0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Have you tried the case $X = \Bbb R$? $\endgroup$ – Omnomnomnom Oct 12 '15 at 18:40
  • $\begingroup$ 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 $\endgroup$ – Jeremy Oct 12 '15 at 19:03
  • $\begingroup$ @Jeremy : $\{ (x,x) \in \Bbb R^2 : x\in \Bbb R\}$ is a straight line, not a single point. $\endgroup$ – Tryss Oct 12 '15 at 19:09
  • $\begingroup$ right, because we are considering $(x_1,x_1),(x_2,x_2)\ldots $ I just drew it out. Thank you. $\endgroup$ – Jeremy Oct 12 '15 at 19:10
1
$\begingroup$

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 $X\times X$.

$\endgroup$
0
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.