# $L=\sum_i t_i L_i$. If L is nonnegative whenever every $L_i$ is nonnegative, we can have all $t_i\ge0$ for every i?

Let $L=\sum_i t_i L_i$, where the $L_i$'s are linear functionals on a real vector space (take $R^n$ if too difficult otherwise). If $Lx\ge 0$ whenever $x$ is such that $L_i x\ge0$ for every $i$ ("Strong Pareto"), then we can redefine the $t_i$'s so that $t_i\ge0$ for every $i$?

This is trivial when the $L_i$'s are linearly independent. I think that I know how to prove it in general, but my proof would be long, and I believe that there is a short or fairly short proof. (Also a reference to a proof would do!)

A not-that-good answer: In that not fairly short proof I sketched I use Theorem 3.4(a) of Walter Rudin's functional analysis to show that the possible $t_i$'s include also positive ones. I just feel that this is probably a known result and hence the proof is not justified.

Theorem 3.4 says that for disjoint nonempty convex sets A and B with A open there is a linear functional F and some $r\in R$ such that $Fx< r\le Fy$ whenever $x\in A$ and $y\in B$. Take A to correspond to $\{L\}$ and B to correspond to positive linear combinations of $L_i$'s. Then clearly r is zero but our above ("strong Pareto") assumption leads to $Fx\ge0$, a contradiction, so $L$ is in $F$, as required.

In $\mathbb{R}^n$, if the $L_i$ are finite, the question can be posed as asking if $L$ is contained in the closed convex cone generated by the $L_i$.

Let $A= \begin{bmatrix} L_1^T &\cdots&L_n^T \end{bmatrix}$, $b=L^T$. Then Farka's lemma states that either there is a solution to $Ax=b$ with $x \ge 0$ or there exists a $y$ such that $y^T A \ge 0$ and $y^T b < 0$. Since the second statement cannot hold (by hypothesis), the first must hold.