I am struggling to work out how to calculate the proximity norm of $ \lambda {\left\| x \right\|}_{2} $.
I believe this should be done using Moreau decomposition $ v = \operatorname{Prox} \left( x \right) + \operatorname{Prox}^{\ast} \left( v \right) $ but I am unsure how this works. Can anyone show me some steps as to what to do?
On
Given $ f \left( x \right) = \left\| x \right\| $ is a norm function its Prox is given by (For any Norm):
$$ \operatorname{Prox}_{\lambda f \left( \cdot \right)} \left( v \right) = v - \lambda \operatorname{Proj}_{ \mathcal{B}_{ \left\| \cdot \right\|_{\ast} } } \left( \frac{v}{\lambda} \right) $$
Where $ \operatorname{Proj}_{ \mathcal{B}_{ \left\| \cdot \right\|_{\ast} } } \left( \cdot \right) $ is the Orthogonal Projection Operator and $ \mathcal{B}_{ \left\| \cdot \right\|_{\ast} } $ is the Norm Unit Ball (Of the Dual Norm).
In your case we're dealing with the $ {L}_{2} $ Norm which is self dual.
Moreover, the Projection onto the $ {L}_{2} $ Unit Ball is given by:
$$ \operatorname{Proj}_{ \mathcal{B}_{ \left\| \cdot \right\|_{2} } } \left( x \right) = \begin{cases} \frac{x}{ \left\| x \right\|_{2} } & \text{ if } \left\| x \right\|_{2} > 1 \\ x & \text{ if } \left\| x \right\|_{2} \leq 1 \end{cases} $$
In summary:
$$ \operatorname{Prox}_{\lambda \left\| \cdot \right\|_{2}} \left( v \right) = v - \lambda \operatorname{Proj}_{ \mathcal{B}_{ \left\| \cdot \right\|_{2} } } \left( \frac{v}{\lambda} \right) = \begin{cases} v - \lambda \frac{ \frac{v}{\lambda} }{ \left\| \frac{v}{\lambda} \right\|_{2} } & \text{ if } \left\| \frac{v}{\lambda} \right\|_{2} > 1 \\ v - \lambda \frac{v}{\lambda} & \text{ if } \left\| \frac{v}{\lambda} \right\|_{2} \leq 1 \end{cases} = \left(1 - \frac{\lambda}{ \left\| v \right\|_{2} } \right)_{+} v $$
Where $ \left( x \right)_{+} = \max \left\{ 0, x \right\} $.
I answered exactly the same question here, using exactly the same Moreau decomposition idea you're thinking about :)