Conjugate of a function $f(x)$ is defined as $$f^*(y)=\sup_{x~\in ~dom~f} \{y^Tx-f(x)\}.$$ If $g(x)=f(Ax+b)$ then how to show that $$g^*(y)=f^*(A^{-T}y)-b^TA^{-T}y,~~~~(1)$$ where $A^{-T}=(A^T)^{-1}=(A^{-1})^T$ and $A$ is positive definite.
My Attempt:
We know that left side of $(1)$ is $$\sup_{x~\in~dom~g}\{y^Tx-g(x)\}=\sup_{Ax+b~\in~dom~f}\{y^Tx-f(Ax+b)\}.~~~(2)$$ Now the right hand side of $(1)$ is $$f^*(A^{-T}y)-b^TA^{-T}y=\sup_{x~\in~dom~f}\{y^TA^{-1}x-f(x)\}-b^TA^{-T}y~~~(3)$$ Now how to show that the right hand sides of $(2)$ and $(3)$ are equal? Any help in this regard will be much appreciated. Thanks in advance.
Assuming $f$ is proper we can simplify life by extending $f$ to have the value $+\infty$ for $x \notin \operatorname{dom} f$.
Then $f^*(y) = \sup_x (y^T x - f(x))$ and so \begin{eqnarray} g^*(x) &=& \sup_x (y^T x - f(Ax+b)) \\ &=& \sup_x (y^T (A^{-1}x) - f(x+b)) \\ &=& \sup_x ((A^{-T}y)^T (x) - f(x+b)) \\ &=& \sup_x ((A^{-T}y)^T (x-b) - f(x)) \\ &=& \sup_x ((A^{-T}y)^T x - f(x)) - (A^{-T}y)^T b\\ &=& f^*(A^{-T}y) - (A^{-T}y)^T b\\ \end{eqnarray}