Let $P$ be the set of all continuous nonnegative functions $[0,1] \to {\mathbb R}$. Let $f_1,f_2,f_3$ and $f_4$ be the elements of $P$ defined by
\begin{align} f_1 (t) &= 1 \\ f_2 (t) &= 301+240899t - 241200t^2 \\ f_3 (t) &= 323612-1224687t + 1221075t^2 \\ f_4 (t) &= 2030167-8100367t+8080200t^2 \end{align}
Note that they are linearly dependent : one has $11390000f_1(t)-335f_2(t)+536f_3(t)-91f_4(t)=0$, so that the vector space $F$ spanned by $f_1,f_2,f_3,f_4$ has dimension three. Let $G=F \cap P$.
Question : are there elements $g_1,g_2,g_3 \in G$ such that each of $f_1,f_2,f_3,f_4$ is a linear combination with nonnegative coefficients of $g_1,g_2,g_3$ ? I conjecture that the answer is no.
Context : I'm interested in the more general case where $f_2,f_3,f_4$ can have arbitrary coefficients (as long as they stay nonnegative on $[0,1]$). I used trial and error on randomly selected coefficients. For small and simple values of the coefficients, I was always able to find a solution $(g_1,g_2,g_3)$. (1) is the smallest set of values for which I was unable to find a solution.