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.
 A: 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
$$
f=\alpha_1f_1+\dots+\alpha_kf_k\in\operatorname{Span}(f_1,\dots,f_k)
$$
A: $(\Leftarrow)$ 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)
A: Probably not the best proof but it was suggested by a colleague :
We will use duality...
Suppose that $\bigcap \limits_{i=1}^{k} \ker{f_i} \subset \ker{f}$.
Using properties of duality we have that $(\ker{f})^{\bot} \subset \left( \bigcap \limits_{i=1}^{k}\ker{f_i}\right)^{\bot} = \displaystyle \sum \limits_{i=1}^{k} (\ker f_i)^{\bot}$.
(Remember that for a set $A\subseteq V$, we have $A^{\bot} :=\{\phi\in V^* \ / \ \forall x \in A, \phi(x)=0 \}$).
If we suppose that all the linear forms are $\not \equiv 0$ then for all $i\in\{1,...,k\}$, all $(\ker{f_i})$ are hyperplanes of $V$ hence all the $(\ker{f_i})^{\bot}$ are of the form $\mathbb{K}a_i$ with $a_1,...,a_k\in \mathbb{K}^{\times}$ and the $(a_i)_i$ can be taken pairwise different.
Then $(\ker{f})^{\bot} \subset \sum \limits_{i=1}^{k} \mathbb{K}a_i$.
A: $(\Rightarrow)$ Suppose $1\leq k\leq n \bigcap_{j\neq k}ker(f_{j})\neq\bigcap_{j=1}^{n}ker(f_{j})$.
Then exists$y_{k}\in\bigcap_{j\neq k}ker(f_{j})$ tal que $y_{k}\notin\bigcap_{j=1}^{n}ker(f_{j})$. then $f_{j}(y_{k})=0$ if $k\neq j$ and $f_{j}(y_{k})\neq0$ if $j=k$.
Let $x_{k}=\frac{yk}{f_{k}(y_{k})}$(Note $f_{k}(y_{k})\neq0)$ Then, $f_{k}(x_{k})=f_{k}(\frac{y_{k}}{f_{k}(y_{k})})=\frac{f_{k}(y_{k})}{f_{k}(y_{k})}=1$.For other way, $f_{j}(x_{k})=f_{j}(\frac{y_{k}}{f_{k}(y_{k})})=\frac{f_{j}(y_{k})}{f_{k}(y_{k})}=0$
Let $f\in V^{*}y x\in\mathbb{R}$ . 
Define $y=x-\sum_{k=1}^{n}f_{k}(x)x_{k}$
Then $f_{j}(y)=f_{j}(x-\sum_{k=1}^{n}f_{k}(x)x_{k})=f_{j}(x)-\sum f_{k}(x)f_{j}(x_{k})=f_{j}(x_{k})-f_{j}(x)f_{j}(x_{j})=0$.
Then, $f_{j}(x_{k})=0\,y\,f_{j}(x_{j})=1$.
By hyphotesis $f(y)=0$,then:
$0=f(x)-\sum_{k=1}^{n}f_{k}(x)f(x_{k})=f(x)-\sum_{k=1}^{n}\alpha_{k}f_{k}(x)\Rightarrow f(x)=\sum_{k=1}^{n}\alpha_{k}f_{k}\Rightarrow f=\sum_{k=1}^{n}\alpha_{k}f_{k}$
In consequence, $f\in span(f_{1},...,f_{n})$
$\smash{}$
