# Corollary of Projection onto a closed convex set and geometric interpretation

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.