Moreau Yosida Approximation Let $f\colon H\to\mathbb{R}\cup \{+\infty\}$ is convex, lower semicontinuous, proper and coercive function. $H$ is a Hilbert space.
$f_{\lambda}:H\to\mathbb{R}\cup \{+\infty\}$ is the Moreau-Yosida approximation with $\lambda>0$: $$f_\lambda(x)=\inf_{y\in X} \left\{ f(y)+\frac{1}{2\lambda}||x-y||^2\right\}$$
Define $J_{\lambda}(x)=y$,  $y$ is the point where the infimum is attained.
Determine $\partial f_{\lambda}(x)$ (the subdifferential of a convex function). I don't know how to do this. Any help would be greatly appreciated!
 A: This can for example be found in "Convex Analysis ans Monotone Operator Theory in Hilbert Spaces" by Bauschke and Combette, Proposition 12.29.
The answer is
$$
\nabla (f_\lambda) = \lambda^{-1}(Id - J_\lambda),
$$
i.e. a gradient step wrt the Moreau-Yosida Regulariztion corresponds to a proximal step of the original function.
Here a sketch of the proof: First you need the result that
$$
p = J_\lambda (x) \quad \Leftrightarrow \quad (\forall y \in H) \quad \langle y-p,x-p \rangle + f(p) \le f(y).
$$
This can be e.g. derived from the optimality conditions (if you know that $Id+\partial f$ has a singlevalued inverse.
Next you derive
$$
f_\lambda (y) - f_\lambda (x) \ge \langle y-x, x - J_\lambda (x) \rangle \gamma^{-1}
$$
and
$$
f_\lambda (y) - f_\lambda (x) \le \langle y-x, y - J_\lambda (y) \rangle \gamma^{-1}.
$$
Combining these two equation with the firmly non-expansiveness of the prox you get that
$$
0 \le f_\lambda (y) - f_\lambda (x) - \langle y-x, x - J_\lambda (x) \rangle \gamma^{-1} \le \lVert y-x \rVert^2 \gamma^{-1}
$$
wihch finishes the proof.
