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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

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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 $$\lambda\psi(\alpha x+s)=\lambda\alpha\psi(x)=\alpha\varphi(x)=\varphi(\alpha x+s)\,.$$

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  • $\begingroup$ Boy I was so close, thanks! $\endgroup$ – Cristian Baeza Oct 18 '20 at 19:30

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