Projection (linear Algebra) - Projections On Normed Vector Spaces

Projections On Normed Vector Spaces

When the underlying vector space X is a (not necessarily finite-dimensional) normed vector space, analytic questions, irrelevant in the finite-dimensional case, need to be considered. Assume now X is a Banach space.

Many of the algebraic notions discussed above survive the passage to this context. A given direct sum decomposition of X into complementary subspaces still specifies a projection, and vice versa. If X is the direct sum X = U ⊕ V, then the operator defined by P(u + v) = u is still a projection with range U and kernel V. It is also clear that P2 = P. Conversely, if P is projection on X, i.e. P2 = P, then it is easily verified that (I − P)2 = (I − P). In other words, (I − P) is also a projection. The relation I = P + (I − P) implies X is the direct sum Ran(P) ⊕ Ran(I − P).

However, in contrast to the finite-dimensional case, projections need not be continuous in general. If a subspace U of X is not closed in the norm topology, then projection onto U is not continuous. In other words, the range of a continuous projection P must be a closed subspace. Furthermore, the kernel of a continuous projection (in fact, a continuous linear operator in general) is closed. Thus a continuous projection P gives a decomposition of X into two complementary closed subspaces: X = Ran(P) ⊕ Ker(P) = Ran(P) ⊕ Ran(I − P).

The converse holds also, with an additional assumption. Suppose U is a closed subspace of X. If there exists a closed subspace V such that X = U ⊕ V, then the projection P with range U and kernel V is continuous. This follows from the closed graph theorem. Suppose x_n → x and Px_n → y. One needs to show Px = y. Since U is closed and {Px_n} ⊂ U, y lies in U, i.e. Py = y. Also, x_n − Px_n = (I − P)x_n → x − y. Because V is closed and {(I − P)x_n} ⊂ V, we have x − y ∈ V, i.e. P(x − y) = Px − Py = Px − y = 0, which proves the claim.

The above argument makes use of the assumption that both U and V are closed. In general, given a closed subspace U, there need not exist a complementary closed subspace V, although for Hilbert spaces this can always be done by taking the orthogonal complement. For Banach spaces, a one-dimensional subspace always has a closed complementary subspace. This is an immediate consequence of Hahn–Banach theorem. Let U be the linear span of u. By Hahn–Banach, there exists a bounded linear functional Φ such that φ(u) = 1. The operator P(x) = φ(x)u satisfies P2 = P, i.e. it is a projection. Boundedness of φ implies continuity of P and therefore Ker(P) = Ran(I − P) is a closed complementary subspace of U.

However, every continuous projection on a Banach space is an open mapping, by the open mapping theorem.