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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