Prove $\bigcap_{i=1}^k\ker(f_i)\subset \ker(f)\iff f\in {\rm span}(f_1,...,f_k) $ Let $V$ a $\mathbb{K}$-vector space of finite dimension $n$, with $\{f_1,...,f_n\}$ a set linearly independent of $V^*$ and $f\in V^*$.
Prove $\bigcap_{i=1}^k\ker(f_i)\subset \ker(f)\iff f\in {\rm span}(f_1,...,f_k).$
($\Leftarrow$) Let $f\in {\rm span}(f_1,...,fk)$ then exists $\alpha_1,...,\alpha_k$ such that $f=\alpha_1f_1+...+\alpha_kf_k$.
Let $f\in\bigcap_{i=1}^k\ker(f_i) $ then $f\in \ker(f_1),...,f\in \ker(f_k)$.
By hypothesis we have if $f\in {\rm span}(f_1,...,fk)$ then exists $\alpha_1,...,\alpha_k$ such that $f=\alpha_1f_1+...+\alpha_kf_k$
Here I'm a little stuck. Can someone help me?
($\Rightarrow$) I do'nt know how to prove this part. Help me, if you can. I will be very grateful.

$(\Leftarrow)$ Let $x\in V$ such that $x\in \bigcap ker(f_i)$ i.e. $x\in ker(f_i)$ for all $i=1,....,n$. 
Hence $f_i(x)=0$ for all $i=1,....,n$.
Now as $f=a_1f_1+.....+a_kf_k$ which gives $f(x)=0$.
$(\Rightarrow)$ for good proof of this part refer to,
J.B.Conway - A Course in Functinal Analysis, 2nd Edition
Proposition 1.4(page no. 371)
