## Abstract

Let This number can be finite or infinite. On the assumption that def .

*F*_{1}be a subspace of the Banach space*F*. The dimension of the orthogonal complement*F*_{1}^{⊥}is called the*defect*or the*codimension*of the subspace*F*_{1}:$$ def{F_1} = \dim F_1^ \bot $$

*F*_{1}= n < ∞, let g_{1},...,g_{n}be a basis for*F*_{1}^{⊥}If u_{1}∉*F*_{1}, then one can construct a linear functional Φ_{1}, ∈*F*_{1}^{*}such that Φ_{1}(u_{1}) = 1 and Φ_{1}(y) = 0 for all y ∈*F*_{1}. Then Φ_{1}∈ F_{1}^{⊥}. Now let u_{2}be an element which is not in the linear span of the subspace*F*_{1}together with the element u_{1}(which is again a subspace of*F*), and choose Φ_{2}∈(F_{1}^{⊥}such that Φ_{2}(u_{2})**=**1, Φ_{2}(u_{1}) = 0. This process may be continued, and clearly it will have at most n steps: the functionals Φ_{1}. are linearly independent (if \( \sum {{c_i}\phi (z) = 0} \) for all z ∈*F*, then taking successively z = u_{1}, z = u_{2},..., we obtain c_{1}= 0, c_{2}= 0,...) and are all in*F*_{1}^{⊥}. In fact, the number of steps is exactly n. Indeed, if the process terminates at the m-th step, then one can write every element z ∈*F*as$$ z = \sum\nolimits_k^m {1{}^\alpha k} {}^uk + y\;\;(y \in {F_1}) $$

## Keywords

Banach Space Orthogonal Complement Linear Span Linear Manifold Null Solution## Preview

Unable to display preview. Download preview PDF.

## Copyright information

© Birkhäuser Boston, Inc. 1982