Basic Analysis – Gently Done Topological Vector Spaces

10: Fréchet Spaces 125
linear subspace of X. Furthermore, for any given v ∈ V, there is x ∈ X such that
Px = v. Hence
Pv = P 2 x = Px = v
and we see that (1l−P)v = 0. Hence V = ker(1l−P) and it follows that V is also
a linear subspace of X. Now, any x ∈ X can be written as x = Px+(1l−P)x with
Px ∈ V = ranP and (1l − P)x ∈ W = kerP. We have seen above that for any
v ∈ V, we have v = Pv. If also v ∈ W = kerP then Pv = 0, so that v = Pv = 0.
It follows that V ∩W = {0} and so X = V ⊕W.
Now suppose that X is a topological vector space and that P : X → X is a
continuous linear operator such that P 2 = P. Then both V = ranP = ker(1l−P)
and W = kerP are closed subspaces of X and X = V ⊕W.
Conversely, suppose that X is a Fréchet space and X = V ⊕ W, where V
and W are closed linear subspaces of X. Define P : X → V as above so that
P 2 = P and V = ranP = ker(1l−P) and W = kerP. We wish to show that P
is continuous. To see this we will show that P is closed and then appeal to the
closed-graph theorem. Suppose, then, that x n → x and Px n → y. Now, Px n ∈ V
for each n and V is closed, by hypothesis. It follows that y ∈ V and so Py = y.
Furthermore, (1l − P)x n = x n − Px n → x − y and (1l − P)x n ∈ W for each n
and W is closed, by hypothesis. Hence x−y ∈ W and so P(x−y) = 0, that is,
Px = Py. Hence we have Px = Py = y and we conclude that P is closed. Thus
P is a closed linear operator from the Fréchet space X onto the Fréchet space V.
By the closed-graph theorem, it follows that P is continuous. We have therefore
proved the following theorem.
Theorem 10.29 Suppose that V is a closed subspace of a Fréchet space X. Then
there is a closed subspace W such that X = V ⊕W if and only if there exists a
continuous idempotent P with ranP = V.
Definition 10.30 We say that a closed subspace V in a topological vector space is
complemented if there is a closed subspace W such that X = V ⊕W.
Theorem 10.31 Suppose that V is a finite-dimensional subspace of a topological
vector space X. Then V is closed and complemented.
Proof Let v 1 ,...,v m be linearly independent elements of X which span V. Define
ℓ i : V → K by linear extension of the rule ℓ i (v j ) = δ ij for 1 ≤ i,j ≤ m. By
Corollary 6.24, each ℓ i is a continuous linear functional on V. By the Hahn-
Banach theorem, we may extend each of these to continuous linear functionals on
X, which wewillalsodenoteby ℓ i . Then, ifv ∈ V isgivenby v = t 1 v 1 +···+t m v m ,
we have ℓ i (v) = t i and so
v = ℓ 1 (v)v 1 +···+ℓ m (v)v m .