Let $P$ an orthogonal projection onto a closed subspace of a hilbert space $X$, and let $x,y \in X$.
I am trying to prove the following two statements hold \begin{align*} &\|x-Py\|^2 \leq \|x-y\|^2 + \|x-Px\|^2, \\ &\|x-y\|^2 \leq \|x-Py\|^2 + \|x-Px\|^2 + 2\|y-Py\|^2. \end{align*} Fred has kindly helped me with the first one, but I am yet to manage the second. One idea I had was to try \begin{equation} \|x-y\|^2 = \|x+y-(Px + Py)+ Px - Py|^2 = \|x+y - Px - Py\|^2 - \|Px + Py\|^2 \end{equation} But it hasn't led me anywhere.
I would appreciate help with this second inequality!
For the first statement observe that $(x-Px) \perp P(x-y)$, since $P$ is an orthogonal projection. We can assume that $P \ne 0$. Then we get with Pythagoras:
$||x-Py||^2=||(x-Px)+P(x-y)||^2=||(x-Px)||^2+||P(x-y)||^2$
$ \le ||x-Px||^2+||P||^2 \cdot||x-y||^2 =||x-Px||^2+||x-y||^2$,
since $||P||=1$.
The second statement is your turn !