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?
Best Answer
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.