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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.

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  • $\begingroup$ Could you explain better exactly what $F$ is? $\endgroup$ – Sergio Parreiras Dec 8 '13 at 3:22
  • $\begingroup$ Thank you for the suggestion, I've done it now. $\endgroup$ – GenH Dec 8 '13 at 5:06
  • $\begingroup$ By which argument do you know $R(\partial f + \lambda \, F) = X^*$? Can you also show this including a translation $R(\partial f(\cdot+x) + \lambda \, F(\cdot))$? $\endgroup$ – gerw Dec 8 '13 at 8:30
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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.

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