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:
- $\norm {h - k_0} = \map d {h, K}$
where $d$ denotes distance to a set.
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:
- $\ds \lim_{n \mathop \to \infty} \norm {k_n} = d$
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:
- $\ds \lim_{n \mathop \to \infty} k_n = k_0$
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 $h_0 = k_0$.
$\blacksquare$
Sources
- 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Theorem $2.5$