Corollary 4', sec 8.2., Functional Analysis, Lax I have some questions concerning this corollary. I will present the statement, the proof, and my issues and finally my thoughts/speculations upon the issues. 

Statement:
Every finite-dimensional subspace Y of a normed linear space X has a closed complement.
Proof:
$y_1, y_2,..., y_n$ a basis for Y.
There exists $n$ linear functionals $l_1, ..., l_n$ such that $l_j(y_k) = \delta_{jk}$. 
The nullspaces $N_1, N_2, ...N_n$ are closed and hence $N = \cap_{k=1}^n N_k$ is closed. It is easy to check that $X = Y \oplus N$, QED.
My issues: 


*

*What is the point of $n$ linear functionals?

*Are all $l_j$ vanishing on the complement of $Y$? It is not clear from the proof how the extensions look like.


My thoughts:


*

*Let $x_k = (l_k(y_1), ..., l_k(y_n))$, for $k = 1,...,n$, be a linearly independent set. Then any other $l|_Y$ can be written as a linear combination of $\{l_k\}_{k=1}^n$. 


Indeed, let $x = (l(y_1), ..., l(y_n))$. This vector can be written uniquely as a linear combination of $\{x_k\}_{k=1}^n$. Suppose $x = \sum b_j x_j$. We claim that $l = \sum_j b_j l_j$. Let $y = \sum a_k y_k$. We have:
$\sum_j b_jl_j (\sum_k a_k y_k) = \sum_j \sum_k b_j a_k l_j(y_k) = \sum_k \sum_j b_j a_k l_j(y_k) = \sum_k a_kl(y_k) = l(y)$.
Any $l|_Y$ has kernel $N$. 


*I think this must be the case. Otherwise it can't be true that $X = Y \oplus N$. 


In summary, we need $n$ linear functionals so that $N$ stays out of $Y$, i.e. $N \cap Y = \{0\}$. We still need that these linear functionals vanishing on the complement of $Y$. 

Many thanks in advance for your responses. 
 A: The construction tells the story. Every $x \in X$ can be written as
$$
          x = \left(x-\sum_{j=1}^{n}l_j(x)y_j\right)+\sum_{n=1}^{n}l_j(x)y_j.
$$
By design of the $l_k$, you have $l_k(y_j)=0$ for $j\ne k$ and $l_k(y_k)=1$. Thus,
\begin{align}
           &l_k(x-\sum_{j=1}^{n}l_j(x)y_j)\\
           &=l_k(x)-\sum_{j=1}^{n}l_j(x)l_k(y_j) \\
           &=l_k(x)-l_k(x) = 0,\;\;\; k=1,2,\cdots,n.
\end{align}
Therefore
$$
       \left(x-\sum_{j=1}^{n}l_j(x)y_j\right) \in N=\bigcap_{j=1}^{n}\mathcal{N}(l_j),\;\;\;\;
          \sum_{j=1}^{n}l_j(x)y_j \in Y.
$$
It's not so hard to check that $X=N\oplus Y$; all that remains is to show that $N\cap Y=\{0\}$.
The importance of the linear functionals $l_j$ is that they are bounded on $L$ and, by the Hahn-Banach theorem, can be chosen so that they are bounded on the whole space. Therefore $\mathcal{N}(l_j)$ is closed, which means that $N$ is closed. So $L$ has a closed complement. It is worth mentioning that $Px=\sum_{j=1}^{n}l_j(x)x_j$ is a projection operator, i.e., $P^{2}=P$. So $Q=I-P$ is also a projection operator. Projection operators and direct sum decompositions are related in this way.
A: My idea is, an element is not in $Y$  iif it is in the kernal for $\ell_{j}$ for all $j$. Each such kernal is closed and hence so is any finite intersection of such kernals. I suppose the functionals has to be zero on the complement for this argument to work. The point of the functionlas would be imo to tell if an element is in $Y$ or not.
