# 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?

Tags :