# Proximal Operator of Weighted ${L}_{2}$ Norm

Define the weighted $${L}_{2}$$ norm $${\left\| x \right\|}_{2,w} = \sqrt{ \sum_{i = 1}^{n} {w}_{i} {x}_{i}^{2} }$$. Find the formula for $$\operatorname{prox}_{\lambda {\left\| \cdot \right\|}_{2,w} }(y)$$, where $$\lambda > 0$$.

By definition we have: $$\operatorname{prox}_{\lambda {\left\| \cdot \right\|}_{2,w}} \left( y \right) = \arg \min_{x} \left( \lambda \sqrt{ \sum_{i = 1}^{n} {w}_{i} {x}_{i}^{2} } + \frac{1}{2} {\left\| x - y \right\|}_{2}^{2} \right)$$

But I have no idea how to proceed from here. Any idea?

It's the usual thing in such situations: set the gradient with respect to $x$ to zero and solve for $x$.
In this case, the $i$-th coordinate of the gradient is $$\frac{w_ix_i}{\sqrt{W}} + \frac{x_i-y_i}{\lambda}$$ where $W=\sum_{i=1}^nw_ix_i^2$ is the weighted squared 2-norm. Setting the above to zero, we get $$x_i = \frac{y_i}{\lambda w_i/\sqrt{W}+1}.$$ So these expressions for $i=1,\ldots,n$ define the minimizer $x$.
• Maybe to clarify the first term can be obtained by applying the chain rule on $f(g(x)), \cases{f(x) = \sqrt{x}\\ g(x) = \sum_{i=1}^{n}w_i{x_i}^2}$. – mathreadler May 3 '17 at 10:15
• I think this is not the closed form answer as $W$ depends on ${x}_{i}$. For example if we set ${w}_{i} = 1$ for all $i$ we should get the result of the Proximal Operator of the vanilla ${L}_{2}$ Norm as in math.stackexchange.com/questions/1681658. But the results are not the same. – Royi Aug 27 '19 at 14:25