Is Inverse Cartesian product of convex set still convex?

Question

Given two compact convex set $X \subset \mathbf{R}^2, Y \subset \mathbf{R}^2 $. The Cartesian product $Z := X \times Y \subset \mathbf{R}^4$ Is again a convex subset.

Is the low dimension projection set $A = \{(x_1, y_1)|(x_1, x_2, y_1, y_2) \in Z\}$, or $B = \{(x_2, y_1)|(x_1, x_2, y_1, y_2) \in Z\}$, or any other projection still convex ?

Answer 1

Yes, convexity is preserved under images of affine functions.

Proof: Suppose $C \subset \mathbb{R}^m$ is convex and $f: \mathbb{R}^m \to \mathbb{R}^n$ is affine, so $f$ can be written as $f(x) = Ax+b$. Then for $y_1, y_2 \in f(C)$ and $\lambda \in [0,1]$, there exist $x_1, x_2$ such that $f(x_1) = y_1$ and $f(x_2) = y_2$.

Thus,

$\begin{align*}\lambda y_1 + (1- \lambda)y_2 &= \lambda f(x_1) + (1-\lambda)f(x_2)\\ &= f(\lambda x_1) + f((1 - \lambda) x_2) \\&= f(\lambda x_1 + (1-\lambda) x_2) \in f(C)\end{align*}$

It follows $f(C)$ is convex. $\square$

Since your projection is linear, its image of a convex set will be convex.

Answer 2

As far as I know a (linear) projection of a convex set is again convex, by writing down the convexity condition and using linearity:

Let $p:K\rightarrow L$ be a linear surjection from a convex set $K\subseteq \Bbb R^m$ to a set $L\subseteq \Bbb R^n$. Then for $z=p(x),w=p(y)$ we find that $$(1-\lambda) z + \lambda w= (1-\lambda)p(x) + \lambda p(y) = p((1-\lambda)x+\lambda y)$$ is contained in $L$, which shows that $L$ is convex.