Given a set $K\subseteq R^d$, we define the projection operator $\pi_K$ as follows for any $x \epsilon R^d$:
$\pi_K(x) = arg min_{y\epsilon K} ||x - y||^2$
That is, $\pi_K(x)$ is the set of closest points in K to x.
(a) Let $K\subseteq R^d$ be a closed and bounded set. Prove that if K is convex, then the projection $\pi_K(x)$ is a singleton (i.e. $|\pi_K(x)| = 1$) for all $x \epsilon R^d$
(b) If $K = \{x : ||x||_2 \leq 1\}$, then for $x \neq K$, show that $\pi_K(x) = \frac{x}{||x||_2} $.
a) $K$ bounded and closed in Euclidean space is equivalent to $K$ compact. Since $K$ is compact, $\pi_K(x)$ is non-empty for any $x$. Now take two elements $y_1, y_2$ of $\pi_K(x)$. Since $K$ is convex, and Euclidean distance is a convex function, then any convex combination $y_3$ of $y_1, y_2$ must be at least as close to $x$ as, say, $y_1$. For example, put $y_3 = \frac{y_1 + y_2}{2}$. But by definition, $y_1, y_2$ achieve the minimum distance, whence the only possibility is that $\| y_1 - x \| = \| y_3 - x \| = \| y_2 - x \|$. The only way that two points and their midpoint are at the same distance from a fixed point $x$ is that they are all the same. We picked two elements of $\pi_K(x)$ and found they must be the same, so it must be a singleton.
b) Fix $x$ and draw the circumference with center at $x$ and radius equal to $\min_{y \in K} \| y - x\|$. By the previous argument, since $K$ is a disk and therefore compact, $\pi_K(x)$ is a singleton and therefore the circumference you just drew is tangent to $K$. Now draw the line joining the origin and $x$. Can you take it from there?