Measure, Integration & Real Analysis, 2021a

Section 8B Orthogonality 229
Orthogonal Complements
8.38 Definition orthogonal complement; U ⊥
Suppose U is a subset of an inner product space V. The orthogonal complement
of U is denoted by U ⊥ and is defined by
U ⊥ = {h ∈ V : 〈g, h〉 = 0 for all g ∈ U}.
In other words, the orthogonal complement of a subset U of an inner product
space V is the set of elements of V that are orthogonal to every element of U.
8.39 Example orthogonal complement
Suppose U is the set of elements of l 2 whose even coordinates are all 0:
U = {(a 1 ,0,a 3 ,0,a 5 ,0,...) : each a k ∈ F and
∞
∑
k=1
|a 2k−1 | 2 < ∞}.
Then U ⊥ is the set of elements of l 2 whose odd coordinates are all 0:
U ⊥ ∞
= {0, a 2 ,0,a 4 ,0,a 6 ,...) : each a k ∈ F and |a 2k | 2 < ∞},
as you should verify.
∑
k=1
8.40 properties of orthogonal complement
Suppose U is a subset of an inner product space V. Then
(a) U ⊥ is a closed subspace of V;
(b) U ∩ U ⊥ ⊂{0};
(c) if W ⊂ U, then U ⊥ ⊂ W ⊥ ;
(d) U ⊥ = U ⊥ ;
(e) U ⊂ (U ⊥ ) ⊥ .
Proof To prove (a), suppose h 1 , h 2 ,...is a sequence in U ⊥ that converges to some
h ∈ V. Ifg ∈ U, then
|〈g, h〉| = |〈g, h − h k 〉|≤‖g‖‖h − h k ‖ for each k ∈ Z + ;
hence 〈g, h〉 = 0, which implies that h ∈ U ⊥ . Thus U ⊥ is closed. The proof of (a)
is completed by showing that U ⊥ is a subspace of V, which is left to the reader.
To prove (b), suppose g ∈ U ∩ U ⊥ . Then 〈g, g〉 = 0, which implies that g = 0,
proving (b).
To prove (e), suppose g ∈ U. Thus 〈g, h〉 = 0 for all h ∈ U ⊥ , which implies that
g ∈ (U ⊥ ) ⊥ . Hence U ⊂ (U ⊥ ) ⊥ , proving (e).
The proofs of (c) and (d) are left to the reader.
Measure, Integration & Real Analysis, by Sheldon Axler