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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.

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This is a fact from linear algebra. No need to use Hahn-Banach, a norm or a Hilbert space structure on $X$. – Martin Apr 30 '13 at 19:48
That's right, but exercise explicitly requires use hahn-banach. Thank you – FASCH Apr 30 '13 at 19:51
But the talk about a reference there. Thank you – FASCH Apr 30 '13 at 19:53
I got that. To spare you the trouble of looking it up in Brézis (Lemma 3.2, page 64): The proof given there is the same as the one given by Tomás. – Martin Apr 30 '13 at 20:06
up vote 4 down vote accepted

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.

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Thanks so much, for the answer – FASCH Apr 30 '13 at 20:04
The last part is not answer, that why I have submitted it as a new question (…) Thanks for the help. – FASCH May 1 '13 at 22:22
@FASCH: It seems to me that the person who asked the question had a very specific argument in mind, so you should ask them what it was. // Since the argument here is only linear algebra taking place in $\mathbb{R}^{N+1}$, the last part of the question simply makes no sense in the context of Tomás's answer. As I wrote in a comment, norms and scalar products on $X$ and any assumptions on the $F_i$ beyond the condition on the kernels are simply not needed (and using them would feel misguided), as the statement in the question is a consequence of a purely algebraic fact. – Martin May 2 '13 at 6:11

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