Can the Proximal Operator Have Discontinuity?

Question

The proximal operator is defined as: $\operatorname{prox}_{\sigma f}(y) = \arg \min_x (f(x) + \frac{1}{2 \sigma} \|x - y\|^2_2), \sigma > 0$.

I was wondering about whether the proximal operator is always continuous or not. My intuition says not, but I can't find any example where it is the case. Any help?

If the proximal operator is always a continuous function, how can I prove?

Answer 1

Let $$ f(x)=\left\{\begin{matrix} 0& x \neq 0 & \\ -1& x =0 & \end{matrix}\right. $$

Take $\sigma = \frac{1}{2} $ then the value of proximal map is $$P(y) =\left\{\begin{matrix} 0& y \neq 0 & \\ -1& y =0 & \end{matrix}\right. $$ which is a discontinuous map.

Answer 2

As Red shoes pointed out, they are not always continuous. However, if $f$ maps from a real Hilbert space to $]-\infty,+\infty]$, is:

• Not identically $+\infty$,
• Convex, and
• lower-semicontinuous,

then its proximal operator is always firmly nonexpansive, hence it is $1$-Lipschitz continuous. See also http://proximity-operator.net/proximityoperator.html and its references