# Unique Point of Minimal Distance

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## 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:

$\norm {h - k_0} = \map d {h, K}$

where $d$ denotes distance to a set.

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

$\norm {h - k_0} = \map d {h, K} \iff \paren {h - k_0} \perp K$

where $\perp$ signifies orthogonality.

## Proof

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

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

$\map d {h, k} = \norm {h - k} = \map d {\mathbf 0_H, k - h}$

it follows that:

$\map d {h, K} = \map d {\mathbf 0_H, K - h}$

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

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

$\norm {k_0} = \map d {\mathbf 0_H, K} = \inf \set {\norm k : k \in K}$

Let $d = \map d {\mathbf 0_H, K}$.

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

$\displaystyle \lim_{n \mathop \to \infty} \norm {k_n} = d$

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

$(1): \quad \norm {\dfrac {k_n - k_m} 2 } = \dfrac 1 2 \paren {\norm {k_n}^2 + \norm {k_m}^2} - \norm {\dfrac {k_n + k_m} 2 }^2$

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

Hence:

$\norm {\dfrac {k_n + k_m} 2 }^2 \ge d^2$

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

$\norm {k_n}^2 < d^2 + \epsilon$

From $(1)$, it follows that:

$\norm {\dfrac {k_n - k_m} 2} < d^2 + \epsilon - d^2 = \epsilon$

and hence that $\sequence {k_n}_{n \mathop \in \N}$ 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:

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

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

This demonstrates existence of $k_0$.

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

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

This implies that $\norm {\dfrac {h_0 + k_0} 2} \ge d$.

Now from the Triangle Inequality:

$\norm {\dfrac {h_0 + k_0} 2} \le \dfrac {\norm {h_0} + \norm {k_0} } 2 = d$

meaning that $\norm {\dfrac {h_0 + k_0} 2} = d$.

Thus, the Parallelogram Law implies that:

$d^2 = \norm {\dfrac {h_0 + k_0} 2}^2 = d^2 - \norm {\dfrac {h_0 - k_0} 2}^2$

from which we conclude that $h_0 = k_0$.

$\blacksquare$