# Bounded linear functionals, normed vector space proof

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?

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 $$g(b) \neq 0$$.