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.
Orthogonal projection to graph of Hilbert space
banach-spacesfunctional-analysishilbert-spaceslinear algebrareal-analysis
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Best Answer
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.