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Let $X$ be a normed space and $L$ a finite dimensional linear subspace.

I need to show that there exists finitely many $l_1,.....l_n \in L$ and $f_1,.....f_n \in X'$ such that $$ l = \sum_{i=1}^{n} f_i(l)l_i$$ for all $l \in L$ Since L is finite it has as basis, so there are $l_1,.....l_n \in L$ with coefficients $f_1,.....f_n$ depending on $l$ such, that we have $$ l = \sum_{i=1}^{n} f_i(l)l_i$$

But how do I show that those $f_i$ are in $X'$??

More I need to show that there exists a continuous projection $P: X \to L$.

So I need to show that $P(X)=L$ and $P^2=P$. Can I define the projection the following? $$ Px = \sum_{i=1}^{n} f_i(x)x_i$$

P continuous: $$||\sum_{i=1}^{n}f_i(x)l_i||_L \leq \sum_{i=1}^{n}||f_i(x)l_i||_L \leq \sum_{i=1}^{n}\|f_i\|_{X'}\|x\|_X \|l_i\|_L \leq .....$$

I am very thankful for help?

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    $\begingroup$ Define your $f_i$ on $L$ and use Hahn-Banach. For the second task you have the right idea, just replace $x_i$ with $l_i$, what should the $x_i$ even be? $\endgroup$
    – Lukas Betz
    Commented Dec 11, 2016 at 13:35
  • $\begingroup$ Thank you! @LeBtz Why can I assume that the $f_i$ are in L'? $\endgroup$
    – Yuhe
    Commented Dec 11, 2016 at 13:48
  • $\begingroup$ Because $L$ is finite-dimensional. $\endgroup$
    – Lukas Betz
    Commented Dec 11, 2016 at 13:54
  • $\begingroup$ AH thanks, can you help me why P is a continuous projection then? $\endgroup$
    – Yuhe
    Commented Dec 11, 2016 at 14:05
  • $\begingroup$ Yeah I tried to show continuous with the definition of bounded operators. Could be correct, but I am not sure, but I dont know how to show that it is a projection? $\endgroup$
    – Yuhe
    Commented Dec 11, 2016 at 14:21

1 Answer 1

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All that is left is, to show that $P^2 = P$.

Since the $f_i$ are defined on $L$ in such a way that $f_i(x)$ is the unique coefficient of $l_i$ in the representation $x = f_1(x)l_1+...+f_n(x)l_n$ and we have $l_i = 0l_1+...+1l_i+...+0l_n$ we have $f_i(l_j) = 1$ if $j=i$ and $f_i(l_j) = 0$ if $i\neq j$.

This yields $$P^2x = P(\sum_{i=1}^nf_i(x)l_i) = \sum_{j=1}^nf_j\left(\sum_{i=1}^nf_i(x)l_i\right)l_j = \sum_{j=1}^n\sum_{i=1}^nf_i(x)f_j(l_i)l_j = \sum_{j=1}^nf_j(x)l_j = Px,$$

where we used that each summand vanishes unless $i=j$ in the second last equality.

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