Theorem in section 2.3.2 of Boyd & Vandenberghe's Convex Optimization:
If $f:R^k \to R^n$ is an affine function and an set $ S \subseteq R^n$ is convex, the inverse image of $S$ under $f$ defined as $$f^{-1}(S)=\{\vec x|f(\vec x)\in S, x \in dom f\}$$ is convex.
$\text{My questions:1 How to prove the theorem? 2 I find an example: if}\ S'=\{(x,y,z)^T|x^2+y^2\le |z|，0\le |z| \lt \infty \} \text{, and if}\ f(\vec x)=[[1,0,0];[0,1,0]]\vec x(\vec x\in S'),\text{then } f(S')=S \text{ is convex. However the } S' \text{ is not convex. dose the example violate the theorem?}$
Let $T$ be an affine transformation, and let $C$ be a convex set. We want to show that if $T(x) \in C$ and $T(y) \in C$ (i.e. $x,y \in T^{-1}(C)$), then $T((1 + t)x + ty) \in C$ holds for any $t \in [0,1]$.
To see that this holds, note that an affine transformation satisfies $$ T((1 - \lambda x) + \lambda y) = (1 - \lambda)T(x) + \lambda T(y) $$ for all $\lambda \in \Bbb R$. Thus, for any $x,y$ with $T(x),T(y) \in C$ and $t \in [0,1]$, we have $$ T((1 + t)x + ty) = (1 + t)T(x) + t T(y). $$ By the convexity of $C$, $(1 + t)T(x) + t T(y) \in C$, as was desired.