Consider the Hilbert product space $X\times X$

Consider the Hilbert product space $X\times X$. In $X\times X$ define the closed convex 'diagonal' set by $$D={(x,x):x\in X}$$ Obtain a formula for projection $P_D$ and rigorously prove it.

I really struggle with finding formulas. I am hopping someone can give me some hints with where to start with finding this projection formula.

Thank you

• Have you tried the case $X = \Bbb R$? – Omnomnomnom Oct 12 '15 at 18:40
• So, if we consider the case where $X=\mathbb{R}$ then we are looking at the product space $\mathbb{R}\times\mathbb{R}$, the cartesian plain. This gives us $$D= (x,x):x\in\mathbb{R},$$ which means we are just looking at a single point? is this correct? Then our projection $P_D=x$? Am I on the right path – Jeremy Oct 12 '15 at 19:03
• @Jeremy : $\{ (x,x) \in \Bbb R^2 : x\in \Bbb R\}$ is a straight line, not a single point. – Tryss Oct 12 '15 at 19:09
• right, because we are considering $(x_1,x_1),(x_2,x_2)\ldots$ I just drew it out. Thank you. – Jeremy Oct 12 '15 at 19:10

Let $(u,v)\in X\times X$ be given, and find the unique $(x,x)\in D$ such that $$\langle(u,v)-(x,x),(y,y)\rangle_{X\times X} = 0,\;\;\; y \in X.$$ The point $(x,x)$ is the orthogonal projection of $(u,v)$ onto $D$. The condition is $$\langle u-x,y\rangle_{X}+\langle v-x,y\rangle_{X}=0, \;\;\; y \in X \\ \langle u+v-2x,y\rangle_{X} = 0,\;\;\; y \in X.$$ Therefore $x = \frac{1}{2}(u+v)$ is the unique solution. So $$P_{D}(u,v) = \frac{1}{2}(u+v,u+v).$$ This function $P_{D}$ is automatically linear, idempotent (i.e., $P_D^{2}=P_D$), and selfadjoint on $X\times X$.

Hint In the case of $X = \Bbb R$, we end up with the linear transformation $$A = \frac 1{2}\pmatrix{1&1\\1&1}$$

• In general, we can naturally think of $P_D$ as the transformation $id_X \otimes A$ – Omnomnomnom Oct 12 '15 at 20:58