[proofplan]
For part (i), we exploit the minimality of $\pi_C(x)$: any convex combination $(1-t)\pi_C(x) + tz$ lies in $C$ by convexity, so the squared distance $\|x - ((1-t)\pi_C(x) + tz)\|_2^2$ must be at least $\|x - \pi_C(x)\|_2^2$. Expanding and sending $t \to 0$ yields the variational inequality. For part (ii), we apply the variational characterisation twice (once for $x$, once for $z$), add the resulting inequalities, and apply the Cauchy--Schwarz inequality to conclude non-expansiveness.
[/proofplan]
[step:Recall that the projection $\pi_C(x)$ exists and is unique]
Since $C$ is closed and convex in $\mathbb{R}^d$ (a finite-dimensional Hilbert space), and the function $z \mapsto \|x - z\|_2^2$ is continuous and coercive (tending to $+\infty$ as $\|z\|_2 \to \infty$), the infimum $\inf_{z \in C} \|x - z\|_2^2$ is attained at some point $\pi_C(x) \in C$. Uniqueness follows from strict convexity of $z \mapsto \|x - z\|_2^2$: if $z_1, z_2 \in C$ both attain the minimum, then for $t \in (0,1)$ the point $(1-t)z_1 + tz_2 \in C$ satisfies
\begin{align*}
\|x - ((1-t)z_1 + tz_2)\|_2^2 < (1-t)\|x - z_1\|_2^2 + t\|x - z_2\|_2^2 = \inf_{z \in C}\|x - z\|_2^2
\end{align*}
(with strict inequality when $z_1 \neq z_2$ by strict convexity), contradicting the definition of the infimum. Hence $\pi_C(x)$ is the unique closest point in $C$ to $x$.
[/step]
[step:Prove the variational characterisation (part (i))]
Write $\pi := \pi_C(x)$ for brevity. Fix any $z \in C$. Since $C$ is convex, for every $t \in [0,1]$ the point $(1-t)\pi + tz$ lies in $C$. By the minimality of $\pi$:
\begin{align*}
\|x - \pi\|_2^2 \leq \|x - ((1-t)\pi + tz)\|_2^2.
\end{align*}
Rewriting the right-hand side:
\begin{align*}
\|x - \pi + t(\pi - z)\|_2^2 = \|x - \pi\|_2^2 + 2t(x - \pi)^\top(\pi - z) + t^2\|\pi - z\|_2^2.
\end{align*}
Substituting and cancelling $\|x - \pi\|_2^2$ from both sides:
\begin{align*}
0 \leq 2t(x - \pi)^\top(\pi - z) + t^2\|\pi - z\|_2^2.
\end{align*}
For $t \in (0, 1]$, divide both sides by $2t > 0$:
\begin{align*}
0 \leq (x - \pi)^\top(\pi - z) + \frac{t}{2}\|\pi - z\|_2^2.
\end{align*}
This holds for all $t \in (0, 1]$. Taking $t \to 0^+$:
\begin{align*}
0 \leq (x - \pi)^\top(\pi - z),
\end{align*}
which is equivalent to $(x - \pi_C(x))^\top(z - \pi_C(x)) \leq 0$. This holds for all $z \in C$.
[guided]
The geometric interpretation is that the vector $x - \pi_C(x)$ (pointing from the projection to the original point) makes an obtuse angle with every vector $z - \pi_C(x)$ (pointing from the projection to any other point in $C$). Equivalently, the hyperplane through $\pi_C(x)$ with normal $x - \pi_C(x)$ is a supporting hyperplane for $C$: all of $C$ lies in the half-space where the inner product with $x - \pi_C(x)$ is non-positive relative to $\pi_C(x)$.
The proof exploits a standard "first-order optimality" technique for constrained minimisation on convex sets. We perturb $\pi$ toward any feasible point $z \in C$ along the line segment, and the minimality of $\pi$ forces the directional derivative to be non-negative. The directional derivative of $\|x - \cdot\|_2^2$ at $\pi$ in the direction $z - \pi$ is $-2(x - \pi)^\top(z - \pi)$, so requiring it to be non-negative gives exactly $(x - \pi)^\top(z - \pi) \leq 0$.
[/guided]
[/step]
[step:Prove non-expansiveness (part (ii))]
Apply the variational characterisation of part (i) to $x$ with the test point $z_0 := \pi_C(z) \in C$:
\begin{align*}
(x - \pi_C(x))^\top(\pi_C(z) - \pi_C(x)) \leq 0.
\end{align*}
Apply the variational characterisation to $z$ with the test point $z_0 := \pi_C(x) \in C$:
\begin{align*}
(z - \pi_C(z))^\top(\pi_C(x) - \pi_C(z)) \leq 0.
\end{align*}
Add these two inequalities:
\begin{align*}
(x - \pi_C(x))^\top(\pi_C(z) - \pi_C(x)) + (z - \pi_C(z))^\top(\pi_C(x) - \pi_C(z)) \leq 0.
\end{align*}
Denote $\Delta := \pi_C(x) - \pi_C(z)$ for brevity. Rewriting:
\begin{align*}
-(x - \pi_C(x))^\top \Delta + (z - \pi_C(z))^\top(-\Delta) &\leq 0,
\end{align*}
which gives
\begin{align*}
\bigl[(x - \pi_C(x)) - (z - \pi_C(z))\bigr]^\top \Delta &\geq 0.
\end{align*}
Since $(x - \pi_C(x)) - (z - \pi_C(z)) = (x - z) - (\pi_C(x) - \pi_C(z)) = (x - z) - \Delta$, we have
\begin{align*}
(x - z)^\top \Delta - \|\Delta\|_2^2 \geq 0,
\end{align*}
so
\begin{align*}
\|\Delta\|_2^2 \leq (x - z)^\top \Delta.
\end{align*}
Applying the Cauchy--Schwarz inequality in $\mathbb{R}^d$ to the right-hand side:
\begin{align*}
\|\Delta\|_2^2 \leq \|x - z\|_2\,\|\Delta\|_2.
\end{align*}
If $\|\Delta\|_2 = 0$ (i.e., $\pi_C(x) = \pi_C(z)$), the bound $\|\pi_C(x) - \pi_C(z)\|_2 \leq \|x - z\|_2$ holds since the right-hand side is non-negative. If $\|\Delta\|_2 > 0$, divide both sides by $\|\Delta\|_2$:
\begin{align*}
\|\pi_C(x) - \pi_C(z)\|_2 \leq \|x - z\|_2,
\end{align*}
which is the non-expansiveness property.
[guided]
The proof of non-expansiveness is a clean application of the variational characterisation. The two variational inequalities -- one for $x$ tested at $\pi_C(z)$, one for $z$ tested at $\pi_C(x)$ -- produce complementary inner-product bounds. Adding them creates a single inequality involving the difference $\Delta = \pi_C(x) - \pi_C(z)$, which after rearrangement gives $\|\Delta\|_2^2 \leq (x - z)^\top \Delta$.
The Cauchy--Schwarz step is the standard way to pass from a bilinear bound $\|\Delta\|_2^2 \leq (x-z)^\top \Delta$ to a norm bound $\|\Delta\|_2 \leq \|x-z\|_2$: one simply estimates the right-hand side by $\|x-z\|_2\|\Delta\|_2$ and cancels one factor of $\|\Delta\|_2$.
Non-expansiveness says that projecting onto a convex set can only bring points closer together, never push them further apart. This is a fundamental property used in the convergence analysis of projected gradient descent: the projection step does not undo the progress made by the gradient step.
[/guided]
[/step]
