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Consider a convex set $S ⊆ ℝ^n$ and an affine function $f(x)=Ax+b$. Then if the image of $S$ under $f$,$f(S)$={$f(x)|x \in S$} , is convex. Then the inverse image of the convex set C,$f^{-1}(C)=\{x|f(x) \in C$}, is convex set.
Hint:Let $y_1$ and $y_2 \in C.$Then there exist $x_1$ and $x_2 \in f^{-1}(C) $.Show that $f^{-1}(C)$ is convex.

How to prove the inverse of affine convex function is still a function?Because for me,i intuitively think the inverse image of the convex is still a convex.

My proof is as below,i am not sure whether it is right or not

For two elements of the preimage $x_1,x_2\in f^{-1}(C)$ and $t\in[0,1]$, you have to show that $tx_1+(1-t)x_2\in f^{-1}(C)$, i.e. that there exists an element $y\in C$, such that $tx_1+(1-t)x_2=f(y)$.

Since $x_1,x_2\in f^{-1}(C)$ there exist $y_1,y_2\in C$, such that $f(y_1)=x_1,f(y_2)=x_2$. Now use the definition of $f$ to find $y$ and you are done.

So \begin{align} f(tx_1+(1-t)x_2) &=A(tx_1+(1-t)x_2)+b\\ & = t(Ax_1+b)+(1-t)(Ax_2+b)\\ & = tf(x_1)+(1-t)f(x_2)\\ & = f(y) \end{align} And the question said $f(x)$ is a convex,that is,$f(tx_1+(1-t)x_2)$ is convex,so $f(y)$ is convex,that is $f^{-1}(C) $ is convex

Is this proof right?

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  • $\begingroup$ If you wish to show curly braces $\{ \ldots \}$ when using MathJax, you can use \{ and \}. $\endgroup$ – Theo Bendit Oct 25 '18 at 7:26
  • $\begingroup$ You don't need $f$ to have an inverse. The inverse image $f^{-1}(S)$ is defined, regardless of whether $f$ is invertible, to be the set of points $x$ such that $f(x) \in S$. For example, if $f : \mathbb{R} \to \mathbb{R}$ is constantly $y$, then $f^{-1}(S)$ will either be $\emptyset$ if $y \notin S$ or $\mathbb{R}$ if $y \in S$. Note that $f^{-1}$ is not a function, as $f$ is very far from being one-to-one! $\endgroup$ – Theo Bendit Oct 25 '18 at 7:29
  • $\begingroup$ @TheoBendit so the proof if this doesn't have relation with $f(x)=Ax+b$?i just said that there exist s $x_1$ and $x_2 \in C$,and C is convex set,so the $f^{-1}(C)$ must be convex too? $\endgroup$ – electronic component Oct 25 '18 at 7:57
  • $\begingroup$ It does relate to $f$, but it doesn't require there to be a function $f^{-1}$. I may have misunderstood your issue with this question, but you seemed to be interested in showing $f^{-1}$ is still a function (which, in general, it won't be). $\endgroup$ – Theo Bendit Oct 25 '18 at 8:00
  • $\begingroup$ not quite. My hint was a bit misleading and there was some mixup with $y$ and $x$, i finished it up in my answer. $\endgroup$ – weee Oct 26 '18 at 20:42
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For two elements of the preimage $x_1,x_2\in f^{-1}(C)$ and $t\in[0,1]$, you have to show that $tx_1+(1-t)x_2\in f^{-1}(C)$, i.e. that $f(tx_1+(1-t)x_2)\in C$. ($S$ is convex so $tx_1+(1-t)x_2\in S$ and we can apply $f$)

Since $x_1,x_2\in f^{-1}(C)$ we have $f(x_1)\in C,f(x_2)\in C$.

Edit: to finish up, we calculate \begin{align} f(tx_1+(1-t)x_2)&=A(tx_1+(1-t)x_2)+b\\ &=t(Ax_1+b)+(1-t)(Ax_2+b)\\ &=tf(x_1)+(1-t)f(x_2). \end{align} Since $C$ is convex, and $f(x_1)\in C,f(x_2)\in C$ we have also $tf(x_1)+(1-t)f(x_2)\in C$ and are done.

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  • $\begingroup$ the definition of $f$ is $Ax+b$,but how can we find $y$ by using this function?how do you know the $f(y_1)=x_1 $ and $f(y_2)=x_2$ from the $tx_1+(1-t)x_2=f(y)$? $\endgroup$ – electronic component Oct 26 '18 at 2:02

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