I am trying to prove the following. I have seen it alluded to in other places of the internet (this site included) but without proof.
Let $L,L_1\ldots L_n$ be linear functionals on a vector space $X$. If $\bigcap_{i=1}^n ker(L_i) \subset ker(L)$ then there exists $t_i$ for $i=1\ldots n \in \mathbb{R}$ such that
$L = \sum_{i=1}^n t_i L_i$.
In other words, if the intersection of kernels of linear functionals is contained by the kernel of another linear functional then they are linearly dependent.
Related:
Intersection of kernels and linear dependence of linear maps
Best Answer
Let $K$ denote the scalar field. Consider $F\colon X \to K^n$ given by
$$F(x) = \begin{pmatrix}L_1(x)\\ L_2(x)\\ \vdots \\ L_n(x)\end{pmatrix}.$$
Let $R = \operatorname{im} F \subset K^n$. We have an induced isomorphism $$\tilde{F}\colon X/\ker F \xrightarrow{\sim} R.$$
Since $\bigcap\limits_{k=1}^n \ker L_k = \ker F \subset \ker L$, we have an induced linear form $\tilde{L} \colon X/\ker F \to K$, and can pull that back to $R$ as $\hat{L} := \tilde{L} \circ \tilde{F}^{-1}$. We can extend $\hat{L}$ to all of $K^n$ (extend a basis of $R$ to a basis of $K^n$, and choose arbitrary values, e.g. $0$, on the basis vectors not in $R$). Thus there is a linear form $\lambda \colon K^n \to K$ with
$$\lambda \circ F = \lambda\lvert_R \circ F = \hat{L}\circ F = \tilde{L}\circ \tilde{F}^{-1}\circ F = \tilde{L} \circ \pi = L,$$
where $\pi \colon X \to X/\ker F$ is the canonical projection.
But every linear form $K^n\to K$ can be written as a linear combination of the component projections, so there are $c_1,\dotsc, c_n$ with
$$\lambda\begin{pmatrix}u_1\\u_2 \\ \vdots \\ u_n \end{pmatrix} = \sum_{k=1}^n c_k\cdot u_k,$$
and that means
$$L(x) = \lambda(F(x)) = \sum_{k=1}^n c_k\cdot L_k(x)$$
for all $x\in X$, or
$$L = \sum_{k=1}^n c_k\cdot L_k.$$