I have a question arising from the Moreau-Yosida regularization in Banach spaces. The real Banach space $X$ and its dual $X^*$ are both reflexive strictly convex, $f:X \rightarrow \mathbb{R} \cup \{+\infty\}$ is lower-semicontinuous, proper and convex. $F$ is the duality mapping of $X$, defined by
$$F(x) := \{ x^* \in X^*: \|x\|_X^2=\|x^*\|_{X^*}^2=x^*(x)\}.$$
For $X$ strictly convex, $F$ is always single-valued.
I know that the subdifferential $\partial f(x)$ is maximal monotone and that $$R(\partial f+\lambda F) = X^*,$$ where $R$ is the range of the operator.
Somehow from this should follow that for every $x \in X, ~\lambda>0$ there is an $x_\lambda \in X$ such that $$ 0 \in\partial f(x_\lambda) + \lambda F(x_\lambda - x) $$.
But I don't see how. Well, if $F$ were linear this would be trivial, but it is not.
Does anyone have an idea how this can be shown? Thank you in advance.
First note that $F$ is the subdifferential of $j(x) = \frac12 \, \|x\|^2$.
Then, your condition is the necessary and sufficient optimality condition of: Minimize (w.r.t. $\tilde x$) $$f(\tilde x) + \frac\lambda2 \, \|\tilde x - x\|^2.$$
Now, it remains to show the existence of a minimizer of this functional. Therefore, you can use the convexity of $f$ and the quadratic growth of $\|\cdot\|^2$ to show that the objective is bounded from below and coercive. Using the lower-semicontinuity of $f$ (and $j$), you can infer the existence of a minimizer using standard arguments.