Measure, Integration & Real Analysis, 2021a

230 Chapter 8 Hilbert Spaces
The results in the rest of this subsection have as a hypothesis that V is a Hilbert
space. These results do not hold when V is only an inner product space.
8.41 orthogonal complement of the orthogonal complement
Suppose U is a subspace of a Hilbert space V. Then
U =(U ⊥ ) ⊥ .
Proof Applying 8.40(a) to U ⊥ , we see that (U ⊥ ) ⊥ is a closed subspace of V. Now
taking closures of both sides of the inclusion U ⊂ (U ⊥ ) ⊥ [8.40(e)] shows that
U ⊂ (U ⊥ ) ⊥ .
To prove the inclusion in the other direction, suppose f ∈ (U ⊥ ) ⊥ . Because
f ∈ (U ⊥ ) ⊥ and P U
f ∈ U ⊂ (U ⊥ ) ⊥ (by the previous paragraph), we see that
f − P U
f ∈ (U ⊥ ) ⊥ .
Also,
by 8.37(a) and 8.40(d). Hence
f − P U
f ∈ U ⊥
f − P U
f ∈ U ⊥ ∩ (U ⊥ ) ⊥ .
Now 8.40(b) (applied to U ⊥ in place of U) implies that f − P U
f = 0, which implies
that f ∈ U. Thus (U ⊥ ) ⊥ ⊂ U, completing the proof.
As a special case, the result above implies that if U is a closed subspace of a
Hilbert space V, then U =(U ⊥ ) ⊥ .
Another special case of the result above is sufficiently useful to deserve stating
separately, as we do in the next result.
8.42 necessary and sufficient condition for a subspace to be dense
Suppose U is a subspace of a Hilbert space V. Then
U = V if and only if U ⊥ = {0}.
Proof
First suppose U = V. Then using 8.40(d), we have
U ⊥ = U ⊥ = V ⊥ = {0}.
To prove the other direction, now suppose U ⊥ = {0}. Then 8.41 implies that
U =(U ⊥ ) ⊥ = {0} ⊥ = V,
completing the proof.
Measure, Integration & Real Analysis, by Sheldon Axler