Finite set of linear functional attaining certain values (Brezis functional analysis exercise 1.12) Problem:
Let $E$ be a vector space. Fix $n$ linear functionals $(f_{i})_{1\leq i\leq n}$
on $E$ and $n$ real numbers $(\alpha_{i})_{1\leq i\leq n}.$ Prove
that the following properties are equivalent.


*

*There exists some $x\in E$ such that $f_{i}(x)=\alpha_{i}$,
$\forall i=1,...,n$.

*For any choice of real numbers $\beta_{1},\beta_{2},...,\beta_{n}$
such that $\sum_{i=1}^{n}\beta_{i}f_{i}=0$, one also has $\sum_{i=1}^{n}\beta_{i}\alpha_{i}=0$.
Attempt 
$(1)\Rightarrow(2)$
This follows from definition. By assumption we can find some $x_{0}\in E$
such that $f_{i}(x)=\alpha_{i}$ for any $i=1,...,n$. Then for any
choice of $\beta\in\mathbb{R}^{n}$ with $\sum_{i=1}^{n}\beta_{i}f_{i}=0,$
we plugin $x_{0}$ and get 
\begin{equation}
\sum_{i=1}^{n}\beta_{i}f_{i}(x_{0})=\sum_{i=1}^{n}\beta_{i}\alpha_{i}=0.
\end{equation}
$(2)\Rightarrow(1)$
I am not sure how to show this. I think the way to go is by argue by contradiction. But I am not getting anywhere but a circular argument. Another way to look at this is that if we assume that if $\cap_{i=1}^{n}\ker{f_i}\subset\ker{g}$, where $g:\mathbb{R^n}\rightarrow\mathbb{R}$ defined map $x\mapsto a^T x$. Then there exists some $x\in E$ such that $f_i(x)=\alpha_i$. Maybe Hahn Banach in its geometric form could be of use here. But I fail to see where to use it. 
 A: Using a geometric form of Hahn-Banach is the right idea. In particular, we will use what Brezis calls the second geometric form of Hahn-Banach.
Let $F = \{ (f_i(x))_{i=1}^n : x \in E\} \subseteq \mathbb{R}^n$. We want to show that $\alpha \in F$. If not, by the second geometric form of Hahn-Banach, $F$ and $\{\alpha\}$ are strictly separated. This implies that there is a $\gamma \in \mathbb{R}^n$ such that for every $x \in E$ we have
$$\sum_{i=1}^n \gamma_i f_i(x) < \sum_{i=1}^n \gamma_i \alpha.$$
This inequality has an asymmetry under rescaling which we can leverage. By replacing $x$ with $\lambda x$ for $\lambda \in \mathbb{R}$ we see that
$$\lambda \sum_{i=1}^n \gamma_i f_i(x) < \sum_{i=1}^n \gamma_i \alpha \qquad \forall \lambda \in \mathbb{R} \quad \forall x \in E.$$
This can only be true if $\sum_{i=1}^n \gamma_i f_i = 0$, else we can pick an $x$ such that the sum on the left hand side is non-zero and let $\lambda \to \pm \infty$. 
Then, by our assumption, we have $\sum_{i=1} \gamma_i \alpha_i = 0$. This contradicts the strict separation so we must have that $\alpha \in F$, as desired.
