Conditions such that norm of matrix vector can be written as the derivative of the norm of the vector for some convex fonction

Question

Problem statement:

Let $A$ be a matrix $\mathbb{R}^{d \times d}$, I want to find some conditions on $A$ such that there exists a differentiable convex function $f: \mathbb{R_{+}} \rightarrow \mathbb{R}$ which verifies:

\begin{equation} \forall x \in \mathbb{R}^{d}\ \|Ax\|_{2}^{2}=f'(\|x\|_{2}^{2}) \end{equation} Some examples:

If $A$ is a orthogonal matrix coming from some isometry then we can take $f(t)=t^{2}/2$, if $A=cI$ for some $c\geq0$ we can take $f(t)=c t^{2}/2$. Are these the only cases ?

I have two strategies so far :

First approach I want to restrict the space of research for $f$. Suppose that $Ker(A)\neq {0}$, then it implies that $f$ must be constant on $[0,\sup_{x\in Ker(A)}(\|x\|_{2}^{2})]$ otherwise it would break the convexity of $f$. Appart from this segment we know that the derivative is bounded: $f'(t)\leq \|A\|_{2}t$. Can we go further on this ?

Second approach, maybe it is useful to decompose $A$ with $A=U\Sigma V^{T}$ where $\Sigma=diag(\lambda_{1},\cdots,\lambda_{d})$, with the change of variable $v=V^{T}x$, the problem is equivalent as finding $f$ satisfying $f'(\|v\|_{2}^{2})=\sum_{i} \lambda_{i}^{2}\|v_{i}\|_{2}^{2}$.

Answer 1

Here is an answer: $\|Ax\|_{2}^{2}=x^{T}A^{T}Ax$, we can write $A^{T}A=P D P^{T}$ where $D$ diagonal and $P$ orthogonal, then it is equivalent to find a $g=f'$ such that:

(E) $||D^{1/2}v ||_{2}^2=\sum_{i} |\lambda_{i} v_{i}|^{2} = g(||v||_{2}^2)$

(with $\lambda_{i}\geq 0$ and we note $v=P^Tx$)

If there is one $\lambda_{i}=0$ (the first one without loss of generality) then for all $t\geq 0$ we can take $v=(\sqrt{t},0,..,0)$ then $g(||v||_{2}^2)=g(t)=0$ which is not possible (unless $A=0$).

In this way all $\lambda_{i}>0$. We will show that they are all equal to some $c$. Indeed let $e_{i}$ be one eigenvector of $A^{T}A$ then $\|Ae_{i}\|_{2}^{2}=e_{i}^{T}A^{T}A e_{i}=\lambda_{i}=g(\|e_{i}\|_{2}^{2})=g(1)$. So $\lambda_{i}=g(1)=c$ for all $i$.

So $A^TA=cI$. Overall it implies that $A$ can be written as $A=c_{0} O$ with $O$ an orthogonal matrix.