Prove That Projection Operator Is Non Expansive I am trying to prove that the projection operator defined as:
\begin{equation}
P(z) := argmin_{x \in \mathcal{C}}  \frac{1}{2}\|x-z\|^2_2
\end{equation}
is non-expansive. Here $\mathcal{C}$ is nonempty closed and convex set. To show this, I proceed as:
\begin{equation}
\|P(z_1) - P(z_2)|| =\|x_1-x_2\|
\end{equation}
, where $x_1, x_2$ are points in the set $\mathcal{C}$. Now, I know that $\|x_1-x_2\| \leq \|z_1-z_2\|$. But, how to prove this last part? Can JL lemma be used someway?
 A: As in your post, let $z_1$, $z_2$ be arbitrary points.
Recall the variational characterization of the projection operator onto nonempty, closed, convex sets:
$$ \langle z_1 - P(z_1),  x- P(z_1) \rangle\leq 0 \; \forall \; x \in C  $$
Now also notice that by definition $P(z_2) \in C$ thus we get:
$$ \langle z_1 - P(z_1),  P(z_2)- P(z_1) \rangle\leq 0 $$
Similarly we also get:
$$ 
\begin{aligned}
&\langle z_2 - P(z_2),  P(z_1)- P(z_2) \rangle\leq 0 \\
\Rightarrow &\langle P(z_2) - z_2,  P(z_2)- P(z_1) \rangle\leq 0
\end{aligned}
$$
Adding these two inequalities, rearranging and finally applying the Cauchy-Schwarz inequality, we get:
$$ 
\begin{aligned}
\langle P(z_2) - P(z_1),  P(z_2)- P(z_1) \rangle &\leq \langle z_2 - z_1,  P(z_2)- P(z_1) \rangle  \\
& \leq \vert\vert z_2 - z_1 \vert\vert \; \vert\vert P(z_2) - P(z_1) \vert\vert
\end{aligned}
$$
Thus:
$$
\begin{aligned}
&\vert\vert P(z_2) - P(z_1) \vert\vert^2 \leq \vert\vert z_2 - z_1 \vert\vert \; \vert\vert P(z_2) - P(z_1) \vert\vert \\
\Rightarrow &\vert\vert P(z_2) - P(z_1) \vert\vert \leq \vert\vert z_2 - z_1 \vert\vert 
\end{aligned}
$$
