Hi: I'm reading a book on optimization and there is an interesting stated theorem but I don't know how to prove it.
Notation: Let $P_{c}(x)$ denote the projection of onto a convex set c which is a subspace of $R^{n}$.
Then, the theorem is the following:
If y and x are 2 points in $R^{n}$, then,
if $|| P_{c}(y) - P_{c}(x) ||= || y - x ||$, then
$x - P_{c}(x) = y - P_{c}(y)$
Thanks to anyone who could provide a reference or a proof. This is not homework. I'm just studying this material for fun.
Since $C$ is a subspace, we have $$ \langle x-P_C(x),P_C(x)-P_C(y)\rangle=0; $$ $$ \langle y-P_C(y),P_C(y)-P_C(x)\rangle=0. $$ Adding two equalities we obtain $$ \langle x-P_C(x)-(y-P_C(y)),P_C(x)-P_C(y)\rangle=0 $$ or equivalently, $$ \langle x-y, P_C(x)-P_C(y)\rangle=\|P_C(x)-P_C(y)\|^2. $$ We have \begin{eqnarray*} \|(x-P_C(x))-(y-P_C(y))\|^2&=&\|(x-y)-(P_C(x)-P_C(y))\|^2\\ &=&\|x-y\|^2-2\langle x-y, P_C(x)-P_C(y)\rangle+\|P_C(x)-P_C(y)\|^2\\ &=&\|P_C(x)-P_C(y)\|^2-2\|P_C(x)-P_C(y)\|^2+\|P_C(x)-P_C(y)\|^2\\ &=&0. \end{eqnarray*} Hence, $x-P_C(x)=y-P_C(y)$.
Remark. We explain why we have $$ \langle x-P_C(x),P_C(x)-P_C(y)\rangle=0, $$ where $C$ is a subspace.
Indeed, by the characterization of the metric projection $$ \langle x-P_C(x), u-P_C(x)\rangle\leq 0 \quad \forall u\in C. $$ Substituting $u=P_C(y)\in C$ into the equality we obtain $$ \langle x-P_C(x), P_C(y)-P_C(x)\rangle\leq 0. $$ Substituting $u=2P_C(x)-P_C(y)\in C$ into the equality we obtain $$ \langle x-P_C(x), P_C(x)-P_C(y)\rangle\leq 0. $$ It follows that $$ \langle x-P_C(x),P_C(x)-P_C(y)\rangle=0. $$
