Integral constraints for positive function Let $C={\cal C}([0,1],(0,\infty))$ denote the set of all continuous
maps $[0,1]\to (0,\infty)$. Let $g_1,g_2 \in C$ ; one can then define
$$
\begin{array}{rcl}
\Phi &: C& \to (0,\infty)^2 \\
f &\mapsto& \bigg(\int_{[0,1]} fg_1,\int_{[0,1]} fg_2\bigg) \\
\end{array}
$$
Obviously, when $g_1$ and $g_2$ are not linearly independent, say $g_1=ag_2$ for some $a>0$, the image of $f$ is the diagonal
$\lbrace (i,ai) | i > 0\rbrace$. I believe that when $g_1$ and $g_2$ are linearly independent,
$\Phi$ is always surjective. Can anyone help me to show this ?

Your conjecture is false.
Let $m$, $M$ be positive numbers such that $m \le g_1$ and $g_2 \le M$. Then, for all $f \in C$, 
$$\frac{\int f g_1}{\int f g_2} \ge \frac{m \int f}{M \int f} = \frac{m}{M}.$$
In particular,
$$\frac{\int f g_1}{\int f g_2} \ge \frac{\min g_1}{\max g_2},$$
and, by the same kind of manipulations,
$$\frac{\int f g_1}{\int f g_2} \le \frac{\max g_1}{\min g_2}.$$
Hence, the couple $(\int fg_1, \int fg_2)$ always live in a cone. What you can show or study is the following:


*

*you can show that the image of $\Phi$ is always a cone: it will be convex, and it has a decomposition in half-lines.

*when is the cone is open, closed, semi-open? This will depend on, among other things, whether the maximum/minimum above are reached on an open set or not.

*find when the estimates above are optimal, i.e. give the boundary of the cone.
The answer are rather easy when the sets where $g_1$ and $g_2$ reach their maximum / minimum are disjoint, but seems to be thorny otherwise (for instance if $g_1$ and $g_2$ reach their maximum at the same point).
