Let $(X,\langle \cdot,\cdot \rangle)$ a Hilbert Space, and $C\subseteq X$ an non empty, closed and convex set.
Show that $\forall x\in X$ and $\alpha \in [0,1]$:
$$\pi_{C}(\pi_{C}(x)+\alpha(x-\pi_{C}(x))=\pi_{C}(x)$$
Where $\pi_C$ is the projection above C.
I tried by contradiction and I get two inequalities:
$$\|x-\pi_C(x)\|<\|x-\pi_C(\pi_C(x)+\alpha(x-\pi_C(x)))\|$$
And
$$\|\pi_C(x)+\alpha(x-\pi_C(x))-\pi_C(\pi_C(x)+\alpha(x-\pi_C(x)))\|<\alpha \|x-\pi_C(x)\|$$
But I don't know how to develop that to reach a contradiction. Any hint?
Here is one way to show this: By the Kolmogorov criterion, $\pi(x)=y$ if and only if $y \in C$ and $\langle y-x,c-y \rangle \ge 0 $ for every $c \in C$. Hence, in order to show that $$ \pi(ax +(1-a) \pi(x)) = \pi(x) , \tag 1$$ it suffices to show that $\langle \pi(x) - ax -(1-a) \pi(x), c- \pi(x) \rangle \ge 0 $ for every $c \in C$. Or that $$ a \langle \pi(x) -x, c- \pi(x) \rangle \ge 0$$ for every $c \in C$. This is true (again, by invoking Kolmogorov's criterion) since $a \ge 0$.
Note that $(1)$ holds for every $a \ge 0$ not just $a \in [0,1]$. Generally, if $C$ is a subset of a Banach space $X$ and $P \colon X \to C$ is a retract of $C$ (that is, $P(c) =c$ for $c \in C$) then $P$ is said to be $\textbf{sunny}$ if for all $t \ge 0$, $$P(tx +(1-t)Px)=Px.$$ Hence, every metric projection on a closed and convex subset of a Hilbert space is sunny.