Let $H$ be a Hilbert space over the field $\mathbb{K}$ and $C$ a closed vector subspace of $H$. My reference defines the orthogonal projection of $u \in H$ onto $C$ as the unique element $P(u) \in C: ||u – P(u)|| \leq ||u – v||$ for all $v \in C$.
Let $v \in H$ be an arbitrary non-zero vector and denote $C = \mathrm{span}(v)$. Take it as granted that $C$ is a complete vector subspace. I am tasked with finding an explicit formula for the orthogonal projection onto $C$. A reasonable guess is the formula $P(u) = \left<u, e_v\right>e_v$ where $e_v \equiv \frac{v}{||v||}$ with which we can verify quite easily that $\left<P(u), (I – P)u\right> = 0$; that is, $P$ divides an element $u$ of $H$ into a collinear and orthogonal part w.r.t. the span of $v$.
(Question:) My question is that how can I argue rigorously that this mapping $P$ is the orthogonal projection onto the subspace $C$ as per the given characterization? In the material I am working with, orthogonal projection is constructed unique closest point in a closed convex subset of a Hilbert space and results such as orthogonality between the images of $P$ and $I – P$ are concluded from the closest point result. Hence it would feel a bit anachronistic to conclude that $P$ is the orthogonal projection solely by the fact that $P(u)$ and $(I – P(u)$ are orthogonal for all $u \in H$, or am I mistaken?
Best Answer
Hint
Use Pythagorean theorem to show that $\lVert u -\left<u, e_v\right>e_v \rVert \le \lVert u - \lambda e_v \rVert$ for any $\lambda \in \mathbb K$.
Note: Pythagorean theorem can be easily deduced from the expansion of $\lVert u -\left<u, e_v\right>e_v) - (\lambda e_v-\left<u, e_v\right>e_v)\rVert^2$