The problem of partial inverse operator

by Ze-Nan Li   Last Updated October 20, 2019 05:20 AM - source

Let $F$ be the proper lower semicontinuous convex function defined on $\mathbb{R}^n$, and $\mathcal{A} = \{X=(x_1,x_2,\dots,x_m) \mid x_1=\dots=x_m\}$. Define the indicate function
\begin{equation} \delta(X \mid \mathcal{A}) = \left\{ \begin{array}{cc} {0} & {\text{if}~X \in \mathcal{A},} \\ {+\infty} & {\text{otherwise}.} \end{array} \right. \end{equation} Our goal is to find a zero of the sum of this two maximal monotone operators $\partial F$ and $\partial \delta(\cdot \mid A)$: \begin{equation} \left\{ \begin{array}{cc} {\text{Find}} & {X^*,} \\ {\text{s.t.}} & {0 \in \partial F(X^*) + \partial \delta(X^* \mid \mathcal{A}).} \end{array} \right. \end{equation} Here the subdifferential of $\delta(\cdot \mid \mathcal{A})$ can be computed as \begin{equation} \partial \delta\left(X^{*} \mid \mathcal{A} \right)= \left\{ \begin{array}{cc} {\mathcal{A}^{\perp}} & {\text{if}~ X^{*} \in \mathcal{A}} \\ {\emptyset} & {\text{otherwise,}} \end{array} \right. \end{equation} where $\mathcal{A}^{\perp}$ is the orthogonal subspace to $\mathcal{A}$ \begin{equation} \mathcal{A}^{\perp}=\left\{Y=\left(y_{1}, \cdots, y_{m}\right) \mid \sum_{i=1}^{m} y_{i}=0\right\} \end{equation} I found that in some literature, this problem is equivalent to the following problem: \begin{equation} \left\{\begin{array}{cc} {\text{Find}} & {X^{*} \in \mathcal{A}, Y^{*} \in \mathcal{A}^{\perp},} \\ {\text{s.t.}} & {Y^* \in \partial F\left(X^{*}\right).} \end{array}\right. \end{equation} I think it is right but I don't know how to prove it, and I guess it has some relations to partial inverse operator?

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