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.

You start well, but soon make a mistake: the statement $f\in\bigcap_{i=1}^k\ker(f_i)$ is wrong, because the kernels are subspaces of $V$ and $f\in V^*$.

What you have to prove is

If $f\in\operatorname{Span}(f_1,\dots,f_k)$ then $\ker(f)\supset\bigcap_{i=1}^k\ker(f_i)$.

Suppose $f\in\operatorname{Span}(f_1,\dots,f_k)$; then $f=\alpha_1f_1+\dots+\alpha_kf_k$ for some scalars $\alpha_1,\dots,\alpha_k$. If $x\in\bigcap_{i=1}^k\ker(f_i)$, then $f_i(x)=0$, for $i=1,\dots,k$ and therefore $$ f(x)=\alpha_1f_1(x)+\dots+\alpha_kf_k(x)=0 $$ proving that $x\in\ker(f)$.

Now let's try the converse:

If $\ker(f)\supset\bigcap_{i=1}^k\ker(f_i)$ then $f\in\operatorname{Span}(f_1,\dots,f_k)$.

Suppose $\ker(f)\supset\bigcap_{i=1}^k\ker(f_i)$. Write $$ f=\alpha_1f_1+\dots+\alpha_kf_k+\alpha_{k+1}f_{k+1}+\dots+\alpha_nf_n $$ which is possible because $\{f_1,\dots,f_n\}$ is a basis of $V^*$.

Lemma. Every basis of $V^*$ is the dual of a basis $\{e_1,\dots,e_n\}$ of $V$.

Once accepted this lemma (for the proof, see https://math.stackexchange.com/a/1772676/62967), we have $\{e_1,\dots,e_n\}$ such that $$ f_i(e_j)=\begin{cases} 1 & i=j \\ 0 & i\ne j \end{cases} $$ In particular, $e_{k+1},\dots,e_n\in\bigcap_{i=1}^k\ker(f_i)$, so $$ f(e_j)=0,\quad j=k+1,\dots,n $$ and therefore, for $j=k+1,\dots,n$, \begin{align} 0=f(e_j)&=\alpha_1f_1(e_j)+\dots+\alpha_kf_k(e_j)+ \alpha_{k+1}f_{k+1}(e_j)+\dots+\alpha_jf_j(e_j)+\dots+\alpha_nf_n(e_j) \\ &=0+\dots+0+0+\dots+\alpha_j\cdot1+\dots+0 \\ &=\alpha_j \end{align} Hence $\alpha_j=0$ for $j=k+1,\dots,n$ and finally $$ f=\alpha_1f_1+\dots+\alpha_kf_k\in\operatorname{Span}(f_1,\dots,f_k) $$

$(\Leftarrow)$ 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)