Measure, Integration & Real Analysis, 2021a

226 Chapter 8 Hilbert Spaces
In the next result, we use for the first time the hypothesis that V is a Hilbert space.
8.28 distance to a closed convex set is attained in a Hilbert space
• The distance from an element of a Hilbert space to a nonempty closed convex
set is attained by a unique element of the nonempty closed convex set.
• More specifically, suppose V is a Hilbert space, f ∈ V, and U is a nonempty
closed convex subset of V. Then there exists a unique g ∈ U such that
‖ f − g‖ = distance( f , U).
Proof First we prove the existence of an element of U that attains the distance to f .
To do this, suppose g 1 , g 2 ,...is a sequence of elements of U such that
8.29 lim ‖ f − g k ‖ = distance( f , U).
k→∞
Then for j, k ∈ Z + we have
8.30
‖g j − g k ‖ 2 = ‖( f − g k ) − ( f − g j )‖ 2
= 2‖ f − g k ‖ 2 + 2‖ f − g j ‖ 2 −‖2 f − (g k + g j )‖ 2
= 2‖ f − g k ‖ 2 + 2‖ f − g j ‖ 2 − 4∥ f − g k + g j ∥
2
≤ 2‖ f − g k ‖ 2 + 2‖ f − g j ‖ 2 − 4 ( distance( f , U) ) 2 ,
where the second equality comes from the parallelogram equality (8.20) and the
last line holds because the convexity of U implies that (g k + g j )/2 ∈ U. Now the
inequality above and 8.29 imply that g 1 , g 2 ,...is a Cauchy sequence. Thus there
exists g ∈ V such that
8.31 lim
k→∞
‖g k − g‖ = 0.
Because U is a closed subset of V and each g k ∈ U, we know that g ∈ U. Now8.29
and 8.31 imply that
‖ f − g‖ = distance( f , U),
which completes the existence proof of the existence part of this result.
To prove the uniqueness part of this result, suppose g and ˜g are elements of U
such that
8.32 ‖ f − g‖ = ‖ f − ˜g‖ = distance( f , U).
Then
8.33
‖g − ˜g‖ 2 ≤ 2‖ f − g‖ 2 + 2‖ f − ˜g‖ 2 − 4 ( distance( f , U) ) 2
= 0,
where the first line above follows from 8.30 (with g j replaced by g and g k replaced
by ˜g) and the last line above follows from 8.32. Now 8.33 implies that g = ˜g,
completing the proof of uniqueness.
∥ 2
Measure, Integration & Real Analysis, by Sheldon Axler