$L$ is a topological complement of $G$
Solution 1:
The inclusion $G + L \subseteq E$ is trivial.
For the other inclusion, let $x \in E$ and consider $y= \sum_{i=1}^n \hat{f}_i(x)e_i.$ Clearly $y \in G.$ Now show that $x-y\in L$ by checking that $\hat{f}_j(x-y)=0$ for all $j=1,\dots,n.$ Then you have $x = y + (x-y) \in G + L.$