Prove that the a linear functional is part of span I am having difficulties proving the following result.
Given a vector space $X$ and the linear functionals $f, f_1, f_2,\dotsc ,f_n$, prove that $f$ belongs to span$\{f_1,f_2,\dotsc,f_n\}$ if and only if $\bigcap\limits_{j=1}^n\ker f_j \subseteq \ker f$.
I managed to prove the statement given $f$ as a linear combination of $f_j$'s, but that part seemed way easier. Could anyone please give me a hint on how to prove the other implication? Thank you!
Note: Question from link makes me think I should use the fact that if $f,g$ are two functionals such that $\ker g\subseteq\ker f$, then $f=\alpha g$ for some $\alpha \in \mathbb{F}$, where $\mathbb{F}$ is the field of $X$, but I fail to see any solutions.

Edit: I also found another solution for exactly the same problem, but it looks more of a trick which does not reveal many mathematical structures. I prefer G. Chiusole's answer as it is more elegant in my opinion and with less technical requirements for a reader to understand.
 A: First assume by passing to a subset that $f_1,...,f_n$ are linearly independent, this is no restriction for the direction you want.
Next check that $X/\bigcap_{i=1}^n \ker(f_i)$ is $n$-dimensional. This works by noting that $\ker(f_i)$ has co-dimension $1$ and choosing an $x_i$ so that $X= \ker(f_i)\oplus \Bbb Fx_i$. Then $X= \bigcap_i\ker(f_i)\oplus\bigoplus_i \Bbb F x_i$ (unless one of the $x_i$ is a linear combination of the other $x_i$, but this contradicts the linear independence of the $f_i$).
Now each $f_i$ induces a linear map $[f_i]:X/\bigcap_{i=1}^n \ker(f_i)\to \Bbb F$. Since the $f_i$ are linearly independent the $[f_i]$ are also independent, then by dimensional reasons they must form a basis of the dual to $X/\bigcap_{i=1}^n\ker(f_i)$. In particular since $\ker(f)\supseteq \bigcap\ker(f_i)$ you also get a well defined induced map $[f]:X/\bigcap\ker(f_i)\to\Bbb F$ so you must have
$$[f]=\sum_i a_i [f_i]$$
for some constants $a_i\in\Bbb F$.
Now make an identification $X\cong \bigcap\ker(f_i) \oplus X/\bigcap\ker(f_i)$ and note that $f$ is $0$ on the first summand and $[f]$ on the second. Hence you get that $f=\sum_i a_i f_i$.
A: Assume that $\{f_1, \ldots, f_n\}$ are linearly independent - otherwise reduce to that case. Now define the function 
$$F: X \rightarrow \mathbb{F}^{n+1}, x \mapsto (f(x), f_1(x), \ldots, f_n(x))$$
Now consider $(1,0, \ldots, 0)$. By assumption, this element is not in the image and thus $\text{Im}(F)$ is a proper linear subspace. Thus we may define a linear functional 
$$\Theta: \mathbb{F}^{n+1} \rightarrow \mathbb{F}, (x,x_1, \ldots, x_n) \mapsto a x + \sum_{i = 1}^n a_i x_i, ~~~~ a, a_1, \ldots, a_n \in \mathbb{F} $$
with kernel $\ker(\Theta) = \text{Im}(F)$. This is possible since $\mathbb{F}^{n+1}$ has finite dimension and thus admits a Hilbert space structure. Since $\text{Im}(F)$ is a proper subspace, we may choose $\Theta$ to be the projection onto the orthogonal complement of $\text{Im}(F)$ and by Riesz Theorem, we have $\forall x \in \mathbb{F}^{n+1}: \Theta(x) = \langle x, \alpha \rangle$ where $\alpha = (a, a_1, \ldots, a_n)$. 
Thus we have 
$$\forall x \in X: a f(x) + \sum_{i = 1}^n a_i f(x) = 0$$
so since the $f_i$ were chosen to be linearly independent, we have 
$$\forall x \in X: f(x) = -\frac{1}{a} \sum_{i = 1}^n a_i f(x)$$
which concludes the proof.
