Measure, Integration & Real Analysis, 2021a

Section 8B Orthogonality 225
8.25 Definition convex
• A subset of a vector space is called convex if the subset contains the line
segment connecting each pair of points in it.
• More precisely, suppose V is a vector space and U ⊂ V. Then U is called
convex if
(1 − t) f + tg ∈ U for all t ∈ [0, 1] and all f , g ∈ U.
Convex subset of R 2 . Nonconvex subset of R 2 .
8.26 Example convex sets
• Every subspace of a vector space is convex, as you should verify.
• If V is a normed vector space, f ∈ V, and r > 0, then the open ball centered at
f with radius r is convex, as you should verify.
The next example shows that the distance from an element of a Banach space to a
closed subspace is not necessarily attained by some element of the closed subspace.
After this example, we will prove that this behavior cannot happen in a Hilbert space.
8.27 Example no closest element to a closed subspace of a Banach space
In the Banach space C([0, 1]) with norm ‖g‖ = sup|g|, let
[0, 1]
{
∫ 1
}
U = g ∈ C([0, 1]) : g = 0 and g(1) =0 .
0
Then U is a closed subspace of C([0, 1]).
Let f ∈ C([0, 1]) be defined by f (x) =1 − x. Fork ∈ Z + , let
g k (x) = 1 2 − x + xk
2 + x − 1
k + 1 .
Then g k ∈ U and lim k→∞ ‖ f − g k ‖ = 1 2 , which implies that distance( f , U) ≤ 1 2 .
If g ∈ U, then ∫ 1
0 ( f − g) = 1 2
and ( f − g)(1) =0. These conditions imply that
‖ f − g‖ > 1 2 .
Thus distance( f , U) = 1 2 but there does not exist g ∈ U such that ‖ f − g‖ = 1 2 .
Measure, Integration & Real Analysis, by Sheldon Axler