Let $C$ be a convex closed cone of a Hilbert space containing $0$. We denote by $P$ the orthogonal projection on $C$ and $x$, $y$ two elements of $H$. My question is: do we have the following inequality:
$\left\|P(x+y)\right\| \leq \left\|P(x)\right\| + \left\|P(y)\right\|$ ?
(The norm of $P(x+y)$ is less than the sum of the norms of $P(x)$ and $P(y)$).
Thanks in advance
The answer is YES.
1) Denote the (closed convex) polar cone $C^\ominus$ by $D$.
2) The Moreau decomposition for cones gives that the identity is equal to $P_C+P_D$.
3) Denote by $f$ the distance function to $D$. Since $D$ is convex, so is $f$.
4) By 2), $f(x) = \|x-P_D(x)\|= \|P_C(x)\|$.
5) By 3) and 4), $$ \|P_C(\tfrac{x+y}{2})\| \leq \tfrac{1}{2}\|P_C(x)\|+\tfrac{1}{2}\|P_C(y)\|.$$
6) Moreover, $C$ is a cone, hence $P_C(\alpha x)=\alpha P_C(x)$ for $\alpha > 0$.
7) Finally, combining 5) and 6) gives the answer.
(Properties used in the above proof can be found in Bauschke-Combettes Convex Analysis and Monotone Operator Theory in Hilbert Spaces.)
