Absolute value function is defined as follows: \begin{equation*} f(x) = |x| = \left\{ \begin{array}{ll} x & \mbox{if } x \geq 0 \\ -x & \mbox{if } x < 0 \end{array} \right. \end{equation*}
Prox of $|x|$ is
\begin{equation*} \text{prox}_{|\cdot|}^{\gamma}(x) = \text{argmin}_{y \in \mathcal{H}} \left\{ \begin{array}{ll} y + \frac{1}{2\gamma} ||y-x||^2 & \mbox{if } y \geq 0 \\ -y + \frac{1}{2\gamma} ||y-x||^2& \mbox{if } y < 0 \end{array} \right. \end{equation*}
Since the objective function of the minization problem is convex, the first order condition of optimality of a convex function is applied to get the minimizing point $y$.
Derivate w.r.t $y$ and using first order condition, $1+ \frac{1}{\gamma} (y -x) = 0 $ implies that the minimizer (proximal point) is $y = x - \gamma$ when $x - \gamma \ge 0$
Similary, $ -1+ \frac{1}{\gamma} (y -x) = 0 $ implies that the proximal point is $y = x + \gamma$ when $x < -\gamma$
\begin{equation*} \text{prox}_{|\cdot|}^{\gamma}(x) = \left\{ \begin{array}{ll} x - \gamma & \mbox{if } x \geq \gamma \\ x + \gamma & \mbox{if } x \leq -\gamma \\ ? & \mbox{if } |x| < \gamma \\ \end{array} \right. \end{equation*}
Question: When $|x| < \gamma $ then what is the proximal point of the absolute value function?.
Let us denote by $g^\gamma(y)$ the function that you are interested in minimizing. If $\vert x \vert < \gamma,$ then the derivative of $g^\gamma$ is always non-zero. However, we know that a proximal point must exist since $g^\gamma$ is convex, so it follows that the minimizing value of $g^\gamma$ must be exactly the point of non-differentiability of $g^\gamma,$ i.e. $\text{prox}^\gamma_{\vert \cdot \vert}(x) = 0$ when $\vert x \vert < \gamma.$ I hope this helps. :)