Hahn-Banach theorem (second geometric form) exercise Let $X$ be a vector normed space and $\{F,F_1,\ldots,F_N\}$ linear functionals over $X$ such that
$$\bigcap_{i=1}^N\mbox{ker}(F_i)\ \subseteq \mbox{ker}(F).$$
Apply the Hahn-Banach theorem (second geometric form)in order to prove that there exists scalars $\alpha_1,\alpha_2,\ldots,\alpha_N$ such that
$$F\ =\ \sum_{i=1}^N\alpha_iF_i.$$
Explain how do you can simplify the above proof, without Hahn-Banach (any form), but using the orthogonality, in the case of $X$ is a Hilbert space and all kernels are not dense on $X$.

Please somebody can help me with this problem? Thanks in advance.
 A: Define $G:X\to\mathbb{R}^{N+1}$ by $$G(x)=(F(x),F_1(x),...,F_N(x))$$
By hypothesis, the point $a=(1,0,...,0)$ does not belong $R(G)$, where $R(G)$ denotes the range of $G$. Now we can separate $a$ and $R(G)$ (note here that we dont need Hanh Banach to separate $a$ and $R(G)$), i.e. there are constants $\alpha,\alpha_1,...,\alpha_N$ and and $\beta\in\mathbb{R}$, such that $$\alpha<\beta<\alpha F(x)+\sum_{i=1}^N\alpha_i F_i(x),\ \forall\ x\in X$$
Fix $x\in X$. If $\alpha F(x)+\sum_{i=1}^N\alpha_i F_i(x)=0$, then we are done. If $\alpha F(x)+\sum_{i=1}^N\alpha_i F_i(x)\neq 0$, then for $\lambda\in\mathbb{R}$, we have that $\alpha F(\lambda x)+\sum_{i=1}^N\alpha_i F_i(\lambda x)=\lambda (\alpha F(x)+\sum_{i=1}^N\alpha_i F_i(x))$. If we let $\lambda\to\infty$ or $\lambda\to-\infty$ we get an absurd, which implies that $\alpha F(x)+\sum_{i=1}^N\alpha_i F_i(x)=0$.
Therefore we have that $\alpha F(x)+\sum_{i=1}^N\alpha_i F_i(x)=0$ for all $x\in X$. Its remains to show that $\alpha\neq 0$, but this is straightforward.
