Let $X$ Hilbert space and $U \subset X$ convex closed. Then we can define the projection operator on $U$ $P \colon X \to X$ such that $P x \in U$ is the orthogonal projection of $x \in X$. Moreover $P$ is a self-adjoint nilpotent operator.
Now define $Q=I-P$ so that $Qx$ is the orthogonal complement of $Px$. Now by the fact that $P$ is a self-adjoint nilpotent operator we also have that $Q$ is a self-adjoint nilpotent operator.
Now every self-adjoint nilpotent operators are projection operators, is it correct? Then if this is true $Q$ is a projection operator on some set $Y$. Is it true that $Y=\overline{ U^C}$?
Best Answer
In case $U$ is a closed subspace, and P is the orthogonal projection onto $U,$ then $(I - P)$ is the orthogonal projection onto the orthogonal complement of $U$ (which is closed, if this is how the question was about.)
I guess, what you mean by "orthogonal projection" onto a closed convex set $U$ is the mapping which to every $x \in X$ assigns the unique $u \in U$ with $\lVert x - u \rVert = \inf_{v \in U}\lVert x - v \rVert$ - note however, that this mapping needs not be linear! For a simple example consider $X = \mathbb{R}^2$ and $U = [-1,1] \times \{0\}.$