Editing Projection (linear algebra) (section)

== Projections on normed vector spaces ==

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

Many of the algebraic results discussed above survive the passage to this context. A given direct sum decomposition of <math>X</math> into complementary subspaces still specifies a projection, and vice versa. If <math>X</math> is the direct sum <math>X = U \oplus V</math>, then the operator defined by <math>P(u+v) = u</math> is still a projection with range <math>U</math> and kernel <math>V</math>. It is also clear that <math>P^2 = P</math>. Conversely, if <math>P</math> is projection on <math>X</math>, i.e. <math>P^2 = P</math>, then it is easily verified that <math>(1-P)^2 = (1-P)</math>. In other words, <math>1 - P</math> is also a projection. The relation <math>P^2 = P</math> implies <math>1 = P + (1-P)</math> and <math>X</math> is the direct sum <math>\operatorname{rg}(P) \oplus \operatorname{rg}(1 - P)</math>.

However, in contrast to the finite-dimensional case, projections need not be [[bounded linear operator|continuous]] in general. If a subspace <math>U</math> of <math>X</math> is not closed in the norm topology, then the projection onto <math>U</math> is not continuous. In other words, the range of a continuous projection <math>P</math> 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 <math>P</math> gives a decomposition of <math>X</math> into two complementary ''closed'' subspaces: <math>X = \operatorname{rg}(P) \oplus \ker(P) = \ker(1-P) \oplus \ker(P)</math>.

The converse holds also, with an additional assumption. Suppose <math>U</math> is a closed subspace of <math>X</math>. If there exists a closed subspace <math>V</math> such that {{nowrap|1=''X'' = ''U'' ⊕ ''V''}}, then the projection <math>P</math> with range <math>U</math> and kernel <math>V</math> is continuous. This follows from the [[closed graph theorem]]. Suppose {{nowrap|''x<sub>n</sub>'' → ''x''}} and {{nowrap|''Px<sub>n</sub>'' → ''y''}}. One needs to show that <math>Px=y</math>. Since <math>U</math> is closed and {{nowrap|{''Px<sub>n</sub>''} ⊂ ''U''}}, ''y'' lies in <math>U</math>, i.e. {{nowrap|1=''Py'' = ''y''}}. Also, {{nowrap|1=''x<sub>n</sub>'' − ''Px<sub>n</sub>'' = (''I'' − ''P'')''x<sub>n</sub>'' → ''x'' − ''y''}}. Because <math>V</math> is closed and {{nowrap|{(''I'' − ''P'')''x<sub>n</sub>''} ⊂ ''V''}}, we have <math>x-y \in V</math>, i.e. <math>P(x-y)=Px-Py=Px-y=0</math>, which proves the claim.

The above argument makes use of the assumption that both <math>U</math> and <math>V</math> are closed. In general, given a closed subspace <math>U</math>, there need not exist a complementary closed subspace <math>V</math>, although for [[Hilbert space]]s 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 <math>U</math> be the linear span of <math>u</math>. By Hahn–Banach, there exists a bounded [[linear functional]] <math>\varphi</math> such that {{nowrap|1=''φ''(''u'') = 1}}. The operator <math>P(x)=\varphi(x)u</math> satisfies <math>P^2=P</math>, i.e. it is a projection. Boundedness of <math>\varphi</math> implies continuity of <math>P</math> and therefore <math>\ker(P) = \operatorname{rg}(I-P)</math> is a closed complementary subspace of <math>U</math>.