# Linear Functionals $\varphi[h]=\lambda\psi[h]$, Variations

From Calculus of Variations, G&F, the problem is: Given two linear functional $$\varphi,\psi$$ over a linear space $$R$$ such that $$\varphi[h]=0\iff\psi[h]=0$$. Show that there is a constant $$\lambda$$ such that $$\varphi[h]=\lambda\psi[h]\tag{1}\label{eq1}$$

Let $$S\subset R$$ the set where $$\varphi,\psi$$ are both zero. What I have so far is: let $$\bar{h}:=\{k\in R:h-k\in S\}$$ then exists $$\lambda_h$$ such that $$\varphi[k]=\lambda_h\psi[k]$$ for every $$k\in\bar{h}$$, thus partitioning $$R$$ into disjoint sets where \eqref{eq1} holds locally. This is what I have. I kinda feel that very close to complete the solution but I just can't see it right away. Thanks

The kernel of a nonzero linear functional always has codimension $$1$$, because of the first isomorphism theorem.
It means that if $$x\in R\setminus S$$ is any element, it together with $$S$$ already spans the whole space, so that $$R={\rm span}(x)\oplus S$$.
Fix such an $$x$$.
Of course, take $$\lambda:=\frac{\varphi(x)}{\psi(x)}$$, as you did 'locally at $$x$$'.
For any scalar $$\alpha$$ and $$s\in S$$ we have $$\lambda\psi(\alpha x+s)=\lambda\alpha\psi(x)=\alpha\varphi(x)=\varphi(\alpha x+s)\,.$$