Proximal Operator of $ f \left( x \right) = {\left\| A x \right\|}_{2} $ Where $ A $ Is Diagonal Matrix (Weighted $ {L}_{2} $ Norm) How to compute the proximal mapping (prox-operator) of  $f(x)=||Ax||_{2}$?
Here $A$ is a diagonal matrix with all positive eigenvalues.
I know how to compute the prox mapping of $f(x)=||x||_{2}$, but I have not found any connection between these two functions. I found someplace saying that for general $A$, with $g(x)=f(Ax)$ the prox-operator of $g(x)$ does not follow easily from prox-operator of $f(x)$. See  L. Vandenberghe - The Proximal Mapping at composition with affine mapping.
 A: I made a mistake in my calculations, and have edited it.
According to the Moreau decomposition, for $t > 0$
\begin{equation}
x = \operatorname{prox}_{t f}(x) + t \operatorname{prox}_{t^{-1} f^{*}}(x / t).
\end{equation}
Now we first compute the conjugate of $f(x)=\|Ax\|$.
\begin{equation}
\begin{aligned}
f^*(u) &= \sup_{x}~ \langle u,x \rangle - f(x) \\
&= \left\{
\begin{array}{cl}
{0} & {\text{if}~\|A^{-1}u\| \leq 1,} \\
{+\infty} & {\text{otherwise.}}
\end{array}
\right.
\end{aligned}
\end{equation}
 And then
\begin{equation}
\begin{aligned}
\operatorname{prox}_{t^{-1}f^*}(x/t) &= \arg\min_{u} f^*(u) + \frac{t}{2}\|u-\frac{x}{t}\|^2 \\
&= \arg\min_{\|A^{-1}u\| \leq 1} \frac{t}{2}\|u-\frac{x}{t}\|^2 \\
&= \left\{
\begin{array}{cl}
{\frac{x}{t}} & {\text{if}~\|A^{-1}x\| \leq t ,} \\
{\frac{A^TAx}{tA^TA + \lambda I}} & {\text{otherwise.}}
\end{array}
\right.
\end{aligned} 
\end{equation}
where $\lambda$ is the solution of 
\begin{equation}
\|\frac{Ax}{tA^TA+ \lambda I}\| = 1.
\end{equation}
Finally, we have
\begin{equation}
\begin{aligned}
\operatorname{prox}_{tf}(x) &= x-t \operatorname{prox}_{t^{-1}f^*}(x/t)  \\
&= \left\{
\begin{array}{cl}
{0} & {\text{if}~\|A^{-1}x\| \leq t ,} \\
{(I - \frac{A^TA}{A^TA + \frac{\lambda}{t}I})x} & {\text{otherwise.}}
\end{array}
\right.
\end{aligned}
\end{equation}
(If $A=I$ and $f$ will be reduced to Euclidean norm.)
Since $A$ is a diagonal matrix with all positive eigenvalues , the Fenchel Conjugate  of $f(x)$ can be computed by the following:
\begin{equation}
\begin{aligned}
f^*(u) &= \sup_{x} \langle u, x \rangle - \|Ax\| \\
&=\sup_{y=Ax} \langle A^{-1}u, y \rangle - \|y\| = g^*(A^{-1}u),
\end{aligned}
\end{equation}
where $g$ is Euclidean norm $g(x)=\|x\|$. Since the conjugate of $g(x)$ is
\begin{equation}
g^*(x^*)= \left\{
\begin{array}{cl}
{0} & {\text{if}~\|x\| \leq 1,} \\
{+\infty} & {\text{otherwise.}}
\end{array}
\right.
\end{equation}
Thus 
\begin{equation}
f^*(u) = g^*(A^{-1}u)= \left\{
\begin{array}{cl}
{0} & {\text{if}~\|A^{-1}u\| \leq 1,} \\
{+\infty} & {\text{otherwise.}}
\end{array}
\right.
\end{equation}
For the computation of $\lambda$, one can refer to @River Li's answer, or solve it easily through Lagrangian multiplier method.
