How to prove $f$ is in the span of $f_{1,\cdots, n}$? Suppose $f, f_{1,\cdots,n}\in X^*$ where $X$ is a normed vector space. How to show that, if $\cap Ker f_i \subset Ker f$, then $f$ is a linear combination of $f_{1,\cdots,n}$?
I'm at a loss even for the classical linear algebra cases (where $X$ is a finite dimensional Euclidean spade with $\ell^2$ norm, so really don't know how to proceed for the more general cases. 

EDIT  It seems I might want to relax the conditions a little here. So let's now assume $X$ is a Banach space. 


EDIT The Banach space assumption can be dropped. Any normed vector space is okay. 
 A: Another proof by my teacher: define $E=\{(f_1(x),\cdots,f_n(x)), x\in X\}$, a subspace of $K^n$ ($K$ denotes the real or complex field), and define $g:E\to K$ as $g(f_1(x),\cdots,f_n(x)):=f(x)$ (well defined by the kernel inclusion). Now we can extend $g$ to $g':K^n\to K$ since $E$ being a finite dimensional subspace is closed, and then 
$$f(x)=g(f_1(x)e_1+\cdots+f_n(x)e_n)=g'(f_1(x)e_1+\cdots+f_n(x)e_n)=\sum f_i(x)g'(e_i).$$
A: Notice that we can reduce the problem to a finite dimensional $X$ by dividing out the closed subspace $K=\bigcap_{i=1}^n \ker f_i$.
Let $\mathbb F$ be the field of real or complex numbers.

*

*Let $\bar f_1, \dotsc, \bar f_n, \bar f : X/K \to \mathbb C$ denote $f_1, \dotsc, f_n, f$ modulo $K$, that is
$$ \bar f_i(x+K) = f_i(x) $$
for every $x\in X$ and so on.
The linear map $g:X/K \to \mathbb F^n$ defined by $g_i = \bar f_i$ has trivial null space and hence is injective.
Thus, $X/K$ must be finite dimensional.
Assume there is $\alpha\in \mathbb F^n$ with
$$ \sum_{i=1}^n \alpha_i \bar f_i = \bar f. $$
Then, for the same $\alpha$ we have
$$ \sum_{i=1}^n \alpha_i f_i = f. $$


*Now the proof for the finite dimensional case:
Let $F = span\{ f_1, \dotsc, f_n \}$. Then, $F$ is closed. Assume the converse, that is $f\notin F$. By hahn banach separation theorem there is a $\lambda\in X^{**}$ (isomorph to $X$) such that
$$ Re\lambda(f) < \inf_{\alpha\in\mathbb F^n} Re\lambda\left(\sum_{i=1}^n \alpha_i f_i \right) = \inf_{\alpha\in\mathbb F^n} \sum_{i=1}^n Re(\alpha_i \lambda(f_i)). $$
That implies $\lambda(f_i) = 0$ for every $i$ (otherwise consider $\alpha_{i,n} = -n\overline{\lambda(f_i)}$).
That is
$$ \lambda\in \bigcap_{i=1}^n \ker f_i \subseteq \ker f. $$
Clearly a contradiction.
