0

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.

$\text{My questions:1 How to prove the theorem? 2 I find an example: if}\ S'=\{(x,y,z)^T|x^2+y^2\le |z|，0\le |z| \lt \infty \} \text{, and if}\ f(\vec x)=[[1,0,0];[0,1,0]]\vec x(\vec x\in S'),\text{then } f(S')=S \text{ is convex. However the } S' \text{ is not convex. dose the example violate the theorem?}$

Rodrigo de Azevedo
  • 1
ACuOoOoO
  • 1
  • 1
    I'm sorry, but I just cannot understand your question. It might help to be more clear about the result. What exactly are the premises, and what is the conclusion? You might find it helpful to see our guide for MathJax formatting so that you can put mathematical symbols into your question. – user804886 Jul 03 '20 at 06:15

1 Answers1

1

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.

Ben Grossmann
  • 225,327