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Let $Y$ be a finite dimensional subspace of a real NLS $X$ and let $Y=span \{y_1,...,y_n\} $ ; then by Hahn-Banach theorem , we can find continuous linear functionals $l_1,...,l_n \in X^*$ such that $l_j(y_k)=\delta_{jk} ; j,k=1,...,n $ ( where $\delta_{ij}$ is the Kronecker delta ) , and consider $Z_j :=\ker l_j$ , then each $Z_j$ is a closed linear subspace , then so is $Z=\cap_{j=1}^n Z_j$ , now I can prove that $Y\cap Z=\{O\}$ my question is , Is it true that $X=Y+Z$ ?

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Yes. We need to express $x\in X$ as a sum $x= y+z$ where $y\in Y$ and $z\in Z$.

Fix $x\in X$. By choice of the $l_j$'s, letting $c_j=l_j (x)$, and writing $y = \sum_j c_j y_j$, it follows that for any $j_0$,

$$l_{j_0} (y) = c_{j_0} l_{j_0}(y_{j_0})+ \sum_{j\ne j_0} l_{j_0} (c_j y_j)= c_{j_0} =l_{j_0} (x).$$

Now $z= x-y$, then from the equality above, we have that $ l_{j_0} (z) =0$ for all $j_0$, that is $z \in Z$.

This completes the proof.

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Let $x\in Y$. Then $x=\langle l_1, x\rangle y_1 + \ldots + \langle l_n, x\rangle y_n$. If $x\in Z$, then $\langle l_1, x\rangle = \ldots = \langle l_n, x\rangle = 0$, so $x=0$. The reverse inclusion is immediate, so $Y\cap Z=\{0\}$. Now, define a map $P\colon X\to X$ by $$Px = \langle l_1, x\rangle y_1 + \ldots + \langle l_n, x\rangle y_n\;(x\in X).$$ Clearly $Y$ is the range of $P$ and $P^2=P$. We have $I=P+(I-P)$ and the range of $I-P$ is precisely $Z$.

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  • $\begingroup$ The norm of P here is n right? Can we get a projection of norm 1 ? $\endgroup$ – Sara Suradi Aug 11 '17 at 18:23
  • $\begingroup$ I mean if Y is finite dimensional normed space of a Banach space , then can we say that Y is complemented by a projection that has norm equal to one ? Can we find such a projection $\endgroup$ – Sara Suradi Aug 11 '17 at 18:25
  • $\begingroup$ @SaraSuradi, no. If every 2-dimensional subspace is 1-complemented, then the space is isometric to a Hilbert space. The best constant you can get in general is $\sqrt{\dim Y}$ (this is the Kadets-Snobar theorem). $\endgroup$ – Tomek Kania Aug 12 '17 at 10:23
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    $\begingroup$ Thanks for your valuable answer 🌷 $\endgroup$ – Sara Suradi Aug 12 '17 at 13:06

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