Small confusion about two orthogonal projection on $L^2$.

Question

Let $ \Omega \subset \mathbb{R}$ be bounded and $ ( \Omega, \mathcal{A}, \mu ) $ be a measure space and . Let $I $ be a closed interval containing $0$ and $$A:=\{f \in L^2(\Omega):f(x) \in I \ \ a.e \}.$$ I have always remembered that the orthogonal projection of $L^{2}(\Omega)$ should be given as $$P_A(f)(x) = f 1_{I}(x).$$Since $$ \langle f - P_{A}(f), g\rangle = 0$$ and $ P_{A}f \in A$ for all $f \in L^{2}(\Omega)$ and $g \in A$. However in an exercise book I have read it claims that this projection is supposed to be $$ P_{A}(f)(x) = P_{I}(f(x)) $$ where $ P_{I} $ is the projection from $\mathbb{R}$ onto $I$. These two are clearly different but I can see where I am wrong. Could anyone give me a quick lift? Thanks!

Answer 1

One possible source of confusion could come from your equality $$\langle f - P_A(f), g\rangle = 0 \qquad \forall g \in A.$$ In fact, this equality characterizes the projection onto a closed subspace. In your case, $A$ is merely a closed convex set. In this case, the projection is characterized by $$\langle f - P_A(f), g - P_A(f) \rangle \le 0 \qquad \forall g \in A.$$ Note that this reduces to the above equality, if $A$ is additionally a subspace.