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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!

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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.

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