(P(0.2))$$ violating the convexity of $q$ over $\Omega\times \mathbb{R}^n_{++}$.