Let $F: \mathbb{R}^n \rightarrow \mathbb{R}^m$ be an affine function, i.e., $F (x) = L(x) + b$, with $L : \mathbb{R}^n \rightarrow \mathbb{R}^m$ linear and $b \in \mathbb{R}^m$ Then for every convex set $C \subset \mathbb{R}^n$ the set $F (C) := \{ F(y) | y \subset C \} \subset \mathbb{R}^n$ is convex. If $G : \mathbb{R}^m \rightarrow \mathbb{R}^n$ is an affine function then
show that the set $G^{-1} (C) := \{z \in \mathbb{R}^m: G(z) \in C\};$
I was wondering that isn't esentially $G^{-1}(C)$ the same as $F(C)$ which is already given as convex? Is there really anything else to do with this proof?
Also what does $F$ being an affine function have to do with anything here. I know what an affine combination looks like, but don't see how this relates to $F$ in this circumstance. Does it tell us anything important?
Any help is very much appreciated.
1) Let $a,b\in F(C)$, let show that $ta+(1-b)t\in F(C)$ for all $t\in[0,1]$.
We have that $a=F(x)=L(x)+b$ and $b=F(y)=L(y)+b$ for certain $x,y\in C$. By linearity of $L$,
$$at+(1-t)b=tF(x)+(1-t)F(y)=tL(x)+(1-t)L(y)+tb+(1-t)b=L(tx+(1-t)y)+b=F(tx+(1-t)y)$$ But $C$ is convexe, therefore $tx+(1-t)y\in C$ and thus $at+(1-t)b\in F(C)$. We conclude that $F(C)$ is convexe.
2) Let $x,y\in G^{-1}(C)$, therefore $G(x),G(y)\in C$ and so, $$t G(x)+(1-t)G(y)=...=G(tx+(1-t)y).$$
By convexity of $C$, for all $t\in[0,1]$, $$t G(x)+(1-t)G(y)\in C$$ and thus $$G(tx+(1-t)y)\in C.$$ Therefore, $$tx+(1-t)y\in G^{-1}(C)$$ and so $G^{-1}(C)$ is convex.