# 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!

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.