Hahn-Banach theorem (second geometric form) exercise #2 Let $X$ be a Hilbert space and $\{F,F_1,\ldots,F_N\}$ linear functionals over $X$ such that
$$\bigcap_{i=1}^N\mbox{ker}(F_i)\ \subseteq \mbox{ker}(F),$$
and any kernel of the involved functionals is not dense on $X$. Using the orthogonality (not Hahn-Banach) in order to prove that there exists scalars $\alpha_1,\alpha_2,\ldots,\alpha_N$ such that
$$F\ =\ \sum_{i=1}^N\alpha_iF_i.$$

This is the unanswered last part of the question
Hahn-Banach theorem (second geometric form) exercise
Thanks in advance.
 A: In a Hilbert space $X$ every (bounded) linear functional $F$ is of the form $\langle x,-\rangle$ for some $x\in X$, and thus $\ker F=x^\perp$. 
(Because if $F\ne 0$, there is an $x_0$ such that $F(x_0)=\|F\|^2$, and project it orthogonally to the (closed subspace) $\ker F$.)
Then we can write $F_i=\langle x_i,-\rangle$ and we have
$$\bigcap_i\,\ker F_i={\rm span}(x_1,..,x_N)^\perp$$
so $x^\perp\supseteq {\rm span}(x_1,..,x_N)^\perp$, therefore $x\in {\rm span}(x_1,..,x_N)$.
A: First of all, note that you can assume $F_i$ to be linearly independent. If not, just argue that removing from the list a linear combination of other functionals does not change things.
Next, show that you can find $x_i \in X$ such that $F_i(x_j) = \delta_{ij}$. This can be done by induction on the number of functionals. Case $N=1$ is simple enough. To get from $N-1$ to $N$, consider the projection $P(x) := x - \sum_{i<N} x_i F_i(x)$. With this definition, $F_i(P(x)) = 0$ for all $x$. I claim that there is $x$ such that $F_N(P(x)) \neq 0$: otherwise we have $0 =  F_N(P(x)) = F_N(x) - \sum_{i<N} F_N(x_i) F_i(x)$, so $F_N$ is linear combination of $F_i$ with $i<N$. So, after rescaling, we can find $x$ such that $F_N(P(x)) = 1$. We can put $x_N = P(x)$, and check it works. This finishes the inductive proof.
Now that we have the sought $x_i$, consider again the projection $P(x) := x - \sum_{i\leq N} x_i F_i(x)$. For any $x$ we have $P(x) \in \ker F_i$, so by the assumption we also have $P(x) \in \ker F$. But now again we have $0 =  F(P(x)) = F(x) - \sum_{i<N} F(x_i) F_i(x)$, so $F$ is a combination of $F_i$, as desired.
Edit: The final step is essentially the same as applying the inductively proved claim + supposition that $F$ is not in linear span of $F_i$, to the sequence of functionals $F_1,F_2,\dots,F_N,F_{N+1}:=F$. The $x_{N+1}$ that comes out is in $\bigcap_{i=1}^N \ker F_i$ but not in $\ker F_{N+1} = \ker F$, which is a contradiction.
