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 $X=Y+Z$ ?
[Math] Does every finite dimensional subspace of any normed linear space have a closed linear complement
functional-analysislinear algebranormed-spaces
Best Answer
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.