I need help with geometric interpretation of this theorem and with the corollary of the theorem:
Theorem: projection onto a closed convex set Let $K \subset H$ be a nonempty closet convex set. Then for every $f \in H$ there exists a unique $u \in K$ such that $|f-u|=mim|f-v|$
Moreover, $u$ is characterized by the property $u \in K \, and \, (f-u, v-u)\leq 0$ $\forall v \in K$
Corolary: Let $M \subset H$ is a closed linear subspace. Let $f \in H$, then $u=P_{M}f$ is characterized by $u\in M$ \, and \, $(f-u,v)=0$ $\forall v \in M$
My attemp for corollary is, like subspace is convex we use the theorem, then i was thinking use. $|(f-u)-t(w-u)|=|f-u|^{2}-2t(f-u,w-u)+t^{2}|w-u|^{2}$ $f \in H, \, w, u \in K$ with $t=1$ , any help with corollary or geometric interpretion of theorem in $\mathbb{R}^{2}$ The angle between these vectors is obtuse, why?. Thanks in advance
Let $v \in M$. $\langle f, v \rangle=\langle (f-u), (v+u)-u \rangle \leq 0$ by the theorem because $v+u \in M$. Since $-v \in M$ we get $\langle (f-u), -v \rangle \leq 0$ or $\langle (f-u), v \rangle \geq 0$. Hence $\langle (f-u), v \rangle=0$.
Conversely the condition obviously implies $\langle (f-u), (v-u) \rangle \leq 0$ .
For a geometric interpretation take $M$ to be a line in $\mathbb R^{2}$, draw a perpendicular from $f$ to $M$ and observe that the perpendicular line meets the space $M$ at the point where the distance form $f$ to $M$ is minimized.
For your last question take $K$ to be a line segment in $\mathbb R^{2}$ and you will easily see why the angle is obtuse.