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Suppose you have a normed vector space $\mathcal{X}$, $n$ bounded linear functionals $f_i \in\mathcal{X}^*$ where $\mathcal{X}^*$ is the space of bounded linear functionals on $\mathcal{X}$ and a vector $b \in \mathbb{R}^n$.

What is a good way to prove that:

For any $\lambda \in \mathbb{R}^n$, $\sum_{i=1}^n \lambda_i f_i = 0 \implies \sum_{i=1}^n \lambda_ib_i = 0$ if and only if there exists a vector $x\in\mathcal{X}$ such that $f_i(x) = b_i$ for all $i \in \{1,...,n\}$.

This statement feels pretty intuitive to me, but I am struggling with the $\Rightarrow$ direction. I wanted to do something like $\sum_{i=1}^n \lambda_i(f_i - b_i) = 0 \implies f_i = b_i$ but I think this approach is wrong. I'm also not sure what the significance of having bounded linear functionals $f_i$ in the question is, could someone please explain how that would be useful in the proof?

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Define $T: X\to \mathbb R^{n}$ by $T(x)=(f_1(x),f_2(x),...,f_n(x))$. If $b =Tx$ for some $x$ then, for any linear map $g:\mathbb R^{n} \to \mathbb R $, $g(Tx)=0$ implies $g(b)=0$. This proves one way. [Just note $g(t_1,t_2,...,t_n)$ is of the form $\sum \lambda_i t_i$].

For the other direction we have to use the fact that if $b$ is not in the range of $T$ then there exists a linear map $g:\mathbb R^{n} \to \mathbb R $ $g$ such that $g=0$ on the range of $T$ but $g(b) \neq 0$.

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