Equivalent condition of linear dependence of linear functionals

**Theorem**. Suppose $f_1,\cdots, f_n$ and $g$ are linear functional on a vector space. Then $g$ is linearly spanned by $f_1,\cdots,f_n$ if and only if $$\bigcap_{i=1}^{n}\ker f_i\subset \ker g.$$

*Proof*. We prove this by induction. If $n=1$ and $g\neq 0$, then there is some $x_1\notin \ker f_1$. Since $$f_1(x)x_1-f_1(x_1)x\in \ker f_1\subset \ker g,$$
$f_1(x)g(x_1)-f_1(x_1)g(x)=0,$
i.e., $$g(x)=\frac{g(x_1)}{f_1(x_1)}f_1(x).$$

Assume the proposition is true for the case of $n-1$, consider the case of $n$.

If there is some $i_0\in \{1,\cdots,n\}$ such that $\cap_{i\neq i_0}\ker f_i\subset\ker f_{i_0},$ then $\cap_{i\neq i_0}\ker f_i=\cap_{i=1}^{n}\ker f_i\subset \ker g$, then the proposition holds for the case of $n$ by assumption.

If for any $j\in \{1,\cdots,n\}$ there is some $x_j\in (\cap_{i\neq j}\ker f_i)\backslash\ker f_j$, then $$x-\sum_{j=1}^n\frac{x_j}{f_j(x_j)}f_j(x)\in \cap_{i=1}^n\ker f_i.$$
so $g(x-\sum_{j=1}^n\frac{x_j}{f_j(x_j)}f_j(x))=0$, i.e., $g(x)=\sum_{j=1}^n\frac{g(x_j)}{f_j(x_j)}f_j(x)$ for all $x$.

Hence the proposition holds for the case of $n$.