Prox Operator of a First Order Perturbation (Adding Linear Term to the Function) Given a function $f$ we can describe its proximal operator as,
$$\mbox{prox}_{\frac{1}{\rho}f}(x) = \arg\min\limits_{u} f(u) + \frac{\rho}{2}\|x-u\|^2$$
How does this change if we introduce a linear perturbation, i.e. we replace $f$ with $f(x) + \langle\mu,x\rangle$? Assuming $f$ is simple, i.e. the prox of $f$ has a closed form, does the perturbation also have a simple prox?
I am curious how easy it is to calculate,
$$\arg\min\limits_{u}f(u)+\langle \mu, u\rangle +\frac{\rho}{2}\|x-u\|^2$$
in particular when $x$ is an $n\times n$ matrix and $f(x) = \lambda \|\mbox{vec}(x)\|_{1}$

[Corrected via comments] Sure. The optimality condition for your original prox function is
$$0 \in \partial f(u) - \rho ( x - u)$$
For the perturbation, it is
$$0 \in \partial f(u) + \mu - \rho ( x - u ) = \partial f(u) - \rho ( x - \rho^{-1} \mu - u)$$
So basically, your perturbation is solved by $$\textstyle\mathop{\textrm{prox}}_{\tfrac{1}{\rho}f}(x-\rho^{-1}\mu)$$
