I want to compute the convex conjugate $f^*$ of $f(x)=\lVert Ax \rVert$, where $x\in \mathbb R^n$ and $A\in \mathbb R^{m\times n}$ such that $\mbox{ran}(A)=\mathbb{R}^m$. Here
$$ f^*(x^*)=\sup_{x\in \mathbb R^n}\{(x^*)^Tx-\lVert Ax \rVert\}$$
I know that $\mathbb R^n=\ker(A)+\ker(A)^{\perp}$, and $A|_{\ker(A)^{\perp}}$ it is invertible, so if $x\in \ker(A)^{\perp}$ here is easy to compute $f^*(x^*)$, but I don't know how to use this fact in my problem and how (if it is posible) to vanish supremum over all $\mathbb{R}^n$ and instead of that calculate convex conjugate over $\ker(A)^{\perp}$. Any help will be appreciated.
Proposition 13.24 from Bauschke & Combettes' book (volume 2) states that the Legendre ("convex") conjugate of $g\circ A$ is written
$$f^* = (g\circ A)^* = A^* \triangleright g^*$$ where $A^*$ is its adjoint (so since it's real you can just use the transpose) and $\triangleright$ is the infimal postcomposition:
$$(\forall y\in\mathbb{R}^m)\quad A^*\triangleright g^*(y)=\inf_{\substack{x\in\mathbb{R}^n\\A^*x=y}}g^*(x).\tag{*}$$
Example 14.5 (same book) states that $$g^*(x)=(\|\cdot\|^*)(x)=\iota_{B(0;1)}(x) = \begin{cases} +\infty &\text{if}\;\;\|x\|>1\\ 0 &\text{if}\;\; \|x\|\leq 1 \end{cases}.$$ Here I use $\iota_C$ to denote the $0$-$\infty$ indicator function and $B(0;1)$ denotes the closed unit ball in $\mathbb{R}^n$. If $B(0;1)$ intersected with the inverse image $(A^*)^{-1}y$ is nonempty, then the infimum in (*) ought to occur when $g^*=0$ i.e. $x\in B(0;1)$. Otherwise, it is identically $+\infty$. Altogether,
$$(\forall y\in \mathbb{R}^m)\quad f^*(y)=\begin{cases} 0 &\text{if}\;\;A^{-1}y\cap B(0;1)\neq \varnothing\\ +\infty &\text{otherwise} \end{cases}.$$
Again, $A^{-1}y=\{x\in\mathbb{R}^n\, |\, Ax=y\}$ is always a well-defined set so here we do not require $A$ to be invertible.