I was trying to prove some special cases of ADMM, but had some trouble getting this one so please help. Thank you in advance
Considering x-update when A=I
$x^+ = argmin_x(f(x)+({\rho}/2){\mid\mid}x-v{\mid\mid}_2^2) = prox_{f,\rho}(v)$.
The special case to be proved is
$f(x)={\lambda}{\mid\mid}.{\mid\mid}_1$, $x_i^+ = S_{\lambda/\rho}(v_i)(S_a(v))= (v-a)_+-(-v-a)_+) $
This is soft-thresholding operator as prox of the $1$-norm. Try to compute it yourself: Start with the case $n=1$, i.e., $f(x) =|x|$. Then observe that the prox of $\|\cdot\|_1$ can be computed component-wise.