Orthogonal projection to graph of Hilbert space Let H be a Hilbert space and A be a bounded linear operator on H.
Let G be the graph of A , i.e a subset of the direct sum of H with itself of the form (h,Ah).
Then we know G is a closed linear subspace (This is one direction of closed graph theorem), so we can define the orthogonal projection P from the direct sum of H with itself onto G.
My question is that do we have an explicit form of P? I have no idea how to express it explicitly.
Any help would be appreciated.
 A: The orthogonal projection of $(y,z)\in\mathcal{H}\times\mathcal{H}$ onto the graph $\mathcal{G}(A)=\{ (x,Ax)\in\mathcal{H}\times\mathcal{H} : x\in\mathcal{H} \}$ is the unique $(x,Ax)\in\mathcal{H}\times\mathcal{H}$ such that the following orthogonality conditions hold in $\mathcal{H}\times\mathcal{H}$:
$$
           ((y,z)-(x,Ax)) \perp (x',Ax'), \;\;\; x'\in\mathcal{H}.
$$
That is, the following hold for all $x'\in\mathcal{H}$:
$$
           (y-x,z-Ax)\perp(x',Ax'),\;\;\; x'\in\mathcal{H}, \\
                \langle y-x,x'\rangle=\langle Ax-z,Ax'\rangle,\;\;\;  x'\in\mathcal{H}, \\
            \langle y-x,x'\rangle=\langle A^*(Ax-z),x'\rangle,\;\;\; x'\in\mathcal{H}, \\
            y-x=A^{*}(Ax-z) \\
            y+A^*z =(A^*A+I)x \\
           x=(A^*A+I)^{-1}(y+A^*z).
$$
Therefore, if $P$ denotes the orthogonal projection of $(y,z)$ onto the graph $\mathcal{G}(A)$, then
$$
        P(y,z) = (x,Ax) \\
   = ((A^*A+I)^{-1}(y+A^*z),A(A^*A+I)^{-1}(y+A^*z)) \in\mathcal{G}(A)\subset\mathcal{H}\times\mathcal{H}.
$$
It's ugly and pretty all at the same time.
