Unique Point of Minimal Distance to Closed Convex Subset of Hilbert Space

Theorem
Let $H$ be a Hilbert space, and let $h \in H$.

Let $K \subseteq H$ be a closed, convex, non-empty subset of $H$.

Then there is a unique point $k_0 \in K$ such that:


 * $\left\|{h - k_0}\right\| = d \left({h, K}\right)$

where $d$ denotes distance to a set.

Furthermore, if $K$ is a linear subspace, this point is characterised by:


 * $\left\|{h - k_0}\right\| = d \left({h, K}\right) \iff \left({h - k_0}\right) \perp K$

where $\perp$ signifies orthogonality.

Proof
Let $\mathbf 0_H$ be the zero of $H$.

Since for every $k \in K$, we have:


 * $d \left({h, k}\right) = \left\|{h - k}\right\| = d \left({\mathbf 0_H, k - h}\right)$

it follows that:


 * $d \left({h, K}\right) = d \left({\mathbf 0_H, K - h}\right)$

Therefore, we may assume without loss of generality that $h = \mathbf 0_H$.

The problem has therefore reduced to finding $k_0 \in K$ such that:


 * $\left\|{k_0}\right\| = d \left({\mathbf 0_H, K}\right) = \inf \left\{{\left\|{k}\right\| : k \in K}\right\}$

Let $d = d \left({\mathbf 0_H, K}\right)$.

By definition of infimum, there exists a sequence $\left({k_n}\right)_{n \in \N}$ such that:


 * $\displaystyle \lim_{n \to \infty} \left\|{k_n}\right\| = d$

By the Parallelogram Law, we have that for all $m, n \in \N$:


 * $(1): \qquad \left\|{ \dfrac {k_n - k_m} 2 }\right\| = \dfrac 1 2 \left({\left\|{k_n}\right\|^2 + \left\|{k_m}\right\|^2}\right) - \left\|{ \dfrac {k_n + k_m} 2 }\right\|^2$

Since $K$ is convex, $\dfrac {k_n + k_m} 2 \in K$.

Hence:


 * $\left\|{ \dfrac {k_n + k_m} 2 }\right\|^2 \ge d^2$

Now given $\epsilon > 0$, choose $N$ such that for all $n \ge N$:


 * $\left\|{k_n}\right\|^2 < d^2 + \epsilon$

From $(1)$, it follows that:


 * $\left\|{ \dfrac {k_n - k_m} 2 }\right\| < d^2 + \epsilon - d^2 = \epsilon$

and hence that $\left({k_n}\right)_{n \in \N}$ is a Cauchy sequence.

Since $H$ is a Hilbert space and $K$ is closed, it follows that there is a $k_0 \in K$ such that:


 * $\displaystyle \lim_{n \to \infty} k_n = k_0$

From Norm is Continuous, we infer that $\left\{{k_0}\right\| = d$.

This demonstrates existence of $k_0$.

For uniqueness, suppose that $h_0 \in K$ has $\left\|{h_0}\right\| = d$.

Since $K$ is convex, it follows that $\dfrac {h_0 + k_0} 2 \in K$.

This implies that $\left\|{\dfrac {h_0 + k_0} 2}\right\| \ge d$.

Now from the Triangle Inequality:


 * $\left\|{\dfrac {h_0 + k_0} 2}\right\| \le \dfrac {\left\|{h_0}\right\| + \left\|{k_0}\right\|} 2 = d$

meaning that $\left\|{\dfrac {h_0 + k_0} 2}\right\| = d$.

Thus, the Parallelogram Law implies that:


 * $d^2 = \left\|{\dfrac {h_0 + k_0} 2}\right\|^2 = d^2 - \left\|{\dfrac {h_0 - k_0} 2}\right\|^2$

from which we conclude that $h_0 = k_0$.